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https://cs.stackexchange.com/questions/43089/reduction-and-decidability | 1,723,416,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00151.warc.gz | 154,435,913 | 46,803 | # Reduction and decidability
Consider the following language: $$L = \{ \langle M \rangle \ |\ M \text { accepts } w \text { whenever it accepts } w^R \}$$
I am trying to understand the following proof that this language $L$ is undecidable.
The proof proceeds by contradiction, by reducing to $L$ the language $A_{TM}=\{\langle M,w \rangle\mid M\text{ accepts }w\}$ known to be undecidable. It goes as follows:
Suppose that $L$ is decidable, then there's a TM $M_L$, that decides $L$
Thus we can build $M_{ATM}$ to decide $A_{TM}$ as follows:
1. Let $\langle M, w \rangle$ be an input for $M_{ATM}$.
2. Construct a machine, $M_1$ which on input $x$: Simulate $M$ on $w$; if $M$ rejects $w$, $reject$. If $M$ accepts $w$, $accept$ if $x=01$, $reject$ otherwise.
3. Simulate $M_L$ on input $\langle M_1 \rangle$.
4. $accept$ if $M_L$ rejects. $reject$ otherwise.
Since $A_{TM}$ is not decidable, we have a contradiction, which implies that $L$ cannot be decidable.
However there is a point of the proof I do not understand.
My question is: What happens if $M$ gets into a loop on some $w$?
• What is $A_{TM}$ ? Can you try to improve the readability? What are you trying to prove? What is the purpose of your construction? Commented May 30, 2015 at 11:55
• Of course. $A_{TM}$ is a common notation for the (undecidable) language $$\{ \langle M,w \rangle \ | \ \text { TM M accepts w } \}$$ Commented May 30, 2015 at 13:22
• I am trying to prove by contradiction (using reduction to $A_{TM}$) that $L$ is undecidable language. Commented May 30, 2015 at 13:38
• Thanks. That is what is needed so that people understand what you are at.It should be in the text of your question, explaining why you are doing all that. Also, do not assume too much in notations when talking to the planet, it is sometimes noted $ATM$, but ATM can also mean Alternating Turing Machine (whatever that is). I personally did not know this notation as standard, and I noticed on the web that many people still give the definition. Commented May 30, 2015 at 13:56
• "In computability theory, a Turing reduction from a problem A to a problem B, is a reduction which solves A, assuming the solution to B is already known". You assume decidability of $L$ is known. The fact that $M$ does not halt, or rejects is immaterial. What you want is that the language accepted by $M_1$ is $\{01\}$ iff $M$ accepts $w$ (and is $\emptyset$ otherwise. If $M$ loops, thus not acceptin $w$, then $M_1$ loops on any input, recognizing $\emptyset$. Hence $M_L$ will accept $\langle M_1 \rangle$, so that $M_{ATM}$ will reject. Commented May 30, 2015 at 15:03
In many problems like this, it's often a bad idea to base your construction on both acceptance and rejection. Try rewriting your $M_1$ in this form
M1(x) =
simulate M on w
if M accepts w
if x = 01
accept
Now the important thing is that
1. If $M$ accepts $w$ (i.e., if $\langle M, w\rangle\in A_{TM}$) then $L(M_1)=\{01\}$.
2. If $M$ fails to accept $w$, either by rejecting or by running forever, then $M_1$ will accept nothing, so $L(M_1)=\varnothing$
Now if $L$ were decidable with decider $M_L$, then you can use the decider and $M_1$ to build a decider, $M_A$, for $A_{TM}$ as you suggested:
MA(<M>, w) =
Construct M1 as above
if ML accepts <M1>
reject
else if ML rejects <M1>
accept
Now we have
1. If $M$ accepts $w$, then $L(M_1)=\{01\}$ so $M_L$ will reject $M_1$ and hence $M_A$ will accept $\langle M, w\rangle$.
2. If $M$ doesn't accept $w$, then $L(M_1)=\varnothing$ so $M_L$ will accept $M_1$ and hence $M_A$ will reject $\langle M, w\rangle$.
In short, $M_A$ is indeed a decider for $A_{TM}$, an undecidable language, meaning that $L$ must not be decidable.
In this reduction of $A_{TM}=\{\langle M,w \rangle\mid M\text{ accepts }w\}$ to $L$, what matters in the construction of the Turing machine $M_1$ is that the language accepted by $M_1$ is $\{01\}$ iff $M$ accepts $w$, and is $\emptyset$ otherwise.
Whether $M$ does not halt on input $w$, or halts rejecting it, does not make any difference. In both cases, the language accepted by $M_1$ is $\emptyset$, thus satisfyiing the definition for $L$. Hence $M_L$ will accept $\langle M_1 \rangle$, so that $M_{ATM}$ will reject.
You are trying to prove that $L$ is undecidable by showing $A_{tm}\le L$.
For this reduction to be valid, you need to convert each input $(\langle M \rangle, w)$ of $A_{tm}$ into an input $\langle M_1 \rangle$ (using your notations) for $L$ so that $$(\langle M \rangle, w) \in A_{tm} \quad\text{ iff }\quad \langle M_1 \rangle \in L$$
Equivalently, such a reduction shows that if you can decide $L$ you will be able to decide $A_{tm}$: simply you convert the input of $A_{tm}$ to an equivalent input of $L$ and run $M_L$ on it.
So the question that hides in your post, is how to convert the inputs. Let's see what happens in each case:
1. case I: $(\langle M \rangle, w) \in A_{tm}$.
In this case $M_1$ will just run $M$ on $w$ (which ends in an accepting state, and specifically, always halts), and then only accepts $x=01$. Therefore $M_1$ is not in $L$.
2. case II: $(\langle M \rangle, w) \notin A_{tm}$.
As you mention, there are 2 cases here: wither $M$ rejects $w$ or it never halts. In both cases, $M_1$ never accept a single word: either it rejects everything, or it never halts (and then also no input is ever accepted). Therefore $M_1$ is in $L$ (vacuously satisfying the requirement).
So this goes the opposite way from the "iff" statement I wrote above, but that's fine because the final decision of your $M_{ATM}$ is the opposite of the decision of $M_L$ on $M_1$, which fixes this negaiton. So everything works fine, and the proof is valid. Makes sense?
• Now, I am curious to see what happens. Commented May 30, 2015 at 15:20
• I spent considerable time trying to explain to the OP how to make a question clearer. But I am afraid it was wasted. The problem is that we can work out what he intends, but the work is useless for further users as understanding what is the issue is to hard. Hence they will not bother reading at all. The question should not have to be explained in answers ,,, Working on question quality is the proper way to say "please" and "thank you" cc @Elimination Commented May 31, 2015 at 14:13
• I appreciate your criticism, @babou Commented May 31, 2015 at 14:40
• Thanks. I edited your question to make it easier for future users. You can thus see what I meant. Good luck. Commented May 31, 2015 at 14:46
Wrong solution ( based on the comment below )
L={⟨M⟩ | M accepts w whenever it accepts $w^R$}. Proof by contradiction. Assume that L is decidable and let $R$ be a decider for it. We will construct decider $S$ for $A_{tm}$, using $R$ as a subrouting. Note that $A_{tm}$ = {<M,w> | M is a TM and M accepts w}.
S = on input <M,w>
constructs M_w = on input x
if x!=010 reject
else run M on w
if M accepts w => accept
else reject
Run R on M_w
if R accepts => accept
if R rejects => reject
Note: if $M$ accepts $w$ then $L(M_w)$ = {010} , $R$ ran on $M_w$ will accept, therefore $S$ will accept. if $M$ rejects or loops on $w$ then $L(M_w)$ = empty set , $R$ ran on $M_w$ will reject, therefore $S$ will reject. We have built a decider for $A_{tm}$, since we know that $A_{tm}$ is undecidable our assumption was wrong => L is undecidable.
Edited Solution>>>
L={⟨M⟩ | M accepts w whenever it accepts $w^R$}. Proof by contradiction. Assume that L is decidable and let $R$ be a decider for it. We will construct decider $S$ for $A_{tm}$, using $R$ as a subrouting. Note that $A_{tm}$ = {<M,w> | M is a TM and M accepts w}.
S = on input <M,w>
constructs M_w = on input x
run M on w
if M accepts w => reject
else => accept
Run R on M_w
if R accepts => accept
if R rejects => reject
Note: if $M$ accepts $w$ then $M_w$ will reject on all input $L(M_w)$ = empty set, $R$ ran on $M_w$ will accept, therefore $S$ will accept. if $M$ rejects or loops on $w$ then $L(M_w)$ = {$\Sigma^*$} , $R$ ran on $M_w$ will reject , therefore $S$ will reject. We have built a decider for $A_{tm}$, since we know that $A_{tm}$ is undecidable our assumption was wrong => L is undecidable.
• This solution doesn't work: R will always accept on $M_w$, since either $L(M) = \emptyset$ or $L(M) = \{ 010 \}$, and in both cases $L(M)$ is closed under string reversal. Commented Aug 11, 2017 at 20:25
• Can you please clarify - $L(M)$ is closed under string reversal ? (Sipser): A collection of objects is closed under some operation if applying that operation to members of the collection returns an object still in the collection. Now empty set contains no elements. Commented Aug 11, 2017 at 20:40
• The empty set is closed under reversal vacuously. You can prove this using the definition – if you apply reversal to a member of the collection, you get another member of the collection. You can easily check that this holds for the empty set. Commented Aug 11, 2017 at 20:44 | 2,681 | 8,932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.908322 |
https://stats.stackexchange.com/questions/550822/endogeneity-and-regression-interpretation-how-does-it-work-is-it-possible | 1,718,874,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861916.26/warc/CC-MAIN-20240620074431-20240620104431-00289.warc.gz | 498,030,312 | 42,622 | # Endogeneity and Regression Interpretation: how does it work? is it possible?
My question is not particularly straightforward. I will explain my reasoning, and then give the question at the end to avoid any confusion.
$$\ Y_{i} = \beta_{0}+\beta_{1}X_{i}+\varepsilon_{i}$$
Suppose that this our population regression function. We note that $$\ \beta_{0}$$ and $$\ \beta_{1}$$ are not arbitrary. This means that the error term $$\ \varepsilon_{i}$$ has some structure to it. In particular, it reflects the deviation of $$Y_{i}$$ from its expected value. To put it even more clearly, we have that $$E[Y_{i}|X_{i}]= \beta_{0}+\beta_{1}X_{i}$$. What I am trying to say here is that our population parameters must take some non-arbitrary value, otherwise we might as well define our errors to make the equation balance and pick them randomly.
Now suppose we want to take a sample and estimate the above population regression function. But before doing so, we want to work out what the true value of $$\beta_{0}$$ and $$\beta_{1}$$ are. To do this we appeal to the mean square error as our guide (I drop the subscripts for simplicity of notation):
$$MSE(b_{0}, b_{1}) = E[(Y - (b_{0}+b_{1}X))^2] = E[Y^2] -2b_{0}E[Y]-2b_{1}E[XY]+E[(b_{0}+b_{1}X)^2]$$
$$MSE(b_{0}, b_{1}) = E[Y^2] -2b_{0}E[Y]-2b_{1}Cov(X,Y) - 2b_{1}E[X]E[Y]+b_{0}^2+2b_{0}b_{1}E[X]+b_{1}^2Var[X]+b_{1}^2(E[X])^2$$
I have skipped some of the algebra because it is not really important at this point. Now I take first order conditions with respect to my two variables:
$$\frac{\partial MSE(b_{0},b_{1})}{\partial b_0}= -2E[Y]+2b_{0}+2b_{1}E[X]$$
$$\frac{\partial MSE(b_{0},b_{1})}{\partial b_1}= -2Cov(X,Y)-2E[X]E[Y]+2b_{0}E[X]+2b_{1}Var[X]+2b_{1}(E[X])^2$$
Setting the partial derivates equal to zero and solving for each parameter gives:
$$\beta_{0}=E[Y]-\beta_{1}E[X]$$
$$\beta_{1}= \frac{Cov[X,Y]}{Var[X]}$$
Now that we have the equations of our true population parameters, we can consider the issue at hand, otherwise known as my question! Let's try to check the correlation between our explanatory variables and the error term defined above:
$$\ E[X\varepsilon] = E[X(Y-(\beta_{0}+\beta_{1}X)] = E[XY - X\beta_{0}-\beta_{1}X^2]$$
Using some simple algebraic manipulations we get that:
$$E[X\varepsilon]= E[XY]- E[X\beta_{0}+\beta_{1}X^2]$$
$$E[X\varepsilon]= E[XY]- E[X(E[Y]-\beta_{1}E[X])+\beta_{1}X^2]$$
$$E[X\varepsilon]= E[XY]- E[XE[Y]-X\beta_{1}E[X]+\beta_{1}X^2]$$
$$E[X\varepsilon]= E[XY]- E[Y]E[X] + E[X\beta_{1}E[X]] - E[\beta_{1}X^2]$$
$$E[X\varepsilon]= E[XY]- E[Y]E[X] + \beta_{1}E[X]^2 - E[\beta_{1}X^2]$$
$$E[X\varepsilon]= E[XY]- E[Y]E[X] + \beta_{1}E[X]^2 - \beta_{1}E[X^2]$$
$$E[X\varepsilon]= E[XY]- E[Y]E[X] -\beta_{1}(E[X^2] - E[X]^2)$$
$$E[X\varepsilon]= Cov[X,Y]-\beta_{1}Var[X]$$
$$E[X\varepsilon]= Cov[X,Y]-\frac{Cov[X,Y]}{Var[X]}Var[X]$$
$$E[X\varepsilon]= 0$$
Now what we have proved is the by construction the parameters of the population regression function are defined in such a way that the covariance between the explanatory variables and errors is equal to zero.
Now bearing all of the above in mind, herein lies my question: how is endogeneity mathematically possible? Is it that my interpretation of the regression is wrong? I have heard others mention that there is some sort of causal interpretation of regression that must be taken, though I am not really sure what that means, or what that is. Any insights?
P.S. Thank you to those that have answered my other questions about regression analysis previously, I am learning a lot!
You are correct. In the linear regression $$\mathbb{E}[X\varepsilon] = 0$$ by construction. I talked about this in another post here
https://stats.stackexchange.com/a/550305/261146
• You need an "externally defined" quantity of interest.
• Based on this external definition you decide whether OLS satisfies exogeneity.
1. An illustrative example
I think the best example is easily selection into treatment, the "causal case".
Suppose that an individual is assigned to a treatment $$D \in \{0,1\}$$. If the individual is part of the control she will have an outcome $$Y_0$$. If she is treated, her outcome is $$Y_1$$. The outcome observed by the researcher is: \begin{align*} Y &= DY_1 + (1-D)Y_0 \\ &= Y_0 + D(Y_1 - Y_0) \end{align*}The variables $$(Y_1,Y_0)$$ are known as "potential outcomes". An individual cannot belong to both groups at the same time, which means that one of the outcomes is never observed. The best the researcher can do is find two groups with similar characteristics and different treatment status to compare their outcomes. The average treatment effect is defined as $$\tau = \mathbb{E}[Y_1 - Y_0]$$ This quantity is important in medical trials and many situations because it indicates the population-level treatment effect.
Can we identify $$\tau$$ from an OLS regression of $$Y$$ on $$D$$?
2. Running an OLS estimator:
The OLS regression takes the form: $$Y = \mu_{OLS} + \tau_{OLS} D + \varepsilon$$
The coefficient $$(\mu_{OLS},\tau_{OLS})$$ are constructed to minimize MSE, and will satisfy the exogeneity condition.
The OLS coefficient of $$\tau$$ is equal to \begin{align*} \tau_{OLS} &= \frac{\mathbb{E}[DY]}{\mathbb{E}[D]} - \frac{\mathbb{E}[(1-D)Y]}{\mathbb{E}[1-D]} \end{align*}
3. Does OLS work, i.e. $$\tau_{OLS} = \tau$$?
• By substituting the definition of $$Y$$ with potential outcomes, this is equal to $$\tau_{OLS} = \frac{\mathbb{E}[DY_1]}{\mathbb{E}[D]} - \frac{\mathbb{E}[(1-D)Y_0]}{\mathbb{E}[1-D]}$$ Case 1: (Random assignment) When $$D$$ is randomly assigned, $$D$$ is independent of $$(Y_1,Y_0)$$. This means that $$\mathbb{E}[DY_1] = \mathbb{E}[D]\mathbb{E}[Y_1]$$ and $$\mathbb{E}[DY_0] = \mathbb{E}[D]\mathbb{E}[Y_0]$$. We can then readily verify that $$\tau_{OLS} = \mathbb{E}[Y_1] - \mathbb{E}[Y_0] = \tau$$.
Case 2: (Self-selection) Suppose that $$D = 1$$ if and only if $$Y_1 > Y_0$$. This is situation where individuals only participate if they (correctly) assess that they will receive some benefit. In this case the terms may not cancel out and $$\tau_{OLS} \ne \tau$$.
3. Insights
• We can always define the potential outcomes as linear model:
$$Y = \mu + \tau D + U$$ where $$U = (Y_0 - \mu) + D(Y_1 - Y_0 - \tau)$$ and $$\mu = \mathbb{E}[Y_0]$$.
• The endogeneity condition in this case requires that
\begin{align*} \mathbb{E}[ D U ] &= \mathbb{E}[D(Y_1 - Y_0 - \tau)] = 0 \\ \mathbb{E}[ U ] &= \mathbb{E}[Y_0 - \mu + D(Y_1 - Y_0 - \tau)] = 0 \\ \end{align*}
• This may fail if $$D$$ is correlated with $$(Y_1,Y_0)$$.
• The OLS minimizes the MSE criterion. It is not guaranteed to produce anything else.
• We sometimes call the parameters $$(\mu_{OLS},\tau_{OLS})$$ "pseudo-parameters" because they solve the equation $$\mathbb{E}[X(Y - \mu_{OLS} - D \tau_{OLS})] = 0$$, where $$X = (1,D)$$.
• However, depending on the "external" model, these may or may not have a "useful" interpretation.
• To make sure I have understood. We distinguish the "true model" (whatever that is) and the OLS population regression function, which we want to estimate using sample data. In other words, there are essentially three models: (1) 'True Model' which encodes causation (2) Hypothesized Regression Function (could be OLS) and (3) Sample Regression Function. The endogeneity problem arises because, due to some external issue (e.g. self-selection above), the 'true' model does not match up with our 'Hypothesized' model. I have been excessively brief, but hopefully in the right direction. Commented Nov 3, 2021 at 21:57
• To put it, perhaps, another way, the population OLS parameters are one way to represent the 'true' causal model, and so we have to make assumptions (external definitions etc.) to make sure that it 'fits'. But this also raises a further question. When trying to show that the sample OLS estimates are unbiased, we get the following expression: $\beta + (X'X)^-1 X'\varepsilon$ This will clearly not refer to our "hypothesized PRF with OLS parameters but the "true" model, right? Commented Nov 3, 2021 at 22:13
• I think you have the right idea. The assumptions are the ones that make sure that the Hypothesized Regression Function (HRF) aligns with the "True model". The sample OLS estimates are unbiased estimators of the HRF coefficients. They may not necessarily recover the parameters of the true model unless the assumptions are satisfied. Commented Nov 3, 2021 at 22:26
• I would also add that in this example, whether $D$ is correlated with $(Y_1,Y_0)$ is not something the analyst can control. Therefore, the same OLS model may or may not recover the "true" model, depending on how the data is produced. This is why running RCTs (randomized controlled trials) is so important in medicine, as opposed to just using convenience sampling, to ensure this holds by design. Commented Nov 3, 2021 at 22:28
• That’s another good example. Commented Nov 4, 2021 at 14:52 | 2,755 | 8,921 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 60, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-26 | latest | en | 0.8602 |
https://mathoverflow.net/questions/443265/encoding-primes-via-ranks-of-sign-matrices | 1,708,851,895,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474594.56/warc/CC-MAIN-20240225071740-20240225101740-00841.warc.gz | 377,910,330 | 28,602 | Encoding primes via ranks of sign matrices
(Reposted from math.SE)
Recently I came across a very simply defined family of matrices: for $$n \in \mathbb{N}$$, set $$A_n := (a_{ij})_{0 \le i, j \le n-1}$$, where
$$\displaystyle a_{ij} := (-1)^{\big\lfloor 2ij/n \big\rfloor}$$
These are normalized $$\pm 1$$ symmetric $$n \times n$$ matrices. The first few are:
$$A_2 = \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix}, \qquad A_3 = \begin{bmatrix} 1&1&1\\ 1&1&-1\\ 1&-1&1 \end{bmatrix}, \qquad A_4 = \begin{bmatrix} 1&1&1&1\\ 1&1&-1&-1\\ 1&-1&1&-1\\ 1&-1&-1&1 \end{bmatrix}, \ldots$$
My original interest was showing that the first standard basis vector $$e_0$$ is always in the column space of $$A_n$$ (which I think I can show). However, computing $$\operatorname{rank}(A_n)$$ for small $$n$$ quickly suggests an intriguing pattern:
$$\operatorname{rank}(A_n) = \sigma_0(n) + \Big\lfloor \frac{n-1}{2} \Big\rfloor$$
where $$\sigma_0(n)$$ is the number (= sum of $$0^\text{th}$$ powers) of divisors of $$n$$. My question is:
Is this formula for $$\operatorname{rank}(A_n)$$ true for all $$n$$?
If so, then since the minimal value of $$\sigma_0$$ is $$2$$, which occurs exactly for prime $$n$$, one would have $$\operatorname{rank}(A_n) = \big\lfloor \frac{n+3}{2} \big\rfloor$$ is minimal $$\iff n$$ is prime. (This would, in my opinion, be an interesting encoding of the primes in a purely linear-algebraic fashion.)
I have tested this up to $$n = 30$$ (and apparently holds up to $$n = 1024$$ even, which is more than enough evidence to convince me personally of its validity). To save some trouble, the proposed formula is A361003 in OEIS. A combinatorial proof e.g. via A361001 would be fine. If anyone knows more about this family of matrices I would be happy to read more.
• Curious! My intuition would have suggested that $A_n$ should have the smaller (not greater) rank the more divisors $n$ has, and yet it is apparently the other way round. Mar 22, 2023 at 19:55
• @darijgrinberg Maybe you were secretly thinking about $\dim \ker$ instead of rank :) Somewhat related, I don't know if there's a good way of viewing $A_n, A_m$ as blocks inside $A_{nm}$ Mar 22, 2023 at 20:25
• No -- I was thinking that divisors of $n$ would lead to equal rows in $A$. Mar 22, 2023 at 21:06
• Small remark : since it has been checked up to 1024, it is in particular true for the first Carmichael number (aka pseudo prime) 561, which adds some weight to the conjecture. Mar 29, 2023 at 10:11
• Side question : if primality is encoded by the number of non-zero eigenvalues, one can wonder what other properties are encoded in the rest of the spectrum. Apr 1, 2023 at 4:15
A modest start, but before doing so, one suggestion: I would index the columns and rows from $$0$$ to $$n-1$$, so $$a_{ij} = (-1)^{\lfloor 2ij/n\rfloor}$$.
We prove that the conjectured value of $$\operatorname{rank}(A_n)$$ is an upper bound. For $$1\le k\le n-1$$ let $$v_k$$ be the row vector with $$1$$ in positions $$k$$ and $$n-k$$ and $$0$$ elsewhere, and for $$0\le j\le n-1$$ let $$a_j$$ be the $$j$$'th column of $$A_n$$.
As $$\lfloor 2kj/n\rfloor$$ and $$\lfloor 2(n-k)j/n\rfloor$$ have different parity if and only if $$n$$ does not divide $$2kj$$, we obtain $$$$v_ka_j= \begin{cases} 0 & \text{if }n\nmid 2kj\\ 2\cdot(-1)^{2kj/n} & \text{if }n\mid 2kj. \end{cases}$$$$ Set $$d=\operatorname{gcd}(n, k)$$ and assume that $$k$$ does not divide $$n$$, hence $$d. We claim that $$w_k=v_k-v_d$$ is a left eigenvector of $$A_n$$. We consider two cases:
• If $$n$$ does not divide $$2kj$$, then $$n$$ does not divide $$2dj$$ either. So $$w_ka_j=v_ka_j-v_da_j=0-0=0$$.
• Now suppose that $$n$$ divides $$2kj$$. Then $$n/d$$ divides $$2jk/d$$, and since $$n/d$$ and $$k/d$$ are relatively prime, we get that $$n/d$$ divides $$2j$$, so $$n\mid 2dj$$. So $$w_ka_j=0$$ once we know that $$2kj/n$$ and $$2dj/n$$ have the same parity. As $$d\mid k$$, this could only fail if $$k/d$$ were even while $$2dj/n$$ were odd. This would imply $$2d\mid k$$ and $$2d\mid n$$, respectively, contrary to the choice of $$d$$.
Thus $$w_k$$ is an eigenvector of $$A_n$$ for $$1\le k if $$k\nmid n$$, and these vectors are linearly independent.
• In fact I did use 0-based indexing in my own calculations (now edited). This is a nice construction of the correct number of linearly independent vectors in $\ker(A_n)$, so it remains to show that they span the kernel - any ideas? Mar 23, 2023 at 18:11
• @math54321 Not really. I guess one again has to write out a sufficiently large set of linearly independent vectors of the row space. The vectors $e_k-e_{n-k}$, $1\le k<n/2$, seem to be in this space. Probably the additional vectors coming from the divisors of $n$ have a similarly simple shape. But at a first glance I did not see a pattern even in how to linearly combine $e_k-e_{n-k}$ them from the rows of $A$. Even if $n$ is prime it might be difficult. Mar 23, 2023 at 19:47
• I tried playing with prime $n$, and it seems like your suggestion of $e_k - e_{n-k}$, $1 \le k < n/2$ being in the row (= column) space checks out - although the coefficients I found for the linear combinations (wrt rows of $A_n$) are not easy to predict, they do seem to be only supported in the latter half of rows (with a single exception for $k = 1$). Also I believe the additional 2 vectors needed in this case can be chosen to be $e_0$ and $\sum_{k=1}^{\lfloor n/2 \rfloor} e_k$ (both of which only involve 2 rows of $A_n$) Mar 23, 2023 at 21:32
Just a long comment: I tried to look at the diagonal of the Smith Normal form of the matrices. I counted the number of occurrences of each number. For example, the row $$\{8,3\} \to 1^1\, 2^4\, 4^1\, 8^1$$ encodes $$n$$, $$\lfloor \frac{n-1}{2} \rfloor$$ and then the non-zero entries on the diagonal. In this case, 1 entry is 1, 4 entries are equal to 2, and then there is a single 4, and a single 8.
$$\begin{array}{l} \{1,0\}\to 1^1 \\ \{2,0\}\to 1^1\, 2^1 \\ \{3,1\}\to 1^1\, 2^2 \\ \{4,1\}\to 1^1\, 2^2\, 4^1 \\ \{5,2\}\to 1^1\, 2^3 \\ \{6,2\}\to 1^1\, 2^3\, 4^2 \\ \{7,3\}\to 1^1\, 2^4 \\ \{8,3\}\to 1^1\, 2^4\, 4^1\, 8^1 \\ \{9,4\}\to 1^1\, 2^5\, 6^1 \\ \{10,4\}\to 1^1\, 2^4\, 4^3 \\ \{11,5\}\to 1^1\, 2^5\, 6^1 \\ \{12,5\}\to 1^1\, 2^6\, 4^1\, 8^3 \\ \{13,6\}\to 1^1\, 2^6\, 10^1 \\ \{14,6\}\to 1^1\, 2^5\, 4^4 \\ \{15,7\}\to 1^1\, 2^9\, 60^1 \\ \{16,7\}\to 1^1\, 2^7\, 4^1\, 8^2\, 16^1 \\ \{17,8\}\to 1^1\, 2^8\, 34^1 \\ \{18,8\}\to 1^1\, 2^7\, 4^4\, 12^2 \\ \{19,9\}\to 1^1\, 2^9\, 54^1 \\ \{20,9\}\to 1^1\, 2^8\, 4^1\, 8^5 \\ \{21,10\}\to 1^1\, 2^{12}\, 504^1 \\ \{22,10\}\to 1^1\, 2^7\, 4^4\, 12^2 \\ \{23,11\}\to 1^1\, 2^{11}\, 534^1 \\ \{24,11\}\to 1^1\, 2^{10}\, 4^3\, 8^1\, 16^4 \\ \{25,12\}\to 1^1\, 2^{12}\, 10^1\, 410^1 \\ \{26,12\}\to 1^1\, 2^8\, 4^5\, 20^2 \\ \{27,13\}\to 1^1\, 2^{13}\, 6^1\, 18^1\, 342^1 \\ \{28,13\}\to 1^1\, 2^{10}\, 4^1\, 8^6\, 16^1 \\ \{29,14\}\to 1^1\, 2^{12}\, 4^2\, 2260^1 \\ \{30,14\}\to 1^1\, 2^{10}\, 4^9\, 60^1\, 120^1 \\ \{31,15\}\to 1^1\, 2^{14}\, 62^1\, 558^1 \\ \{32,15\}\to 1^1\, 2^{11}\, 4^3\, 8^2\, 16^3\, 32^1 \\ \{33,16\}\to 1^1\, 2^{16}\, 6^1\, 66^1\, 2904^1 \\ \{34,16\}\to 1^1\, 2^{10}\, 4^7\, 68^2 \\ \{35,17\}\to 1^1\, 2^{18}\, 18^1\, 81900^1 \\ \{36,17\}\to 1^1\, 2^{14}\, 4^1\, 8^6\, 24^4 \\ \{37,18\}\to 1^1\, 2^{18}\, 524290^1 \\ \{38,18\}\to 1^1\, 2^{11}\, 4^8\, 108^2 \\ \{39,19\}\to 1^1\, 2^{20}\, 520^1\, 32760^1 \\ \{40,19\}\to 1^1\, 2^{13}\, 4^5\, 8^1\, 16^6\, 48^1 \\ \end{array}$$
The Mathematica code:
aa[n_] :=
aa[n] = Table[(-1)^Floor[2 i j/n], {i, 0, n - 1}, {j, 0,
n - 1}]; {#, Floor[(# - 1)/2]} ->
Row[(Superscript[#1, #2] & @@@
Tally[DeleteCases[Diagonal[SmithDecomposition[aa[#]][[2]]],
0]]), " "] & /@ Range[40] // Column
• Thanks for the data! Any guesses as to the product of the nonzero invariant factors? Mar 27, 2023 at 18:00
• Powers of 2 seems to give particularly nice Smith normal forms, perhaps start with a conjecture there. Mar 28, 2023 at 5:24 | 3,270 | 7,939 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 79, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.851077 |
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# Brief Introduction to Vectors and Matrices by kylemangan
VIEWS: 76 PAGES: 18
• pg 1
``` CHAPTER 1
Brief Introduction to Vectors and Matrices
In this chapter, we will discuss some needed concepts found in in-
troductory course in linear algebra. We will introduce matrix, vector,
vector-valued function, and linear independency of a group of vectors
and vector-valued functions.
1. Vectors and Matrices
A matrix is a group of numbers(elements) that are arranged in
rows and columns. In general, an m × n matrix is a rectangular array
of mn numbers (or elements) arranged in m rows and n columns. If
m = n the matrix is called a square matrix. For example a 2×2 matrix
is
a11 a12
a21 a22
and an 3 × 3 matrix is
a11 a12 a13
a21 a22 a23
a31 a32 a33
Generally, we use bold phase letter, like A, to denote a matrix,
and lower case letters with subscripts, like aij , to denote element of
a matrix. Here aij would be the element at ith row and j th column.
So a11 is an element at 1st row and column. Sometime we use the
abbreviation A = (aij ) for a matrix with elements aij .
1.1. Special matrices. 0 denotes the zero matrix whose elements
are all zeroes. So 2 × 2 and 3 × 3 zero matrices are
0 0 0
0 0
and 0 0 0
0 0
0 0 0
Another special matrix is the identity matrix, denoted by I, a iden-
tity matrix is an matrix whose main diagonal elements are 1, and all
1
2 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
other elements are 0. So 2 × 2 and 3 × 3 zero matrices are
1 0 0
1 0 0 1
and 0
0 1
0 0 1
A vector is a matrix with one row or one column. In this chapter,
a vector is always a matrix with one column as
x1
x2
for a two-dimensional vector and
x1
x2
x3
for a three dimensional vector. Here the element has only one index
that denotes the row position (Sometimes we use different variable to
denote number in different position such as using
x
y
for a 2-dimensional vector). We use bold lower case, such as v, to
denote a vector.
1.2. Operations on Matrices. Arrange number in rectangular
fashion, as a matrix, itself is not something terribly interesting. The
most important advantage from that kind arrangement is that we can
define matrix addition, multiplication, and scalar multiplication.
Definition 1.1.
(i) Equality: Two matrix A = (aij ) and B = (bij ) are equal
if corresponding elements are equal, i.e. aij = bij .
(ii) Addition: If A = (aij ) and B = (bij ) and the sum of Aand
B is A + B = (cij ) = aij + bij .
(iii) Scalar Product: If A = (aij ) is matrix and k is num-
ber(scalar), the kA = (kaij ) is product of k and A.
From the above definition, we see that, to multiply a matrix by
a number k, we simply multiply each of its entries by k; to add two
matrices we just add their corresponding entries; A−B = A+(−1)B.
Example 1.1. Let
2 3
A=
−1 4
1. VECTORS AND MATRICES 3
and
0 5
B= ,
3 −4
find (a) A + B, (b) 3A, (c) 4A − B.
Solution
(a)
2 3 0 5
A+B = +
−1 4 3 −4
2+0 3+5 2 8
= =
−1 + 3 4 + (−4) 2 0
(b)
2 3 6 9
3A = 3 =
−1 4 −3 12
(c)
8 12 0 5 8 7
4A − B = − =
−4 16 3 −4 −7 20
The following fact lists all properties of matrix addition and scalar
product.
Theorem 1.1. Let A, bB, and C be matrices. Let a, b be scalars
(numbers). We have
(1) A + 0 = 0 + A = A, A − A = 0;
(2) A + B = B + A (commutativity);
(3) A + (B + C) = (A + B) + C, (ab)A = a(bA) (associa-
tivity);
(4) a(A+B) = aA+aB, (a+b)A = aA+bA (distributivity)
When we have a row vector and a column vector with the same
number of elements, we can define the dot product as
Definition 1.2. Dot Product:
y1
• in 2-dimension: Let x = x1 x2 and y = , the
y2
dot product of x and y is,
x · y = x1 y1 + x2 y2
4 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
y1
• in 3-dimension: Let x = x1 x2 x3 and y = y2 ,
y3
the dot product of x and y is,
x · y = x1 y1 + x2 y2 + x3 y3
Definition 1.3. Matrix product
Let A = (aij ) and B = (bij ), if the number of columns of A
is the same as number of rows of B, then the product of A and B is
given by AB = (cij ) where cij is dot product of ith row of A with j th
column of B.
Example 1.2. Let
2 3
A=
−1 4
and
0 5
B= ,
3 −4
find AB
Solution
2 3 0 5
AB =
−1 4 3 −4
2 · (0) + 3 · (3) 2 · 5 + 3 · (−4) 9 −2
= =
(−1) · (0) + 4 · 3 −1 · 5 + 4 · (−4) 12 −21
Notice, the first element of AB is 2 · (0) + 3 · (3) which is the dot
0
product of first row of A, 2 3 and first column of B,
3
The following fact gives properties of matrix product,
Theorem 1.2. Let A, B, C be three matrices and r be a scalar,
we have
• A(BC) = (AB)C, r(AB) = A(rB) (associativity)
• A(B + C) = AB + AC (distributivity)
Notice, in general AB = BA, that is for most of the times, AB
is not equal to BA.
Using the matrix notation and matrix product, we can write the
following system of equations
ax1 + bx2 = y1
cx1 + dx2 = y2
1. VECTORS AND MATRICES 5
as Ax = y with
a b
A= ,
c d
x1 y1
x= , and y = .
x2 y2
Definition 1.4. A square (ex. 2 × 2 or 3 × 3) matrix A is
invertible if there is a matrix A-1 such that AA-1 = A-1 A = I.
Theorem 1.3. Let
a b
A=
c d
be a 2 × 2 matrix, if A is invertible, we have
1 d −b
A-1 =
So if A is invertible, to solve Ax = y, we need to simply multiply
both sides with A-1 , that is
x = A-1 y.
Example 1.3. Solve the system of equation
3x1 − 4x2 = 2
−2x1 + 5x2 = 7
Solution The equation can be rewrite Ax = y with
3 −4
A= ,
−2 5
x1 2
x = , and y = . So in matrix form the system of
x2 7
equation is
3 −4 x1 2
= .
−2 5 x2 7
Now the inverse of A is
1 5 4
A-1 = ,
3(5) − (−2)(−4) 2 3
so the solution is
1 38
5 4 2
x = A-1 y = = 7
25
7 2 3 7 7
6 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
Example 1.4. Solve the system of equation
3x1 − 4x2 + 5x3 = 2
−2x1 + 5x2 = 7
x − 5x + 8x = −1
1 2 3
Solution The equation can be rewrite Ax = y with
3 −4 5
A = −2 5 0 ,
1 −5 8
x1 2
x = x2 , and y = 7 . So in matrix form the system of
x3 −1
equation, Ax = y, is
3 −4 5 x1 2
−2 5 0 x2 = 7 .
1 −5 8 x3 −1
It is a little harder to compute the inverse of a 3 × 3 matrix, we
will use Mathcad to solve the equation. Here is how to do it,
• Type A:[Ctrl][M] at a blank area to bring up the matrix
definition screen, put 3 in the both input boxes and click OK,
you will get a 3 × 3 matrix place holder like
A :=
Fill the entries of A in the corresponding position, using [Tab]
key to navigate among the place holders(or just click each one).
• Type b:[Ctrl][M] in another blank area, the matrix defini-
tion screen is up again. This time put 3 in the number of row
box, and 1 in the number of column box and click OK. You
will get b:= put the values of y in the corresponding
position.
• Type A^-1 *b= you will get the solution, which is,
154
81
175
81
80
81
1. VECTORS AND MATRICES 7
• Notice, by default, Mathcad will display the results as dec-
imal, you can double click on the result vector to change it
to fraction, after you double click the result you will have a
Figure 1. Format Result
The next example shows how we can determine the unknown con-
stants typically found in the initial value problems of system of differ-
ential equations.
Example 1.5. Let x1 (t) = C1 et + C2 e 2t and x2 = 2C1 et −
C2 e 2t . If x1 (0) = 2, x2 (0) = 3, find C1 and C2
Solution From x1 (t) = C1 et + C2 e 2t , set t = 0 we have
x1 (0) = C1 e0 + C2 e 2(0) = C1 + C2 .
Similarly, x2 (t) = 2C1 et − C2 e 2t , gives x2 (0) = 2C1 e( 0) −
C2 e 2(0) = 2C1 − C2 .
Together with x1 (0) = 2, x2 (0) = 3 we have the following system
of equations,
C1 + C2 = 1
2C1 − C2 = 3
Rewrite the equation in matrix form
1 1 C1 1
= ,
2 −1 C2 3
and using Mathcad we find the solution is
4
C1
= 3
C2 −13
8 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
So x1 (t) = 4 et − 1 e 2t and x2 = 8 et + 1 e
3 3 3 3
2t
. They are solution of
the following system of differential equations,
x1 (t) = −x1 (t) + x2 (t)
x2 (t) = 2x1
a b
1.3. Eigenvalues and Eigenvectors. If A = the de-
c d
a b
termined of A is defined as |A| = = ad − bc. For a 3 × 3
c d
matrix
a11 a12 a13
a21 a22 a23
a31 a32 a33
we can compute the matrix as
a11 a12 a13
a22 a23 a21 a23 a21 a22
a21 a22 a23 = a11 −a12 +a13 .
a32 a33 a31 a33 a31 a32
a31 a32 a33
In Mathcad , type the vertical bar — to bring up the absolute
evaluator | |, put the matrix in the place holder and press = to compute
the determinant. The following screen shot shows an example,
Figure 2. Compute determinant in Mathcad
The concepts of eigenvalue and eigenvector play an important role
in find solutions to system of differential equations.
Definition 1.5. We say λ is an eigenvalue of a matrix A (2 × 2
or 3 × 3) if the determinant
|A − λI| = 0.
An nonzero vector v is an eigenvector associated with λ if
Av = λv.
Remark 1.1.
1. VECTORS AND MATRICES 9
- The above definition of eigenvector and eigenvalue is valid for
any square matrix with n rows and columns.
- p(λ) = |A−λI| is a polynomial of degree n for n×n matrix
A, which is called the characteristic polynomial of A.
- If we view A as an transform that maps a vector x to Ax,
an eigenvector v defines a straight line passing origin that is
invariant under A.
- If v is an eigenvector then for and number s = 0, sv is also
an eigenvector. This is especially useful when using Mathcad
might need to remove the common factor of the component of
the vector to make it ”better.”
Computing eigenvalues and eigenvectors of a given matrix is quite
tedious, Mathcad provides two functions eigenvals() and eigenvecs()
to compute eigenvalues and eigenvectors of a matrix.
In Mathcad , eigenvecs(M) Returns a matrix containing the eigenvectors. The
nth column of the matrix returned is an eigenvector corresponding to the nth
eigenvalue returned by eigenvals.
The results of these functions by default is in decimal, you can
change it by using simplify key word as shown in the following dia-
gram.
(a) Find eigenvalue (b) Find eigenvector
Figure 3. Compute eigenvalue and eigenvector in Mathcad
√
√ 3
Notice, in the diagram, the eigenvalues are listed as vector
− 3
and the eigenvectors are listed in a matrix
√ √
1+ 3 -( 3-1)
√ 1 √ 1
(8+2 3) 2 (8-2 3) 2 ,
2 2
√ 1 √ 1
(8+2 3) 2 (8-2 3) 2
10 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
each column represents a eigenvector. Since multiplying an eigenvector
by a nonzero constant you still get an eigenvector, so we can simplify
√ √
1+ 3 1− 3
the eigenvectors as v 1 = , and v 2 =
2 2
Here is how to use Mathcad ,
• Define the matrix by type A:[Ctrl][M] and specify the row
and column number, fill the entries.
• type eigenvals(, you will get eigenvals( ) and in the place
holder type A.
• Click at end of the eigenvals(A) and press [Shift][Ctrl][.],
you will get eigenvals(A) → . In the place holder type in
key word simple. And click any area outside the box to get
result.
• Using the same procedure for find eigenvector using eigenvecs()
function.
2. Vector-valued functions
A vector-valued function over [a, b] is a function whose value is a
vector or matrix. For example the following functions are vector-valued
functions,
t
Example 2.1. (1) v(t) =
t2
1
(2) x = t2
et
1 t3 − 4t + 5
(3) A(t) =
0 sin(t)
2.1. Arithmetics of vector-valued function.
ing components.
• To multiply a vector-valued function by a scalar function to
to multiply each entry by the scalar function.
• To multiply a vector(matrix) valued function to another vector-
valued function is same as multiply a matrix with a vector.
The following example illustrate how to add/subtract two vector-valued
functions and how to multiply a vector-valued function by a scalar
function and how to apply a vector-valued function that is matrix to a
vector value function.
2. VECTOR-VALUED FUNCTIONS 11
t 1
Example 2.2. Suppose v(t) = t2 , x = t2 , and
t3 − 2 et
1 t3 − 4t + 5 1
A(t) = 0 2 sin(t) .
2 0 1
(a) Find v(t) + x(t);
(b) Let f (t) = et , find f (t)x(t);
(c) Find A(t)x(t)
Solution
(a)
t 1 t+1
v(t) + x(t) = t2 + t2 == 2t2 ;
t3 − 2 et 3
t −2+e t
(b)
1 et
f (t)x(t) = et t2 = t e
2 t ;
et e2t
(c)
1 t3 − 4t + 5 1 1
A(t)x(t) == 0 2 sin(t) t2
2 0 1 et
1 + t2 (t3 − 4t + 5) + et
= 2t2 + et sin(t)
2 + sin(t)
2.2. derivative and integrations of vector-valued functions.
• A vector-valued function is continuous if each of its entries
are continuous.
• A vector-valued function is differentiable if each of its entries
are differentiable.
• If v(t) is an vector-valued function, then the derivative dv (t) =
dt
v (t) of v(t) is a vector-valued function whose entries are the
derivative of corresponding entries of v(t). That is to find
derivative of a vector-valued function we just need to find de-
rivative of each of its component.
12 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
• The antiderivative v(t) dt of an vector-valued function v(t)
is a vector-valued function whose entries are the antiderivative
of corresponding entries of v(t).
3t2 − 5
Example 2.3. Find derivative of x(t) =
sin(t)
Solution
dx(t) d d
3t2 − 5 (3t2 − 5) 6t
x (t) = = = dt
d =
dt dt sin(t) dt
(sin(t)) cos(t)
3t2 − 5
Example 2.4. Find antiderivative of x(t) =
sin(t)
Solution
3t2 − 5 (3t2 − 5) dt
x(t) dt = dt =
sin(t) sin(t) dt
3 3
t − 5t + C1 t − 5t C1
= = +
− cos(t) + C2 − cos(t) C2
Theorem 2.1. Suppose v(t), x(t), A(t) are differentiable vector-
valued functions (A(t) is matrix), and f (t) is differentiable scalar
function. We have,
(1) Sum and Difference rule:
- [v(t) ± x(t)] = v (t) ± x (t),
- v(t) ± x(t) dt = v(t) dt ± x(t) dt.
(2) Product rule:
- [f (t)v(t)] = f (t)v(t) + f (t)v (t),
- [A(t)x(t)] = A (t)x(t) + A(t)x (t),
Using Mathcad to find derivative or antiderivative of a vector-
valued function using Mathcad , you need to find derivative or anti-
derivative component wise as shown in the following screen shot,
2. VECTOR-VALUED FUNCTIONS 13
Figure 4. Differentiate and integrate vector-valued function
14 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
Notice:
- Press [Shift][/]to get the derivative operator and press [Ctrl][I]
to get the antiderivative operator.
- To get dx(t)simplif y → you type dx(t) and press [Shift][Ctrl][.]
and type the key word simplify in the place holder before → .
- To execute symbolically (→ operator), just press [Ctrl][.]
3. Linearly independency
3.1. Linearly independency of vectors. Let x1 , x2 , · · · , xn
be n vectors, C1 , C2 , · · · , Cn are n scalars(numbers), the expression
C1 x1 + C2 x2 + · · · + Cn xn
is called a linear combination of vectors x1 , x2 , · · · , xn .
Definition 3.1. n vectors x1 , x2 , · · · , xn is linearly independent
if
C1 x1 + C2 x2 + · · · + Cn xn = 0
leads to C1 = 0, C2 = 0, · · · , Cn = 0.
A set of vectors are linearly dependent if they are not linearly in-
dependent.
• If 0 is one of x1 , x2 , · · · , xn , then they linearly dependent.
• Two nonzero vectors x and y are linearly dependent if and
only if x = sy for some s = 0.
• n nonzero vectors are linearly independent if one can be rep-
resented as linear combination of the others.
• Any three or more 2-dimensional vectors (vectors with two
entries) are linear dependent.
• Any four or more 3-dimensional(vectors with three entries)
vectors are linear dependent.
To determine if a given set of vectors are linearly independent,
create a matrix so that the row of the matrix are given vectors. Using
Mathcad function rref( ) to find the reduced echelon form of the
matrix, if the result contains one or more rows that are entirely zero
the vectors are linearly dependent, otherwise the vectors are linearly
independent.
3. LINEARLY INDEPENDENCY 15
2 0
Example 3.1. For x1 = 3 , x2 = 1 , and x3 =
4 −4
4
8 , we can form a matrix,
0
2 3 4
A = 0 1 −4 ,
4 8 0
apply rref(type rref and in the place holder type A, then press =),
1 0 8
rref (A) = 0 1 −4
0 0 0
.
We see that the vectors are linearly dependent as the last row is
entirely zero.
3.2. Linearly independency of functions. We can also define
linearly independency for a group of functions over an given inter-
val [a, b]. Let f1 , f2 , · · · , fn be n functions defined over [a, b],
C1 , C2 , · · · , Cn are n scalars(numbers), the expression
C1 f1 + C2 f2 + · · · + Cn fn
is called a linear combination of functions f1 , f2 , · · · , fn .
Definition 3.2. n functions f1 , f2 , · · · , fn is linearly indepen-
dent over [a, b]if
(1) C1 f1 + C2 f2 + · · · + Cn fn = 0 for all a≤t≤b
leads to C1 = 0, C2 = 0, · · · , Cn = 0.
A set of function are linearly dependent if they are not linearly
independent.
• If 0 function is one of f1 , f2 , · · · , fn , then they linearly de-
pendent.
• Two nonzero functions f (t) and g(t) are linearly dependent
over [a, b] if and only if f (t) = sg(t) for a constant s = 0
and all a ≤ t ≤ b, for example, f (t) = t and g(t) = 4t
are linearly dependent but f (t) = t and g(t) = 4t2 are not,
even f (0) = 4g(0) and f (1) = 4g(1).
• There are exists infinite many functions that are linearly in-
dependent. For example the set {1, t, t2 , t3 , · · · , tn , · · · } is
a linearly independent set.
16 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
Here are some sets of linearly independent functions that we en-
counter in solving a system of differential equations, assume k1 , k2 , · · · , kn
are different numbers,
- {tk1 , tk2 , · · · , tkn }.
- {ek1 t , ek1 t , ·, ek1 t }.
- {sin(k1 t), sin(k2 t), ·, sin(kn t)}.
- {cos(k1 t), cos(k2 t), ·, cos(kn t)}.
- The mixing of above sets.
- For each above set, when multiplying each element by an com-
mon nonzero factor, we get another linearly independent set.
The following screen shot displays a heuristic Mathcad function that
tries to determine if a given set of functions are linearly independent.
Figure 5. Calculus tool bar
One warning, the result of the program is not very reliable, the user
should check the result manually to confirm the result.
To manually check if an set of functions are linearly independent
on [a, b], one need to show that the only solutions are C1 = 0, C2 =
0, · · · , Cn = 0. if equation (1) holds for all t in [a, b], which requires
strong algebraic skill.
One method is to choose n different numbers {t1 , t2 , · · · , tn }
from [a, b] and using the functions to create an matrix, the compute
the determinant of the matrix A = (fi (tj )), if the determinant is not
zero, the functions are linearly independent, but if the determinant is
zero, it is inconclusive(most likely are linearly dependent).
Example 3.2. Determine if f1 (t) = t2 − 2t + 3, f2 (t) = 2t2 −
5t − 6, and f3 (t) = 5t2 − 11t + 4 are linearly independent.
3. LINEARLY INDEPENDENCY 17
Solution Choose t1 = −1, t2 = 0, and t3 = 1,
f1 (t1 ) f1 (t2 ) f1 (t3 ) 7 3 2
f2 (t1 ) f2 (t2 ) f2 (t3 ) = 1 −6 −9
f3 (t1 ) f3 (t2 ) f3 (t3 ) 20 4 −2
Compute the determinant,
7 3 2
1 −6 −9
= 50,
20 4
−2
so the functions are linearly independent.
Project
At beginning you should enter: Project title, your name, ss#, and
due date in the following format
Project One: Define and Graph Functions
John Doe
SS# 000-00-0000
Due: Mon. Nov. 23rd, 2003
You should format the text region so that the color of text is different
than math expression. You can choose color for text from Format–
>Style select normal and click modify, then change the settings for
font. You can do this for headings etc.
(1) Independent of functions as vectors
Goal: Familiar your self with the concept of linearly indepen-
dency.
• Use the Mathcad code provided at at the website www.unf.edu/∼mzhan/linear
to check if given set of functions are linearly independent
or not.
{sin(x), sin(2x), sin(3x)}
{t2 , 2t2 − 2t + 4, 3t, 6}
{et , tet , t3 et }
{e2t , e t , e 3t }
• Using algebraic arguments or reasoning to verify the con-
18 1. BRIEF INTRODUCTION TO VECTORS AND MATRICES
(2) Condition Number In solving Ax = b, one number is very
important, it is called the condition number, which can be de-
fined as C(A) = | s , where λs is the eigenvalue with smallest
l
absolute value and lambdal is the eigenvalue with largest ab-
solute value, if C(A) is too large or too small, a little change
in b will result in a large in the solution x. We say the system
1 1 1
2 3
Ax = b is not stable. Now if A = 1 3 1 2
1
4
1 1 1
3 4 5
• Find all eigenvalues, all eigenvectors, and C(A).
1
• Find solution of Ax = b if b = 1
1
1
• Change b a little to b = 1 we get different solu-
1.1
tion, which component of the new solution change most?
The change of the third component if 10% what is the
percentage change of the most changed component?
Note:
• Our definition of condition number is not accurate, the
1
true definition is C(A) = where · is a
A A-1
given norm (metric).
• Mathcad provides three functions cond1(A), cond2(A)
and condi(A) in compute condition number for A in
different metric.
```
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4.14. Timing Cryptographic Primitives
Problem
You want to compare the efficiency of two similar cryptographic primitives and would like to ensure that you do so in a fair manner.
Solution
Time operations by calculating how many cycles it takes to process each byte, so that you can compare numbers across processors of different speeds more fairly.
Focus on the expected average case, best case, and worst case.
Discussion
When you’re looking at timing information, you usually have one of two motivations: either you’re interested in comparing the performance of two algorithms, or you’d like to get a sense of how much data you’ll actually be able to pump through a particular machine.
Measuring bytes per second is a useful thing when you’re comparing the performance of multiple algorithms on a single box, but it gives no real indication of performance on other machines. Therefore, cryptographers prefer to measure how many processor clock cycles it takes to process each byte, because doing so allows for comparisons that are more widely applicable. For example, such comparisons will generally hold fast on the same line of processors running at different speeds.
If you’re directly comparing the speed of an algorithm on a 2GHz Pentium 4 against the published speed of the same algorithm run on a 800 MHz Pentium 3, the first one will always be faster when measured in bytes per second. However, if you convert the numbers from bytes per second to cycles per byte, you’ll see that, if you run the same implementation of an algorithm on a P3 and a P4, the P3 will generally be faster by 25% or so, just because instructions on a P4 take longer to execute on average than they do on a P3.
If you know the speed of an algorithm in bytes per second, you can calculate the number of cycles per byte simply by dividing by the clock speed in hertz (giving you bytes per cycle) and taking the reciprocal (getting cycles per byte). If you know the speed measured in gigabytes per second, you can divide by the clock speed in gigahertz, then take the reciprocal. For example, you can process data at 0.2 gigabytes per second on a 3 GHz CPU as follows:
.2/3 = 0.066666666666666666 (bytes processed per cycle) 1/0.066666666666666666 = 15.0 cycles per byte
For many different reasons, it can be fairly difficult to get timing numbers that are completely accurate. Often, internal clocks that the programmer can read are somewhat asynchronous from the core processor clock. More significantly, there’s often significant overhead that can be included in timing results, such as the cost of context switches and sometimes timing overhead.
Tip
Some CPUs, such as AMD’s Athlon, are advertised such that the actual clock speed is not obvious. For example, the Athlon 2000 runs at roughly 1666 MHz, significantly less than the 2000 MHz one might suspect.
Generally, you’ll want to find out how quickly a primitive or algorithm can process a fixed amount of data, and you’d like to know how well it does that in a real-world environment. For that reason, you generally shouldn’t worry much about subtracting out things that aren’t relevant to the underlying algorithm, such as context switches and procedure call overhead. Instead, we recommend running the algorithm many times and averaging the total time to give a good indication of overall performance.
In the following sections we’ll discuss timing basics, then look at the particulars of timing cryptographic code.
Timing basics
You need to be able to record the current time with as much precision as possible. On a modern x86 machine, it’s somewhat common to see people using inline assembly to call the RDTSC instruction directly, which returns the number of clock cycles since boot as a 64-bit value. For example, here’s some inline assembly for GCC on 32-bit x86 platforms (only!) that reads the counter, placing it into a 64-bit unsigned ```long long``` that you pass in by address:
```#define current_stamp(a) asm volatile("rdtsc" : "=a"(((unsigned int *)(a))[0]),\
"=d"(((unsigned int *)a)[1]))```
The following program uses the above macro to return the number of ticks since boot:
```#include <stdio.h>
int main(int argc, char *argv[ ]) {
spc_uint64_t x;
current_stamp(&x);
printf("%lld ticks since boot (when I read the clock).\n", x);
return 0;
}```
RDTSC is fairly accurate, although processor pipelining issues can lead to this technique’s being a few cycles off, but this is rarely a big deal.
On Windows machines, you can read the same thing using ```QueryPerformanceCounter( )``` , which takes a pointer to a 64-bit integer (the `LARGE_INTEGER` or `_ _int64` type).
You can get fairly accurate timing just by subtracting two subsequent calls to ```current_stamp( )``` . For example, you can time how long an empty `for` loop with 10,000 iterations takes:
```#include <stdio.h>
int main(int argc, char *argv[ ]) {
spc_uint64_t start, finish, diff;
volatile int i;
current_stamp(&start);
for (i = 0; i < 10000; i++);
current_stamp(&finish);
diff = finish - start;
printf("That loop took %lld cycles.\n", diff);
return 0;
}```
On an Athlon XP, compiling with GCC 2.95.4, the previous code will consistently give 43-44 cycles without optimization turned on and 37-38 cycles with optimization turned on. Generally, if `i` is declared volatile, the compiler won’t eliminate the loop, even when it can figure out that there are no side effects.
Note that you can expect some minimal overhead in gathering the timestamp to begin with. You can calculate the fixed timing overhead by timing nothing:
```int main(int argc, char *argv[ ]) {
spc_uint64_t start, finish, diff;
current_stamp(&start);
current_stamp(&finish);
diff = finish - start;
printf("Timing overhead takes %lld cycles.\n", diff);
return 0;
}```
On an Athlon XP, the overhead is usually reported as 0 cycles and occasionally as 1 cycle. This isn’t really accurate, because the two store operations in the first timestamp call take about 2 to 4 cycles. The problem is largely due to pipelining and other complex architectural issues, and it is hard to work around. You can explicitly introduce pipeline stalls, but we’ve found that doesn’t always work as well as expected. One thing to do is to time the processing of a large amount of data. Even then, you will get variances in timing because of things not under your control, such as context switches. In short, you can get within a few cycles of the truth, and beyond that you’ll probably have to take some sort of average.
A more portable but less accurate way of getting timing information on Unix-based platforms is to ask the operating system for the clock using the `gettimeofday( )` function. The resolution varies depending on your underlying hardware, but it’s usually very good. It can be implemented using RDTSC but might have additional overhead. Nonetheless, on most operating systems, `gettimeofday( )` is very accurate.
Here’s a macro that will use ```gettimeofday( )``` to put the number of microseconds since January 1, 1970 into an unsigned 64-bit integer (if your compiler does not support a 64-bit integer type, you’ll have to store the two 32-bit values separately and diff them properly; see below).
```#include <sys/time.h>
#define current_time_as_int64(a) { \
struct timeval t; \
gettimeofday(&t, 0); \
*a = (spc_uint64_t)((t.tv_sec * 1000000) + t.tv_usec); \
}```
Attackers can often force the worst-case performance for functionality with well-chosen inputs. Therefore, you should always be sure to determine the worst-case performance characteristics of whatever it is you’re doing, and plan accordingly.
Warning
The `gettimeofday( )` -based macro does not compute the same thing the RDTSC version does! The former returns the number of microseconds elapsed, while the latter returns the number of cycles elapsed.
You’ll usually be interested in the number of seconds elapsed. Therefore, you’ll need to convert the result of the `gettimeofday( )` call to a number of cycles. To perform this conversion, divide by the clock speed, represented as a floating-point number in gigahertz.
Because you care about elapsed time, you’ll want to subtract the starting time from the ending time to come up with elapsed time. You can transform a per-second representation to a per-cycle representation once you’ve calculated the total running time by subtracting the start from the end. Here’s a function to do both, which requires you to define a constant with your clock speed in gigahertz:
```#define MY_GHZ 1.6666666666666667 /* We're using an Athlon XP 2000 */
spc_uint64_t get_cycle_count(spc_uint64_t start, spc_uint64_t end) {
return (spc_uint64_t)((end - start) / (doublt)MY_GHZ);
}```
Timing cryptographic code
When timing cryptographic primitives, you’ll generally want to know how many cycles it takes to process a byte, on average. That’s easy: just divide the number of bytes you process by the number of cycles it takes to process. If you wish, you can remove overhead from the cycle count, such as timing overhead (e.g., a loop).
One important thing to note about timing cryptographic code is that some types of algorithms have different performance characteristics as they process more data. That is, they can be dominated by per-message overhead costs for small message sizes. For example, most hash functions such as SHA1 are significantly slower (per byte) for small messages than they are for large messages.
You need to figure out whether you care about optimal performance or average-case performance. Most often, it will be the latter. For example, if you are comparing the speeds of SHA1 and some other cryptographic hash function such as RIPEMD-160, you should ask yourself what range of message sizes you expect to see and test for values sampled throughout that range.
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###### How does this sample mean compare to the distribution of sample means? In other words, how does the height of the girls in the sample compare to the height of girls in th general population? Answer
January 17, 2018
PSYC 354 Homework 1 Introduction to Statistics
When submitting this file, be sure the filename includes your full name, course and section.
Be sure you have reviewed this module/week’s lesson and presentations before proceeding to the
homework exercises. Number all responses. Review the “Homework Instructions: General”
document for an example of how homework assignments must look.
All Questions
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Fill in the blank with the appropriate word or words.
1. A statistic that uses sample data to make general estimates about the larger population is statistic.
2. A statistic that summarizes a group of numbers is a(n) statistic.
3. A(n) is a set of observations drawn from the larger group of interest.
4. The large group about which you want to know something is called .
5. A five-star rating system for movies is a variable at the level of measurement.
6. A person’s religious preference is a variable at the level of measurement.
7. A(n) variable is a variable that meets the criteria for an interval and ratio variable.
8. Discrete values that the independent variable can take on are called the of the independent variable.
9. A variable that is manipulated to determine its effects on another variable is a(n)
variable.
10. A(n) variable makes it impossible to determine whether the independent variable is the cause of changes in the dependent variable.
11. is the process of drawing conclusions about whether or not a particular relation between variables is supported by the data.
12. The hallmark of a(n) is the random assignment of participants to conditions
so that cause and effect can be discovered.
1. When each participant has the same chance of being assigned to the various levels of the manipulation, they are said to be to conditions.
2. Each participant experiences all levels of the independent variable in a(n) design.
3. Participants experience only one level of the independent variable in a(n) design.
Question 1a-1b
A health psychologists decides to study the effects of living region (rural, suburban, urban) on the number of extra-curricular activities of residents.
1-a) What would the independent variable be in this study?
1-b) What would the dependent variable be in this study?
Question 2 (a-d)
Hsee and Tang (2007) reported the results of a study in which 195 college students completed a happiness
scale (from 1 to 7) one day in class. On this scale, 1 corresponded to very unhappy and 7 to very happy. On average, the students rated their happiness as 6.18. Identify each of the following for this study.
Questions 3-6
3) In which window do you enter data in SPSS?
4) Which window displays the results of your analysis in SPSS?
5) Which SPSS main menu would you use to choose a particular statistical test?
6) If you wanted to custom define a variable in SPSS, which window would you open?
Question 7
Suppose you have administered a personality inventory to 12 people and have recorded their score on
each item in SPSS. The personality inventory has 8 items, so the SPSS file has 8 variables representing each item (item1, item2, etc.). You want to compute a total score for each participant that represents the
sum of items 1–8.
7a) Which main SPSS menu contains the option for computing a variable like this?
7b) According to the presentation and to Lesson 19 in Green and Salkind, what could you type into the “Numeric Expression” text box in order to have SPSS add items 1–8 together? (There is more than one way to do this.)
Submit Homework 1 by 11:59 p.m. (ET) on Monday of Module/Week 1. Remember to name file appropriately.
PSYC 354 Homework 2 Frequency Tables and Graphs
All Questions
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Use the following table to answer Question 1. This table depicts the scores of 83 students on an exam worth 65 points.
1) Use the information in the table to determine the percentages for each interval. Depending on your rounding decisions, these may or may not add up to exactly 100% but should be very close.
Table: Grouped Frequency Table
Exam score Frequency
60–62 3
57–59 9
54–56 21
51–53 18
48–50 14
45–47 10
42–44 8
TOTAL 83
2) When constructing a histogram and labeling the x- and y-axes, the lowest number on each axis should ideally be ……
3) A frequency distribution that is bell-shaped, symmetrical, and unimodal is
4) A frequency distribution that has a tail trailing off to the right of the distribution is
5) A frequency distribution of ages of residents at a senior citizen home is clustered around 83 with a long tail to the left. This distribution is
6) When a variable cannot take on values above a certain level, this is known as a(n) effect.
7) A grouped frequency table has the following intervals: 30–44, 45–59, and 60–74. If converted into a histogram, what would the midpoints be?
8) Do the data below show a linear relation, non-linear relation, or no relation at all?
9) Do the data below show a linear relation, non-linear relation, or no relation at all?
Part I:
Question 10a- 10e
• Read the introduction and click on different “subway lines” to see how the interactive graph works.
10-a) In which of the four boroughs is the median household income highest?
10-b) Click on the “A” line. Does the line graph for Manhattan show high or low variability? What does this
level of variability tell us about the household income in this area of Manhattan?
10-c) Click on line 2. Which borough (not a street) shows the least variability in median household income?
10-d) On line 2, find the following two subway stops: Park Place (the first of the highest Manhattan stops) and E 180 St. (one of the lowest Bronx stops, located about halfway across the BRX section). What is the difference (calculate) between the median household incomes of the two areas?
10-e) Click on the “D” line. Which subway stop in Brooklyn seems to be an outlier?
Part II: SPSS Analysis
Green and Salkind, Lesson 20
• Open the “Lesson 20 Exercise File 1” document (found in the course’s Assignment Instructions folder) in order
to complete these exercises.
• Always use the Blackboard files instead of the files on the Green and Salkind website as some files have been modified for the purposes of this course.
• Reminder: For Exercise 1, be sure to paste in the SPSS output and write out the answers for A, B, and C beneath it.
Part III: SPSS Data Entry and Analysis
The steps will be the same in Part III as the ones you have been practicing in Part I of the assignment; the only difference is that you are now responsible for creating the data file as well. Remember to do the following:
• Name and define your variables under the “Variable View,” then return to the “Data View” to enter the data;
and
• Paste all SPSS output and graphs into your homework file at the appropriate place.
Part III: Questions 1a-1c
PSYC 354 Homework 3 Central Tendency and Variability
Questions 1–7
Questions 1-4
Fill in the highlighted blanks to answer/complete the statements.
The arithmetic average of a set of numbers is the mean.
2) What is an outlier?
3) Which widely-used measure describes the typical amount or distance a score deviates from the mean in any given distribution?
3. Standard Deviation (-2) – correction from professor
4) What is the mathematical relationship between variance and standard deviation?
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Question 5
For the following terms, write the equivalent mathematical symbol (letter or letters) for the sample statistic.
Question 6a-6c
These are the winning percentages for 11 baseball players for each one’s best 4-year pitching performance:
6-c ) Compare the mean and median. Does the difference between them suggest that the data are skewed very much?
6-a) What is the mean of the scores? (Compute this using a calculator):
6-b) What is the median of these scores?
Explain.
Question 7
Recall the interactive graph from Homework 2 depicting household income and location in New
York City.
Based on this module/week’s reading, why do you think the author of this graph chose the
median household income instead of the mean household income to describe central tendency?
Part II: SPSS Analysis
Green and Salkind Course Text, Lesson 21: Exercises 1, 4, 7, and 8
• Questions 1 and 4
o Green and Salkind Text, Lesson 21
o Lesson 21 Exercise File 1 (located under course’s Assignment Instructions folder)
Part II:
Exercises 1a-1d and Exercise 4
Green and Salkind, Lesson 21, Exercises 1 and 4
Open “Lesson 21 Exercise File 1” in order to complete these exercises.
Part III: SPSS Data Entry and Analysis
The steps will be the same as the ones you have been practicing in Part I of the assignment; the only difference is that you are now responsible for creating the data file, as well. Remember to do the following:
a) Name and define your variables under the Variable View, then return to the Data View to enter the data; and
b) Paste all SPSS output and graphs into your homework file at the appropriate place.
Part III: Questions 1-4
The following data are taken from the Bureau of Labor Statistics surveys from the years 2009–2013. They represent the average weekly pay for wage and salary earners measured at 4 different quarters each year and broken down by gender.
……………………………………………..
PSYC 354 Homework 4
Sampling and Probability
When submitting this file, be sure the filename includes your full name, course and section. Example: HW4_JohnDoe_354B01
Be sure you have reviewed this module/week’s lesson and presentations along with the practice
data analysis before proceeding to the homework exercises. Complete all analyses in SPSS, then copy and paste your output and graphs into your homework document file. Answer any written questions (such as the text-based questions or the APA Participants section) in the appropriate place within the same file.
Questions 1–15
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Questions 1-10
Fill in the highlighted blanks to answer/complete the statements.
1) A(n) random sample occurs when everyone in the population has the same chance of being selected.
2) The ability to apply research findings to contexts or samples other than the one studied is called generalizability.
3) The duplication of the results of a research study in a different context or with a different sample is called replication.
4) Your own estimate of the likelihood that you will uphold your New Year’s resolution is known as a(n) personal probability.
5) Expected relative-frequency probability is computed as the number of successes divided by the number of trials.
6) The experimental group is the group receiving the intervention or treatment of interest.
7) A statement that 2 populations are different from one another is a(n) research hypothesis.
8) If you find that the groups you studied differed from each other more than would be expected by chance alone, you reject the null hypothesis.
9) When you reject the null hypothesis but the null hypothesis is in fact true, you have made a(n) Type I error.
10) When you fail to reject the null hypothesis but the null hypothesis is false, you have made a(n) Type II error.
Question 11
Researchers were interested in whether touch therapy improves the weight gain of preterm infants and
compared the weight gain over a 3-week period of infants receiving touch therapy to the weight gain of infants not receiving touch therapy.
11-a) What is the likely null hypothesis for this experiment?
Touch therapy does not improve the weight gain of preterm infants.
11-b) What is the likely research hypothesis for this experiment?
Touch therapy does improve the weight gain of preterm infants.
Question 12-15
Complete the Nolan and Heinzen end-of-chapter Exercises 5.26, 5.27, 5.28, and 5.29.
12) Exercises 5.26
71/489=0.1452
13) Exercises 5.27
8/266=0.0301
14) Exercises 5.28
a)
1.73%
b)
80%
c)
37.19%
15) Exercises 5.29
a)
0.627
b)
0.003
c)
0.042
Part II & Part III
There is no new SPSS material this module/week. No questions for Part II or III
.
Part IV: Cumulative
Data provided below for Questions 1 and 2.
The final grades for students in a freshman seminar are shown in the left-hand column. Enter the data into a new SPSS data file.
• Create the appropriate SPSS graph to show the distribution of students’ grades.
• Compute descriptive statistics for this data.
• Which measure of central tendency would you use to describe this data?
• Why?
99.00
93.00
89.00
86.00
82.00
79.00
78.00
76.00
75.00
75.00
73.00
72.00
69.00
65.00
63.00
55.00
50.00
1-a) Create the appropriate SPSS graph to show the distribution of students’ grades.
Page 3 of 6
1-b) Compute descriptive statistics for this data.
Which measure of central tendency would you use to describe this data? Why?
Page 4 of 6
Descriptives
Statistic
Std. Error
Mean
95% Confidence Interval for Lower Bound
Mean Upper Bound
5% Trimmed Mean
Median
Variance
Grade Std. Deviation Minimum Maximum Range
Interquartile Range
Skewness
Kurtosis
75.24
3.097
68.67
81.80
75.32
75.00
163.066
12.770
50
99
49
17
-.146
.550
.037
1.063
Mean
Explanation/Justification:
You generally use the mean unless there are outliers or a small amount of observations. Neither is the case for this. Also, the mean and median are only 0.24 points different and the mode is the same as the median.
Page 5 of 6
Explanation/Justification: You would use a bar graph because the independent variable is nominal and the dependent variable is scale.
2-b) What level of measurement (nom., ord., scale) is the variable “City”?
What level of measurement is the variable “Livability Rating”?
Nominal
Scale 2.b. ratings are ordinal (-2) – correction from professor
Submit Homework 4 by 11:59 p.m. (ET) on Monday of Module/Week 4. Remember to name file appropriately.
PSYC 354 Homework 5
Z-Scores
Questions 1–9
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
What are always the mean and standard deviation of the z-distribution? (2 pts)
2) Define the central limit theorem. (1 pt)
3)
Fill in the blanks (1 pt): A z-score is based on a distribution of equally likely events, while a z- statistic is based on a distribution of sample percentage or average.
Part I: Questions 4-8
Module 5 Lesson 21 Exercise File 1
Part II: SPSS Analysis
Open the “Lesson 21 Exercise File 1” document (found in the course’s Assignment Instructions folder) in order to complete these exercises.
………………………………………………………………..
PSYC 354 Homework 6 Percentiles and Hypothesis Testing with Z-Tests
Questions 1–4
Part I: Concepts
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Part I: Questions 1-8
End-of-chapter problems:
Exercise 7.8 (1 pt)
2) Exercise 7.18 (1 pt each)
3) Exercise 7.20 (1 pt each)
More Questions Included
PSYC 354 Homework 7
Confidence Intervals, Effect Size, and Statistical Power
Part I: Concepts
Questions 1–8
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Part I: Questions 1a-1e
Fill in the highlighted blanks with the best word or words.
Part I: Questions 2-5
End-of-chapter problems:
Complete the following Nolan and Heinzen end-of-chapter exercises for Chapter 8: 8.15, 8.22, 8.26, and 8.40.
If applicable, remember to show work in your homework document to receive partial credit.
More Questions Included
PSYC 354 Homework 8
Single-Sample T-Test
When submitting this file, be sure the filename includes your full name, course and section. Example:
Be sure you have reviewed this module/week’s lesson and presentations along with the practice data analysis before proceeding to the homework exercises. Complete all analyses in SPSS, then copy and paste your output and graphs into your homework document file. Answer any written questions (such as the text-based questions or the APA Participants section) in the appropriate place within the same file.
Part I: Concepts
Questions 1–3
These questions are based on the Nolan and Heinzen reading and end-of-chapter questions.
Part I: Questions 1a-1g
Fill in the highlighted blanks with the best word or words.
Part II: SPSS Analysis
Module/Week 8 Exercise File 1
Open the “Module/Week 8 Exercise File 1” document (found in the course’s Assignment Instructions folder) in order to complete these exercises.
Part II:
Exercises 1-3
Use file: Module/Week 8 Exercise File 1
End-of-chapter exercise 9.37 in Nolan and Heinzen text
Part III: SPSS Data Entry and Analysis
Data provided below.
Part IV: Cumulative
Data provided below for respective questions.
PSYC 354 Review Test Exam 4
Attempt Score 97.5 points scored
Instructions
Time limit: 1 hour and 30 minutes
40 multiple¬choice questions
Open¬book/open¬notes
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1 Imagine that the confidence interval around a group’s mean is [980, 1160], what is the mean?
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2 A negatively skewed distribution would most likely violate which assumption?
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3 A researcher conducts a single¬sample t test and finds statistical significance at the 0.01 level.
Question
4 The formula “ ” is used to represent the:
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5 Cohen’s d is one measure of:
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6 The single¬sample t test compares a sample mean to a population mean when:
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do we use to compare a sample to the population?
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8 Why are effect sizes rather than test statistics used when comparing study results?
Question
9 What falls within the 95% confidence interval?
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10 We calculate a statistical power and find that it is 0.93. This means that if the null hypothesis is , we have a % chance of rejecting the null hypothesis.
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11 A confidence interval is a(n. that includes the population mean after repeatedly sampling.
Question
12 One of the roles of the researcher performing a meta¬analysis is to:
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13 The statement “The findings based on a sample of 1000 participants were statistically significant, providing evidence for our hypothesis” would be strengthened by:
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14 We can increase statistical power with each of the following except:
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Question
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15 Imagine that a study of memory and aging finds that younger participants correctly recall 55% of studied words, older participants correctly recall 42% of studied words, and the size of this effect is Cohen’s d = 0.49. This effect size indicates that the memory performance of:
Question
16 That a results is statistically significant does not mean that it is
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17 When we report that something is statistically significant, it means that
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18 Increasing sample size:
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19 Many companies that manufacture light bulbs advertise their 60¬watt bulbs as having an average life of 1000 hours. A cynical consumer bought 30 bulbs and burned them until they failed. He found that they burned for an average of M = 1233, with a standard deviation of s =
232.06. What statistical test would this consumer use to determine whether the average burn time of light bulbs differs significantly from that advertised?
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20 The correct formula for effect size using Cohen’s d for a single¬sample t test is:
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21 In statistics, what does “homogeneous” mean?
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22 According to Cohen’s convention, a value of is a small effect size.
Question
23 Meta¬analysis involves:
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Question
24 The practical use of statistical power is that it informs you the researcher:
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25 Alpha refers to:
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26 The larger the effect size, the:
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27 Candidate #2 received 32% of the votes and candidate #3 received 14% of the votes, with a margin of error of 6%. What does this information tell us about the popularity of these candidates?
Question
28 Statistical convention for the minimal acceptable power is:
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29 The numerator (top portion) of the ratio for calculating all the t statistics contains:
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30 Identify the formula for the single¬sample t test.
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31 The second step in conducting the single¬sample t test involves
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32 Which of the following reports of statistical results are in appropriate APA format?
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33 According to Cohen’s convention, a value of is a large effect size.
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34 Effect size assesses the degree to which two:
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35 The formula for the degrees of freedom for the single¬sample t test is:
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36 Using the following information, convert the z score to a raw score: z = ¬2.3, mu = 15.5, sigma = 4.
Question
37 In a distribution of Z scores, the mean is always
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38 A population is normally distributed with μ = 30, σ =
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10, and our sample N = 25. What are the mean and standard error of the distribution of
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39 A population is normally distributed with μ = 50, σ =
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40 If the mean of a test is 13 and its standard deviation is 3, the Z score for a person with a raw score of 7 is | 5,396 | 22,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-40 | latest | en | 0.924077 |
https://us.metamath.org/mpeuni/coshalfpip.html | 1,721,748,908,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00235.warc.gz | 509,145,030 | 8,058 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > coshalfpip Structured version Visualization version GIF version
Theorem coshalfpip 25087
Description: The cosine of π / 2 plus a number. (Contributed by Paul Chapman, 24-Jan-2008.)
Assertion
Ref Expression
coshalfpip (𝐴 ∈ ℂ → (cos‘((π / 2) + 𝐴)) = -(sin‘𝐴))
Proof of Theorem coshalfpip
StepHypRef Expression
1 coshalfpi 25062 . . . . 5 (cos‘(π / 2)) = 0
21oveq1i 7145 . . . 4 ((cos‘(π / 2)) · (cos‘𝐴)) = (0 · (cos‘𝐴))
3 coscl 15472 . . . . 5 (𝐴 ∈ ℂ → (cos‘𝐴) ∈ ℂ)
43mul02d 10827 . . . 4 (𝐴 ∈ ℂ → (0 · (cos‘𝐴)) = 0)
52, 4syl5eq 2845 . . 3 (𝐴 ∈ ℂ → ((cos‘(π / 2)) · (cos‘𝐴)) = 0)
6 sinhalfpi 25061 . . . . 5 (sin‘(π / 2)) = 1
76oveq1i 7145 . . . 4 ((sin‘(π / 2)) · (sin‘𝐴)) = (1 · (sin‘𝐴))
8 sincl 15471 . . . . 5 (𝐴 ∈ ℂ → (sin‘𝐴) ∈ ℂ)
98mulid2d 10648 . . . 4 (𝐴 ∈ ℂ → (1 · (sin‘𝐴)) = (sin‘𝐴))
107, 9syl5eq 2845 . . 3 (𝐴 ∈ ℂ → ((sin‘(π / 2)) · (sin‘𝐴)) = (sin‘𝐴))
115, 10oveq12d 7153 . 2 (𝐴 ∈ ℂ → (((cos‘(π / 2)) · (cos‘𝐴)) − ((sin‘(π / 2)) · (sin‘𝐴))) = (0 − (sin‘𝐴)))
12 halfpire 25057 . . . 4 (π / 2) ∈ ℝ
1312recni 10644 . . 3 (π / 2) ∈ ℂ
14 cosadd 15510 . . 3 (((π / 2) ∈ ℂ ∧ 𝐴 ∈ ℂ) → (cos‘((π / 2) + 𝐴)) = (((cos‘(π / 2)) · (cos‘𝐴)) − ((sin‘(π / 2)) · (sin‘𝐴))))
1513, 14mpan 689 . 2 (𝐴 ∈ ℂ → (cos‘((π / 2) + 𝐴)) = (((cos‘(π / 2)) · (cos‘𝐴)) − ((sin‘(π / 2)) · (sin‘𝐴))))
16 df-neg 10862 . . 3 -(sin‘𝐴) = (0 − (sin‘𝐴))
1716a1i 11 . 2 (𝐴 ∈ ℂ → -(sin‘𝐴) = (0 − (sin‘𝐴)))
1811, 15, 173eqtr4d 2843 1 (𝐴 ∈ ℂ → (cos‘((π / 2) + 𝐴)) = -(sin‘𝐴))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2111 ‘cfv 6324 (class class class)co 7135 ℂcc 10524 0cc0 10526 1c1 10527 + caddc 10529 · cmul 10531 − cmin 10859 -cneg 10860 / cdiv 11286 2c2 11680 sincsin 15409 cosccos 15410 πcpi 15412 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2770 ax-rep 5154 ax-sep 5167 ax-nul 5174 ax-pow 5231 ax-pr 5295 ax-un 7441 ax-inf2 9088 ax-cnex 10582 ax-resscn 10583 ax-1cn 10584 ax-icn 10585 ax-addcl 10586 ax-addrcl 10587 ax-mulcl 10588 ax-mulrcl 10589 ax-mulcom 10590 ax-addass 10591 ax-mulass 10592 ax-distr 10593 ax-i2m1 10594 ax-1ne0 10595 ax-1rid 10596 ax-rnegex 10597 ax-rrecex 10598 ax-cnre 10599 ax-pre-lttri 10600 ax-pre-lttrn 10601 ax-pre-ltadd 10602 ax-pre-mulgt0 10603 ax-pre-sup 10604 ax-addf 10605 ax-mulf 10606 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2598 df-eu 2629 df-clab 2777 df-cleq 2791 df-clel 2870 df-nfc 2938 df-ne 2988 df-nel 3092 df-ral 3111 df-rex 3112 df-reu 3113 df-rmo 3114 df-rab 3115 df-v 3443 df-sbc 3721 df-csb 3829 df-dif 3884 df-un 3886 df-in 3888 df-ss 3898 df-pss 3900 df-nul 4244 df-if 4426 df-pw 4499 df-sn 4526 df-pr 4528 df-tp 4530 df-op 4532 df-uni 4801 df-int 4839 df-iun 4883 df-iin 4884 df-br 5031 df-opab 5093 df-mpt 5111 df-tr 5137 df-id 5425 df-eprel 5430 df-po 5438 df-so 5439 df-fr 5478 df-se 5479 df-we 5480 df-xp 5525 df-rel 5526 df-cnv 5527 df-co 5528 df-dm 5529 df-rn 5530 df-res 5531 df-ima 5532 df-pred 6116 df-ord 6162 df-on 6163 df-lim 6164 df-suc 6165 df-iota 6283 df-fun 6326 df-fn 6327 df-f 6328 df-f1 6329 df-fo 6330 df-f1o 6331 df-fv 6332 df-isom 6333 df-riota 7093 df-ov 7138 df-oprab 7139 df-mpo 7140 df-of 7389 df-om 7561 df-1st 7671 df-2nd 7672 df-supp 7814 df-wrecs 7930 df-recs 7991 df-rdg 8029 df-1o 8085 df-2o 8086 df-oadd 8089 df-er 8272 df-map 8391 df-pm 8392 df-ixp 8445 df-en 8493 df-dom 8494 df-sdom 8495 df-fin 8496 df-fsupp 8818 df-fi 8859 df-sup 8890 df-inf 8891 df-oi 8958 df-card 9352 df-pnf 10666 df-mnf 10667 df-xr 10668 df-ltxr 10669 df-le 10670 df-sub 10861 df-neg 10862 df-div 11287 df-nn 11626 df-2 11688 df-3 11689 df-4 11690 df-5 11691 df-6 11692 df-7 11693 df-8 11694 df-9 11695 df-n0 11886 df-z 11970 df-dec 12087 df-uz 12232 df-q 12337 df-rp 12378 df-xneg 12495 df-xadd 12496 df-xmul 12497 df-ioo 12730 df-ioc 12731 df-ico 12732 df-icc 12733 df-fz 12886 df-fzo 13029 df-fl 13157 df-seq 13365 df-exp 13426 df-fac 13630 df-bc 13659 df-hash 13687 df-shft 14418 df-cj 14450 df-re 14451 df-im 14452 df-sqrt 14586 df-abs 14587 df-limsup 14820 df-clim 14837 df-rlim 14838 df-sum 15035 df-ef 15413 df-sin 15415 df-cos 15416 df-pi 15418 df-struct 16477 df-ndx 16478 df-slot 16479 df-base 16481 df-sets 16482 df-ress 16483 df-plusg 16570 df-mulr 16571 df-starv 16572 df-sca 16573 df-vsca 16574 df-ip 16575 df-tset 16576 df-ple 16577 df-ds 16579 df-unif 16580 df-hom 16581 df-cco 16582 df-rest 16688 df-topn 16689 df-0g 16707 df-gsum 16708 df-topgen 16709 df-pt 16710 df-prds 16713 df-xrs 16767 df-qtop 16772 df-imas 16773 df-xps 16775 df-mre 16849 df-mrc 16850 df-acs 16852 df-mgm 17844 df-sgrp 17893 df-mnd 17904 df-submnd 17949 df-mulg 18217 df-cntz 18439 df-cmn 18900 df-psmet 20083 df-xmet 20084 df-met 20085 df-bl 20086 df-mopn 20087 df-fbas 20088 df-fg 20089 df-cnfld 20092 df-top 21499 df-topon 21516 df-topsp 21538 df-bases 21551 df-cld 21624 df-ntr 21625 df-cls 21626 df-nei 21703 df-lp 21741 df-perf 21742 df-cn 21832 df-cnp 21833 df-haus 21920 df-tx 22167 df-hmeo 22360 df-fil 22451 df-fm 22543 df-flim 22544 df-flf 22545 df-xms 22927 df-ms 22928 df-tms 22929 df-cncf 23483 df-limc 24469 df-dv 24470 This theorem is referenced by: sincosq2sgn 25092 sincosq3sgn 25093 sincosq4sgn 25094 1cubrlem 25427
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#### Resources tagged with Factors and multiples similar to The Number Crunching Machine:
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### There are 143 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Zios and Zepts
##### Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Being Resilient - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
### Abundant Numbers
##### Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Being Collaborative - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level to work on with others.
### Scoring with Dice
##### Age 7 to 11 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### Two Primes Make One Square
##### Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
### Making Pathways
##### Age 7 to 11 Challenge Level:
Can you find different ways of creating paths using these paving slabs?
### A Mixed-up Clock
##### Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Sweets in a Box
##### Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Sets of Numbers
##### Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Path to the Stars
##### Age 7 to 11 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Tiling
##### Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### What Is Ziffle?
##### Age 7 to 11 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### What Two ...?
##### Age 7 to 11 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### In the Money
##### Age 7 to 11 Challenge Level:
One quarter of these coins are heads but when I turn over two coins, one third are heads. How many coins are there?
### What Do You Need?
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Tom's Number
##### Age 7 to 11 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### Round and Round the Circle
##### Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Mystery Matrix
##### Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Fractions in a Box
##### Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
### Ducking and Dividing
##### Age 7 to 11 Challenge Level:
Your vessel, the Starship Diophantus, has become damaged in deep space. Can you use your knowledge of times tables and some lightning reflexes to survive?
### Factor-multiple Chains
##### Age 7 to 11 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Divide it Out
##### Age 7 to 11 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Fitted
##### Age 7 to 11 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
### Number Tracks
##### Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Which Is Quicker?
##### Age 7 to 11 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Sets of Four Numbers
##### Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### Multiplication Squares
##### Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Ip Dip
##### Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
### Multiply Multiples 3
##### Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
### Becky's Number Plumber
##### Age 7 to 11 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Multiply Multiples 1
##### Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
### Surprising Split
##### Age 7 to 11 Challenge Level:
Does this 'trick' for calculating multiples of 11 always work? Why or why not?
### Give Me Four Clues
##### Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Pebbles
##### Age 7 to 11 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Number Detective
##### Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
### Multiply Multiples 2
##### Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
### Seven Flipped
##### Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Crossings
##### Age 7 to 11 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Multiples Grid
##### Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
##### Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Multiplication Series: Number Arrays
##### Age 5 to 11
This article for teachers describes how number arrays can be a useful representation for many number concepts.
### Flashing Lights
##### Age 7 to 11 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
### Three Dice
##### Age 7 to 11 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Money Measure
##### Age 7 to 11 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes? | 2,235 | 9,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-43 | longest | en | 0.898929 |
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# Elementary Statistics - PowerPoint PPT Presentation
## Elementary Statistics
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1. Elementary Statistics Q: What is data? Q: What does the data look like? Q: What conclusions can we draw from the data? Q: Where is the middle of the data? Q: Why is the spread of the data important? Q: Can we model the data? Q: How do we know if we have a good model? Q: Is our data affected by other variables?
2. Definitions Individuals : Objects described by a set of data. Individuals may be people, but they may also be animals or things. Variable : Any characteristic of an individual. A variable can take on different values for different individuals. Categorical and Quantitative Variables Categorical variable : Places an individual into one or several categories. Quantitative variable : Takes numerical values for which arithmetic operations make sense. Distribution : Tells what values the data takes and how often it takes these values.
3. Homework 1, 2, 4, 6
4. Exploring Data Two Basic Strategies : 1) Begin by examining each variable by itself. Then move on to study the relationships among variables. 2) Begin with a graph or graphs. Then add numerical summaries of specific aspects of data. Different types of graphs : Bar graph, Pie chart, Stemplot, back-to-back Stemplot, Histogram, Time plot
5. Bar Graphs Grade A B C D Other Count Bar graph - A graph which displays the data using heights of bars to represent the counts of the variables. Example : Consider the following grade distribution : 6 12 15 9 3 How could we display the data using a bar graph ?
6. Bar Graphs 15 12 9 Grade A B C D Other 6 Count 6 12 15 9 3 3 A B C D F
7. Pie Charts Pie Chart : 1) A chart which represents the data using percentages. 2) Break up a circle (pie) into the respected percentages.
8. Pie Charts Percent Grade A B C D Other Count 6 12 15 9 3 13 27 33 20 7 B A C D F
9. Homework 13, 14, 16
10. Stemplot How to make a Stemplot : 1) Separate each observation into a stem consisting of all but the final (rightmost) digit, and a leaf, the final digit. Stems may have as many digits as needed, but each leaf contains only a single digit. 2) Write the stems in a vertical column with the smallest at the top, and draw a vertical line at the right of this column. 3) Write each leaf in a row to the right of the stem, in increasing order out from the stem.
11. Stemplot 4 4 5 5 Steps 1 and 2 : 6 6 Step 3 : 7 7 8 8 9 9 10 10 Example: Here are the grades Max achieved while in school his first two years. Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66 5 8 6 2 7 7 6 2 8 3 3 6 2 3 8 1 0 4 1 0
12. Stemplot 4 4 5 5 5 8 Steps 1 and 2 : 6 6 6 Step 3 : 7 7 2 2 6 7 7 8 8 2 3 3 3 6 8 8 9 9 0 1 1 4 10 10 0 Example: Here are the grades Max achieved while in school his first two years. Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66
13. Back-To-Back Stemplot • This is a stemplot which allows you to see and compare the • distribution of two related data sets Example : Here are the grades Lulu received during her first two years at college : Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 • To make a Back-To-Back Stemplot, you make the stem, and the • stems going off to the right and the left. You want the smaller • values closer to the stem.
14. Back-To-Back Stemplot 4 5 5 8 6 6 7 2 2 6 7 7 8 2 3 3 3 6 8 8 9 0 1 1 4 10 0 Lulu’s Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 Max’s Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66 5 6 6 5 6 3 1 8 8 7 2 7 2 6 4 3 5 3 0 2
15. Back-To-Back Stemplot 4 5 5 5 8 6 6 6 6 8 8 7 6 5 3 1 7 2 2 6 7 7 7 6 4 2 2 8 2 3 3 3 6 8 8 5 3 3 2 0 9 0 1 1 4 10 0 Lulu’s Grades: 66, 77, 78, 84, 92, 90, 86, 78, 71, 93, 82, 55, 73, 95, 87, 76, 93, 82, 66, 75 Max’s Grades: 88, 72, 91, 83, 77, 90, 45, 83, 94, 91, 86, 77, 82, 100, 58, 76, 83, 88, 72, 66
16. Splitting Stems 7 8 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 Normally we would set up the stems as follows :
17. Splitting Stems 7 8 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 Normally we would set up the stems as follows : 1, 1, 2, 4, 5, 5, 6, 7, 9 0, 1, 1, 2, 3, 3, 3, 3, 4, 5, 5, 8, 9 0, 0, 0, 1, 3, 5, 6 , 7
18. Splitting Stems 7 7 This stem gets scores 70 - 74 8 8 This stem gets scores 75 - 79 9 9 • If you have a large data set (leaves), then sometimes a stemplot • will not work very well. For instance, if you have a large amount • of leaves, and only a few stems, you might want to split the stems. Example : Consider the following test scores : 71, 71, 72, 74, 75, 75, 75, 76, 77, 79, 80, 81, 81, 82, 83, 83, 83, 83, 84, 85, 85, 88, 89, 90, 90, 90, 91, 93, 95, 96, 97 However, we could set up the stems as follows : 1 1 2 4 5 5 5 6 7 9 0 1 1 2 3 3 3 3 4 5 5 8 9 0 0 0 1 3 5 6 7
19. Rounding Stems 29.1 29.0 5.7 5.6 5.5 Q: What if we have a lot of stems, but not a lot of leaves? A: One might want to join the stems into larger stems by rounding. Example: Consider the following charges for filling a car with gas : 9.73 10.12 8.72 6.53 12.89 15.67 5.50 16.97 11.38 10.77 7.77 9.00 10.50 8.00 17.12 13.00 21.00 18.11 9.99 25.12 22.57 15.00 23.00 29.11 What would this stem look like ?
20. Rounding Stems 2 1 0 Q: What if we have a lot of stems, but not a lot of leaves? A: One might want to join the stems into larger stems by rounding. Example: Consider the following charges for filling a car with gas : 9.73 10.12 8.72 6.53 12.89 15.67 5.50 16.97 11.38 10.77 7.77 9.00 10.50 8.00 17.12 13.00 21.00 18.11 9.99 25.12 22.57 15.00 23.00 29.11 We could round the stems to be \$10 stems : 1 5 2 3 9 0 2 5 6 1 0 0 7 3 8 5 9 8 6 5 7 9 8 9
21. Homework 20, 22, 23, 26
22. Histograms A histogram breaks the range of variables up into (equal) intervals, and displays only the count or percent of the observations which fall into the particular intervals. Notes: • You can choose the intervals (usually equal) • Slower to construct than stemplots • Histograms do not display the individual observations • In case a score falls on an interval point, you must decide in • advance which interval in which the point will go.
23. Histograms Steps to drawing a histogram : 1) Divide the range into classes of equal width. 2) Count the number of observations in each class. These are called frequencies. 3) Draw the histogram.
24. Histograms Grade Amount Percent 90 - 100 8 20 80 - 90 10 25 70 - 80 10 25 60 - 70 8 20 50 - 60 4 10 Frequency Table 10 10 8 8 4 Example : Suppose the final breakdown in grades looks like this : 50 60 70 80 90 100
25. Histograms Grade Amount Percent 90 - 100 8 20 80 - 90 10 25 70 - 80 10 25 60 - 70 8 20 50 - 60 4 10 25% 25% 20% 20% 10% Example : Suppose the final breakdown in grades looks like this : 50 60 70 80 90 100
26. Homework 31, 32
27. Time Plot Variable Time A Time Plot is a graph with two axis. One axis represents time ,and the other axis represents the variable being measured.
28. Time Plot 89 90 91 92 93 94 95 96 97 98 Year HR 33 39 22 42 9 9 39 52 58 70 Example : The following are homerun totals for a certain baseball player the last 10 years : Construct a timeplot for this data set.
29. Time Plot 89 90 91 92 93 94 95 96 97 98 Year HR 33 39 22 42 9 9 39 52 58 70 Home Run Year
30. Time Plot 89 90 91 92 93 94 95 96 97 98 Year 70 HR 33 39 22 42 9 9 39 52 58 70 60 50 40 30 20 10 89 90 91 92 93 94 95 96 97 98
31. Homework 35, 36 | 3,245 | 8,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2019-51 | latest | en | 0.760555 |
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# A set of numbers has the property that for any number t in
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A set of numbers has the property that for any number t in [#permalink]
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17 Oct 2005, 06:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
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17 Oct 2005, 07:06
Ans is D
If -1 is in set, then t+2 = 1 can be in set (Hence II)
If 1 can be in set then t+2 = 3 can be in set.
If 3 can be in set then t+2 = 5 can be in set. (Hence III)
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17 Oct 2005, 07:14
E.
-3 + 2 = -1
- 1
and 5.
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17 Oct 2005, 07:16
jainvineet wrote:
E.
-3 + 2 = -1
- 1
and 5.
The question did not mention anything about -3. How did you get that?
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17 Oct 2005, 08:02
D is correct in my opinion
You can not assume that the first number of the set has a realtion with -1
Maybe the set begins with -2 and the numbers are then higher and higher. So you can not be sure that -3 is inside it.
It doesn't say that all numbers have the property to have a relation with other numbers by doing t-2. So you can only go in the way t+2.
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17 Oct 2005, 08:50
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
Logically, just because -1 is in the set doesn't mean that -3 is in the set. Yes, -3 COULD be in the set but the question stem asks for what numbers MUST be in the set. You must always start with the numbers that are given to you and not extrapolate backwards on a problem like this (IMHO). I will go with D.
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19 Oct 2005, 06:42
Another D. Agree with explanations above about not being E.
OA?
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Re: p/s - # [#permalink]
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19 Oct 2005, 10:00
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
If t then t+2 doesn't mean that if t+2 then t, or if t then t-2.
_________________
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keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
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Re: p/s - # [#permalink]
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19 Oct 2005, 15:30
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
B is my answer.
1 must be in the set because we are given -1 as one of the number. I didn't choose 5 because we don't know how many numbers are in the set.
5 COULD be in the set of numbers but it doesn't have to be.
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Re: p/s - # [#permalink]
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19 Oct 2005, 15:36
TeHCM wrote:
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
B is my answer.
1 must be in the set because we are given -1 as one of the number. I didn't choose 5 because we don't know how many numbers are in the set.
5 COULD be in the set of numbers but it doesn't have to be.
I hope this trickygmat guy posts the OA; hopefully in the next millenium at the very least.
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Re: p/s - # [#permalink]
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19 Oct 2005, 17:32
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
this is one of the most discussed question.
should be D. if t, then t+2 means 1 and 5 are in the set since -1 is in the set but not -3 (=-1-2).
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19 Oct 2005, 18:10
If t = -1, then t+2 = 1. This number 1 will then become another t, which gives t+2 of 3, and the proess is repeated to give us 5.
If t+2 = -1, then t = -3. -1 will then give 1, 3, 5 as sequences of t and t+2.
I'll take E.
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19 Oct 2005, 18:26
Ywilfried
you know that T+2 is in the set
Moreover it is not said that ONLY T+2 are in the set
so let's say T=-1, T could be in the set and not comes from a T-2 number, however T will bring T+2 into the set
if a number is in the set, it will bring this number+2
Now imagine the set begins with 2 as the smallest value for example. So you will not find 0 inside this set (it will 2,4,6,8,10,...). However you will find 4 and 6
if there is a disscussion it should be between B and D
can we consider t+2 as t ?
i stick with D, if you have T, you have T+2 in the set and it is said that any number in the set then has T+2...
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19 Oct 2005, 20:21
Antmavel wrote:
Ywilfried
you know that T+2 is in the set
Moreover it is not said that ONLY T+2 are in the set
so let's say T=-1, T could be in the set and not comes from a T-2 number, however T will bring T+2 into the set
if a number is in the set, it will bring this number+2
Now imagine the set begins with 2 as the smallest value for example. So you will not find 0 inside this set (it will 2,4,6,8,10,...). However you will find 4 and 6
if there is a disscussion it should be between B and D
can we consider t+2 as t ?
i stick with D, if you have T, you have T+2 in the set and it is said that any number in the set then has T+2...
Your explanation was for the case if -1 is the first number in the series. However, you picked D. How do you know 3 is not the last in the series? If 3 is the last number in the series, there's no 5.
The question says for every T, there is a T+2. This T and T+2 value is not fixed at a certain number. The number -1 is the T value of 1, and the T+2 value of -3. Similarly, 5 is the T+2 value of 3, but 3 is the T value of 5 and T+2 value of 1. You get the picture? I think this is an infinite series, since each value has a relationship with another number.
I'll stick with E
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19 Oct 2005, 20:30
I understand your way of reasonning now
OA will decide our fate
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19 Oct 2005, 20:30
Antmavel wrote:
I understand your way of reasonning now
OA will decide our fate
I like your light-sabre
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19 Oct 2005, 20:51
trickygmat wrote:
A set of numbers has the property that for any number t in the set t+2 is also in the set. If -1 is in the set, which of the following must also be in the set?
I. -3
II. 1
III. 5
A. I only
B. II only
C. I & II only
D. II & III Only
E. I, II & III
My answer is B. The question stem says that for any t there is a t+2, and not the opposite. So if -1 is there then 1 should be there. -3 would belong to this set only when we say for every t+2 there is a t.
I would say B for this.
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19 Oct 2005, 22:24
Antmavel wrote:
I understand your way of reasonning now
OA will decide our fate
Your fates may never come my friends. This guy (or girl i should say) trickygmat is very bad about posting the OA.
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20 Oct 2005, 06:19
I dont know how reliable the OA is.... but it is D.
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20 Oct 2005, 06:29
It must be D.
The reasoning for why -3 is not is the set is well explained.
B is wrong because 5 must be in the set if -1 is part of it. The stem states that if t then t+2 is there as well. This creates an infinite set of numbers. Size is not a factor. If -1 is part of the set you know that 1, 3, 5, 7, 9, .... etc. every positive odd number is part of the set.
Brendan.
20 Oct 2005, 06:29
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# A set of numbers has the property that for any number t in
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# luca-deltodesco
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Offline Last Active Sep 07 2013 03:50 PM
### #5043030Confused on Big Oh
Posted by on 14 March 2013 - 07:11 AM
I think the best example of that is triangulation:
Triangulating a polygon 'can' be done in O(n) time, but the algorithm to do this is so incredibly complex that unless you have a polygon with 100000000000000 (*number pulled out of my arse) vertices, an O(n.log(n)) or O(n^2) algorithm is going to do it faster.
### #5027342Collision Physics Question
Posted by on 30 January 2013 - 04:21 PM
It is not that there is less trig involved when using x-y vectors, but that there is NO trig involved.
I already showed you earlier how to do bouncing.
That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*
The solution is incredibly simple, using simple vector operations.
You have the slope normal vector (As opposed to its angle)
You have the incident velocity.
The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)
where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.
If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x). The normal is a unit-length vector that points 'out' (or 'in' for these purposes, makes no difference to the result) of the surface, so the left side of a box will have normal (-1, 0) etc.
### #5026170Collision Physics Question
Posted by on 27 January 2013 - 02:17 PM
That you say you have done bouncing, but cannot figure out this, plus that you talk about the 'angle' of the slope tells me you are using trigonometry to do the bouncing... *sigh*
The solution is incredibly simple, using simple vector operations.
You have the slope normal vector (As opposed to its angle)
You have the incident velocity.
The outgoing velocity is then: newVelocity = velocity - normal*(1+e)*dot(velocity, normal)
where e is the coefficient of restitution. For a perfectly elastic bounce e = 1, for a perfectly inelastic bounce (your 'slide') e = 0.
If you're struggling to compute the normal vector. Assuming you have two positions for the slope 'a' and 'b' that you're using to compute your angle like atan2(b.y-a.y, b.x-a.x), then the normal for this slope is instead: unit(a.y-b.y, b.x-a.x). The normal is a unit-length vector that points 'out' (or 'in' for these purposes, makes no difference to the result) of the surface, so the left side of a box will have normal (-1, 0) etc.
### #5015554How do I create a vector tangent to each point in a non-function defined surf...
Posted by on 29 December 2012 - 04:40 PM
sounds like an XY situation: http://www.perlmonks.org/index.pl?node_id=542341
Why don't you describe what your actually trying to do?
You mention lighting, you mean that you want to write a shader to emulate how opengl would do its lighting?
You're talking about 'vector tangent to a point', do you mean the surface normal? You need to specify those for opengl lighting in the first place, so lighting is not really the question here, it seems to me you have a mesh, and you want to find the normals for each vertex of the mesh's triangles.
### #5006619Caustics - simple questions
Posted by on 03 December 2012 - 09:11 AM
It seems to simply mean that, given the triangle of a model which is specular, it casts a caustic; the caustic it casts when projected onto the plane is also a triangle (the caustic triangle) and the volume that the original specular triangle, and the projected caustic make is the caustic volume.
### #5006584New is SLOW!
Posted by on 03 December 2012 - 07:09 AM
Well if you think about what is actually happening:
In the first case, it is in the worst scenario doing a loop that has nothing to do; the local int is going to have been assigned to a register and at worst a space on the stack. In either case setting it to 0 is like nothing at all. Most likely your compiler is not even bothering to set it since it's unused, and may even be removing the loop altogether as it's not doing any work.
In the second case, each time you call 'new int' it's calling the operating system to ask it for space and performing lots of computations to choose the best place to allocate the int, with perhaps further logic to maintain the heap, same with delete in reverse.
### #5006468I need to remove these interior lines.
Posted by on 02 December 2012 - 08:07 PM
Further examples shown with results in red (additional number being number of times in total that vertex appears in output)
http://www.zimagez.com/zimage/screenshot-031212-021000.php
### #5006465I need to remove these interior lines.
Posted by on 02 December 2012 - 07:55 PM
For fun, I implemented my solution in Haxe and checked it worked with a somewhat degenerate case as well of having a dangling vertex connected outside the main exterior set
http://pastebin.com/uyu2ZehA
### #5006450I need to remove these interior lines.
Posted by on 02 December 2012 - 06:24 PM
Your step doesn't really need a triangulation. Eitherway given purely a set of lines you need to determine what the triangles are in the first place; and that same step can be used to simply determine the sub-polygons which would then lead to the same rule of an internal line has more than one sub-polygon using it and the triangulation step is pointless.
Eitherway, I don't see how you could determine those internal sub-polygons any faster than just traversing the exterior directly.
### #5006422I need to remove these interior lines.
Posted by on 02 December 2012 - 04:51 PM
Paradigm shifter, that only works if the lines always form triangles..
eg: http://www.zimagez.com/zimage/screenshot-021212-225045.php
here, your solution would choose to remove the crossed-out red line. My algorithm would give us the green polygon.
### #5006421I need to remove these interior lines.
Posted by on 02 December 2012 - 04:44 PM
One way to identify the exterior shell of your set of points (so we're assuming there are not any 'holes' that should be considered exterior to the volume).
a) Identify a point on the exterior of the line set.
b) Walk around the exterior of the line set till you get back to start.
This identifies all the lines required to define the exterior; and then ones that you can remove are all those not belonging to the exterior.
For a) you can choose the lexicographically minimal point (a, b) < (c, d) iff a < c || (a == c &amp;&amp; b < d)
For b) you need to be able to; from any given point of the exterior identify the next point of the exterior. This comes in two cases; the initial case where you have 2 exterior lines to choose and you have to pick one of them (for this, abuse that your start point is lexicographically minimal), and in the other case you can make use of the point of the exterior you have just came from to order the outgoing edges.
Presume that we choose, as a result of the first case of b) the point that is clockwise around the exterior (We'll come back to this after).
If you label the previous point 'p' and the current point 'c'; then you can order the outgoing edges based on their relative rotation from the vector (c - p) http://www.zimagez.com/zimage/screenshot-021212-223758.php here, we want to pick vertex v0.
We can do this by computing the perp-dot products of (v[i] - c) with (c - p) normalised by the length of (v[i] - c); and choosing the minimal (or maximal depending on coordinate system); since the perp-dot of two vectors u,v is |u||v|sin(theta). We can also avoid any division if sorting of the outgoing edges uses a comparison function instead of a key.
Back to the first case of b); choosing the second vertex of exterior we can use the same technique, and use the vector (1, 0) instead of (c - p) above. This is valid given that our first vertex was lexicographically minimal since the clockwise direction of the next vertex can in the worst case be in the direction (1, 0) and there cannot exist any vertex of the set that would give us a 'negative' perp-dot since that point would need to have a lower y-coordinate and would have been chosen as the lexicographically minimal one.
^^ this is very similar to gift wrapping algorithm for computing a convex hull and is exactly equivalent if we join every vertex to every other one using this algorithm.
### #5005040LPD3DXMESH to btConeShape
Posted by on 28 November 2012 - 01:26 PM
.... Assuming that the box is axis aligned, and the cone is aligned with y-axis.
### #4996291Calculating angles of rotation
Posted by on 01 November 2012 - 01:43 PM
Previous two responses: I think you missed his response that the API he's using permits ONLY setting z-x-y euler angles.
### #4996120Calculating angles of rotation
Posted by on 01 November 2012 - 03:01 AM
Use a better SDK?
We need to know what order the final transform gets applied, does it do X followed by Y, followed by Z or some other order?
### #4995979Calculating angles of rotation
Posted by on 31 October 2012 - 04:56 PM
You only need 2 angles to specify a direction in 3 dimensions, and your question really depends on what axis those rotations should be about, and in what order the rotations are supposed to occur.
Even specified properly, there's no total solution, since you can end up with gimbal lock, eg:
Say you define the rotations to be: a rotation about the y-axis, then a rotation about the x-axis; if AB starts at (0, 1, 0) then it is only possible to define the rotations if the new AB happens to be in the yz plane. No matter how you specify the two rotations you will end up with this problem. I 'guess' you could add a third rotation, that is only used in this corner case; but really you can probably do much better with a different representation of rotation, whether it be a quaternion or axis-angle
[Edit: Bob's suggestion is to use axis-angle]
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1 Chapter 4: Polynomials I. Exponents & Their Properties II. Negative Exponents III. Scientific Notation IV. Polynomials V. Addition & Subtraction of Polynomials VI. Multiplication of Polynomials VII. Greatest Common Factor (GCF) VIII. Factoring Trinomials IX. Solving Quadratic Equations by Factoring 1
2 VII. Greatest Common Factor (GCF) a polynomial is to find an equivalent expression that is a product. An equivalent expression of this type is called a factorization of the polynomial. In the last sections, we spent much time in multiplying polynomials. up coming sections we want to reverse that process and factor. 2
3 What are the factors of 24? What is the GCF of 24 & 32? What are the factors of 32? At this point in the course, we have use the Distributive Law for multiplication. It can also be used to factor out the GCF. 24x + 32 = 3
4 4
5 5
6 6
7 1) x(3x - 2) - 8(3x - 2) 2) x(7x + 1) + (7x + 1) 7
8 What is the GCF? In this form, there is no common factor to factor out. There is method that could possible work, called factoring by grouping 8
9 Factor by Grouping. 9
10 VIII. Factoring Trinomials What is the pattern? 10
11 ( )( ) "Check if O + I = bx 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 Going back over the Pattern! What part of the multiplication does this term come from? 20
21 Possible Factorizations (3x - 5)(x + 1) Which one is correct? (3x - 1)(x + 5) (3x + 5)(x - 1) (3x + 1)(x - 5) 21
22 22
23 23
24 Factor Completely 24
25 Factor Completely 25
26 26
27 27
28 IX. Solving Quadratic Equations by Factoring Standard Form of a Quadratice Equation The strategy we applied to solve linear equations will NOT work on this type of equation. The Principle of Zero Products (x - 3)(x + 5) = 0 28
29 Note: It is difficult to see solutions while equation is in this form! Strategy to solve quadratic (or higher degree polynomial) : Set one side to zero (Write equation in standard form). : Set each factor to zero & solve. 29
30 30
31 31
32 32
33 Homework Chapter 4 Part I: Exponents & Their Properties Simplify the following Expressions 33
34 Part II: Negative Exponents Simplify the following expressions. Your final answer should NOT include negative exponents. Part III: Scientific Notation Multiply or divide as appropriate. Write final answer in scientific notation. 34
35 Part IV: Polynomials For each polynomial below (a) classify it as a monomial, binomial, trinomial, or a polynomial with no special name; (b) give the degree of the polynomial; (c) determine the leading term and leading coefficient. Part V: Addition & Subtraction of Polynomials Add or Subtract as appropriate 35
36 Part VI: Multiplication of Polynomials Multiply the following polynomials 36
37 Part VII: Greatest Common Factor (GCF) Factor out the GCF from each of the following polynomials Factor the following polynomials by grouping (if possible) 37
38 Part VIII: Factoring Trinomials Factor the following trinomials. If it is not possible to factor the trinomial then state that it is prime 38
39 Part IX: Solving Quadratic Equations by Factoring Solve the following equations by factoring 39
### MathB65 Ch 4 IV, V, VI.notebook. October 31, 2017
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## Presentation on theme: "1 Chapter 2: Time Value of Money Future value Present value Rates of return Amortization."— Presentation transcript:
1 Chapter 2: Time Value of Money Future value Present value Rates of return Amortization
2 Time lines show timing of cash flows. CF 0 CF 1 CF 3 CF 2 0123 i% Tick marks at ends of periods, so Time 0 is today; Time 1 is the end of Period 1; or the beginning of Period 2.
3 Time line for a \$100 lump sum due at the end of Year 2. 100 012 Year i%
4 Time line for an ordinary annuity of \$100 for 3 years 100 0123 i%
5 Time line for uneven CFs 100 50 75 0123 i% -50
6 FV of an initial \$100 after 3 years (i = 10%) FV = ? 0123 10% Finding FVs (moving to the right on a time line) is called compounding. 100
7 After 2 years FV 1 = PV + INT 1 = PV + PV (i) = PV(1 + i) = \$100(1.10) = \$110.00. FV 2 = FV 1 (1+i) = PV(1 + i)(1+i) = PV(1+i) 2 = \$100(1.10) 2 = \$121.00.
8 After 3 years FV 3 = FV2(1+i)=PV(1 + i) 2 (1+i) = PV(1+i) 3 = \$100(1.10) 3 = \$133.10 In general, FV n = PV(1 + i) n.
9 Three Ways to Find FVs Solve the equation with a regular calculator. Use a financial calculator. Use a spreadsheet.
10 Financial calculator: HP10BII Adjust display brightness: hold down ON and push + or -. Set number of decimal places to display: Orange Shift key, then DISP key (in orange), then desired decimal places (e.g., 3). To temporarily show all digits, hit Orange Shift key, then DISP, then =
11 HP10BII (Continued) To permanently show all digits, hit ORANGE shift, then DISP, then. (period key) Set decimal mode: Hit ORANGE shift, then./, key. Note: many non-US countries reverse the US use of decimals and commas when writing a number.
12 HP10BII: Set Time Value Parameters To set END (for cash flows occurring at the end of the year), hit ORANGE shift key, then BEG/END. To set 1 payment per period, hit 1, then ORANGE shift key, then P/YR
13 Financial calculators solve this equation: FV n + PV (1+i) n = 0. There are 4 variables. If 3 are known, the calculator will solve for the 4th. Financial Calculator Solution
14 310-100 0 NI/YR PV PMTFV 133.10 Clearing automatically sets everything to 0, but for safety enter PMT = 0. Set:P/YR = 1, END. INPUTS OUTPUT Heres the setup to find FV
15 Spreadsheet Solution Use the FV function: see spreadsheet in FM11 Ch 02 Mini Case.xls = FV(Rate, Nper, Pmt, PV) = FV(0.10, 3, 0, -100) = 133.10
16 10% Whats the PV of \$100 due in 3 years if i = 10%? Finding PVs is discounting, and its the reverse of compounding. 100 0123 PV = ?
17 Solve FV n = PV(1 + i ) n for PV PV = FV n (1 + i) n = FV n () 1 1 + i n PV= \$100 1 1.10 = \$1000.7513 = \$75.13. 3
18 3 10 0100 N I/YR PV PMTFV -75.13 Either PV or FV must be negative. Here PV = -75.13. Put in \$75.13 today, take out \$100 after 3 years. INPUTS OUTPUT Financial Calculator Solution
19 Spreadsheet Solution Use the PV function: see spreadsheet in FM11 Ch 02 Mini Case.xls = PV(Rate, Nper, Pmt, FV = PV(0.10, 3, 0, 100) = -75.13
20 20% 2 012? FV= PV(1 + i) n \$2= \$1(1 + 0.20) n (1.2) n = \$2/\$1 = 2 nLN(1.2)= LN(2) n= LN(2)/LN(1.2) n= 0.693/0.182 = 3.8. Finding the Time to Double
21 20 -1 0 2 NI/YR PV PMTFV 3.8 INPUTS OUTPUT Financial Calculator Solution
22 Spreadsheet Solution Use the NPER function: see spreadsheet in FM11 Ch 02 Mini Case.xls = NPER(Rate, Pmt, PV, FV) = NPER(0.10, 0, -1, 2) = 3.8
23 ?% 2 0123 FV= PV(1 + i) n \$2= \$1(1 + i) 3 (2) (1/3) = (1 + i) 1.2599= (1 + i) i= 0.2599 = 25.99%. Finding the Interest Rate
24 3 -1 0 2 NI/YR PV PMTFV 25.99 INPUTS OUTPUT Financial Calculator
25 Spreadsheet Solution Use the RATE function: = RATE(Nper, Pmt, PV, FV) = RATE(3, 0, -1, 2) = 0.2599
26 Ordinary Annuity PMT 0123 i% PMT 0123 i% PMT Annuity Due Ordinary Annuity vs. Annuity Due
27 Whats the FV of a 3-year ordinary annuity of \$100 at 10%? 100 0123 10% 110 121 FV= 331
28 FV Annuity Formula The future value of an annuity with n periods and an interest rate of i can be found with the following formula: = PMT (1+i) n -1 i = 100 (1+0.10) 3 -1 0.10 = 331
29 Financial Calculator Formula for Annuities Financial calculators solve this equation: FV n + PV(1+i) n + PMT (1+i) n-1 i = 0. There are 5 variables. If 4 are known, the calculator will solve for the 5th.
30 3 10 0 -100 331.00 NI/YRPVPMTFV Have payments but no lump sum PV, so enter 0 for present value. INPUTS OUTPUT Financial Calculator Solution
31 Spreadsheet Solution Use the FV function: see spreadsheet. = FV(Rate, Nper, Pmt, Pv) = FV(0.10, 3, -100, 0) = 331.00
32 Whats the PV of this ordinary annuity? 100 0123 10% 90.91 82.64 75.13 248.69 = PV
33 PV Annuity Formula The present value of an annuity with n periods and an interest rate of i can be found with the following formula: = PMT 1 - 1 (1+i) n i = 100 1 - 1 (1+0.10) 3 0.10 = 248.69
34 Have payments but no lump sum FV, so enter 0 for future value. 3 10 100 0 NI/YRPVPMTFV -248.69 INPUTS OUTPUT Financial Calculator Solution
35 Spreadsheet Solution Use the PV function: see spreadsheet. = PV(Rate, Nper, Pmt, Fv) = PV(0.10, 3, 100, 0) = -248.69
36 Find the FV and PV if the annuity were an annuity due. 100 0123 10% 100
37 PV and FV of Annuity Due vs. Ordinary Annuity PV of annuity due: = (PV of ordinary annuity) (1+i) = (248.69) (1+ 0.10) = 273.56 FV of annuity due: = (FV of ordinary annuity) (1+i) = (331.00) (1+ 0.10) = 364.1
38 310 100 0 -273.55 NI/YRPVPMTFV INPUTS OUTPUT PV of Annuity Due: Switch from End to Begin Then enter PV = 0 and press FV to find FV = \$364.10.
39 Excel Function for Annuities Due Change the formula to: =PV(10%,3,-100,0,1) The fourth term, 0, tells the function there are no other cash flows. The fifth term tells the function that it is an annuity due. A similar function gives the future value of an annuity due: =FV(10%,3,-100,0,1)
40 What is the PV of this uneven cash flow stream? 0 100 1 300 2 3 10% -50 4 90.91 247.93 225.39 -34.15 530.08 = PV
41 Financial calculator: HP10BII Clear all: Orange Shift key, then C All key (in orange). Enter number, then hit the CFj key. Repeat for all cash flows, in order. To find NPV: Enter interest rate (I/YR). Then Orange Shift key, then NPV key (in orange).
42 Financial calculator: HP10BII (more) To see current cash flow in list, hit RCL CFj CFj To see previous CF, hit RCL CFj – To see subsequent CF, hit RCL CFj + To see CF 0-9, hit RCL CFj 1 (to see CF 1). To see CF 10-14, hit RCL CFj. (period) 1 (to see CF 11).
43 Input in CFLO register: CF0 = 0 CF1 = 100 CF2 = 300 CF3 = 300 CF4 = -50 Enter I = 10%, then press NPV button to get NPV = 530.09. (Here NPV = PV.)
44 Excel Formula in cell A3: =NPV(10%,B2:E2) ABCDE 101234 2100300 -50 3530.09
45 Nominal rate (i Nom ) Stated in contracts, and quoted by banks and brokers. Not used in calculations or shown on time lines Periods per year (m) must be given. Examples: 8%; Quarterly 8%, Daily interest (365 days)
46 Periodic rate (i Per ) i Per = i Nom /m, where m is number of compounding periods per year. m = 4 for quarterly, 12 for monthly, and 360 or 365 for daily compounding. Used in calculations, shown on time lines. Examples: 8% quarterly: i Per = 8%/4 = 2%. 8% daily (365): i Per = 8%/365 = 0.021918%.
47 Will the FV of a lump sum be larger or smaller if we compound more often, holding the stated I% constant? LARGER! If compounding is more frequent than once a year--for example, semiannually, quarterly, or daily--interest is earned on interest more often.
48 FV with Different Compounding Periods (e.g., \$100 at a 12% nominal rate with semiannual compounding for 5 years) = \$100(1.06) 10 = \$179.08. FV = PV1.+ i m n Nom mn FV = \$1001+ 0.12 2 5S 2x5
49 FV of \$100 at a 12% nominal rate for 5 years with different compounding FV(Annual)= \$100(1.12) 5 = \$176.23. FV(Semiannual)= \$100(1.06) 10 =\$179.08. FV(Quarterly)= \$100(1.03) 20 = \$180.61. FV(Monthly)= \$100(1.01) 60 = \$181.67. FV(Daily)= \$100(1+(0.12/365)) (5x365) = \$182.19.
50 Effective Annual Rate (EAR = EFF%) The EAR is the annual rate which causes PV to grow to the same FV as under multi-period compounding. Example: Invest \$1 for one year at 12%, semiannual: FV = PV(1 + iNom/m) m FV = \$1 (1.06) 2 = 1.1236. EFF% = 12.36%, because \$1 invested for one year at 12% semiannual compounding would grow to the same value as \$1 invested for one year at 12.36% annual compounding.
51 Comparing Rates An investment with monthly payments is different from one with quarterly payments. Must put on EFF% basis to compare rates of return. Use EFF% only for comparisons. Banks say interest paid daily. Same as compounded daily.
52 EFF% = - 1 ( 1 + ) i Nom m m = - 1.0 ( 1 + ) 0.12 2 2 = (1.06) 2 - 1.0 = 0.1236 = 12.36%. EFF% for a nominal rate of 12%, compounded semiannually
53 Finding EFF with HP10BII Type in nominal rate, then Orange Shift key, then NOM% key (in orange). Type in number of periods, then Orange Shift key, then P/YR key (in orange). To find effective rate, hit Orange Shift key, then EFF% key (in orange).
54 EAR (or EFF%) for a Nominal Rate of of 12% EAR Annual = 12%. EAR Q =(1 + 0.12/4) 4 - 1= 12.55%. EAR M =(1 + 0.12/12) 12 - 1= 12.68%. EAR D(365) =(1 + 0.12/365) 365 - 1= 12.75%.
55 Can the effective rate ever be equal to the nominal rate? Yes, but only if annual compounding is used, i.e., if m = 1. If m > 1, EFF% will always be greater than the nominal rate.
56 When is each rate used? i Nom :Written into contracts, quoted by banks and brokers. Not used in calculations or shown on time lines.
57 i Per :Used in calculations, shown on time lines. If i Nom has annual compounding, then i Per = i Nom /1 = i Nom. When is each rate used? (Continued)
58 When is each rate used? (Continued) EAR (or EFF%): Used to compare returns on investments with different payments per year. Used for calculations if and only if dealing with annuities where payments dont match interest compounding periods.
59 Amortization Construct an amortization schedule for a \$1,000, 10% annual rate loan with 3 equal payments.
60 Step 1: Find the required payments. PMT 0123 10% -1,000 3 10 -1000 0 INPUTS OUTPUT NI/YRPVFV PMT 402.11
61 Find interest charge and principal repayment for Year 1. Step 1: Find interest INT t = Beg bal t (i) INT 1 = \$1,000(0.10) = \$100. Step 2: Find repaid princ. Repmt = PMT - INT = \$402.11 - \$100 = \$302.11.
62 Step 4: Find ending balance after Year 1. End bal= Beg bal - Repmt = \$1,000 - \$302.11 = \$697.89. Repeat these steps for Years 2 and 3 to complete the amortization table.
63 Amortization Table YEAR BEG BALPMTINT PRIN PMT END BAL 1\$1,000\$402\$100\$302\$698 269840270332366 3 402373660 TOT1,206.34206.341,000
64 Interest declines because outstanding balance declines.
65 Amortization tables are widely used--for home mortgages, auto loans, business loans, retirement plans, and more. They are very important! Financial calculators (and spreadsheets) are great for setting up amortization tables.
66 Fractional Time Periods On January 1 you deposit \$100 in an account that pays a nominal interest rate of 11.33463%, with daily compounding (365 days). How much will you have on October 1, or after 9 months (273 days)? (Days given.)
67 i Per = 11.33463%/365 = 0.031054% per day. FV=? 012273 0.031054% -100 Convert interest to daily rate
68 273-100 0 108.85 INPUTS OUTPUT N I/YRPVFV PMT i Per =i Nom /m =11.33463/365 =0.031054% per day. Calculator Solution
69 Whats the value at the end of Year 3 of the following CF stream if the quoted interest rate is 10%, compounded semiannually? 01 100 23 5% 45 6 6-mos. periods 100
70 Non-matching rates and periods Payments occur annually, but compounding occurs each 6 months. So we cant use normal annuity valuation techniques.
71 1st Method: Compound Each CF 01 100 23 5% 456 100100.00 110.25 121.55 331.80 FVA 3 = \$100(1.05) 4 + \$100(1.05) 2 + \$100 = \$331.80.
72 2nd Method: Treat as an annuity, use financial calculator Find the EAR for the quoted rate: EAR = ( 1 + ) - 1 = 10.25%. 0.10 2 2
73 3 10.25 0 -100 INPUTS OUTPUT N I/YR PVFV PMT 331.80 Use EAR = 10.25% as the annual rate in calculator.
74 Whats the PV of this stream? 0 100 1 5% 23 100 90.70 82.27 74.62 247.59
75 Comparing Investments You are offered a note which pays \$1,000 in 15 months (or 456 days) for \$850. You have \$850 in a bank which pays a 6.76649% nominal rate, with 365 daily compounding, which is a daily rate of 0.018538% and an EAR of 7.0%. You plan to leave the money in the bank if you dont buy the note. The note is riskless. Should you buy it?
76 i Per =0.018538% per day. 1,000 0365456 days -850 Daily time line
77 Method 1: Greatest Future Wealth Find FV of \$850 left in bank for 15 months and compare with notes FV = \$1,000. FV Bank =\$850(1.00018538) 456 =\$924.97 in bank. Buy the note: \$1,000 > \$924.97.
78 456-850 0 924.97 INPUTS OUTPUT NI/YRPVFV PMT i Per =i Nom /m =6.76649%/365 =0.018538% per day. Calculator Solution to FV
79 Find PV of note, and compare with its \$850 cost: PV=\$1,000/(1.00018538) 456 =\$918.95. Method 2: Greatest Present Wealth
80 456.018538 0 1000 -918.95 INPUTS OUTPUT NI/YRPVFV PMT 6.76649/365 = PV of note is greater than its \$850 cost, so buy the note. Raises your wealth. Financial Calculator Solution
81 Find the EFF% on note and compare with 7.0% bank pays, which is your opportunity cost of capital: FV n = PV(1 + i) n \$1,000= \$850(1 + i) 456 Now we must solve for i. Method 3: Rate of Return
82 456-850 0 1000 0.035646% per day INPUTS OUTPUT NI/YRPV FV PMT Convert % to decimal: Decimal = 0.035646/100 = 0.00035646. EAR = EFF%= (1.00035646) 365 - 1 = 13.89%. Calculator Solution
83 P/YR=365 NOM%=0.035646(365)= 13.01 EFF%=13.89 Since 13.89% > 7.0% opportunity cost, buy the note. Using Interest Conversion
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# New constants and real representation of new spaces(part one)
```Title : New constants and real representation of new spaces
Elia T .Alabbas
Title : New constants and real representation of new spaces( Part one)
Author : Elia T . Alabbas
Comments 8 Pages, 4 figures.
Tishreen University
Subject-class : Algebraic Geometry
[email protected]
ABSTRACT
This is a continuation of what the researcher started from previous
research, This research aims to reach an infinite number of Pythagoras
triads, By proportioning the lengths of the right-angled triangle, This will
help to represent really many spaces, The researcher was able to discover
New constants in the right-angled triangle: They are (0.28), (0.96), (1),
These New constants capable of generating an infinite number of
Pythagorean triads,(0.28,0.96,1) It is the source of the Pythagorean triad
( 7,24,25).
Also the researcher was able to discover New constants in the rightangled triangle: They are (1), (2.4), (2.6), These New constants capable of
generating an infinite number of Pythagorean triads,(1,2.4,2.6) It is the
source of the Pythagorean triad ( 5,12,13).
Title : New constants and real representation of new spaces
Elia T .Alabbas
Depending on the Pythagorean theorem" a right angled triangle:the square of the
hypotenuse is equal to the sum of the squares of the other two sides" and
previous research. the researcher works to find New constants in the rightangled triangle.
A2 + B2 = C2
C
A
B
Figure 1
The research paper symbols:
A: The first side length.
B: The second side length
C: Hypotenuse length
IN previous research (Alabbas,2017,6)The researcher was able to
A = (0.6) C
discover constants in the right-angled triangle: They are (0.6), (0.8).
A = (0.6)C
B= (0.8) C
C
B= (0.8) C
Figure 2
Title : New constants and real representation of new spaces
Elia T .Alabbas
A new start:
First:
By proportioning the lengths of the right-angled triangle:
New Observation:
(7,24,25)
A=7
B = 24
C = 25
A / C = 7 / 25 = 0.28
A = (0.28)C
B / C = 24 /25 = 0.96
B = (0,96)C
C / C = 25 /25 =1
From the above, the researcher reaches the new constants:
(0.28) ,(0.96), (1)
[A, B ,C]
[0.28, 0.96, 1]
A2 + B2 = C2
(0.28)2 + (0.96)2 = (1)2
0.0784 + 0.9216= 1
example:
A=?
B=?
C=2
The solution:
A = (0.28)C
A = 0.56
B = (0.96)C
Title : New constants and real representation of new spaces
Elia T .Alabbas
B = 1.92
solution Validation by using the Pythagorean theorem:
A2 + B2 = C2
(0.56)2 + (1.92)2 = (2)2
0.3136 + 3.6864= 4
Pythagorean trilogy
(0.56,1.92,2)
The researcher was able to discover constants in the right-angled
triangle: They are(0.28), (0.96), (1).
[0.28, 0.96, 1]
C
A = (0.28) C
A = (0.28)C
B = (0,96)C
B= (0.96) C
A: The first side length.
Figure 3
B: The second side length
C: Hypotenuse length
[0.28, 0.96, 1] These New constants capable of generating an
infinite number of Pythagorean triads
[0.28, 0.96,1], [0.56,1.92, 2] , [0.84, 2.88, 3] ,[1.12, 3.84, 4] , [1.4, 4.8, 5]
[1.68, 5.76, 6]………………..(1)
IN previous research (Alabbas,2017,) [ 0.6, 0.8, 1] These constants
capable of generating an infinite number of Pythagorean triads
[ 0.6, 0.8, 1] , [1.2, 1.6, 2] , [1.8, 2,4, 3] , [2.4, 3.2, 4] , [3, 4, 5]
[3.6, 4.8,6]…………………(2)
{
From (1) & (2) The Researcher notes [ 0.6, 0.8, 1] ,[0.28, 0.96,1]
}
Title : New constants and real representation of new spaces
Elia T .Alabbas
Hypotenuse length is represented In two different places in space in a
very simple way, Thus enabling the researcher to represent six spaces
in a real way.
Second:
By proportioning the lengths of the right-angled triangle:
New Observation:
(5,12,13)
A=5
B = 12
C = 13
A/A=5/5=1
B / A = 12 / 5 = 2.4
B = (2.4)A
C / A = 13 / 5 = 2.6
C = (2.6)A
From the above, the researcher reaches the new constants:
(1) ,(2.4), (2.6)
[A, B ,C]
[1, 2.4, 2.6]
A2 + B2 = C2
(1)2 + (2.4)2 = (2.6)2
1 + 5.76=6.76
First example:
A=2
B=?
C=?
The solution:
A= 2
B = (2.4)(2)
Title : New constants and real representation of new spaces
Elia T .Alabbas
B = 4.8
C = (2.6)(2)
C = 5.2
solution Validation by using the Pythagorean theorem:
A2 + B2 = C2
(2)2 + (4.8)2 = (5.2)2
4 + 23.04=27.04
Pythagorean trilogy
(2,4.8,5.2)
Second example:
A=3
B=?
C=?
The solution:
A= 2
B = (2.4)(3)
B = 7.2
C = (2.6)(3)
C = 7.8
solution Validation by using the Pythagorean theorem:
A2 + B2 = C2
(3)2 + (7.2)2 = (7.8)2
9 + 51.84=60.84
Pythagorean trilogy
(3,7.2,7.8)
Third example:
A=4
B=?
C=?
The solution:
A= 4
B = (2.4)(4)
B = 9.6
C = (2.6)(4)
C = 10.4
Title : New constants and real representation of new spaces
Elia T .Alabbas
solution Validation by using the Pythagorean theorem:
A2 + B2 = C2
(4)2 + (9.6)2 = (10.4)2
16 + 92.16=108.16
Pythagorean trilogy
(4,9.6,10.4)
Fifth example:
A=5
B=?
C=?
The solution:
A= 5
B = (2.4)(5)
B = 12
C = (2.6)(5)
C = 13
solution Validation by using the Pythagorean theorem:
A2 + B2 = C2
(5)2 + (12)2 = (13)2
25+144=169
Pythagorean trilogy
(5,12,13)
The researcher was able to discover constants in the right-angled
triangle: They are(1), (2.4), (2.6).
[1, 2.4, 2.6]
C = (2.6)A
A
B = (2.4)A
C = (2.6)A
A: The first side length.
B: The second side length
C: Hypotenuse length
B= (2.4) A
Figure 4
Title : New constants and real representation of new spaces
Elia T .Alabbas
[1, 2.4, 2.6] These New constants capable of generating an
infinite number of Pythagorean triads
[1, 2.4,2.6], [2,4.8,5.2] , [3, 7.2, 7.8] , [4, 9.6, 10,4]……………………….
Thus enabling the researcher to represent new spaces in a real way.
``` | 2,177 | 5,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-24 | latest | en | 0.850966 |
http://fractionslearningpathways.ca/digitalPaper/equivFrac | 1,579,500,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250597458.22/warc/CC-MAIN-20200120052454-20200120080454-00083.warc.gz | 70,640,785 | 8,009 | ## Unit Fraction: Day 1: Representing Fractions Using Manipulatives Jr/Int
MO 5 min
A 45 min
C/D 25 min
75 min
Math Learning Goals
Students will:
• represent fractions as parts of a whole using a variety of models
• reason about meaning of a fraction and the relationship between numerator and denominator
• communicate strengths of different representations for different students and in certain contexts e.g. , use of benchmarks to support and refine the meaning of fractions
Materials
• a variety of manipulatives
• sticky notes
Minds On...
Independent Math Log
Students respond to the prompt: What is a fraction?
Action!
Ask students to select one of the following fractions: 410 or 25, and represent it in as many ways as they can. Pairs display all of their representations in their workspace.
Whole Group Gallery Walk
Students circulate around the room and review the different representations. Students consider which representation they think most clearly shows the fraction, and indicate their preference by placing a sticky note with their name by their first choice. Ask students to be prepared to discuss any similarities or differences they notice between the representations of 410 and 25.
Teachers can provide scaffolding by suggesting different manipulatives for different pairs.
Students may notice that the two fractions are equivalent. Allow them to reason and explore to reach this conclusion.
Consolidate/ Debrief
Whole Group Discussion
Organize and name the different types of representations students preferred. Ask students who put their name on a sticky note by each particular representation to explain why they think that one is the most effective. Use prompts such as:
• What did you see?
• What did you like?
• I am noticing a lot of students selected (this) model. Why do you think so many of us chose that representation?
• Why didn’t you pick (the most popular)?
• I am interested in this representation…. (least picked)
• One of my favourite representations of 25 is (show a representation that students did not show e.g., position on a number line). Why do you think I like this type of representation?
Push their thinking for each representation by using some of the following questions:
#### Key Questions:
• So what does the 4 (2) represent?
• What does the 10 (5) represent?
• How are the 4 and the 10 (2 and 5) related?
• Why is it important for this to be partitioned into equal parts?
• What do equal parts mean in this model?
Independent Math Log
Use another colour of ink to build on your note in your Math Log. Be sure to include at least one example of each type of representation that shows your understanding of a fraction.
Include the following types of representations in the discussion:
• Part-whole: measure
• length (1-D)
• area (2-D) Include discussion of strengths of rectangular representations over circular representations
• volume (3-D)
• Part-whole: sets with either identical or non-identical items
• Part-Part: a ratio representation which shows numerator : denominator
• Position (point) on a number line
• Segment of a number line
Home Activity or Further Classroom Consolidation
Find at least two different representations of fractions. You may consider looking in the kitchen, garage, or newspapers and magazines.
## Unit Equivalency in Fractions: Day 2: Thinking Proportionally Junior/In
MO 15 min A 25 min C/D 35 min 75 min Math Learning Goals Students will: represent fractions as parts of a whole using a variety of models reason about meaning of a fraction and the relationship between numerator and denominator communicate strengths of different representations for different students and in certain contexts e.g., use of benchmarks to support and refine the meaning of fractions Materials Minds On... Pairs Exploration Ask students to use their multiplication chart (BLM 2.1) to identify equivalent fractions for 2⁄5, 6⁄7, and 3⁄10. Students list their observations. Possible observations include: you are counting by 2’s at the top and by 2’s at the bottom the equivalent fractions are found by looking across the chart you could extend the pattern beyond the chart the numerator and denominator grow at the same rate (proportionally) Action! Pairs Activity Ask students to show how they know that 2⁄3 and 8⁄12 are equivalent on chart paper. They should provide enough detail to support their classmates in understanding their work during the Gallery Walk. This could be differentiated by using the fractions 1⁄2 and 2⁄4 for some students. Consolidate/ Debrief Whole Group Gallery Walk Students go on a Gallery Walk affix their sticky note on the representation they think best helps them to understand equivalent fractions. Facilitate an Elmo presentation and post examples Bansho style. Point out the different representations (pictures, fraction circles, fraction towers, number lines, numerical operations). Discuss the different ways to represent. Post the charts that contain important learning – with labels – so that the students can refer back. For example, two area models that are accurate / same size or two line models that are overlaid, or a numerical explanation 2x4=8 … 3x4=12 Home Activity or Further Classroom Consolidation Choose A or B. A: Name a fraction that is equivalent to 1⁄3 and show how you know. B: 1⁄3 and 2⁄6 are equivalent. How do you know?
## Unit Equivalency in Fractions: Day 3: Representing Fractions on Number Lines Junior/Int
MO 15 min A 45 min C/D 20 min 75 min Math Learning Goals Students will: represent fractions on a number line reason about meaning of a fraction and the relationship between numerator and denominator communicate strengths of different representations for different students and in certain contexts e.g., use of benchmarks to support and refine the meaning of fractions Materials large number line copies of BLM 3.1 Minds On... Whole Class Brainstorm Here is a number line. (get a large “wall” number line from a primary classroom) What do you notice? How is this like a ruler? How can these help us learn math? Create a list. [measuring, counting, adding on, subtracting, skip counting] How could a number line help us with fractions? Write on a chart. Action! Pair-Share Investigation In groups label each partition on BLM 3.1. Compare the number lines. What do you notice? What equivalent fractions do you see? How do you know they are equivalent? Write questions based on your number lines. Pairs rotate and answer another group’s questions. Consolidate/ Debrief Whole Class Guided Discussion Look back at the chart we created about number lines and fractions. What else have you learned about the ways that number lines can help us learn fractions? Home Activity or Further Classroom Consolidation
#### BLM 3.1 Equivalence on the Number Line
Our group’s questions:
## Unit Equivalency in Fractions: Day 4: Determining Equivalency Junior/Int
MO 15 min A 40 min C/D 20 min 75 min Math Learning Goals Students will: represent fractions on a number line reason about meaning of a fraction and the relationship between numerator and denominator communicate strengths of different representations for different students and in certain contexts e.g. , use of benchmarks to support and refine the meaning of fractions Materials envelopes with fraction pieces, one per pair chart paper Minds On... Think-Pair Reflection Individually, students write a journal entry about the number line. What is it? What are the rules? Make a number line that you would use to show 2⁄8. Could you use this number line to show 2⁄5? Pairs share their journal entries. Action! Small Groups Exploration Distribute envelopes containing Set A or Set B below of eight fractions in numerical form to pairs. Students decide how they will place their set of fractions on a number line. Remind students that they may need to make more than one number line to help show fractions that are equivalent. [Set A: 2⁄4, 1⁄2, 1⁄3, 2⁄6, 3⁄4, 6⁄8, 5⁄10, 3⁄9] [Set B: 8⁄6, 1 2⁄6, 1 1⁄3, 4⁄3, 5⁄8, 2⁄8, 1⁄4, 1⁄3] Consolidate/ Debrief Whole Class Discussion How did you decide whether to do one / two / three number lines? How are your number lines like the fraction towers? Which numbers were easy to place? What did you find challenging? What questions do you still have about equivalent fractions? Home Activity or Further Classroom Consolidation
[Choose from these for students who need more practice, or an extension.]
Choose a fraction. Show as many equivalent fractions as you can, using the manipulatives provided.
What do you notice about the numerator and denominator in the fractions 35 and 1220?
You have a bag of marbles. There are 5 black and 5 white marbles in the bag. What is the probability of getting a black? Now, if you remove one black and one white (4 of each), what is the probability of pulling out a black? What about 3 of each? What do you notice?
Look at the two granola bars. Sue says that the two bars both show 14 of the granola bar is shaded. Mitchell says that the fractions are different. Who is right?
Jian threw his paper airplane 0.66 m and Sylvain threw his 23 of a meter. Whose airplane went the farthest?
(Other examples: 0.75 and 34; 0.2 and 210; 1.25m and 54) | 2,074 | 9,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-05 | latest | en | 0.942141 |
https://solvedlib.com/n/starting-with-an-initial-value-of-p-0-25-the-population,4450282 | 1,696,116,590,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00558.warc.gz | 579,003,510 | 17,593 | # Starting with an initial value of P(0) = 25, the population of prairie dog community grows at a rate of
###### Question:
Starting with an initial value of P(0) = 25, the population of prairie dog community grows at a rate of P'(t)= 10 - (in units of prairie dogsimonth) , for 0sts50 What is the population months later? Find the population P(t) for 0 sts50.
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company has such datas, what can you say about these all? what do you see? the condition of the company is good or bad? if good why if bad why? Net Profit Total operating revenue Total Equity Total Assets Net profit margin ROA ROE 2014 2015 % change 2016 481860 351320 -27 0996174775 4815177151545...
Explain what is meant by the term Cold War. (Make sure your answer names the two superpowers that were involved in the Cold War and explain why they were ‘fighting’.) ... | 3,344 | 11,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-40 | latest | en | 0.751249 |
http://nrich.maths.org/public/leg.php?code=-99&cl=3&cldcmpid=670 | 1,506,378,831,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693459.95/warc/CC-MAIN-20170925220350-20170926000350-00512.warc.gz | 243,543,216 | 10,052 | # Search by Topic
#### Resources tagged with Working systematically similar to All Square:
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### There are 128 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### Where Can We Visit?
##### Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Maths Trails
##### Stage: 2 and 3
The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails.
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Colour Islands Sudoku
##### Stage: 3 Challenge Level:
An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Twinkle Twinkle
##### Stage: 2 and 3 Challenge Level:
A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour.
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Product Sudoku
##### Stage: 3 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### American Billions
##### Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
##### Stage: 3 Challenge Level:
You need to find the values of the stars before you can apply normal Sudoku rules.
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### M, M and M
##### Stage: 3 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
##### Stage: 3 Challenge Level:
Rather than using the numbers 1-9, this sudoku uses the nine different letters used to make the words "Advent Calendar".
### Crossing the Bridge
##### Stage: 3 Challenge Level:
Four friends must cross a bridge. How can they all cross it in just 17 minutes?
### Multiples Sudoku
##### Stage: 3 Challenge Level:
Each clue in this Sudoku is the product of the two numbers in adjacent cells.
### Coins
##### Stage: 3 Challenge Level:
A man has 5 coins in his pocket. Given the clues, can you work out what the coins are?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Medal Muddle
##### Stage: 3 Challenge Level:
Countries from across the world competed in a sports tournament. Can you devise an efficient strategy to work out the order in which they finished?
### You Owe Me Five Farthings, Say the Bells of St Martin's
##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Rectangle Outline Sudoku
##### Stage: 3 and 4 Challenge Level:
Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.
### Integrated Product Sudoku
##### Stage: 3 and 4 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
### Tea Cups
##### Stage: 2 and 3 Challenge Level:
Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.
### Colour Islands Sudoku 2
##### Stage: 3, 4 and 5 Challenge Level:
In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
### Cinema Problem
##### Stage: 3 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### Tetrahedra Tester
##### Stage: 3 Challenge Level:
An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
### When Will You Pay Me? Say the Bells of Old Bailey
##### Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Twin Chute-swapping Sudoku
##### Stage: 4 and 5 Challenge Level:
A pair of Sudokus with lots in common. In fact they are the same problem but rearranged. Can you find how they relate to solve them both?
### Peaches Today, Peaches Tomorrow....
##### Stage: 3 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### LCM Sudoku II
##### Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Rainstorm Sudoku
##### Stage: 4 Challenge Level:
Use the clues about the shaded areas to help solve this sudoku | 2,242 | 9,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-39 | latest | en | 0.909057 |
https://math.libretexts.org/Courses/Mount_Royal_University/MATH_3200%3A_Mathematical_Methods/2%3A_Ordinary_differential_equations/2.4%3A_The_Method_of_Undetermined_Coefficient/2.4E%3A_Exercises | 1,586,328,469,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810617.95/warc/CC-MAIN-20200408041431-20200408071931-00059.warc.gz | 568,629,240 | 24,238 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 2.4E: Exercises
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In Exercises $$(2.4E.1)$$ to $$(2.4E.14)$$, find a particular solution.
## Exercise $$\PageIndex{1}$$
$$y''-3y'+2y=e^{3x}(1+x)$$
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## Exercise $$\PageIndex{2}$$
$$y''-6y'+5y=e^{-3x}(35-8x)$$
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## Exercise $$\PageIndex{3}$$
$$y''-2y'-3y=e^x(-8+3x)$$
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## Exercise $$\PageIndex{4}$$
$$y''+2y'+y=e^{2x}(-7-15x+9x^2)$$
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## Exercise $$\PageIndex{5}$$
$$y''+4y=e^{-x}(7-4x+5x^2)$$
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## Exercise $$\PageIndex{6}$$
$$y''-y'-2y=e^x(9+2x-4x^2)$$
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## Exercise $$\PageIndex{7}$$
$$y''-4y'-5y=-6xe^{-x}$$
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## Exercise $$\PageIndex{8}$$
$$y''-3y'+2y=e^x(3-4x)$$
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## Exercise $$\PageIndex{9}$$
$$y''+y'-12y=e^{3x}(-6+7x)$$
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## Exercise $$\PageIndex{10}$$
$$2y''-3y'-2y=e^{2x}(-6+10x)$$
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## Exercise $$\PageIndex{11}$$
$$y''+2y'+y=e^{-x}(2+3x)$$
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## Exercise $$\PageIndex{12}$$
$$y''-2y'+y=e^x(1-6x)$$
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## Exercise $$\PageIndex{13}$$
$$y''-4y'+4y=e^{2x}(1-3x+6x^2)$$
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## Exercise $$\PageIndex{14}$$
$$9y''+6y'+y=e^{-x/3}(2-4x+4x^2)$$
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In Exercises $$(2.4E.15)$$ to $$(2.4E.19)$$, find the general solution.
## Exercise $$\PageIndex{15}$$
$$y''-3y'+2y=e^{3x}(1+x)$$
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## Exercise $$\PageIndex{16}$$
$$y''-6y'+8y=e^x(11-6x)$$
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## Exercise $$\PageIndex{17}$$
$$y''+6y'+9y=e^{2x}(3-5x)$$
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## Exercise $$\PageIndex{18}$$
$$y''+2y'-3y=-16xe^x$$
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## Exercise $$\PageIndex{19}$$
$$y''-2y'+y=e^x(2-12x)$$
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In Exercises $$(2.4E.20)$$ to $$(2.4E.23)$$, solve the initial value problem and plot the solution.
## Exercise $$\PageIndex{20}$$
$$y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10$$
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## Exercise $$\PageIndex{21}$$
$$y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8$$
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## Exercise $$\PageIndex{22}$$
$$y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2$$
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## Exercise $$\PageIndex{23}$$
$$y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17$$
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In Exercises $$(2.4E.24)$$ to $$(2.4E.29)$$, use the principle of superposition to find a particular solution.
## Exercise $$\PageIndex{24}$$
$$y''+y'+y=xe^x+e^{-x}(1+2x)$$
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## Exercise $$\PageIndex{25}$$
$$y''-7y'+12y=-e^x(17-42x)-e^{3x}$$
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## Exercise $$\PageIndex{26}$$
$$y''-8y'+16y=6xe^{4x}+2+16x+16x^2$$
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## Exercise $$\PageIndex{27}$$
$$y''-3y'+2y=-e^{2x}(3+4x)-e^x$$
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## Exercise $$\PageIndex{28}$$
$$y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)$$
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## Exercise $$\PageIndex{29}$$
$$y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)$$
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## Exercise $$\PageIndex{30}$$
(a) Prove that $$y$$ is a solution of the constant coefficient equation
\label{eq:2.4E.1}
ay''+by'+cy=e^{\alpha x}G(x)
if and only if $$y=ue^{\alpha x}$$, where $$u$$ satisfies
\label{eq:2.4E.2}
au''+p'(\alpha)u'+p(\alpha)u=G(x)
and $$p(r)=ar^2+br+c$$ is the characteristic polynomial of the complementary equation
\begin{eqnarray*}
ay''+by'+cy=0.
\end{eqnarray*}
For the rest of this exercise, let $$G$$ be a polynomial. Give the requested proofs for the case where
\begin{eqnarray*}
G(x)=g_0+g_1x+g_2x^2+g_3x^3.
\end{eqnarray*}
(b) Prove that if $$e^{\alpha x}$$ isn't a solution of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $$u_p=A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example $$(2.4.4)$$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $$y_p=e^{\alpha x}A(x)$$.
(c) Show that if $$e^{\alpha x}$$ is a solution of the complementary equation and $$xe^{\alpha x}$$ isn't, then \eqref{eq:2.4E.2} has a particular solution of the form $$u_p=xA(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example $$(2.4.5)$$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $$y_p=xe^{\alpha x}A(x)$$.
(d) Show that if $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are both solutions of the complementary equation then \eqref{eq:2.4E.2} has a particular solution of the form $$u_p=x^2A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, and $$x^2A(x)$$ can be obtained by integrating $$G/a$$ twice, taking the constants of integration to be zero, as in Example $$(2.4.6)$$. Conclude that \eqref{eq:2.4E.1} has a particular solution of the form $$y_p=x^2e^{\alpha x}A(x)$$.
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Exercises $$(2.4E.31)$$ to $$(2.4E.36)$$ treat the equations considered in Examples $$(2.4.1)$$ to $$(2.4.6)$$. Substitute the suggested form of $$y_p$$ into the equation and equate the resulting coefficients of like functions on the two sides of the resulting equation to derive a set of simultaneous equations for the coefficients in $$y_p$$. Then solve for the coefficients to obtain $$y_p$$. Compare the work you've done with the work required to obtain the same results in Examples $$(2.4.1)$$ to $$(2.4.6)$$.
## Exercise $$\PageIndex{31}$$
Compare with Example $$(2.4.1)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{32}$$
Compare with Example $$(2.4.2)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{33}$$
Compare with Example $$(2.4.3)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{34}$$
Compare with Example $$(2.4.4)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{35}$$
Compare with Example $$(2.4.5)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{36}$$
Compare with Example $$(2.4.6)$$:
\begin{eqnarray*}
\end{eqnarray*}
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## Exercise $$\PageIndex{37}$$
Write $$y=ue^{\alpha x}$$ to find the general solution.
(a) $$y''+2y'+y=\displaystyle{e^{-x}\over\sqrt x}$$
(b) $$y''+6y'+9y=e^{-3x}\ln x$$
(c) $$y''-4y'+4y=\displaystyle{e^{2x}\over1+x}$$
(d) $$4y''+4y'+y=\displaystyle{4e^{-x/2}\left({1\over x}+x\right)}$$
Add texts here. Do not delete this text first.
## Exercise $$\PageIndex{38}$$
Suppose $$\alpha\ne0$$ and $$k$$ is a positive integer. In most calculus books integrals like $$\int x^k e^{\alpha x}\,dx$$ are evaluated by integrating by parts $$k$$ times. This exercise presents another method. Let
\begin{eqnarray*}
y=\int e^{\alpha x}P(x)\,dx
\end{eqnarray*}
with
\begin{eqnarray*}
P(x)=p_0+p_1x+\cdots+p_kx^k, \mbox{\quad (where $p_k\ne0$)}.
\end{eqnarray*}
(a) Show that $$y=e^{\alpha x}u$$, where
\label{eq:2.4E.3}
u'+\alpha u=P(x).
(b) Show that \eqref{eq:2.4E.3} has a particular solution of the form
\begin{eqnarray*}
u_p=A_0+A_1x+\cdots+A_kx^k,
\end{eqnarray*}
where $$A_k$$, $$A_{k-1}$$, $$\dots$$, $$A_0$$ can be computed successively by equating coefficients of $$x^k,x^{k-1}, \dots,1$$ on both sides of the equation
\begin{eqnarray*}
u_p'+\alpha u_p=P(x).
\end{eqnarray*}
(c) Conclude that
\begin{eqnarray*}
\int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,
\end{eqnarray*}
where $$c$$ is a constant of integration.
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## Exercise $$\PageIndex{39}$$
Use the method of Exercise $$(2.4E.38)$$ to evaluate the integral.
(a) $$\int e^x(4+x)\,dx$$
(b) $$\int e^{-x}(-1+x^2)\,dx$$
(c) $$\int x^3e^{-2x}\,dx$$
(d) $$\int e^x(1+x)^2\,dx$$
(e) $$\int e^{3x}(-14+30x+27x^2)\,dx$$
(f) $$\int e^{-x}(1+6x^2-14x^3+3x^4)\,dx$$
## Exercise $$\PageIndex{40}$$
Use the method suggested in Exercise $$(2.4E.38)$$ to evaluate $$\int x^ke^{\alpha x}\,dx$$, where $$k$$ is an arbitrary positive integer and $$\alpha\ne0$$. | 3,816 | 10,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-16 | latest | en | 0.337151 |
https://www.mathsai.com/worksheet/equivalent-algebraic-expression.htm | 1,580,140,399,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00472.warc.gz | 974,908,311 | 14,960 | ## Find the Equivalent Algebraic Expression
Name
Date
1
9(1+10) =
1. 23+(3x8)
2. 19+(30-8)
3. 2(3+9)
4. 3(14+19)
2
2(5+12) =
1. 2x11
2. 2x17
3. 2x47
4. 32+(3x13)
3
3x19 =
1. 2(14-4)
2. 3(8+11)
3. 5(6+13)
4. 26+(43-10)
4
16+(3x7) =
1. 18+(23-4)
2. 2(2+12)
3. 4(5+6)
4. 26+(41-6)
5
3x7 =
1. 5x7
2. 3(38-6)
3. 3(3+4)
4. 4(13-8)
6
5x11 =
1. 9(0+9)
2. 5(1+10)
3. 2(19+24)
4. 8(2+6)
7
2(5+6) =
1. 2(2+4)
2. 26+(3x9)
3. 2x11
4. 5+(2x7)
8
35+(2x19) =
1. 17+(30-4)
2. 4(7+15)
3. 2(21+24)
4. 35+(46-8)
9
3x9 =
1. 2(0+10)
2. 3(0+9)
3. 4(4+7)
4. 7(12-3)
10
2(18+23) =
1. 3(13+16)
2. 2(18+26)
3. 16+(5x5)
4. 2x41
11
46+(3x17) =
1. 3x7
2. 3(12+20)
3. 46+(58-7)
4. 37+(3x14)
12
2(6+16) =
1. 7+(2x5)
2. 2(36-4)
3. 4(4+7)
4. 3(1+4)
13
4+(14-5) =
1. 4(27-6)
2. 16+(5x5)
3. 4+(3x3)
4. 2(6+8)
14
2(4+5) =
1. 3(1+5)
2. 2(-2+7)
3. 6(4+7)
4. 38+(5x9)
15
2(3+7) =
1. 2(16-4)
2. 5(3+9)
3. 4(8-3)
4. 2(10+17)
16
3x5 =
1. 19+(27-3)
2. 5x19
3. 3(1+4)
4. 27+(39-5)
17
2(35-7) =
1. 2(24-10)
2. 4(18-4)
3. 32+(41-2)
4. 2(41-9)
18
17+(35-9) =
1. 2(9+16)
2. 17+(2x13)
3. 4(12+13)
4. 7(5+8)
19
2(8+11) =
1. 2(3+6)
2. 4(2+3)
3. 5(8+9)
4. 2x19
20
3x11 =
1. 5(1+5)
2. 2(15-5)
3. 2(7+14)
4. 3(2+9)
Show All Workout
1
9(1+10) = 3(14+19)
Solution
9(1+10) = 9 x 11 = 99
3(14+19) = 3 x 33 = 99
Hence 9(1+10) = 3(14+19)
2
2(5+12) = 2x17
Solution
2(5+12) = 2 x 17 = 34
2 x 17 = 34
Hence 2(5+12) = 2x17
3
3x19 = 3(8+11)
Solution
3 x 19 = 57
3(8+11) = 3 x 19 = 57
Hence 3x19 = 3(8+11)
4
16+(3x7) = 18+(23-4)
Solution
16+(3x7) = 16 + 21 = 37
18+(23-4) = 18 + 19 = 37
Hence 16+(3x7) = 18+(23-4)
5
3x7 = 3(3+4)
Solution
3 x 7 = 21
3(3+4) = 3 x 7 = 21
Hence 3x7 = 3(3+4)
6
5x11 = 5(1+10)
Solution
5 x 11 = 55
5(1+10) = 5 x 11 = 55
Hence 5x11 = 5(1+10)
7
2(5+6) = 2x11
Solution
2(5+6) = 2 x 11 = 22
2 x 11 = 22
Hence 2(5+6) = 2x11
8
35+(2x19) = 35+(46-8)
Solution
35+(2x19) = 35 + 38 = 73
35+(46-8) = 35 + 38 = 73
Hence 35+(2x19) = 35+(46-8)
9
3x9 = 3(0+9)
Solution
3 x 9 = 27
3(0+9) = 3 x 9 = 27
Hence 3x9 = 3(0+9)
10
2(18+23) = 2x41
Solution
2(18+23) = 2 x 41 = 82
2 x 41 = 82
Hence 2(18+23) = 2x41
11
46+(3x17) = 46+(58-7)
Solution
46+(3x17) = 46 + 51 = 97
46+(58-7) = 46 + 51 = 97
Hence 46+(3x17) = 46+(58-7)
12
2(6+16) = 4(4+7)
Solution
2(6+16) = 2 x 22 = 44
4(4+7) = 4 x 11 = 44
Hence 2(6+16) = 4(4+7)
13
4+(14-5) = 4+(3x3)
Solution
4+(14-5) = 4 + 9 = 13
4+(3x3) = 4 + 9 = 13
Hence 4+(14-5) = 4+(3x3)
14
2(4+5) = 3(1+5)
Solution
2(4+5) = 2 x 9 = 18
3(1+5) = 3 x 6 = 18
Hence 2(4+5) = 3(1+5)
15
2(3+7) = 4(8-3)
Solution
2(3+7) = 2 x 10 = 20
4(8-3) = 4 x 5 = 20
Hence 2(3+7) = 4(8-3)
16
3x5 = 3(1+4)
Solution
3 x 5 = 15
3(1+4) = 3 x 5 = 15
Hence 3x5 = 3(1+4)
17
2(35-7) = 4(18-4)
Solution
2(35-7) = 2 x 28 = 56
4(18-4) = 4 x 14 = 56
Hence 2(35-7) = 4(18-4)
18
17+(35-9) = 17+(2x13)
Solution
17+(35-9) = 17 + 26 = 43
17+(2x13) = 17 + 26 = 43
Hence 17+(35-9) = 17+(2x13)
19
2(8+11) = 2x19
Solution
2(8+11) = 2 x 19 = 38
2 x 19 = 38
Hence 2(8+11) = 2x19
20
3x11 = 3(2+9)
Solution
3 x 11 = 33
3(2+9) = 3 x 11 = 33
Hence 3x11 = 3(2+9)
# Find the Equivalent Algebraic Expressions Worksheet
Find the equivalent algebraic expressions worksheet with answers for 6th grade math curriculum is available online for free in printable and downloadable (pdf & image) format. Tap on PRINT, PDF or IMAGE button to print or download this grade-6 algebra worksheet to practice finding an equivalent algebraic expression from the given options.
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In this sixth grade algebra worksheet, students are required to select an equivalent algebraic expression from the given options for the given expression. The selected expression must be equal to the given expression to form the algebraic equation. To choose the correct expression, find the numerical value by solving the given expression and for the expressions in the options. Match the expressions having the same numerical values.
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https://en.m.wikipedia.org/wiki/Jensen%27s_inequality | 1,632,075,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00293.warc.gz | 281,386,034 | 39,110 | # Jensen's inequality
In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function. It was proved by Jensen in 1906.[1] Given its generality, the inequality appears in many forms depending on the context, some of which are presented below. In its simplest form the inequality states that the convex transformation of a mean is less than or equal to the mean applied after convex transformation; it is a simple corollary that the opposite is true of concave transformations.[2]
Jensen's inequality generalizes the statement that a secant line of a convex function lies above the graph.
Visualizing Convexity and Jensen's Inequality
Jensen's inequality generalizes the statement that the secant line of a convex function lies above the graph of the function, which is Jensen's inequality for two points: the secant line consists of weighted means of the convex function (for t ∈ [0,1]),
${\displaystyle tf(x_{1})+(1-t)f(x_{2}),}$
while the graph of the function is the convex function of the weighted means,
${\displaystyle f\left(tx_{1}+(1-t)x_{2}\right).}$
Thus, Jensen's inequality is
${\displaystyle f\left(tx_{1}+(1-t)x_{2}\right)\leq tf(x_{1})+(1-t)f(x_{2}).}$
In the context of probability theory, it is generally stated in the following form: if X is a random variable and φ is a convex function, then
${\displaystyle \varphi \left(\operatorname {E} [X]\right)\leq \operatorname {E} \left[\varphi (X)\right].}$
The difference between the two sides of the inequality, ${\displaystyle \operatorname {E} \left[\varphi (X)\right]-\varphi \left(\operatorname {E} [X]\right)}$, is called the Jensen gap.[3]
## Statements
The classical form of Jensen's inequality involves several numbers and weights. The inequality can be stated quite generally using either the language of measure theory or (equivalently) probability. In the probabilistic setting, the inequality can be further generalized to its full strength.
### Finite form
For a real convex function ${\displaystyle \varphi }$ , numbers ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ in its domain, and positive weights ${\displaystyle a_{i}}$ , Jensen's inequality can be stated as:
${\displaystyle \varphi \left({\frac {\sum a_{i}x_{i}}{\sum a_{i}}}\right)\leq {\frac {\sum a_{i}\varphi (x_{i})}{\sum a_{i}}}}$
(1)
and the inequality is reversed if ${\displaystyle \varphi }$ is concave, which is
${\displaystyle \varphi \left({\frac {\sum a_{i}x_{i}}{\sum a_{i}}}\right)\geq {\frac {\sum a_{i}\varphi (x_{i})}{\sum a_{i}}}.}$
(2)
Equality holds if and only if ${\displaystyle x_{1}=x_{2}=\cdots =x_{n}}$ or ${\displaystyle \varphi }$ is linear on a domain containing ${\displaystyle x_{1},x_{2},\cdots ,x_{n}}$ .
As a particular case, if the weights ${\displaystyle a_{i}}$ are all equal, then (1) and (2) become
${\displaystyle \varphi \left({\frac {\sum x_{i}}{n}}\right)\leq {\frac {\sum \varphi (x_{i})}{n}}}$
(3)
${\displaystyle \varphi \left({\frac {\sum x_{i}}{n}}\right)\geq {\frac {\sum \varphi (x_{i})}{n}}}$
(4)
For instance, the function log(x) is concave, so substituting ${\displaystyle \varphi (x)=\log(x)}$ in the previous formula (4) establishes the (logarithm of the) familiar arithmetic-mean/geometric-mean inequality:
${\displaystyle \log \!\left({\frac {\sum _{i=1}^{n}x_{i}}{n}}\right)\geq {\frac {\sum _{i=1}^{n}\log \!\left(x_{i}\right)}{n}}\quad {\text{or}}\quad {\frac {x_{1}+x_{2}+\cdots +x_{n}}{n}}\geq {\sqrt[{n}]{x_{1}\cdot x_{2}\cdots x_{n}}}}$
A common application has x as a function of another variable (or set of variables) t, that is, ${\displaystyle x_{i}=g(t_{i})}$ . All of this carries directly over to the general continuous case: the weights ai are replaced by a non-negative integrable function f (x), such as a probability distribution, and the summations are replaced by integrals.
### Measure-theoretic and probabilistic form
Let ${\displaystyle (\Omega ,A,\mu )}$ be a probability space, i.e. ${\displaystyle \mu (\Omega )=1}$ . If ${\displaystyle g}$ is a real-valued function that is ${\displaystyle \mu }$ -integrable, and if ${\displaystyle \varphi }$ is a convex function on the real line, then:
${\displaystyle \varphi \left(\int _{\Omega }g\,d\mu \right)\leq \int _{\Omega }\varphi \circ g\,d\mu .}$ [4]
In real analysis, we may require an estimate on
${\displaystyle \varphi \left(\int _{a}^{b}f(x)\,dx\right),}$
where ${\displaystyle a,b\in \mathbb {R} }$ , and ${\displaystyle f\colon [a,b]\to \mathbb {R} }$ is a non-negative Lebesgue-integrable function. In this case, the Lebesgue measure of ${\displaystyle [a,b]}$ need not be unity. However, by integration by substitution, the interval can be rescaled so that it has measure unity. Then Jensen's inequality can be applied to get[5]
${\displaystyle \varphi \left({\frac {1}{b-a}}\int _{a}^{b}f(x)\,dx\right)\leq {\frac {1}{b-a}}\int _{a}^{b}\varphi (f(x))\,dx.}$
The same result can be equivalently stated in a probability theory setting, by a simple change of notation. Let ${\displaystyle (\Omega ,{\mathfrak {F}},\operatorname {P} )}$ be a probability space, X an integrable real-valued random variable and φ a convex function. Then:
${\displaystyle \varphi \left(\operatorname {E} [X]\right)\leq \operatorname {E} \left[\varphi (X)\right].}$ [6]
In this probability setting, the measure μ is intended as a probability ${\displaystyle \operatorname {P} }$ , the integral with respect to μ as an expected value ${\displaystyle \operatorname {E} }$ , and the function ${\displaystyle g}$ as a random variable X.
Note that the equality holds if and only if φ is a linear function on some convex set ${\displaystyle A}$ such that ${\displaystyle \mathrm {P} (X\in A)=1}$ (which follows by inspecting the measure-theoretical proof below).
### General inequality in a probabilistic setting
More generally, let T be a real topological vector space, and X a T-valued integrable random variable. In this general setting, integrable means that there exists an element ${\displaystyle \operatorname {E} [X]}$ in T, such that for any element z in the dual space of T: ${\displaystyle \operatorname {E} |\langle z,X\rangle |<\infty }$ , and ${\displaystyle \langle z,\operatorname {E} [X]\rangle =\operatorname {E} [\langle z,X\rangle ]}$ . Then, for any measurable convex function φ and any sub-σ-algebra ${\displaystyle {\mathfrak {G}}}$ of ${\displaystyle {\mathfrak {F}}}$ :
${\displaystyle \varphi \left(\operatorname {E} \left[X\mid {\mathfrak {G}}\right]\right)\leq \operatorname {E} \left[\varphi (X)\mid {\mathfrak {G}}\right].}$
Here ${\displaystyle \operatorname {E} [\cdot \mid {\mathfrak {G}}]}$ stands for the expectation conditioned to the σ-algebra ${\displaystyle {\mathfrak {G}}}$ . This general statement reduces to the previous ones when the topological vector space T is the real axis, and ${\displaystyle {\mathfrak {G}}}$ is the trivial σ-algebra {∅, Ω} (where is the empty set, and Ω is the sample space).[7]
### A sharpened and generalized form
Let X be a one-dimensional random variable with mean ${\displaystyle \mu }$ and variance ${\displaystyle \sigma ^{2}\geq 0}$ . Let ${\displaystyle \varphi (x)}$ be a twice differentiable function, and define the function
${\displaystyle h(x)\triangleq {\frac {\varphi \left(x\right)-\varphi \left(\mu \right)}{\left(x-\mu \right)^{2}}}-{\frac {\varphi '\left(\mu \right)}{x-\mu }}.}$
Then[8]
${\displaystyle \sigma ^{2}\inf {\frac {\varphi ''(x)}{2}}\leq \sigma ^{2}\inf h(x)\leq E\left[\varphi \left(X\right)\right]-\varphi \left(E[X]\right)\leq \sigma ^{2}\sup h(x)\leq \sigma ^{2}\sup {\frac {\varphi ''(x)}{2}}.}$
In particular, when ${\displaystyle \varphi (x)}$ is convex, then ${\displaystyle \varphi ''(x)\geq 0}$ , and the standard form of Jensen's inequality immediately follows for the case where ${\displaystyle \varphi (x)}$ is additionally assumed to be twice differentiable.
## Proofs
A graphical "proof" of Jensen's inequality for the probabilistic case. The dashed curve along the X axis is the hypothetical distribution of X, while the dashed curve along the Y axis is the corresponding distribution of Y values. Note that the convex mapping Y(X) increasingly "stretches" the distribution for increasing values of X.
This is a proof without words of Jensen's inequality for n variables. Without loss of generality, the sum of the positive weights is 1. It follows that the weighted point lies in the convex hull of the original points, which lies above the function itself by the definition of convexity. The conclusion follows.[9]
Jensen's inequality can be proved in several ways, and three different proofs corresponding to the different statements above will be offered. Before embarking on these mathematical derivations, however, it is worth analyzing an intuitive graphical argument based on the probabilistic case where X is a real number (see figure). Assuming a hypothetical distribution of X values, one can immediately identify the position of ${\displaystyle \operatorname {E} [X]}$ and its image ${\displaystyle \varphi (\operatorname {E} [X])}$ in the graph. Noticing that for convex mappings Y = φ(X) the corresponding distribution of Y values is increasingly "stretched out" for increasing values of X, it is easy to see that the distribution of Y is broader in the interval corresponding to X > X0 and narrower in X < X0 for any X0; in particular, this is also true for ${\displaystyle X_{0}=\operatorname {E} [X]}$ . Consequently, in this picture the expectation of Y will always shift upwards with respect to the position of ${\displaystyle \varphi (\operatorname {E} [X])}$ . A similar reasoning holds if the distribution of X covers a decreasing portion of the convex function, or both a decreasing and an increasing portion of it. This "proves" the inequality, i.e.
${\displaystyle \varphi (\operatorname {E} [X])\leq \operatorname {E} [\varphi (X)]=\operatorname {E} [Y],}$
with equality when φ(X) is not strictly convex, e.g. when it is a straight line, or when X follows a degenerate distribution (i.e. is a constant).
The proofs below formalize this intuitive notion.
### Proof 1 (finite form)
If λ1 and λ2 are two arbitrary nonnegative real numbers such that λ1 + λ2 = 1 then convexity of φ implies
${\displaystyle \forall x_{1},x_{2}:\qquad \varphi \left(\lambda _{1}x_{1}+\lambda _{2}x_{2}\right)\leq \lambda _{1}\,\varphi (x_{1})+\lambda _{2}\,\varphi (x_{2}).}$
This can be generalized: if λ1, ..., λn are nonnegative real numbers such that λ1 + ... + λn = 1, then
${\displaystyle \varphi (\lambda _{1}x_{1}+\lambda _{2}x_{2}+\cdots +\lambda _{n}x_{n})\leq \lambda _{1}\,\varphi (x_{1})+\lambda _{2}\,\varphi (x_{2})+\cdots +\lambda _{n}\,\varphi (x_{n}),}$
for any x1, ..., xn.
The finite form of the Jensen's inequality can be proved by induction: by convexity hypotheses, the statement is true for n = 2. Suppose the statement is true for some n, so
${\displaystyle \varphi \left(\sum _{i=1}^{n}\lambda _{i}x_{i}\right)\leq \sum _{i=1}^{n}\lambda _{i}\varphi \left(x_{i}\right)}$
for any λ1, ..., λn such that λ1 + ... + λn = 1.
One needs to prove it for n + 1. At least one of the λi is strictly smaller than ${\displaystyle 1}$ , say λn+1; therefore by convexity inequality:
{\displaystyle {\begin{aligned}\varphi \left(\sum _{i=1}^{n+1}\lambda _{i}x_{i}\right)&=\varphi \left((1-\lambda _{n+1})\sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}x_{i}+\lambda _{n+1}x_{n+1}\right)\\&\leq (1-\lambda _{n+1})\varphi \left(\sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}x_{i}\right)+\lambda _{n+1}\,\varphi (x_{n+1}).\end{aligned}}}
Since λ1 + ... +λn + λn+1 = 1,
${\displaystyle \sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}=1}$ ,
applying the induction hypothesis gives
${\displaystyle \varphi \left(\sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}x_{i}\right)\leq \sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}\varphi \left(x_{i}\right)}$
therefore
{\displaystyle {\begin{aligned}\varphi \left(\sum _{i=1}^{n+1}\lambda _{i}x_{i}\right)&\leq (1-\lambda _{n+1})\sum _{i=1}^{n}{\frac {\lambda _{i}}{1-\lambda _{n+1}}}\varphi \left(x_{i}\right)+\lambda _{n+1}\,\varphi (x_{n+1})=\sum _{i=1}^{n+1}\lambda _{i}\varphi \left(x_{i}\right)\end{aligned}}}
We deduce the equality is true for n + 1, by the principle of mathematical induction it follows that the result is also true for all integer n greater than 2.
In order to obtain the general inequality from this finite form, one needs to use a density argument. The finite form can be rewritten as:
${\displaystyle \varphi \left(\int x\,d\mu _{n}(x)\right)\leq \int \varphi (x)\,d\mu _{n}(x),}$
where μn is a measure given by an arbitrary convex combination of Dirac deltas:
${\displaystyle \mu _{n}=\sum _{i=1}^{n}\lambda _{i}\delta _{x_{i}}.}$
Since convex functions are continuous, and since convex combinations of Dirac deltas are weakly dense in the set of probability measures (as could be easily verified), the general statement is obtained simply by a limiting procedure.
### Proof 2 (measure-theoretic form)
Let g be a real-valued μ-integrable function on a probability space Ω, and let φ be a convex function on the real numbers. Since φ is convex, at each real number x we have a nonempty set of subderivatives, which may be thought of as lines touching the graph of φ at x, but which are at or below the graph of φ at all points (support lines of the graph).
Now, if we define
${\displaystyle x_{0}:=\int _{\Omega }g\,d\mu ,}$
because of the existence of subderivatives for convex functions, we may choose a and b such that
${\displaystyle ax+b\leq \varphi (x),}$
for all real x and
${\displaystyle ax_{0}+b=\varphi (x_{0}).}$
But then we have that
${\displaystyle \varphi \circ g(x)\geq ag(x)+b}$
for all x. Since we have a probability measure, the integral is monotone with μ(Ω) = 1 so that
${\displaystyle \int _{\Omega }\varphi \circ g\,d\mu \geq \int _{\Omega }(ag+b)\,d\mu =a\int _{\Omega }g\,d\mu +b\int _{\Omega }d\mu =ax_{0}+b=\varphi (x_{0})=\varphi \left(\int _{\Omega }g\,d\mu \right),}$
as desired.
### Proof 3 (general inequality in a probabilistic setting)
Let X be an integrable random variable that takes values in a real topological vector space T. Since ${\displaystyle \varphi :T\to \mathbb {R} }$ is convex, for any ${\displaystyle x,y\in T}$ , the quantity
${\displaystyle {\frac {\varphi (x+\theta \,y)-\varphi (x)}{\theta }},}$
is decreasing as θ approaches 0+. In particular, the subdifferential of ${\displaystyle \varphi }$ evaluated at x in the direction y is well-defined by
${\displaystyle (D\varphi )(x)\cdot y:=\lim _{\theta \downarrow 0}{\frac {\varphi (x+\theta \,y)-\varphi (x)}{\theta }}=\inf _{\theta \neq 0}{\frac {\varphi (x+\theta \,y)-\varphi (x)}{\theta }}.}$
It is easily seen that the subdifferential is linear in y[citation needed] (that is false and the assertion requires Hahn-Banach theorem to be proved) and, since the infimum taken in the right-hand side of the previous formula is smaller than the value of the same term for θ = 1, one gets
${\displaystyle \varphi (x)\leq \varphi (x+y)-(D\varphi )(x)\cdot y.}$
In particular, for an arbitrary sub-σ-algebra ${\displaystyle {\mathfrak {G}}}$ we can evaluate the last inequality when ${\displaystyle x=\operatorname {E} [X\mid {\mathfrak {G}}],\,y=X-\operatorname {E} [X\mid {\mathfrak {G}}]}$ to obtain
${\displaystyle \varphi (\operatorname {E} [X\mid {\mathfrak {G}}])\leq \varphi (X)-(D\varphi )(\operatorname {E} [X\mid {\mathfrak {G}}])\cdot (X-\operatorname {E} [X\mid {\mathfrak {G}}]).}$
Now, if we take the expectation conditioned to ${\displaystyle {\mathfrak {G}}}$ on both sides of the previous expression, we get the result since:
${\displaystyle \operatorname {E} \left[\left[(D\varphi )(\operatorname {E} [X\mid {\mathfrak {G}}])\cdot (X-\operatorname {E} [X\mid {\mathfrak {G}}])\right]\mid {\mathfrak {G}}\right]=(D\varphi )(\operatorname {E} [X\mid {\mathfrak {G}}])\cdot \operatorname {E} [\left(X-\operatorname {E} [X\mid {\mathfrak {G}}]\right)\mid {\mathfrak {G}}]=0,}$
by the linearity of the subdifferential in the y variable, and the following well-known property of the conditional expectation:
${\displaystyle \operatorname {E} \left[\left(\operatorname {E} [X\mid {\mathfrak {G}}]\right)\mid {\mathfrak {G}}\right]=\operatorname {E} [X\mid {\mathfrak {G}}].}$
## Applications and special cases
### Form involving a probability density function
Suppose Ω is a measurable subset of the real line and f(x) is a non-negative function such that
${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=1.}$
In probabilistic language, f is a probability density function.
Then Jensen's inequality becomes the following statement about convex integrals:
If g is any real-valued measurable function and ${\textstyle \varphi }$ is convex over the range of g, then
${\displaystyle \varphi \left(\int _{-\infty }^{\infty }g(x)f(x)\,dx\right)\leq \int _{-\infty }^{\infty }\varphi (g(x))f(x)\,dx.}$
If g(x) = x, then this form of the inequality reduces to a commonly used special case:
${\displaystyle \varphi \left(\int _{-\infty }^{\infty }x\,f(x)\,dx\right)\leq \int _{-\infty }^{\infty }\varphi (x)\,f(x)\,dx.}$
This is applied in Variational Bayesian methods.
### Example: even moments of a random variable
If g(x) = x2n, and X is a random variable, then g is convex as
${\displaystyle {\frac {d^{2}g}{dx^{2}}}(x)=2n(2n-1)x^{2n-2}\geq 0\quad \forall \ x\in \mathbb {R} }$
and so
${\displaystyle g(\operatorname {E} [X])=(\operatorname {E} [X])^{2n}\leq \operatorname {E} [X^{2n}].}$
In particular, if some even moment 2n of X is finite, X has a finite mean. An extension of this argument shows X has finite moments of every order ${\displaystyle l\in \mathbb {N} }$ dividing n.
### Alternative finite form
Let Ω = {x1, ... xn}, and take μ to be the counting measure on Ω, then the general form reduces to a statement about sums:
${\displaystyle \varphi \left(\sum _{i=1}^{n}g(x_{i})\lambda _{i}\right)\leq \sum _{i=1}^{n}\varphi (g(x_{i}))\lambda _{i},}$
provided that λi ≥ 0 and
${\displaystyle \lambda _{1}+\cdots +\lambda _{n}=1.}$
There is also an infinite discrete form.
### Statistical physics
Jensen's inequality is of particular importance in statistical physics when the convex function is an exponential, giving:
${\displaystyle e^{\operatorname {E} [X]}\leq \operatorname {E} \left[e^{X}\right],}$
where the expected values are with respect to some probability distribution in the random variable X.
The proof in this case is very simple (cf. Chandler, Sec. 5.5). The desired inequality follows directly, by writing
${\displaystyle \operatorname {E} \left[e^{X}\right]=e^{\operatorname {E} [X]}\operatorname {E} \left[e^{X-\operatorname {E} [X]}\right]}$
and then applying the inequality eX ≥ 1 + X to the final exponential.
### Information theory
If p(x) is the true probability density for X, and q(x) is another density, then applying Jensen's inequality for the random variable Y(X) = q(X)/p(X) and the convex function φ(y) = −log(y) gives
${\displaystyle \operatorname {E} [\varphi (Y)]\geq \varphi (\operatorname {E} [Y])}$
Therefore:
${\displaystyle -D(p(x)\|q(x))=\int p(x)\log \left({\frac {q(x)}{p(x)}}\right)\,dx\leq \log \left(\int p(x){\frac {q(x)}{p(x)}}\,dx\right)=\log \left(\int q(x)\,dx\right)=0}$
a result called Gibbs' inequality.
It shows that the average message length is minimised when codes are assigned on the basis of the true probabilities p rather than any other distribution q. The quantity that is non-negative is called the Kullback–Leibler divergence of q from p.
Since −log(x) is a strictly convex function for x > 0, it follows that equality holds when p(x) equals q(x) almost everywhere.
### Rao–Blackwell theorem
If L is a convex function and ${\displaystyle {\mathfrak {G}}}$ a sub-sigma-algebra, then, from the conditional version of Jensen's inequality, we get
${\displaystyle L(\operatorname {E} [\delta (X)\mid {\mathfrak {G}}])\leq \operatorname {E} [L(\delta (X))\mid {\mathfrak {G}}]\quad \Longrightarrow \quad \operatorname {E} [L(\operatorname {E} [\delta (X)\mid {\mathfrak {G}}])]\leq \operatorname {E} [L(\delta (X))].}$
So if δ(X) is some estimator of an unobserved parameter θ given a vector of observables X; and if T(X) is a sufficient statistic for θ; then an improved estimator, in the sense of having a smaller expected loss L, can be obtained by calculating
${\displaystyle \delta _{1}(X)=\operatorname {E} _{\theta }[\delta (X')\mid T(X')=T(X)],}$
the expected value of δ with respect to θ, taken over all possible vectors of observations X compatible with the same value of T(X) as that observed. Further, because T is a sufficient statistics, ${\displaystyle \delta _{1}(X)}$ does not depend on θ, hence, becomes a statistics.
This result is known as the Rao–Blackwell theorem.
## Notes
1. ^ Jensen, J. L. W. V. (1906). "Sur les fonctions convexes et les inégalités entre les valeurs moyennes". Acta Mathematica. 30 (1): 175–193. doi:10.1007/BF02418571.
2. ^ Dekking, F.M.; Kraaikamp, C.; Lopuhaa, H.P.; Meester, L.E. (2005). A Modern Introduction to Probability and Statistics: Understanding Why and How. London: Springer.
3. ^ Gao, Xiang; Sitharam, Meera; Roitberg, Adrian (2019). "Bounds on the Jensen Gap, and Implications for Mean-Concentrated Distributions" (PDF). The Australian Journal of Mathematical Analysis and Applications. 16 (2). arXiv:1712.05267.
4. ^ p. 25 of Rick Durrett (2019). Probability: Theory and Examples (5th ed.). Cambridge University Press. ISBN 978-1108473682.
5. ^ Niculescu, Constantin P. "Integral inequalities", P. 12.
6. ^ p. 29 of Rick Durrett (2019). Probability: Theory and Examples (5th ed.). Cambridge University Press. ISBN 978-1108473682.
7. ^ Attention: In this generality additional assumptions on the convex function and/ or the topological vector space are needed, see Example (1.3) on p. 53 in Perlman, Michael D. (1974). "Jensen's Inequality for a Convex Vector-Valued Function on an Infinite-Dimensional Space". Journal of Multivariate Analysis. 4 (1): 52–65. doi:10.1016/0047-259X(74)90005-0.
8. ^ Liao, J.; Berg, A (2018). "Sharpening Jensen's Inequality". American Statistician. arXiv:1707.08644. doi:10.1080/00031305.2017.1419145.
9. ^ Bradley, CJ (2006). Introduction to Inequalities. Leeds, United Kingdom: United Kingdom Mathematics Trust. p. 97. ISBN 978-1-906001-11-7. | 6,895 | 22,602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 104, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-39 | latest | en | 0.903327 |
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SULIT3472(2Additional MathematicsKertas 1September, 20082 ja m 30 minit
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH
NEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2008
Dua ja m tiga puluh mlnlt
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. This question paper consists o f three sections: Section A, Section B and Section C.
2. Answer all questions in Section A, four questions from Section B and two questionsfrom Section C.
3. Give only one answer/solution to each question.
5. The diagrams provided are not drawn according to scale unless stated.
6. The marks allocatedfor each question and sub - part of a question are shown inbrackets.
7. You may use a non-programmable scientific calculator.
8. A list offormulae is provided in page 3 and 4.
This question paper consists of 17 printed pages and 1 blank page.
3472/2 [Lihat sebelah
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Section ABahagian A[4 0 marks]
[4 0 markah]
Answer al l questions.Jawab semua soalan.
1. Solve the simultaneous equations 2x - Y = 3 and JI- xy + IGive your answers correct to three decimal places.
= 10.[5 marks]
Selesaikan persamaan serentak 2x - Y = 3 dan JI- xy + Y.Beri jawapan anda betul kepada tiga tempat perpuluhan.
= 10.[5 markah]
2.
Th e diagram shows the first three of an infinite series of circles. The
first circle has a radius o f 8 em, the second circle has a radius of 4 em,the third circle has a radius of2 em and so on.Find, in terms of'n,
[3 marks](a) the su m o f the circumferences, in em, of the first seven circles,
(b) the area, in cm 2, of this infinite series of circles. [3 marks]
Rajah menunjukkan tiga bulatan daripada satu siri bulatan yangtakterhingga. Bulatan pertama mempunyai jejari 8 em, bulatan keduamempunyai jejari 4 em , bulatan ketiga mempunyai jejari 2 em danseterusnya.
Cari, dalam sebutan n,
(a) hasil tambah panja ng lilitan, dalam em, bagi tujuh bulatanpertama, [3 markah]
2(b) luas, dalam cm , bagi siri bulatan yang takterhingga ini. [3 markah]
Additional Mathematics Paper 2 [Lihat sebelahSULIT
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3. The quadratic function j{x) =i + px + q has a minimum point at(3 , - 5).
(a) Without using differentiation. find the values ofp and q. [3 marks](b) Hence, find the range of the values of x if f (x) -31 ~ o . [2 marks]
Fungsi kuadratik j{x) = i + px + q mempunyai titik minimum di(3 , - 5).
(a) Tanpa menggunakan kaedah pembezaan, carikan nilai p dan q. [3 markah](b) Seterusnya, cari jul at nilai x jika f (x) -31 ~ o . [2 markah]
4. the identi cosxa) Prove e 1 entity tanx + sec x [3 marks]l+sinx
(b) (i) Sketch the graph of y =-3
sin 2x forO:s
x:s 21t .
[3 marks]
(ii) By using the same axes, sketch a suitable straight line to findthe number of solutions for the equation
2 - 3 sin 2x = 5 x for 0 < x < 21t- - .1r
State the number of solutions. [3 marks]
kosx(a) Buktikan identiti tan x + sek x [3 markah]l+sinx
(b) (i) Lakarkan graf bagi y =- 3 sin 2x untuk O:s x :s 21t . [3 markah]
(ii) Dengan menggunakan paksi yang sama, lakarkan satu garislurus yang sesuai untuk mencari bilangan penyelesaian bagi
persamaan 2 - 3sin 2x = 5 x untuk 0 :s x :s 21t.1r
Nyatakan b ilangan penyelesaian itu. [3 markah]
Addition al Mathematics Paper 2 [Lihat sebelahSULIT
3472/2
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SULIT 7 S ~ p t ~ m b ~ r , 1008
5. SetA has 20 numbers with mean 12 and standard deviation 3. Thenumbers in se t B are 5, 8, 10, 11 and 14. All the numbers in set A andset B are combined to form a ne w set, C.Calculate(a) the mea n,(b) the standard deviationfor set C. [7 marks]
Set A mengandungi 20 nombor dengan min 12 don sisihanpiawai 3.Nombor-nombor dalam set B ialah 5, 8, 10, 11 don 14. Kesemuanombor dart set A dan set B dicantum untuk membentuk satu setbaru, C.Hitungkan(a) min,(b) sisihanpiawaibagi set C. [7 marlcah]
Addltlona/ Mat1lDlUllics Paper Z [Ubat sebelabSULIT
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6.
o RTh e diagram shows a trapezium OPQR. The straight lines OQ andPR intersect at point S.Given OP=6g ,OR=15Q WId PQ : OR=3: 5.
(a) Express, in terms of g and lo r Q,
(i) PR(ii) OQ [3 markr]
(b) Given OS = hO Q an d PS = kPR ,express OS in terms o f(i) h, g and Q,(ii) k, g and ~ .Hence, find the value o f h and k . [5 markr]
Rajah menunjukkan satu trapezium OPQR. Garis lurus OQ dan PRbersilang di ti tik S.Diberi OP =6g , OR =15Q dan PQ: OR = 3 : 5.
(a) Ungkapkan, dalam sebutan q dan I atau Q,(i) PR(ii) OQ [3 markah]
(b) Dlberi OS = hO Q dan PS = kPR, ungkapkan OS dalamsebutan(i) h, g dan ~ ,(ii) k, g dan QSeterusnya, cari nilai h dan k . [5 markah]
Additional Mathematics Paper 2 [Libat sebelabSULIT
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SULlT 9 September, 1008
Section BBahagianB
[4 0 marks][ 40 markahs ]
7. The table below shows the values of two variables, x and y,obtained from an experiment. The variables x and y are related bythe equation y" =4kx , where n and k are constants.
x 1.5 2.0 2.5 4.0 5.5y 3.0 3.5 3.9 4.9 5.7
(a) Plot loglo y against loglo x , using a seale of 2 em to 0.1unit on both axes.
Hence, draw the line of best fit.
(b) Use your graph from (a) to find the value of
(i) n,(ii) k.
x dan y dihubungkan oleh persamaan y"=
[6 marks]
[4 marks]
x 1.5 2.0 2.5 4.0 5.5y 3.0 3.5 3.9 4.9 5.7
(a) Plot loglo y melawan loglo x , dengan menggunakan skala2 em kepada 0.1 unit pada kedua-dua paksi.
Seterusnya, lukis garis lurus penyuaian terbaik: [6 markah]
(b) Gunakan graf anda dari (a) untuk mencari nilai
(i)(ii)
n,k. [4markah]
3472/2 Additional MatheltUltics Paper1 [ l ihat sebelahSULIT
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SULIT 10 SeptembeT. 2008
8.
2y=x 2y
y = x + 4
x
The diagram shows the straight line y =x + 4 intersecting the curve2y = ; at the points P and Q.
Find(a) the coordinates of P and Q. [3 marks)
(b) the area of the shaded region A. [4 marks]
(c) the volume generated. in terms of 7 t. when the shaded regionB is revolved through 360 0 about the x-axis. [3 marks]
Rajah menunjukkan garis lurus y =x + 4 yang menyilang/engkung2y = ~ pada titik P dan Q.
Cari(a) koordinat P dan Q. [3 markah]
(b) /uas rantau berlorekA. [4 markah]
(c) isipadujanaan, da/amsebutan 7 t, apabilarantau berlorek B
dikisarkan melalui 360 0 pada paksi-x. [3 markah]
Addi tional MaJhematics Paper 2 [Lib at sebelahSULIT
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9. Solutions by scale drawing will not be accepted.
Penyelesaian seeara lukisan berskala tidak diterima.
y
o x
In the diagram. OABC is a parallelogram and A is the point (2. 6).
The equations of OC and BC are x - 2y = 0 and y = px + qrespectively. where p and q are constants. The perpendicular fromA to OC meets at the point D.
Given that D is the mid-point of OCt find
(b) the coordinates of D.
(c)
(d)
the values o fp and q.
the area, in unir.of the parallelogram OABC. [10 marks]
Dalam rajah. OABC adalah sebuah segiempat selari dan A ialahtltik (2. 6). Persamaan-persamaan bagi OC dan BC adalahmasing-masing x - 2y = 0 dan y = px + q dengan keadaan pdan q ialah pemalar. Garis serenjang dari A ke OC bertemu dititik D.
Diberi bahawa D ialah titik tengah OC, earikan
(b) koordinat tit ik D,
(e) nilai p dan nilai q,
(d) luas, dalam unit", bagi segiempat selari OABC. [10 markah]
3472/2
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10.
Water levelParas air
The diagram shows the cross-section of a spherical solid objectwith centre 0 and radius 10 em partially submerged in water.Points P and Q lie on the circumference of the cross-section andPQ=12cm.Calculate .
[3 marks](a) L POQ in radian,(b) the length, in em, of the major arc PQ, [3 marks]
(c) the area, in cm 2 , o f the shaded region. [4 marks]
Rajah menunjukkan keratan rentas sebuah pepejal sfera denganpusat 0 dan jejari 10 em yang sebahagian tenggelam dalamair. Titik P dan Q terletak di lilitan keratan rentas itu danPQ = 12 em.
Hitungkan
[3 markah](a) L POQ dalam radian,(b) panjang, dalam ern, lengkok major PQ, [3 markah]
(c) luas, dalam cm2
.kawasan berlorek. [4markah]
3472/2 Additional Mathematics Paper2 [Lihat sebelahSULIT
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SULIT ] 3 September, 2008
11(a) The school football team held a training session on penalty kicks.Each player was given 5 trials. After the session, it was found thaton the average, the mean for number of goals scored by a player
is 3. If a player is chosen at random, find the probability that he(i) fails to score any goal,
(ii) scores at least 4 goals. [5 marks]
(b) The body mass of 483 pupils in a primary school follows a normaldistribution with a mean of 35 kg and a standard deviation of10kg.
(i) If a pupil is chosen at random, find the probability that hisbody mass is between 30 kg an,d-69 kg.
(ii) A pupil will be placed under the obesity list ifhis body mass
exceeds 60 kg. Estimate the number of pupils whose names
(a)
will appear in the list.
Pasukan bola sepak sekolah mengadakan satu sesi latihantendangan penalti. Setiap pemain diberi 5 percubaan. Selepas sesiitu, didapati secara purata, min bilangan gol yang berjaya dijaringoleh seseorang .pemain adalah 3. Jika seorang pemain dipilihsecara rawak; carikan kebarangkalian bahawa pemain itu
[5 marks]
(i)
(ii)
gagal menjaringkan sebarang gol,
menjaring sekurang-kurangnya 4 gol.[5 markah]
(b) Jisim ba d an 483 murid sebuah sekolah rendah adalah mengikuttaburan normal dengan min 35 kg dan sisihan piawai 10 kg.
(i) Jika seorang murid dipilih secara rawak; carikankebarangkalian bahawa jisim badannya berada di antara30 kg dan 60 kg.
(ii) Seseorang akan disenaraikan dalam senarai murid obes jikaj isim badannya melebihi 60 kg. Anggarkan bilangan murid
yang namanya akan muncul dalam senarai itu: [5 markah]
3472/2 Additional Mathematics Paper 1 [Lihat sebelahSULIT
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SULIT 14
Section CBahagian C
[2 0 marks]
[2 0 markah]Answer two questions from this section.
Jawab dna soa/an daripada bahagian ini.
12. A particle moves along a straight line from a fixed point 0. Itsvelocity, v rns", is given by v = 15t - 3r ,where t is the time, inseconds, after leaving the point O.[Assume motion to the right is positive.]
Find
(a) the acceleration, in ms 2, when the particle starts to move,
(b) the maximum velocity of the particle, in ms",
(c) the value of t when the particle passes the point 0 again,
(d) the distance travelled, in m, by the particle during the fourthsecond.
Suatu zarah bergerak di sepanjang suatu garis lurus dari satu titiktetap 0. Halajunya, vms ! , diberi o/eh v = 15t - 3r . dengankeadaan t ialah masa, da/am saat, se/epas meningga/kan titik 0.[Anggapkan gerakan Ire arah kanan sebagai positif.]
Cari
(a) pecutan, da/am ms 2 ,apabila zarah itu mula bergerak,
(b) ha/aju maksimum zarah itu, da/am ms! ,
(c) nilai bagi t apabila zarah itu melalui titik 0 semula,
(d) jarak yang dilalui, da/am m, o/eh zarah itu da/am saat keempat.
September, 2008
[2 marks]
[2 marks]
[3 marks]
[3 marks]
[2 markah]
[2 markah]
[3 markah]
[3 markah]
3472/2
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13. The table shows the prices and price indices o f four components,P, Q, Rand S. used in the manufacture o f a certain product T.
Component
,
Price (RM)for the year
Price index forthe year 2007
based on,the year 2005Harga indekspacla tahun
Komponen2005
,2007
P RM 40.50 x 120
Q RM46.00 RM 59.80 130R RM 58.00 RM 78.30 YS z RM 11 1.20 139
(a) Find the value of x, y and z.
(b) Given that the composite index for the production cost of
product T in the year 2007 based on the year 2005 is 132.1,calculate
(i) the value of m if the quantities o f components P, Q. Rand S used are in the ratio 25 : m : 80 : 30.
(ii) the production cost o f each product T in the year 2005 ifthe corresponding production cost in the year 2007 isRM 150.00.
(iii) the composite index for the year 2008 based on the year2005 if the price of each component increases by 30 %from the year 2007 to the year 2008.
Jadual menunjukkan harga dan indeks harga bagi empatkomponen, P, Q. R dan S. yang digunakan untuk membuat sejenisproduk T.
(a) Cari nilai x, y dan z.
(b) Diberi bahawa indeks gubahan bagi kos penghasilanproduk T dalam tahun 2007 berasaskan tahun 2005 ialah132.1, hitungkan
(i) nilai m jika kuantiti komponen P, Q. R dan S yang
digunakan adalah dalam nisbah 25 : m : 80 : 30.
(ii) kos penghasilan setiap produk T dalam tahun 2005 jikakos penghasilan sepadan pada tahun 2007 ialahRMI50.00.
(iii) indeks gubahan bagi tahun 2008 berasaskan tahun2005 jika harga setiap komponen menaik sebanyak30 % dari tahun 2007 ke tahun 2008.
[3 marks]
[3 marks]
[2 marks]
[2 marks]
[3 markah]
[3 markah]
[2 markah]
[2 markah]
[Libat sebelahSULlT
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SULIT 16 September, 2008
14. Use graph paper to answer this question.
A dress making shop produces x baju kurung and y baju kebayaper week. A baju kurung takes 1 hour to complete while a bajukebaya takes 4 hours. The production of the dresses is based on thefollowing constraints:
I The total number of baju kurung and boju kebaya producedis not more than 80.
I I The number of baju kebaya produced is at most four timesthe number of baju kurung.
III The total time used for making the dresses pe r week is notless than 120 hours.
(a) Write three inequalities, other than x ~ 0 and y ~ 0, which [3 marks]satisfy all of the above constraints.
(b) Using a scale of 2 em to 10 dresses on both axes, constructand shade the region R which satisfies all of the above
[3 marks]constraints.
(c) Use your graph in 14(b), to find
(i) the minimum number of baju kebaya that can beproduced if the shop keeper fixed the number of bajukurung to be produced as 30.
(ii) the maximum total profit per week if the profit from abaju kurung is RM20 and from a baju kebaya isRM40. [4 marks]
3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT
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SULIT 17 September, 2008
Gunakan kertas graf untuk menjawab soalan ini.
Sebuah kedai jahitan menghasilkan x helai baju kurung dan yhelai baju kebaya seminggu. Masa uruuk menyiapkan sehelai baju/curung ialah 1 ja m manakala masa untuk menyiapkan sehelai bajukebaya ialah 4 jam. Penghasilan pakaian tersebut dalam semingguadalah berdasarkan kekangan berikut:
I : Jumlah bilangan baju kurung dan baju kebaya tidak melebihi80 helai.
I I Bilangan baju kebaya yang dihasilkan tidak melebihi empatkali ganda bilangan baju /curung.
I I I : Jumlah masa dalam seminggu yang digunakan untukmenjahit baju tidak kuran g daripada 120 jam.
(a) Tuliskan tiga ketaksamaan, selain x ~ 0 dan y ~ 0, yang
[3 march]memenuhi semua kekangan di atas.
(b) Menggunakan skala 2 em kepada 10 helai baju padakedua-dua paksi, bina dan lorek rantau R yang memenuhisemua kekangan di atas. [3 march]
(e) Gunakan grof anda di 14(b), untuk mencari
(i) bilangan minimum baju kebaya yang dapat dihasilkan jikapekedai itu menetapkan bilangan baju kurung yangdihasilkan ialah 30 helai.
(ii) jumlah keuntungan maksimum seminggu jika keuntungansehelai baju kurung ialah RM20 dan sehelai baju kebayaialahRM40. [4 markah]
3472/2 Addition al Mathematics Paper 1 [Libat sebelabSULIT
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SULIT 18 September, 2008
15. The diagram shows a triangle ACD and B is a point on the side ACwhere CB = 6.0 em. The area of triangle BCD is 20 cm 2 andL CBD is obtuse.
Rajah menunjukkan satu segitiga ACD dan B ialah satu titik padasis; AC di mana CB = 6.0 em. Luas segitiga BCD ia/ah 20 em 2
dan L CBD ialah cakah.
D
AC 6.0 em B
Calculate
(a) LBCD, [2 marks]
(b) the length, in em, o f BD, [2 marks]
[2 marks](c) LCBD.
(d) the area, in em 2, of triangle ACD. [4 marks]
Hitung
[2 markah](a) LBCD.
[2 markah](b) panjang, da/am em, bagi BD,
[2 markah](c) LCBD.
(d) luas, dalam em 2, bag! segitiga ACD. [4 markah]
END OF QUESTION PAPERKERTAS SOALAN TAMAT
3472/2 Additiona l Mathematics Paper 1SULIT
2k | 5,353 | 15,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.467507 |
https://stemez.com/subjects/maths/AAlgebra/AAlgebra/AAlgebra/A00-Ch17.htm | 1,675,661,485,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00036.warc.gz | 556,776,073 | 7,645 | CHAPTER 17: SOLUTION BY QUADRATIC FORMULA
PROBLEM 17 - 471: Solve for x2 : 4x2 – 7 = 0 .
PROBLEM 17 - 472: Obtain the quadratic equation in standard form that is
equivalent to 4x − 3 = 5x2 .
PROBLEM 17 - 473: Find the roots of the equation x2 + 12 85 = 0 .
PROBLEM 17 - 474: Use the quadratic formula to solve for x in the equation
x2 − 5x + 6 = 0 .
PROBLEM 17 - 475: Solve the equation x2 + 5x + 6 = 0 by the quadratic
formula.
PROBLEM 17 - 476: Solve 6x2 − 7x − 20 = 0 .
PROBLEM 17 - 477: Solve the equation 2x2 − 5x + 3 = 0 .
PROBLEM 17 - 478: Solve x2 − 7x + 10 = 0 .
PROBLEM 17 - 479: Solve the equation 3x2 – 5x +2 = 0 by means of the quadratic
formula .
PROBLEM 17 - 480: Solve for the roots of the equation
6x2 + 5x − 2 = 0 and for the roots of the equation
3x2 + 4x − 4 = 0 .
PROBLEM 17 - 481: Solve the quadratic equation 6x2 – x – 35 = 0 .
PROBLEM 17 - 482: Solve x2 + 2x − 5 = 0.
PROBLEM 17 - 483: Use the Quadratic Formula to solve the following equation:
x2 − 7x − 7 = 0 .
PROBLEM 17 - 484: Solve the equation 3x2 + 5x – 7 = 0 .
PROBLEM 17 - 485: Use the quadratic formula to solve the equation
3x2 + 4x − 5 = 0 .
PROBLEM 17 - 486: Solve t2 – 8t + 3 = 0 by the quadratic formula .
PROBLEM 17 - 487: Solve the equation x2 = 4x − 1,
PROBLEM 17 - 488: Use the quadratic formula to solve equation
8z (z + 1) = 1 for z .
PROBLEM 17 - 489: Solve 1/ x + 1/ (x + 2) = 2 .
PROBLEM 17 - 490: Solve the equation x2 − x + 1 = 0.
PROBLEM 17 - 491: Solve the equation √(x + 1) + √(2x + 3) – √(8x + 1) = 0.
PROBLEM 17 - 492: Solve for y if 6x2 + 9y2 + x − 6y = 0.
PROBLEM 17 - 493: Solve for y if 2x2 + y2 + 2xy − 2x = 0.
PROBLEM 17 - 494: Solve for x by using the quadratic formula,
(a) 3x2 = x + 6 (b) 5x2 − 6x + 7 = 0.
PROBLEM 17 - 495: Solve the equation 4√(3 − x / 3 + x) – √(3 + x / 3 – x) = √2
PROBLEM 17 - 496: Solve for x: 3x2 +5 = 0 .
PROBLEM 17 - 497: Use the quadratic formula to solve 2x2 − 5x + 8 = 0.
PROBLEM 17 - 498: Use the quadratic formula to solve the following equation for
x: x2 + 2x + 4 = 0
PROBLEM 17 - 499: Solve x2 + 2x + 5 = 0 .
PROBLEM 17 - 500: Solve the equation 2x2 + 5x + 8 = 0.
PROBLEM 17 - 501: Solve x2 + 2x + 5 = 0, by the quadratic formula .
PROBLEM 17 - 502: Solve the equation 4x2 = 8x – 7 by means of the
PROBLEM 17 - 503: Find the roots of the function F whose rule of
Correspondence is F(x) = 2x2 + 8x + 4 .
PROBLEM 17 - 504: Solve 3x2 − 5x + 4 = 0 by the quadratic formula.
PROBLEM 17 - 505: Determine the quadratic equation whose roots are
1/2 and (2/3)
PROBLEM 17 - 506: Show that the roots of the quadratic equation
x2 – x – 3 = 0 are x1 = (1 + √13)/ 2 and x2 = (1 − √13)/ 2
PROBLEM 17 - 507: Find the sum and product of the roots of the equation
3x2 – 2x + 1 = 0 .
PROBLEM 17 - 508: Find the sum and product of the roots of the equation
3x2 + 13x – 10 = 0 .
PROBLEM 17 - 509: Without solving, find the sum and product of the roots of
8x2 = 2x + 3.
PROBLEM 17 - 510: Find the sum and the product of the roots in each of the
following equations: x2 − 3x + 2 = 0, 2x2 + 8x – 5 = 0, and
√2x2 + 5x – √8 = 0
PROBLEM 17 - 511: Determine the quadratic equation whose roots are
x = 2 + (√3) and x = 2 – (√3).
PROBLEM 17 - 512: Find the equation whose roots are 3 + √2 and 3 − √2 .
PROBLEM 17 - 513: Find a quadratic equation whose roots are 3 + 2√3 and
3 – 2√3 .
PROBLEM 17 - 514: Form the equation whose roots are 2 + (√3) and 2 – (√3) .
PROBLEM 17 - 515: Find the equation whose roots are α/β , β/α .
PROBLEM 17 - 516: Find the equation whose roots are the negatives of the
roots of x2 + 7x − 2 = 0 .
PROBLEM 17 - 517: Find the value of k If one root is twice the other.
x2kx + 18 = 0.
PROBLEM 17 - 518: Find the values of the constant k in the equation
2x2kx + 3k = 0 if the difference of the roots is 5/2 .
PROBLEM 17 - 519: Find the value of k if, in the equation
2x2 – kx2 + 4x + 5k = 0, one root is the reciprocal of the
other.
PROBLEM 17 - 520: If α and β are the roots of x2px + q = 0, find the value of
(1) α2 + β2 , (2) α3 + β3 .
PROBLEM 17 - 521: Solve (1 – a2 ) (x + a) – 2a (1 – x2 ) = 0 .
PROBLEM 17 - 522: Find the condition that the roots of ax2 + bx + c = 0 ,
may be (1) both positive, (2) opposite in sign, but the
greater of them negative .
PROBLEM 17 - 523: Find the discriminant of 3x2 − 7x + 5 = 0. Then solve.
PROBLEM 17 - 524: Compute the value of the discriminant and then determine
the nature of the roots of each of the following four
equations:
4x2 − 12x +9 = 0 ,
3x2 − 7x − 6 = 0
5x2 + 2x − 9 = 0
and x2 + 3x + 5 = 0 .
PROBLEM 17 - 525: Discuss the nature of the roots of
(a) 3x2 − 7x + 3 = 0 (b) 5x2 + 3x + 1 = 0
PROBLEM 17 - 526: Determine the character of the roots of the equation
2x2 − x + 5 = 0.
PROBLEM 17 - 527: Determine the character of the roots of the equation
4x2 − 12x + 9 = 0.
PROBLEM 17 - 528: Determine the character of the roots of the equation
x2 − 5x + 6 = 0.
PROBLEM 17 - 529: Without solving the equation
2x2 − 3x + 5 = 0,
determine the nature of its roots.
PROBLEM 17 - 530: Compute the value of the discriminant and determine the
nature of the roots in each of the following three equations:
4x2 4 √(5x) + 5 = 0, √(3x2) − 6x + √12 = 0, and
√(2x2) + 3x + √5 = 0 .
PROBLEM 17 - 531: Can the expression 16x2 − 76x + 21 be factored into
rational factors ?
PROBLEM 17 - 532: Show that the graph of y = − x2 + x − 1 has no real zeros.
PROBLEM 17 - 533: Show that the equation 2x2 − 6x + 7 = 0 cannot be satisfied
by any real values of x.
PROBLEM 17 - 534: If the equation x2 + 2(k + 2)x + 9k = 0 has equal roots, find k. | 2,390 | 6,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-06 | latest | en | 0.377905 |
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# Constrained data model - can you solve it?
By Murray Bourne, 21 Jul 2012
## Constrained data model - can you solve it?
Statisticians are often asked to predict the future. Their job is to look at historical trends and to make their best guess what will happen tomorrow, or maybe next year. This involves a process called extrapolation, where we project into the future based on past data.
For example, here is a set of data representing company sales (variable s) for a period of 10 days (t):
t s 1 2 3 4 5 6 7 8 9 10 2 4 3 5 9 6 8 9 8 10
Let's graph the data to get a better idea what it looks like:
To extrapolate intelligently, we need to consider the shape of the data curve, and then fit a curve as best we can through the points.
At the beginning, there is a steep rise in sales, but then it flattens out (perhaps the salespeople were getting tired). It appears to be a logarithmic curve, which in its standard form, starts fairly steeply, and then flattens out.
There are various formulas for fitting a curve to data (see the bottom of the page), but we'll use software.
We'll get Excel to draw a curve through the data points and then extrapolate for us. After entering the data in Excel, we highlight it, then choose the "XY(Scatter)" option. Once we see the chart, we select the graph and right click, then choose "Add trendline". We choose "logarithmic" because that's the shape of our data curve. Under "Options", we can choose to "Forecast" by as many steps as we like.
So in this next graph, you'll see the logarithmic growth model passing through the given data. I've forecast the next 20 days, so we can get an idea how our sales team will perform for the rest of the month, assuming their historical performance (for the first 10 days) continues.
(The above set of data and curve is just to illustrate the problem. We need a formula that will work with any set of data, within the constraint mentioned next.)
Constraint
Now, (and here's where the challenge comes) say we need to produce a curve such that the sum of the values of the model curve is exactly the same as the sum of the values (for the first 10 days) in the original data set?
For example, here is the data again, but with a new row (m) containing values that fit the above constraint. That is, the sum of the first 10 sales values (variable s) is the same as the sum of the first 10 model values (using variable m).
t s m 1 2 3 4 5 6 7 8 9 10 Σ 2 4 3 5 9 6 8 9 8 10 64 2.17 3.53 4.69 5.68 6.52 7.23 7.84 8.36 8.8 9.18 64
The sum (Σ) is given in the final column. Both rows s and m add to 64.
Here is the graph of the m-values given in the above table. You'll notice it is similar to the Excel logarithmic graph, but not exactly the same. I have once again extrapolated another 20 steps in the chart below. We observe using this constrained model that the sales per day at the end of the month is less (around 11.5) than Excel's logarithm curve (which gave almost 13).
So, your challenge is to develop a formula to produce a curve that goes through any set of data, given the above constraint. Good luck! (This is quite challenging, but give it a go.)
### 10 Comments on “Constrained data model - can you solve it?”
1. Peter Hunter says:
There is an underlying assumption concerning this problem that was not explicity stated: sales are reported as sales/salesperson. This cancels the effect of an expanding (or shrinking) sales force. Given that, the model suggests a logistic form. Why logistic? Presumably there is a learning curve for sales people to effectively promote a new product. Once they are proficient, there is a maximum number of sales that can be made per day (hours spent selling/time per sales pitch, assuming each sales pitch is successful!).
The logistic in its general form can be expressed as:
sales = A + (B - A)/(1 + (C/x)^D)
where
A = is the minimum number of sales per person per day
B = maximum number of sales per person per day
C,D determine slope and inflection point of the learning curve
This formula can be simplified by noting that A=0 (can't sell anything before you start).
I placed this equation and the data in Excel and used the equation solver as follows:
1. I defined the target cell as then squared error between the actual sales and predicted and minimized it.
2. Added the constraint that the sum of sales of the predicted had to equal the sum of sales of the raw data
3. Based on the original log-linear regression, I set the initial conditions as:
A=0
B=13.5
C=5
D=1
Furthermore, I instructed the solver to only alter parameters B and C, leaving A and D as specified. The resulting solution was:
A=0
B=15.2676679048944
C=6.63383395244721
D=1
The sum of the predicted sales was indeed 64, consistent with the stated constraint.
A plot of the predicted data closely matched your predicted values. The main difference is that at day 30, the logistic prediction was 12.5 sales vs your 11.5
I would be happy to provide the working spreadsheet on request.
Great question and challenge!
2. Murray says:
Great solution, Peter. Your logistic equation makes a lot of sense, given the description of the problem.
I guess I should have said "keeping all other things equal" (so the sales force is the same throughout, their skill level is consistent and all those other unrealistic things that may or may not be going on in such a problem). However, my "sales" story and data was just to illustrate the kind of problem that we aim to model.
I realized after I posted the challenge that I hadn't explicitly stated that our formula must apply for any set of data. That is, we aim to find a model that will fit any shaped data while maintaining the constraint that the sum of the data values equals the sum of the model output values up to a known time value.
I amended the post to make that more clear. Very sorry to have led you up the garden path.
3. Mathblogging.org Weekly Picks « Mathblogging.org — the Blog says:
[...] squareCircleZ, Murray Bourne challenges you to predict the [...]
4. Alan says:
Let there be n data values, d(t), where t is time. Let the function you are fitting be f(a,t), where a is a vector of the constants to be found. Let g(f,a,t) be the function that represents the sum of f(a,t) over all the times for which you have data. Then minimise the differences between f(a,t)+w*(g(f,a,t) and d(t)+w*(sum of d(t)), using whatever is your favourite minimisation technique, where w is a weighting constant. Increase w until you are satisfied that the sum of the n fitted values adequately matches that of the data points.
Of course, the end result will still depend on the applicability of the form of function, f. In general, if the fitted function isn’t based on some underlying model, but just on guesswork, I’d be very wary of extrapolating.
5. Peter Hunter says:
This may well drive me batty! I now understand that you wanted a generic solution, not one that depended on the logistic model, even thought that might make sense for this particular problem.
After plotting your predicted data against the logarithmic with a log scale along with the trending line, I noticed that the plot of your predicted values followed an S shape starting above the Excel trend line crossing it at day 2 reaching a maximum below it between days 3 and 4, approaches the Excel trending line at day 10, then falls below it further on.
This suggests two things: first, there is probably a weighting function related to the time variable, and second, the equality of the sum of dependent y’s and the sum of the predicted y’s has to be part of the error sum.
With respect to the weighting factor, the closer one gets to the end of the observed the weighting factor should trend to 1. Thus, a linear weighting could be expressed as
2 – x(i)/x(n)
Thus the beginning of the time series would have the greatest effect (2 – 1/10 = 19/10’s) and the last data point the least (2 – 10/10 = 1).
The fitted y’s are calculated based on the model:
yHat = m*log10(x)^(2 – x/10) + b
I think this remains generic because we are still only estimating the two parameters, m and b.
I then modified the minimization function to include the squared difference of total and predicted sales:
SSE = sum[( y(i) – yHat(i))^2] + [sum(y) – sum(yHat)]^2
Upon solving this, m=7.15 and b=2.27. The sum of yHat’s equals 64 and what is really interesting is the sum of squared error using this model(12.62) IS LESS than the sum of squared error from your solution(12.97). At day 30, my solution yields 12.8 vs your value of 11.5.
So, my solution doesn’t match your’s, but it does insure the equality of the observed and predicted values as well as having a slightly lower SSE.
6. Rene Grothmann says:
See
http://euler.rene-grothmann.de/renegrothmann/Predictions.html
7. Murray says:
@Rene - Great discussion - thanks for your inputs.
Yes, the whole thing is very arbitrary. It's basically there as a thought piece - and your article indicates that objective was met!
8. Rene Grothmann says:
Did you notice, that it solves your problem? Well not quite. It does not provide a formula for the minimization, just a nonlinear system to be solved.
9. Murray says:
@Rene: Indeed I did! As you said, "the best fit does this automatically".
I enjoyed your Black Swan article.
10. Vorhersagen | Observations says:
[...] Intmath Newsletter (SquareCircleZ Blog) [...]
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Science > You are Here
#### Example – 01:
• Check the correctness of physical equation, s = ut +1/2at2, Where u is the initial velocity, v is the final velocity, a is the acceleration, s is the displacement and t is the time in which the change occurs.
• Solution:
Given equation is s = ut +1/2at2
L.H.S. = v, hence [L.H.S.] = [s] = [L1M0T0] ………………. (1)
R.H.S = ut +1/2at2 , hence [R.H.S] = [u][t]+ [a][t]2
= [L1M0T-1][L0M0T1] + [L1M0T-2][L0M0T1]2
= [L1M0T0] + [L1M0T-2][L0M0T2]
= [L1M0T0] + [L1M0T0]
= [L1M0T0] ……………. (2)
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
#### Example – 02:
• Check the correctness of physical equation, F = mv²/r, Where F is the centripetal force acting on a body of mass m performing uniform circular motion along a circle of radius r with linear speed v.
• Solution:
Given equation is F = mv²/r
L.H.S. = F, hence [L.H.S.] = [F] = [L1M1T-2] ………………. (1)
R.H.S = mv²/r , hence [R.H.S] = [m][v]²/[r]
= [L0M1T0][L1M0T-1]2 / [L1M0T0]
= [L0M1T0][L2M0T-2] [L-1M0T0]
= [L1M1T-2] ……………. (2)
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
#### Example – 03:
• Check the correctness of physical equation, v2 = u2 + 2as2
• Solution:
Given equation is v2 = u2 + 2as2
L.H.S. = v2, hence [L.H.S.] = [v]2 = [L1M0T-1]2 = [L2M0T-2] ………………. (1)
R.H.S = u2 + 2as2
hence [R.H.S] = [u]2 + 2[a][s]2
[R.H.S] = [L1M0T-1]2 + [L1M0T-2][L1M0T0]2
[R.H.S] = [L2M0T-2] + [L1M0T-2][L2M0T0]
[R.H.S] = [L2M0T-2] + [L3M0T-2] ……………. (2)
From (1) and (2) we have [L.H.S.] ≠ [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally not correct.
#### Example – 03:
• Check the correctness of equation, when the frequency of vibration ‘n’ of a string of length ‘l’ having mass per unit length ‘m’ kept under tension ‘F’ is given by
• Solution:
Given equation is
L.H.S. = n, hence [L.H.S.] = [n] = [L0M0T-1] ………………. (1)
R.H.S = mv²/r , hence [R.H.S] = [m][v]²/[r]
= [L0M1T0][L1M0T-1]2 / [L1M0T0]
= [L0M1T0][L2M0T-2] [L-1M0T0]
= [L1M1T-2] ……………. (2)
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
#### Example – 04:
• Check the homogeneity of equation, when the rate of flow of a liquid having coefficient of viscosity’η’ through a capillary tube of length ‘l’ and radius ‘a’ under pressure head ‘p’ is given by
• Solution:
Given equation is
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.
#### Example – 05:
• Check the homogeneity of equation, when the periodic time ‘T’ of vibration of magnet of moment of inertia ‘I’, magnetic moment ‘M’ vibrating in magnetic induction ‘B’ is given by
• Solution:
Given equation is
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.
#### Example – 06:
• Check the homogeneity of equation, when the terminal velocity ‘v’ of a small sphere of radius ‘a’ and density ‘ρ’ falling through a liquid of density ‘σ’ and coefficient of viscosity ‘η’ is given by
• Solution:
Given equation is
From (1) and (2) we have [L.H.S.] = [R.H.S.]
Hence by the principle of homogeneity, the given equation is dimensionally homogeneous.
### Problems Based on Dimensionally Correct Physical Equations
#### Example – 07:
• A force F is given by F = at + bt2, where t is the time. Find the dimensional formula of ‘a’ and ‘b’.
• Solution:
• By the principle of homogeneity the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same.
Dimensions of F = Dimensions of at
[L1M1T-2] = [a] [t]
[a] =[L1M1T-2] / [T]
[a]=[L1M1T-3]
Dimensions of F = Dimensions of bt2
[L1M1T-2] = [b] [T] 2
[b]= [L1M1T-2] / [T2]
[b]= [L1M1T-4]
Ans: The dimensional formula of ‘a’ and ‘b’ are [M1L1T-3 ] and [M1L1T-4].
#### Example – 08:
• The distance covered by a body in time t is given by the relation S = a + bt + ct2. What are the dimension of a,b,c ? Also write the quantities they represent.
• Solution:
• By the principle of homogeneity the dimensions on either side of the physical equation must be the same. Now two physical quantities can be added or subtracted if and only if their dimensions are the same.
Dimensions of S = Dimensions of a
[L1M0T0] = [a]
Dimensions of ‘a’ are that of displacement. Hence ‘a’ represent the displacement.
Dimensions of S = Dimensions of bt
[L1M0T0] = [b] [t]
[b] =[L1M0T0] / [T]
[b]=[L1M0T-1]
Dimensions of ‘b’ are that of velocity. Hence ‘b’ represent the velocity.
Dimensions of S = Dimensions of ct2
[L1M0T0] = [c] [T] 2
[c]= [L1M0T0] / [T2]
[b]= [L1M0T-2]
Dimensions of ‘c’ are that of acceleration. Hence ‘c’ represent the acceleration.
Ans: The dimensional formula of ‘a’, ‘b’ and ‘c’ are [L1M0T0], [L1M0T-1], and [L1M0T-2].
The quantities represented by a, b and c are displacement, velocity and acceleration.
#### Example – 09:
• According to Laplace’s formula, the velocity ( V) of sound in a gas is given by v = (γP / d)1/2, where P is the pressure, d is the density of a gas. What is the dimensional formula for γ?
• Solution:
Ans: The dimensional formula for γ is [ M0 L0 T0 ]
#### Example – 10:
• In an equation { P + a / V2 } (V – b) = RT, where P is the pressure, V is the volume, T is the temperature and a, b, R are constants. What is the dimensional formula of a /b?
• Solution:
Two physical quantities can be added or subtracted if and only if their dimensions are the same.
Thus the dimensions of P and a / V2 are same
Dimensions of b = Dimensions of V
[b] = [L3M0T0]
Now [a]/[b] = [L5M1T-2]/ [L3M0T0] = [L2M1T-2]
Ans: The dimensional formula of a /b is [ M1 L2 T-2 ]
#### Example – 11:
• A force is given in terms of distance x and time t by F = A sin Ct + B cos Dx. Then what are the dimensions of A/B and C/D?
• Solution:
• By the principle of homogeneity the dimensions on either side of a physical equation must be the same. The trigonometric ratios sin and cos are pure ratios of length, hence are dimensionless quantity.
Thus [A] = [B] =[F] = [ M1 L1 T-2 ] Thus dimensions of A/B are [ M0 L0 T0 ]
Now the angle is dimensionless quantity
[C][t] = [ M0 L0 T0 ]
[C] = [ M0 L0 T0 ]/ [T] = [ M0 L0 T-1 ]
[D][x] = [ M0 L0 T0 ]
[D] = [ M0 L0 T0 ]/ [L] = [ M0 L-1 T0 ]
[C]/[D] = [ M0 L0 T-1 ]/ [ M0 L-1 T0 ] = [ M0 L1 T-1 ]
Ans: The dimensions of A/B and C/D are [M0L0T0] [M0L1 T-1]
#### Example – 12:
• In a equation { P + a / V2 } ( V – b) = constant , where P is the pressure, V is the volume and a, b are constants. Find the unit of ‘a’.
• Solution:
Two physical quantities can be added or subtracted if and only if their units are the same.
Thus the units of P and a / V2 are same
Unit of P = Unit of a /(unit of V)2
dyne cm-2 = unit of a / (cm3) 2
dyne cm-2 = unit of a / cm6
unit of a = dyne cm-2 x cm6 = dyne cm4
Ans: The unit of a is dyne cm4
#### Example – 13:
• What are the dimensions of constant C in the equation P + ½ρv2 + hρg = C? Where P is a pressure, ρ is a density, v is the velocity of liquid flow, h is the height and g is the acceleration due to gravity.
• Solution:
Two physical quantities can be added or subtracted if and only if their units are the same.
Thus the dimension of P, ½ρv2, hρg, and C should be that of pressure..
Dimensions of C = Dimensions of P = [M1L-1 T-2]
Ans: The dimensions of C are[M1L-1 T-2]
#### Example – 14:
• The velocity v, the acceleration a and displacement s of a body in motion are related by the expression v2 ∝ axsy. Find x and y.
• Solution:
Given v2 ∝ axsy
∴ v2 = k axsy
∴ [M0L1 T-1]2 = [M0L1 T-2]x[M0L1 T0]y
∴ [M0L2 T-2]2 = [M0Lx T-2x] [M0Ly T0]
∴ [M0L2 T-2]2 = [M0Lx + y T-2x]
∴ x + y = 2 and -2x = -2 i.e. x = 1 and y = 1
Ans: x = 1 and y = 1
Science > You are Here
### One Comment
1. F. Murairwa
A very good update to the internet. Keep on doing the same. This is good staff for reading of learners | 2,927 | 8,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-13 | latest | en | 0.791979 |
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|n|+3=7
Steve and Reiny
A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when the water is 7 meters deep WELL I HAVE THIS v=...
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A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when the water is 7 meters deep
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A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when the water is 7 meters deep
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WELL I HAVE THIS v=1/3pi*r^2h v=1/3pi*r2(4r) v=4/3pi*r^3 dv/dt=4pi*r^2 2=4pi(7/4)^2 2=49pi/4 * dr/dt dr/dt= 8/49pi = 0.052 meters per minute
is the answer 8/49pi= 0.052 meters per minute
A tank has the shape of an inverted circular cone with a base radius of 5 meters and a height of 20 meters. If water is being pumped into the tank at 2 cubic meters per minute, find the rate at which the water level is rising when the water is 7 meters deep
Help Math
Thanks Steve
Help Math
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8/12 and 9/12
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help ss !!!!!!!!!!!!!!!
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Oh okay. I understand that, i was just not understand where the 9 came from, but i see now. Thank you both
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Sorry about that Reiny. I have a quick question though, where does the 9 come from?
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ohh i finally get it!
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Puzzle 929. Six primes such that the sum of any five is a prime number The set of six non-consecutive primes {5, 7, 11, 19, 29, 37} has the interesting property that the the sum of any five of these is a prime number. Q1: Find a set of six consecutive primes with the same property than the set above? Q2: Redo Q1 using 8, 10, ... consecutive primes with the corresponding property similar to the set above?
Contributions came from Seiji Tomita, Giovanni Resta, Paul Cleary, Simon Cavegn and Emmanuel Vantieghem.
***
Seiji wrote on Oct 28, 2018
Q1:
Search condition: {a,b,c,d,e,f}<10^10
Min: {9733, 9739, 9743, 9749, 9767, 9769}
Max: {996712309, 996712331, 996712333, 996712349, 996712361, 996712369}
Q2:
Eight consecutive primes such that the sum of any seven is a prime number.
Min:{69398759, 69398761, 69398779, 69398789, 69398803, 69398807, 69398821, 69398831}
***
Giovanni wrote on Oct 29, 2018:
The starts of 6, 8, 10. and 12 consecutive primes such that the sum of all
apart one gives a prime are 9733, 69398759, 422701985389 and 28999134198739.
***
On October 31, 2018 Paul Cleary wrote:
I have found the following for 6 consecutive primes. Format {consecutive primes}, {sum of any 5 primes}.
{9733,9739,9743,9749,9767,9769} , {48731,48733,48751,48757,48761,48767}
{970217,970219,970231,970237,970247,970259} , {4851151,4851163,4851173,4851179,4851191,4851193}
{3218471,3218473,3218477,3218483,3218503,3218533} , {16092407,16092437,16092457,16092463,16092467,16092469}
{5241937,5241959,5241961,5241983,5241989,5242021} , {26209829,26209861,26209867,26209889,26209891,26209913}
{5691893,5691899,5691913,5691919,5691929,5691943} , {28459553,28459567,28459577,28459583,28459597,28459603}
{8445251,8445259,8445271,8445343,8445377,8445419} , {42226501,42226543,42226577,42226649,42226661,42226669}
{8788079,8788081,8788121,8788141,8788147,8788151} , {43940569,43940573,43940579,43940599,43940639,43940641}
{11268497,11268503,11268511,11268527,11268541,11268571} , {56342579,56342609,56342623,56342639,56342647,56342653}
{11881901,11881913,11881927,11881931,11881957,11882071} , {59409629,59409743,59409769,59409773,59409787,59409799}
And 8 consecutive primes
{69398759,69398761,69398779,69398789,69398803,69398807,69398821,69398831} , {485791519,485791529,485791543,485791547,485791561,485791571,485791589,485791591}
{120009433,120009541,120009551,120009553,120009563,120009619,120009629,120009641} , {840066889,840066901,840066911,840066967,840066977,840066979,840066989,840067097}
{130095619,130095701,130095731,130095733,130095739,130095803,130095877,130095887} , {910670203,910670213,910670287,910670351,910670357,910670359,910670389,910670471}
Still looking for 10 consecutive primes.
***
Simon wrote of Nov 1, 2018
Using the algorithm described below, tested all consecutive prime sets of length 4,6,8,10 up to the prime 218775961, sets of length 12 up to the prime 195194897
Found no consecutive set of primes of length 10 or 12.
Sum any 7 of 8 to get a prime:
{69398759, 69398761, 69398779, 69398789, 69398803, 69398807, 69398821, 69398831}
{120009433, 120009541, 120009551, 120009553, 120009563, 120009619, 120009629, 120009641}
{130095619, 130095701, 130095731, 130095733, 130095739, 130095803, 130095877, 130095887}
Sum any 5 of 6 to get a prime:
{9733, 9739, 9743, 9749, 9767, 9769}
{970217, 970219, 970231, 970237, 970247, 970259}
{3218471, 3218473, 3218477, 3218483, 3218503, 3218533}
{5241937, 5241959, 5241961, 5241983, 5241989, 5242021}
{5691893, 5691899, 5691913, 5691919, 5691929, 5691943}
{8445251, 8445259, 8445271, 8445343, 8445377, 8445419}
{8788079, 8788081, 8788121, 8788141, 8788147, 8788151}
{11268497, 11268503, 11268511, 11268527, 11268541, 11268571}
{11881901, 11881913, 11881927, 11881931, 11881957, 11882071}
{16697419, 16697489, 16697497, 16697501, 16697543, 16697581}
{19604623, 19604633, 19604657, 19604677, 19604693, 19604701}
{22057961, 22057963, 22057969, 22058039, 22058051, 22058077}
{22926473, 22926487, 22926517, 22926583, 22926587, 22926593}
{26027723, 26027737, 26027741, 26027747, 26027773, 26027779}
{26939197, 26939201, 26939239, 26939243, 26939249, 26939251}
{38187463, 38187467, 38187493, 38187607, 38187671, 38187683}
{38938153, 38938181, 38938187, 38938243, 38938259, 38938267}
{39901963, 39901997, 39902029, 39902047, 39902063, 39902081}
{45190247, 45190253, 45190261, 45190279, 45190333, 45190337}
{52489691, 52489699, 52489721, 52489729, 52489747, 52489751}
{54887597, 54887627, 54887629, 54887647, 54887689, 54887711}
{58296113, 58296149, 58296167, 58296169, 58296181, 58296211}
{61909753, 61909781, 61909787, 61909807, 61909819, 61909823}
{62686369, 62686387, 62686409, 62686427, 62686447, 62686469}
{68142289, 68142299, 68142301, 68142311, 68142329, 68142343}
{69567359, 69567401, 69567439, 69567457, 69567461, 69567481}
{69799033, 69799061, 69799063, 69799091, 69799109, 69799123}
{72085687, 72085697, 72085711, 72085721, 72085771, 72085781}
{72973723, 72973763, 72973777, 72973781, 72973799, 72973807}
{79517741, 79517743, 79517777, 79517783, 79517827, 79517863}
{82464511, 82464517, 82464539, 82464553, 82464563, 82464587}
{82792109, 82792117, 82792141, 82792169, 82792181, 82792183}
{88902937, 88902953, 88902997, 88903007, 88903049, 88903063}
{91374029, 91374043, 91374047, 91374061, 91374071, 91374079}
{102004219, 102004249, 102004283, 102004297, 102004319, 102004349}
{106966463, 106966487, 106966507, 106966511, 106966513, 106966543}
{121639181, 121639183, 121639201, 121639247, 121639277, 121639321}
{122052719, 122052727, 122052757, 122052767, 122052779, 122052787}
{124626977, 124627043, 124627049, 124627091, 124627109, 124627121}
{128438207, 128438227, 128438257, 128438293, 128438327, 128438393}
{130755571, 130755607, 130755617, 130755641, 130755643, 130755671}
{139131367, 139131389, 139131397, 139131401, 139131407, 139131427}
{139651189, 139651199, 139651231, 139651261, 139651271, 139651307}
{154908697, 154908731, 154908737, 154908739, 154908763, 154908773}
{160032949, 160032967, 160032989, 160032997, 160033007, 160033031}
{166920179, 166920181, 166920199, 166920203, 166920251, 166920319}
{175310021, 175310059, 175310101, 175310119, 175310129, 175310159}
{175827607, 175827623, 175827637, 175827647, 175827649, 175827677}
{189126473, 189126503, 189126517, 189126527, 189126559, 189126607}
{191001277, 191001287, 191001311, 191001359, 191001367, 191001397}
{196047721, 196047739, 196047749, 196047779, 196047809, 196047853}
{198009953, 198010009, 198010049, 198010081, 198010121, 198010129}
{198621271, 198621281, 198621289, 198621329, 198621337, 198621341}
{199825291, 199825313, 199825319, 199825331, 199825333, 199825363}
{201612769, 201612791, 201612793, 201612811, 201612833, 201612863}
{207464149, 207464161, 207464171, 207464197, 207464219, 207464261}
Description of the algorithm in C#/Linq:
Load the first 105000000 primes from drive into ram.
Make a hashset of the primes to easily check if a number is contained in the hashset.
for (int i = 0; i < 10829000; i++) // i means index to primes hashset. When summing up 11 primes, do not go above the largest loaded prime in the hashset.
{
foreach (int setLength in new int[] { 4, 6, 8, 10, 12 })
{
List<int> set = _primes.Skip(i).Take(setLength).ToList();
if (GetOrderedSubSets(set, setLength - 1).All(subSet => _primes.Contains(subSet.Sum())))
{
Log(string.Join(", ", set));
}
}
}
public static IEnumerable<IEnumerable<T>> GetOrderedSubSets<T>(IEnumerable<T> list, int length) where T : IComparable
{
if (length == 1) return list.Select(t => new T[] { t });
return GetOrderedSubSets(list, length - 1).SelectMany(t => list.Where(e => t.All(g => g.CompareTo(e) == -1)), (t1, t2) => t1.Concat(new T[] { t2 }));
}
***
Emmanuel wrote on Oct 2, 2018:
Q1.
This is the smallest example :
{9733, 9739, 9743, 9749, 9767, 9769}
The six sums of five of them are :
{48731, 48733, 48751, 48757, 48761, 48767}.
They happen to be consecutive too !
Q2.
This is the smallest example :
{69398759, 69398761, 69398779, 69398789, 69398803, 69398807, 69398821, 69398831}.
The eight sums of seven of them are :
{485791519, 485791529, 485791543, 485791547, 485791561, 485791571, 485791589, 485791591}
These primes are not consecutive.
I could not find an example below 4.6*10^9 in which the eight sums were consecutive.
For an example with ten consecutive primes with prime sums of nine of them we should search among primes greater than 10^10.
***
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Engineering Mathematics SEM – 1Concept
#### LOGARITHMS
Logarithm: For ant two positive real numbers a, b, and a ≠ 1. If the real number x such then ax = b, then x is called logarithm of b to the base a. it is denoted by
#### Standard formulae of logarithms:
##### Logarithmic Function:
Let a be a positive real number and a ≠ 1. The function f: (o, ∞) → R Defined by f(x) =
PARTIAL FRACTIONS
Fractions:
If f(x) and g(x) are two polynomials, g(x) ≠ 0, then is called rational fraction.
Ex:
etc. are rational fractions.
##### Proper Fraction:
A rational fraction is said to be a Proper fraction if the degree of g(x) is greater than the degree of f(x).
Ex:
etc. are the proper fractions.
Improper Fraction:
A rational fraction is said to be an Improper fraction if the degree of g(x) is less than the degree of f(x).
Ex:
etc. are the Improper fractions.
##### Partial Fractions:
Expressing rational fractions as the sum of two or more simpler fractions is called resolving a given fraction into a partial fraction.
∎ If R(x) = is proper fraction, then
Case(i): – For every factor of g(x) of the form (ax + b) n, there will be a sum of n partial fractions of the form:
Case(ii): – For every factor of g(x) of the form (ax2 + bx + c) n, there will be a sum of n partial fractions of the form:
∎ If R(x) = is improper fraction, then
Case (i): – If degree f(x) = degree of g(x), where k is the quotient of the highest degree term of f(x) and g(x).
Case (ii): – If f(x) > g(x)
R(x) =
### MATRICES AND DETERMINANTS
Matrix: A set of numbers arranged in the form of a rectangular array having rows and columns is called Matrix.
•Matrices are generally enclosed by brackets like
•Matrices are denoted by capital letters A, B, C, and so on
•Elements in a matrix are real or complex numbers; real or complex real-valued functions.
Oder of Matrix: A matrix having ‘m’ rows and ‘n’ columns is said to be of order m x n read as m by n.
Ex:
##### Types Of Matrices
Rectangular Matrix: A matrix in which the no. of rows is not equal to the no. of columns is called a rectangular matrix.
Square Matrix: A matrix in which the no. of rows is equal to no. of columns is called a square matrix.
Principal diagonal (diagonal) Matrix: If A = [a ij] is a square matrix of order ‘n’ the elements a11, a22, a33, ………. An n is said to constitute its principal diagonal.
Trace Matrix: The sum of the elements of the principal diagonal of a square matrix A is called the trace of the matrix. It is denoted by Tr (A).
Diagonal Matrix: If each non-diagonal element of a square matrix is ‘zero’ then the matrix is called a diagonal matrix.
Scalar Matrix: If each non-diagonal element of a square matrix is ‘zero’ and all diagonal elements are equal to each other, then it is called a scalar matrix.
Identity Matrix or Unit Matrix: If each of the non-diagonal elements of a square matrix is ‘zero’ and all diagonal elements are equal to ‘1’, then that matrix is called unit matrix
Null Matrix or Zero Matrix: If each element of a matrix is zero, then it is called a null matrix.
Row matrix & column Matrix: A matrix with only one row s called a row matrix and a matrix with only one column is called a column matrix.
###### Triangular matrices:
A square matrix A = [aij] is said to be upper triangular if aij = 0 ∀ i > j
A square matrix A = [aij] is said to be lower triangular matrix aij = 0 ∀ i < j
###### Equality of matrices:
matrices A and B are said to be equal if A and B are of the same order and the corresponding elements of A and B are equal.
###### Addition of matrices:
If A and B are two matrices of the same order, then the matrix obtained by adding the corresponding elements of A and B is called the sum of A and B. It is denoted by A + B.
Subtraction of matrices:
If A and B are two matrices of the same order, then the matrix obtained by subtracting the corresponding elements of A and B is called the difference from A to B.
##### Product of Matrices:
Let A = [aik]mxn and B = [bkj]nxp be two matrices, then the matrix C = [cij]mxp where
Note: Matrix multiplication of two matrices is possible when no. of columns of the first matrix is equal to no. of rows of the second matrix.
A m x n . Bp x q = AB mx q; n = p
Transpose of Matrix: If A = [aij] is an m x n matrix, then the matrix obtained by interchanging the rows and columns is called the transpose of A. It is denoted by AI or AT.
Note: (i) (AI)I = A (ii) (k AI) = k . AI (iii) (A + B )T = AT + BT (iv) (AB)T = BTAT
Symmetric Matrix: A square matrix A is said to be symmetric if AT =A
If A is a symmetric matrix, then A + AT is symmetric.
Skew-Symmetric Matrix: A square matrix A is said to be skew-symmetric if AT = -A
If A is a skew-symmetric matrix, then A – AT is skew-symmetric.
Minor of an element: Consider a square matrix
the minor element in this matrix is defined as the determinant of the 2×2 matrix obtained after deleting the rows and the columns in which the element is present.
Ex: – minor of a3 is = b1c2 – b2c1
Minor of b2 is = a1c3 – a3c1
Cofactor of an element: The cofactor of an element in i th row and j th column of A3×3 matrix is defined as its minor multiplied by (- 1) i+j.
###### Properties of determinants:
If each element of a row (column) f a square matrix is zero, then the determinant of that matrix is zero.
Ex: – A = ⇒ det(A) = 0
If A is a square matrix of order 3 and k is scalar then.
If two rows (columns) of a square matrix are identical (same), then Det. Of that matrix is zero.
Ex: – A = ⇒ det(A) = 0
If each element in a row (column) of a square matrix is the sum of two numbers then its determinant can be expressed as the sum of the determinants.
Ex: –
If each element of a square matrix are polynomials in x and its determinant is zero when x = a, then (x-a) is a factor of that matrix.
For any square matrix A Det(A) = Det (AI).
Det (AB) = Det(A). Det(B).
For any positive integer n Det (An) = (DetA)n.
###### Singular and non-singular matrices:
A Square matrix is said to be singular if its determinant is is zero, otherwise it is said to be non-singular matrix.
Ex: – A = det(A) = 4 – 4 = 0
∴ A is singular matrix
B =
Det(B) = 4 + 4 = 8≠ 0
∴ B is non-singular
Adjoint of a matrix: The transpose of the matrix formed by replacing the elements of a square matrix A with the corresponding cofactors is called the adjoint of A.
Let A = and cofactor matrix of A =
Then adj (A) =
Invertible matrix: Let A be a square matrix, we say that A is invertible if there exists a matrix B such that AB =BA = I, where I is a unit matrix of the same order as A and B.
1. (A– I)– I = A
2. (AI)– I = (A-I)I
3. (AB)-I = B-I A-I
4. A-I =
#### Compound Angles
The algebraic sum of two or more angles is called a ‘compound angle’. Thus, angles A + B, A – B, A + B + C etc., are Compound Angles
For any two real numbers A and B
sin (A + B) = sin A cos B + cos A Cos B
sin (A − B) = sin A cos B − cos A Cos B
cos (A + B) = cos A cos B − sin A sin B
cos (A − B) = cos A cos B + sin A sin B
tan (A + B) =
tan (A − B) =
cot (A + B) =
cot (A − B) =
tan ( + A) =
tan ( − A) =
cot (+ A) =
cot (− A) =
sin (A + B + C) = ∑sin A cos B cos C − sin A sin B sin C
cos (A + B + C) = cos A cos B cos C− ∑cos A sin B sin C
tan (A + B + C) =
⋇ cot (A + B + C) =
⋇ sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A
⋇ cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A
##### Multiple and Sub Multiple Angles
If A is an angle, then its integral multiples 2A, 3A, 4A, … are called ‘multiple angles ‘of A and the multiple of A by fraction like are called ‘submultiple angles.
∎ If is not an add multiple of
#### PROPERTIES OF TRIANGLES
In ∆ABC,
Lengths AB = c; BC = a; AC =b
The area of the triangle is denoted by ∆.
Perimeter of the triangle = 2s = a + b + c
A = ∠CAB; B = ∠ABC; C = ∠BCA.
R is circumradius.
Sine rule:
In ∆ABC,
⟹ a = 2R sin A; b = 2R sin B; c = 2R sin C
Where R is the circumradius and a, b, c, are lengths of the sides of ∆ABC.
Cosine rule:
In ∆ABC,
###### Projection rule:
In ∆ABC,
a = b cos C + c cos B
b = a cos C + c cos A
c = a cos B + b cos A
###### Tangent rule (Napier’s analogy):
Area of the triangle:
In ∆ABC, a, b, and c are sides
S = and area of the triangle
### HYPERBOLIC FUNCTIONS
The function f: R→R defined by f(x) = ∀ x ∈ R is called the ‘hyperbolic sin’ function. It is denoted by Sinh x.
∴Sinh x =
Similarly,
###### Identities:
cosh2x – sinh2 x = 1
cosh2x = 1 + sinh2 x
sinh2 x = cosh2 x – 1
sech2 x = 1 – tanh2 x
tanh2 x = 1 – sesh2 x
cosech2 x = coth2 x – 1
coth2 x = 1 + coth2 x
###### Addition formulas of hyperbolic functions:
Sinh (x + y) = Sinh x Cosh y + Cosh x Sinh y
Sinh (x − y) = Sinh x Cosh y − Cosh x Sinh y
Cosh (x + y) = Cosh x Cosh y + Sinh x Sinh y
Cosh (x − y) = Cosh x Cosh y − Sinh x Sinh y
tanh (x + y) =
tanh (x − y) =
coth (x + y) =
sinh 2x = 2 sinh x cosh 2x =
cosh 2x = cosh2x + sinh2 x = 2 cosh2x – 1 = 1 + 2 sinh2x =
tanh 2x =
###### Inverse hyperbolic functions:
Sinh−1x = ∀ x ∈ R
Cosh−1x = ∀ x ∈ (1, ∞)
Tanh−1x = ∀ < 1
#### COMPLEX NUMBERS
The equation x2 + 1 = 0 has no roots in real number system.
∴ scientists imagined a number ‘i’ such that i2 = − 1.
#### Complex number:
if x, y are any two real numbers then the general form of the complex number is
z = x + i y; where x real part and y is the imaginary part.
3 + 4i, 2 – 5i, – 3 + 2i are the examples for Complex numbers.
• z = x +i y can be written as (x, y).
• If z1 = x1 + i y1, z2 = x2 + i y2, then
• z1 + z2 = (x1 + x2, y1 + y2) = (x1 + x2) + i (y1 + y2)
• z1 − z2 = (x1 − x2, y1 − y2) = (x1 − x2) + i (y1 − y2)
• z1∙ z2 = (x1 x2 −y1 y2, x1y2 + x2y1) = (x1x2 −y1 y2) + i (x1y2 +x2 y1)
• z1/ z2 = (x1x2 + y1 y2/x22 +y22, x2 y1 – x1y2/ x22 +y22)
= (x1x2 + y1 y2/x22 +y22) + i (x2 y1 – x1y2/ x22 +y22)
###### Multiplicative inverse of complex number:
The multiplicative inverse of the complex number z is 1/z.
z = x + i y then 1/z = x – i y/ x2 + y2
Conjugate complex numbers:
The complex numbers x + i y, x – i y are called conjugate complex numbers.
Conjugate of z is denoted by
The sum and product of two conjugate complex numbers are real.
If z1, z2 are two complex numbers then
###### Modulus and amplitude of complex numbers:
Modulus: – If z = x + i y, then the non-negative real number is called modulus of z and it is denoted by or ‘r’.
Amplitude: – The complex number z = x + i y represented by the point P (x, y) on the XOY plane. ∠XOP = θ is called amplitude of z or argument of z.
x = r cos θ, y = r sin θ
x2 + y2 = r2 cos2θ + r2 sin2θ = r2 (cos2θ + sin2θ) = r2(1)
⇒ x2 + y2 = r2
⇒ r = and = r.
• Arg (z) = tan−1(y/x)
• Arg (z1.z2) = Arg (z1) + Arg (z2) + nπ for some n ∈ { −1, 0, 1}
• Arg(z1/z2) = Arg (z1) − Arg (z2) + nπ for some n ∈ { −1, 0, 1}
Note:
∎ e = cos θ + i sin θ
∎ e−iθ = cos θ − i sin θ
##### De- Moiver’s theorem
For any integer n and real number θ, (cos θ + i sin θ) n = cos n θ + i sin n θ.
cos α + i sin α can be written as cis α
cis α.cis β= cis (α + β)
1/cisα = cis(-α)
cisα/cisβ = cis (α – β)
(cos θ + i sin θ) -n = cos n θ – i sin n θ
(cos θ + i sin θ) (cos θ – i sin θ) = cos2θ – i2 sin2θ = cos2θ + sin2θ = 1.
cos θ + i sin θ = 1/ cos θ – i sin θ and cos θ – i sin θ = 1/ cos θ + i sin θ
(cos θ – i sin θ) n = (1/ (cos θ –+i sin θ)) n = (cos θ + i sin θ)-n = cos n θ – i sin n θ
nth root of a complex number: let n be a positive integer and z0 ≠ 0 be a given complex number. Any complex number z satisfying z n = z0 is called an nth root of z0. It is denoted by z01/n or
let z = r (cos θ + i sin θ) ≠ 0 and n be a positive integer. For k∈ {0, 1, 2, 3…, (n – 1)}
let Then a0, a1, a2, …, an-1 are all n distinct nth roots of z and any nth root of z is coincided with one of them.
nth root of unity: Let n be a positive integer greater than 1 and
Note:
• The sum of the nth roots of unity is zero.
• The product of nth roots of unity is (– 1) n – 1.
• The nth roots of unity 1, ω, ω2, …, ωn-1 are in geometric progression with common ratio ω.
##### Cube root of unity:
x3 – 1 = 0 ⇒ x3 = 1
x =11/3
cube roots of unity are: 1
ω = , ω2 =
ω2 +ω + 1 = 0 and ω3 = 1
### TRANSFORMATIONS
For A, B∈ R
⋇ sin (A + B) + sin (A – B) = 2sin A cos B
⋇ sin (A + B) −sin (A – B) = 2cos A sin B
⋇ cos (A + B) + cos (A – B) = 2 cos A cos B
⋇ cos (A + B) − cos (A – B) = − 2sin A sin B
For any two real numbers C and D
⋇ sin C + sin D = 2sin cos
⋇ sin C −sin D= 2cos sin
⋇ cos C + cos D = 2 cos cos
⋇ cos C − cos D = − 2sin sin
If A + B + C = π or 1800, then
⋇ sin (A + B) = sin C; sin (B + C) = sin A; sin (A + C) = sin B
⋇ cos (A + B) = − cos C; cos (B + C) = −cos A; cos (A + C) = − cos B
If A + B + C = 900 or then
⋇ sin = cos; sin = cos ; sin = cos
⋇ cos = sin; cos = sin; cos = sin
If then
⋇ sin (A + B) = cos C; sin (B + C) = cos A; sin (A + C) = cos B
⋇ cos (A + B) = sin C; cos (B + C) = sin A; cos (A + C) = sin B
### INVERSE TRIGONOMETRIC RATIOS
If A, B are two sets and f: A→ B is a bijection, then f-1 is existing and f-1: B → A is an inverse function.
### Solutions of Simultaneous Equations
#### Matrix Inversion Method:
Let a system of simultaneous equations be
a1 x + b1 y + c1z = d1
a2 x + b2 y + c2z = d2
a3 x + b3 y + c3z = d3
The matrix form of the above equations is
Therefore, the matrix equation is AX = B
If Det A ≠ 0, A-1 is exists
X = A-1 B
By using above Condition, we get the values of x, y and z
This Method is called as Matrix Inversion Method
#### Cramer’s Method:
Let system of simultaneous equations be
a1 x + b1 y + c1z = d1
a2 x + b2 y + c2z = d2
a3 x + b3 y + c3z = d3
1 is obtained by replacing the coefficients of x (1st column elements of ∆) by constant values
2 is obtained by replacing the coefficients of y (2nd column elements of ∆) by constant values
3 is obtained by replacing the coefficients of z (3rd column elements of ∆) by constant values
Now
This method is called Cramer’s Method
#### Gauss-Jordan Method:
Let a system of simultaneous equations be
a1 x + b1 y + c1z = d1
a2 x + b2 y + c2z = d2
a3 x + b3 y + c3z = d3
Augmented matrix: The coefficient matrix (A) augmented with the constant column matrix (B) is called the augmented matrix. It is denoted by [AD].
[AD] =
This Matrix is reduced to the standard form ofby using row operations
1. Interchanging any two rows
2. Multiplying the elements of any two elements by a constant.
3. Adding to the elements of one row with the corresponding elements of another row multiplied by a constant.
∴ The solution of a given system of simultaneous equations is x = α, y = β, and z = γ.
##### Procedure to get the standard form:
1. Take the coefficient of x as the unity as a first equation.
2. If 1 is there in the first-row first column, then make the remaining two elements in the first column zero.
3. After that, if one element in R2 or R3 is 1, then make the remaining two elements in that column C2 or C3 as zeroes.
4. If any row contains two elements as zeros and only non-zero divide that row elements with the non-zero element to get unity and make the remaining two elements in that column as zeros. | 5,107 | 15,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-06 | longest | en | 0.858316 |
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# 8-3 - PowerPoint PPT Presentation
8-3. Perimeter and Circumference. Course 2. Warm Up. Problem of the Day. Lesson Presentation. 8-3. Perimeter and Circumference. Course 2. Warm Up Add or multiply. 1. 8.2 + 5.6 + 8.2 + 5.6 2. 12.4 + 15.8 + 9.3 3. 4 · 8.5 4. 22 · 3.14. 27.6. 37.5. 34. 69.08. 8-3.
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Presentation Transcript
Perimeter and Circumference
Course 2
Warm Up
Problem of the Day
Lesson Presentation
Perimeter and Circumference
Course 2
Warm Up
1.8.2 + 5.6 + 8.2 + 5.6
2. 12.4 + 15.8 + 9.3
3. 4 · 8.5
4. 22 · 3.14
27.6
37.5
34
69.08
Perimeter and Circumference
Course 2
Problem of the Day
A rectangular piece of sheet metal measures 5 in. by 6 in. A rectangular notch 1 in. deep and in. wide is cut
7
64
out of the middle of each side. What is
the perimeter of the piece after the
cuts have been made? (Hint: You don’t
have to use fractions.)
30 in. (the original 6 in. by 5 in. sheet
plus 8 in. for the eight 1 in. cuts)
Perimeter and Circumference
Course 2
Learn to find the perimeter of a polygon and the circumference of a circle.
Perimeter and Circumference
Course 2
Insert Lesson Title Here
Vocabulary
perimeter
circumference
Perimeter and Circumference
Course 2
The distance around a geometric figure is its perimeter. To find the perimeter P of a polygon, you can add the lengths of its sides.
Perimeter and Circumference
Course 2
Additional Example 1A: Finding the Perimeter of a Polygon
Find the perimeter of the polygon.
A.
4 in.
4 in.
5 in.
5 in.
4 in.
Use the side lengths.
P = 4 + 4 + 5 + 4 + 5
P = 22
The perimeter of the pentagon is 22 in.
Perimeter and Circumference
Course 2
Additional Example 1B: Finding the Perimeter of a Polygon
Find the perimeter of the polygon.
B.
25 cm
25 cm
15 cm
P = 25 + 25 + 15
Use the side lengths.
P = 65
The perimeter of the triangle is 65 cm.
Perimeter and Circumference
Course 2
Insert Lesson Title Here
Try This: Example 1A
Find the perimeter of the polygon.
5 in.
A.
4 in.
6 in.
8 in.
Use the side lengths.
P = 5 + 6 + 8 + 4
P = 23
The perimeter of the polygon is 23 in.
Perimeter and Circumference
Course 2
Insert Lesson Title Here
Try This: Example 1B
Find the perimeter of the polygon.
20 cm
B.
25 cm
30 cm
P = 25 + 20 + 30
Use the side lengths.
P = 75
The perimeter of the triangle is 75 cm.
Perimeter and Circumference
Course 2
You can find the perimeter of a rectangle by adding the length of its sides. Or, since opposite sides of a rectangle are equal in length, you can find the perimeter by using the formula P = 2l+ 2w.
w
l
Perimeter and Circumference
Course 2
Additional Example 2: Using Properties of a Rectangle to Find Perimeter
Find the perimeter of a rectangle.
14 ft
18 ft
P = 2l+ 2w
Use the formula.
P = (2 ·18) + (2 ·14)
Substitute for l and w.
P = 36 + 28
Multiply.
P = 64
The perimeter of the rectangle is 64 ft.
Perimeter and Circumference
Course 2
Insert Lesson Title Here
Try This: Example 2
Find the perimeter of a rectangle.
16 ft
8 ft
P = 2l + 2w
Use the formula.
P = (2 · 8) + (2 · 16)
Substitute for l and w.
P = 16 + 32
Multiply.
P = 48
The perimeter of the rectangle is 48 ft.
Perimeter and Circumference
Course 2
The distance around a circle is
called circumference. For
every circle, the ratio of
circumference C to diameter d
is the same. This ratio,
C
d
, is
represented by the symbol ,
called pi. Pi is approximately
equal to 3.14 or . By
22
7
Diameter
multiplying both sides of the
equation
C
d
= by d, you get
Circumference
the formula for circumference,
C = d, or C = 2r.
Perimeter and Circumference
Course 2
Additional Example 3A: Finding the Circumference of a Circle
Find the circumference of the circle to the nearest tenth. Use 3.14 for .
A.
C = d
You know the diameter.
12 in.
C 3.14 · 12
Substitute for and d.
C 37.68
Multiply.
The circumference of the circle is about 37.7 in.
Perimeter and Circumference
Course 2
Additional Example 3B: Finding the Circumference of a Circle
Find the circumference of the circle to the nearest tenth. Use 3.14 for .
B.
C = 2r
You know the radius.
18 cm
C 2 · 3.14 · 18
Substitute for and r.
C 113.04
Multiply.
The circumference of the circle is about 113.0 cm.
Perimeter and Circumference
Course 2
Try This: Example 3A
Find the circumference of the circle to the nearest tenth. Use 3.14 for .
A.
C = d
You know the diameter.
7 in.
C 3.14 · 7
Substitute for and d.
C 21.98
Multiply.
The circumference of the circle is about 22 in.
Perimeter and Circumference
Course 2
Try This: Example 3B
Find the circumference of the circle to the nearest tenth. Use 3.14 for .
B.
C = 2r
You know the radius.
11 cm
C 2 · 3.14 · 11
Substitute for and r.
C 69.08
Multiply.
The circumference of the circle is about 69.1 cm.
Perimeter and Circumference
Course 2
Additional Example 4: Application
The diameter of a circular pond is 42 m. What is its circumference? Use
for .
22
7
You know the diameter.
C = d
22
7
22
7
· 42
Substitute
for and 42 for d.
C
22
7
42
1
Write 42 as a fraction.
C
·
6
22
7
42
1
·
Simplify.
C
1
C 132
Multiply.
The circumference of the pond is about 132 m.
Perimeter and Circumference
Course 2
Try This: Example 4
The diameter of a circular spa is 14 m. What is its circumference? Use
for .
22
7
You know the diameter.
C = d
22
7
22
7
· 14
Substitute
for and 14 for d.
C
22
7
14
1
Write 14 as a fraction.
C
·
2
22
7
14
1
·
Simplify.
C
1
C 44
Multiply.
The circumference of the spa is about 44 m.
Perimeter and Circumference
Course 2
Insert Lesson Title Here
Lesson Quiz
Find the perimeter of each polygon.
1.2.
3. What is the circumference of a circle with a
a radius of 8 inches? Use 3.14 for .
4. If a guardrail is to be placed around the edge
of a circular swimming pool with a diameter of
21 ft, how many feet of railing will be needed?
Use
8 m
12 m
8 in.
10 m
45 m
36 in.
9 m
10 in.
6 m
50.24 in.
for .
66 ft
22
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# Special Right Triangles
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### Special Right Triangles
1. 1. Two Special Right Triangles 45°- 45°- 90° 30°- 60°- 90°
2. 2. 45°- 45°- 90° The 45-45-90 triangle is based on the square with sides of 1 unit. 1 1 1 1
3. 3. 45°- 45°- 90° If we draw the diagonals we form two 45-45-90 triangles. 45° 45° 45° 45° 1 1 1 1
4. 4. 45°- 45°- 90° Using the Pythagorean Theorem we can find the length of the diagonal. 45° 45° 45° 45° 1 1 1 1
5. 5. 45°- 45°- 90° 1 2 + 1 2 = c 2 1 + 1 = c 2 2 = c 2 2 = c 45° 45° 45° 45° 2 1 1 1 1
6. 6. 45°- 45°- 90° Conclusion: the ratio of the sides in a 45-45-90 triangle is 1-1- 2 1 1 2 45° 45°
7. 7. 45°- 45°- 90° Practice 4 4 2 SAME leg* 2 4 45° 45°
8. 8. 45°- 45°- 90° Practice 9 9 2 SAME leg* 2 9 45° 45°
9. 9. 45°- 45°- 90° Practice 2 2 2 SAME leg* 2 2 45° 45°
10. 10. 45°- 45°- 90° Practice 14 SAME leg* 2 7 7 45° 45°
11. 11. 45°- 45°- 90° Practice Now Let's Go Backward
12. 12. 45°- 45°- 90° Practice 3 2 hypotenuse 2 45° 45°
13. 13. 45°- 45°- 90° Practice = 3 3 2 2
14. 14. 45°- 45°- 90° Practice 3 2 hypotenuse 2 3 SAME 3 45° 45°
15. 15. 45°- 45°- 90° Practice 6 2 hypotenuse 2 45° 45°
16. 16. 45°- 45°- 90° Practice = 6 6 2 2
17. 17. 45°- 45°- 90° Practice 6 2 hypotenuse 2 6 SAME 6 45° 45°
18. 18. 45°- 45°- 90° Practice 11 2 hypotenuse 2 45° 45°
19. 19. 45°- 45°- 90° Practice = 11 11 2 2
20. 20. 45°- 45°- 90° Practice 11 2 hypotenuse 2 11 SAME 11 45° 45°
21. 21. 45°- 45°- 90° Practice 8 hypotenuse 2 45° 45°
22. 22. 45°- 45°- 90° Practice = 4 2 8 2 2 2 * = 8 2 2
23. 23. 45°- 45°- 90° Practice 8 hypotenuse 2 4 2 SAME 4 2 45° 45°
24. 24. 45°- 45°- 90° Practice 4 hypotenuse 2 45° 45°
25. 25. 45°- 45°- 90° Practice = 2 2 4 2 2 2 * = 4 2 2
26. 26. 45°- 45°- 90° Practice 4 hypotenuse 2 2 2 SAME 2 2 45° 45°
27. 27. 45°- 45°- 90° Practice 6 Hypotenuse 2 45° 45°
28. 28. 45°- 45°- 90° Practice = 3 2 6 2 2 2 * = 6 2 2
29. 29. 45°- 45°- 90° Practice 6 hypotenuse 2 3 2 SAME 3 2 45° 45°
30. 30. 30°- 60°- 90° The 30-60-90 triangle is based on an equilateral triangle with sides of 2 units. 2 2 2 60 ° 60 °
31. 31. 30°- 60°- 90° The altitude (also the angle bisector and median) cuts the triangle into two congruent triangles. 1 1 30 ° 30 ° 2 2 2 60 ° 60 °
32. 32. 30°- 60°- 90° This creates the 30-60-90 triangle with a hypotenuse a short leg and a long leg. hypotenuse Short Leg Long Leg 30 ° 60 °
33. 33. 30°- 60°- 90° Practice 1 2 We saw that the hypotenuse is twice the short leg. We can use the Pythagorean Theorem to find the long leg. 60° 30°
34. 34. 30°- 60°- 90° Practice 1 2 3 A 2 + B 2 = C 2 A 2 + 1 2 = 2 2 A 2 + 1 = 4 A 2 = 3 A = 3 60° 30°
35. 35. 30°- 60°- 90° Conclusion: the ratio of the sides in a 30-60-90 triangle is 1- 2 - 3 3 1 2 60° 30°
36. 36. 30°- 60°- 90° Practice 4 8 Hypotenuse = short leg * 2 4 3 The key is to find the length of the short side. Long Leg = short leg * 3 60° 30°
37. 37. 30°- 60°- 90° Practice 5 10 Hypotenuse = short leg * 2 5 3 Long Leg = short leg * 3 60° 30°
38. 38. 30°- 60°- 90° Practice 7 14 Hypotenuse = short leg * 2 7 3 Long Leg = short leg * 3 60° 30°
39. 39. 30°- 60°- 90° Practice 3 2 3 Hypotenuse = short leg * 2 3 Long Leg = short leg * 3 60° 30°
40. 40. 30°- 60°- 90° Practice 10 2 10 Hypotenuse = short leg * 2 30 Long Leg = short leg * 3 60° 30°
41. 41. 30°- 60°- 90° Practice Now Let's Go Backward
42. 42. 30°- 60°- 90° Practice 11 22 Short Leg = Hypotenuse 2 11 3 Long Leg = short leg * 3 60° 30°
43. 43. 30°- 60°- 90° Practice 2 4 Short Leg = Hypotenuse 2 2 3 Long Leg = short leg * 3 60° 30°
44. 44. 30°- 60°- 90° Practice 9 18 Short Leg = Hypotenuse 2 9 3 Long Leg = short leg * 3 60° 30°
45. 45. 30°- 60°- 90° Practice 15 30 Short Leg = Hypotenuse 2 15 3 Long Leg = short leg * 3 60° 30°
46. 46. 30°- 60°- 90° Practice 23 46 Hypotenuse = Short Leg * 2 23 3 Short Leg = Long leg 3 60° 30°
47. 47. 30°- 60°- 90° Practice 14 28 Hypotenuse = Short Leg * 2 14 3 Short Leg = Long leg 3 60° 30°
48. 48. 30°- 60°- 90° Practice 16 32 Hypotenuse = Short Leg * 2 16 3 Short Leg = Long leg 3 60° 30°
49. 49. 30°- 60°- 90° Practice 3 3 6 3 Hypotenuse = Short Leg * 2 9 Short Leg = Long leg 3 60° 30°
50. 50. 30°- 60°- 90° Practice 4 3 8 3 Hypotenuse = Short Leg * 2 12 Short Leg = Long leg 3 60° 30°
51. 51. 30°- 60°- 90° Practice 9 3 18 3 Hypotenuse = Short Leg * 2 27 Short Leg = Long leg 3 60° 30°
52. 52. 30°- 60°- 90° Practice 7 3 14 3 Hypotenuse = Short Leg * 2 21 Short Leg = Long leg 3 60° 30°
53. 53. 30°- 60°- 90° Practice 11 3 22 3 Hypotenuse = Short Leg * 2 33 Short Leg = Long leg 3 60° 30°
54. 54. THE END | 2,860 | 5,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-43 | latest | en | 0.458071 |
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# If x and y are integers greater than 1, is x a multiple of y
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If x and y are integers greater than 1, is x a multiple of y [#permalink]
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29 Jan 2012, 16:21
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If x and y are integers greater than 1, is x a multiple of y?
(1) 3y^2+7y=x
(2) x^2-x is a multiple of y
[Reveal] Spoiler:
For me its D. Unless someone else thinks otherwise. This is how I solved this:
Statement 1
y is a multiple of 3 and 7
So when y = 2 then x will be 26 and YES x is a multiple of y.
when y = 5 then x will be 110 and YES x is a multiple of y.
Therefore, statement 1 is sufficient to answer this questions.
Statement 2
x^2-x is a multiple of y
x(x-1) is a multiple of y. Sufficient to answer.
Therefore D for me i.e. both statements alone are sufficient to answer this question. Any thoughts guys?
[Reveal] Spoiler: OA
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Last edited by Bunuel on 30 Oct 2012, 01:44, edited 2 times in total.
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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
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x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x
(2): x^2 - x is multiple of y
The question is asking whether $$x/y$$ is an integer or not.
Statement 1 can be written as $$y(3y +7)=x$$ or $$3y+7=x/y$$. Since $$y$$ is an integer, therefore 3y+7 must also be an integer. Hence $$x/y$$ will be an integer or x is a multiple of y. Sufficient.
Statement 2 can be written as $$x(x-1)/y$$ is an integer. Notice that we can't deduce that x is a multiple of y because it is quite possible that the product is a multiple of y, but not the individual entities.
ex. 2*3/6 is a mltiple of 6 BUT neither 2 nor 3 is a multiple of 6.
Insufficient.
+1A
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Re: Is x a multiple of y [#permalink]
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If x and y are integers great than 1, is x a multiple of y?
(1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient.
(2) $$x^2-x$$ is a multiple of $$y$$ --> $$x(x-1)$$ is a multiple of $$y$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient.
Hope it helps.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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03 May 2014, 12:26
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FACTS ESTABLISHED BY STEMS
• $$x$$ is an integer
• $$y$$ is an integer
• $$x > 1$$
• $$y > 1$$
QUESTION TRANSLATED
Since both $$x$$ and $$y$$ are integers the question "is $$x$$ a multiple of $$y$$?" really asks if
$$x = q \cdot y + r$$
where $$q$$ is an integer (the quotient) and $$r$$ is the remainder when you divide $$x$$ by $$y$$ so that $$r = 0$$ which means that
$$x = q \cdot y + 0$$
which implies that
$$x = q \cdot y$$
So the question translates to:
is $$x = q \cdot y$$ when $$x$$, $$q$$ and $$y$$ are all integers, and when $$x>1$$ and $$y > 1$$?
STATEMENT 1
$$3y^2 + 7y = x$$
can be rewritten as
$$y(3y + 7) = x$$
however, from the question stems we know that $$y$$ is an integer, which implies $$3y$$ is an integer and that $$3y+7$$ is also an integer:
$$y \cdot integer = x$$
which can be rewritten as
$$x = y \cdot integer$$ or better yet, as
$$x = integer \cdot y$$
which if you notice is an equation of the form $$x = q \cdot y$$ since $$q$$ is an integer which implies that $$x$$ is indeed a multiple of $$y$$.
Therefore, Statement A is sufficient.
STATEMENT 2
the statement "$$x^2 - x$$ is a multiple of $$y$$" must be rewritten as "$$x (x - 1)$$ is a multiple of $$y$$" which implies that
either $$x$$ is a multiple of $$y$$ or $$x-1$$ is a multiple of $$y$$
the sub-statement $$x$$ is a multiple of $$y$$ implies that $$x = q \cdot y$$ which would answer the question, but we still need to figure out if the second sub-statement answers the question with the same answer as the first sub-statement.
the second sub-statement $$x-1$$ is a multiple of $$y$$ implies that
$$x-1 = q \cdot y$$
which can be rewritten as
$$x = q \cdot y + 1$$
which is an equation of the form $$x = q \cdot y + r$$ when $$r=1$$ which implies that when $$x$$ is divided by $$y$$ we get a remainder of 1, meaning that $$x$$ is NOT a multiple of $$y$$.
Since both sub-statements are contradictory, this means that when $$x^2 - x$$ is a multiple of $$y$$, $$x$$ is not always a multiple of $$y$$. meaning that we cannot determine whether $$x$$ is always a multiple of $$y$$ according to Statement 2 alone.
Therefore, Statement 2 is not sufficient.
(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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for(2)
x(x-1) is multiple of y.
so either x is divisible by y or x-1 is divisible by y.
if x-1 is divisible by y, then x is not divisble by y.
plug in y=8, x=16--->16*15 is divisible by 8 --> yes x is a multiple of y.
plug in y=8,x=17----> 17*16 is still divisible by 8 --> But x is not a multiple of y.
So insufficient.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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18 Jun 2012, 07:07
Hi Bunuel,
Can you please explain this problem. I did not understand the explanation above for Point 2.
Sorry for the inconvenience.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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18 Jun 2012, 07:10
harshavmrg wrote:
Hi Bunuel,
Can you please explain this problem. I did not understand the explanation above for Point 2.
Sorry for the inconvenience.
Can you please be a little bit more specific? What exactly didn't you understand in the second statement?
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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05 Sep 2012, 08:00
Hi Bunuel,
I have not understood the answer to the (2) statement, infact we have that x(x-1) = Y*F (because is a multiple), doesn't it depend on the number F if X is a multiple of y? For example:
If x=2, then y could be either 2 or 1; so the answer would be yes;
If x=3, then y could be either 2 or 3; so the answer would be no;
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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
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31 Dec 2012, 22:41
x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x
(2): x^2 - x is multiple of y
The Questions asks whether x/y is an integer or not.
from St 1 we have 3y^2 +7y=x ---> x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer.
If y=7, then x/y=4 which is an integer.
So St1 alone not sufficient
From St 2 we have x^2-x is multiple of y ----> x(x-1)/y= Integer value
Now we know x-1 and x are consecutive integers
X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x is even. Now x can be odd or even and hence alone not sufficient.
Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7.
But we are not sure from above statement so ans should be E.
What is OA??
Thanks
Mridul
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Re: Is x multiple of y? If: 3y^2 + 7y = x? If: x^2-x is mult? [#permalink]
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31 Dec 2012, 22:59
mridulparashar1 wrote:
x and y are integers > 1. Is x multiple of y?
(1): 3y^2 + 7y = x
(2): x^2 - x is multiple of y
The Questions asks whether x/y is an integer or not.
from St 1 we have 3y^2 +7y=x ---> x/y= 3y+7 if y=2, x=13 and 13/2 is not an integer.
If y=7, then x/y=4 which is an integer.
So St1 alone not sufficient
From St 2 we have x^2-x is multiple of y ----> x(x-1)/y= Integer value
Now we know x-1 and x are consecutive integers
X2-x is even because if x is odd then x2-x (odd-odd)=even and if x is even then x2-x is even. Now x can be odd or even and hence alone not sufficient.
Combining both statement,we get for x/y to be an integer y has to be 7 and x will be have to be multiple of 7.
But we are not sure from above statement so ans should be E.
What is OA??
Thanks
Mridul
if x/y=3y+7, then taking y as 2 will yield x/y as 13 and not 13/2.
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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18 Sep 2014, 18:36
Bunuel wrote:
If x and y are integers great than 1, is x a multiple of y?
(1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient.
(2) $$x^2-x$$ is a multiple of $$y$$ --> $$x(x-1)$$ is a multiple of $$y$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient.
Hope it helps.
Hi Bunuel,
Two questions:
1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc?
2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest?
3) Lastly, the way I solved statement 1 was -- I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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19 Sep 2014, 02:31
russ9 wrote:
Bunuel wrote:
If x and y are integers great than 1, is x a multiple of y?
(1) $$3y^2+7y=x$$ --> $$y(3y+7)=x$$ --> as $$3y+7=integer$$, then $$y*integer=x$$ --> $$x$$ is a multiple of $$y$$. Sufficient.
(2) $$x^2-x$$ is a multiple of $$y$$ --> $$x(x-1)$$ is a multiple of $$y$$ --> $$x$$ can be multiple of $$y$$ ($$x=2$$ and $$y=2$$) OR $$x-1$$ can be multiple of $$y$$ ($$x=3$$ and $$y=2$$) or their product can be multiple of $$y$$ ($$x=3$$ and $$y=6$$). Not sufficient.
Hope it helps.
Hi Bunuel,
Two questions:
1) If x = 2 and y = 2, does that mean that x is a multiple of y? I always thought that x needed to be 1 multiple greater, meaning, x=4 and y=2 etc?
2) In situations like these, I usually stumble with either starting off with factoring(like you did above) or some other method. What do you suggest?
3) Lastly, the way I solved statement 1 was -- I saw it as (y)(multiple of y) = x, and the rule states that if i multiply two numbers that are multiples of y, then the result will be a multiple of y. Is that correct to assume?(I know that we can assume that with addition, right?)
1. The least multiple of a positive integer is that integer itself. So, the least multiple of 2 is 2.
2. I suggest the way I used.
3. 3y+7 is not necessarily a multiple of y but it does not need to be. y*integer=x implies that x is a multiple of y.
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If x and y are integers greater than 1, is x a multiple of y [#permalink]
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20 Aug 2016, 19:46
x,y are integers>1, is x multiple of y?
x multiple of y is same as x/y=integer
1. 3y^2+7y=x
y(3y+7)=x
dividing both sides by y, we get:
(3y+7)=x/y
x/y=int as question stem says y is integer
statement 1. is sufficient
2. x^2-x is a multiple of y
x(x-1) is a multiple of y
i.e {x(x-1)}/y must be integer
x/y say 4/2 =integer or 4-1/2=3/2 not integer
Thus, statement 2 is insufficient.
Hence, A
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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21 Aug 2016, 01:59
Awesome Question .
Here we need to prove if x=y*I for some integer I
Statement 1=> x=y[3y+y]=y*I => Sufficient
Statement 2 => x(x-1) is a multiple of y let x=7 x-1=6 and y =6 here clearly the answer is NO
now let y=7 => the answer will be YES
hence Insufficient
Smash that A
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
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26 Jan 2017, 19:34
Statement 1: x = y(3y + 7) = y*(Some integer)
Therefore, x is a multiple of y
Sufficient
Statement 2: x(x - 1) is a multiple of y.
Therefore, either x or (x - 1) or none of them is multiple of y.
Not sufficient
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink]
### Show Tags
16 Oct 2017, 06:13
What if we change the question stem to factors... is B the answer?
If x and y are integers greater than 1, is x a factor of y?
(1) 3y^2+7y=x
(2) x^2-x is a multiple of y
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Re: If x and y are integers greater than 1, is x a multiple of y [#permalink] 16 Oct 2017, 06:13
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1 ELCT201: DIGITAL LOGIC DESIGN Dr. Eng. Haitham Omran, Dr. Eng. Wassim Alexan, Lecture 3 Following the slides of Dr. Ahmed H. Madian ذو الحجة 1438 ه Winter 2017
2 COURSE OUTLINE 1. Introduction 2. Gate-Level Minimization 3. Combinational Logic 4. Synchronous Sequential Logic 5. Registers and Counters 6. Memories and Programmable Logic 2
3 4-VARIABLE MAP Notice the order of the minterms Remember that the cells in the top row are adjacent to the cells in the bottom row Remember that cells in the most left column are adjacent to the cells in the most right column Remember that the cells in the four corners are adjacent to each 3
4 NOTES ON A 4-VARIABLE MAP The number of adjacent squares that may be combined must always represent a number that is a power of two, such as 1, 2, 4, 8 and 16 As more adjacent squares are combined, we obtain a product term with fewer literals One square represents one minterm, giving a term with 4 literals Two adjacent squares represent a term with 3 literals Four adjacent squares represent a term with 2 literals Eight adjacent squares represent a term with 1 literal Sixteen adjacent squares encompass the entire map and produce a function that is always equal to logic 1 4
5 4-VARIABLE MAP: EXAMPLE I Simplify the Boolean expression: F A, B, C, D = Σ(0,1,2,4,5,7,8,9,10,12,13) 5
6 4-VARIABLE MAP: EXAMPLE I Simplify the Boolean expression: F A, B, C, D = Σ(0,1,2,4,5,7,8,9,10,12,13) F A, B, C, D = C + B D + A BD 6
7 4-VARIABLE MAP: EXAMPLE II Simplify the Boolean expression: F w, x, y, z = Σ(0,1,2,4,5,6,8,9,12,13,14) 7
8 4-VARIABLE MAP: EXAMPLE II Simplify the Boolean expression: F w, x, y, z = Σ(0,1,2,4,5,6,8,9,12,13,14) F w, x, y, z = y + w z + xz 8
9 CHOICE OF BLOCKS We can simplify a function by using larger blocks Do we really need all blocks? Can we leave some out to further simplify an expression? Any function needs to contain a special type of blocks These are called Essential Prime Implicants We need to define some new terms: Implicant Prime implicant Essential prime implicant 9
10 TERMINOLOGY Implicant (I) Any product term in the SOP form A block of 1s in a K-map Prime implicant (PI) Block of 1s that cannot be further increased Product term that cannot be further reduced Essential prime implicant (EPI) A prime implicant on a K-map which covers at least one 1 which is not covered by any other prime implicant is called an Essential Prime Implicant Is C an essential prime implicant? 10
11 THE SYSTEMATIC PROCEDURE FOR SIMPLIFYING BOOLEAN FUNCTIONS 1. Generate all Prime Implicants of the function 2. Include all Essential Prime Implicants 3. For the remaining minterms not included in the Essential Prime Implicants, select a set of other Prime Implicants to cover them, with minimal overlap in the set 4. The resulting simplified function is the logical OR of the product terms selected above 11
12 ILLUSTRATING THE TERMS: EXAMPLE I The Prime Implicants are: A D gray, AC rose, BC D pink, CD purple, ABD green, A BC (yellow). Of which only three are Essential: A D gray, AC rose and BC D pink. 12
13 ILLUSTRATING THE TERMS: EXAMPLE II The Prime Implicants are: BD gray, A BC yellow, AC D purple, ABC green, A CD rose. Of which only four are Essential: A BC yellow, AC D purple, ABC green and A CD rose. 13
14 PRODUCT OF SUMS SIMPLIFICATION USING K-MAPS Use the SOP simplification on the zeros of the function in the K-map to get F Find the complement of F, i.e. F = F Recall that the complement of a Boolean function can be obtained by (1) taking the dual and (2) complementing each literal Or by using DeMorgan s theorem 14
15 PRODUCT OF SUMS MINIMIZATION How to generate a POS from a K- map? Use duality of Boolean algebra (DeMorgan s law) Look at the 0s in map instead of the 1s Generate blocks around the 0s This gives the inverse of F Use duality to generate POS Example: F = (0,1,2,5,8,9,10) F = AB + CD + BD F = (A + B )(C + D )(B + D) 15
16 GATE IMPLEMENTATION 16
17 EXAMPLE ON POS MINIMIZATION Given the K-map below, produce the F from the zeros in the map and then obtain F from it F = AB +AC +A BCD F = (AB )(AC )(A BCD ) F = (A + B)(A + C)(A + B + C + D) 17
18 DON T CARE CONDITIONS There may be a combination of input values which: Will never occur, If they do occur, the output is of no concern The function value for such combinations is called a don t care They are usually denoted with an X Each X may be arbitrarily assigned the value 0 or 1 in an implementation Don t cares can be used to further simplify a function 18
19 MINIMIZATION USING DON T CARES Treat don t cares as if they are 1s to generate Prime Implicants Delete Prime Implicants that cover only don t care minterms Treat the covering of the remaining don t care minterms as optional in the selection process (i.e. they maybe, but need not be covered) 19
20 MINIMIZATION EXAMPLE F w, x, y, z = (1,3,7,11,15) and d w, x, y, z = (0,2,5) What are the possible solutions? 20 20
21 EXAMPLE INVOLVING X Simplify the function whose K-map is shown at the right F = A C D + AB + CD + A BC or F = A C D + AB + CD + A BD 21
22 ANOTHER EXAMPLE Simplify the function whose K-map is shown at the right F = A C + AB or F = A C + BD 22
23 NAND AND NOR IMPLEMENTATIONS Digital circuits are frequently constructed with NAND or NOR gates rather than AND & OR gates. NAND and NOR gates are easier to fabricate with electronic components and are the basic gates used in all IC digital logic families A NAND gate has the smallest propagation time delay among the other gates, except for the inverter! 23
24 NAND AND NOR IMPLEMENTATIONS Each NAND or NOR gate requires only 4 transistors Each AND gate requires 6 transistors 24
25 LOGIC OPERATION WITH NAND GATE NOT, AND, and OR can be implemented with NAND! 25
26 CONVERSION TO NAND IMPLEMENTATION Minimized expressions are AND-OR combinations There are two illustrations for NAND gates Key observation: two bubbles eliminate each other Two bubbles equal a straight wire How to generate a sum of minterms using NAND? Use AND-invert for minterms Use invert-or for sum 26
27 CONVERSION TO NAND IMPLEMENTATION Sum of minterms Replace AND with AND-invert and OR with invert-or Still the same circuit! Replace AND-invert and invert-or with NAND F = (A B )(C D ) = AB + CD 27
28 NAND EXAMPLE IMPLEMENTATION Minimize and implement the function F(x, y, z) = (1,2,3,4,5,7), using only NAND gates 28
29 MULTILEVEL NAND CIRCUITS Multilevel circuits conversion rules: 1. Convert all AND gates to NAND with AND-invert symbols 2. Convert all OR gates to NAND with invert-or symbols 3. Check all bubbles in the diagram. For every bubble that is not compensated by another bubble, insert an inverter. Example: 29
30 MULTILEVEL NAND CIRCUITS: AN EXAMPLE F = (AB + A B)(C + D ) 30
31 LOGIC OPERATION WITH NOR GATE NOR can also replace NOT, AND & OR There are two representations of the NOR gate: 31
32 CONVERTING TO NOR IMPLEMENTATIONS Same rules as for NAND implementations F = (AB + A B)(C + D ) With NOR F = (AB + A B)(C + D ) 32
33 DESIGN PROBLEM Design a digital system whose output is defined as logically low if the 4-bit input binary number is a multiple of 3; otherwise, the output will be logically high. The output is defined if and only if the input binary number is greater than 2. 33
34 DESIGN PROBLEM Design a digital system whose output is defined as logically low if the 4-bit input binary number is a multiple of 3; otherwise, the output will be logically high. The output is defined if and only if the input binary number is greater than 2. 34
35 DESIGN PROBLEM 35
36 DESIGN PROBLEM SOP POS Y SOP = B D + A C + A BD + BC D + AB C + ACD Y POS = (A + B)(B + C + D )(A + C + D)(A + B + C + D)(A + B + C + D ) 36
37 DESIGN PROBLEM What if you are asked to implement these digital circuits using only NAND gates? Is that possible? 37
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Unit-4: Introduction to Boolean Algebra Learning Objective At the end of the chapter students will: Learn Fundamental concepts and basic laws of Boolean algebra. Learn about Boolean expression and will | 8,534 | 33,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-49 | latest | en | 0.839041 |
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## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Product formulas
These are derived from the four addition formulas:
\begin{aligned} &\cos (A+B)=\cos A \cos B-\sin A \sin B \ &\cos (A-B)=\cos A \cos B+\sin A \sin B \ &\sin (A+B)=\sin A \cos B+\cos A \sin B \ &\sin (A-B)=\sin A \cos B-\cos A \sin B \end{aligned}
(a) $+(\mathrm{b})$ gives:
$\cos A \cos B=\frac{1}{2}(\cos (A+B)+\cos (A-B)) \quad$ Product Formula $\mathbf{p}(\mathbf{i})$
(b) – (a) gives:
$\sin A \sin B=\frac{1}{2}(\cos (A-B)-\cos (A+B)) \quad$ Product Formula $p(i i)$
(c) $+(\mathrm{d})$ gives:
$\sin A \cos B=\frac{1}{2}(\sin (A+B)+\sin (A-B)) \quad$ Product Formula p $($ iii)
(c) $-$ (d) gives:
$\cos A \sin B=\frac{1}{2}(\sin (A+B)-\sin (A-B)) \quad$ Product Formula $\mathbf{p}($ iv $)$
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Sum formulas
This set of formulas allows us to achieve the opposite of what the product formulas do, namely, to express the sum or difference of two cosines, or of two sines, in terms of products.
If, in formulas $\mathrm{p}(\mathrm{i})$ through $\mathrm{p}$ (iv) above, we multiply both sides by 2 , reverse sides, and let $x=A+B$ and $y=A-B$, so that $A=\frac{x+y}{2}$, and $B=\frac{x-y}{2}$, then we have:
$\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) \quad$ Sum Formula s(i)
$\cos y-\cos x=2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) \quad$ Sum Formula s(ii)
$\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$
Sum Formula s(iii)
$\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$
Sum Formula s(iv)
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Product formulas
$$\cos (A+B)=\cos A \cos B-\sin A \sin B \quad \cos (A-B)=\cos A \cos B+\sin A \sin B \sin (A+B)=\sin A \cos B+\cos A \sin B \quad \sin (A-B)=\sin A \operatorname{co}$$
(一个) $+(\mathrm{b})$ 给出:
$\cos A \cos B=\frac{1}{2}(\cos (A+B)+\cos (A-B)) \quad$ 产品配方 $\mathbf{p}(\mathbf{i})$
(b) – (a) 给出:
$\sin A \sin B=\frac{1}{2}(\cos (A-B)-\cos (A+B)) \quad$ 产品配方 $p(i i)$
(C) $+(\mathrm{d})$ 给出:
$\sin A \cos B=\frac{1}{2}(\sin (A+B)+\sin (A-B)) \quad$ 产昭配方 $\mathrm{p}($ 三 $)$
(c) $-(\mathrm{d})$ 给出:
$\cos A \sin B=\frac{1}{2}(\sin (A+B)-\sin (A-B)) \quad$ 产品配方 $\mathbf{p}(\mathrm{iv})$
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Sum formulas
$B=\frac{x-y}{2}$ ，那么我们有:
$\cos x+\cos y=2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$ 求和公式 $\mathrm{s}(\mathrm{i})$
$\cos y-\cos x=2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$ 我是公式 $\mathrm{s}(\mathrm{ii})$
$\sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$
$\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)$
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
Posted on Categories:CEMC代考, 数学代写, 数学竞赛代写, 数学竞赛代考
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Consider the points $P, Q, R, S$ and $T$ lying on the unit circle above. Observe that the arc-lengths from $S$ to $P$ and from $T$ to $Q$ are equal. Since equal arcs in a circle subtend equal chords, it follows from Pythagoras’ theorem that:
\begin{aligned} \sqrt{\left(\cos \left(s_1+s_2\right)-1\right)^2+\left(\sin \left(s_1+s_2\right)-0\right)^2} \ &=\sqrt{\left(\cos s_2-\cos s_1\right)^2+\left(\sin s_2-\left(-\sin s_1\right)\right)^2}, \ \cos ^2\left(s_1+s_2\right)+\sin ^2\left(s_1+s_2\right) &-2 \cos \left(s_1+s_2\right) \ &=\cos ^2 s_2+\sin ^2 s_2-2 \cos s_1 \cos s_2+\cos ^2 s_1 \ &+\sin ^2 s_1+2 \sin s_1 \sin s_2, \end{aligned}
hence $1-2 \cos \left(s_1+s_2\right)=1-2 \cos s_1 \cos s_2+2 \sin s_1 \sin s_2$, and therefore $\cos \left(s_1+s_2\right)=\cos s_1 \cos s_2-\sin s_1 \sin s_2$.
Thus, for any $A, B \in \mathbb{R}$,
$\cos (A+B)=\cos A \cos B-\sin A \sin B \quad$ Addition Formula $\mathbf{f}(\mathbf{i})$
It follows, by replacing $B$ with $-B$ and using the identities $\cos (-x)=\cos x$, $\sin (-x)=-\sin x$, that:
$\cos (A-B)=\cos A \cos B+\sin A \sin B \quad$ Addition Formula $\mathbf{f}(i i)$
We showed earlier on that $\cos \left(\frac{\pi}{2}-\alpha\right)=\sin \alpha$. Letting $\alpha=A+B$, we have
$$\cos \left(\frac{\pi}{2}-(A+B)\right)=\sin (A+B)$$
therefore $\sin (A+B)=\cos \left(\left(\frac{\pi}{2}-A\right)-B\right)$
\begin{aligned} &=\cos \left(\frac{\pi}{2}-A\right) \cos B+\sin \left(\frac{\pi}{2}-A\right) \sin B \ &=\sin A \cos B+\cos A \sin B . \end{aligned}
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Double angle formulas
These are deduced by setting $A=B$ in the addition formulas above: from $f(i)$ $\cos 2 A=\cos ^2 A-\sin ^2 A$ Double Angle Formula d(i)
Now, since $\cos ^2 A+\sin ^2 A=1$,
\begin{aligned} \cos 2 A &=\left(1-\sin ^2 A\right)-\sin ^2 A \ &=1-2 \sin ^2 A . \end{aligned}
Hence $\cos 2 A=1-2 \sin ^2 A \quad$ Double Angle Formula d(ii)
Similarly, we can show:
$\cos 2 A=2 \cos ^2 A-1$ Double Angle Formula d(iii)
from $\mathrm{f}($ iii) $\sin 2 A=2 \sin A \cos A$ Double Angle Formula d(iv)
from $g(i) \quad \tan 2 A=\frac{2 \tan A}{1-\tan ^2 A} \quad$ Double Angle Formula $d(\mathbf{v})$
## 滑铁卢数学竞赛代考
$$\sqrt{\left(\cos \left(s_1+s_2\right)-1\right)^2+\left(\sin \left(s_1+s_2\right)-0\right)^2}=\sqrt{\left(\cos s_2-\cos s_1\right)^2+\left(\sin s_2-\left(-\sin s_1\right)\right)^2}, \cos ^2\left(s_1+s_2\right)+\sin ^2\left(s_1+s_2\right) \quad-2 \cos \left(s_1+s_2\right)=\operatorname{co}$$
$\cos (A+B)=\cos A \cos B-\sin A \sin B$ 加法公式 $f(i)$
$\cos (A-B)=\cos A \cos B+\sin A \sin B \quad$ 加法公式 $f(i i)$
$$\cos \left(\frac{\pi}{2}-(A+B)\right)=\sin (A+B)$$
$$=\cos \left(\frac{\pi}{2}-A\right) \cos B+\sin \left(\frac{\pi}{2}-A\right) \sin B \quad=\sin A \cos B+\cos A \sin B$$
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|Double angle formulas
$$\cos 2 A=\left(1-\sin ^2 A\right)-\sin ^2 A \quad=1-2 \sin ^2 A .$$
$\cos 2 A=2 \cos ^2 A-1$ 双角公式 d(iii)
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
Posted on Categories:CEMC代考, 数学代写, 数学竞赛代写, 数学竞赛代考
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## 滑铁卢数学竞赛代考Waterloo Math Contest代考|The Unique Factorization Theorem
Here is a problem to help you to fully appreciate the tools of this section. You are invited to try it now. Its solution is given at the end of the section.
Appetizer Problem: $N$ is a natural number such that $\frac{N}{5}$ is a perfect square and $\frac{N}{2}$ is a perfect cube. The smallest value of $N$ for which $\frac{N}{3^3}$ is another natural number is:
(A) 2916000
(B) 729000
(C) 250000
(D) 1458000
(E) 364500 .
The fundamental theorem of arithmetic is that every natural number can be factored in essentially one way into a product of prime numbers. Here is the precise statement:
Theorem 5 Unique Factorization Theorem Any natural number $N$ can be written in precisely one way in the form
$$N=p_1^{z_1} p_2^{z_2} p_3^{z_3} \cdots p_k^{z_k}$$
where $p_1, p_2, \ldots, p_k$ are prime, $p_1<p_2<\cdots<p_k$, and $z_1, z_2, \ldots, z_k$ are positive integers.
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|The Chinese Remainder Theorem
We start by considering a typical problem that can be solved easily with the help of the Chinese Remainder Theorem. You are invited to try it before looking at the methods of solution, one given immediately afterwards, and another (using the theorem) at the end of the section.
Appetizer Problem: Brilliant Duck in Bertrand Carroll’s fiction book ‘Mathland’ was indeed brilliant. She devised a quick way of counting her ducklings which was basically a residue evaluation process after division (separately) by 5, 3, and 11. She knew that:
(i) When counted by fives, 2 ducklings remained.
(ii) When counted by threes, 2 ducklings remained.
(iii) When counted by elevens, 3 ducklings remained.
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|The Unique Factorization Theorem
(A) 2916000
(B) 729000
(C) 250000
(D) 1458000
(E) 364500 。
$$N=p_1^{z_1} p_2^{z_2} p_3^{z_3} \cdots p_k^{z_k}$$
## 滑铁卢数学竞赛代考Waterloo Math Contest代考|The Chinese Remainder Theorem
(i) 当按 5 计数时，剩下 2 只小鸭。
(ii) 按三点数，剩下 2 只小鸭。
(iii) 按 11 计数时，剩下 3 只小鸭。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 4,708 | 10,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | longest | en | 0.296652 |
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## College Algebra 8Th Edition by Ziegler, Byleen Barnett -Test Bank A+
\$35.00
College Algebra 8Th Edition by Ziegler, Byleen Barnett -Test Bank A+
1. A parabola has its vertex at the origin, and the equation of its directrix is . Find the coordinates of the focus.
2. A) B) C) D)
Ans: C Section: 6.1
1. A parabola has its vertex at the origin, and the coordinates of its focus are . Find the equation of the directrix.
2. A) B) C) D)
Ans: D Section: 6.1
Use the following to answer questions 3-5:
y2 = 28x
1. Find the vertex.
2. A) (0, 0) B) (0, –6) C) (0, –24) D) (–24, 0)
Ans: A Section: 6.1
1. Find the focus.
2. A) (0, –1) B) (0, –4) C) (–1, 0) D) (–4, 0)
Ans: C Section: 6.1
1. Find the directrix.
2. A) y = –3 B) y = 3 C) x = –3 D) x = 3
Ans: C Section: 6.1
Use the following to answer questions 6-8:
y2 = 12x
1. Find the focus.
Ans: (3, 0)
Section: 6.1
1. Find the directrix.
Ans: x = –3
Section: 6.1
1. Sketch the graph.
Ans:
Section: 6.1
Use the following to answer questions 9-11:
x2 = 16y
1. Find the vertex.
Ans: (0, 0)
Section: 6.1
1. Find the focus.
Ans: (0, 6)
Section: 6.1
1. Find the directrix.
Ans: y = 5
Section: 6.1
Use the following to answer questions 12-14:
y2 = –8x
1. Find the focus.
2. A) (–2, 0) B) (2, 0) C) (0, –2) D) (0, 2)
Ans: A Section: 6.1
1. Find the directrix.
2. A) x = –2 B) y = –2 C) x = 2 D) y = 2
Ans: C Section: 6.1
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: B Section: 6.1
Use the following to answer questions 15-17:
x2 = –16y
1. Locate the focus.
Ans: (0, –4)
Section: 6.1
1. Locate the directrix.
Ans: y = 4
Section: 6.1
1. Sketch the graph.
Ans:
Section: 6.1
Use the following to answer questions 18-20:
x2 = 14y
1. Locate the focus.
2. A) (7/2, 0) B) (–7/2, 0) C) (0, 7/2) D) (0, –7/2)
Ans: C Section: 6.1
1. Locate the directrix.
2. A) x = –7/2 B) y = –7/2 C) x = 7/2 D) y = 7/2
Ans: B Section: 6.1
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: D Section: 6.1
1. Find the coordinates of the focus to two decimal places.
y2 = 34x
Ans: (8.50, 0)
Section: 6.1
1. Find the coordinates of the focus to two decimal places.
x2 = –79y
1. A) (–19.75, 0) B) (0, –19.75) C) (0, 19.75) D) (–79, 0)
Ans: B Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the x– or y-axis, and directrix y = 10.
2. A) y2 = –40x B) y2 = 40x C) x2 = –40y D) x2 = 40y
Ans: C Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the x– or y-axis, and focus (0, –1).
Ans: x2 = –4y
Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the x– or y-axis, and directrix x = –10.
Ans: y2 = 40x
Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the x– or y-axis, and focus (–16, 0).
2. A) y2 = –64x B) y2 = –4x C) x2 = –64y D) x2 = –4y
Ans: A Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the y-axis, and passing through (4, 2).
2. A) y2 = 8x B) y2 = 2x C) x2 = 8y D) x2 = 2y
Ans: C Section: 6.1
1. Find the equation of the parabola with vertex at the origin, axis of symmetry the y-axis, and passing through (–2, 4).
Ans: y2 = –8x
Section: 6.1
1. Find the first-quadrant points of intersection for the pair of parabolas to three decimal places.
x2 = 5y
y2 = 7x
Ans: (0, 0) and (5.593, 6.257)
Section: 6.1
1. Use the definition of a parabola and the distance formula to find the equation of a parabola with directrix y = –3 and focus (–4, 1).
2. A) y2 + 8y – 8x + 24 = 0 C) x2 + 8x – 8y + 24 = 0
3. B) y2 – 8y – 8x + 8 = 0 D) x2 – 8x – 8y + 8 = 0
Ans: C Section: 6.1
1. Find the distance between the foci of an ellipse that has a major axis of length 10 and a minor axis of length 8.
2. A) 3 B) 6 C) 12 D) 9
Ans: B Section: 6.2
1. Find the length of the major axis of an ellipse that has the distance between the foci equal to 10 and the length of the minor axis equal to 24.
2. A) 13 B) 18 C) 52 D) 26
Ans: D Section: 6.2
1. Sketch the graph. Find the coordinates of the foci and the lengths of the major and minor axes.
Ans: Foci at (–3, 0) and (3, 0)
Major axis length = 10, minor axis length = 8
Section: 6.2
1. Sketch the graph. Find the coordinates of the foci and the lengths of the major and minor axes.
Ans: Foci at
Major axis length = 8, minor axis length = 6
Section: 6.2
Use the following to answer questions 35-37:
9x2 + y2 = 36
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: B Section: 6.2
1. Find the coordinates of the foci
2. A) C)
3. B) D)
Ans: A Section: 6.2
1. Find the lengths of the major and minor axes.
2. A) Major axis length = 6, minor axis length = 3
3. B) Major axis length = 12, minor axis length = 6
4. C) Major axis length = 12, minor axis length = 4
5. D) Major axis length = 9, minor axis length = 4
Ans: C Section: 6.2
Use the following to answer questions 38-40:
x2 + 4y2 = 36
1. Sketch the graph.
Ans:
Section: 6.2
1. Find the coordinates of the foci.
Ans:
Section: 6.2
1. Find the lengths of the major and minor axes.
Ans: Major axis = 12, Minor axis length = 6
Section: 6.2
1. Sketch the graph. Find the coordinates of the foci and the lengths of the major and minor axes.
25x2 + 16y2 = 400
Ans: Foci at (0, –3) and (0, 3)
Major axis length = 10, Minor axis length = 8
Section: 6.2
Use the following to answer questions 42-44:
25x2 + y2 = 100
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: B Section: 6.2
1. Find the coordinates of the foci.
2. A) C)
3. B) D)
Ans: C Section: 6.2
1. Find the lengths of the major and minor axes.
2. A) Major axis length = 10, minor axis length = 2
3. B) Major axis length = 20, minor axis length = 2
4. C) Major axis length = 10, minor axis length = 4
5. D) Major axis length = 20, minor axis length = 4
Ans: D Section: 6.2
1. Sketch the graph. Find the coordinates of the foci and the lengths of the major and minor axes.
5x2 + 2y2 = 10
Ans: Foci at
Major axis length = , minor axis length =
Section: 6.2
1. Write the equation of the ellipse in the form whose graph is shown.
Ans:
Section: 6.2
1. Write the equation of the ellipse in the form whose graph is shown.
1. A) B) C) D)
Ans: A Section: 6.2
1. Find the equation of an ellipse in the form with major axis along the x-axis, major axis length 14 and minor axis length 6.
2. A) B) C) D)
Ans: C Section: 6.2
1. Write the equation of an ellipse in the form with major axis along the x-axis, major axis length 16, and distance of foci from center = 4.
Ans:
Section: 6.2
1. Write the equation of an ellipse in the form with major axis along the y-axis, major axis length 16, and distance of foci from center = 7.
2. A) B) C) D)
Ans: B Section: 6.2
1. Find the equation of an ellipse in the form with major axis along the y-axis, minor axis length 8, and distance of foci from center = 3.
2. A) B) C) D)
Ans: B Section: 6.2
Use the following to answer questions 52-53:
A landscaper wishes to create an elliptical garden 20 m long and 12 m across with decorative fountains located at the foci.
1. How far from the center of the ellipse should the fountains be located? (Round to the nearest 100th of a meter.)
2. A) 18 m B) 2.30 m C) 3.32 m D) 4.66 m
Ans: C Section: 6.2
1. How far apart are the fountains?
2. A) 94 m B) 6.90 m C) 11.62 m D) 12.10 m
Ans: A Section: 6.2
1. Find the distance between the foci of a hyperbola that has a transverse axis of length 6 and a conjugate axis of length 8.
2. A) 5 B) 20 C) 10 D) 7
Ans: C Section: 6.3
1. Match each equation with its graph.
2. , B.
3. , D.
III.
Ans: A matches II, B matches IV, C matches III, D matches I
Section: 6.3
Use the following to answer questions 56-58:
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: A Section: 6.3
1. Find the coordinates of the foci.
2. A) C)
3. B) D)
Ans: C Section: 6.3
1. Find the lengths of the transverse and conjugate axes.
2. A) Transverse axis length = 8, conjugate axis length = 4
3. B) Transverse axis length = 16, conjugate axis length = 4
4. C) Transverse axis length = 4, conjugate axis length = 2
5. D) Transverse axis length = 8, conjugate axis length = 2
Ans: A Section: 6.3
Use the following to answer questions 59-61:
1. Sketch the graph.
Ans:
Section: 6.3
1. Find the coordinates of the foci.
Ans:
Section: 6.3
1. Find the lengths of the transverse and conjugate axes.
Ans: Transverse axis length = 4, conjugate axis length = 8
Section: 6.3
1. Find the coordinates of the foci.
Ans:
Section: 6.3
Use the following to answer questions 63-65:
1. Sketch the graph.
Ans:
Section: 6.3
1. Find the coordinates of the foci.
Ans:
Section: 6.3
1. Find the lengths of the transverse and conjugate axes.
Ans: Transverse axis length = 4, conjugate axis length = 8
Section: 6.3
1. Sketch the graph of the equation. Find the coordinates of the foci, and find the lengths of the transverse and conjugate axes.
4x2 – 9y2 = 36
Ans: Foci at
Transverse axis length = 6, conjugate axis length = 4
Section: 6.3
Use the following to answer questions 67-69:
y2 – 4x2 = 16
1. Sketch the graph.
2. A) C)
3. B) D)
Ans: C Section: 6.3
1. Find the coordinates of the foci.
2. A) C)
3. B) D)
Ans: D Section: 6.3
1. Find the lengths of the transverse and conjugate axes.
2. A) Transverse axis length = 8, conjugate axis length = 4
3. B) Transverse axis length = 4, conjugate axis length = 8
4. C) Transverse axis length = 2, conjugate axis length = 4
5. D) Transverse axis length = 4, conjugate axis length = 2
Ans: A Section: 6.3
1. Sketch the graph of the equation. Find the coordinates of the foci, and find the lengths of the transverse and conjugate axes.
25y2 – 4x2 = 100
Ans: Foci at
Transverse axis length = 4, conjugate axis length = 10
Section: 6.3
1. Sketch the graph of the equation. Find the coordinates of the foci, and find the lengths of the transverse and conjugate axes.
9y2 – 4x2 = 36
Ans: Foci at
Transverse axis length = 4, conjugate axis length = 6
Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin and whose graph is shown below.
1. A) B) C) D)
Ans: D Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin and whose graph is shown below.
Ans:
Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin, whose transverse axis is on the x-axis and has length 14, and whose conjugate axis has length 10.
1. A) B) C) D)
Ans: D Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin, whose transverse axis is on the x-axis and has length 8, and whose foci are a distance of 6 units from the center.
Ans:
Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin, whose transverse axis is on the y-axis and has length 6, and whose foci are a distance of 4 units from the center.
1. A) B) C) D)
Ans: C Section: 6.3
1. Find an equation of a hyperbola in the form
whose center is at the origin, whose conjugate axis is on the x-axis and has length 12, and whose foci are a distance of from the center.
Ans:
Section: 6.3
1. Find the equations of the asymptotes.
Ans:
Section: 6.3
1. Find the equations of the asymptotes.
1. A) B) C) D)
Ans: B Section: 6.3
1. Find the equations of the asymptotes.
Ans:
Section: 6.3
1. Find the equations of the asymptotes.
16x2 – 25y2 = 400
1. A) B) C) D)
Ans: A Section: 6.3
1. A ship is traveling on a course parallel to and 70 miles from a straight shoreline. Two transmitting stations, S1 and S2, are located 300 miles apart on the shoreline. By timing radio signals from the stations, this ship’s navigator determines that the ship is between the two stations and 40 miles closer to S2 than to S1. Find the distance from the ship to each station. Round answers to one decimal place.
Ans: 185.8 mi from S1 and 145.8 mi from S2
Section: 6.3
1. A ship is traveling on a course parallel to and 90 miles from a straight shoreline. Two transmitting stations, S1 and S2, are located 400 miles apart on the shoreline. By timing radio signals from the stations, this ship’s navigator determines that the ship is between the two stations and 30 miles closer to S2 than to S1. Find the distance from the ship to station S2. Round to one decimal place.
2. A) 2 mi B) 202.6 mi C) 204.4 mi D) 206.8 mi
Ans: C Section: 6.3
Chapter 7
1. Solve the system by graphing.
x + y = 4
x + y = 2
Ans: (1, 3)
Section: 7.1
1. Solve the system by graphing.
x – 2y = 8
x + y = –1
Ans: (2, –3)
Section: 7.1
1. Solve the system by graphing.
2x + y = 4
x + y = 3
Ans: (1, 2)
Section: 7.1
1. Solve the system by graphing.
3x + y = 6
3xy = 0
Ans: (1, 3)
Section: 7.1
1. Solve the system by graphing.
xy = 2
3x – 3y = 6
Ans: Infinitely many solutions (dependent system)
Section: 7.1
1. Solve the system of equations.
x + y = 4
xy = 2
1. A) (1, 3) B) (–1, 5) C) (3, 1) D) (5, –1)
Ans: C Section: 7.1
1. Solve the system of equations.
x + y = 9
xy = 9
Ans: (9, 0)
Section: 7.1
1. Solve the system of equations.
2x + 5y = –12
7x – 5y = 3
1. A) (–1, –3) B) (–1, –2) C) (0, –2) D) (0, –3)
Ans: B Section: 7.1
1. Solve the system of equations.
x – 2y = –3
5x – 10y = –10
Ans: No solution (parallel lines)
Section: 7.1
1. Solve the system of equations.
xy = 4
x + 2y = –14
1. A) (–2, –6) B) (–3, –6) C) (–2, –5) D) (–3, –5)
Ans: A Section: 7.1
1. Solve the system of equations.
y = x + 3
y = 5x – 5
Ans: (2, 5)
Section: 7.1
1. Solve the system of equations.
7x – 2y = 5
6x + 5y = 11
Ans: (1, 1)
Section: 7.1
1. Solve the system of equations.
3x – 4y = 8
6x + 3y = 5
Ans:
Section: 7.1
1. Solve the system of equations.
Ans: (–20, 12)
Section: 7.1
1. Solve the system of equations.
Ans: (–1, –3, 3)
Section: 7.1
1. Solve the system using elimination by addition.
xyz = 2
2x + yz = –3
3x + yz = –2
Ans: (1, –3, 2)
Section: 7.1
1. Solve the system of equations.
xyz = 5
xy + 3z = –7
2x + yz = 1
1. A) (1, –3, –2) B) (0, –2, –2) C) (0, –2, –3) D) No solution
Ans: C Section: 7.1
1. Solve the system of equations.
3x + 2y + 3z = 6
4x + 3y + 2z = 5
6x + 4y + 5z = 11
Ans: (3, –3, 1)
Section: 7.1
1. Solve the system of equations.
2x – 3y + 4z = –19
5x – 2y + z = –19
–7x + 5y – 5z = 39
1. A) (–4, –3, –5) B) (–4, –4, –4) C) (–4, –2, –6) D) No solution
Ans: D Section: 7.1
1. Solve the system of equations.
1. A) (–1, –7, 0)
2. B) (3, 3, 2)
3. C) {(2s – 3, 5s – 7, s) | s is any real number}
4. D) No solution
Ans: C Section: 7.1
1. Solve using elimination by addition.
Ans:
Section: 7.1
1. Solve the system of equations.
2x + 3y – 4z = –9
–6xy + z = –18
4x – 2y + 3z = 28
Ans: No solution
Section: 7.1
1. A boat traveled 48 mi up a river in 4 hours. Returning downstream, the boat took 3 hours. What is the boat’s rate in still water, and what is the rate of the river’s current?
Ans: Boat: 14 mi/h; current: 2 mi/h
Section: 7.1
1. A chemist wants to combine a 30% alcohol solution with a 50% alcohol solution to form 400 mL of a 35% alcohol solution. How much of each solution should the chemist use to form the mixture?
Ans: 300 mL of 30% solution and 100 mL of 50% solution
Section: 7.1
1. A coin jar contains nickels, dimes, and quarters. There are 30 coins in all. There are 11 more nickels than quarters. The value of the dimes is \$1.70 less than the value of the quarters. How many coins of each type are in the jar?
Ans: nickels: 19, dimes: 3, quarters: 8
Section: 7.1
1. Angus invested \$12,000, part at 15% and part at 9%. If the total interest at the end of the year is \$1,440, how much did he invest at each rate?
Ans: \$6,000 at 15% and \$6,000 at 9%
Section: 7.1
1. Janet invested \$10,000, part at 2% and part at 12%. If the total interest at the end of the year is \$600, how much did she invest at 2%?
2. A) \$4,000 B) \$7,000 C) \$6,000 D) \$5,000
Ans: C Section: 7.1
1. A company manufactures three products, tables, chairs, and bookcases. The labor, material and shipping costs for manufacturing one unit of each product are given in the table. The weekly allocations for labor, materials, and shipping are \$62,800, \$56,700, and \$26,300, respectively. How many of each type of product should be manufactured each week in order to exactly use the weekly allocations?
_____________________________________________
Product Table Chair Bookcase
Labor \$40 \$65 \$50
Materials \$85 \$45 \$60
Shipping \$40 \$20 \$30
Ans: 120 tables, 700 chairs, and 250 bookcases
Section: 7.1
1. Is the matrix in reduced form?
1. A) Yes B) No
Ans: A Section: 7.2
1. Is the matrix in reduced form?
1. A) Yes B) No
Ans: B Section: 7.2
1. Write the linear system corresponding to the reduced augmented matrix and solve.
Ans: x1 = 1, x2 = –1, x3 = –4
Section: 7.2
1. Write the linear system corresponding to the reduced augmented matrix and solve.
1. A) x1 = –7t + 4, x2 = –t + 5, x3 = t, t any real number
2. B) x1 = 7t + 4, x2 = –t + 5, x3 = t, t any real number
3. C) x1 = –7t + 4, x2 = –t – 5, x3 = t, t any real number
4. D) x1 = 7t – 4, x2 = –t + 5, x3 = t, t any real number
Ans: A Section: 7.2
1. Perform the indicated row operation, then write the new matrix.
–2R1 + R2 → R2
Ans:
Section: 7.2
1. Perform the indicated row operations, then write the new matrix.
R1 + R2 → R2; –2R1 + R3 → R3
1. A) C)
2. B) D)
Ans: B Section: 7.2
1. Use row operations to change the matrix to reduced form.
Ans:
Section: 7.2
1. Use row operations to change the matrix to reduced form.
1. A) C)
2. B) D)
Ans: D Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
x1 – 4x2 = –17
4x1 + x2 = –17
1. A) x1 = –5, x2 = 2 B) x1 = –4, x2 = 3 C) x1 = –5, x2 = 3 D) No solution
Ans: C Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
–10x1 – 5x2 = 35
2x1 + x2 = 7
1. A) x1 = 5, x2 = –4 B) x1 = 6, x2 = –3 C) x1 = 5, x2 = –3 D) No solution
Ans: D Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
–3x1 + 3x2 = 9
2x1 – 2x2 = –6
Ans: x1 = s – 3, x2 = s, s any real number
Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
x1x2 + x3 = –4
x1 + x2 – 3x3 = 2
5x1x2 + x3 = 14
1. A) x1 = 3, x2 = 2, x3 = 1 C) x1 = 4, x2 = 1, x3 = 2
2. B) x1 = 3, x2 = 2, x3 = 2 D) No solution
Ans: A Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
x1x2x3 = 0
x1 + x2 + 5x3 = –4
–2x1x2 + x3 = –5
Ans: x1 = 3, x2 = –2, x3 = –1
Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
3x1 – 9x2 + 2x3 = –7
–6x1 + 18x2 + x3 = –11
9x1 – 27x2 + 5x3 = –16
1. A) x1 = 4, x2 = 1, x3 = –5
2. B) x1 = 3s + 1, x2 = s + 2, x3 = s, s any real number
3. C) x1 = 3s + 1, x2 = s, x3 = –5, s any real number
4. D) No solution
Ans: C Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
x1 + x2 + 5x3 = 7
–5x1 + x2x3 = –17
–20x1 + 4x2 – 4x3 = –69
1. A) x1 = 3, x2 = 0, x3 = 2 C) x1 = 3, x2 = 1, x3 = 1
2. B) x1 = 3, x2 = 0, x3 = 1 D) No solution
Ans: D Section: 7.2
1. Solve the system using Gauss-Jordan elimination.
3x1 + 6x2 + 10x3 + 6x4 = 30
x1 + 2x2 + 3x3 + x4 = 9
Ans: x1 = –2s + 8t, x2 = s, x3 = –3t + 3, x4 = t, s and t any real numbers
Section: 7.2
1. Find a, b, and c so that the graph of the parabola with equation y = ax2 + bx + c passes through the points (1, 6), (–2, 21), and (3, 16).
Ans: a = 2, b = –3, c = 7
Section: 7.2
1. Find a, b, and c so that the graph of the circle with equation x2 + y2 + ax + by + c = 0 passes through the points (0, 7), (–6, –1), and (1, 0).
Ans: a = 6, b = –6, c = –7
Section: 7.2
Ans:
Section: 7.3
Ans:
Section: 7.3
1. A) B) C) D) Not defined
Ans: D Section: 7.3
1. Subtract, if possible.
Ans:
Section: 7.3
1. Multiply.
Ans:
Section: 7.3
1. Multiply, if possible.
1. A) B) C) D) Not defined
Ans: C Section: 7.3
1. Find AB, if possible.
Ans:
Section: 7.3
1. Find A2, if possible.
1. A) B) C) D) Not defined
Ans: D Section: 7.3
1. Find A2, if possible.
Ans:
Section: 7.3
1. Find the value of a, b, c, and d if
+ = .
Ans: a = –3, b = –2, c = 9, d = –5
Section: 7.3
1. Find the value of a, b, c, and d if
.
1. A) C)
2. B) D)
Ans: D Section: 7.3
1. A company with two different plants makes shoes and hiking boots. The production costs for each item are given in the following matrices.
Find the matrix and explain what information it provides.
Ans:
This is the average cost of materials and labor for each product at the two plants.
Section: 7.3
1. A car dealer sells two models of a car. Current dealer invoice price (cost) and the retail price for the basic models and the indicated options are given in the following two matrices.
The markup matrix is defined to be Economic conditions have resulted in an across-the-board increase of 15% in costs to the dealer for next year’s models. To stay competitive, the dealer will only increase the retail price by 10%. Calculate the markup matrix for next year’s models and the indicated options. (Compute results to the nearest dollar.)
1. A)
2. B)
3. C)
4. D)
Ans: B Section: 7.3
1. A company with manufacturing plants located in different parts of the country has labor-hour and wage requirements for the manufacturing of two types of inflatable boats as given in the following matrices.
Find the labor costs for a one-person boat manufactured at plant I.
1. A) \$17.50 B) \$18.60 C) \$23.70 D) \$13.80
Ans: D Section: 7.3
1. Multiply.
Ans:
Section: 7.4
1. Multiply.
1. A) B) C) D)
Ans: D Section: 7.4
1. Use matrix products to determine if the matrices are inverses of each other.
and
1. A) Yes B) No
Ans: A Section: 7.4
1. Write the matrix equation as a system of linear equations.
Ans:
Section: 7.4
1. Write the system in the form of a matrix equation.
Ans:
Section: 7.4
1. Write the system in the form of a matrix equation.
1. A) C)
2. B) D)
Ans: A Section: 7.4
1. Find x1 and x2.
Ans: x1 = 2, x2 = 5
Section: 7.4
1. Find x1 and x2.
1. A) x1 = 12, x2 = 8 B) x1 = 8, x2 = 12 C) x1 = 10, x2 = 6 D) x1 = 6, x2 = 10
Ans: C Section: 7.4
1. Find the inverse of the matrix, if it exists.
Ans:
Section: 7.4
1. Find the inverse of the matrix, if it exists.
1. A) B) C) D) The inverse does not exist.
Ans: D Section: 7.4
1. Find A–1, if it exists.
Ans:
Section: 7.4
1. Write the system as a matrix equation and solve using inverses.
x1 + 2x2 = 18
4x1 + x2 = 16
Ans: x1 = 2, x2 = 8
Section: 7.4
1. Write the system as a matrix equation and solve using inverses.
x1 – 2x2 + x3 = –4
–2x1 + x2 + 5x3 = –1
2x1 + x2x3 = 7
1. A) x1 = 3, x2 = 2, x3 = 1 C) x1 = 2, x2 = 3, x3 = 0
2. B) x1 = 2, x2 = 3, x3 = 1 D) x1 = 2, x2 = 2, x3 = 1
Ans: C Section: 7.4
1. Write the system as a matrix equation and solve using inverses.
x1x2x3 = 0
–2x1 + x2 – 5x3 = 5
5x1 – 5x2 + x3 = –6
Ans: x1 = 1, x2 = 2, x3 = –1
Section: 7.4
1. Adult tickets for a play cost \$8 and child tickets cost \$6. If there were 20 people at a performance and the theatre collected \$152 from ticket sales, how many adults and how many children attended the play?
Ans: 16 adults and 4 children
Section: 7.4
1. Evaluate the second-order determinant.
1. A) –10 B) 30 C) 10 D) 8
Ans: B Section: 7.5
1. Solve the system using Cramer’s rule.
1. A) B) C) D)
Ans: A Section: 7.5
1. Solve the system using Cramer’s rule.
1. A) , C) ,
2. B) , D) ,
Ans: D Section: 7.5
1. Consider the following determinant.
Write the minor of the element . Leave the answer in determinant form.
1. A) B) C) D)
Ans: C Section: 7.5
1. Consider the following determinant.
Evaluate the cofactor of the element .
1. A) 52 B) –52 C) 28 D) –28
Ans: A Section: 7.5
1. Evaluate the following determinant using cofactors.
1. A) 20 B) –20 C) 70 D) –70
Ans: A Section: 7.5
1. Solve the system using Cramer’s rule.
1. A) , C) ,
2. B) , D) ,
Ans: D Section: 7.5
1. Solve the system using Cramer’s rule.
1. A) C)
2. B) D)
Ans: C Section: 7.5
1. Solve the system using Cramer’s rule.
1. A) , , C) , ,
2. B) , , D) , ,
Ans: D Section: 7.5
1. Discuss the number of solutions for the system below where a and b are real numbers. Use Cramer’s rule where appropriate and Gauss-Jordan elimination otherwise.
Ans: If and there are an infinite number of solutions. If and there are no solutions. If there is one solution.
Section: 7.5
1. Use Cramer’s rule to solve for x only.
1. A) B) C) D)
Ans: C Section: 7.5
1. Use Cramer’s rule to solve for y only.
1. A) B) C) D)
Ans: B Section: 7.5
1. Use Cramer’s rule to solve for z only.
1. A) B) C) D)
Ans: A Section: 7.5
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https://juxbiketer.web.app/1074.html | 1,726,237,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651523.40/warc/CC-MAIN-20240913133933-20240913163933-00222.warc.gz | 311,335,528 | 7,027 | # Nfinding domain and range of functions pdf
I can determine the domain and range given a graph andor an equation. The range of a function is the list of all possible outputs. The range of a function is the set of values that the function assumes. To get an idea of the domain and range of the combined function, you simply break down the problem and look at the individual domains and ranges. All functions are relations, but not all relations are functions. In order to grasp domain and range, students must understand how to determine if a relation is a function and interpreting graphs. Using the function model describe the appropriate domain and range. Test skills acquired with this printable domain and range revision worksheets that provide a mix of absolute, square root, quadratic and reciprocal functions fx. Thus, this particular rational function has two numbers excluded from its domain but no restrictions on its range. In particular we will discuss finding the domain of a function of several variables as well as level curves, level surfaces and traces. Function worksheets for high school students comprises a wide variety of subtopics like domain and range of a function, identifying and evaluating functions, completing tables, performing arithmetic operations on functions, composing functions, graphing linear and quadratic functions, transforming linear and quadratic functions and a lot more in a nutshell. Determine the domain x and plug in the possible xvalues to find the range y. Describe the domain and range of each function below as discrete or continuous 7. Domain and range of a quadratic function onlinemath4all.
Use the mapping diagrams to i write the set of ordered pairs of the relation ii state if the relation is a function since every value in the domain maps to exactly one value in the range, this relation is a function. To find the domain and range of a rational function, set the denominator equal to zero and solve. Domain and range center for academic program support. The domain and range of this function are both the set of real numbers, and the relation is a function because for any value of x there is a unique value of y. This set is the values that the function shoots out after we plug an x value in. The easiest way is to look at the graph, examine the yvalues from bottom to top. That is, it is the set of all y values for which there is an x value such that y. In cases like this,the domain of a function is the largest set of real numbers for which the value of is a real number. The table shows the number of adult and child tickets sold for a school concert. Domain and range of a function 7th and 8th grade math.
Nov 02, 2016 domain and range interactive notebook pages here are this years interactive notebook pages for domain and range in algebra 1. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the xaxis. The domain of a function is the set of all the numbers you can substitute into the function x values. Snyder 2014 algebra unit 2 functions domainrangefunctions day 1 previous vocabdefinitions. However, one strategy that works most of the time is to find the domain of the inverse function if it exists. Wolframalpha is a great tool for finding the domain and range of a function. Find the domain and range of the quadratic function given below.
What are some ways to tell if a relation is a function. Then find the domain and range in set builder notation. In this applet, you can change the domain and see the effect on the range of several different functions. Ixl domain and range of radical functions precalculus. We will take a look at two 2 examples on how to find the domain and range of radical functions, and also two 2 examples of rational functions. How to find domain and range of a rational equation using. Graphically speaking, the domain is the portion of the. Tinspire cas finding range and domain of any function the. The domain is the set of all possible xvalues which will make the function work, and will output real yvalues. Really clear math lessons prealgebra, algebra, precalculus, cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. In this video you will learn find the domain and range of functions fully explained in urduhindi domain range in hindi mathematics foundation. Write a rule in function notation for each situation.
If there is a requirement that a yvalue produced by a function. The domain of a function is the complete set of possible values of the independent variable. The vertex of the function is at 1,1 and therfore the range of the function is all real y. This is because the set may contain any element which doesnt have an image in the right set. Find the domain of this new equation and it will be the range of the original. The domain and range you find for a combined function depend on the domain and range of each of the original functions individually. The collection of all the outputs of a function finding the domain of functions finding the range of functions skills practiced. Apr 09, 20 use your tinspire cas to find the range and domain of any function.
In algebra, when we deal with points on a graph, you may be asked to find its domain and range. Domain and range of quadratic functions video khan academy. Remember that domain refers to the xvalues that are represented in a problem and range refers to the yvalues that are represented in a problem. This is not a function because a united states citizen can be a citizen of more than one country. The curve of y sin x shows the range to be between. An understanding of toolkit functions can be used to find the domain and range of related functions.
This easy to print worksheetprintable pdf of 4 pages contains graphs of twelve functions and space given for writing domain and range for each graph. Functions a function is a relation where each x goes to only one y no x values are repeated among ordered pairs a graph would pass the vertical line test any vertical line only crosses graph once. Finding the domain of the inverse function is done the same way. Domain and range the domain of a function is the set of values that we are allowed to plug into our function. There is no set way to find the range algebraically. The range of a function is the list of all possible outputs yvalues of the function. An important part of understanding functions is understanding their domain and range. Determining the domain of a function from its equation. How to find domain and range of a rational equation using inverse duration. The range of a function f is the set of all values that fx takes on as x runs through the domain of f. The domain of a function f consists of all values of x for which the value fxis defined. The table shows the number of adult and child tickets. Find the x and y intercepts of the following functions. Domain and range of radical and rational functions.
When functions are first introduced, you will probably have some simplistic functions and relations to deal with, usually being just sets of points. Notes,whiteboard,whiteboard page,notebook software,notebook, pdf,smart,smart technologies ulc,smart board interactive whiteboard. Improve your math knowledge with free questions in domain and range of radical functions and thousands of other math skills. Learn vocabulary, terms, and more with flashcards, games, and other study tools. I can write and graph piecewisedefined functions to represent realworld situations. If a function does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of. In order to find the domain of a function, if it isnt stated to begin with, we need to look at the function definition to determine what values are not allowed. Iii identifying functions give the domain and range of each relation. We determined that the yvalue of the inner function in a composite function becomes the xvalue of the outer function.
The effect of ph on the action of a certain enzyme is shown on the accompanying graph. Finding the domain and range of radical and rational. Choose from 500 different sets of domain range function flashcards on quizlet. First, swap the x and y variables everywhere they appear in the equation and then solve for y.
Finding the domain of a function, without graphing. Domain and range are all the possible xvalues and yvalues of the function, and can often be described easily by looking at a graph. Pdf a computer program that plots functions from algebraic input should be. But in case of functions, the domain will always be equal to the first set. Finding the domain of a function cool math has free online cool math lessons, cool math games and fun math activities. If youre seeing this message, it means were having trouble loading external resources on our website.
These pictures provide an example of something similar that you could do. Find the domain and range of functions fully explained. What is the range of the joggers heart rate during this interval. You are asked about the range of the sine, along with the domain of a. Domains and ranges of functions and their inverses are the subjects of this quiz and worksheet. If youre behind a web filter, please make sure that the domains. It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. Find the domain of a function defined by an equation in functions and function notation, we were introduced to the concepts of domain and range. Sometimes it isnt possible to list all the values that x or y can be because the graph. The domain has to do with the values of x in your function. In this section, we will practice determining domains and ranges for specific functions.
Common core functions common core for mathematics examples, solutions, videos, and lessons to help high school students understand that a function from one set called the domain to another set called the range assigns to each element of the domain exactly one element of the range. Another way to identify the domain and range of functions is by using graphs. Functions domain, range examples, solutions, videos. Examples of domains and ranges from graphs important notes about domains and ranges from graphs. Finding the domain and range from the graph of a given function is one of the important standard in algebra 2 eoc. Let y fx be a function with an independent variable x and a dependent variable y.
For many functions, the domain and range can be determined from a graph. The set of all first components of the ordered pairs. For a function defined by a table, its domain consists of numbers in the first row. Determine the domain of functions whose expressions are a combination of different types of assignment rules. Use the vertical line test to check if it is a relation or a function. Domain and range worksheets math worksheets 4 kids. Domain and range of a function definitions of domain and range domain. Finding the domain of a f unction, without graphing. This time we will tackle how to find the domain and range of more interesting functions, namely, radical functions and rational functions. Finding an informative domain for a given function depends.
This quiz and worksheet can help students to test the following skills. Domain and range exercises studysmarter question 1 find the domain and range of each of the following, where y is a function of x. This applet lets you explore the domain and range examples discussed on the previous page, domain and range of a function. The domain of a function combinations of functions. Grieser 4 domain and range determine if the following relations are functions. Now, any integer when squared will generated a positive perfect square.
A piecewise function is described by more than one formula. Before we tackle the idea of the domain and range of a function, we first need to spend a few minutes talking about how to picture what any particular function does. Domain and range of a functions domain and range meaning. We will graph the function and state the domain and range of each function. The domain of f has already been stated in the question. The range of a function is the complete set of all possible resulting values of the dependent variable y, usually, after we have substituted the domain. Nov 30, 2008 finding the domain of a function made easy. A vending company realized a relationship between the number of people. Domain and range of trig and inverse trig functions. However, for most functions, only an equation is given and the domain is not specified.
Finding a function s domain if a function does not model data or verbal conditions,its domain is the largest set of real numbers for which the value of is a real number. Domain of advanced functions practice khan academy. Jul 05, 2017 this would require the students to not only be able to look at a graph of a function or a function and find the domain and range, but also look at a domain and range and be able to identify the function or graph that fits for that domain and range. How to adjust the domain and range of combined functions. The range of a function f consists of all values fxit assumes when x ranges over its domain. Find a reasonable domain and range for the function.
If a function f provides a way to successfully produce a single value y using for that purpose a value for x then that chosen xvalue is said to belong to the domain of f. Domain and range concept algebra video by brightstorm. Find the domain of a function defined by an equation. How to find domain and range of a rational equation using inverse brian mclogan. Domain and range of inverse functions pike page 1 of 2 domain and range of inverse functions finding the domain and range of an inverse function finding the domain of a function is nothing new. Determining the domain of advanced functions algebra 2 level domain of advanced functions. More than just an online function properties finder. Learn domain range function with free interactive flashcards. We are dealing with real numbers only in this work. What is the range of the function represented by the table. We have found the domain of several different functions up to this point.
Remember that in case of a relation, the domain might not be the same as the left set in the arrow diagram. I can write the equation of a piecewisedefinied function, given the graph. Mar 02, 2020 how to find the domain and range of a function. For instance, we know that you cannot take the square root of a negative number, and you cannot divide by latex0latex. Graphs of basic functions there are six basic functions that we are going to explore in this section. These wont be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. Domain and range of general functions the domain of a function is the list of all possible inputs xvalues to the function.
Functions are rules that take inputs from one set and produce outputs in another. Functions that model data often have their domains explicitly given with the functions equation. Then describe the domain and range of the relations in set builder and interval notations. Exclude from a function s domain real numbers that cause division by zero and real numbers that result in an even root of a negative number. This one has no domain exclusions because no value of x will cause its denominator to be zero. Pdf algorithms for choosing the domain and range when. Domain and range of inverse functions finding the domain. A piecewise function can be graphed using each algebraic formula on its assigned subdomain. The domain of this function cannot include 1, but includes all real numbers except 1.
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https://www.thelaunchpadtech.net/fiu-football-cdlkuks/hw9u4ao.php?tag=determine-whether-or-not-matrix-multiplication-is-commutative-2b2e54 | 1,627,514,872,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153803.69/warc/CC-MAIN-20210728220634-20210729010634-00015.warc.gz | 1,076,253,703 | 21,194 | *Response times vary by subject and question complexity. &= B \cdot A \blacksquare Lv 7. … Support your answer using at least one (1) real-work example or scenario. \begin{pmatrix} One of the most common questions we get on Analytics Vidhya is,Even though the question sounds simple, there is no simple answer to the the question. Is It Important? \end{pmatrix}$$,$$B := \begin{pmatrix} The product BA is defined (that is, we can do the multiplication), but the product, when the matrices are multiplied in this order, will be 3×3, not 2×2. Problem 23. Here is an example: You might note that (I) is the same as (IV). Assess the importance of causality in estimating cost functions/relationships. Determine whether or not matrix multiplication is commutative. HR Assignment Help From UK Native Writers. Then look no further. III) & ce + dg &= ag + ch & \Leftrightarrow g (a - d) = c (e - h) \end{pmatrix} = B \cdot A$$,$$\begin{pmatrix} This is … 31 & 46 We have step-by-step solutions for your textbooks written by Bartleby experts! Determine whether or not matrix multiplication is commutative. I) & bg &= cf \\ \end{eqnarray*}, \begin{eqnarray*} \end{align},\begin{pmatrix}0 & 1 \\ Support your answer using at least one (1) real-world example or scenario. Answer to Determine whether or not matrix multiplication is commutative. Now determine whether the array elements of A are a matrix. (d) Discuss inverses. Matrix multiplication is always commutative if ... 1. As A and B are simultaneous diagonalizable, a matrix $$T \in \mathbb{R}^{n \times n}$$ exists, such that $$D_A = S^{-1} \cdot A \cdot S$$ and $$D_B = S^{-1} \cdot B \cdot S$$ with $$D_A$$ and $$D_B$$ are diagonal matrices. Support your answer using at least one (1) real-work example or scenario. \begin{pmatrix}1 & 0 \\ When is matrix multiplication commutative? This is the perfect way you can prepare your own unique academic paper and score the grades you deserve. That is good to start.But, once you have covered the basic concepts in machine learning, you will need to learn some more math. Explain the importance of providing employee benefit plans to employees working in the chosen positions. ... one matrix is the Identity matrix. If we have two matrix A and B, multiplication of A and B not equal to multiplication of B and A. II) & 0 &= b (e - h)\\ 13 -6. A*B != B*A This c program is used to check whether order of matrix multiplication is commutative or not. Relevance. Are you looking for a similar paper or any other quality academic essay? So we’re being asked whether we get the same result when we multiply the two matrices together in different orders. I) & bg &= cf \\ 0 0. \begin{pmatrix} e & f \\ Describe what you have learned from at least two specific courses (e.g., philosophy, history, or psychology) that illustrated usefulness in your daily life. If you want matrix multiplication, use. \begin{pmatrix}1 & 0 \\ Usually, we say that you need to know basic descriptive and inferential statistics to start. \end{pmatrix} \neq Answer to Determine whether the statement is true: Matrix multiplication is not commutative. So we only demand that $$bg = cf$$ and $$a \neq d$$ and $$e \neq h$$ for commutative matrix multiplication of $$2 \times 2$$ matrices. Determine whether or not matrix multiplication is commutative. Prove your answers. (You should expect to see a "concept" question relating to this fact on your next test.) Check whether the second page of the 3-D array is a matrix. Multipliers of Zero Problem 2: For real numbers, the zero-product property states that if the product of two factors is zero, then at least one of the factors is zero. c & d Add a Comment. 19 & 22 \\ EXAMPLE 10. 0 & 0\end{pmatrix} \cdot 1 Answer. TomV. There is another difference between the multiplication of scalars and the multiplication of matrices. It is so too equal! Answer Save. Use the order calculator below and get started! Give at least one (1)example What's an example of two matrices and show your work So to show that matrix multiplication is NOT commutative we simply need to give one example where this is not the case. The only sure examples I can think of where it is commutative is multiplying by the identity matrix, in which case B*I = I*B = B, or by the zero matrix, that is, 0*B = B*0 = 0. It is a fundamental property of many binary operations, and many mathematical proofs depend on it. Matrix multiplication in general is not commutative. each of these are discussions, no formal outline or format just need a couple paragraphs for each one. If we multiply A x B we get a 1 x 1 matrix, which is just a number. Matrix Multiplication. Show that multiplication of matrices is not commutative by determining the product matrices ST and TS.? 2010-04-20 11:25:32 2010-04-20 11:25:32. \Rightarrow A \cdot B &= S \cdot D_A S^{-1} \cdot S \cdot D_B \cdot S^{-1} \\ \begin{pmatrix} II) & af + bh &= be + df\\ ($$a = d$$ and $$bg = cf$$) or ($$f = g = 0$$). Acc350 cost accounting. \end{eqnarray*}, \begin{align} 3 & 4 III) & ce + dg &= ag + ch\\ II) & \frac{f}{g} &= \frac{b}{c} \Leftrightarrow cf = bg each of these are discussions, no formal outline or format just need a couple paragraphs for each one. (b) Determine whether the operation is associative and/or commutative. 1 Let be a binary operation on the set M 2(R) of all 2 2 matrices de ned by 8A 1;A 2 2M 2(R); A 1 A 2 = A 1 + A 2: (a) Prove that the operation is binary. Contact our live support team for any assistance or inquiry. III) & 0 &= c (e - h) PHI 210 “Take a Look at Ethics Around You:” (half page) Using what you’ve read through the WebText on ethics as a guide, examine the world around you. 0 & 1\end{pmatrix} = Get your answers by asking now. IV) & cf + dh &= bg + dh &\Leftrightarrow cf = bg Determine whether or not matrix multiplication is commutative. a & b \\ \end{pmatrix}$$,$$A \cdot B = \begin{pmatrix} (I.e. TF = ismatrix(A(:,:,2)) TF = logical 1 Check whether the second row of the 3-D array is a matrix. support your answer using at least one (1) real-world example of scenario. Matrix multiplication is NOT commutative. $$A := \begin{pmatrix} g & h Support your answer using at least one (1) real-work example or scenario. numpy.dot(A, B) You can also use matrix objects instead of ndarrays, but the inconsistencies they cause can be annoying to work with. If a and b are real numbers, then the equation ab = 0 implies that a = 0 or b = 0. Our team of experienced writers is on standby to deliver to you an original paper as per your specified instructions with zero plagiarism guaranteed. It is not for rectangular matrices of different sizes as it's not even defined for both "directions"! 34 = 12 and 43 = 12). 1 & 2 \\ 7 & 8 ae + cf & be + df \\ Is It Important? \end{pmatrix} \cdot Assess the importance of causality in estimating cost functions/relationships . If we multiply B x A, we get a new 3 x 3 matrix. II) & f (a - d) &= 0\\ You need it to understand how these algorithms work. Support your answer using at least one (1) real-world example or scenario Matrix multiplication is NOT commutative. When is 2x2 matrix multiplication commutative. Determine whether or not matrix multiplication is commutative. QuickBooks provides many tools for managing and securing your QuickBooks system (e.g., the QuickBooks Audit Report, and the QuickBooks’ Accountant Copy). However, matrix multiplication is not defined if the number of columns of the first factor differs from the number of rows of the second factor, and it is non-commutative, even when the product … In particular, matrix multiplication is not "commutative"; you cannot switch the order of the factors and expect to end up with the same result. (c) Determine whether the operation has identities. For example, multiplication of real numbers is commutative since whether we write ab or ba the answer is always the same. Join. We know that order matrix multiplication is important and matrix multiplication is not commutative. 2 months ago. If A is an m × p matrix, B is a p × q matrix, and C is a q × n matrix, then A (B C) = (A B) C. This important property makes simplification of many matrix expressions possible. Unless A and B are matrices. Statement: $$A \cdot B = B \cdot A \nRightarrow A, B \in \mathbb{R}^{n \times n}$$ are simultaneous diagonalizable. In general, we know that matrix multiplication is not commutative, which means that we get a different result if we multiply the matrices together in a different order. Statement: $$A, B \in \mathbb{R}^{n \times n}$$ are simultaneous diagonalizable $$\Rightarrow A \cdot B = B \cdot A$$. \end{pmatrix}$$, $$\begin{eqnarray*} (basically case #2) 4. Evaluate the effectiveness of these and other tools that are used to manage and secure QuickBooks. a & b \\ ST =-24 3. II) & af + bh &= be + df & \Leftrightarrow f (a - d) = b (e - h)\\ 2. Enter your Email id used at the time of registration and hit "Recover Password". Question 1. Acc350 cost accounting . So you have those equations: Now (I) and (II) are essentially the same. \end{eqnarray*}$$, $$\begin{eqnarray*} \end{pmatrix} = III) & g (a - d) &= 0 ... both matrices are Diagonal matrices. Assess the importance of causality in estimating cost functions/relationships Proof: When we change order of matrix multiplication, usally result is not same mostly. g & h So you end up with: TS =-20 -5-19 -10. Multiplication is commutative. Either way, if A and B are matrices, then, no, it is not equal because matrix multiplication is NOT commutative. Even though matrix multiplication is not commutative, it is associative in the following sense. 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Plagiarism guaranteed to multiplication of a and B are real numbers, then, formal. | 4,565 | 19,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-31 | latest | en | 0.80526 |
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+ + + =
# Examples of Fractional Calculus Computer Algebra System 例题
## Content
• Arithmetic 算术
• Numeric math 数值数学
• Algebra 代数
• Function 函数
• Calculus 微积分
• Equation 方程
• Transform 转换
• Discrete Math 离散数学
• Definition 定义式
• Number Theory 数论
• Probability 概率
• Multi elements 多元
• Statistics 统计
• Linear Algebra 线性代数
• programming 编程
• Graphics
1. Classification by plotting function 按制图函数分类
2. Classification by dimension 按维数分类
3. Classification by appliaction 按应用分类
4. Classification by objects 按物体分类
5. Classification by function 按函数分类
6. Classification by equation 按方程分类
7. Classification by domain 按领域分类
8. Classification by libary 按文库分类
9. Classification by calculator 按计算器分类
10. Classification by platform 按平台分类
• Plot 制图
1. Interactive plot 互动制图
2. parametric plot, polar plot
3. solve equation graphically
4. area plot with integral
5. complex plot
6. Geometry 几何
• plane graph 平面图
1. plane graph 平面图 with plot2D
2. function plot with funplot
3. differentiate graphically with diff2D
4. integrate graphically with integrate2D
5. solve ODE graphically with odeplot
• 3D graph 立体图
1. surface in 3D with plot3D
2. contour in 3D with contour3D
3. wireframe in 3D with wirefram3D
4. complex function in 3D with complex3D
5. a line in 3D with parametric3D
6. a column in 3D with parametric3D
7. the 4-dimensional object (x,y,z,t) in 3D with implicit3D
• Drawing 画画
Do exercise and learn from example. If something wrong, please clear meomery by clicking the AC button, then do again.
## Arithmetic 算术 >>
### Exact computation
1. Fraction 1E2-1/2
2. mod operation:
input mod(3,2) for 3 mod 2
#### big number in Java
3. Add prefix "big" to number for Big integer:
big1234567890123456789-1
4. Add prefix "big" to number for Big decimal:
big1.234567890123456789-1
5. #### big number in JavaScript
computation in JavaScript should to click the numeric button or expression ending with equal sign =
The default precision for BigNumber is 64 digits, and can be configured with the option precision.
6. bignumber(1.23456789123456789)=
#### Round-off errors
Calculations with BigNumber are much slower than calculations with Number, but they can be executed with an arbitrary precision. By using a higher precision, it is less likely that round-off errors occur:
// round-off errors with numbers
7. math.add(0.1, 0.2) // Number, 0.30000000000000004
8. math.divide(0.3, 0.2) // Number, 1.4999999999999998
// no round-off errors with BigNumbers :)
9. math.add(math.bignumber(0.1), math.bignumber(0.2)) // BigNumber, 0.3
10. math.divide(math.bignumber(0.3), math.bignumber(0.2)) // BigNumber, 1.5
### Numerical approximations
There are two types of numeric computation: JavaScript and Java numeric computation:
#### JavaScript numeric computation
11. numeric computation with the ≈ button :
acos(bignumber(-1))
12. numeric computation end with the equal sign = :
acos(bignumber(-1))=
#### Java numeric computation
13. numeric computation with the ≈≈ button :
sin(pi/4)
14. Convert back with numeric computation function n(x) :
n(polar(2,45degree))
n( sin(pi/4) )
n( sin(30 degree) )
15. sin^((0.5))(1) is the 0.5 order derivative of sin(x) at x=1 :
n( sin(0.5,1) )
16. sin(1)^(0.5) is the 0.5 power of sin(x) at x=1 :
n( sin(1)^0.5 )
17. more are in numeric math
### Complex 复数
math handbook chapter 1.1.2 complex
There are two types of complex number: JavaScript complex and Java complex:
#### input complex numbers
18. input complex numbers in the complex plane:
1+2i
19. in small letters of complex(1,2) is complex number:
complex(1,2)
20. in first upper case letter of Complex(1,2) is object in JavaScript complex:
Complex(1,2)
21. input complex number in polar(r,theta) coordinates:
polar(1,pi)
22. input complex number in polar(r,theta*degree) coordinates:
polar(1,45degree)
23. input complex number in polar(r,theta) coordinates for degree by polard(r,degree):
polard(1,45)
24. input complex number in trig format of r*cis(theta*degree) :
2cis(45degree)
25. #### Convert complex to polar(r,theta) coordinates
26. Convert complex a+b*i to polar(r,theta) coordinates in Java:
convert 1-i to polar
topolar(1-i)
convert back to a+b*i format by to remove last letter s or click the simplify button
27. Convert complex a+b*i to polar(r,theta*degree) coordinates:
topolard(1-i)
28. convert back to a+b*i format by to remove last letter s or click the simplify button
29. Convert complex a+b*i to polar(r,theta) coordinates in JavaScript:
math.Complex(1,2).toPolar()=
math.complex(1,2).toPolar()=
#### convert to vector
complex(1,2) number is special vector, i.e. the 2-dimentional vector, so it can be converted to vector.
30. to Java vector(1,2)
convert 1-i to vector
tovector(1-i)
31. to JavaScript vector [1,2]
math.Complex(1,2).toVector()=
math.complex(1,2).toVector()=
#### complex number plot
32. in order to auto plot complex number as vector, input complex(1,-2) for 1-2i, or convert 1-2i to complex(1,-2) by
convert 1-2i to complex
convert(1-2i to complex)
tocomplex(1-2i)
#### complex function plot
33. complex animation(z) show animation of complex function in complex domain of complex variable z.
34. complexplot(z) show phrase and/or modulus of complex function in complex domain of complex variable z.
35. re2D plot: real curve (blue) and imag cureve (red) on real domain
re2D(x^x)
36. im2D plot: real curve (blue) and imag cureve (red) on imag domain
im2D(x^x)
37. put imagary i with variable x for plot on imag domain:
exp(i*x)
more are in complex2D(x) show 2 curves of real and imag parts in realand imag domains.
38. complex 3D plot:
complex3D(pow(x,x))
more are in complex function with complex3D(x) in complex domain of complex variable x.
39. ## Numeric math 数值数学 >>
数值数学 = 计算数学
math handbook chapter 3.4
40. computation in JavaScript should to click the numeric button or expression ending with equal sign =
sin(pi/4)=
41. numeric computation with the n(x) ≈≈ button:
n( sin(30 degree) )
n sin(30 degree)
42. numeric solve equation:
nsolve( x^2-5*x+6=0 )
nsolve( x^2-5*x+6 )
43. find_root(x,-10,10) between -10 and 10 for numeric equation:
find_root( x^2-5*x+6==0 )
find_root( x^2-5*x+6 )
44. JavaScript numeric calculator with the ≈ button can calculate numeric, number theory, Probability, Statistics, matrix, solve equation.
more example in JavaScript mathjs
45. numeric integrate, by default x from 0 to 1:
nint( x^2-5*x+6,x,0,1 )
nint x^2-5*x+6 as x from 0 to 1
nint sin(x)
46. numeric computation with the funplot ≈ button:
integrate(x=>sin(x),[1,2])
more calculus operation in JavaScript calculus
47. ## Algebra 代数 >>
math handbook chapter 1 algebra
48. simplify:
taylor( (x^2 - 1)/(x-1) )
49. expand:
expand( (x-1)^3 )
50. factorizing:
factor( x^2+3*x+2 )
51. factor high order polymonial by factor(x)== :
factor( x^3-1 )==
52. factorization:
factor( x^4-1 )==
53. combine two terms:
combine( log(a)+log(b) )
54. ### tangent
55. tangent equation at x=0 by default
tangent( sin(x) )
56. tangent equation at x=1
tangent( sin(x),x=1 )
57. tangentplot(x) show dynamic tangent line when your mouse over the curve.
tangentplot( sin(x) )
### convert
convert( sin(x) to exp) is the same as toexp(sin(x))
58. convert to exp:
toexp( cos(x) )
59. convert to trig:
convert exp(x) to trig
60. convert sin(x) to exp(x):
convert sin(x) to exp = toexp( sin(x) )
61. Convert to exp(x):
toexp(Gamma(2,x))
62. ### inverse function
63. input sin(x), click the inverse button
inverse( sin(x) )
check its result by clicking the inverse button again.
In order to show multi-value, use the inverse equation instead function.
### inverse equation
64. inverse equation to show multivalue if it has:
inverse( sin(x)=y )
check its result by clicking the inverse button again.
### polynomial
math handbook chapter 20.5 polynomial
65. the unit polynomial:
Enter poly(3,x) = poly(3) for the unit polynomial with degree 3: x^3+x^2+x+1.
66. Hermite polynomial:
hermite(3,x) gives the Hermite polynomial while hermite(3) gives Hermite number.
67. harmonic polynomial:
harmonic(-3,1,x) = harmonic(-3,x)
68. the zeta( ) polynomial:
zeta(-3,x)
69. simplify(x):
taylor( (x^2 - 1)/(x-1) )
70. expand(x) polynomial:
expand(hermite(3,x))
71. topoly(x) convert polynomial to polys(x) as holder of polynomial coefficients,
convert x^2-5*x+6 to poly = topoly( x^2-5*x+6 )
72. simplify polys(x) to polynomial:
simplify( polys(1,-5,6,x) )
73. polyroots(x) is holder of polynomial roots, topolyroot(x) convert a polynomial to polyroots.
convert (x^2-1) to polyroot = topolyroot(x^2-1)
74. polysolve(x) numerically solve polynomial for multi-roots:
polysolve(x^2-1)
75. nsolve(x) numerically solve for a single root:
nsolve(x^2-1)
76. solve(x) for sybmbloic and numeric roots:
solve(x^2-1)
77. construct polynomial from roots, activate polyroots(x) to polynomial, and reduce roots to polynomial equation = 0 by click on the simplify(x) button.
simplify( polyroots(2,3) )
### Number
When the variable x of polynomial is numnber, it becomes polynomial number, please see Number_Theory section.
78. ## Function 函数 >>
math handbook chapter 1 Function 函数
#### Trigonometry 三角函数
79. expand Trigonometry by expandtrig(x) :
expandtrig( sin(x)^2 )
80. inverse function :
inverse( sin(x) )
81. plot a multivalue function by the inverse equation :
inverse( sin(x)=y )
82. expand trig function :
expandtrig( sin(x)^2 )
83. expand special function :
expand( gamma(2,x) )
84. factor(x) :
factor( sin(x)*cos(x) )
85. #### Complex Function 复变函数
math handbook chapter 10 Complex Function 复变函数
complex2D(x) shows 2 curves of the real and imag parts in real domain x, and complex3D(x) shows complex function in complex domain x, for 20 graphes in one plot.
86. re2D plot: real curve (blue) and imag cureve (red) on real domain
re2D(x^x)
87. im2D plot: real curve (blue) and imag cureve (red) on imag domain
im2D(x^x)
88. put imagary i with variable x for plot on imag domain:
exp(i*x)
more are in complex2D(x) for 2 curves of real and imag parts in complex domain of complex variable x.
89. complex animation(z) show animation of complex function in complex domain of complex variable z.
90. complexplot(z) show phrase and/or modulus of complex function in complex domain of complex variable z.
91. complex 3D plot :
complex3D(pow(x,x)) in complex domain of complex variable x.
## Complex
1. complex - complex function - complex math
2. complex animate(z) or complex2D(x) for phase animation in complex plane, the independent variable must be z.
3. complex plot(z) for phase and/or modulus in complex plane, the independent variable must be z.
4. plot complex(z) for phase and/or modulus in complex plane, the independent variable must be z.
5. complex2D show re2D(x) and im2D(x) for complex 2 curves of real and imag parts in real and imag domain, the independent variable must be x.
6. complex3D(x) for 3 dimensional graph, the independent variable must be x.
7. color WebXR surface of complex function on complex plane
8. Riemann surface - Complex Branches - complex coloring
## References
1. math handbook content 2 chapter 10 complex function
2. math handbook content 3 chapter 10 complex function
3. math handbook content 4 chapter 10 complex function
4. Complex analysis
more are in complex function
#### special Function
math handbook chapter 12 special Function
92. ## Calculus 微积分 >>
### Limit
math handbook chapter 4.1 limit
93. click the lim(x) button for lim at x->0 :
lim_(x->0) sin(x)/x = lim sin(x)/x as x->0 = lim(sin(x)/x)
94. click the nlim(x) button for numeric limit at x->0 :
nlim(sin(x)/x)
95. click the limit(x) button for limit at x->oo :
limit_(x->oo) log(x)/x = limit( log(x)/x as x->inf )
= limoo( log(x)/x )
96. one side limit, left or right side :
lim(exp(-x),x,0,right)
### Derivatives
Math Handbook chapter 5 differential calculus
97. Differentiate
d/dx sin(x) = d(sin(x))
98. Second order derivative :
d^2/dx^2 sin(x) = d(sin(x),x,2) = d(sin(x) as x order 2)
99. sin(0.5,x) is inert holder of the 0.5 order derivative sin^((0.5))(x), it can be activated by simplify(x):
simplify( sin(0.5,x) )
100. Derivative as x=1 :
d/dx | _(x->1) x^6 = d( x^6 as x->1 )
101. Second order derivative as x=1 :
d^2/dx^2| _(x->1) x^6 = d(x^6 as x->1 order 2) = d(x^6, x->1, 2)
#### Fractional calculus
Fractional calculus
102. semiderivative :
d^(0.5)/dx^(0.5) sin(x) = d(sin(x),x,0.5) = d( sin(x) as x order 0.5) = semid(sin(x))
103. input sin(0.5,x) as the 0.5 order derivative of sin(x) for
sin^((0.5))(x) = sin^((0.5))(x) = sin(0.5,x)
104. simplify sin(0.5,x) as the 0.5 order derivative of sin(x) :
sin^((0.5))(x) = simplify(sin(0.5,x))
105. 0.5 order derivative again :
d^(0.5)/dx^(0.5) d^(0.5)/dx^(0.5) sin(x) = d(d(sin(x),x,0.5),x,0.5)
106. Minus order derivative :
d^(-0.5)/dx^(-0.5) sin(x) = d(sin(x),x,-0.5)
107. inverse the 0.5 order derivative of sin(x) function :
f(-1)( sin(0.5)(x) ) = inverse(sin(0.5,x))
108. Derive the product rule :
d/dx (f(x)*g(x)*h(x)) = d(f(x)*g(x)*h(x))
109. … as well as the quotient rule :
d/dx f(x)/g(x) = d(f(x)/g(x))
110. for derivatives :
d/dx ((sin(x)* x^2)/(1 + tan(cot(x)))) = d((sin(x)* x^2)/(1 + tan(cot(x))))
111. Multiple ways to derive functions :
d/dy cot(x*y) = d(cot(x*y) ,y)
112. Implicit derivatives, too :
d/dx (y(x)^2 - 5*sin(x)) = d(y(x)^2 - 5*sin(x))
113. the nth derivative formula :
d^n/dx^n (sin(x)*exp(x)) = nthd(sin(x)*exp(x))
114. #### differentiate graphically
some functions cannot be differentiated or integrated symbolically, but can be semi-differentiated and integrated graphically in diff2D.
### Integrals
Math Handbook chapter 6 integral calculus
115. indefinite integrate : int sin(x) dx = integrate(sin(x))
116. enter a function sin(x), then click the ∫ button to integrate :
int(cos(x)*e^x+sin(x)*e^x)\ dx = int(cos(x)*e^x+sin(x)*e^x)
int tan(x)\ dx = integrate tan(x) = int(tan(x))
117. Exact answers for integral :
int (2x+3)^7 dx = int (2x+3)^7
118. Multiple integrate :
int int (x + y)\ dx dy = int( int(x+y, x),y)
int int exp(-x)\ dx dx = integrate(exp(-x) as x order 2)
119. Definite integration :
int _1^3 (2*x + 1) dx = int(2x+1,x,1,3) = int(2x+1 as x from 1 to 3)
120. Improper integral :
int _0^(pi/2) tan(x) dx =int(tan(x),x,0,pi/2)
121. Infinite integral :
int _0^oo 1/(x^2 + 1) dx = int(1/x^2+1),x,0,oo)
122. Definite integration :
int_0^1 sin(x) dx = integrate( sin(x),x,0,1 ) = integrate sin(x) as x from 0 to 1
#### integrator
If integrate(x) cannot do, please try integrator(x) :
123. integrator(sin(x))
124. enter sin(x), then click the ∫ dx button to integrator
#### fractional integrate
125. semi integrate, semiint(x) :
int sin(x) \ dx^(1/2) = int(sin(x),x,1/2) = int sin(x) as x order 1/2 = semiint(sin(x)) = d(sin(x),x,-1/2)
126. indefinite semiintegrate :
int sin(x)\ dx^0.5 = d^(-0.5)/dx^(-0.5) sin(x) = int(sin(x),x,0.5) = semiint(sin(x))
127. Definite fractional integration :
int_0^1 sin(x) (dx)^0.5 = integrate( sin(x),x,0.5,0,1 ) = semiintegrate sin(x) as x from 0 to 1
#### numeric computation
128. numeric computation by click on the "~=" button :
n( int _0^1 sin(x) dx )
#### numeric integrate
If numeric computation ail, please try numeric integrate nintegrate(x) or nint(x) :
nint(sin(x),x,0,1) = nint(sin(x))
#### integrate graphically
some functions cannot be differentiated or integrated symbolically, but can be semi-differentiated and integrated graphically in integrate2D.
#### vector calculus
129. differentiate vector(x,x) :
d(vector(x,x))
130. differentiate sin(vector(x,x)) :
d(sin(vector(x,x)))
131. ## Equation 方程 >>
equation world
#### Your operation step by step
132. solve x^2-1=0 step by step
(x^2-1=0)+1
2. add ( ) to equation, then power by 0.5:
(x^2=1)^0.5
### inverse an equation
133. inverse an equation to show multivalue curve.
inverse( sin(x)=y )
check its result by clicking the inverse button again.
### polynomial equation
134. polyroots(x) is holder of polynomial roots, topolyroot(x) convert a polynomial to polyroots.
convert (x^2-1) to polyroot = topolyroot(x^2-1)
135. polysolve(x) numerically solve polynomial for multi-roots.
polysolve(x^2-1)
136. construct polynomial from roots, activate polyroots(x) to polynomial, and reduce roots to polynomial equation = 0 by click on the simplify(x) button.
simplify( polyroots(2,3) )
137. solve(x) for sybmbloic and numeric roots :
solve(x^2-1)
solve( x^2-5*x-6 )
138. solve equation and inequalities, by default, equation = 0 for default unknown x if the unknown omit.
solve( x^2+3*x+2 )
139. Symbolic roots :
solve( x^2 + 4*x + a )
140. Complex roots :
solve( x^2 + 4*x + 181 )
141. solve equation for x.
solve( x^2-5*x-6=0,x )
142. numerically root :
nsolve( x^3 + 4*x + 181 )
143. nsolve(x) numerically solve for a single root.
nsolve(x^2-1)
144. find_root(x) numerically find for a single root.
find_root(x^2-1)
### Algebra Equation f(x)=0
math handbook chapter 3 algebaic Equation
solve(x) also solve other algebra equation, e.g.
145. nonlinear equations:
solve(exp(x)+exp(-x)=4)
solve cos(x)+sin(x)=1
if solve(x) cannot solve, then click the numeric button to solve numerically.
146. solve(cos(x)-sin(x)=1)
then click the numeric button
#### absolute equation
147. solve(x) absolute equation for the unknown x inside the abs(x) function :
solve abs(x-1)+abs(x-2)=3 for
|x-1|+|x-2|=3
#### Modulus equation
solve(x) Modulus equation for the unknown x inside the mod(x) function :
##### first order equation
148. solve mod(3x,5)=1 for
3x mod 5 = 1
click the solve button. it is solved by inversemod(3,5)
##### second order equation
149. solve mod(x^2-5x+7,2)=1 for
(x^2-5x+7) mod 2 = 1
150. solve mod(x^2-5x+6,2)=0 for
(x^2-5x+6) mod 2 = 0
#### congruence equation
a x ≡ b* (mod m)
math handbook chapter 20.3 congruence
a x ≡ b* (mod m) means two remindars in both sides of equation are the same, i.e. congruence, it is the same as the modular equation mod(a*x,m)=mod(b,m). if b=1, the modular equation mod(a*x,m)=1 can be solved by inversemod(a*x,m). By definition of congruence, a x ≡ b* (mod m) if a x − b is divisible by m. Hence, a x ≡ b (mod m) if a x − b = m y, for some integer y. Rearranging the equation to the equivalent form of Diophantine equation a x − m y = b, which can be solved by solve(a*x-m*y=b,x,y).
##### first order equation:
151. solve 3x=1*(mod 5)
solve mod(3x,5)=mod(1,5)
##### second order equation:
152. solve x^2+3x+2=1*(mod 11)
solve x^2+3x+2=1 mod(11)
solve x^2+3x+2=mod(11)
### Probability_equation
153. solve(x) Probability equation for the unknown k inside the Probability function P(x),
solve( P(x>k)=0.2, k)
### recurrence_equation
154. rsolve(x) recurrence and functional and difference equation for y(x)
y(x+1)+y(x)+x=0
y(x+1)+y(x)+1/x=0
155. fsolve(x) recurrence and functional and difference equation for f(x)
f(x+1)=f(x)+x
f(x+1)=f(x)+1
### functional_equation
156. rsolve(x) recurrence and functional and difference equation for y(x)
y(a+b)=y(a)*y(b)
y(a*b)=y(a)+y(b)
157. fsolve(x) recurrence and functional and difference equation for f(x)
f(a+b)=f(a)*f(b)
f(a*b)=f(a)+f(b)
### difference equation
158. rsolve(x) recurrence and functional and difference equation for y(x)
y(x+1)-y(x)=x
y(x+2)-y(x+1)-y(x)=0
159. fsolve(x) recurrence and functional and difference equation for f(x)
f(x+1)-f(x)=x
f(x+2)- f(x+1)-f(x)=0
see vector
### 2 variables equations f(x,y) = 0
#### Diophantine equation f(x,y) = 0
math handbook chapter 20.5 polynomial
It is that number of equation is less than number of the unknown, e.g. one equation with 2 unknowns x and y.
160. One 2D equation f(x,y) = 0 for 2 integer solutions x and y
solve( 3x-2y-2=0, x,y )
solve( x^2-3x-2y-2=0, x,y )
#### 2 variables equation f(x,y) = 0 solved graphically on 2D
One equation for 2 unknowns x and y, f(x,y) = 0, solved graphically by implicitplot(x)
161. solve x^2-y^2=1 graphically
x^2-y^2-1=0
### 3D equations
#### 3 variables equation f(x,y,z) = 0
One equation for 3 unknowns x and y and z, f(x,y,z) = 0, solved graphically :
162. implicit3D( x-y-z )
163. plot3D( x-y-z )
### 4D equations
One equation with 4 variables,
164. f(x,y,z,t) = 0, solved graphically :
implicit3D( x-y-z-t )
165. f(x,y,n,t) = 0, solved graphically :
plot2D( x-y-n-t )
### system of equations
math handbook chanpter 4.3 system of equations
#### system of 2 equations
166. system of 2 equations f(x,y)=0, g(x,y)=0 for 2 unknowns x and y by default if the unknowns omit with the solve() button :
solve( 2x+3y-1=0, 3x+2y-1=0 )
167. system of 2 equations f(x,y)=0 and g(x,y)=0 for 2 unknowns x and y by default if the unknowns omit. On First graph it is solved graphically, where their cross is solution:
( 2x+3y-1=0 and 3x+2y-1=0 )
168. system of 2 equations f(x,y)=0, g(x,y)=0 for 2 unknowns x and y by default if the unknowns omit with the solver() button :
solver( 2*x+3*y-1=0, 3*x+2*y-1=0 )
#### One variable parametric equations x=f(t), y=g(t) on 2D
A system of 2 equations with a parameter t for 2 unknowns x and y, x=f(t), y=g(t), solved graphically :
169. parametricplot( x=cos(t), y=sin(t) )
170. parametric3D( cos(t),sin(t) )
171. parametric2D( cos(t),sin(t) )
#### 2 variables parametric equations x=f(u,v), y=f(u,v) on 2D
A system of 3 equations with 2 parameters u and v for 3 unknowns x and y and z, x=f(u,v), y=f(u,v) solved graphically :
172. parametric3D( cos(u*v),sin(u*v),u*v )
173. wireframe3D( cos(x*y),sin(x*y) )
174. parametric3D( cos(t),sin(t) )
#### system of 3 equations
175. system of 3 equations f(x,y,z)=0, g(x,y,z)=0 h(x,y,z)=0 for 3 unknowns x and y and z by default if the unknowns omit with the solver() button :
solve([x-y-z==0,x+y+z==6,x+y==5],x,y,z )
#### one variable parametric equations x=f(t), y=f(t), z=f(t) in 3D
A system of 3 equations with a parameter t for 3 unknowns x and y and z, x=f(t), y=f(t), z=f(t), solved graphically :
176. parametric3D( t,cos(t),sin(t) )
#### 2 variables parametric equations x=f(u,v), y=f(u,v), z=f(u,v) in 3D
A system of 3 equations with 2 parameters u and v for 3 unknowns x and y and z, x=f(u,v), y=f(u,v), z=f(u,v), solved graphically :
177. parametric3D( u,u-v,u*v )
178. parametric surface
### Inequalities
179. solve(x) Inequalities for x.
solve( 2*x-1>0 )
solve( x^2+3*x+2>0 )
### differential equation
Math handbook chapter 13 differential equation.
There are two types of differential equations: a single variable is ordinary differential equation (ODE) and multi-variables is partial differential equation (PDE).
#### ordinary differential equation (ODE)
ODE(x) and dsolve(x) and lasove(x) solve ordinary differential equation (ODE) for unknown y.
#### Your operation step by step
180. solve dy/dx=x/y step by step
1. add ( ) to equation, then time by y:
(dy/dx=x/y)*y
2. add ( ) to equation, then time by dx:
(dy/dx*y=x)*dx
3. integrate by click the integrate button
int(dy*y=dx*x)
4. add ( ) to equation, then time by 2:
(1/2y^2=1/2x^2)*2
5. add ( ) to equation, then power by 0.5:
(y^2=x^2)^0.5
181. linear ordinary differential equations:
dsolve y'=x*y+x
ode y'= 2y
ode y'-y-1=0
182. nonlinear ordinary differential equations:
ode (y')^2-2y^2-4y-2=0
dsolve( y' = sin(x-y) )
dsolve( y(1,x)=acos(y)-sin(x)-x )
dsolve( ds(y)-cos(x)=asin(y)-x )
dsolve( ds(y)=exp(y)-exp(x) )
dsolve( ds(y)-exp(x)=log(y)-x )
ode y'-exp(y)-1/x-x=0
ode y'-sinh(y)+x-1/sqrt(1+x^2)=0
ode y'-tan(y)+x-1/(1+x^2)=0
183. second order nonlinear ordinary differential equations:
dsolve( ds(y,x,2)=asin(y)-sin(x)-x )
dsolve( ds(y,x,2)-exp(x)=log(y)-x )
184. 2000 examples of Ordinary differential equation (ODE)
more examples in bugs
#### solve graphically
The odeplot(x) can be used to visualize individual functions, First and Second order Ordinary Differential Equation over the indicated domain. Input the right hand side of Ordinary Differential Equations, y"=f(x,y,z), where z for y', then click the checkbox. by default it is first order ODE.
185. second order ODE
odeplot y''=y'-y
### integral equation
Math handbook chapter 15 integral equation.
#### indefinite integral equation
indefinite integral equation
186. linear equation
input ints(y) -2y = exp(x) for
ode int y dx - 2y = exp(x)
187. nonlinear equation
input ints(y) -2y^2 = 0 for
ode int y dx - 2y^2 = 0
#### double integral equation
input ints(y,x,2) for double integral
188. linear equation
ode( int int y dx -y = exp(x) )
189. nonlinear equation
ode( int int y dx *y= exp(x) )
#### definite integral equation
definite integral equation
190. linear equation
input integrates(y(t)/sqrt(x-t),t,0,x) = 2y for
ode int_0^x (y(t))/sqrt(x-t) dt = 2y
191. nonlinear equation
input integrates(y(t)/sqrt(x-t),t,0,x) = 2y^2 for
ode int_0^x (y(t))/sqrt(x-t) dt = 2y^2
### differential integral equation
192. input ds(y)-ints(y) -y-exp(x)=0 for
ode dy/dx-int y dx -y-exp(x)=0
### fractional differential equation
dsolve(x) also solves fractional differential equation
193. linear equations:
ode d^0.5/dx^0.5 y = 2y
ode d^0.5/dx^0.5 y -y -exp(4x) = 0
ode (d^0.5y)/dx^0.5 -y=x
ode d^0.5/dx^0.5 y -y - exp(x)*x = 0
194. nonlinear equations:
ode (d^0.5y)/dx^0.5 = y^2*exp(x)
ode (d^0.5y)/dx^0.5 = sin(y)*exp(x)
ode (d^0.5y)/dx^0.5 = exp(y)*exp(x)
ode (d^0.5y)/dx^0.5 = log(y)*exp(x)
ode (d^0.5y)/dx^0.5 - a*y^2-b*y-c = 0
ode (d^0.5y)/dx^0.5 - log(y) - exp(x) + x=0
ode (d^(1/2)y)/dx^(1/2)-asin(y)+x-sin(x+pi/4)=0
#### linear fractional integral equation
195. ode d^-0.5/dx^-0.5 y = 2y
#### nonlinear fractional integral equation
196. ode d^-0.5/dx^-0.5 y = 2y^2
#### fractional differential integral equation
197. ds(y,x,0.5)-ints(y,x,0.5) -y-exp(x)=0
ode (d^0.5y)/(dx^0.5)-int y (dx)^0.5 -y-exp(x)=0
#### complex order differential equation
198. ode (d^(1-i) y)/dx^(1-i)-2y-exp(x)=0
#### variable order differential equation
199. ode (d^sin(x) y)/dx^sin(x)-2y-exp(x)=0
### system of differential equations
system of 2 equations with 2 unknowns x and y with a variable t :
200. linear equations:
dsolve( ds(x,t)=x-2y,ds(y,t)=2x-y )
201. nonlinear equations:
dsolve( ds(x,t)=x-2y^2,ds(y,t)=2x^2-y )
202. the second order system of 2 equations with 2 unknowns x and y with a variable t :
dsolve( x(2,t)=x,y(2,t)=2x-y )
203. the 0.5th order system of 2 equations with 2 unknowns x and y with a variable t :
dsolve( x(0.5,t)=x,y(0.5,t)=x-y )
### partial differental equation
Math handbook chapter 14 partial differential equation.
PDE(x) and pdsolve(x) solve partial differental equation with two variables t and x for y, then click the plot2D button to plot solution, pull the t slider to change the t value. click the plot3D button for 3D graph.
204. linear equation:
pde dy/dt = dy/dx-2y
205. nonlinear equation:
pde dy/dt = dy/dx*y^2
pde dy/dt = 2dy/dx-y^2
pde (d^2y)/(dt^2) -2* (d^2y)/(dx^2)-y^2-2x*y-x^2=0
#### partial differential integral equation
206. ds(y,t)-ints(y,x)-y-exp(x)=0
pde (dy)/(dt)-int y (dx) -y-exp(x)=0
#### fractional partial differental equation
PDE(x) and pdsolve(x) solve fractional partial differental equation.
207. linear equations:
pde (d^0.5y)/dt^0.5 = dy/dx-2y
208. nonlinear equations:
pde (d^0.5y)/dt^0.5 = 2* (dy)/dx*y^2
pde (d^0.5y)/dt^0.5 = 2* (d^0.5y)/dx^0.5-y^2
pde (d^1.5y)/(dt^1.5) + (d^1.5y)/(dx^1.5)-2y^2-4x*y-2x^2 =0
More examples are in Analytical Solution of Fractional Differential Equations
#### fractional partial differential integral equation
209. ds(y,t)-ints(y,x,0.5)-exp(x)=0
pde (dy)/(dt)-int y (dx)^0.5 -exp(x)=0
210. ds(y,t)-ints(y,x,0.5)+2y-exp(x)=0
pde (dy)/(dt)-int y (dx)^0.5 +2y-exp(x)=0
211. ds(y,t)-ds(y,x)-ints(y,x,0.5)+3y-exp(x)=0
pde (dy)/(dt)-dy/dy-int y (dx)^0.5 +3y-exp(x)=0
### system of partial differential equations
system of 2 equations with two variables t and x for 2 unknowns y and z:
212. linear equations:
pde( ds(y,t)-ds(y,x)=2z-2y,ds(z,t)-ds(z,x)=4z-4y )
213. the second order system of 2 equations :
pde( ds(y,t)-ds(y,x,2)=2z-2y,ds(z,t)-ds(z,x,2)=4z-4y )
214. the 0.5th order system of 2 equations :
pde( ds(y,t)-ds(y,x,0.5)=2z-2y,ds(z,t)-ds(z,x,0.5)=4z-4y )
### test solution
#### test solution for algebaic equation
test solution for algebaic equation to the unknown x by test( solution,eq, x) or click the test button :
215. test( -1, x^2-5*x-6, x )
216. test( -1, x^2-5*x-6 )
#### test solution for differential equation
test solution for differential equation to the unknown y by test( solution, eq ) or click the test button :
217. test( exp(2x), dy/dx=2y )
218. test( exp(4x), (d^0.5y)/dx^0.5=2y )
#### test solution for recurrence equation to the unknown y
by rtest( solution, eq ) or click the rtest button.
#### test solution for recurrence equation to the unknown f
by ftest( solution, eq ) or click the ftest button.
### test equation
219. tests over 1000 Differential Equations
### bugs
220. There are over 1000 bugs
221. ## Transform 转换 >>
Math handbook chapter 11 integral transform
### laplace transform
222. First graph is in real domain, second graph is in Laplace domain by Lapalce transform
laplace(x)
223. Input your function, click the laplace button :
laplace(sin(x))
### inverse laplace transform
224. First graph is in Laplace domain , second graph is in real domain by inverse Lapalce transform
inverselaplace(1/x^2)
### Fourier transform
225. First graph is in real domain, second graph is in Fourier domain by Fourier transform
fourier(x)
Input your function, click the Fourier button :
226. fourier(exp(x))
227. sine wave
228. Weierstrass function animation
### convolution transform
First graph is in real domain, second graph is in convolution domain by convolution transform convolute(x) with x by default:
229. Input your function, click the convolute button :
convolute(exp(x))
230. convoute sin(x) with exp(x) :
convolute(sin(x),exp(x))
231. convolution of two lists
convolution([1,2],[3,4])
## Discrete Math 离散数学 >>
The default index variable in discrete math is k.
232. Input harmonic(2,x), click the defintion(x) button to show its defintion, check its result by clicking the simplify(x) button, then click the limoo(x) button for its limit as x->oo.
### Difference
233. Δ(k^2) = difference(k^2)
Check its result by the sum(x) button
### Summation ∑
#### Indefinite sum
234. ∑ k = sum(k)
235. Check its result by the difference(x) button
Δ sum(k) = difference( sum(k) )
236. In order to auto plot, the index variable should be x :
sum_x x = sum(x,x)
#### definite sum
237. Definite sum = Partial sum x from 1 to x :
1+2+ .. +x = sum _(k=1) ^x k = sum(k,k,1,x)
238. Definite sum, sum x from 1 to 5 :
1+2+ .. +5 = ∑(x,x,1,5) = sum(x,x,1,5)
sum(x^k,k,1,5)
#### Definite sum with parameter x as upper limit
sum(k^2, k,1, x)
239. Check its result by the difference(x) button, and then the expand(x) button.
240. convert to sum series definition :
tosum( exp(x) )
241. expand above sum series by the expand(x) button :
expand( tosum(exp(x)) )
#### Indefinite sum
242. ∑ k
243. sum( x^k/k!,k )
244. partial sum of 1+2+ .. + k = ∑ k = partialsum(k)
245. Definite sum of 1+2+ .. +5 = ∑ k
#### partial sum with parameter upper limit x
246. sum(1/k^2,k,1,x)
#### infinite sum
247. sum from 1 to oo:
Infinite sum of 1/1^2+1/2^2+1/3^2 .. +1/k^2+... = sum( 1/k^2,k,1,oo )
248. sum from 0 to oo:
Infinite sum of 1/0!+1/1!+1/2! .. +1/k!+... = sum( 1/k!,k,0,oo )
249. Infinite sum x from 0 to inf :
1/0!+1/1!+1/2!+ .. +1/x! = sum 1/(x!) as x->oo
### Series 级数
250. convert to sum series definition :
tosum( exp(x) ) = toseries( exp(x) )
251. check its result by clicking the simplify(x) button :
simplify( tosum( exp(x) ))
252. expand above sum series :
expand( tosum(exp(x)) )
253. compare to Taylor series with numeric derivative:
taylor( exp(x), x=0, 8)
254. compare to series with symbolic derivative:
series( exp(x) )
255. Taylor series expansion as x=0, by default x=0.
taylor( exp(x) as x=0 ) = taylor(exp(x))
256. series expand not only to taylor series:
series( exp(x) )
257. but aslo to other series expansion:
series( zeta(2,x) )
#### the fractional order series expansion at x=0 for 5 terms and the 1.5 order
258. series( sin(x),x,0,5,1.5 )
### Product ∏
259. prod(x,x)
260. prod x
## Definition 定义式 >>
261. definition of function :
definition( exp(x) )
262. check its result by clicking the simplify(x) button :
simplify( def(exp(x)) )
263. #### series definition
264. convert to series definition :
toseries( exp(x) )
265. check its result by clicking the simplify(x) button :
simplify( tosum(exp(x)) )
266. #### integral definition
267. convert to integral definition :
toint( exp(x) )
268. check its result by clicking the simplify(x) button :
simplify( toint(exp(x)) )
269. ## Number Theory 数论 >>
math handbook chapter 20 Number Theory.
When the variable x of polynomial is numnber, it becomes polynomial number :
270. poly number:
poly(3,2)
271. Hermite number:
hermite(3,2)
272. harmonic number:
harmonic(-3,2)
harmonic(-3,2,4)
harmonic(1,1,4) = harmonic(1,4) = harmonic(4)
273. Bell number:
bell(5)
274. double factorial 6!!
275. binomial number ((4),(2))
276. combination number C_2^4
277. harmonic number H_4
#### prime
278. is prime number? isprime(12321)
279. is prime number? is_prime(12321)
280. Calculate the 4nd prime by prime(4)
281. Calculate the 4nd prime by nth_prime(4)
282. next prime greater than 4 by nextprime(4)
283. next prime greater than 4 by next_prime(4)
284. prime in range 4 to 9 by prime_range(4,9)
285. prime in range 1 to 9 by prime_range(9)
286. #### integer equation
287. congruence equation:
3x-1 = 2*(mod 2)
x^2-3x-2 = 2mod( 2)
288. modular equation:
solve mod(3x,5)=1 for
3x mod 5 = 1
solve mod(x^2-5x+7,2)=1 for
(x^2-5x+7) mod 2 = 1
solve mod(x^2-5x+6,2)=0 for
(x^2-5x+6) mod 2 = 0
289. Diophantine equation:
number of equation is less than number of the unknown, e.g. one equation with 2 unkowns x and y,
solve( 3x-2y-2=0, x,y )
solve( x^2-3x-2y-2=0, x,y )
more example is number theory in function and in JavaScript javascsript math
290. ## Probability 概率 >>
math handbook chapter 16 Probability
291. standard normal probability density function
Guassian function gauss(x)
292. standard normal distribution function gaussi(x)
Gaussian integral gaussi(x)
293. standard normal distribution function Phi(x):
Phi(x)
294. Probability of standard normal distribution P(range(-1,1)) between -1 and 1:
P(range(-1,1))
295. All probability of standard normal distribution P(x) between -oo and oo:
P(range(-oo,oo))
296. probability of standard normal distribution P(x<0.8):
P(x<0.8)
297. solve Probability equation for x by default :
solve(P(t>x)=0.2)
298. the binomial coefficient or choice number, combination(4,x) = C(4,2) = bimonial(4,2)
C(4,2)
299. arrangement number, permutation(4,x) = P(4,2),
P(4,2)
more example in JavaScript mathjs
## Multi elements 多元 >>
We can put multi elements together with list(), vector(), and. Most operation in them is the same as in one element, one by one. e.g. +, -, *, /, ^, differentiation, integration, sum, etc. We count its elements with size(), as same as to count elements in function. But vector operation may be not.
### and
Its position of the element is fix so we cannot sort it. We can plot multi curves with the and. e.g.
300. differentiate :
d(x and x*x)
301. integrate :
int(x and x*x)
### list
There are 2 types of list: [1,2,3] is numeric list in JavaScript, and the list(1,2,3) function is symbolic list in mathHand. The symbolic list element can be symbol, formula and function.
#### symbolic list
302. sort list value :
sort(list(4,2,1))
303. two lists added value :
list(2,3)+list(3,4)
#### numeric list
304. two lists join together
[1,2]+[3,4]
305. convolution of two lists
convolution([1,2],[3,4])
306. sort with JavaScript calculator:
[1,2,3].sort()=
307. two lists added value by the ending with =
[1,2]+[3,4]=
308. JavaScript calculation on the ≈ button by input:
[1,2]+[3,4]
click the ≈ button
more example in JavaScript mathjs
## Statistics 统计
math handbook chapter 16 Statistics
#### symbolic list
309. sort(list(x)), add numbers together by total(list()), max(list()), min(list()), size(list()) with mathHand calculator on the = button. e.g.
total(list(1,2,3))
#### numeric list
310. with JavaScript numeric calculator on the ≈ button :
sum([1,2])
more example in JavaScript mathjs
## Linear Algebra 线性代数 >>
### vector
math handbook chapter 8 vector
It has direction. the position of the element is fix so we cannot sort it. numeric vector is number with direction. the system auto plot the 2-dimentional vector. two vector(x) in the same dimention can be operated by +, -, *, /, and ^, the result can be checked by its reverse operation.
There are two types of vector: symbolic vector(a,b) and numeric vector([1,2])
#### symbolic vector
311. vector(1,2)+vector(3,4)
##### vector equation
312. solve vector(1,2)+x=vector(2,4) is as same as x=vector(2,4)-vector(1,2)
313. solve 2x-vector(2,4)=0 is as same as x=vector(2,4)/2
314. solve 2/x-vector(2,4)=0 is as same as x=2/vector(2,4)
315. solve vector(1,2)*x-vector(2,4)=0 is as same as x=vector(2,4)/vector(1,2)
316. solve vector(1,2)*x-20=0 is as same as x=20/vector(1,2)
317. solve vector(2,3)*x+vector(3,2)*y=vector(1,1),x,y is as same as solve(-1+2*x+3*y=0,-1+3*x+2*y=0)
##### vector calculus
318. differentiate vector(x,x) :
d(vector(x,x))
319. differentiate sin(vector(x,x)) :
d(sin(vector(x,x)))
#### numeric vector
320. vector([1,2])+vector([3,4])
### Array 数组
math handbook chapter 4 matrix
Complex array [[1,2],[3,4]] can be operated by +,-,*,/,^,
321. with array calculator 数组计算器
322. with JavaScript numeric calculator
more example in JavaScript mathjs
### Matrix 矩阵
math handbook chapter 4 matrix
Complex matrix([[1,2],[3,4]]) can be operated by +,-,*,/,\,^,
323. with matrix calculator 矩阵计算器
324. matrix([[1,2],[3,4]])
## programming 编程 >>
There are many coding :
325. math coding 数学编程
326. HTML + JavaScript coding 网页编程
327. cloud computing = web address coding 云计算 = 网址编程 = 网址计算器
## Graphics >>
328. Classification by plotting function 按制图函数分类
329. Classification by dimension 按维数分类
330. Classification by appliaction 按应用分类
331. Classification by objects 按物体分类
332. Classification by function 按函数分类
333. Classification by equation 按方程分类
334. Classification by domain 按领域分类
335. Classification by libary 按文库分类
336. Classification by calculator 按计算器分类
337. Classification by platform 按平台分类
## Plot 制图 >>
338. plane curve 2D
339. surface 2D
## 3D graph 立体图 plot 3D >>
340. space curve 3D
341. surface 3D
342. surface 4D
## Drawing 画画 >>
343. drawing
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In the last math post, we dealt with exponents and really big numbers in quantitative comparison. Let’s now try the same but in the problem solving context.
The general approach is not too different—you want to look for a pattern. However this time you will have to be exact, instead of simply saying which side is bigger. One tactic the GRE employs is to make a problem look impossible to solve in 2 minutes. But don’t despair. All problems are solvable once you unlock them. So if your approach is taking too much time, step back and look at the problem from a new angle.
Try the following and see if you can solve it in fewer than 2 minutes.
What is the sum of the digits of integer x, where x = 4^10 x 5^13?
(A) 13
(B) 11
(C) 10
(D) 8
(E) 5
If you spotted the pattern early on, you may have only required a minute. But if you couldn’t unlock the problem, here is how you do so. Notice that 4^10 can be rewritten as 2^20. We can now express x as 2^20 x 5^13. The logic here is that 2 x 5 = 10. That is, 10 to any integer power greater than 1,will be a 1 followed by zeroes
So now let’s rewrite the problem again so we get 2^7 x 2^13 x 5^13. Combine 2^13 x 5^13 and we get 10^13. That is, we get 1 followed by 13 zeroes. If you are taking the sum, it’s straightforward: 1 plus 13 zeroes is 1. We are not done yet as we have the 2^7. When you multiply this out, you get 128. 128 x 1,000 thirteen zeroes is equal to 128 followed by the thirteen zeroes. Ignore the zeroes and we get 1 + 2 + 8, which equals 11. Answer B
### 25 Responses to Advanced Exponents–Problem Solving
1. Azam April 24, 2018 at 3:58 am #
In the practice question Why power of y equals 17?
Shouldn’t it be y to 12-(+5)=7
Y has -5 in denominator. Will it not become +5 in Numerator?
• Magoosh Test Prep Expert April 24, 2018 at 12:38 pm #
Hi Azam,
I’m not sure if I understand this question–are you sure you wrote it on the correct blog post? The practice question in this blog post uses the variable ‘x’ and doesn’t have a fraction in it. I’d be happy to answer your question if you let me know where it came from!
2. aditya September 13, 2016 at 10:09 am #
can we solve this problem by unit digit?
4^10*5^13
unit digit of 4^10 is 6
unit digit of 5^13 is 5
6+5=11
• Magoosh Test Prep Expert September 17, 2016 at 5:17 am #
Unfortunately, focusing only on the units digit for this type of problem will not work. In this specific situation, the answer worked out by coincidence. The question asks us for the “sum of digits of integer x”. Technically, if “x = 4^10 x 5^13”, we should not be adding the units digit of “4^10” and “5^13”. Within “x” we are told those two numbers (therefore, the units digits) must be multiplied together. As such, this approach is not valid, but I like your thinking!
3. Rania June 26, 2016 at 11:49 am #
Hi Magooshers,
If I have a small circle in a large circle, both circles share a 90 degree sector. If no other information is given. How can I compare the circumference of the small circle to the length of Arc of the bigger circle?
Are the quantities equal, one is smaller, larger or can’t be decided based on information given?
Thank you
• Magoosh Test Prep Expert June 27, 2016 at 6:00 am #
Hi Raina,
Because you are a premium member and your questions are rather intricate, I have sent them over for our team of test prep experts to help you with! You should be hearing from them soon. 🙂
• Rania June 27, 2016 at 7:46 am #
Hi Magoosh Expert,
Thank you so much:)
4. Karan August 4, 2014 at 5:32 am #
Hey Chris,
I noticed that you can make this question slightly trickier by adding 12 as an answer choice.
I noticed that while solving the question I accidentally counted the 10^13 as 1(as 1 + 13 zeros = 1) and then counted 128 separately as 11. So I got a wrong answer of sum of digits = 12, but since 12 was not part of the answers I quickly realized my mistake.
Good question though 🙂
• Chris Lele August 4, 2014 at 12:03 pm #
That’s true! Good eye :). I’ll make the change now.
• Prashanth October 1, 2015 at 10:18 pm #
How to solve something like
Find the greatest integer
a) 10^10 + 2^100
b) 100^10 + 2^10
c) (100 + 2 ) ^10
• Rasitha July 27, 2016 at 6:31 am #
b) can be re written as 10^100 +2^10, which you can intuitively see that is bigger than a)
(because a. has the same powers but and the same bases but switched up)
So a<b
Then, looking at c) we see that it is definitely bigger than b) because,
using FOIL method, c can be written as 100 ^10 + 2^10 + other terms
So, b<c
Combining both results above we get a<b<c
5. Richard July 23, 2014 at 8:34 pm #
Hi chris, I tried a crude method and had the same answer.
this is how my method worked:
4 raised to an even number will always yield 6 as its unit number and 5 raised to any number yields 5 as its unit number, add them up and you will have 11. Is it a perfect method?
• Chris Lele August 4, 2014 at 12:06 pm #
Hmm..actually, I think that you got luck in this case :). The question is actually asking for the sum of all of the digits that make up ‘x’, hot the sum of the units digit of the two numbers. That said, I might have to rewrite this question, because I’m pretty sure you are not the first person who used this approach 🙂
6. Remya May 31, 2014 at 11:27 pm #
I loved this question 🙂
• Chris Lele June 4, 2014 at 11:59 am #
7. Sandeep September 9, 2013 at 2:56 am #
What a great Problem !!! I have to appreciate you !!! You are blessed with great Mind.
8. Cecilia September 19, 2011 at 6:24 pm #
You’re so right!!!!
Thanks!
9. Cecilia September 19, 2011 at 12:49 pm #
Hi there,
I got a slightly different answer and I am not being able to see what am I doing wrong…. (…. starting to think it could be the effect of studying too much!….)
I solved the problem as follows:
x=4^10 * 5^13
x= 4^10 * 5^10 * 5^3
x= 20^10 * 5^3 = 2 + 1+ 2 + 5 = 10 … not 11…
• Chris September 19, 2011 at 3:41 pm #
This one is tricky – at the very end you have 20^10. This is an astronomical number and therefore it is difficult to realize that the sum of it’s digits does not equal 2. If you break up 20^10 to 2^10 x 10^10 you get a number that has the first four digits of 1,024 (followed by a massive string of zeroes). If you try to multiply this number times 5^3 things get even messier.
So therefore, we want to simply this massive number at the very beginning by changing 4^10 to 2^20. Then we can share 13 of those zeroes with 5^13: 2^7 x (2^13 x 5^13), which give us 2^7 x 10^13. Now all we have to do is multiply out 2^7 getting 128. 10^13 gives us 13 zeroes, which we can discount when finding the sum of the digits. So we have 1+2+8 = 11.
This is a tough problem that only about 20% of test takers would be able to answer in less than 3 minutes.
• Albert October 19, 2013 at 2:40 pm #
The problem is that 20^10 is not 2 and ten zeroes (as if it was 10^10), it’s 1024 and ten zeroes.
But still, your method looks alright and the sum would be 1+2+4+1+2+5=15, not 11, which puzzles me…
• Chris Lele October 21, 2013 at 1:00 pm #
Hi Albert,
The thing is if you just look at 20^10, then, yes, you get 1024 followed by 10 zeroes. But you have to multiply that result by the remaining 5^3. Don’t forget that 4^10 x 5^13 equals 20^10 x 5^3. Let me know if that makes sense :).
• Akhil June 15, 2014 at 3:25 am #
20^10 x 5^3 can be simplified to 2^10 x 10^10 x 5^3 right?
So that is 1024 x 10^10 x 5^3
Which is 1024 x 625 x 10^10
Which is 640000 x 10^10
So the answer is 6+4 = 10 right??
So where did I go wrong?
• Chris Lele June 16, 2014 at 1:28 pm #
Akhil,
It looks like you made a slight mistake in the second line, where you translate 5^3 as 625.
Hope that helps!
• Akhil June 17, 2014 at 12:50 am #
Thanks Chris. Silly Me!
• Chris Lele June 17, 2014 at 1:55 pm #
No prob 🙂
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11 months ago
# CLASS_11_MATHS_SOLUTIONS_NCERT
## Class XI Chapter 13 –
Class XI Chapter 13 – Limits and Derivatives Maths ______________________________________________________________________________ pr qr pr ps 2 x x x qr ps 2 x Question 4: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed nonzero constants and m and n are integers): (ax + b) (cx + d) 2 Solution 4: Let f ' x ax b cx d 2 By Leibnitz product rule, d 2 2 d f ' x ax b cx d cx d ax b dx dx d 2 2 2 2 d ax b c x 2cdx d cx d ax b dx dx d 2 2 d d 2 2 d d ax b c x 2cdx d cx d ax b dx dx dx dx dx 2 2 2 ax b c x cd cx d a 2c ax b cx d a cx d 2 2 Question 5: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non zero constants and m and n are integers): Solution 5: Let f(x) = ax b cx d By quotient rule, d f ' dx x ax b cx d cx d ax bax b cx d cx d 2 d dx Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class XI Chapter 13 – Limits and Derivatives Maths ______________________________________________________________________________ cx daax dc 2 cx d acx ad acx bc 2 cx d ad bc cx d 2 Question 6: Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s 1 1 are fixed nonzero constants and m and n are integers): x 1 1 x Solution 6: 1 x 1 1 1 Let x x x f x ,where x 0 1 x 1 1 x 1 x x By quotient rule, d d x 1 x 1 x 1 x 1 f ' x dx dx , x 0,1 2 x-1 x11 x11 2 x-1 = , x 0,1 x1 x1 2 , x 0,1 x-1 2 = , x 0,1 2 x-1 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
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#### Resources tagged with Area similar to Fence It:
Filter by: Content type:
Stage:
Challenge level:
### There are 83 results
Broad Topics > Measures and Mensuration > Area
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Dissect
##### Stage: 3 Challenge Level:
It is possible to dissect any square into smaller squares. What is the minimum number of squares a 13 by 13 square can be dissected into?
### Framed
##### Stage: 3 Challenge Level:
Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Isosceles Triangles
##### Stage: 3 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Square Areas
##### Stage: 3 Challenge Level:
Can you work out the area of the inner square and give an explanation of how you did it?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### Take Ten
##### Stage: 3 Challenge Level:
Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube made from 27 unit cubes so that the surface area of the remaining solid is the same as the surface area of the original 3 by 3 by 3. . . .
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Hallway Borders
##### Stage: 3 Challenge Level:
A hallway floor is tiled and each tile is one foot square. Given that the number of tiles around the perimeter is EXACTLY half the total number of tiles, find the possible dimensions of the hallway.
### Rati-o
##### Stage: 3 Challenge Level:
Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle?
### Cylinder Cutting
##### Stage: 2 and 3 Challenge Level:
An activity for high-attaining learners which involves making a new cylinder from a cardboard tube.
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Inscribed in a Circle
##### Stage: 3 Challenge Level:
The area of a square inscribed in a circle with a unit radius is, satisfyingly, 2. What is the area of a regular hexagon inscribed in a circle with a unit radius?
### Lying and Cheating
##### Stage: 3 Challenge Level:
Follow the instructions and you can take a rectangle, cut it into 4 pieces, discard two small triangles, put together the remaining two pieces and end up with a rectangle the same size. Try it!
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Tiling Into Slanted Rectangles
##### Stage: 2 and 3 Challenge Level:
A follow-up activity to Tiles in the Garden.
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Extending Great Squares
##### Stage: 2 and 3 Challenge Level:
Explore one of these five pictures.
### Areas of Parallelograms
##### Stage: 4 Challenge Level:
Can you find the area of a parallelogram defined by two vectors?
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Great Squares
##### Stage: 2 and 3 Challenge Level:
Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
### Gutter
##### Stage: 4 Challenge Level:
Manufacturers need to minimise the amount of material used to make their product. What is the best cross-section for a gutter?
### Towers
##### Stage: 3 Challenge Level:
A tower of squares is built inside a right angled isosceles triangle. The largest square stands on the hypotenuse. What fraction of the area of the triangle is covered by the series of squares?
### Exploration Versus Calculation
##### Stage: 1, 2 and 3
This article, written for teachers, discusses the merits of different kinds of resources: those which involve exploration and those which centre on calculation.
### Trapezium Four
##### Stage: 4 Challenge Level:
The diagonals of a trapezium divide it into four parts. Can you create a trapezium where three of those parts are equal in area?
### Isosceles
##### Stage: 3 Challenge Level:
Prove that a triangle with sides of length 5, 5 and 6 has the same area as a triangle with sides of length 5, 5 and 8. Find other pairs of non-congruent isosceles triangles which have equal areas.
### Kite
##### Stage: 3 Challenge Level:
Derive a formula for finding the area of any kite.
### Maths Filler
##### Stage: 4 Challenge Level:
Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?
### Carpet Cuts
##### Stage: 3 Challenge Level:
You have a 12 by 9 foot carpet with an 8 by 1 foot hole exactly in the middle. Cut the carpet into two pieces to make a 10 by 10 foot square carpet.
### Making Rectangles
##### Stage: 2 and 3 Challenge Level:
A task which depends on members of the group noticing the needs of others and responding.
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
### Growing Rectangles
##### Stage: 3 Challenge Level:
What happens to the area and volume of 2D and 3D shapes when you enlarge them?
### Areas and Ratios
##### Stage: 4 Challenge Level:
What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Appearing Square
##### Stage: 3 Challenge Level:
Make an eight by eight square, the layout is the same as a chessboard. You can print out and use the square below. What is the area of the square? Divide the square in the way shown by the red dashed. . . .
### Kissing Triangles
##### Stage: 3 Challenge Level:
Determine the total shaded area of the 'kissing triangles'. | 2,190 | 9,352 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2016-40 | longest | en | 0.886942 |
https://zooquizzes.com/quiz/only-geniuses-can-pass-this-logic-quiz-can-you | 1,580,206,203,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778168.77/warc/CC-MAIN-20200128091916-20200128121916-00551.warc.gz | 1,155,092,631 | 41,952 | # Only Geniuses Can Pass This Logic Quiz. Can You?
EDUCATION
Olivia Cantor
6 Min Quiz
Image: shutterstock
Are you great at logic? This quiz can help determine how logical your thought process is and how your brain handles a challenge. Yes, it's also a game for the geniuses among us -- and that might be you.
Logic has been defined as "the science that investigates the principles governing correct or reliable inference." That means a highly logical person can see the truth of things when presented with a scenario or challenged with a situation. That's why logic tests are often formulated in storytelling fashion. In this one, we'll provide the premise of the story, leave a question hanging, and then it's up to you to figure out the answer -- the next step in the story.
Another definition of logic is "a particular method of reasoning or argumentation." While it closely resembles the storytelling mode of some logic quizzes, this is where a more "linear" or straightforward type of questioning occurs. So here, you'll also see questions that relate to math or allude to numbers, as well as abstract reasoning.
Whether it's a combination of inference, argument, or reasoning, this quiz tests multiple skills to discover if your logical reasoning is on par with geniuses. Are you up for the challenge? Let's find out!
# On my way to the cinema house, I saw Hannah and her 3 sisters, the 12 brothers Karamazov, and the 7 brides for the 7 brothers. How many people were going to the cinema house?
The answer is 1. Since it was only the speaker "I" who was going to the cinema house, then it's just 1!
# Jane’s father has 4 daughters. The eldest child is Joy, the second's name is Happy, and the third's name is Glee. What is the name of the youngest child?
The youngest should be Jane. Since her father has 4 daughters, it's only her name that's not mentioned in the "roll call."
# Hidden inside the name Tom Marvolo Riddle is an anagram of which Harry Potter villain?
An anagram forms a new word/phrase when you rearrange it. Therefore, "I am Lord Voldemort" is hidden within that anagram of a name of Tom Marvolo Riddle.
# All spiders are poisonous. A tarantula is a spider. Therefore, ___?
If we follow the syllogism of this question, the answer would be "A tarantula is poisonous." It's the valid conclusion proposed from the two valid premises given.
# Fill in the blank: 1C 2D 3E 4F __
Follow the numbers and you get 1-4 while you get C-F when you follow the letters. Therefore, the next number is 5 and the next letter is G -- 5G.
# "Fork" is to "road" as "silver lining" is to "___"
"A fork in the road" is an idiomatic expression. "Every cloud has a silver lining" is also an idiomatic expression
# Fill in the blank: May the __ be with you!
May the Force be with you! No amount of confusing homonyms would steer a Star Wars fan away from that quote!
# Two cars crash along the border of North and South Korea. Where should they bury the survivors?
When a car crash has survivors, that means those people are still alive and should most definitely not be buried anywhere.
# Father = Male. Mother = Female. Niece = Female. Nephew = __?
A nephew is the male counterpart of niece.
# Fill in the number that comes next in this series: 11, 22, 44, 55, __, 88
It's 77. See how 33 is omitted, so 66 will be omitted as well, before reaching 88. Or you could notice the pattern seems to be alternating between numbers that are 11 and 22 apart.
# Chris is your mother’s child. But Chris is not your brother. What is Chris’ relation to you?
Chris is not an uncommon female nickname. It could be short for Christina, Christine, or Christy, for example.
# What word am I? My 1st letter is the same as that of Harry Potter’s female best friend's name; my 2nd letter is the 1st letter of the name for the U.S. president’s office in the West Wing; my 3rd letter is the 13th letter of the alphabet; and my 4th letter is the most common letter in the English language (and this paragraph).
Home is where the heart is. It's also the answer to this question.
# From a dozen roses, you take 3 out of 12. How many roses do you now have?
You now have 3 roses at hand, because that's what you took!
# What is the palindrome of "live evil"?
A palindrome is a word or phrase that reads just the same if you flip it from end to beginning. Therefore, the palindrome is the same forwards or backwards -- live evil.
# Which town building has the most stories - the mayor’s office, the bank, the school, or the town library?
The library should have the most number of stories in a given town. We're talking of stories found in books, both fiction and nonfiction, not the number of floors of a building.
# Which is heavier, a pound of wet leaves or a pound of dry branches?
A pound is a pound is a pound. Whatever it is measuring, it's the same weight if it's just the same unit of measurement!
# Upon God’s orders in the Bible, how many pairs of 4-legged land animals did Abraham take on the ark?
Noah, not Abraham, was said to have built the ark and loaded it up with animals. Therefore Abraham took zero animals.
# What number comes next in the sequence: 12, 13, 15, 17, 21, __
If you take prime numbers starting from 2 and add 10, you get the numbers in the sequence. Therefore, the last prime number here is 17, which becomes 27 if you add 10.
# Martha’s pink house is pink while Cheryl’s blue house is blue. What is the color of Mark’s greenhouse?
A greenhouse is made of glass, therefore it is transparent and has no color. It's not like a real house, if you know your gardening science.
# How many eggs can a rooster lay in one day?
A rooster doesn't lay eggs at all. It's the hen that does that.
# January, May, October and December each have 31 days, while June, September and November all have 30 days. Which months have 28 days?
All months have 28 days. Only February doesn't exceed this, but the others reach either 30 or 31.
# If it rains outside, then the ground will be wet. The ground is wet. Therefore...
Based only on the premises, it's impossible to know whether or not rain is the cause of the wet ground. Sometimes human intuition tells us that this argument is valid (sometimes written as "If A, then B. / B [is true], therefore A [is true]"), but it's invalid. The correct logic is "If A, then B. A is true, therefore B must also be true."
# If you were born in a leap year, how many birthdays could you have?
Everyone only has 1 birthday, regardless of when you were born. Yep, we're only born once!
# Tony the librarian has 12 rare print books. All but 3 burned in a fire. How many were left?
All but 3 books burned. Therefore, only those 3 were left!
# No dogs have scales. Archie is a dog. Therefore, __.
"Archie doesn’t have scales" should be the logical conclusion to this argumentation. Yep, it's logical!
# If you take away 1 from 19, you will have 20 left. How can this be logical?
If you use Roman numerals, this statement is logical. In Roman numerals, you write 19 as XIX, and if you take away 1, or I, then you're left with XX, which is the Roman number for 20.
# Say silk silk silk silk silk! Now tell me: What does a cow drink?
The "silk" portion should redirect your brain to say "milk." But what a cow drinks is water.
# Reese Witherspoon's is quite long, but Megan Fox's is quite short. Meanwhile Madonna doesn't seem to have one, officially, while performing. What is it?
The question pertains to the surnames of the ladies. Officially, Madonna doesn't really use her last name, but of course legally she has one!
# Which animal doesn’t belong in the series: Fish, frog, turtle, pigeon
A pigeon is warm-blooded, which makes it the odd one out on the list full of coldblooded creatures.
# Mary Poppins sings a happy song with a very long title. How do you spell it?
Did you miss "it"?
# Johnny was standing outside his house when it suddenly rained, but his hair didn't get wet. He wasn't carrying an umbrella, and he didn't have a hat, too. How can this be logical?
Johnny is bald. That's why there's no hair to get drench in the rain.
# Fill in the blank: Go __ and multiply.
Forth should be the reply. Always be careful of homonyms or words that sound the same -- or almost!
# If a = 1 and b = 5, what is the answer to a + b - ab?
In mathematical notation, if you put two variables next to each other, they're multiplied. So "ab" would equal 5, and 6 minus 5 equals 1.
# Sheila's parents have 5 children: Sam, Sem, Sim, Som, and __. What's the name of the 5th child?
Sheila should be the 5th child. The other 4 are her siblings.
# If silver : spoon, then __ : hot.
Being born with a silver spoon and being red hot are two examples of hyperbole. It could also be symbolism, depending on usage. | 2,165 | 8,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-05 | longest | en | 0.960254 |
https://en.academic.ru/dic.nsf/enwiki/994171 | 1,585,559,737,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496901.28/warc/CC-MAIN-20200330085157-20200330115157-00485.warc.gz | 467,732,466 | 17,704 | # Principal curvature
Principal curvature
Saddle surface with normal planes in directions of principal curvatures
In differential geometry, the two principal curvatures at a given point of a surface are the eigenvalues of the shape operator at the point. They measure how the surface bends by different amounts in different directions at that point.
## Discussion
At each point p of a differentiable surface in 3-dimensional Euclidean space one may choose a unit normal vector. A normal plane at p is one that contains the normal, and will therefore also contain a unique direction tangent to the surface and cut the surface in a plane curve. This curve will in general have different curvatures for different normal planes at p. The principal curvatures at p, denoted k1 and k2, are the maximum and minimum values of this curvature.
Here the curvature of a curve is by definition the reciprocal of the radius of the osculating circle. The curvature is taken to be positive if the curve turns in the same direction as the surface's chosen normal, and otherwise negative. The directions of the normal plane where the curvature takes its maximum and minimum values are always perpendicular, a result of Euler (1760), and are called principal directions. From a modern perspective, this theorem follows from the spectral theorem because they can be given as the principal axes of a symmetric tensor—the second fundamental form. A systematic analysis of the principal curvatures and principal directions was undertaken by Gaston Darboux, using Darboux frames.
The product k1k2 of the two principal curvatures is the Gaussian curvature, K, and the average (k1+k2)/2 is the mean curvature, H.
If at least one of the principal curvatures is zero at every point, then the Gaussian curvature will be 0 and the surface is a developable surface. For a minimal surface, the mean curvature is zero at every point.
## Formal definition
Let M be a surface in Euclidean space with second fundamental form II(X,Y). Fix a point pM, and an orthonormal basis X1, X2 of tangent vectors at p. Then the principal curvatures are the eigenvalues of the symmetric matrix
$\left[I\!I_{ij}\right] = \begin{bmatrix} I\!I(X_1,X_1)&I\!I(X_1,X_2)\\ I\!I(X_2,X_1)&I\!I(X_2,X_2) \end{bmatrix}.$
If X1 and X2 are selected so that the matrix [IIij] is a diagonal matrix, then they are called the principal directions. If the surface is oriented, then one often requires that the pair (X1, X2) to be positively oriented with respect to the given orientation.
Without reference to a particular orthonormal basis, the principal curvatures are the eigenvalues of the shape operator, and the principal directions are its eigenvectors.
### Generalizations
For hypersurfaces in higher dimensional Euclidean spaces, the principal curvatures may be defined in a directly analogous fashion. The principal curvatures are the eigenvalues of the matrix of the second fundamental form II(Xi,Xj) in an orthonormal basis of the tangent space. The principal directions are the corresponding eigenvectors.
Similarly, if M is a hypersurface in a Riemannian manifold N, then the principal curvatures are the eigenvalues of its second-fundamental form. If k1, ..., kn are the n principal curvatures at a point pM and X1, ..., Xn are corresponding orthonormal eigenvectors (principal directions), then the sectional curvature of M at p is given by
K(Xi,Xj) = kikj.
## Classification of points on a surface
• At elliptical points, both principal curvatures have the same sign, and the surface is locally convex.
• At umbilic points, both principal curvatures are equal and every tangent vector can be considered a principal direction. These typically occur in isolated points.
• At hyperbolic points, the principal curvatures have opposite signs, and the surface will be locally saddle shaped.
• At parabolic points, one of the principal curvatures is zero. Parabolic points generally lie in a curve separating elliptical and hyperbolic regions.
• At flat umbilic points both principal curvatures are zero. A generic surface will not contain flat umbilic points. The Monkey saddle is one surface with an isolated flat umbilic.
## Lines of curvature
The lines of curvature or curvature lines are curves which are always tangent to a principal direction (they are integral curves for the principal direction fields). There will be two lines of curvature through each non-umbilic point and the lines will cross at right angles.
In the vicinity of an umbilic the lines of curvature form one of three configurations star, lemon and monstar (derived from lemon-star)[1]. These points are also called Darbouxian Umbilics, in honor to Gaston Darboux, the first to make a systematic study in Vol. 4, p455, of his Leçons (1896).
In these figures, the red curves are the lines of curvature for one family of principal directions, and the blue curves for the other.
When a line of curvature has a local extremum of the same principal curvature then the curve has a ridge point. These ridge points form curves on the surface called ridges. The ridge curves pass through the umbilics. For the star pattern either 3 or 1 ridge line pass through the umbilic, for the monstar and lemon only one ridge passes through.[2]
## References
• Darboux, Gaston (1887,1889,1896). Leçons sur la théorie génerale des surfaces: Volume I, Volume II, Volume III, Volume IV. Gauthier-Villars.
• Guggenheimer, Heinrich (1977). "Chapter 10. Surfaces". Differential Geometry. Dover. ISBN 0-486-63433-7.
• Kobayashi, Shoshichi and Nomizu, Katsumi (1996 (New edition)). Foundations of Differential Geometry, Vol. 2. Wiley-Interscience. ISBN 0471157325.
• Spivak, Michael (1999). A Comprehensive introduction to differential geometry (Volume 3). Publish or Perish. ISBN 0-914098-72-1.
1. ^ Berry, M V, & Hannay, J H, 'Umbilic points on Gaussian random surfaces', J.Phys.A 10, 1977, 1809-21, .
2. ^ Porteous, I. R. (1994). Geometric Differentiation. Cambridge University Press. ISBN 0-521-39063-X
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# Among the more effective kinds of publicity
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22 Jul 2013, 19:06
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Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher.
Which of the following conclusions is best supported by the information above?
A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book.
B) Because the financial advantage of excerpting a new book in a magazine usually accrues to the book's publisher, magazine editors are unwilling to publish excerpts from new books.
C) In calculating the total number of copies that a book has sold, publishers include sales of copies of magazines that featured an excerpt of the book.
D) The effectiveness of having excerpts of a book published in a magazine, measured in terms of increased sales of a book, is proportional to the circulation of the magazine in which the excerpts are published.
E) Books that are suitable for excerpting in high-circulation magazines sell more copies than book that are not suitable for excerpting.
[Reveal] Spoiler: OA
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Re: Among the more effective kinds of publicity [#permalink]
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22 Jul 2013, 19:11
I agree that A is the most relevant one among the 5 options.
But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage.
Any explanation would be appreciated!!
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Re: Among the more effective kinds of publicity [#permalink]
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22 Jul 2013, 23:00
joonhy wrote:
I agree that A is the most relevant one among the 5 options.
But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage.
Any explanation would be appreciated!!
Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published.
The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher.
Which of the following conclusions is best supported by the information above?
The benefits include increase in sales
A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book.==> What does this mean?
If the number of people stimulated to buy the book is greater compared to the number of people discouraged to buy the book (because consider "excerpt of a book an adequate substitute for reading") the sales will increase.
This is supported by the passage, because correspond to the one of the benefits of the publicity.
Hope it helps.
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Re: Among the more effective kinds of publicity [#permalink]
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22 Jul 2013, 23:29
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joonhy wrote:
I agree that A is the most relevant one among the 5 options.
But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage.
Any explanation would be appreciated!!
Actually (A) is an inference from the passage. You can infer/conclude it if what is given in the argument is correct.
The author says that publishing an excerpt is effective. Sales increases and you get a fee.
Since we are given that sales increases, this means that more people buy the book if you publish an excerpt. So even if there are people who do not buy the book after publishing an excerpt because the excerpt is enough for them, there are more people who buy the book because they get tantalized by the excerpt. Only then will the sales rise when you publish an excerpt.
I would like to point out that there is an assumption here in (A): The assumption is that the people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book would have bought the book had they not read the excerpt. But since no other option comes close to being a conclusion, (A) is the best choice.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 14 Jul 2013 Posts: 3 Followers: 0 Kudos [?]: 13 [0], given: 2 Re: Among the more effective kinds of publicity [#permalink] ### Show Tags 24 Jul 2013, 19:50 Thanks a lot! I think I was too focused on the term "conclusion". Verbal Forum Moderator Joined: 16 Jun 2012 Posts: 1132 Location: United States Followers: 278 Kudos [?]: 3112 [2] , given: 123 Re: Among the more effective kinds of publicity [#permalink] ### Show Tags 25 Jul 2013, 00:35 2 This post received KUDOS joonhy wrote: Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher. Which of the following conclusions is best supported by the information above? A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book. B) Because the financial advantage of excerpting a new book in a magazine usually accrues to the book's publisher, magazine editors are unwilling to publish excerpts from new books. C) In calculating the total number of copies that a book has sold, publishers include sales of copies of magazines that featured an excerpt of the book. D) The effectiveness of having excerpts of a book published in a magazine, measured in terms of increased sales of a book, is proportional to the circulation of the magazine in which the excerpts are published. E) Books that are suitable for excerpting in high-circulation magazines sell more copies than book that are not suitable for excerpting. ANALYZE THE STIMULUS: Fact: Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. Fact: The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher. ANALYZE EACH ANSWER: A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book. Correct. KEY words in the stimulus are “increase in sales”. ==> when people read the excerpts, they desire to read a whole book ==> They will buy the book ==> sales increase. B) Because the financial advantage of excerpting a new book in a magazine usually accrues to the book's publisher, magazine editors are unwilling to publish excerpts from new books. Wrong. Totally wrong, the stimulus does not say “magazine editor unwilling to publish excerpts”. C) In calculating the total number of copies that a book has sold, publishers include sales of copies of magazines that featured an excerpt of the book. Wrong. Nothing about the formula to calculate the total number of copies sold. D) The effectiveness of having excerpts of a book published in a magazine, measured in terms of increased sales of a book, is proportional to the circulation of the magazine in which the excerpts are published. Wrong. Nothing about the relationship “effectiveness of having excerpts is proportional to the circulation of the magazine”. E) Books that are suitable for excerpting in high-circulation magazines sell more copies than book that are not suitable for excerpting. Wrong. TEMPTING. Even though the books that are suitable for excerpting in high-circulation magazine can increase sales, the stimulus DOES NOT say they sell more copies than other books. Hope it helps. _________________ Please +1 KUDO if my post helps. Thank you. "Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong." Chris Bangle - Former BMW Chief of Design. Intern Joined: 11 Aug 2013 Posts: 5 Concentration: Finance, Marketing GMAT Date: 09-12-2013 Followers: 0 Kudos [?]: 1 [0], given: 14 Re: Among the more effective kinds of publicity [#permalink] ### Show Tags 12 Aug 2013, 12:27 Zarrolou wrote: joonhy wrote: I agree that A is the most relevant one among the 5 options. But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage. Any explanation would be appreciated!! Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher. Which of the following conclusions is best supported by the information above? The benefits include increase in sales A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book.==> What does this mean? If the number of people stimulated to buy the book is greater compared to the number of people discouraged to buy the book (because consider "excerpt of a book an adequate substitute for reading") the sales will increase. This is supported by the passage, because correspond to the one of the benefits of the publicity. Hope it helps. i m confused between A and E.............. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7376 Location: Pune, India Followers: 2288 Kudos [?]: 15127 [1] , given: 224 Re: Among the more effective kinds of publicity [#permalink] ### Show Tags 12 Aug 2013, 22:37 1 This post received KUDOS Expert's post anilvb wrote: Zarrolou wrote: joonhy wrote: I agree that A is the most relevant one among the 5 options. But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage. Any explanation would be appreciated!! Among the more effective kinds of publicity that publishers can get for a new book is to have excerpts of it published in a high-circulation magazine soon before the book is published. The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher. Which of the following conclusions is best supported by the information above? The benefits include increase in sales A) The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book is smaller than the number for whom the excerpt stimulates a desire to read the book.==> What does this mean? If the number of people stimulated to buy the book is greater compared to the number of people discouraged to buy the book (because consider "excerpt of a book an adequate substitute for reading") the sales will increase. This is supported by the passage, because correspond to the one of the benefits of the publicity. Hope it helps. i m confused between A and E.............. The reason (E) is not correct: (E) Books that are suitable for excerpting in high-circulation magazines sell more copies than book that are not suitable for excerpting. - Incorrect The argument tells you that actual excerption (not excerption suitability) increases sales. Also the comparison "higher" is of the sales of the same book "after excerpt" with "before excerpt". The comparison is not with books whose excerpt is not published or which are not suitable for excerption. Say, a J K Rowling book which is not suitable for excerption will have higher sales than most other books even after they publish excerpts wildly. Note that the argument only says that if you publish an excerpt, the sales will be higher. Just being suitable for excerption does not increase sales. The point is that books suitable for excerpting COULD sell more copies (compared to what they will sell if they don't publish excerpts). _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Among the more effective kinds of publicity [#permalink]
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20 Jul 2014, 22:18
VeritasPrepKarishma wrote:
joonhy wrote:
I agree that A is the most relevant one among the 5 options.
But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage.
Any explanation would be appreciated!!
Actually (A) is an inference from the passage. You can infer/conclude it if what is given in the argument is correct.
The author says that publishing an excerpt is effective. Sales increases and you get a fee.
Since we are given that sales increases, this means that more people buy the book if you publish an excerpt. So even if there are people who do not buy the book after publishing an excerpt because the excerpt is enough for them, there are more people who buy the book because they get tantalized by the excerpt. Only then will the sales rise when you publish an excerpt.
I would like to point out that there is an assumption here in (A): The assumption is that the people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book would have bought the book had they not read the excerpt. But since no other option comes close to being a conclusion, (A) is the best choice.
Thanks for pointing it out. I was stuck at the same point. My analysis was that let's say 10 people are wiling to buy the book in the beginning, and the excerpt is released to a magazine that is read by 100 people. 60 people are discouraged to buy and 40 are encouraged --> this fact refutes option A but still increases the sale.
My question here is, what if questions like these (with an underlying assumption not so evident) appear in GMAT? is there a way around or do we just rely on our instinct?
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Re: Among the more effective kinds of publicity [#permalink]
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21 Jul 2014, 03:05
Karishma , although you have explained before but still I fail to understand how come A which is clearly an assumption for the argument be the conclusion ? I would have expected D to be a logical conclusion which relates the circulation of the magazine in which article is appearing to the increased sales. Extrapolating this relation to the logical extremities , if the circulation of magazine is zero ( i.e. no excerpts published in any magazine) , then no increased sale ( i.e. the usual default sale). On the other hand say if the circulation of magazine is high , then highly increased sale.
Clearly the ASSUMPTION over here is that people will be interested to buy the book after reading the excerpt (pt A) . So for e.g. if its a suspense book and the excerpts give away some pointers to the climax , not many people might be interested to buy the book
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Re: Among the more effective kinds of publicity [#permalink]
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21 Jul 2014, 18:13
I went for D as well. Option A does not talk about circulation. How do we know whether the question is asking about the desirability of the book or the effectiveness of circulation?
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Re: Among the more effective kinds of publicity [#permalink]
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21 Jul 2014, 22:52
gauravkaushik8591 wrote:
Thanks for pointing it out. I was stuck at the same point. My analysis was that let's say 10 people are wiling to buy the book in the beginning, and the excerpt is released to a magazine that is read by 100 people. 60 people are discouraged to buy and 40 are encouraged --> this fact refutes option A but still increases the sale.
My question here is, what if questions like these (with an underlying assumption not so evident) appear in GMAT? is there a way around or do we just rely on our instinct?
The number analysis is not correct. If 10 people were willing to buy the book, the excerpt cannot discourage 60 people since they were not interested in buying in the first place. Only 10 were interested in the first place!
This group - "The number of people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book" has to come from the people who were interested in buying the book in the first place.
The number analysis would be something like this: If 100 people were interested in buying the book initially and then the excerpt was released, 30 of those 100 people were not interested in buying anymore because the excerpt was a good enough substitute for them. But the excerpt has to bring in another '31 or more' people who were not interested initially but are interested now in buying the book. That is when the sales will increase due to the excerpt.
Now, I hope you see that the underlying assumption is clear.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7376 Location: Pune, India Followers: 2288 Kudos [?]: 15127 [0], given: 224 Re: Among the more effective kinds of publicity [#permalink] ### Show Tags 21 Jul 2014, 23:04 himanshujovi wrote: Karishma , although you have explained before but still I fail to understand how come A which is clearly an assumption for the argument be the conclusion ? I would have expected D to be a logical conclusion which relates the circulation of the magazine in which article is appearing to the increased sales. Extrapolating this relation to the logical extremities , if the circulation of magazine is zero ( i.e. no excerpts published in any magazine) , then no increased sale ( i.e. the usual default sale). On the other hand say if the circulation of magazine is high , then highly increased sale. Clearly the ASSUMPTION over here is that people will be interested to buy the book after reading the excerpt (pt A) . So for e.g. if its a suspense book and the excerpts give away some pointers to the climax , not many people might be interested to buy the book (A) is not an assumption. It is a clear inference. You are given "The benefits of such excerption include not only a sure increase in sales" The sure increase in sales is given as a fact. It is not the author's opinion. From this fact, you can say that definitely number of people getting encouraged to buy the book is more than the number of people getting discouraged. So from the given data in the argument, you can INFER option (A). When you have data given and you can infer something from it, it is called an inference/conclusion. When you have author's opinion (conclusion of the argument) and you need something to be true for the opinion to hold, that is an assumption. For Example: Argument 1 All A are B. All B are C. You can conclude that: All A are C. This must be true. Argument 2 All A are B. All B are C. If you conclude that 'All C are A' (your opinion, not necessarily a fact), you are assuming that A, B and C overlap. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Among the more effective kinds of publicity [#permalink]
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16 Sep 2014, 13:32
I eliminated choice A just because of the underlying assumption that some one who is tantalized by the excerpt in the book will be motivated enough to buy the book. Generally in Inference Questions there are no assumptions affecting the validity of an answer choice, but this official questions serves as an exception I believe.
Experts, any opinion?
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Re: Among the more effective kinds of publicity [#permalink]
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26 Oct 2014, 10:22
Logical answer is A. If more people who see the excerpt then want to buy the actual book, then posting the excerpt in high circulation magazines is indeed a good idea.
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Re: Among the more effective kinds of publicity [#permalink]
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13 Dec 2015, 16:29
VeritasPrepKarishma wrote:
joonhy wrote:
I agree that A is the most relevant one among the 5 options.
But, I still don't understand the relationship between the passage and A because A seems like assumption or evidence to support the passage, not the conclusion supported by the passage.
Any explanation would be appreciated!!
Actually (A) is an inference from the passage. You can infer/conclude it if what is given in the argument is correct.
The author says that publishing an excerpt is effective. Sales increases and you get a fee.
Since we are given that sales increases, this means that more people buy the book if you publish an excerpt. So even if there are people who do not buy the book after publishing an excerpt because the excerpt is enough for them, there are more people who buy the book because they get tantalized by the excerpt. Only then will the sales rise when you publish an excerpt.
I would like to point out that there is an assumption here in (A): The assumption is that the people for whom seeing an excerpt of a book in a magazine provides an adequate substitute for reading the whole book would have bought the book had they not read the excerpt. But since no other option comes close to being a conclusion, (A) is the best choice.
http://www.beatthegmat.com/effective-ki ... 32609.html
This is a must be true question.. i.e assumption question..
going through the explaination in link above.. i understood that opt A is MUST BE TRUE.. however, at first i thought of this as conclusion question and was looking for restatement of conclusion mentioned in the argument
I figured out below as conclusion
Conclusion - The benefits of such excerption include not only a sure increase in sales but also a fee paid by the magazine to the book's publisher.
and hence felt opt D close the conclusion..
i.e. opt D somewhere specifies excerption in magazines increases sales of the book..
however, i understood how D falls apart.. from the explaination in that link..
my confusion is
1. how to get must be true/conclusion question when it is clearly specified in the stimulus as 'conclusion'
2. if i have though of stimulus as conclusion question is my pick as D wrong?? because i feld D was closest, though was confused btw A and D..
Re: Among the more effective kinds of publicity [#permalink] 13 Dec 2015, 16:29
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### A brief talk on how to start from Q_ Learning to dqn
DRL（Deep Reinforcement Learning） For the first time , Should be DeepMind stay 2013 It was first applied to Atari In the game DQN（Deep Q Network） Algorithm . In today's （2017 year ）,DRL It's still one of the most cutting-edge areas of research . But in just four years ,DRL Has been playing from Atari, Evolved into go （Alphago）、 Playing video games （Dota AI、StarCraft AI）, Refresh your three views again and again .
### 1. What is? Q-Learning
Q-Learning The algorithm is a method to solve the reinforcement learning control problem by using time series difference . Through the current state $S$, action $A$, Instant rewards $R$, Attenuation factor $γ$, Exploration rate $ϵ$, The best action value function $Q$ And the most strategic $π$.
• $S$： Represents the state of the environment , stay $t$ The state of the environment $S_t$
• $A$：agent The action of , stay $t$ Actions taken all the time $A_t$
• $R$： Environmental rewards , stay $t$ moment agent In state St Take action $A_t$ The corresponding reward $R_{t+1}$ Will be in $t+1$ Always get
• $\gamma$： The discount factor , The weight of the current delay Award
• $\epsilon$： Exploration rate , stay Q-learning We will select the most valuable action in the current iteration , It may lead to some actions that have never been performed again , stay agent When choosing an action , There is a small probability that it is not to select the most valuable action in the current iteration .
#### 1.1 Q-Learning Introduction to the algorithm
First, we're based on state $S$, use $ϵ−greed$( greedy ) Select to action $A$, And then perform the action $A$, Get a reward $R$, And get into the State $S'$, $Q$ The update formula of the value is as follows ：
$Q(S,A)=Q(S,A)+\alpha(R+\gamma maxQ(S',a)-Q(S,A))$
#### 1.2 Q-learning Algorithm flow of
• The value corresponding to the random initialization state and action value .（ initialization $Q$ form ）
• for i from 1 to TT： The total number of iterations ）
a） initialization $S$ Is the first state in the sequence of the current state
b） use $ϵ$− Greedy law in the current state $S$ Choose the action $A$
c） In state $S$ Perform the current action $A$, Get a new state $S′$ And rewards $R$
d） Update value function $Q(S,A)$： $Q(S,A)=Q(S,A)+\alpha(R+\gamma maxQ(S',a)-Q(S,A))$
e） $S=S'$
f） if $done$ Complete the current iteration
#### 1.3 About Q_table for instance
（1） Game map
• The black frame is a trap
• The yellow box is the exit （ Reward points ）
（2） This is after a training model Q form
（3） A simple example
• If agent stay “1” Enter the maze from the right position , It will be better Q form , Going down Q The maximum value is 0.59, that agent It's going to be “5” The location of .
• agent stay “5” After the position of , More Q form , Going down Q The maximum value is 0.66, Still going down , that agent It's time to “9” The location of .
• agent stay “9” After the position of , More Q form , On the right Q The maximum value is 0.73, Still going down , that agent It's time to “10” The location of .
• agent stay “10” After the position of , More Q form , Going down Q The maximum value is 0.81, Still going down , that agent It's time to “14” The location of .
• agent stay “14” After the position of , More Q form , On the right Q The maximum value is 0.9, Still going down , that agent It's time to “15” The location of .
• agent stay “15” After the position of , More Q form , On the right Q The maximum value is 1, Still going down , that agent It's time to “16” The location of , Reach the end point .
Last agent The course of action is 1-->5-->9-->10-->14-->15-->16
Every run , $Q$ The values of the table will vary change , But the principle is the same .
If you want to see more intuitive vision Poke it here
### 2. DQN(Deep Q Network)
I talked about it before. Q-Learning Our decision is based on Q The values of the table , You get more rewards for that action , Just select the action to execute . The state space and action space mentioned above are very small , If the state space and action space become very large , Then we can use one more Q Table to show ？ Obviously not , It's introduced The value function approximates .
#### 2.1 The value function approximates
Because in practical problems , The scale of a problem is very large , One possible solution is to use the value function approximation . We introduce a state value function $\hat v$, By weight $\omega$ describe , In state $s$ As input , Calculate the State $s$ The value of ： $\hat v(s,w)\approx v_\pi(s)$
As we mentioned above $\omega$ It's equivalent to the parameters in our neural network , Through the state of the input $s$, use MC( Monte Carlo )/TD( Temporal difference ) Calculate the value function as the output , And then the weight $\omega$ Training , Until it converges . in fact , So-called DQN That is to combine neural networks with Q-Learning combination , take Q The form becomes Q The Internet .
#### 2.2 Deep Q-Learning Algorithm ideas
DQN It's a kind of Off-Policy Algorithm , In the words of Mr. Li Hongyi , You can watch others learn , that DQN Why can I watch others learn ？DQN It adopts a way of experience playback to learn . Every time agent Rewards for interacting with the environment , The current state and the next state are saved , For the back Q Network updates .
So let's see Nature DQN, Actually Nature DQN by DQN The second generation ,DQN NIPS For the most primitive DQN, There's a lot more on top of that DQN Version of , such as Double DQN,Dueling DQN wait . I'm here to introduce Nature DQN Well ！ I think this version of DQN, It should be the most classic . Now let's see DQN How to carry out reinforcement learning .
#### 2.3 algorithm flow chart
Input : The total number of iterations $T$, State feature dimension $n$, Action dimension $A$, step $a$, Attenuation factor $\gamma$, Exploration rate $\epsilon$, At present Q The Internet $Q$, The goal is Q The Internet $Q'$, Sample size of batch gradient descent $m$, The goal is Q Network parameter update frequency $P$.
Output ：Q Network parameters
• Randomly initializes the values of all states and actions $Q$, Randomly initialize the current $Q$ All parameters of the network $\omega$, Initialize target Q The Internet $Q'$ Parameters of $\omega$'= $\omega$, Clear the experience pool $D$
• for i from 1 to T( Iterating on and on )
a） Initialization environment , Get the first state $s$, Get eigenvectors \phi$$(S) b） stay $Q$ Network usage \phi$$(S) As input , obtain $Q$ All the actions of the Internet Q value , Use $\epsilon$- The greedy law in the current Q Select the corresponding action in the value output $A$
c） In state $S$ Perform the action below $A$, Get a new state $S'$, And the corresponding \phi$$(S') And rewards $R$, Whether it is in the end state $isdone$ d） take { $\phi(S)$, $A$, $R$, $\phi(S')$, $isdone$ } Will this 5 Put elements into the experience pool $D$ e） $S=S'$ f） From the experience pool $D$ Adopted in $m$ Samples ,{ $\phi(S_j)$, $A_j$, $R_j$, $\phi(S'_j$ $isdone_j$), $j=1,2,3,4....m$, Calculate current $Q$ value $y_j$: g） Use the mean square error loss function \left(\frac{1}{m}\right)$$\sum_{j=1}^m( $y_j-Q(\phi (S_j),A_j,\omega))^2$ Update parameters by neural network gradient descent back propagation $\omega$
h） If i%P=0, With new goals $Q$ Network parameters $\omega'=\omega$
i） If $S'$ Is the termination state , Then the current iteration is completed , Otherwise jump to step (2)
#### 2.4 DQN Implementation code
（1） Network structure
class Net(nn.Module):
def __init__(self, ):
super(Net, self).__init__()
self.fc1 = nn.Linear(N_STATES, 50)
self.fc1.weight.data.normal_(0, 0.1) # initialization
self.out = nn.Linear(50, N_ACTIONS)
self.out.weight.data.normal_(0, 0.1) # initialization
def forward(self, x):
x = self.fc1(x)
x = F.relu(x)
actions_value = self.out(x)
return actions_value
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• Two networks Same structure .
• There are differences in parameter weights , One is Real time updates , One is After a while Updating .
（2） The choice of action
def choose_action(self, x): #x For the current state of 4 It's worth
x = torch.unsqueeze(torch.FloatTensor(x), 0) # At the end of the data 0 Add one dimension to the dimension
# input only one sample
if np.random.uniform() < EPSILON: # greedy # Greedy
actions_value = self.eval_net.forward(x) ## Pass in eval_net Get the next action
action = torch.max(actions_value, 1)[1].data.numpy() ## Returns the index of the largest value in this row
action = action[0] if ENV_A_SHAPE == 0 else action.reshape(ENV_A_SHAPE) # return the argmax index
else: # random
action = np.random.randint(0, N_ACTIONS)
# action = random.sample(N_ACTIONS)
action = action if ENV_A_SHAPE == 0 else action.reshape(ENV_A_SHAPE)
return action
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Added a Exploration value $(\epsilon)$, The small possibility is to randomly choose actions .
（3） Experience pool
def store_transition(self, s, a, r, s_): #s and s_ All for 4 It's worth , Respectively Location Movement speed angle Moving angle
transition = np.hstack((s, [a, r], s_))
# replace the old memory with new memory # Update the experience
index = self.memory_counter % MEMORY_CAPACITY
self.memory[index, :] = transition # Will be the first index Experience is replaced by transition
self.memory_counter += 1
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（4） Update network parameters
def learn(self):
# target parameter update Target parameter update
if self.learn_step_counter % TARGET_REPLACE_ITER == 0:
self.target_net.load_state_dict(self.eval_net.state_dict()) ## Every time I study 200 Step by step eval_net The parameters of are assigned to target_net
self.learn_step_counter += 1
# sample batch transitions # Select transition
sample_index = np.random.choice(MEMORY_CAPACITY, BATCH_SIZE) # from MEMORY_CAPACITY Random selection BATCH_SIZE individual
b_memory = self.memory[sample_index, :]
b_s = torch.FloatTensor(b_memory[:, :N_STATES]) # First state
b_a = torch.LongTensor(b_memory[:, N_STATES:N_STATES+1].astype(int)) # action
print("--------")
print(b_a)
print("-----")
b_r = torch.FloatTensor(b_memory[:, N_STATES+1:N_STATES+2]) # score
b_s_ = torch.FloatTensor(b_memory[:, -N_STATES:]) # Next state
# q_eval w.r.t the action in experience
q_eval = self.eval_net(b_s).gather(1, b_a) # shape (batch, 1) The current state of Q Value usage eval_net Calculation
# print("++++++")
# print(self.eval_net(b_s))
# print(self.eval_net(b_s).gather(1,b_a))
# print("+++++++")
q_next = self.target_net(b_s_).detach() # Use target_net Calculate the next step Q value # detach from graph, don't backpropagate detach prevent targent——net Back propagation
q_target = b_r + GAMMA * q_next.max(1)[0].view(BATCH_SIZE, 1) # shape (batch, 1)
loss = self.loss_func(q_eval, q_target)
loss.backward() # Back propagation
self.optimizer.step() # Perform the next optimization
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DQN It's the threshold of deep reinforcement learning , Just step into the gate , The later study will be very easy .
PS： More technical dry goods , Quick attention 【 official account | xingzhe_ai】, Discuss it with the traveler ！
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Inverse trigonometric functions
In mathematics, the inverse trigonometric functions (occasionally called cyclometric functions[1]) are the inverse functions of the trigonometric functions with suitably restricted domains .
The notations sin−1, cos−1, etc. are often used for arcsin, arccos, etc., but this convention logically conflicts with the common semantics for expressions like sin2(x), which refer to numeric power rather than function composition, and therefore may result in confusion between multiplicative inverse and compositional inverse.
In computer programming languages the functions arcsin, arccos, arctan, are usually called asin, acos, atan. Many programming languages also provide the two-argument atan2 function, which computes the arctangent of y / x given y and x, but with a range of (−π, π].
## Principal values
Since none of the six trigonometric functions are one-to-one, they must be restricted in order to have inverse functions. Therefore the ranges of the inverse fuctions are subsets of the domains of the original functions
For example, just as the square root function $y = \sqrt{x}$ is defined such that y2 = x, the function y = arcsin(x) is defined so that sin(y) = x. There are multiple numbers y such that sin(y) = x; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. It follows that the arcsine function is multivalued: arcsin(0) = 0, but also arcsin(0) = π, arcsin(0) = 2π, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain the expression arcsin(x) will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.
The principal inverses are listed in the following table.
Name Usual notation Definition Domain of x for real result Range of usual principal value
Range of usual principal value
(degrees)
arcsine y = arcsin x x = sin y −1 ≤ x ≤ 1 −π/2 ≤ y ≤ π/2 −90° ≤ y ≤ 90°
arccosine y = arccos x x = cos y −1 ≤ x ≤ 1 0 ≤ y ≤ π 0° ≤ y ≤ 180°
arctangent y = arctan x x = tan y all real numbers −π/2 < y < π/2 −90° < y < 90°
arccotangent y = arccot x x = cot y all real numbers 0 < y < π 0° < y < 180°
arcsecant y = arcsec x x = sec y x ≤ −1 or 1 ≤ x 0 ≤ y < π/2 or π/2 < y ≤ π 0° ≤ y < 90° or 90° < y ≤ 180°
arccosecant y = arccsc x x = csc y x ≤ −1 or 1 ≤ x −π/2 ≤ y < 0 or 0 < y ≤ π/2 -90° ≤ y < 0° or 0° < y ≤ 90°
If x is allowed to be a complex number, then the range of y applies only to its real part.
## Relationships among the inverse trigonometric functions
The usual principal values of the arcsin(x) (red) and arccos(x) (blue) functions graphed on the cartesian plane.
The usual principal values of the arctan(x) and arccot(x) functions graphed on the cartesian plane.
Principal values of the arcsec(x) and arccsc(x) functions graphed on the cartesian plane.
Complementary angles:
$\arccos x = \frac{\pi}{2} - \arcsin x$
$\arccot x = \frac{\pi}{2} - \arctan x$
$\arccsc x = \frac{\pi}{2} - \arcsec x$
Negative arguments:
$\arcsin (-x) = - \arcsin x \!$
$\arccos (-x) = \pi - \arccos x \!$
$\arctan (-x) = - \arctan x \!$
$\arccot (-x) = \pi - \arccot x \!$
$\arcsec (-x) = \pi - \arcsec x \!$
$\arccsc (-x) = - \arccsc x \!$
Reciprocal arguments:
$\arccos (1/x) \,= \arcsec x \,$
$\arcsin (1/x) \,= \arccsc x \,$
$\arctan (1/x) = \tfrac{1}{2}\pi - \arctan x =\arccot x,\text{ if }x > 0 \,$
$\arctan (1/x) = -\tfrac{1}{2}\pi - \arctan x = -\pi + \arccot x,\text{ if }x < 0 \,$
$\arccot (1/x) = \tfrac{1}{2}\pi - \arccot x =\arctan x,\text{ if }x > 0 \,$
$\arccot (1/x) = \tfrac{3}{2}\pi - \arccot x = \pi + \arctan x,\text{ if }x < 0 \,$
$\arcsec (1/x) = \arccos x \,$
$\arccsc (1/x) = \arcsin x \,$
If you only have a fragment of a sine table:
$\arccos x = \arcsin \sqrt{1-x^2},\text{ if }0 \leq x \leq 1$
$\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}$
Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real).
From the half-angle formula $\tan \frac{\theta}{2} = \frac{\sin \theta}{1+\cos \theta}$, we get:
$\arcsin x = 2 \arctan \frac{x}{1+\sqrt{1-x^2}}$
$\arccos x = 2 \arctan \frac{\sqrt{1-x^2}}{1+x},\text{ if }-1 < x \leq +1$
$\arctan x = 2 \arctan \frac{x}{1+\sqrt{1+x^2}}$
## Relationships between trigonometric functions and inverse trigonometric functions
$\sin (\arccos x) = \cos(\arcsin x) = \sqrt{1-x^2}$
$\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}}$
$\cos (\arctan x) = \frac{1}{\sqrt{1+x^2}}$
$\tan (\arcsin x) = \frac{x}{\sqrt{1-x^2}}$
$\tan (\arccos x) = \frac{\sqrt{1-x^2}}{x}$
## General solutions
Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent begins its period at 2πk − π/2, finishes it at 2πk + π/2, and then repeats it (forward) over 2πk + π/2 to 2πk + 3π/2. Cotangent begins its period at 2πk, finishes it at 2πk + π, and then repeats it (forward) over 2πk + π to 2πk + 2π.
This periodicity is reflected in the general inverses where k is some integer:
$\sin(y) = x \ \Leftrightarrow\ y = \arcsin(x) + 2k\pi \text{ or } y = \pi - \arcsin(x) + 2k\pi$
$\cos(y) = x \ \Leftrightarrow\ y = \arccos(x) + 2k\pi \text{ or } y = 2\pi - \arccos(x) + 2k\pi$
$\tan(y) = x \ \Leftrightarrow\ y = \arctan(x) + k\pi$
$\cot(y) = x \ \Leftrightarrow\ y = \arccot(x) + k\pi$
$\sec(y) = x \ \Leftrightarrow\ y = \arcsec(x) + 2k\pi \text{ or } y = 2\pi - \arcsec (x) + 2k\pi$
$\csc(y) = x \ \Leftrightarrow\ y = \arccsc(x) + 2k\pi \text{ or } y = \pi - \arccsc(x) + 2k\pi$
## Derivatives of inverse trigonometric functions
Simple derivatives for real and complex values of x are as follows:
\begin{align} \frac{d}{dx} \arcsin x & {}= \frac{1}{\sqrt{1-x^2}}\\ \frac{d}{dx} \arccos x & {}= \frac{-1}{\sqrt{1-x^2}}\\ \frac{d}{dx} \arctan x & {}= \frac{1}{1+x^2}\\ \frac{d}{dx} \arccot x & {}= \frac{-1}{1+x^2}\\ \frac{d}{dx} \arcsec x & {}= \frac{1}{x\,\sqrt{x^2-1}}\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{x\,\sqrt{x^2-1}} \end{align}
Only for real values of x:
\begin{align} \frac{d}{dx} \arcsec x & {}= \frac{1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{|x|\,\sqrt{x^2-1}}; \qquad |x| > 1 \end{align}
For a sample derivation: if $\theta = \arcsin x \!$, we get:
$\frac{d \arcsin x}{dx} = \frac{d \theta}{d \sin \theta} = \frac{1} {\cos \theta} = \frac{1} {\sqrt{1-\sin^2 \theta}} = \frac{1}{\sqrt{1-x^2}}$
## Expression as definite integrals
Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:
\begin{align} \arcsin x &{}= \int_0^x \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arccos x &{}= \int_x^1 \frac {1} {\sqrt{1 - z^2}}\,dz,\qquad |x| \leq 1\\ \arctan x &{}= \int_0^x \frac 1 {z^2 + 1}\,dz,\\ \arccot x &{}= \int_x^\infty \frac {1} {z^2 + 1}\,dz,\\ \arcsec x &{}= \int_1^x \frac 1 {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1\\ \arcsec x &{}= \pi + \int_x^{-1} \frac 1 {z \sqrt{z^2 - 1}}\,dz, \qquad x \leq -1\\ \arccsc x &{}= \int_x^\infty \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \geq 1\\ \arccsc x &{}= \int_{-\infty}^x \frac {1} {z \sqrt{z^2 - 1}}\,dz, \qquad x \leq -1 \end{align}
When x equals 1, the integrals with limited domains are improper integrals, but still well-defined.
## Infinite series
Like the sine and cosine functions, the inverse trigonometric functions can be calculated using infinite series, as follows:
\begin{align} \arcsin z & {}= z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots\\ & {}= \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}
\begin{align} \arccos z & {}= \frac {\pi} {2} - \arcsin z \\ & {}= \frac {\pi} {2} - (z + \left( \frac {1} {2} \right) \frac {z^3} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^5} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^7} {7} + \cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{2n+1}} {(2n+1)} ; \qquad | z | \le 1 \end{align}
\begin{align} \arctan z & {}= z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \qquad z \neq i,-i \end{align}
\begin{align} \arccot z & {}= \frac {\pi} {2} - \arctan z \\ & {}= \frac {\pi} {2} - ( z - \frac {z^3} {3} +\frac {z^5} {5} -\frac {z^7} {7} +\cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \frac {(-1)^n z^{2n+1}} {2n+1} ; \qquad | z | \le 1 \qquad z \neq i,-i \end{align}
\begin{align} \arcsec z & {}= \arccos {(1/z)} \\ & {}= \frac {\pi} {2} - (z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {z^{-5}} {5} + \left( \frac{1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6 } \right) \frac{z^{-7}} {7} + \cdots ) \\ & {}= \frac {\pi} {2} - \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{-(2n+1)}} {(2n+1)} ; \qquad \left| z \right| \ge 1 \end{align}
\begin{align} \arccsc z & {}= \arcsin {(1/z)} \\ & {}= z^{-1} + \left( \frac {1} {2} \right) \frac {z^{-3}} {3} + \left( \frac {1 \cdot 3} {2 \cdot 4 } \right) \frac {z^{-5}} {5} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {z^{-7}} {7} +\cdots \\ & {}= \sum_{n=0}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {z^{-(2n+1)}} {2n+1} ; \qquad \left| z \right| \ge 1 \end{align}
Leonhard Euler found a more efficient series for the arctangent, which is:
$\arctan z = \frac{z}{1+z^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k z^2}{(2k+1)(1+z^2)}.$
(Notice that the term in the sum for n= 0 is the empty product which is 1.)
Alternatively, this can be expressed:
$\arctan z = \sum_{n=0}^\infty \frac{2^{\,2n}\,(n!)^2}{\left(2n+1\right)!} \; \frac{z^{\,2n+1}}{\left(1+z^2\right)^{n+1}}$
## Continued fractions for arctangent
Two alternatives to the power series for arctangent are these generalized continued fractions:
$\arctan z=\cfrac{z} {1+\cfrac{(1z)^2} {3-z^2+\cfrac{(3z)^2} {5-3z^2+\cfrac{(5z)^2} {7-5z^2+\cfrac{(7z)^2} {9-7z^2+\ddots}}}}} =\cfrac{z} {1+\cfrac{(1z)^2} {3+\cfrac{(2z)^2} {5+\cfrac{(3z)^2} {7+\cfrac{(4z)^2} {9+\ddots\,}}}}}\,$
The second of these is valid in the cut complex plane. There are two cuts, from −i to the point at infinity, going down the imaginary axis, and from i to the point at infinity, going up the same axis. It works best for real numbers running from −1 to 1. The partial denominators are the odd natural numbers, and the partial numerators (after the first) are just (nz)2, with each perfect square appearing once. The first was developed by Leonhard Euler; the second by Carl Friedrich Gauss utilizing the Gaussian hypergeometric series.
## Indefinite integrals of inverse trigonometric functions
For real and complex values of x:
\begin{align} \int \arcsin x\,dx &{}= x\,\arcsin x + \sqrt{1-x^2} + C\\ \int \arccos x\,dx &{}= x\,\arccos x - \sqrt{1-x^2} + C\\ \int \arctan x\,dx &{}= x\,\arctan x - \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arccot x\,dx &{}= x\,\arccot x + \frac{1}{2}\ln\left(1+x^2\right) + C\\ \int \arcsec x\,dx &{}= x\,\arcsec x - \ln\left(x\left(1+\sqrt{{x^2-1}\over x^2}\right)\right) + C\\ \int \arccsc x\,dx &{}= x\,\arccsc x + \ln\left(x\left(1+\sqrt{{x^2-1}\over x^2}\right)\right) + C \end{align}
For real x ≥ 1:
\begin{align} \int \arcsec x\,dx &{}= x\,\arcsec x - \ln\left(x+\sqrt{x^2-1}\right) + C\\ \int \arccsc x\,dx &{}= x\,\arccsc x + \ln\left(x+\sqrt{x^2-1}\right) + C \end{align}
All of these can be derived using integration by parts and the simple derivative forms shown above.
### Example
Using $\int u\,\mathrm{d}v = u v - \int v\,\mathrm{d}u$, set
\begin{align} u &{}=&\arcsin x &\quad\quad\mathrm{d}v = \mathrm{d}x\\ \mathrm{d}u &{}=&\frac{\mathrm{d}x}{\sqrt{1-x^2}}&\quad\quad{}v = x \end{align}
Then
$\int \arcsin(x)\,\mathrm{d}x = x \arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x$
Substitute
$k = 1 - x^2.\,$
Then
$\mathrm{d}k = -2x\,\mathrm{d}x$
and
$\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x = -\frac{1}{2}\int \frac{\mathrm{d}k}{\sqrt{k}} = -\sqrt{k}$
Back-substitute for x to yield
$\int \arcsin(x)\, \mathrm{d}x = x \arcsin x + \sqrt{1-x^2}+C$
## Two-argument variant of arctangent
The two-argument atan2 function computes the arctangent of y / x given y and x, but with a range of (−π, π]. In other words, atan2(yx) is the angle between the positive x-axis of a plane and the point (xy) on it, with positive sign for counter-clockwise angles (upper half-plane, y > 0), and negative sign for clockwise angles (lower half-plane, y < 0). It was first introduced in many computer programming languages, but it is now also common in other fields of science and engineering.
In terms of the standard arctan function, that is with range of (−π/2, π/2), it can be expressed as follows:
$\operatorname{atan2}(y, x) = \begin{cases} \arctan(\frac y x) & \qquad x > 0 \\ \pi + \arctan(\frac y x) & \qquad y \ge 0 , x < 0 \\ -\pi + \arctan(\frac y x) & \qquad y < 0 , x < 0 \\ \frac{\pi}{2} & \qquad y > 0 , x = 0 \\ -\frac{\pi}{2} & \qquad y < 0 , x = 0 \\ \text{undefined} & \qquad y = 0, x = 0 \end{cases}$
It also equals the principal value of the argument of the complex number x + iy.
This function may also be defined using the tangent half-angle formulae as follows:
$\operatorname{atan2}(y, x)=2\arctan \frac{y}{\sqrt{x^2 + y^2} + x}$
provided that either x > 0 or y ≠ 0. However this fails if given x ≤ 0 and y = 0 so the expression is unsuitable for computational use.
The above argument order (y, x) seems to be the most common, and in particular is used in ISO standards such as the C programming language, but a few authors may use the opposite convention (x, y) so some caution is warranted. These variations are detailed at Atan2.
## Logarithmic forms
These functions may also be expressed using complex logarithms. This extends in a natural fashion their domain to the complex plane.
\begin{align} \arcsin x &{}= -i\,\ln\left(i\,x+\sqrt{1-x^2}\right) &{}= \arccsc \frac{1}{x}\\[10pt] \arccos x &{}= -i\,\ln\left(x+i\,\sqrt{1-x^2}\right) = \frac{\pi}{2}\,+i\ln\left(i\,x+\sqrt{1-x^2}\right) = \frac{\pi}{2}-\arcsin x &{}= \arcsec \frac{1}{x}\\[10pt] \arctan x &{}= \tfrac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) &{}= \arccot \frac{1}{x}\\[10pt] \arccot x &{}= \tfrac{1}{2}i\left(\ln\left(1-\frac{i}{x}\right)-\ln\left(1+\frac{i}{x}\right)\right) &{}= \arctan \frac{1}{x}\\[10pt] \arcsec x &{}= -i\,\ln\left(i\,\sqrt{1-\frac{1}{x^2}}+\frac{1}{x}\right) = i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right)+\frac{\pi}{2} = \frac{\pi}{2}-\arccsc x &{}= \arccos \frac{1}{x}\\[10pt] \arccsc x &{}= -i\,\ln\left(\sqrt{1-\frac{1}{x^2}}+\frac{i}{x}\right) &{}= \arcsin \frac{1}{x} \end{align}
Elementary proofs of these relations proceed via expansion to exponential forms of the trigonometric functions.
### Example proof
$\arcsin x = \theta \,$
$\frac{e^{i\theta} - e^{-i\theta}}{2i} = x$
(exponential definition of sine)
Let
$k=e^{i\,\theta}. \,$
Then
$\frac{k-\frac{1}{k}}{2i} = x$
$k^2-2\,i\,k\,x-1\,=\,0$
$k = ix \pm \sqrt{1-x^2} = e^{i\theta} \,$
(the positive branch is chosen)
$\theta = \arcsin x = -i \ln \left(ix + \sqrt{1-x^2}\right) \,$
arcsin(z) arccos(z) arctan(z) arccot(z) arcsec(z) arccsc(z)
## Arctangent addition formula
$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi, \qquad u v \ne 1 \,.$
This is derived from the tangent addition formula
$\tan ( \alpha + \beta ) = \frac{\tan \alpha + \tan \beta} {1 - \tan \alpha \tan \beta} \,,$
by letting
$\alpha = \arctan u \,, \quad \beta = \arctan v \,.$
## Application: finding the angle of a right triangle
A right triangle.
Inverse trigonometric functions are useful when trying to determine the remaining two angles of a right triangle when the lengths of the sides of the triangle are known. Recalling the right-triangle definitions of sine, for example, it follows that
$\theta = \arcsin \left( \frac{\text{opposite}}{\text{hypotenuse}} \right).$
Often, the hypotenuse is unknown and would need to be calculated before using arcsine or arccosine using the Pythagorean Theorem: a2 + b2 = h2 where h is the length of the hypotenuse. Arctangent comes in handy in this situation, as the length of the hypotenuse is not needed.
$\theta = \arctan \left( \frac{\text{opposite}}{\text{adjacent}} \right).$
For example, suppose a roof drops 8 feet as it runs out 20 feet. The roof makes an angle θ with the horizontal, where θ may be computed as follows:
$\theta = \arctan \left(\frac{\text{opposite}}{\text{adjacent}} \right) = \arctan \left( \frac{\text{rise}}{\text{run}} \right) = \arctan \left( \frac{8}{20} \right) = 21.8^{\circ}.$
## Practical considerations
For angles near 0 and π, arccosine is ill-conditioned and will thus calculate the angle with reduced accuracy in a computer implementation (due to the limited number of digits). Similarly, arcsine is inaccurate for angles near −π/2 and π/2. To achieve full accuracy for all angles, arctangent or atan2 should be used for the implementation.
## References
1. ^ For example Dörrie, Heinrich (1965). Triumph der Mathematik. Trans. David Antin. Dover. p. 69. ISBN 0486613488.
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• List of integrals of inverse trigonometric functions — The following is a list of integrals (antiderivative formulas) for integrands that contain inverse trigonometric functions (also known as arc functions ). For a complete list of integral formulas, see lists of integrals.Note: There are three… … Wikipedia
• Trigonometric functions — Cosine redirects here. For the similarity measure, see Cosine similarity. Trigonometry History Usage Functions Generalized Inverse functions … Wikipedia
• inverse trigonometric function — noun : the inverse of any of the six trigonometric functions … Useful english dictionary
• Differentiation of trigonometric functions — Trigonometry History Usage Functions Generalized Inverse functions Further reading … Wikipedia
• inverse function — Math. the function that replaces another function when the dependent and independent variables of the first function are interchanged for an appropriate set of values of the dependent variable. In y = sin x and x = arc sin y, the inverse function … Universalium
• Inverse Winkelfunktion — Eine Arkusfunktion (von lat. arcus „Bogen“) oder inverse Winkelfunktion ist die Umkehrfunktion einer trigonometrischen Funktion. Sie berechnet die Bogenlänge eines Sektors des Einheitskreises x2 + y2 = 1. Ihr Funktionswert ist ein Winkel. Die… … Deutsch Wikipedia
• Inverse function — In mathematics, if fnof; is a function from A to B then an inverse function for fnof; is a function in the opposite direction, from B to A , with the property that a round trip (a composition) from A to B to A (or from B to A to B ) returns each… … Wikipedia
• Inverse hyperbolic function — The inverses of the hyperbolic functions are the area hyperbolic functions. The names hint at the fact that they compute the area of a sector of the unit hyperbola x^{2} y^{2} = 1 in the same way that the inverse trigonometric functions compute… … Wikipedia
• List of trigonometric identities — Cosines and sines around the unit circle … Wikipedia
• List of mathematical functions — In mathematics, several functions or groups of functions are important enough to deserve their own names. This is a listing of pointers to those articles which explain these functions in more detail. There is a large theory of special functions… … Wikipedia | 7,430 | 20,224 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 74, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-43 | latest | en | 0.858258 |
https://helperscript.com/category/python/ | 1,642,524,372,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00581.warc.gz | 347,943,908 | 29,865 | ## hacker-rank | min-max sum of array
find the minimum and maximum values that can be calculated by summing exactly four of the five integers. Then print the respective minimum and maximum values as a single line of two space-separated long integers.
Example:
arr = [1,3,5,7,9]
The minimum sum is 1+3+5+7 = 16 and the maximum sum is 3+5+7+9 = 24. The function prints
16 24
Output Format
Print two space-separated long integers denoting the respective minimum and maximum values that can be calculated by summing exactly four of the five integers. (The output can be greater than a 32-bit integer.)
Sample Input
1 2 3 4 5
Sample Output
10 14
Explanation:
The numbers are 1, 2, 3, 4, and 5. Calculate the following sums using four of the five integers:
• Sum everything except 1, the sum is 1+2+3+4+5 = 14.
• Sum everything except 2, the sum is 1+3+4+5 = 13.
• Sum everything except 3, the sum is 1+2+4+5 = 12.
• Sum everything except 4, the sum is 1+2+3+5 = 11.
• Sum everything except 5, the sum is 1+2+3+4 = 10.
From the above calculation, 10 is the minimum and 14 is maximum.
Code language => Python
Solutions:
```arr = [1,3,5,7,9]
def sumOfarray(arry):
s = 0
l = len(arry)
for x in range(0,l):
s =s+ arry[x]
return s
def findminandmax(arr):
sumOfarr = []
for i in range(0, len(arr)):
a = arr[:]
a.pop(i)
sumOfarr.append(sumOfarray(a))
return print(str(min(sumOfarr)) + " "+ str(max(sumOfarr)))
findminandmax(arr)
###output###
# 16 24
```
Note: we use a deep clone([:]) or deep copy for changing an array without affecting previous array values.
Example for deep clone:
1st case without deep clone
```# Example for deep clone:
a = [5, 2]
b = a
b[0] = 100
print(a)
###output###
#[100 ,2]
```
2nd case with a deep clone [:]
```# if we dont want to change variable a, then
a = [5,2]
b = a[:]
b[0] = 100
print(a)
###output###
#[5,2]
```
## hacker-rank | print staircase right aligned with space and hash
Staircase detail
This is a staircase of size: n = 4
#
##
###
####
Note: Right-aligned #
Its base and height are both equal to n. It is drawn using # symbols and spaces. The last line is not preceded by any spaces.
Query: Write a program that prints a staircase of size n.
Function Description
The staircase has the following parameter(s):
int n: an integer
Print a staircase as described above.
Input Format
A single integer, n, denoting the size of the staircase.
Code language => Python
Solutions:
```def staircase(n):
l = n
for x in range(0,l):
ns = l-x -1
print(ns * " "+("#")*(x+1))
staircase(6)
# output
#
##
###
####
#####
######
```
Note:
In python, we can multiply a string. eg,
=> “ab” * 2
=> abab
From above we can multiply the “ab” 2 times. So, we use the same concept to print space.
## hacker-rank | calculate ratios of positive, negative, and zero elements
Given an array of integers, calculate the ratios of its elements that are positive, negative, and zero. Print the decimal value of each fraction on a new line with 6 places after the decimal.
Note: This challenge introduces precision problems. The test cases are scaled to six decimal places, though answers with an absolute error of up to 10^4 are acceptable.
Example
arr = [1, 1, 0, -1, -1]
There are n= 5 elements, two positive, two negative, and one zero. Their ratios are (2/5) = 0.400000 , (2/5) = 0.400000 and (1/5) = 0.200000. Results are printed as each in new lines:
0.400000
0.400000
0.200000
Coding language => Python
Coding logic => own logic
Solutions:
```arr = [-4, 3, -9, 0, 4, 1]
# iterating each number in list
def findcount(numbers):
pos_count, neg_count, zero_count = 0, 0, 0
for num in numbers:
# checking condition
if num > 0:
pos_count += 1
elif num == 0:
zero_count += 1
elif num < 0:
neg_count += 1
return [pos_count, neg_count, zero_count]
def plusMinus(arr):
l = len(arr)
ratiolist = findcount(arr)
ratioOfPositive = ratiolist[0] / l
ratioOfNegative = ratiolist[1] / l
ratioOfZero = ratiolist[2] / l
print(format(ratioOfPositive,'.6f' )+"\n"+ format(ratioOfNegative,'.6f' )+"\n"+ format(ratioOfZero,'.6f' ))
plusMinus(arr)
```
OUTPUT:
0.400000
0.400000
0.200000
NOTE: python print float digit with 6 decimal point
=> format(‘2′,’.6f’)
=> 2.000000
## find factorial of a number in python simple way
find factorial in python easy way only 5 lines
## Coding Style = my style of coding
Coding language = python
```def findFactorial(num):
a = 1
for x in range(0, num):
a = a * (num -x)
return a
print(findFactorial(5))
# 5! * 4! * 3! * 2! * 1!
# output :
# 120
```
Explanation:
for x in range(0, 5)
Already set a = 1, so
First loop
=> a = 1 * (5 – 0)
=> a = 5
Second loop
=> a = 5 * (5 – 1)
=> a = 5 * 4
=> a = 20
Third loop
=> a = 20 * (5 – 2)
=> a = 20 * 3
=> a = 60
Fourth loop
=> a = 60 * (5 – 3)
=> a = 60 * 2
=> a = 120
Fourth loop
=> a = 120 * (5 – 4)
=> a = 120 * 1
=> a = 120
After that x range is now 4, we set only in the range between 0 to 5
Now loop ended. So we print( findFactorial(5) ) => its return answer 120.
Hope this logic will help you to understand the factorial concept in programming. Thank you for reading!
## Python Strings | tutorial 4
How to add single quotes, double quotes, or both quotes in the python strings?
Example:
```course = "Python's for everyone"
print(course)
new_course = 'Python for "EveryOne"'
print(new_course)
email = '''
Hi John!,
Welcome to helperscript to learn Python's for "Everyone"
Thank you,
Support Team
'''
print(email)
```
Triple quotes (”’Python for “Everyone””’) – Three single quotes from the start and three single quotes from the end. This type of string escapes all the symbols in the string.
Split string into an array (or) Split the string in python
Python string starts index with zero(0).
Example: course variable has string “Python for Everyone”
Course = ‘Python for Everyone’
Get the first character of a string in python
course[0]
Get the second character of a string in python
course[1]
Get the last character of a string in python
course[-1]
Get the last second character of a string in python
course[-2]
Get the first three character of a string in python
course[0:3]
Remove the first character and remove the last 3 characters of a string in python
course[1:-3]
Remove the character after 6 index
course[:6]
Remove the first 6 characters in the string
course[6:]
Clone or copy the string
course[:]
Code:
```# strings
course = 'Python for Everyone'
print(course);
print('course[0] = ' + course[0])
print('course[1] = ' + course[1])
print('course[-1] = ' + course[-1])
print('course[-2] = ' + course[-2])
print('course[0:3] = ' + course[0:3])
print('course[1:-3] = ' + course[1:-3])
print('course[:5] = ' + course[:6])
print('course[6:] = ' + course[6:])
print('course[:] = ' + course[:])
```
OUTPUT:
String Concatenation
Two ways to concat the string dynamically.
Normal way – firstname +’ ’+ lastname +’ is a coder’
Format way – f‘{firstname} {lastname} is a code’
f – prefix with f define a formatted string
Code:
Python strings method
Let,
course = ‘Python for Beginners’
# find the length of string
len(course)
# convert uppercase
course.upper()
# convert lowercase
course.lower()
# convert first character uppercase in each word
course.title()
# find an index of character or sequence of character.
Python find() method is case sensitive.
print(course.find(‘P’)) # gives index of capital P
print(course.find(‘p’)) # return -1 if not present.
print(course.find(‘Beginners’))
# Replace a character
course.replace(‘P’, ‘J’)
# Replace a sequence of character
course.replace(‘Python’, ‘Angular’)
# find a string contains in the existing string(case sensitive)
‘Python’ in course
App.py
```course = 'Python for Beginners'
# find the length of string
print(len(course))
# convert uppercase
print(course.upper())
# convert lowercase
print(course.lower())
# convert first character uppercase in each word
print(course.title())
# find index of character or sequence of character
print(course.find('P')) # capital
print(course.find('p')) # small letter
print(course.find('Beginners'))
# Replace a character
print(course.replace('P', 'J'))
# Replace a sequence of character
print(course.replace('Python', 'Angular'))
# find a string contains in existing string
print('Python' in course) # gives boolean output either True or False
```
In Next article, we see about Operators in Python.
## PYTHON variables | tutorial 3
Python is a case sensitive language. So when you define a variable you should consider the uppercase and lowercase.
eg: price = 10 is different to PRICE = 10
Important Variable Types and Declaration
```price = 10
# price variable type is INTEGER
# price is an identifier for variable and initialised with the value 10
update_price = 20.5
# update_price variable type is FLOAT
is_published = True
# is_published variable type is BOOLEAN
# either True or False (case sensitive first letter should be capital)
name = 'Jasim'
# name variable type is STRING
```
# – is a comment line. Python interpreted not read this line.
Let’s do one small task.
Variable Declaration Task: We check in a patient named John Toe. He’s 20 years old and is a new patient.
```name = 'John Toe'
Age = 20
is_new_patient = True
```
As of now we successfully learn how to declare the variable in python. Next, we learn about how to get input from users.
Instead of print() method use input() method to get the input from the user.
Filename: app.py
```name = input('What is your name? ')
print('Hey! ' + name)
```
Task 2: Ask two questions – Person’s name and his favorite color. Print the output like “John Toe like violet”.
```name = input('What is your name? ')
Color = input('what is your favourite color? ')
print(name + ' like ' + color);
```
Type conversion
Type conversion is used to define a variable for calculation. If the number is in string format we need to convert the number string into the integer type.
Example:
```birth_year = input('Your birth year? ')
print(type(birth_year))
age = 2020 - birth_year
print('Your age is ' + age)
```
You get input from the terminal 1994 and the type of 1994 is string identified with the use of type() method.
You get an error in line 20 in the app.py file. It will be shown in the terminal due to the type differentiation. So how to solve this problem. We need to convert the type from string to integer.
How to convert the string to integer
int() – method is used to convert the typed string to an integer.
str() – method is used to convert the integer type to string type.
Solution:
```birth_year = input('Your birth year? ')
print(type(birth_year))
age = 2020 - int(birth_year)
print('Your age is ' + str(age))
```
In the next article let see the strings in python. Go the next continuation on the menu.
## Basics of python tutorial 2
Open your PyCharm or your favorite IDE. Click to create a new project and name your project.
Give your project name and check the base interpreter has the current version. If you need another version, you can select the other version from the list.
Create a first python program
Name your file. In my case app.py
Now let’s start to learn the code first.
How to console or print the value in the python
file name: app.py
```print('Hi! welcome to helperscript');
```
print(data) => print function helps to print or console in terminal output. It is used to identify whether the data passed are correct or incorrect.
data => you can pass data that you want to output. When your data is a string. String declaration with a single quote or either double quote.
Eg: print(“hello”) is equal to print(‘hello’)
Then Run the app.py file from the menu Run. It will show configuration settings. As of now we just run the file. So just click on the app.
After Run the app.py file. The below terminal will open and show the output like below.
Python code gets executed line by line from the top.
Print the string with the multiplication
file name: app.py
```print('Hi! welcome to helperscript')
print('*' * 5)
```
In python, the String data type can be generated and printed multiple times with an asterisk.
So the python is powerful for AI, Automation, desktop application, and web application.
Let us see more new topics in python in the next article. Go to the menu of Python and find the next tutorial for continuation…
## Python 3 tutorial step by step learning through free courses
INTRODUCTION TO PYTHON 3
Python is one of the most popular languages nowadays.
Python has the ability to do
1. Automation
2. AI (Artificial Intelligence)
3. Applications
4. Websites
We talk about all the core concepts in python and then we build three(3) python projects together.
BUILD WEB APPLICATION
Django administration
HOW TO USE 1000 OF SPREADSHEET IN MIN 100 SECONDS WITH CHARTS
Let’s start the course for the absolute beginners.
INSTALL PYTHON
HOW TO DOWNLOAD AND INSTALL PYTHON ON YOUR COMPUTER
1. Go to python.org
2. Click on downloads
3. Click the Download button for the latest version
In windows simply click and open the downloaded file. Make sure you check Add path to the variable.
For Mac, Just continue and Agree with the license agreement and install it.
Next, we are going to install a code editor. PyCharm (https://jetbrains.com/pycharm) is one of the best code editors for python. It fully depends on you. You can use another editor too.
Download and install pyCharm on your computer depends on your OS.
Windows => click next => next => and install
Mac => drag and drop the pyCharm to the application to install.
Choose your theme, Most of them use Dracula or blacky themes to protect their eyes. Now open your pyCharm application.
Congratulations, you are now successfully downloading and installing everything for python to learn.
Go to the python menu on top and see the continuation tutorial in the python. | 3,684 | 13,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-05 | longest | en | 0.712464 |
https://handwiki.org/wiki/Degree_of_a_polynomial | 1,670,068,015,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00396.warc.gz | 326,346,016 | 16,813 | # Degree of a polynomial
Short description: Mathematical concept
In mathematics, the degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer. For a univariate polynomial, the degree of the polynomial is simply the highest exponent occurring in the polynomial.[1][2] The term order has been used as a synonym of degree but, nowadays, may refer to several other concepts (see order of a polynomial (disambiguation)).
For example, the polynomial $\displaystyle{ 7x^2y^3 + 4x - 9, }$ which can also be written as $\displaystyle{ 7x^2y^3 + 4x^1y^0 - 9x^0y^0, }$ has three terms. The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5, which is the highest degree of any term.
To determine the degree of a polynomial that is not in standard form, such as $\displaystyle{ (x+1)^2 - (x-1)^2 }$, one can put it in standard form by expanding the products (by distributivity) and combining the like terms; for example, $\displaystyle{ (x+1)^2 - (x-1)^2 = 4x }$ is of degree 1, even though each summand has degree 2. However, this is not needed when the polynomial is written as a product of polynomials in standard form, because the degree of a product is the sum of the degrees of the factors.
## Names of polynomials by degree
The following names are assigned to polynomials according to their degree:[3][4][5][2]
• Special case – zero (see Degree of the zero polynomial below)
• Degree 0 – non-zero constant[6]
• Degree 1 – linear
• Degree 3 – cubic
• Degree 4 – quartic (or, if all terms have even degree, biquadratic)
• Degree 5 – quintic
• Degree 6 – sextic (or, less commonly, hexic)
• Degree 7 – septic (or, less commonly, heptic)
For higher degrees, names have sometimes been proposed,[7] but they are rarely used:
• Degree 8 – octic
• Degree 9 – nonic
• Degree 10 – decic
Names for degree above three are based on Latin ordinal numbers, and end in -ic. This should be distinguished from the names used for the number of variables, the arity, which are based on Latin distributive numbers, and end in -ary. For example, a degree two polynomial in two variables, such as $\displaystyle{ x^2 + xy + y^2 }$, is called a "binary quadratic": binary due to two variables, quadratic due to degree two.[lower-alpha 1] There are also names for the number of terms, which are also based on Latin distributive numbers, ending in -nomial; the common ones are monomial, binomial, and (less commonly) trinomial; thus $\displaystyle{ x^2 + y^2 }$ is a "binary quadratic binomial".
## Examples
The polynomial $\displaystyle{ (y - 3)(2y + 6)(-4y - 21) }$ is a cubic polynomial: after multiplying out and collecting terms of the same degree, it becomes $\displaystyle{ - 8 y^3 - 42 y^2 + 72 y + 378 }$, with highest exponent 3.
The polynomial $\displaystyle{ (3 z^8 + z^5 - 4 z^2 + 6) + (-3 z^8 + 8 z^4 + 2 z^3 + 14 z) }$ is a quintic polynomial: upon combining like terms, the two terms of degree 8 cancel, leaving $\displaystyle{ z^5 + 8 z^4 + 2 z^3 - 4 z^2 + 14 z + 6 }$, with highest exponent 5.
## Behavior under polynomial operations
The degree of the sum, the product or the composition of two polynomials is strongly related to the degree of the input polynomials.[8]
The degree of the sum (or difference) of two polynomials is less than or equal to the greater of their degrees; that is,
$\displaystyle{ \deg(P + Q) \leq \max\{\deg(P),\deg(Q)\} }$ and $\displaystyle{ \deg(P - Q) \leq \max\{\deg(P),\deg(Q)\} }$.
For example, the degree of $\displaystyle{ (x^3+x)-(x^3+x^2)=-x^2+x }$ is 2, and 2 ≤ max{3, 3}.
The equality always holds when the degrees of the polynomials are different. For example, the degree of $\displaystyle{ (x^3+x)+(x^2+1)=x^3+x^2+x+1 }$ is 3, and 3 = max{3, 2}.
### Multiplication
The degree of the product of a polynomial by a non-zero scalar is equal to the degree of the polynomial; that is,
$\displaystyle{ \deg(cP)=\deg(P) }$.
For example, the degree of $\displaystyle{ 2(x^2+3x-2)=2x^2+6x-4 }$ is 2, which is equal to the degree of $\displaystyle{ x^2+3x-2 }$.
Thus, the set of polynomials (with coefficients from a given field F) whose degrees are smaller than or equal to a given number n forms a vector space; for more, see Examples of vector spaces.
More generally, the degree of the product of two polynomials over a field or an integral domain is the sum of their degrees:
$\displaystyle{ \deg(PQ) = \deg(P) + \deg(Q) }$.
For example, the degree of $\displaystyle{ (x^3+x)(x^2+1)=x^5+2x^3+x }$ is 5 = 3 + 2.
For polynomials over an arbitrary ring, the above rules may not be valid, because of cancellation that can occur when multiplying two nonzero constants. For example, in the ring $\displaystyle{ \mathbf{Z}/4\mathbf{Z} }$ of integers modulo 4, one has that $\displaystyle{ \deg(2x) = \deg(1+2x) = 1 }$, but $\displaystyle{ \deg(2x(1+2x)) = \deg(2x) = 1 }$, which is not equal to the sum of the degrees of the factors.
### Composition
The degree of the composition of two non-constant polynomials $\displaystyle{ P }$ and $\displaystyle{ Q }$ over a field or integral domain is the product of their degrees:
$\displaystyle{ \deg(P \circ Q) = \deg(P)\deg(Q) }$.
For example:
• If $\displaystyle{ P = (x^3+x) }$, $\displaystyle{ Q = (x^2+1) }$, then $\displaystyle{ P \circ Q = P \circ (x^2+1) = (x^2+1)^3+(x^2+1) = x^6+3x^4+4x^2+2 }$, which has degree 6.
Note that for polynomials over an arbitrary ring, this is not necessarily true. For example, in $\displaystyle{ \mathbf{Z}/4\mathbf{Z} }$, $\displaystyle{ \deg(2x) \deg(1+2x) = 1\cdot 1 = 1 }$, but $\displaystyle{ \deg(2x\circ(1+2x)) = \deg(2+4x)=\deg(2) = 0 }$.
## Degree of the zero polynomial
The degree of the zero polynomial is either left undefined, or is defined to be negative (usually −1 or $\displaystyle{ -\infty }$).[9]
Like any constant value, the value 0 can be considered as a (constant) polynomial, called the zero polynomial. It has no nonzero terms, and so, strictly speaking, it has no degree either. As such, its degree is usually undefined. The propositions for the degree of sums and products of polynomials in the above section do not apply, if any of the polynomials involved is the zero polynomial.[10]
It is convenient, however, to define the degree of the zero polynomial to be negative infinity, $\displaystyle{ -\infty, }$ and to introduce the arithmetic rules[11]
$\displaystyle{ \max(a,-\infty) = a, }$
and
$\displaystyle{ a + (-\infty) = -\infty. }$
These examples illustrate how this extension satisfies the behavior rules above:
• The degree of the sum $\displaystyle{ (x^3+x)+(0)=x^3+x }$ is 3. This satisfies the expected behavior, which is that $\displaystyle{ 3 \le \max(3, -\infty) }$.
• The degree of the difference $\displaystyle{ (x)-(x) = 0 }$ is $\displaystyle{ -\infty }$. This satisfies the expected behavior, which is that $\displaystyle{ -\infty \le \max(1,1) }$.
• The degree of the product $\displaystyle{ (0)(x^2+1)=0 }$ is $\displaystyle{ -\infty }$. This satisfies the expected behavior, which is that $\displaystyle{ -\infty = -\infty + 2 }$.
## Computed from the function values
A number of formulae exist which will evaluate the degree of a polynomial function f. One based on asymptotic analysis is
$\displaystyle{ \deg f = \lim_{x\rarr\infty}\frac{\log |f(x)|}{\log x} }$;
this is the exact counterpart of the method of estimating the slope in a log–log plot.
This formula generalizes the concept of degree to some functions that are not polynomials. For example:
• The degree of the multiplicative inverse, $\displaystyle{ \ 1/x }$, is −1.
• The degree of the square root, $\displaystyle{ \sqrt x }$, is 1/2.
• The degree of the logarithm, $\displaystyle{ \ \log x }$, is 0.
• The degree of the exponential function, $\displaystyle{ \exp x }$, is $\displaystyle{ \infty. }$
The formula also gives sensible results for many combinations of such functions, e.g., the degree of $\displaystyle{ \frac{1 + \sqrt{x}}{x} }$ is $\displaystyle{ -1/2 }$.
Another formula to compute the degree of f from its values is
$\displaystyle{ \deg f = \lim_{x\to\infty}\frac{x f'(x)}{f(x)} }$;
this second formula follows from applying L'Hôpital's rule to the first formula. Intuitively though, it is more about exhibiting the degree d as the extra constant factor in the derivative $\displaystyle{ d x^{d-1} }$ of $\displaystyle{ x^d }$.
A more fine grained (than a simple numeric degree) description of the asymptotics of a function can be had by using big O notation. In the analysis of algorithms, it is for example often relevant to distinguish between the growth rates of $\displaystyle{ x }$ and $\displaystyle{ x \log x }$, which would both come out as having the same degree according to the above formulae.
## Extension to polynomials with two or more variables
For polynomials in two or more variables, the degree of a term is the sum of the exponents of the variables in the term; the degree (sometimes called the total degree) of the polynomial is again the maximum of the degrees of all terms in the polynomial. For example, the polynomial x2y2 + 3x3 + 4y has degree 4, the same degree as the term x2y2.
However, a polynomial in variables x and y, is a polynomial in x with coefficients which are polynomials in y, and also a polynomial in y with coefficients which are polynomials in x. The polynomial
$\displaystyle{ x^2y^2 + 3x^3 + 4y = (3)x^3 + (y^2)x^2 + (4y) = (x^2)y^2 + (4)y + (3x^3) }$
has degree 3 in x and degree 2 in y.
## Degree function in abstract algebra
Given a ring R, the polynomial ring R[x] is the set of all polynomials in x that have coefficients in R. In the special case that R is also a field, the polynomial ring R[x] is a principal ideal domain and, more importantly to our discussion here, a Euclidean domain.
It can be shown that the degree of a polynomial over a field satisfies all of the requirements of the norm function in the euclidean domain. That is, given two polynomials f(x) and g(x), the degree of the product f(x)g(x) must be larger than both the degrees of f and g individually. In fact, something stronger holds:
$\displaystyle{ \deg(f(x)g(x)) = \deg(f(x)) + \deg(g(x)) }$
For an example of why the degree function may fail over a ring that is not a field, take the following example. Let R = $\displaystyle{ \mathbb{Z}/4\mathbb{Z} }$, the ring of integers modulo 4. This ring is not a field (and is not even an integral domain) because 2 × 2 = 4 ≡ 0 (mod 4). Therefore, let f(x) = g(x) = 2x + 1. Then, f(x)g(x) = 4x2 + 4x + 1 = 1. Thus deg(fg) = 0 which is not greater than the degrees of f and g (which each had degree 1).
Since the norm function is not defined for the zero element of the ring, we consider the degree of the polynomial f(x) = 0 to also be undefined so that it follows the rules of a norm in a Euclidean domain.
## Notes
1. For simplicity, this is a homogeneous polynomial, with equal degree in both variables separately.
1. Weisstein, Eric W.. "Polynomial Degree" (in en).
2. "Names of Polynomials". November 25, 1997.
3. Mac Lane and Birkhoff (1999) define "linear", "quadratic", "cubic", "quartic", and "quintic". (p. 107)
4. King (2009) defines "quadratic", "cubic", "quartic", "quintic", "sextic", "septic", and "octic".
5. Shafarevich (2003) says of a polynomial of degree zero, $\displaystyle{ f(x)=a_0 }$: "Such a polynomial is called a constant because if we substitute different values of x in it, we always obtain the same value $\displaystyle{ a_0 }$." (p. 23)
6. James Cockle proposed the names "sexic", "septic", "octic", "nonic", and "decic" in 1851. (Mechanics Magazine, Vol. LV, p. 171)
7. Lang, Serge (2005). Algebra (3rd ed.). Springer. pp. 100. ISBN 978-0-387-95385-4.
8. Shafarevich (2003) says of the zero polynomial: "In this case, we consider that the degree of the polynomial is undefined." (p. 27)
Childs (1995) uses −1. (p. 233)
Childs (2009) uses −∞ (p. 287), however he excludes zero polynomials in his Proposition 1 (p. 288) and then explains that the proposition holds for zero polynomials "with the reasonable assumption that $\displaystyle{ -\infty }$ + m = $\displaystyle{ -\infty }$ for m any integer or m = $\displaystyle{ -\infty }$".
Axler (1997) uses −∞. (p. 64)
Grillet (2007) says: "The degree of the zero polynomial 0 is sometimes left undefined or is variously defined as −1 ∈ $\displaystyle{ \mathbb{Z} }$ or as $\displaystyle{ -\infty }$, as long as deg 0 < deg A for all A ≠ 0." (A is a polynomial.) However, he excludes zero polynomials in his Proposition 5.3. (p. 121)
9. Barile, Margherita. "Zero Polynomial".
10. Axler (1997) gives these rules and says: "The 0 polynomial is declared to have degree $\displaystyle{ -\infty }$ so that exceptions are not needed for various reasonable results." (p. 64) | 3,828 | 13,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-49 | latest | en | 0.949254 |
http://athometuition.com/linear-equation-system.aspx | 1,571,316,530,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675316.51/warc/CC-MAIN-20191017122657-20191017150157-00397.warc.gz | 19,066,869 | 8,367 | ## Linear Equations System
Until now, we have solved linear equations with two variables. Let's see how we can solve two linear equations having two variables x and y.
Solve x + y = 5
x - y = 1
If an equation contains two variables, there are a number of ordered pairs that satisfy the given equation.
To solve an equation having two variables, we need two independent equations, i.e., to determine the ordered pairs that satisfy both equations.
Each ordered pair that satisfies a pair of equations is called the solution of that pair of equations.
Method of elimination using addition or subtraction
Example 1
Solve x + 2y = 5
x - 2y = 1
In this method, we eliminate one variable by addition or subtraction. The equation consists of the other variable after elimination.
Step 1:
x + 2y = 5
x - 2y = 1
The coefficient is the numerical value assigned to the variables. In this case, the coefficient of y is 2 in both the equations but with different signs.
Step 2:
If we add the two equations, the variable y will be eliminated.
x + 2y = 5 ------------- (1)
x - 2y = 1 ------------(2)
Add equations (1) and (2)
to get 2x = 6
x = 6/2 = 3
Step 3:
Substituting the value of x, i.e., 3, in equation (1) or (2)
x + 2y = 5
3 + 2y = 5
2y = 5 - 3 = 2
2y = 2
y = 1
Step 4:
If x = 3 then y = 1
Solution set = { (3,1) }
Verification:
We have to verify whether or not the solution set (3,1) satisfies both equations.
Substitute x = 3 and y = 1 into both equations
x + 2y = 5 x - 2y = 1
3 + 2 = 5 ∗3 - 2 = 1
The solution (3, 1) is correct.
Example 2
Solve x + 3y = 7, 3x + 4y = 11
In this example, we cannot eliminate x and y by addition or subtraction because the coefficients of x and y are different.
If every term in an equation is multiplied with a real number, the value of the equation remains unaltered.
Therefore, to make the coefficients of either x or y the same, we should multiply each equation with a number so that the value of the coefficient in both equations becomes the same.
Solution:
Step 1:21
x + 3y = 7 ------------ (1)
3x + 4y = 11 ------------ (2)
Multiply all the terms of the first equation with 3 to make the coefficients of x equal in both the equations.
x + 3y = 7 ----------- (1)
3x + 4y = 11 ---------- (2)
Equation (1) gives
3x + 9y = 21 ------------(3)
Subtracting the equation (3) from (2), we get
- 5y = - 10
y = -10/ -5 =2
y = 2
Substitute y = 2 in any one of the given equations to find x.
x + 3y = 7
x + 3 (2) = 7
x + 6 = 7
Therefore x = 1 and y = 2
Solution set = { (1,2) }
Verification:
Substitute x = 1 and y = 2 in the given equations
x + 3y = 7 3x + 4y = 11
1 + 3 (2) = 7 3 (1) + 4 (2) = 11
The solution (1,2) satisfies both the equations.
Therefore, the solution is correct.
#### Method of substitution
In this method, we express one variable (say x) in terms of the other variable (say y) in one of the equations and solve the resulting equation for the value of x.
Example 1
Solve 3x + 2y = 8 ---------------(1)
2x + 3y = 7 ----------------(2)
Solution:
Step 1:
Find the value of x in terms of y
2x + 3y = 7
2x = 7 - 3y
x = (7-3y)/2 ------------------(3)
Step 2:
The equation is a simultaneous linear equation. So the values of x and y are the same in both equations.
Substitute the values of x in equation (1)
3[ 7-3y / 2 ] + 2y = 8
Step 3:
Multiplying on both sides by 2
3(7 - 3y) + 2 ∗ 2y = 2 ∗8
21 - 9y + 4y = 16
-5y = 16 - 21
-5y = -5
y = -5/-5 =1
Step 4:
Substituting the value of y in equation (3) we get the value of x
x = 7-3 ∗ 1 / 2 = 7-3/ 2 = 4/2 = 2
x = 2; y = 1
Therefore, the solution set = {(2,1)}
Example 2:
Solve 1/x + 1/y = 9 -------------- (1)
3/x +2/y = 22 -------------- (2)
Solution:
Let 1/x = a and 1/y = b
The given equations can be written as
a + b = 9 -------------- (3)
3a + 2b = 22 ----------------(4)
Multiply the equation (3) by 2
2a + 2b = 18 --------------- (5)
Subtracting (5) from (4)
2a + 2b = 18 ----------------(5)
+3a + 2b = 22 ------------------(4)
Subtract (4) from (5), we get
-a = -4
a = 4
Substitute a = 4 in equation (3)
4 + b = 9
b = 9 - 4 = 5
But a = 1/x
therefore, 1/x = 4
and x =1/4
b = 1/y
therefore, 1/y = 5
and y = 1/5.
Solution set = {(1/4, 1/5)}
Verify whether or not { 1/4, 1/5 } is a solution set.
Example 3
Solve (x + y)/2 - (x - y )/3 = 6 -------------- (1)
(x - 3y)/2 + (x + 2y)/6 = 1 -------------- (2)
Solution:
Write both equations in the form ax + by + c = 0
(x + y)/2 - (x - y)/3 - 6 = 0
The L.C.M. of 2 and 3 is 6
Multiply the equation by 6
(x + y)/2 ∗ 6 - (x - y)/3 ∗ 6 - 6 ∗ 6 = 0
3( x + y ) - 2 ( x - y ) - 36 = 0
x + 5y - 36 = 0 ---------------- (3)
(x - 3y)/2 + (x + 2y)/6 = 1
The L.C.M. of 2 and 6 is 6
Multiply the equation by 6
(x - 3y)/2 ∗ 6 + (x + 2y)/6 ∗ 6 - 1∗ 6 = 0
3( x - 3y ) + x + 2y - 6 = 0
3x - 9y + x + 2y - 6 = 0
4x - 7y - 6 = 0 ---------------- (4)
Solve equations (3) and (4) for the values of x and y
Solution set = { (12,6) }
#### Dependent Equations
If the equation ax + by + c = 0 is multiplied by a constant k, the resulting equation is k ax + k by + kc = 0.
The equation ax + by + c = 0 and k ax + k by + kc = 0 are known as dependent equations.
Example 1
Solve 8x - 5y = 3 ---------------(1)
16x - 10y = 6 ------------------ (2)
In the above example, the first equation is multiplied by 2 to get the second equation.
Therefore, 8x - 5y = 8 and 16x - 10y = 6 are dependent equations.
8x - 5y = 3----------------(1) 16x - 10y = 6 -------------- (2)
8x = 3 + 5y 16x = 6 + 10y
x = (3+5y)/8 -------------- (3) x = (6+ 10y)/16 -------------(4)
Solving the equation for the value of y
(3+ 5y) / 8 = (6 + 10y) / 16
Cross-multiplying
48 + 80y = 48 + 80y
80y = 80y
The equation is true for any value of y.
For every value of y, we can determine the corresponding value of x from equation (1) or (2). The ordered pair that satisfies equation (1) will also satisfy equation (2).
The solution set for the above equations is infinite.
#### Try these questions
Find the solution sets of the following systems of equations
1. x + y = 6
x - y = 2
Answer: x + y = 6 ________ (1)
x - y = 2 ________ (2)
Adding (1) & (2) 2x = 8
Therefore x =8/2= 4
Substituting x = 4 in (1) we get 4 + y = 6
Therefore y = 6 - 4 = 2
Therefore solution set = { (4, 2) }
2. x + y = 7
x - y = 5
Answer: x + y = 7 ________ (1)
x - y = 5 ________ (2)
Adding (1) & (2) 2x = 12
Therefore x = 12/2= 6
Substituting x = 6 in (1) we get
6 + y = 7
Therefore y = 7 -6 = 1
Therefore solution set = { (6, 1) }
3. 4x - y = 5
4x + 4y = 20 Answer: 4x - y = 5 ________ (1)
4x + 4y = 20 ________ (2)
Subtract 1 & 2
-5y = -15
y = 3
Substituting y = 3 in equation 1
4x - 3 = 5
4x = 8
x =8/4
x =2
Therefore solution set = {( 2, 3)}
4. a - b = 5
a + b = 9
Answer: a - b = 5 ________ (1)
a + b = 9 ________ (2)
Adding (1) & (2) 2a = 14
Therefore a = 14/2= 7
Substituting a = 7 in (1) we get
7 - b = 5
Therefore b = 7 - 5 = 2
Therefore solution set = { (7, 2) }
5. 2x + 2y = 10
3x - 2y = 5 Answer: 2x + 2y = 10 ________ (1)
3x - 2y = 5 ________ (2)
Adding (1) & (2) 5x = 15
Therefore x = 15/5= 3
Substituting x = 3 in (1) we get
2 ∗3 + 2y = 10
Therefore 2y = 4, y = 2
Therefore solution set = { (3, 2) } | 2,749 | 7,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5 | 5 | CC-MAIN-2019-43 | latest | en | 0.899173 |
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1 6.1 Systems of Linear Equations An equation is a linear equation in two variables x and y. Similarly, is a linear equation in three variables x, y, and z. We may also consider linear equations in four, five, or any number of variables. A system of linear equations is collection of two or more linear equations each containing one or more variables. In this section we shall consider only systems of two linear equations in two variables. Systems involving more than two variables are discussed in the next section. We can view the problem of solving a system of two linear equation containing two variables as a geometry problem. The graph of each equation in such as system is a line. So, a system of two equations containing two variables represents a pair of lines. 1. intersecting lines; system 2. Parallel lines; system. Lines coincide; system has exactly one solution has no solution. has infinitely many (system is inconsistent) solutions (system is dependent) If the system has infinitely many solutions (dependent system) the system is solved by finding the general form of the solution set. The two methods of solving a system of linear equations that will be discussed in this section are the substitution method and the elimination method. Steps For Solving By Substitution 1. Pick one of the equations and solve for one of the variables in terms of the other variable. 2. Substitute this quantity in the remaining equation and solve. Find the value of the remaining variable by back Solve by Substitution Solution: Step 1 The easiest to solve is y in equation Step 2. Substitute 56 in 6 5 equation
2 Step 2. Substitute 56 in equation The solution is 1, Step 1 The easiest to solve is x in equation Step 2. Substitute 5 in equation The solution is 1, Step 1 The easiest to solve is y in equation Step 2. Substitute 42 in equation The System has no solution. The system is inconsistent.
3 Steps For Solving By Elimination 1. Multiply both sides of one (or both) equation(s) by the appropriate nonzero numbers so that when the equations are added together one of the variables will be eliminated. 2. Solve this equation for the remaining variable.. Find the value of the remaining variable by back substitution. Solve by Elimination Solution: We multiply both sides of equation (2) by 2 so that the coefficient of x in the two Now find the value of the remaining variable by back substitution The solution is 2, Solution: The smallest common multiple between 6 and 4 is 12, so we multiply both sides of equation (1) by 2 and both sides of equation (2) by so that the coefficient of y in the two Now find the value of the remaining variable by back substitution The Solution to the system is 1,1 6.
4 Solution: We multiply both sides of equation (1) by 2 so that the coefficient of x in the two The system is a dependent system of linear equations and thus has infinitely many solutions. To determine the general form of the solution set, we solve for one of the variables in either equation The General Form of the solution set is 4 2,: 7. The population y in year x of Long Beach and new Orleans is approximated by the equations : : Where x = 0 corresponds to 1980 and y in thousands. In what year do the two cities have the same population? Solution: We multiply both sides of equation (2) by 2 so that the coefficient of y in the two The year = 2002 the two cities will have the same population. 8. An apparel shop sells skirts for \$45 and blouses for \$5. Its entire stock is worth \$51,750, but sales are slow and only half the skirts and two thirds of the blouses are sold, for a total of \$0,600. How many skirts and blouses are left in the store? x = number of skirts in the store y = number of blouses in the store , , , ,600
5 Multiply equation (1) by , , , ,50 51, , Half the skirts are left (they sold half) and one third of the blouses are left (they sold two thirds) Therefore, there are 260 skirts left and 270 blouses left in the store. 9. A company purchases two models of bicycles: model 201 and model 01. Model 201 requires 2 hours of assembly time, and model 01 requires hours of assembly time. The parts for model 201 cost \$25 per bike, and the parts for model 01 cost \$0 per bike. If the company has a total of 4 hours of assembly time and \$65 available per day for these two models, how many of each can be made in a day? x = number of 201 models y = number of 01 models models and 8 01 models can be produced in one day.
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2. System of linear equations can be solved by graphing, substitution, or eliminating a variable.
1 Subject: Algebra 1 Grade Level: 9 th Unit Plan #: 6 UNIT BACKGROUND Unit Title: Systems of Equations and Inequalities Grade Level: 9 Subject/Topic: Algebra 1 Key Words: Graphing, Substitution, Elimination | 7,124 | 30,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2018-47 | longest | en | 0.919569 |
https://considerlearning.com/2013/04/27/teaching-log-0-9999-1/?replytocom=268 | 1,590,374,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00164.warc.gz | 305,176,055 | 32,508 | # Teaching Log: 0.9999… = 1
It worries me that as I move into leadership roles, I get fewer and fewer opportunities to teach. Hence, I was excited to give a talk at the Sonya Kovalevsky Day at Barnard College. The entire 9th grade of the Urban Assembly Institute of Math and Science for Young Women was at the event, and about a quarter were in my class. It was a challenging talk for a number of reasons. First, the students had not chosen to be there, and had no particular interest in math. Moreover, I felt that it was important to do more than give a fun math talk. I wanted the students to really learn something that would help them in their mathematics in school.
The topic I chose was “Does 0.9999… = 1?” I did this for a few reasons:
• I could pose an interesting question at the beginning and let the students think about it and come up with ideas.
• I could vary the level of the discussion based on student background. At one end I could always retreat to 1/3 = 0.3333…, multiply both sides by 3, and get 1 = 0.9999… (ignoring issues of how you multiply an infinite decimal); at the other end I could talk about infinite geometric series and convergence (although that was very unlikely). In the middle, I could talk about how you prove things in general, give the overall idea of a series, and talk about what convergence means.
• I would be able to emphasize ideas of “math makes sense”—that different pieces of math fit together and work in the same way—and I would also be able to reinforce basic mathematics skills. This is as opposed to a topic like, say, combinatorial game theory, where I could talk about logical reasoning but wouldn’t be able to tie it into what they’ve already seen or build their ability to understand numbers.
I began by introducing myself and having all of the girls tell me their name. I asked about their subway ride to Barnard and generally tried to bond a bit and be friendly. Then I put up the question “Does 0.9999… = 1?” I took some questions (“what does the bar over the 9 mean?”) and took a poll. A bunch thought no, a bunch thought yes, several said they were unsure. Great!
I next asked students to explain why they were or were not equal. I got a few ideas but nothing very deep. I knew that underneath, students were struggling with what this question even means, so I asked them straight up, what does this question mean? I got some not terribly enlightening answers.
Now it was time to actually help them understand the question. “What does = mean?” I asked (and wrote on the board). We discussed the notion of equality for a while, and gave examples of things that were equal (1 = 1, 1/4 = 0.25, 2 + 3 = 1 + 4); part of my goal was to emphasize that = means “are the same as”, not “is the outcome of an operation”. I asked them if x + 3 = 2x, and we had some debate before deciding that it is only true if x = 3 (I’m still not sure if they got this); then we had some more debate about if x + x = 2x before deciding yes. Although I had hoped to doubly address the meaning of = and the hidden quantifiers in most algebra problems, I realize in retrospect that this probably muddled the picture more than it helped.
Once I felt like they understood =, I asked what 1 means (briefly, although I wish I’d had time to talk about it more) and then I asked what 0.9999… means. I ended up getting drawn off-topic in the discussion, which is OK, but I wish I’d gotten more of a bead on “it’s 9/10 + 9/100 + 9/1000 + …”.
Instead, we ended up talking about what 0.3333… is. None of them recognized it as 1/3, which took me rather by surprise. OK, time to talk about 0.3333…
We spent some time discussing how to turn decimals into fractions in general, but quickly concluded that you couldn’t straight-up turn 0.3333… into a fraction; 333333…/100000… makes no sense. So instead I suggested looking at 0.3, 0.33, 0.333, 0.3333, and so forth. I was again a bit surprised: they seemed to have no intuition that these numbers are actually very close together. So it was time to take out a number line and draw them.
Where, I asked, is 0.3 on the number line? Only one girl knew how to put it up; she knew that 0.1, 0.2, …, 0.9 were equally spaced and so she found 0.3. But when asked about 0.33, she didn’t know. So we spent some time talking about how 0.3 is 3/10, and how to find 3/10 (divide the number line into 10ths; there’s 1/10, now we want three of them, so we go over here). Then we decided that 0.33 is 33/100, and we talked about dividing the number line into 100 pieces. Then we saw that 0.333 is 333/1000 and we talked about dividing the number line into 1000 pieces. (I actually liked this part a lot, because it required a certain amount of abstraction to visualize dividing the number line into so many pieces!)
Then I asked, “OK, where is 0.3333333333 on the number line?” The students really had no idea that it would be right by the other numbers. Some thought it would be very far out indeed! Others thought it would be close to 0; an original hypothesis had been that as you add 3s, the number gets closer and closer to 0. Eventually I got them to write it out as a fraction: 3,333,333,333/10,000,000,000. I asked them where this fraction was on the number line, and again, no idea. (Although they were amused, perhaps even impressed, at my ability to rattle off “three billion, three hundred and thirty-three million, three hundred and thirty-three thousand, three hundred and thirty-three.”) Eventually we realized that this fraction was close to the others, although I’m still not sure all of them were convinced. I finally explained that if I offered to give you that many dollars, I’d give you 34 cents and then you’d owe me some money! This seemed to help put it in perspective for them, and we briefly discussed how you might scam someone with this “trick”. Note to my future self: give the example of 5,000,000,000/10,000,000,000; it should be much easier to see that it is just 1/2 despite the large numbers.
At some point we noted that these were getting close to 1/3, although many of them didn’t have an intuition for where 1/3 is on the number line. We talked then about how to convert 1/3 into a decimal and I divided 3 into 1, at which point some of them realized that they had seen this before. That said, I realize in retrospect that I didn’t emphasize that fractions are division and so something of a learning opportunity was lost.
We’d also gotten that 1/9 = 0.1111… in this discussion, but still no insight on 0.9999…. I had them discover the relationship between 1/9 = 0.1111… and 1/3 = 0.3333…, that in both cases you can multiply by 3 to go from one to the other. Then we saw that 1/3 times 3 is 1, and that 0.3333… times 3 is 0.9999…. At this point I tried not to hint further, and despite having everything right there, the class went off on a wild tangent, thinking they could “add” something to 0.9999… to get 1. (A perfectly good theory, if they were different, and I wish I’d emphasized that more. We spent a while talking about what happens if you add 1, if you add 1/2, if you add 0.1, and 0.01, and 0.001, and so forth; I wish I’d had more time to allow them to do this themselves.) Somewhat flummoxed, I did a poll. Nearly everyone was now convinced that 0.9999… is not equal to 1!
Eventually I brought us back on track, and someone made the magical connection, and a student realized that since 1/3 times 3 is both 1 and 0.9999…, they must be equal. This part could have used more discussion, relating it back to equality, but we were really quite short on time. I ended the class by hinting at a graphical argument, and tantalizing them with the sum 1 + 1/2 + 1/4 + 1/8 + 1/16 + …, and then we were done.
In the end, the class was engaged and had very high energy, but had such a deficit of basic facts that they really struggled. I don’t think I managed to bring everyone with me as we went forward, nor was the class nearly as student-led as I’d hoped because it was hard to get them moving in this context. I’m actually happy with the presentation itself, and I hope to continue to refine it and find ways to make it accessible to everyone.
I am reminded just how much students in schools lack, however. The insistence on moving students forward to algebra and beyond when they don’t fully have the number sense to understand how things go on the number line seems to be doing them a great disservice, and I wish we could find a way to enable teachers to address these basic challenges with them. Otherwise, they’ll be continually forced to memorize procedures and they’ll never understand why they do what they do.
## Author: danzaharopol
I am a math geek. I love doing math, learning math, and teaching math. Nothing excites me more than working with young people who are discovering new and amazing things. Professionally, I founded Bridge to Enter Advanced Mathematics (BEAM), a program that makes it possible for low-income and underserved students to become scientists, mathematicians, engineers, and programmers. That's where I spend most of my time geeking out about math these days. Prior to BEAM, I was a math graduate student (studying algebraic topology) and taught math in places all around the country. I also co-founded and served as the founding CEO of Learning Unlimited, an organization that mentors college students to create enrichment programs for local middle and high school students. In my non-existent free time, I love board games, great plays, frisbee, and reading.
## 33 thoughts on “Teaching Log: 0.9999… = 1”
1. Reena says:
++ on 5,000,000,000/10,000,000,000. For my Splash “calculus for middle school” kids who tend to be very advantaged math students, I give “what is 0.00000002/0.00000001” as a warm-up problem, and get them to tell me that if you keep inserting zeroes forever it will still be 2 (because later with introducing derivatives, they need to understand that an arbitrarily small number divided by an arbitrarily small number is not necessarily also small, and though they realize that in a snap when it’s explicitly brought up, somehow it isn’t by default in their conscious).
2. Reena says:
this also, in general, looks like it will be super fantastic next time you do it because you have a detailed reflection to look back on. I should write up thoughts like this after teaching, it would probably help me improve a lot.
3. I really don’t like the 3 times .33333… “argument.” First it’s logically circular, as in order to know that 1/3 = .33333… you need to know that 3 x .33333… = .99999… = 1. If you try to dodge this by saying 1/3 = .33333… follows from long division, you could also just get 1/1 = .99999… by doing long division. Finally any philosophical issues that come up with .9999… = 1 already come up in understanding why 1/3 is .3333… rather than infinitely many 3’s followed by a 4.
1. I agree with you, although for different reasons, and it was my backup plan! However, I would claim that the class was valuable for students even if the argument was not the best one, because it provides a good foil against which you can test your knowledge and understanding of fractions.
How do you get 1/1 = .999999… by long division? That is, how do you justify not simply getting 1.000000…?
1. Certainly you can also get 1.000000… by long division if you want to, but you could just as well get .999999999… by doing it slightly differently.
2. What do you mean by slightly differently? Do you mean using some other algorithm for division? Do you mean that when you divide 1 into 1 (or 9 into 9) you don’t write a “1” at the top even though you can, and instead write a 0 and see what happens? (Surely that needs more justification!) I’d be very curious about this.
1. It’s a little hard to explain in a blog post, but I just meant do ordinary long division but put 0.999999… at the top instead. So each step in the long division looks like 10-9 = 1, then bring down the 0 to get 10, subtract 9, etc. Keeps working just fine.
1. In this case it works, but you’re not following the algorithm and you need to prove that it works! This is the part I feel needs more justification. I agree that it works, because the numbers at the bottom stay bounded, but I think this example would be more confusing than helpful in a classroom unless there’s a good and easy reason why this kind of division is OK.
4. David Jao says:
This is a great write-up and very useful for anyone involved in math education in any way. For cutting-edge researchers, it’s sometimes easy to forget just how much basic preparedness is missing in students today. I find it disheartening that students either don’t have any idea what fractions represent, or lack the ability to apply their ideas in unfamiliar situations such as 3,333,333,333/10,000,000,000. Key concepts and ideas such as these need constant emphasis and reinforcement or else they’ll just fall through. The teachers themselves also need to know the concepts and recognize which ones are fundamental. I suspect that the understanding deficit among students starts from the very beginning of the curriculum and grows each year until it becomes impossible to correct.
I’ve also had my share of Internet Arguments about 0.99999… = 1. I find that this is a very hard question to answer satisfactorily for the student. Congratulations to you for attempting it! The fundamental obstacle is that the building blocks in the definition of 0.99999…, such as 1/3, 1/9, and even 0.3, 0.33, 0.333, etc. are algebraic quantities: they are defined from fractions. But the final step, that of constructing the infinite decimal, is an analytic concept. The typical student who asks or considers this question is still struggling with the algebra, and is not prepared to make the leap to analysis. At best, the students might be able to take baby steps towards analysis, such as recognizing the concept of distance: 0.3 is close to 0.33 which is close to 0.333 etc. (although it seems that your class needed considerable assistance even to get this far). Still, it’s a big jump from that to the least upper bound property, which is what you need to show that 0.99999… = 1. Yes, you really do need the LUB property, since there exist number systems without this property (hyperreals, surreals) in which infinitesimals exist, and where 0.99999… can reasonably be defined to equal something different from 1. Of course, you would never be able to mention the LUB property to this kind of audience, but you cannot avoid its implicit presence in the background; although the two are not formally logically equivalent, I am not aware of any rigorous way to show that 0.99999… = 1 without invoking the LUB property (other than by fiat), and I have thought about this for a while. I find that even students with no ability whatsoever to reason logically about mathematics will still have some intuitive sense that “0.99999… = 1” is a deep equation requiring a powerful new concept for a full understanding. It can be frustrating, to say the least, when students recognize (even subconsciously) that they need something, but cannot understand what it is!
I wonder how long your talk was, and how many were in attendance? These variables would certainly affect what kind of talk I could give.
1. Actually, we’re saved a little bit here: I’m pretty sure that all you really need for the core issues is the Archimedean property, which is weaker than the LUB property, and much more intuitive to believe. (I’ve also wrestled extensively with how to address 0.99999… = 1 without the LUB property.)
I think the core argument is that if 0.9999… is not 1, then there must be a number between them. What is the average of 0.99999… and 1? What number could you find between them? What would you get if you subtracted one from the other if not 0? Ultimately students are forced to add extra digit after infinitely many digits, and a socratic discussion of what those extra digits might mean is usually enough to demonstrate that it’s not the number system that we usually work with, and the exercise of thinking about it with guidance can be very valuable.
Ideally, if the students are comfortable enough with fractions and decimals, you’d establish much earlier in the class what decimals really mean. In particular, the idea that 0.99999… = 9/10 + 9/100 + 9/1000 + … is going to help tremendously with the above explanation. Then if students try to add on extra digits on the end, you can ask them what that would mean as a number. They won’t be able to come up with anything once they’ve established the meanings of decimals in general, although you can discuss how there are other number systems with infinitesimals briefly without confusing them too much, and then you’ve even seeded the idea that we can create other number systems by setting up sufficient rules at the beginning.
Another way to deal with this is to talk about geometric sequences. From the formula for finite geometric sequences, you can derive the formula for infinite geometric sequences. Then you have a formula showing you plain as day that as you add more 9s, you get arbitrarily close to 1.
One other technique I’ve seen, although I haven’t used it much myself, is to do a graphical argument. For example, you can intuitively get the idea of 1/2 + 1/4 + 1/8 + … = 1 by taking a square and dividing it in half; then taking the remaining half and dividing it in half; and so forth. It’s not hard to see that every point in the square will eventually be covered by one of the rectangles you’re drawing. You can do the same thing with 9/10 + 9/100 + 9/1000 + ….
2. Are there self-consistent theories of infinite decimals for the hyperreals or surreals according to which .9999… is less than 1? In the hyperreals, you could cut off the sum at some randomly chosen infinite decimal place, but that’s pretty arbitrary. In the surreals, it seems like the most natural definition of an infinite decimal would be a cut defined by its finite approximations above and below, but since those are all rational, the cut will be the ordinary real number.
1. If we look at .9999… as the sequence of its truncations, i.e. {.9, .99, .999, …}, 1. as the constant sequence
{1., 1., …}, and our hyperreals as real sequences modulo sequences that vanish on some element of an ultrafilter that is an extension of the filter of subsets of natural numbers with finite complements we will get 1. – .999… = an infinitely small number represented by the sequence {.1, .01, .001, …}. See my article http://en.wikipedia.org/wiki/Hyperreals#An_intuitive_approach_to_the_ultrapower_construction for more details.
1. Oh, you want to drag in the *construction* of the hyperreals as an ultraproduct. I find that inelegant, and essentially just as arbitrary, since it amounts to picking the particular infinite natural number represented by the sequence (1,2,3,4,5,…) and cutting off the decimal expansion there.
2. It may be not too elegant, but at least it is understandable. Whatever way you do it, you need to postulate the existence of an ultrafilter or introduce a similar assumptions to get a hyperreal field. Anyhow, you asked a question and you got an answer, sorry you didn’t like it.
5. suevanhattum says:
This is a beautiful description of what sounds very much like a math circle. Thank you for sharing the details.
How many girls were you working with?
1. I think it was in the range of 12-15, although I don’t exactly remember.
Also, the style of math circles was very much on my mind when I was planning this class! I wanted to capture the spirit of inquiry about interesting questions that seems to be a common thread among them.
6. On my first day teaching eighth grade, we hit a major speed bump when we got to the sequence 1, 4, 9, 16, 25… and not a single one of them knew what the pattern was.
I blame calculators.
1. Calculators are absolutely a factor; the beautiful thing about doing calculations by hand is that it forces you to meditate on the numbers, at least if you’re the sort to naturally think about what you’re doing. Unfortunately, I am not really convinced the situation was any better before calculators.
2. I was wondering the other day (when I was teaching an upper level college class on why every real number has a decimal expansion and related issues like .99999… = 1) whether students typically knew that 1/3 = .3333… before the invention of calculators. It seemed plausible to me that it’s only because the calculator turns 1/3 into .3333… that student know it. On the other hand, maybe before calculators people did enough long division by hand that they also knew this then.
1. Clearly, the solution is to redesign calculators so that when you divide 1/1 they give the answer .9999…
7. I wonder whether students would be less confused if we never taught them the concept of ‘infinite decimal’ at all. As David said, the meaning of an infinite decimal is a fundamentally analytic one, essentially involving limits, which we don’t try to teach students about in general until calculus. Of course, that’s not something we can change, but are we really explaining an important concept, or just mitigating the damage done by teaching people to trust symbols that they don’t have the conceptual background to really understand?
8. Ari Nieh says:
Well, we do have to teach them the decimal expansions for 1/2, 1/4, and 1/5. I don’t see a good way to get around mentioning the decimal expansion of 1/3.
1. What I was trying to say is that 1/3 doesn’t _have_ a decimal expansion in the same way that 1/2, 1/4, and 1/5 do. Maybe we are doing students a disservice by implying that it does.
1. Don’t students need decimal approximations? I feel like knowing that 33% is roughly 1/3 is important. And once you’ve gone there, it’s hard to avoid the question of why it’s only approximate.
1. Of course they need to know that 1/3 is approximately .33. I was wondering whether it would be possible to tell them that .33, .333, .3333, and so on are successively better approximations to 1/3 without telling them that it’s meaningful to write “.3333….”.
2. Maybe we are doing students a disservice by not explaining approximations in general on an elementary level, without dragging in the whole abstract elephant of the real numbers.
1. Ah yes! I’ve read that post before, and it’s really good. Now I’ve actually scrolled down to the comments, and noted them as fodder for future classes. :)
9. And why would you want to do that, Mike?
1. Oh, sorry, I got it, you don’t want to confuse them…
10. I remember approaching this exact same problem as a student in 7th grade, when the teacher stated with no explanation that .9 repeating = 1. Between classes the other students and I came to the conclusion that it wasn’t actually true (that it must’ve been something mathematicians decided because practically you had to eventually round /somewhere/; ah, the days when we all thought math was only for practical purposes!), because if you go 9/10 of the way to somewhere, and do it again, and again, no matter how many times you would never actually get there. You couldn’t give a distance between it and 1, but that was like how you can write x < 1, and x can be any infinitesimal amount smaller than 1 to satisfy the inequality; .9 repeating was on one side of that line and 1 on the other.
It came into a little bit better perspective when I got to infinite series a year later (where it was replaced by something unsatisfying like "it's true because inexplicably weird things happen when you add infinity things"), but I don't think I had an "aha" moment until I read Anders's answer to this post: http://www.quora.com/Is-it-possible-to-count-in-base-pi . I realized I had never even considered the possibility of 2 representations meaning the same number until I had to think about irrational bases. Don't know if that's useful to you as a teaching tool, but thought you might find it interesting.
11. DH says:
I am an engineer who made a mid-life career change to teaching. I’m currently teaching chemistry in a public high school, and I am continually amazed by the lack of basic number sense many of my high school students exhibit. They all must have algebra I in order to take chemistry, but the majority of my students have extreme difficulties with math. Take the problem “round 299854 to the thousands place” as an example. We did this as part of a year-end review (we had covered place value and rounding early in the year). At least a third of the class had no idea where the thousands place was. What surprised me, however, was that even after a review of place value and rounding rules, some students rounded to 299 or even to 300 in addition to the usual mistake of 299000. When I talked to them about their thought process I found the same lack of general comprehension of what rounding meant and what place value meant that you found in your lesson. Unfortunately, I did not think of putting up the number line, but I am certain that these students had the same issues with the number line. Incidentally, this same issue occurred when rounding 144 to the tens place, so it was not because of the comma between 299,854.
I find myself going back to manipulatives frequently when explaining what should be a simple math concept, such as last week when we did a lab measuring the energy content of snack foods with calorimetry and I found that many of the students could not understand that to convert from 1/2 serving to 1 serving they should multiply by two. I had to get out beakers and fill two of them with 1/2 a serving of Cheetos and another with a full serving of Cheetos and show them how two half-full beakers was the same as 1 full beaker. I also took out an empty beaker and poured the two half beakers together. After that, they were able to understand that 1/2 + 1/2 = 1 whole, but a few were still unable to see that 1/2 + 1/2 = 2 x 1/2. These students seem to lack the fundamental understanding of what multiplication means, which they should have grasped in elementary school.
Re-teaching math is a significant part of my chemistry teaching job. So far in my non-quantitative observations I have to say that using real things to demonstrate math concepts works with this group of students better than anything. I suspect that many students lack a bridge from concrete math to abstract math. | 6,423 | 26,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-24 | latest | en | 0.986355 |
https://areavolumecalculator.com/centimeters-to-kilometers/ | 1,685,827,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00471.warc.gz | 127,835,301 | 15,110 | Created By : Abhinandan Kumar
Reviewed By : Rajashekhar Valipishetty
Last Updated : Jun 03, 2023
1 Centimeter is equal to 0.0000100001087999999 Kilometers. To convert a Centimeter measurement to a KiloMeter measurement, multiply the number of Centimeters by 0.0000100001087999999 and change the units to Kilometers.
Enter the value in Centimeters
## How to Convert Centimeters to Kilometer?
To Convert Foot to Kilometer multiply the number of centimeters by kilometers and change the units to a kilometer. Let's have a detailed explanation of cm to km conversionby the following lines so that you can easily find the calculations on how to convert Centimeter to Kilometer. Or else, make use of our free online calculator and get the calculation done faster.
The formula for Unit conversion of 1 Centimeter = 0.00001 Kilometers.
So, if you want to convert any centimeter input into kilometer then substitute the value into this formula and calculate the cm to km conversion.
In order to give proper clarity for you, we have provided an example of how to convert centimeters to a kilometer below. Have a look at the solved example and understand the calculations involved in converting from Cm to Km.
Solved Example:
Convert 5000 Centimeters to Kilometers?
Solution:
We know the formula to convert centimeter to kilometers as
1 Centimeter = 1/100000 Kilometer
To convert Centimeters to Kilometers, divide the centimeter value by 100000.
Here, in this case, the centimeter value given is 5,000
so on substituting in the conversion formula we will have the results as such
km=5000/100000 km
= 0.05 kilometers
Result in Kilometers: 0.05 meters
Therefore, 5000 Centimeter = 0.05 Kilometers.
### Steps to Use Free Online Centimeter to Kilometer Conversion Calculator
Our handy and free calculator is very simple and easy to use. It converts centimeters to kilometers within a fraction of sections and displays output on the screen along with an explanation. In order to know how to use and how it works just follow the simple steps given below:
1. Enter the value of centimeters in the input field.
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### Centimeter to Kilometer Conversion Table
Centimeter [cm]Kilometer [km]
0.01 cm1.0E-7 km
0.1 cm1.0E-6 km
1 cm1.0E-5 km
2 cm2.0E-5 km
3 cm3.0E-5 km
5 cm5.0E-5 km
10 cm0.0001 km
20 cm0.0002 km
50 cm0.0005 km
100 cm0.001 km
1000 cm0.01 km
### FAQs on Convert Cm to Km Calculator - Centimeter Conversion Calculator
1. How do you convert cm to KM?
As there are 100000 centimeters in a kilometer, to convert centimeters to kilometers, divide the centimeter value by 100000.
2. What is the formula to convert Centimeters (cm) to Kilometer (Km)?
You can convert centimeters into kilometers with a simple formula i.e.,Kilometers = Centimeters รท 100,000
3. Why we use centimeters and kilometers units?
Centimeters and kilometers are both units used to measure the length.
4. What is the unit conversion of 5 cm to km?
Converting into km. (1km=100000cm). So 5 cm=5/100000=0.00005km or 5.0E-5 km.
### More Examples
(some results rounded)
cm km
0.0001 1.00 x 10-09
0.0006 6.00 x 10-09
0.001 1.00 x 10-08
0.006 6.00 x 10-08
0.01 1.00 x 10-07
0.06 6.00 x 10-07
0.1 0.000001
0.6 0.000006
1 0.00001
2 0.00002
3 0.00003
4 0.00004
5 0.00005
6 0.00006
7 0.00007
8 0.00008
9 0.00009
10 0.0001
11 0.00011
12 0.00012
13 0.00013
14 0.00014
15 0.00015
16 0.00016
17 0.00017
18 0.00018
19 0.00019
20 0.0002
21 0.00021
22 0.00022
23 0.00023
24 0.00024
25 0.00025
26 0.00026
27 0.00027
28 0.00028
29 0.00029
30 0.0003
31 0.00031
32 0.00032
33 0.00033
34 0.00034
35 0.00035
36 0.00036
37 0.00037
38 0.00038
39 0.00039
40 0.0004
41 0.00041
42 0.00042
43 0.00043
44 0.00044
45 0.00045
46 0.00046
47 0.00047
48 0.00048
49 0.00049
50 0.0005
51 0.00051
52 0.00052
53 0.00053
54 0.00054
55 0.00055
56 0.00056
57 0.00057
58 0.00058
59 0.00059
60 0.0006
61 0.00061
62 0.00062
63 0.00063
64 0.00064
65 0.00065
66 0.00066
67 0.00067
68 0.00068
69 0.00069
70 0.0007
71 0.00071
72 0.00072
73 0.00073
74 0.00074
75 0.00075
76 0.00076
77 0.00077
78 0.00078
79 0.00079
80 0.0008
81 0.00081
82 0.00082
83 0.00083
84 0.00084
85 0.00085
86 0.00086
87 0.00087
88 0.00088
89 0.00089
90 0.0009
91 0.00091
92 0.00092
93 0.00093
94 0.00094
95 0.00095
96 0.00096
97 0.00097
98 0.00098
99 0.00099
100 0.001
101 0.00101
102 0.00102
103 0.00103
104 0.00104
105 0.00105
106 0.00106
107 0.00107
108 0.00108
109 0.00109
110 0.0011
111 0.00111
112 0.00112
113 0.00113
114 0.00114
115 0.00115
116 0.00116
117 0.00117
118 0.00118
119 0.00119
120 0.0012
121 0.00121
122 0.00122
123 0.00123
124 0.00124
125 0.00125
126 0.00126
127 0.00127
128 0.00128
129 0.00129
130 0.0013
131 0.00131
132 0.00132
133 0.00133
134 0.00134
135 0.00135
136 0.00136
137 0.00137
138 0.00138
139 0.00139
140 0.0014
141 0.00141
142 0.00142
143 0.00143
144 0.00144
145 0.00145
146 0.00146
147 0.00147
148 0.00148
149 0.00149
150 0.0015
151 0.00151
152 0.00152
153 0.00153
154 0.00154
155 0.00155
156 0.00156
157 0.00157
158 0.00158
159 0.00159
160 0.0016
161 0.00161
162 0.00162
163 0.00163
164 0.00164
165 0.00165
166 0.00166
167 0.00167
168 0.00168
169 0.00169
170 0.0017
171 0.00171
172 0.00172
173 0.00173
174 0.00174
175 0.00175
176 0.00176
177 0.00177
178 0.00178
179 0.00179
180 0.0018
181 0.00181
182 0.00182
183 0.00183
184 0.00184
185 0.00185
186 0.00186
187 0.00187
188 0.00188
189 0.00189
190 0.0019
191 0.00191
192 0.00192
193 0.00193
194 0.00194
195 0.00195
196 0.00196
197 0.00197
198 0.00198
199 0.00199
200 0.002
202 0.00202
204 0.00204
206 0.00206
208 0.00208
210 0.0021
212 0.00212
214 0.00214
216 0.00216
cm km
218 0.00218
220 0.0022
222 0.00222
224 0.00224
226 0.00226
228 0.00228
230 0.0023
232 0.00232
234 0.00234
236 0.00236
238 0.00238
240 0.0024
242 0.00242
244 0.00244
246 0.00246
248 0.00248
250 0.0025
252 0.00252
254 0.00254
256 0.00256
258 0.00258
260 0.0026
262 0.00262
264 0.00264
266 0.00266
268 0.00268
270 0.0027
272 0.00272
274 0.00274
276 0.00276
278 0.00278
280 0.0028
282 0.00282
284 0.00284
286 0.00286
288 0.00288
290 0.0029
292 0.00292
294 0.00294
296 0.00296
298 0.00298
300 0.003
302 0.00302
304 0.00304
306 0.00306
308 0.00308
310 0.0031
312 0.00312
314 0.00314
316 0.00316
318 0.00318
320 0.0032
322 0.00322
324 0.00324
326 0.00326
328 0.00328
330 0.0033
332 0.00332
334 0.00334
336 0.00336
338 0.00338
340 0.0034
342 0.00342
344 0.00344
346 0.00346
348 0.00348
350 0.0035
352 0.00352
354 0.00354
356 0.00356
358 0.00358
360 0.0036
362 0.00362
364 0.00364
366 0.00366
368 0.00368
370 0.0037
372 0.00372
374 0.00374
376 0.00376
378 0.00378
380 0.0038
382 0.00382
384 0.00384
386 0.00386
388 0.00388
390 0.0039
392 0.00392
394 0.00394
396 0.00396
398 0.00398
400 0.004
402 0.00402
404 0.00404
406 0.00406
408 0.00408
410 0.0041
412 0.00412
414 0.00414
416 0.00416
418 0.00418
420 0.0042
422 0.00422
424 0.00424
426 0.00426
428 0.00428
430 0.0043
432 0.00432
434 0.00434
436 0.00436
438 0.00438
440 0.0044
442 0.00442
444 0.00444
446 0.00446
448 0.00448
450 0.0045
452 0.00452
454 0.00454
456 0.00456
458 0.00458
460 0.0046
462 0.00462
464 0.00464
466 0.00466
468 0.00468
470 0.0047
472 0.00472
474 0.00474
476 0.00476
478 0.00478
480 0.0048
482 0.00482
484 0.00484
486 0.00486
488 0.00488
490 0.0049
492 0.00492
494 0.00494
496 0.00496
498 0.00498
500 0.005
505 0.00505
510 0.0051
515 0.00515
520 0.0052
525 0.00525
530 0.0053
535 0.00535
540 0.0054
545 0.00545
550 0.0055
555 0.00555
560 0.0056
565 0.00565
570 0.0057
575 0.00575
580 0.0058
585 0.00585
590 0.0059
595 0.00595
600 0.006
605 0.00605
610 0.0061
615 0.00615
620 0.0062
625 0.00625
630 0.0063
635 0.00635
640 0.0064
645 0.00645
650 0.0065
655 0.00655
660 0.0066
665 0.00665
670 0.0067
675 0.00675
680 0.0068
685 0.00685
690 0.0069
695 0.00695
700 0.007
705 0.00705
710 0.0071
715 0.00715
720 0.0072
725 0.00725
730 0.0073
735 0.00735
740 0.0074
745 0.00745
750 0.0075
755 0.00755
760 0.0076
765 0.00765
770 0.0077
775 0.00775
780 0.0078
785 0.00785
790 0.0079
795 0.00795
800 0.008
805 0.00805
810 0.0081
815 0.00815
820 0.0082
825 0.00825
830 0.0083
835 0.00835
840 0.0084
845 0.00845
850 0.0085
855 0.00855
860 0.0086
865 0.00865
870 0.0087
cm km
875 0.00875
880 0.0088
885 0.00885
890 0.0089
895 0.00895
900 0.009
905 0.00905
910 0.0091
915 0.00915
920 0.0092
925 0.00925
930 0.0093
935 0.00935
940 0.0094
945 0.00945
950 0.0095
955 0.00955
960 0.0096
965 0.00965
970 0.0097
975 0.00975
980 0.0098
985 0.00985
990 0.0099
995 0.00995
1000 0.01
1050 0.0105
1100 0.011
1150 0.0115
1200 0.012
1250 0.0125
1300 0.013
1350 0.0135
1400 0.014
1450 0.0145
1500 0.015
1550 0.0155
1600 0.016
1650 0.0165
1700 0.017
1750 0.0175
1800 0.018
1850 0.0185
1900 0.019
1950 0.0195
2000 0.02
2050 0.0205
2100 0.021
2150 0.0215
2200 0.022
2250 0.0225
2300 0.023
2350 0.0235
2400 0.024
2450 0.0245
2500 0.025
2550 0.0255
2600 0.026
2650 0.0265
2700 0.027
2750 0.0275
2800 0.028
2850 0.0285
2900 0.029
2950 0.0295
3000 0.03
3050 0.0305
3100 0.031
3150 0.0315
3200 0.032
3250 0.0325
3300 0.033
3350 0.0335
3400 0.034
3450 0.0345
3500 0.035
3550 0.0355
3600 0.036
3650 0.0365
3700 0.037
3750 0.0375
3800 0.038
3850 0.0385
3900 0.039
3950 0.0395
4000 0.04
4050 0.0405
4100 0.041
4150 0.0415
4200 0.042
4250 0.0425
4300 0.043
4350 0.0435
4400 0.044
4450 0.0445
4500 0.045
4550 0.0455
4600 0.046
4650 0.0465
4700 0.047
4750 0.0475
4800 0.048
4850 0.0485
4900 0.049
4950 0.0495
5000 0.05
5500 0.055
6000 0.06
6500 0.065
7000 0.07
7500 0.075
8000 0.08
8500 0.085
9000 0.09
9500 0.095
10000 0.1
11000 0.11
12000 0.12
13000 0.13
14000 0.14
15000 0.15
16000 0.16
17000 0.17
18000 0.18
19000 0.19
20000 0.2
21000 0.21
22000 0.22
23000 0.23
24000 0.24
25000 0.25
26000 0.26
27000 0.27
28000 0.28
29000 0.29
30000 0.3
31000 0.31
32000 0.32
33000 0.33
34000 0.34
35000 0.35
36000 0.36
37000 0.37
38000 0.38
39000 0.39
40000 0.4
41000 0.41
42000 0.42
43000 0.43
44000 0.44
45000 0.45
46000 0.46
47000 0.47
48000 0.48
49000 0.49
50000 0.5
51000 0.51
52000 0.52
53000 0.53
54000 0.54
55000 0.55
56000 0.56
57000 0.57
58000 0.58
59000 0.59
60000 0.6
61000 0.61
62000 0.62
63000 0.63
64000 0.64
65000 0.65
66000 0.66
67000 0.67
68000 0.68
69000 0.69
70000 0.7
71000 0.71
72000 0.72
73000 0.73
74000 0.74
75000 0.75
76000 0.76
77000 0.77
78000 0.78
79000 0.79
80000 0.8
81000 0.81
82000 0.82
83000 0.83
84000 0.84
85000 0.85
86000 0.86
87000 0.87
88000 0.88
89000 0.89
90000 0.9
91000 0.91
92000 0.92
93000 0.93
94000 0.94
95000 0.95
96000 0.96
97000 0.97
98000 0.98
99000 0.99
100000 1
200000 2
300000 3
400000 4
500000 5
600000 6
700000 7
800000 8
900000 9
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```EF 507
QUANTITATIVE METHODS FOR ECONOMICS
AND FINANCE
FALL 2008
Chapter 6
Continuous Random Variables and Probability
Distributions
Chap 6-1
Probability Distributions
Probability
Distributions
Ch. 5
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Uniform
Hypergeometric
Normal
Poisson
Exponential
Ch. 6
Chap 6-2/62
Continuous Probability Distributions
A continuous random variable is a variable
that can assume any value in an interval
thickness of an item
time required to complete a task
temperature of a solution
height in inches
These can potentially take on any value,
depending only on the ability to measure
accurately.
Chap 6-3/62
Cumulative Distribution Function
The cumulative distribution function, F(x), for
a continuous random variable X expresses the
probability that X does not exceed the value of x
F(x) P(X x)
Let a and b be two possible values of X, with
a < b. The probability that X lies between a
and b is
P(a X b) F(b) F(a)
Chap 6-4/62
Probability Density Function
The probability density function, f(x), of random variable X has the
following properties:
1. f(x) > 0 for all values of x
2. The area under the probability density function f(x) over all values of the
random variable X is equal to 1.0
3. The probability that X lies between two values is the area under the
density function graph between the two values
4. The cumulative density function F(x0) is the area under the probability
density function f(x) from the minimum x value up to x0
x0
f(x 0 ) f(x)dx
xm
where xm is the minimum value of the random variable x
Chap 6-5/62
Probability as an Area
Shaded area under the curve is the
probability that X is between a and b
f(x)
P (a ≤ x ≤ b)
= P (a < x < b)
(Note that the
probability of any
individual value is
zero)
a
b
x
Chap 6-6/62
The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Exponential
Chap 6-7/62
The Uniform Distribution
The uniform distribution is a probability
distribution that has equal probabilities for all
possible outcomes of the random variable
f(x)
Total area under the
uniform probability
density function is 1.0
xmin
xmax x
Chap 6-8/62
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
f(x) =
1
if a x b
ba
0
otherwise
where
f(x) = value of the density function at any x value
a = minimum value of x
b = maximum value of x
Chap 6-9/62
Properties of the
Uniform Distribution
The mean of a uniform distribution is
ab
μ
2
The variance is
2
(b
a)
σ2
12
Chap 6-10/62
Uniform Distribution Example
Example: Uniform probability distribution
over the range 2 ≤ x ≤ 6:
1
f(x) = 6 - 2 = 0.25 for 2 ≤ x ≤ 6
f(x)
μ
0.25
ab 26
4
2
2
(b - a)2 (6 - 2)2
σ
1.333
12
12
2
2
6
x
Chap 6-11/62
Expectations for Continuous
Random Variables
The mean of X, denoted μX , is defined as the
expected value of X
μX E(X)
The variance of X, denoted σX2 , is defined as the
expectation of the squared deviation, (X - μX)2, of a
random variable from its mean
σ 2X E[(X μX )2 ]
Chap 6-12/62
Linear Functions of Variables
Let W = a + bX , where X has mean μX and
variance σX2 , and a and b are constants
Then the mean of W is
μW E(a bX) a bμX
the variance is
σ Var(a bX) b σ
2
W
2
2
X
the standard deviation of W is
σW b σX
Chap 6-13/62
Linear Functions of Variables
(continued)
An important special case of the previous results is the
standardized random variable
X μX
Z
σX
which has a mean 0 and variance 1
Chap 6-14/62
The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Uniform
Normal
Exponential
Chap 6-15/62
The Normal Distribution
(continued)
‘Bell Shaped’
Symmetrical
f(x)
Mean, Median and Mode
are Equal
Location is determined by the
mean, μ
σ
μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+ to
x
Mean
= Median
= Mode
Chap 6-16/62
The Normal Distribution
(continued)
The normal distribution closely approximates the
probability distributions of a wide range of random
variables (Central Limit Theorem)
Distributions of sample means approach a normal
distribution given a “large” sample size
Computations of probabilities are direct and elegant
The normal probability distribution has led to good
business decisions for a number of applications
Chap 6-17/62
Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
Chap 6-18/62
The Normal Distribution
Shape
f(x)
Changing μ shifts the
distribution left or right.
σ
Changing σ increases
or decreases the
μ
x
Given the mean μ and variance σ we define the normal
distribution using the notation
X ~ N(μ,σ 2 )
Chap 6-19/62
The Normal Probability
Density Function
The formula for the normal probability density
function is
1
f(x)
e
2π
Where
(x μ)2
2σ 2
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
x = any value of the continuous variable, < x <
Chap 6-20/62
Cumulative Normal Distribution
For a normal random variable X with mean μ and
variance σ2 , i.e., X~N(μ, σ2), the cumulative
distribution function is
F(x 0 ) P(X x 0 )
f(x)
P(X x 0 )
0
x0
x
Chap 6-21/62
Finding Normal Probabilities
The probability for a range of values is
measured by the area under the curve
P(a X b) F(b) F(a)
a
μ
b
x
Chap 6-22/62
Finding Normal Probabilities
(continued)
F(b) P(X b)
a
μ
b
x
a
μ
b
x
a
μ
b
x
F(a) P(X a)
P(a X b) F(b) F(a)
Chap 6-23/62
The Standardized Normal
Any normal distribution (with any mean and variance
combination) can be transformed into the
standardized normal distribution (Z), with mean 0
and variance 1
f(Z)
Z ~ N(0 ,1)
1
0
Z
Need to transform X units into Z units by subtracting the
mean of X and dividing by its standard deviation
X μ
Z
σ
Chap 6-24/62
Example
If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
X μ 200 100
Z
2.0
σ
50
This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Chap 6-25/62
Comparing X and Z units
100
0
200
2.0
X
Z
(μ = 100, σ = 50)
(μ = 0, σ = 1)
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (X) or in standardized units (Z)
Chap 6-26/62
Finding Normal Probabilities
b μ
a μ
P(a X b) P
Z
σ
σ
b μ a μ
F
F
σ σ
f(x)
a
a μ
σ
µ
b
x
0
b μ
σ
Z
Chap 6-27/62
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(X) P( X μ) 0.5
0.5
P(μ X ) 0.5
0.5
μ
X
P( X ) 1.0
Chap 6-28/62
Appendix Table 1
The Standardized Normal table in the textbook
(Appendix Table 1) shows values of the
cumulative normal distribution function
For a given Z-value a , the table shows F(a)
(the area under the curve from negative infinity to a)
F(a) P(Z a)
0
a
Z
Chap 6-29/62
The Standardized Normal Table
Appendix Table 1 gives the probability F(a) for
any value a (A)
0.9772
Example:
P(Z < 2.00) = 0.9772
0
2.00
Z
Chap 6-30/62
The Standardized Normal Table
(continued)
For negative Z-values, use the fact that the
distribution is symmetric to find the needed
probability:
0.9772
0.0228
Example:
P(Z < -2.00) = 1 – 0.9772
= 0.0228
0
2.00
Z
0.9772
0.0228
-2.00
0
Z
Chap 6-31/62
General Procedure for
Finding Probabilities
To find P(a < X < b) when X is
distributed normally:
1- Draw the normal curve for the problem
in terms of X
2- Translate X-values to Z-values
3- Use the Cumulative Normal Table
Chap 6-32/62
Finding Normal Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0
Find P(X < 8.6)
X
8.0
8.6
Chap 6-33/62
Finding Normal Probabilities
(continued)
Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(X < 8.6)
X μ 8.6 8.0
Z
0.12
σ
5.0
μ=8
σ = 10
8 8.6
P(X < 8.6)
μ=0
σ=1
X
0 0.12
Z
P(Z < 0.12)
Chap 6-34/62
Solution: Finding P(Z < 0.12)
Standardized Normal Probability
Table (Portion) (B)
z
F(z)
P(X < 8.6)
= P(Z < 0.12)
F(0.12) = 0.5478
0.10 0.5398
0.11
0.5438
0.12 0.5478
Z
0.13 0.5517
0.00
0.12
Chap 6-35/62
Upper Tail Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now Find P(X > 8.6)
X
8.0
8.6
Chap 6-36/62
Upper Tail Probabilities
(continued)
Now Find P(X > 8.6)…
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - 0.5478 = 0.4522
0.5478
1.000
1.0 - 0.5478
= 0.4522
Z
0
0.12
Z
0
0.12
Chap 6-37/62
Finding the X value for a Known
Probability
Steps to find the X value for a known
probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
Z
X μ
σ
X μ Zσ
Chap 6-38/62
Finding the X value for a
Known Probability
(continued)
Example:
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now find the X value so that only 20% of all
values are below this X
0.20
?
?
8.0
0
X
Z
Chap 6-39/62
Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known probability
Standardized Normal Probability 20% area in the lower
Table (Portion) (C)
tail is consistent with a
z
F(z)
Z value of -0.84
0.80
0.82 0.7939
0.83 0.7967
0.20
0.84 0.7995
0.85 0.8023
?
8.0
-0.84 0
X
Z
Chap 6-40/62
Finding the X value
2. Convert to X units using the formula:
X μ Zσ
8.0 ( 0.84)5.0
3.80
So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80
Chap 6-41/62
Assessing Normality
Not all continuous random variables are
normally distributed
It is important to evaluate how well the data is
approximated by a normal distribution
Chap 6-42/62
Normal Distribution Approximation
for Binomial Distribution
Recall the binomial distribution:
n independent trials
probability of success on any given trial = P
Random variable X:
Xi =1 if the ith trial is “success”
Xi =0 if the ith trial is “failure”
E(X) μ nP
Var(X) σ nP(1- P)
2
Chap 6-43/62
Normal Distribution Approximation
for Binomial Distribution
(continued)
The shape of the binomial distribution is
approximately normal if n is large
The normal is a good approximation to the binomial
when nP(1 – P) > 9
Standardize to Z from a binomial distribution:
X E(X)
X np
Z
Var(X)
nP(1 P)
Chap 6-44/62
Normal Distribution Approximation
for Binomial Distribution
(continued)
Let X be the number of successes from n independent
trials, each with probability of success P.
If nP(1 - P) > 9,
a nP
b
nP
P(a X b) P
Z
nP(1 P)
nP(1
P)
Chap 6-45/62
Binomial Approximation Example
40% of all voters support ballot proposition A. What
is the probability that between 76 and 80 voters
indicate support in a sample of n = 200 ?
E(X) = µ = nP = 200(0.40) = 80
Var(X) = σ2 = nP(1 – P) = 200(0.40)(1 – 0.40) = 48
( note: nP(1 – P) = 48 > 9 )
76 80
80 80
P(76 X 80) P
Z
200(0.4)(1 0.4)
200(0.4)(1 0.4)
P(0.58 Z 0)
F(0) F(0.58)
0.5000 0.2810 0.2190
Chap 6-46/62
The Exponential Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Exponential
Chap 6-47/62
The Exponential Distribution
Used to model the length of time between two
occurrences of an event (the time between
arrivals)
Examples:
Time between trucks arriving at an unloading dock
Time between transactions at an ATM Machine
Time between phone calls to the main operator
Chap 6-48/62
The Exponential Distribution
(continued)
The exponential random variable T (t>0) has a
probability density function
f(t) λ e
λ t
for t 0
Where
is the mean number of occurrences per unit time
t is the number of time units until the next occurrence
e = 2.71828
T is said to follow an exponential probability distribution
Chap 6-49/62
The Exponential Distribution
Defined by a single parameter, its mean (lambda)
The cumulative distribution function (the probability that
an arrival time is less than some specified time t) is
F(t) 1 e
where
λt
e = mathematical constant approximated by 2.71828
= the population mean number of arrivals per unit
t = any value of the continuous variable where t > 0
Chap 6-50/62
Exponential Distribution
Example
Example: Customers arrive at the service counter at
the rate of 15 per hour. What is the probability that the
arrival time between consecutive customers is less
than three minutes?
The mean number of arrivals per hour is 15, so = 15
Three minutes is 0.05 hours
P(arrival time <0.05) = 1 – e- X = 1 – e-(15)(0.05) = 0.5276
So there is a 52.76% probability that the arrival time
between successive customers is less than three
minutes
Chap 6-51/62
Joint Cumulative Distribution
Functions
Let X1, X2, . . .Xk be continuous random variables
Their joint cumulative distribution function,
F(x1, x2, . . .xk)
defines the probability that simultaneously X1 is less
than x1, X2 is less than x2, and so on; that is
F(x1, x2,, xk ) P(X1 x1 X2 x2 Xk xk )
Chap 6-52/62
Joint Cumulative Distribution
Functions
(continued)
The cumulative distribution functions
F(x1), F(x2), . . .,F(xk)
of the individual random variables are called their
marginal distribution functions
The random variables are independent if and only if
F(x1, x2,, xk ) F(x1)F(x 2 )F(xk )
Chap 6-53/62
Covariance
Let X and Y be continuous random variables, with
means μx and μy
The expected value of (X - μx)(Y - μy) is called the
covariance between X and Y
Cov(X, Y) E[(X μx )(Y μy )]
An alternative but equivalent expression is
Cov(X, Y) E(XY) μxμy
If the random variables X and Y are independent, then the
covariance between them is 0. However, the converse is not true.
Chap 6-54/62
Correlation
Let X and Y be jointly distributed random variables.
The correlation between X and Y is
Cov(X, Y)
ρ Corr(X, Y)
σ Xσ Y
Chap 6-55/62
Sums of Random Variables
Let X1, X2, . . .Xk be k random variables with
means μ1, μ2,. . . μk and variances
σ12, σ22,. . ., σk2. Then:
The mean of their sum is the sum of their
means
E(X1 X2 Xk ) μ1 μ2 μk
Chap 6-56/62
Sums of Random Variables
(continued)
Let X1, X2, . . .Xk be k random variables with means μ1,
μ2,. . . μk and variances σ12, σ22,. . ., σk2. Then:
If the covariance between every pair of these random
variables is 0, then the variance of their sum is the
sum of their variances
Var(X1 X2 Xk ) σ12 σ 22 σk2
However, if the covariances between pairs of random
variables are not 0, the variance of their sum is
K 1 K
Var(X1 X2 Xk ) σ12 σ 22 σk2 2 Cov(X i , X j )
i1 ji1
Chap 6-57/62
Differences Between Two
Random Variables
For two random variables, X and Y
The mean of their difference is the difference of their
means; that is
E(X Y) μX μY
If the covariance between X and Y is 0, then the
variance of their difference is
Var(X Y) σ 2X σ 2Y
If the covariance between X and Y is not 0, then
the variance of their difference is
Var(X Y) σ 2X σ 2Y 2Cov(X, Y)
Chap 6-58/62
Linear Combinations of
Random Variables
A linear combination of two random variables, X and Y,
(where a and b are constants) is
W aX bY
The mean of W is
μW E[W] E[aX bY] aμX bμY
Chap 6-59/62
Linear Combinations of
Random Variables
(continued)
The variance of W is
σ 2W a2σ 2X b2σ 2Y 2abCov(X, Y)
Or using the correlation,
σ 2W a2σ 2X b2σ 2Y 2abCorr(X, Y)σ Xσ Y
If both X and Y are joint normally distributed
random variables, then the linear combination, W, is
also normally distributed
Chap 6-60/62
Example
Two tasks must be performed by the same worker.
X = minutes to complete task 1; μx = 20, σx = 5
Y = minutes to complete task 2; μy = 30, σy = 8
X and Y are normally distributed and independent
What is the mean and standard deviation of the time to
Chap 6-61/62
Example
(continued)
X = minutes to complete task 1; μx = 20, σx = 5
Y = minutes to complete task 2; μy = 30, σy = 8
What are the mean and standard deviation for the time to complete | 6,261 | 16,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-21 | latest | en | 0.839728 |
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2021-01-04, 19:41 #1 Dr Sardonicus Feb 2017 Nowhere 24·32·31 Posts An old puzzle for the new year The following problem is not difficult. Any Forumite should be able to determine the answer. I don't know any really elegant solution. The answer seems curious to me, but I am not aware of it being anything more than a curiosity. Which positive integers can not be expressed as the sum of two or more consecutive positive integers?
2021-01-04, 20:00 #2 Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 22·2,371 Posts 1 and the evens.
2021-01-04, 20:07 #3
bsquared
"Ben"
Feb 2007
22×23×37 Posts
The powers of 2
I don't have an elegant solution either - just starting writing down sums and observed a pattern.
Quote:
Originally Posted by Uncwilly 1 and the evens.
The puzzles states two or more consecutive integers.
2021-01-04, 20:13 #4
Uncwilly
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
22·2,371 Posts
Quote:
Originally Posted by bsquared The puzzles states two or more consecutive integers.
At least I was partially right.
2021-01-04, 20:50 #5 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 2×36 Posts We need for a given x>0 that x=a+..+(a+b)=(2*a+b)*(b+1)/2 with a,b>0 So: 2*x=(2*a+b)*(b+1) Notice that 2*a+b and b+1 has different parity, hence if x is a power of two then this equation has no solution [because 2*a+b>1 and b+1>1 and of them is odd]. Let 2*x=2^u*(2*k+1) with u>0 and k>0 (because x is not power of two). Even there could be lots of such decompositions of 2*x to (2*a+b)*(b+1) we can regard only one of them and its pair to get a "good" solution: 2*x=2^u * (2*k+1) 2*x=(2*k+1) * 2^u solving this: a1=(2^u-2*k)/2 ; b1=2*k>0 a2=(2*k-2^u+2)/2 ; b2=2^u-1>0 notice that a1+a2=1 so one will be a good solution, giving a1>0 or a2>0. In code [if x is a power of two then it will return the fake trivial solution]. F(x)={tmp=2*x;u=0;while(tmp%2==0,u+=1;tmp/=2);k=(tmp-1)/2; a1=(2^u-2*k)/2;a2=(2*k-2^u+2)/2;if(a1>0,print(a1"+...+"a1+2*k),print(a2"+...+"a2+2^u-1))}
2021-01-04, 21:54 #6 charybdis Apr 2020 2·5·23 Posts Not sure if this counts as elegant: Suppose n = ab where b>1 is odd. Then when we write n = a+...+a, one of the a's is in the middle, so we can easily rewrite this as a sum of consecutive integers, eg 7+7+7+7+7 becomes 9+8+7+6+5; the sums are the same because 9+5 = 7+7 etc. If the resulting sum contains negative numbers, that isn't a problem because they just cancel out the smallest positive numbers, eg 14 = 2+2+2+2+2+2+2 = 5+4+3+2+1+0+(-1) = 5+4+3+2. We can also run the above process in reverse to go from an expression of n as a sum of consecutive positive integers to an odd factor of n. If the sum contains an odd number of terms, then n = (number of terms)*(middle term). If it contains an even number of terms, then if k is the smallest term in the sum, we can add (k-1)+(k-2)+....+(-(k-2))+(-(k-1)) to get a new sum with an odd number of terms. Some of these terms are negative but that doesn't matter. The only numbers without nontrivial odd factors are the powers of 2, so we're done.
2021-01-05, 00:12 #7
uau
Jan 2017
10110002 Posts
Quote:
Originally Posted by charybdis Not sure if this counts as elegant:
The cancellation trick is nice and gives a short solution to most of the problem. I think the other case is a bit simpler without reducing to that though:
That is, the sum is always the mean of the numbers times their count. If the count is odd, the mean is an integer. If it's even, the mean is m+1/2, and the product is divisible by 2m+1.
2021-01-05, 03:18 #8 LaurV Romulan Interpreter Jun 2011 Thailand 33·347 Posts We inadvertently put the mouse on bb's spoiler and spoiled all the fun... But we read the answers and consider that if we put all together and cut out complicate parts, you all solved the puzzle in an elegant way.
2021-01-05, 07:03 #9 Citrix Jun 2003 1,579 Posts Sum of n consecutive numbers==> (n / 2) *(first number + last number) = sum Last number=first number +n-1 n*(2*first number +n-1)-2*sum=0 first number =f n^2+(2*f-1)*n-2*sum=0 This is a quadratic equation. We want to find integer roots and positive solutions. This will have integer roots if (2*f-1)^2+8*sum=y^2 y will be odd as 2*f-1 is odd We also have n>=2 and y>=(2*f-1)+4 for positive solutions greater than 2 or y-(2*f-1)>=4 -- Concept 1 Or 8*sum needs to be a difference of 2 odd squares; with each even factor pair giving a unique solution. y^2-(2*f-1)^2=8*sum (y-(2*f-1)) * (y+(2*f-1)) = 8*sum Both (y-(2*f-1)) & (y+(2*f-1)) will be even Both (y-(2*f-1)) & (y+(2*f-1)) cannot be 0 (mod 4) otherwise y and 2f-1 will be even For odd y and (2*f-1) only solution would be one of the set of factors1 =2 (mod 4) and other factors2=0 (mod 4) - Concept 2 If sum is a power of 2; 8*sum=2^x with only factors =2 and 2^(x-1) in accordance with concept 2 then y=2^(x-2)+1 and (2*f-1)=2^(x-2)-1 But y-(2*f-1)>=4 so there is no solution for powers of 2 Combining concept 1 and 2 For all other values of sum; 8*sum can be factored into a solution where the smaller factor >=4 and one of the factors =2 (mod 4) and other factor=0 (mod 4) For all other values, other than powers of base 2, their will be a solution as explained above Simpler solution from google https://nrich.maths.org/summingconsecutive/solution Last fiddled with by Citrix on 2021-01-05 at 07:09
2021-01-06, 15:17 #10 Dr Sardonicus Feb 2017 Nowhere 24·32·31 Posts Here's my solution - considerably refined, thanks to the postings to this thread! I believe what I first saw about this curiosity was the assertion that any positive integer could be expressed as a sum of two or more consecutive integers, unless it was a power of 2. This, of course, was a great advantage in analyzing the situation, since I already knew what was true, and only needed to determine why it was true. The obvious commonality of positive integers that are not powers of two is, they have an odd factor greater than 1. So the obvious thing to look for was an algebraic factorization of a sum of consecutive integers. It occurred to me that a sum of consecutive integers is the difference of two triangular numbers. And I knew that "8*triangle plus 1 = square," so a factorization as a difference of two squares was in sight. s = Tm - Tn with m - n > 1 8*s = 8*Tm - 8*Tn = (8*Tm + 1) - (8*Tn + 1) = (2*m + 1)^2 - (2*n + 1)^2 8*s = (2*m - 2*n)*(2*m + 2*n + 2) 2*s = (m - n)*(m + n + 1) where, again, m - n > 1 Both algebraic factors of 2*s are greater than 1 by hypothesis, and the difference of the two factors is 2*n + 1, an odd number, so one of the factors is an odd number greater than 1. Obviously, given a positive integer s, the factorizations of 2*s as above which produce m and n with m - n > 1, correspond exactly to the odd divisors d > 1 of s, and the cofactor 2*s/d of 2*s. Ta-daaaah! One irksome thing about this analysis is that Tm - Tn is the sum of the m - n integers from n+1 to m, not the integers from n to m. It is also less "elegant" than the "trapezoidal" factorizations of a sum of consecutive integers. One satisfying thing is, in the difference of two squares above, both squares are odd. It is this fact that leads to the algebraic factors of 2*s differing by an odd number (the smaller odd root). One pitfall of allowing non-positive integers is, it allows ANY integer to be expressed as a sum of two or more consecutive integers. Zero is the sum of the integers from -k to k for any k > 0. Any positive integer s can be expressed as the sum of the integers from -(s-1) to s. A negative integer s is the sum of the integers from s to -s-1. Last fiddled with by Dr Sardonicus on 2021-01-06 at 17:59 Reason: xigfin posty
2021-01-07, 05:02 #11 LaurV Romulan Interpreter Jun 2011 Thailand 33×347 Posts If n contains no odd factor, Robert gave the demonstration why it can't be written as so. If n contains an odd factor, charybdis gave the simplest algorithm to write it so. (edit: Robert also solved this, but charybdis' proof is simpler and constructive). The rest are complications (and only partially solve the problem, you all show how to write n if it have odd factors, but nobody except Robert show that no power of two can be written so. Maybe there are some HIGH powers of two which could be accidentally written so, therefore the second half of the proof is important too). Last fiddled with by LaurV on 2021-01-07 at 05:07
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#### Resources tagged with Working systematically similar to LOGO Challenge 10 - Circles:
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##### Other tags that relate to LOGO Challenge 10 - Circles
Circles. Investigations. Programming. Radius (radii) & diameters. Recursion. STEM - General. Logo. Tangents. Pythagoras' theorem. Practical Activity.
### There are 129 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### LOGO Challenge - Sequences and Pentagrams
##### Stage: 3, 4 and 5 Challenge Level:
Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?
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##### Stage: 3 and 4 Challenge Level:
Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?
### LOGO Challenge - Pentagram Pylons
##### Stage: 3, 4 and 5 Challenge Level:
Pentagram Pylons - can you elegantly recreate them? Or, the European flag in LOGO - what poses the greater problem?
### The Best Card Trick?
##### Stage: 3 and 4 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### LOGO Challenge - Triangles-squares-stars
##### Stage: 3 and 4 Challenge Level:
Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.
### LOGO Challenge - Following On
##### Stage: 3, 4 and 5 Challenge Level:
Remember that you want someone following behind you to see where you went. Can yo work out how these patterns were created and recreate them?
### Making Maths: Double-sided Magic Square
##### Stage: 2 and 3 Challenge Level:
Make your own double-sided magic square. But can you complete both sides once you've made the pieces?
### Bent Out of Shape
##### Stage: 4 and 5 Challenge Level:
An introduction to bond angle geometry.
### One Out One Under
##### Stage: 4 Challenge Level:
Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
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##### Stage: 3 and 4 Challenge Level:
Can you use your powers of logic and deduction to work out the missing information in these sporty situations?
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##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
##### Stage: 3 and 4 Challenge Level:
This is a variation of sudoku which contains a set of special clue-numbers. Each set of 4 small digits stands for the numbers in the four cells of the grid adjacent to this set.
### Triangles to Tetrahedra
##### Stage: 3 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Star Product Sudoku
##### Stage: 3 and 4 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### Corresponding Sudokus
##### Stage: 3, 4 and 5
This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.
### Constellation Sudoku
##### Stage: 4 and 5 Challenge Level:
Special clue numbers related to the difference between numbers in two adjacent cells and values of the stars in the "constellation" make this a doubly interesting problem.
##### Stage: 3 and 4 Challenge Level:
Four small numbers give the clue to the contents of the four surrounding cells.
### The Naked Pair in Sudoku
##### Stage: 2, 3 and 4
A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article.
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##### Stage: 4 and 5 Challenge Level:
A pair of Sudokus with lots in common. In fact they are the same problem but rearranged. Can you find how they relate to solve them both?
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##### Stage: 2 and 3 Challenge Level:
Arrange 9 red cubes, 9 blue cubes and 9 yellow cubes into a large 3 by 3 cube. No row or column of cubes must contain two cubes of the same colour.
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This Sudoku combines all four arithmetic operations.
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##### Stage: 2 and 3 Challenge Level:
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##### Stage: 3 and 4 Challenge Level:
This Sudoku, based on differences. Using the one clue number can you find the solution?
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##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
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##### Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
### Ratio Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
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##### Stage: 3 and 4 Challenge Level:
A Sudoku that uses transformations as supporting clues.
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##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
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##### Stage: 3 Challenge Level:
The letters of the word ABACUS have been arranged in the shape of a triangle. How many different ways can you find to read the word ABACUS from this triangular pattern?
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##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
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The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .
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##### Stage: 3 and 4 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
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##### Stage: 3 and 4 Challenge Level:
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##### Stage: 3, 4 and 5 Challenge Level:
In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
### Sandwiches
##### Stage: 2, 3, 4 and 5 Challenge Level:
Arrange the digits 1, 1, 2, 2, 3 and 3 so that between the two 1's there is one digit, between the two 2's there are two digits, and between the two 3's there are three digits.
### Alphabetti Sudoku
##### Stage: 3 and 4 Challenge Level:
This Sudoku requires you to do some working backwards before working forwards.
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This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
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##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
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##### Stage: 3 Challenge Level:
Move your counters through this snake of cards and see how far you can go. Are you surprised by where you end up?
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##### Stage: 3 and 4 Challenge Level:
Solve the equations to identify the clue numbers in this Sudoku problem.
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Intersection Sums Sudoku
##### Stage: 2, 3 and 4 Challenge Level:
A Sudoku with clues given as sums of entries.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### Magnetic Personality
##### Stage: 2, 3 and 4 Challenge Level:
60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
### Pole Star Sudoku
##### Stage: 4 and 5 Challenge Level:
A Sudoku based on clues that give the differences between adjacent cells.
### Diagonal Sums Sudoku
##### Stage: 2, 3 and 4 Challenge Level:
Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells. | 2,263 | 9,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-34 | latest | en | 0.867898 |
https://graphicallinearalgebra.net/2017/07/31/determinants-and-the-lindstrom-gessel-vienot-lemma/?like_comment=3621&_wpnonce=44d4b675a2 | 1,638,867,581,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00069.warc.gz | 354,501,953 | 35,546 | # Determinants and the Lindström-Gessel-Vienot Lemma
Solomon is currently a student at the University of Pennsylvania where he is pursuing a Ph.D in Computer Science. He has a keen interest in all things Computer Science and Maths, so he decided to specialize in Programming Languages, an area which applies ideas from diverse areas of Computer Science, Mathematics and Philosophy. One of his favourite things is when connections are made between seemingly disparate areas of inquiry, so he has enjoyed reading the Graphical Linear Algebra blog for the connections between string diagrams and linear algebra. As the Lindström-Gessel-Vienot Lemma shows us, there is also a surprising combinatorial connection. Such occurrences make Solomon very happy!
Consider the familiar diagram for the $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$:
Suppose that we wanted to push some flow from the input nodes $X_1$ and $X_2$ to the output nodes $Y_1$ and $Y_2$. Moreover, we want to do this in such a way that every output node receives some flow without any congestion in the network. More specifically, want to push flow from the input nodes to the output nodes in such a way that:
1. no two different paths that are part of the flow intersect at
a vertex (not even endpoints), and
2. for each output node $Y_j$, there is a path from some input
node $X_i$ to $Y_j$.
For example, in the diagram above, we can push $a$ units of flow from $X_1$ to $Y_1$ and $d$ units of flow from $X_2$ to $Y_2$.
We can also push $c$ units of flow from $X_1$ to $Y_2$ and $b$ units of flow from $X_2$ to $Y_1$.
We are not allowed to push $a$ units of flow from $X_1$ to $Y_1$ and $b$ units of flow from $X_2$ to $Y_1$, since then we would have two different paths intersecting at $Y_1$.
Also, we are not allowed to push only $a$ units of flow from $X_1$ to $Y_1$, since then there is no path from some $X_i$ to $Y_2$.
We call a flow that satisfies conditions (1) and (2) a viable flow. Now we will measure the signed weight a viable flow. The weight of a viable flow is simply the product of all weights on the edges that are part of the flow. To get the signed weight, we multiply the result by -1 if the flow links $X_1$ to $Y_2$. Otherwise, we multiply the result by +1.
For example, in our $2 \times 2$ matrix example, the first viable flow has weight $+ad$. The unsigned weight of this flow is $ad$ since we push $a$ units of flow from $X_1$ to $Y_1$ and $d$ units of flow from $X_2$ to $Y_2$. To get the signed weight of this flow, we multiply the unsigned weight $ad$ by +1 since $X_1$ is linked to $Y_1$. A similar computation shows that the signed weight of the second viable flow is $-bc$.
Interestingly, when we add signed weights of all the viable flows in the diagram, we get the determinant of the matrix associated with the diagram i.e. $ad-bc$. Coincidence?
The Lindström-Gessel-Vienot Lemma (LGVL) tells us that this in fact is not a
coincidence. Specifically, LGVL associates to each finite directed acyclic
graph (dag) $G$ a matrix whose determinant is exactly equal to the
sum of the signed weights of the viable flows in $G$. To make this relationship precise we shall give the statement of LGVL, but first we need to introduce some definitions and notation that we will need in order to give the statement of the theorem.
First, we introduce permutations. For any sets $A=\{a_1, \ldots, a_n\}$ and $B=\{b_1, \ldots, b_n\}$ we call any bijection $\sigma : A \longrightarrow B$ a permutation from $A$ to $B$. Now assume that $\sigma$ sends $a_1, \ldots, a_n$ to $\sigma(a_1), \ldots, \sigma(a_n)$ respectively. For example, let $A = B = \{1,2,3,4,5\}$, and let $\sigma$ send $1,2,3,4,5$ to $3,5,2,1,4$ respectively. Write $a_1, \ldots, a_n$ in a row in that order, skip a line, then write $b_i$ below $a_i$ for each $b_i$. Draw a line from $a_i$ to $\sigma(a_i)$, and count the number $m$ of crossings in between the ${a_i}'$s and the ${b_i}'$s. Define the sign of $\sigma$ be $(-1)^m$. For the permutation $\sigma$ we have just described, there are six crossings, so the sign of the permutation is $(-1)^6 = 1$.
The sign of a permutation $\sigma$ can also be computed by raising -1 to the
number $inv(\sigma)$ of inversions of $\sigma$, where $inv(\sigma)$ is the number of ordered pairs $(i, j)$ such that $i < j$ but $\sigma(i) > \sigma(j)$. In the example above, the number of inversions of $\sigma$ is 6 = $|\{(1,3), (1,4), (2,3), (2,4), (2,5), (3,4)\}|$, which gives the same result as before.
For the next few pieces of notation as well as the statement of LGVL, assume that $G$ is a finite directed acyclic graph. Assume further that $G$ has input nodes $X_1, \ldots, X_n$ and output nodes $Y_1, \ldots Y_n$.
As before, we are interested in viable flows, i.e. flows that have the property that:
1. no two different paths that are part of the flow intersect at
a vertex (not even endpoints), and
2. for each output node $Y _j$, there is a path from some input
node $X_i$ to $Y_j$.
Each viable flow can described by an $n$-tuple $P = (P_1, \ldots, P_n)$ of paths , where $P_i$ is a path from the input node $X_i$ to the output node $Y_{\sigma(i)}$ for some permutation $\sigma$. Let $\sigma(P)$ be this permutation.
For each directed path $P$ between any two vertices in $G$, let $\omega (P)$ be the product of the weights of the edges of the path. For any two vertices $X_i$ and $Y_j$, write $e(X_i, Y_j)$ for the sum
$e(X_i,Y_j)=\sum\limits _{P:X_i\to Y_j} \omega (P),$
where $P$ ranges over all paths from $X_i$ to $Y_j$.
We are finally in a position to give the statement of LGVL, which is that
$\left \vert {\begin{pmatrix}e(X_{1},Y_{1})&e(X_{1},Y_{2})&\cdots &e(X_{1},Y_{n})\\e(X_{2},Y_{1})&e(X_{2},Y_{2})&\cdots &e(X_{2},Y_{n})\\\vdots &\vdots &\ddots &\vdots \\e(X_{n},Y_{1})&e(X_{n},Y_{2})&\cdots &e(X_{n},Y_{n})\end{pmatrix}} \right \vert = \sum\limits _{{P = (P_{1},\ldots ,P_{n})}}{\mathrm {sign}}(\sigma (P))\prod\limits _{{i=1}}^{n}\omega (P_{i}),$
where $P$ ranges over all viable flows in $G$!
LGVL holds for any finite directed acyclic graph, not just graphs that look like the $2 \times 2$ example. For example consider the following graph:
or equivalently, the following string diagram:
The matrix for this is
$\begin{pmatrix} ag+bh & bi & 0\\ ch & ci + dj & dk\\ 0 & ej & ek + fl \end{pmatrix}$.
It has determinant $agciek + agcifl + agdjfe + bhdjfe$, and it is easy to check that each of these terms indeed gives a viable flow from the ${s_i}'$s to the ${t_i}'$s. Note also that the sign of each term is +1 since $s_i$ is sent to $t_i$ in each of the viable flows.
Now on the one hand, LGVL gives us a way of computing the sum of the signed weights of viable flows in a dag as the determinant of a certain matrix. On the other hand, given a matrix $M$, we may compute the determinant of $M$ by doing the following. First, we find a special dag $G$. Next, we find a set of input vertices $\{X_i\}$ and output vertices $\{Y_j\}$ in $G$. We check to see that $M_{ij} = e(X_i, Y_j)$. Then, we (easily!) compute the sum of the signed weights of viable flows in $G$ from $\{X_i\}$ to $\{Y_j\}$. Finally we conclude that the result is equal to the determinant of $M$ by LGVL. This compositional approach to computing the determinant often leads to useful and surprising results, as we shall see below.
Consider the determinant of the following matrix:
$A = \begin{pmatrix} (1,1) & (1,2) & ... & (1,n) \\ (2,1) & (2,2) & ... & (2,n) \\ \vdots & \vdots & \ddots & \vdots \\(n-1,1) & (n-1,2) & ... & (n-1,n)\\(n,1) & (n,2) & ... & (n,n)\end{pmatrix},$
where $(k, \ell)$ denotes the greatest common divisor of $k$ and $\ell$.
It is difficult to come up with a closed formula for this determinant by using say row-reductions or the co-factor expansion. However, using the strategy involving LGVL gives a simple and surprising result.
But first, a few more definitions! Given any positive integer $k$, let $\phi(k)$ denote the number of positive integers less than or equal to $k$ which are relatively prime to $k$. For example, $\phi(6) = 2$ since the only numbers less than 6 that are relatively prime to 6 are 1 and 5. On the other hand $\phi(7)=6$ since all the positive integers less than 7 are relatively prime to 1. In fact, it is easy to see that $\phi(p)=p-1$ for any prime $p$.
Also, for the purposes of the proof, we shall need to use the fact that
$\sum\limits_{d|k} \phi(d) = k,$
which is a property first established by Gauss. For example
$6 = \phi(1) +\phi(2) + \phi(3) + \phi(6) = 1 + 1 + 2 + 2$.
Now we proceed to the actual computation. Consider the graph $G$ which has $3n$ vertices: $\{s_1, \ldots, s_n\}$ are the input vertices, $\{t_1, \ldots, t_n\}$ are the output vertices and $\{v_1, \ldots, v_n\}$ are intermediate vertices. If $d$ divides $k$, put the edge $s_k \to v_d$ with weight $\phi(d)$ and the edge $v_d \to t_k$ with weight 1.
What is the weight $e(s_k, t_{\ell})$ of all the paths from $s_k$ to $t_{\ell}$? Well each path from $s_k$ to $t_{\ell}$ can have only two edges: an edge $s_k \rightarrow v_d$ of weight $\phi(d)$ and an edge $v_d \rightarrow t_{\ell}$ of weight 1, for some $v_d$, where $d$ divides $k$ and $d$ divides $\ell$. Of course the weight of this path is $\phi(d)$ multiplied by 1, which is just $\phi(d)$. Therefore,
$e(s_k, t_{\ell}) = \sum\limits_{\substack{d|k \\ d|\ell}}\phi(d) = \sum\limits_{d|(k, \ell)}\phi(d) = (k, \ell)$
The last equality is justified by Gauss’ result.
By the LGVL, it follows that
$\det(A) = \sum\limits_{P = (P_1,\ldots,P_n)} \mathrm{sign}(\sigma(P)) \prod\limits_{i=1}^n \omega(P_i),$
where $P = (P_1, \ldots P_n)$ ranges over all viable flows from the input vertices $S = \{s_1, \ldots, s_n\}$ to the output vertices $T = \{t_1, \ldots, t_n\}$.
Now the only path from $s_1$ to $T$ is $s_1 \to v_1 \to t_1$ since 1 has only one divisor, namely 1. As a result, for each prime $1 \leq p \leq n$, any path from $s_p$ to $T$ in a viable flow must pass through $v_p$. This is because $p$ has only two divisors, namely 1 and $p$, but the path through 1 has already been blocked by 1.
Consequently, for any product $pq$ of two primes, any path from $s_{pq}$ to $T$ must pass through $v_{pq}$. Repeating this argument for all the other numbers less than or equal to $n$, we conclude that all the viable flows from $S$ to $T$ link $s_k$ to $v_k$.
Now if there is a viable flow which links $v_k$ to $t_{\ell}$ for some $k \neq \ell$, then there must be some $k', \ell'$ such that $v_{k'}$ is linked to $t_{\ell}'$ with $k' > \ell'$ (why?). But this means that $k'$ divides $\ell'$, a contradiction since $k' > \ell$! So we must have $v_k = t_k$, for all $k$.
Thus, there is a unique viable flow from $S$ to $T$ i.e. the flow for which $s_k$ is sent to $t_k$ via $v_k$. Thus, we conclude that the determinant of $A$ is $\phi(1) \cdot \ldots \cdot \phi(n)$, since the weight of each path $P_k$ in this flow is $\phi(k)$ as we showed earlier:
$\left \vert \begin{pmatrix}(1,1) & (1,2) & ... & (1,n) \\ (2,1) & (2,2) & ... & (2,n) \\ \vdots & \vdots & \ddots & \vdots \\(n-1,1) & (n-1,2) & ... & (n-1,n)\\(n,1) & (n,2) & ... & (n,n)\end{pmatrix} \right \vert = \phi(1) \cdot \ldots \cdot \phi(n)$
## 5 thoughts on “Determinants and the Lindström-Gessel-Vienot Lemma”
1. Sweet. Thank you for sharing this!
I noticed two minor typos (omitted words) and had a few questions.
“signed weight a viable flow” –> “signed weight of a viable flow”
“Each viable flow can described by” –> “Each viable flow be can described by”
I think the variable P is overloaded between this and the following paragraphs (as both an arbitrary viable flow and an arbitrary path between two given endpoints). Am I missing something, or are these actually distinct objects?
You pass lightly over a potentially interesting point: “…given a matrix M, we may compute the determinant of M by doing the following. First, we find a special dag G…” Is this inverse operation always possible?
Like
• markusq says:
Oops. I messed up. “be can” –> “can be”. 🙂
Like
2. . says:
”It has determinant agciek + agcifl + agdjfe + bhdjfe, and it is easy to check that each of these terms indeed gives a viable flow from the {s_i}’s to the {t_i}’s.“
should be
”It has determinant agciek + agcifl + agdjfl + bhdjfl, and it is easy to check that each of these terms indeed gives a viable flow from the {s_i}’s to the {t_i}’s.“
Like | 3,882 | 12,436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 214, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-49 | latest | en | 0.9332 |
https://maker.pro/forums/threads/hysteresis-needed-in-741-op-amp-circuit.22556/ | 1,680,289,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00475.warc.gz | 437,460,406 | 81,246 | # Hysteresis needed in 741 Op-Amp circuit
J
#### John Fields
Jan 1, 1970
0
1. THE THERMISTOR
For a good discussion of how thermistors work, refer to:
http://www.epcos.com/inf/50/db/ntc_02/00170027.pdf
For the purpose at hand, (determining how the change in resistance
affects the voltage into the comparator) refer to the "1006" column
in:
http://www.epcos.com/inf/50/db/ntc_02/01470178.pdf
and you'll notice that at 15°C the value of Rt/R25 for 15°C is 1.4883,
and at 35°C it's 0.686.
What that means is that if the thermistor has a resistance of 10000
ohms at 25°C, then at 15°C it'll have a resistance of 14,883 ohms, and
at 35°C it'll have a resistance of 6860 ohms.
2. THE VOLTAGE DIVIDER
Now, if we connect the thermistor in series with a resistor and apply
a fixed voltage across them, like this:
E1
|
[R1]
|
+---->E2
|
[RT1]
|
GND
Then when temperature increases and the resistance of the thermistor
(RT1) decreases, E2 will also decrease. Conversely, when the
temperature to which RT1 is exposed decreases, E2 will increase.
The exact relationship is:
E1 RT1
E2 = ------
R1+RT1
so, if we make E1 = 12V and R1 = 100kohms, then at 15°C,
12*14883
E2 = ------------- ~ 1.555V
100000+14883
and the power being dissipated in the thermistor will be:
E² 1.555²
P = --- = ----- = 162µW
R 10000
At 35°C, the potput voltage will be:
12*6860
E2 = ------------- ~ 0.7703V
100000+6860
and the power dissipated by the thermistor:
E² 0.7703²
P = --- = ------- = 86µW
R 6860
The power dissipated in the thermistor may be enough to cause a
self-heating error to occur, but you'd need to look at the data sheet
for the particular thermistor you're using to make sure.
In any case, if the self-heating error is negligible with respect to
the accuracy you're looking for, with 1.555V out at 15°C and 0.770V
out at 35°C, that's a spread of 780mV from one end of the temp range
to the other, which is pretty good.
3. THE REFERENCE DIVIDER.
If we now connect our thermistor circuitry to the + input of a voltage
comparator, like this,
+12V>---------+---------+----+
| | |
[100K] | [RL]
| | |
+--------|+\ |
| | >--+----->OUT
| |-/
| |
[RT1] |
| |
GND>----------+---------+---------->GND
and we arrange things so that the voltage on the - input is 0.770V,
then when the voltage on the + input rises to 0.770V or higher, the
comparator's output transistor will switch off and the voltage on OUT
will be pulled up to +V through RL.
We can do that by using another voltage divider, like this:
+V>---+-------+-----+----+
| | | |
E1--+ [100K] | [RL]
| | | |
[R1] +----|+\ |
| | | >--+----->OUT
E2--+-------|----|-/
| | |
[R2] [RT1] |
| | |
GND>--+-------+-----+---------->GND
and selecting R1 and R2 so that the voltage at their junction will be
0.770V. An easy way to do this is to look at the comparator data
sheet to find out what the required input current is and making the
resistances small enough to swamp out that current. For the LM311,
that comes out to about 350nA, so if we make the divider current 350µA
the LM311's input currents will contribute about a 1% error to the
divider's output voltage. Better yet, let's make the divider current
about 1mA and use 1% resistors for the divider. That way, most of the
error will be in the resistors, where we can easily account for it.
Since we've got a 12V supply, and 10kohms is a standard 1% value, and
12V into 10k is about 1.2mA, let's make R1 = to 10k and figure out
what R2 needs to be to get us to 0.770V.
Since our new divider is the same as the old one, we can write:
E1 R2
E2 = ------
R1+R2
and rearrange to solve for R2:
E2 R1
R2 = -------
E1-E2
which gives us:
0.770*10000
R2 = ------------- = 685.66 ohms
12-0.770
681 ohms is a standard 1% value, so plugging that into
E1 R2
E2 = ------
R1+R2
to get the voltage into the comparator's - input gives us:
E1 R2 12*681
E2 = ------- = ----------- = 0.765V
R1+R2 10000+681
That's a voltage error of 5 parts out of 770, which is just a little
over 1/2 a percent, so since the resistor tolerance error will
dominate at this point we have a tentative solution, and our circuit
now looks like this:
+V>---+-------+-----+----+
| | | |
| [100.0K] | [RL]
| | | |
[10.0K] +----|+\ |
| | | >--+----->OUT
+-------|----|-/
| | |
[681] [RT1] |
| | |
GND>--+-------+-----+---------->GND
4. THE VARIABLE REFERENCE.
Now, we could use the same procedure to figure out a new R2 for a temp
of 15°C, and switch between the two, or figure out a whole bunch of
different R2's and switch them in and out for the different temps we
wanted to set to between 15 and 35°, but a much nicer way to do it
would be to use a pot.
In order to be able to do that, our circuit changes to:
+V>---+-------+-----+----+
| | | |
[R1] [100.0K] | [RL]
| | | |
| +----|+\ |
| | | >--+----->OUT
[R2]<-----|----|-/
| | |
[R3] [RT1] |
| | |
GND>--+-------+-----+---------->GND
Where R1 is a pot and R1 and R3 are 1% resistors used to set the ends
of the pot's resistive element to the voltages corresponding to 15°
and 35°C.
Just to get a feel for things, if we want the end of the pot connected
to R1 to be at the higher endpoint voltage (1.555V) and we keep R1 at
10k,then the series combination of R2 and R3 has to be:
E2 R1 1.555*10000
R2+R3 = ------- = ------------ ~ 1489 ohms
E1-E2 12-1.555
and if we want the lower end of the pot to be at 0.77V, then since
there's
E1 12
I = ----------- = ------------ = 1.04mA
R1+(R2+R3) 10000+1489
flowing in the circuit, and
E
R = ---
I
0.77V
then R3 = ------- = 740 ohms.
1.04mA
Since 750 ohms is a standard 1% value, that leaves
1489ohms - 750ohms = 739 ohms for the pot.
739 ohms is _not_ a standard end-to-end resistance for a pot, so we'll
have to do something to fix that. Since R1R2R3 is a voltage divider,
all we're really concerned with is the ratios of resistances, so let's
scale up the 739 ohms to 1000 ohms, which is a standard end-to-end
resistance for pots, and then scale up everything else and see what
happens.
1000/739 = 1.353, so R1 needs to go to 13530 ohms, R2 will go to 1000
ohms and R3 will need to go to 1000 ohms as well. So now, since the
closest we can get to 13530 ohms with 1% resistors is 13700 ohms, our
string now looks like this:
+12V
|
[13.7K]
|
|
[1000]<----Vout
|
|
[1000]
|
GND
With the wiper at the top of the pot, Vout will be:
12*2000
Vout = --------- = 1.528V
15700
and at the bottom of the pot it will be:
12*1000
Vout = --------- = 0.764V
15700
(1.555-1.528)*100
So, we'll have an error of ----------------- ~ -1.7%
1.555
(0.770-0.764)*100
on the high end, and ------------------- = + 0.78% on the high end.
0.764
Not too shabby, and now, after adding all the resistor values and
introducing the hysteresis resistor, our circuit looks like this:
+12>---+--------+-----+-------+
| | | |
[13.7K] [100.0K] | [RL]
| | | |
| +-------[Rh]--+
| | | |
| +----|+\ |
| | | >-----+----->OUT
[1000]<-----|----|-/
| | |
| | |
[1000] [RT1] |
| | |
GND>---+--------+-----+---------->GND
5. HYSTERESIS
Since a comparator has what amounts to essentially infinite gain, if
its input signals are allowed to move through the switching region
slowly, any noise present on the inputs during that time _will_ cause
the output to swing wildly between its limits. If slowly moving
signals _must_ be accommodated, the cure for the problem is to cause
the input(s) to quickly move out of the sensitive region when the
output changes state. This is done by connecting the output to the
non-inverting input so that when the non-inverting input goes more
positive than the inverting input, causing the output to go high, a
little of that high-going output signal is coupled back to the
non-inverting input, causing it to quickly go even higher, moving it
quickly out of the noisy region, thus preventing the "chatter" which
would otherwise surely occur. In this instance, that positive feedback
is called "hysteresis".
The amount of hysteresis intoduced into the circuit can be controlled
by selecting the resistance of the feedback resistor, which is part of
yet another voltage divider.
Looking at the output of the circuit above in a little more detail
shows us that the output stage is a transistor with an open collector
which pulls the putput down to ground when it is turned on and allows
the output to float when it's turned off, and what's important to
remember is that when the non-inverting (+) input is more positive
than the inverting (-) input the output will be turned off, allowing
current to flow from the supply and through RL into whatever is
connected to the comparator's output.
+12>---+--------+-------+-----+
| | | |
[13.7K] [100.0K] | [RL]
| | | |
| +--[Rh]-------+
| | | |
| | +------+ |
| +----|+ C|-+----->OUT
| | |LM311 |
[1000]<-----|----|- E|-+
| | +------+ |
| | | |
[1000] [RT1] | |
| | | |
GND>---+--------+-------+-----+----->GND
Since the LM311 can sink 50mA, there's no reason why it can't be used
to dirctly drive a 12V relay with a coil resistance of
12V
R = ----- = 240 ohms
50mA
or greater, so adding that circuitry yields:
+12>---+--------+-------+-----+-------+
| | | |K | O--> |
[13.7K] [100.0K] | [1N4001] [COIL]- | - -|
| | | | | | O------>COM
| +--[Rh]-------+-------+ |
| | | | +----------->NO
| | +------+ |
| +----|+ C|-+
| | |LM311 |
[1000]<-----|----|- E|-+
| | +------+ |
| | | |
[1000] [RT1] | |
| | | |
GND>---+--------+-------+-----+----->GND
Checking the sense of the output, we find that as the thermistor heats
up its resistance will decrease until it causes the comparator's +
input to go more negative than its - input, which will cause the
output transistor to turn on, energizing the relay, so this a cooling
thermostat. That is, if it gets hot enough to turn on the relay, the
relay will turn on the A/C, cooling the thermistor until the voltage
on the + input goes more positive than the voltage on the - input,
turning off the output transistor, turning off the relay, then
turning off the AC until it gets hot enough to turn it back on...
Now that we have that all figured out, if we simplify the circuit by
replacing the comparator's input with a resistor and its output with a
switch we'll have what amounts to:
+12>---+---------+
| |
[100.0K] [240]
| |
+--[Rh]---+-+-------->Eout
| |
| +-> |
Ein---+----+ |
| | O
[RT1] [Rin] |
| | |
GND>---+----+----------+---->GND
If 240 ohms (the relay coil resistance) << than Rh and Rin is >> RT1,
(which they will be) then the circuit further simplifies to:
+12>---+-------------+
| |
[100.0K] |
| |
+--[Rh]---O |
| | |
| O--> |<--O
Ein---+ |
| |
[RT1 |
| |
GND>---+----+
So that with the switch in the rightmost position we have:
+12>---+------+
| |
[100.0K] [Rh]
| |
+------+--->Ein
|
[RT1]
|
GND>---+
And with the switch in the leftmost position we have:
+12>---+
|
[100.0K]
|
+------+--->Ein
| |
[RT1] [Rh]
| |
GND>---+------+
Since, at 15°C, (without Rh in the circuit) we'll have an output
voltage of 1.55V from the thermistor and, at 35°C, an output voltage
of 0.77V, we'll have a 780mV change for a 20° change in temperature,
which is a sensitivity of 39mV/°C. That won't be quite true over
every increment of that range, since the thermistor is a non-linear
device but, for now, it's close enough.
Next, let's say that the accuracy spec's for the thermostat will
allow, after everything else has been accounted for, an extra degree C
to be used up to get the hysteresis we need. That means, since we've
got a sensitivity of 39mV/°C, that we've got 39mV to play around with
to get some hysteresis.
Now, below 15°C our thermistor will exhibit a resistance > 14883 ohms,
and if it warms up to 15°C, our circuit will look like this:
+12>---+
|
[100.0K]
|
+---->1.55V
|
[14883]
|
GND>---+
When this happens and the comparator's - input is set to 1.55V, the
output will go low and turn on the relay, and what we'd like to do is
to fool the thermostat into thinking that when that happens it's
actually 16° outside so that it'll move the input switching point a
little closer to ground to get away from the noise-susceptible input
transition region.
Since the thermistor resistance decreases when it gets hotter, what we
can do when the output goes low is to parallel the thermistor with
enough resistance to lower the 1.55 volts by one degree, to 1.51V
(rounding the 39mV/°C sensitivity to 40mV/°C).
Looking at the thermistor circuitry like this:
E1
|
[R1]
|
+---->E2
|
[R2]
|
GND
And setting E1 = 12V, E2 = 1.51V, and R1 = 100kohms,
we can solve for R2, the total resistance we need to get 1.51V out by
writing:
E2 R1 1.51*100000
R2 = ------- = ------------- ~ 14395 ohms
E1-E2 12.0-1.51
Since, at 15°, the thermistor will be at 14883 ohms, we'll need to
parallel it with a resistor to make the combination look like 14395
ohms, like this:
E1
|
[R1]
|
+-----+--->E2
| |
[R2] [R3]
| |
+-----+
|
GND
Knowing that the equation for the total resistance of two parallel
resistors is:
R2 R3
Rt = -------
R2+R3
and that R2 = 14883 ohms and Rt = 14395 ohms,
we can rewrite it to solve for R3 like this:
Rt R2 14395*14883
R3 = -------- = ------------- ~ 439kohms
R2-Rt 14883-14395
The closest 1% resistor we can get which won't exceed our 1°C error
budget is 432k, but that'll still give us nearly the 40mV of
hysteresis we're looking for, so let's choose that value tentatively.
Now, let's say the temp starts falling from that point. Since it's
really 15° outside and we've fooled the comparator into thinking that
it's 16°, it's going to have to fall to 14° for the comparator to
think that it's 15° outside, but when it gets there the comparator's
output will go open-collector and the feedback resistor will now be
connected to +12V through the relay coil. If the coil looks like 240
ohms its resistance will be negligible compared to the feedback
resistor's 439kohm resistance, so since the thermistor's sensitivity
is -401 ohms/°C, (assuming it's linear for the sake of this
discusion...) its resistance will rise to 14482 ohms and the circuit
will now look like this:
12V 12V
| |
[100K] [439K]
| |
+-------+--->E2
[14482]
|
GND
100k in parallel reduces to a total resistance of
100000*439000
Rt = --------------- = 81447 ohms.
100000+439000
So, the circuit further reduces to
12V
|
[81447]
|
+---->E2
|
[14482]
|
GND
And the voltage into the comparator's + input, E2, will now be
12.0*14482
E2 = ------------- ~ 1.812V
81447+14482
Since the [15°C] reference voltage on the comparator's -input will be
sitting at 1.555V, that 1.812V represents a hysteresis of about 257mV,
which represents the introduction of about a 6.6°C error. If the
error spec is +/- 1°, that's very, very, bad.
Fortunately, since the reason for the error is the asymmetry of the
thermistor string, we can fix it by making the string symmetrical.
That is, by making the series resistor and the thermistor the same
value at the center of the thermistor's resistance range for the
temperature range it will be sensing.
For our thermistor, with an R of 14883 ohms at 15°C and an R of 6860
ohms at 35°C, that comes out to 10,872 ohms for the series resistor,
so with the nearest 1% resistor installed, our new circuit will look
like:
12V
|
[10700]
|
+---->E2
|
[Rth]
|
GND
Now E2 will be 6.98V at 15°C (Rth = 14883 ohms) and 4.688V at 35°C
(Rth = 6860 ohms), so the new voltage sensitivity will will be about
115mV/°C.
115mV is a lot, so if we limit our hysteresis to, say, 50mV that'll
cause less than a 0.5° error, but let's see how it works out.
Just like last time, if we want 50 mV of hysteresis we'll have to make
the parallel combination of the 10.7k resistor and and the feeback
resistor change the + input voltage to the comparator by the amount we
want, which will be 50mV in this case.
We really only have four cases to consider:
The hysteresis when the temp is rising and falling through 15°C, and
the hysteresis when the temp is rising and falling through 35°C.
Looking at falling through 15°C first, we'll have:
12V 12V
| |
[10.7K] [Rf]
| |
+-------+--->E2
|
[14883]
|
GND
Since the resistance of the thermistor is increasing as it gets
colder, the voltage at E2 will be increasing also. Without feedback
E2 would be 6.98V when the thermistor got to 15°C (14482 ohms), and
what we want to do with the feedback is to make it seem like it's a
little bit colder than it really is when E2 gets to 6.98V, so that
would mean making E2 jump a little more positive when the comparator
output goes open collector.
We can do that by paralleling the 10.7kohm resistor with a resistor of
a suitable value and, using the procedure outlined earlier, we find
that for a 50mV jump (from 6.98V to 7.03V) the parallel combination of
the 10.7kohm resistor and the feedback resistor (Rf) needs to be 10522
ohms, with Rf being equal to 632502 ohms. The closest 1% resistor is
634kohms, which will give us an E2 of 7.02V as the temp falls through
15°C, about 40mV of hysteresis over the non-feedback case.
With the temp rises through 15°C and the comparator's output
transistor turns on, our circuit becomes:
12V
|
[10.7K]
|
+-------+--->E2
| |
[14883] [634K]
| |
GND GND
and E2 will now become 6.91V, giving us 70mV of hysteresis.
When the temp falls through 35°C, the circuit will become:
12V 12V
| |
[10.7K] [634k]
| |
+-------+--->E2
|
[6808]
|
GND
Without feedback, E2 would be 4.666V, but with feedback it'll be
4.714V, giving us about 48mV of hysteresis.
Finally, When the temp rises through 35°C, the circuit will become:
12V
|
[10.7K]
|
+-------+--->E2
| |
[6808] [634K]
| |
GND GND
and E2 will go to 4.636V, giving us about 30mV of hysteresis.
So, here's the final schematic, and barring any fuzzy thinking, math
errors, trypos, etc., all that's left to do now is to recalculate the
padding resistors around the pot for the new voltage range and trim
things up if you want/need more or less hysteresis, and it should be
pretty much done.
+--- >= 240 ohms
+12>---+--------+--------+-----+-------+ /
| | | |K | / O--> |
[R1] [10.7K] | [1N4001] [COIL]- | - -|
| | | | | | O------>COM
| +-[634K]-------+-------+ |
| | | | +----------->NO
| | +------+ |
| +----|+ C|--+
| | |LM311 |
[1000]<-----|----|- E|--+
| | +------+ |
| | | |
[R2] [RT1] | |
| | | |
GND>---+--------+--------+-----+----->GND
Simple, huh?^)
PS...
I've got some schematics for heating-cooling thermostats using
LM393's, and if you'd like I can post them to
alt.binaries.schematics.electronic.
R
#### Robert Baer
Jan 1, 1970
0
------------- SNIPped ------------
.... a *VERY* long-winded paper that might be called a thesis (or
mini-novel?).
If you had a ?questiion? concerning hysteresis, it would have been
nice to make a *BRIEF* synopsis of the problem.
Else, do not waste bandwidth!
W
#### Winfield Hill
Jan 1, 1970
0
Robert Baer wrote...
------------- SNIPped ------------
... a *VERY* long-winded paper that might be called a thesis (or
mini-novel?).
If you had a ?questiion? concerning hysteresis, it would have been
nice to make a *BRIEF* synopsis of the problem.
Else, do not waste bandwidth!
Hah! That's some of the best bandwidth wasting we've seen here
in some time.
Thanks,
- Win
(email: use hill_at_rowland-dot-org for now)
T
#### Tim Auton
Jan 1, 1970
0
Winfield Hill said:
Robert Baer wrote...
Hah! That's some of the best bandwidth wasting we've seen here
in some time.
I concur. I found it a most interesting insight into how I would like
to be able to design circuits. I'd probably have chucked a uC input at
it long ago
Perhaps Robert Baer missed the thread in s.e.basics that prompted John
Fields' post. I think it's on-topic here too.
Tim
J
#### Jim Thompson
Jan 1, 1970
0
------------- SNIPped ------------
... a *VERY* long-winded paper that might be called a thesis (or
mini-novel?).
If you had a ?questiion? concerning hysteresis, it would have been
nice to make a *BRIEF* synopsis of the problem.
Else, do not waste bandwidth!
washed out with soap.
...Jim Thompson
J
#### John Fields
Jan 1, 1970
0
------------- SNIPped ------------
... a *VERY* long-winded paper that might be called a thesis (or
mini-novel?).
If you had a ?questiion? concerning hysteresis, it would have been
nice to make a *BRIEF* synopsis of the problem.
Else, do not waste bandwidth!
---
It wasn't a question. As Tim Auton noted, this was a reply, which I
originated on seb.
Sorry you didn't find it at least interesting...
S
#### Spehro Pefhany
Jan 1, 1970
0
Robert Baer wrote...
Hah! That's some of the best bandwidth wasting we've seen here
in some time.
Yup. A lot of effort went into that.
Best regards,
Spehro Pefhany
P
#### Product developer
Jan 1, 1970
0
John Fields said:
---
It wasn't a question. As Tim Auton noted, this was a reply, which I
originated on seb.
Sorry you didn't find it at least interesting...
I sure did as with all your posts. Absolutely brilliant. I love these
exercises and I benefit from nearly all them. This NG is one of the
most valuable sources I have ever found. Many thanks to you and the
few other Titans in here that take the time to educate and advance the
Thanks again!
J
#### John Fields
Jan 1, 1970
0
I sure did as with all your posts. Absolutely brilliant. I love these
exercises and I benefit from nearly all them. This NG is one of the
most valuable sources I have ever found. Many thanks to you and the
few other Titans in here that take the time to educate and advance the
Thanks again!
T
#### Terry Given
Jan 1, 1970
0
Dont thank us John, just keep posting such excellent work
Cheers
Terry
J
Jan 1, 1970
0
W
#### Winfield Hill
Jan 1, 1970
0
John Fields wrote...
What should I do then, ignore a compliment?
I don't think so...
We don't want you to get a big head. :>)
Thanks,
- Win
(email: use hill_at_rowland-dot-org for now)
T
#### Terry Given
Jan 1, 1970
0
Winfield Hill said:
John Fields wrote...
We don't want you to get a big head. :>)
Thanks,
- Win
I printed it to a PDF - it will be useful in the future, as I often find
myself working with people who dont understand this sort of thing, and John
kindly saved me from writing it myself
- besides, it will frighten the hell out of my year 12 electronics students
cheers
Terry
F
#### Fred Bloggs
Jan 1, 1970
0
What was the question this "paper" answers?
John said:
1. THE THERMISTOR
For a good discussion of how thermistors work, refer to:
http://www.epcos.com/inf/50/db/ntc_02/00170027.pdf
For the purpose at hand, (determining how the change in resistance
affects the voltage into the comparator) refer to the "1006" column
in:
http://www.epcos.com/inf/50/db/ntc_02/01470178.pdf
and you'll notice that at 15°C the value of Rt/R25 for 15°C is 1.4883,
and at 35°C it's 0.686.
What that means is that if the thermistor has a resistance of 10000
ohms at 25°C, then at 15°C it'll have a resistance of 14,883 ohms, and
at 35°C it'll have a resistance of 6860 ohms.
2. THE VOLTAGE DIVIDER
Now, if we connect the thermistor in series with a resistor and apply
a fixed voltage across them, like this:
E1
|
[R1]
|
+---->E2
|
[RT1]
|
GND
Then when temperature increases and the resistance of the thermistor
(RT1) decreases, E2 will also decrease. Conversely, when the
temperature to which RT1 is exposed decreases, E2 will increase.
The exact relationship is:
E1 RT1
E2 = ------
R1+RT1
so, if we make E1 = 12V and R1 = 100kohms, then at 15°C,
12*14883
E2 = ------------- ~ 1.555V
100000+14883
and the power being dissipated in the thermistor will be:
E² 1.555²
P = --- = ----- = 162µW
R 10000
At 35°C, the potput voltage will be:
12*6860
E2 = ------------- ~ 0.7703V
100000+6860
and the power dissipated by the thermistor:
E² 0.7703²
P = --- = ------- = 86µW
R 6860
The power dissipated in the thermistor may be enough to cause a
self-heating error to occur, but you'd need to look at the data sheet
for the particular thermistor you're using to make sure.
In any case, if the self-heating error is negligible with respect to
the accuracy you're looking for, with 1.555V out at 15°C and 0.770V
out at 35°C, that's a spread of 780mV from one end of the temp range
to the other, which is pretty good.
3. THE REFERENCE DIVIDER.
If we now connect our thermistor circuitry to the + input of a voltage
comparator, like this,
+12V>---------+---------+----+
| | |
[100K] | [RL]
| | |
+--------|+\ |
| | >--+----->OUT
| |-/
| |
[RT1] |
| |
GND>----------+---------+---------->GND
and we arrange things so that the voltage on the - input is 0.770V,
then when the voltage on the + input rises to 0.770V or higher, the
comparator's output transistor will switch off and the voltage on OUT
will be pulled up to +V through RL.
We can do that by using another voltage divider, like this:
+V>---+-------+-----+----+
| | | |
E1--+ [100K] | [RL]
| | | |
[R1] +----|+\ |
| | | >--+----->OUT
E2--+-------|----|-/
| | |
[R2] [RT1] |
| | |
GND>--+-------+-----+---------->GND
and selecting R1 and R2 so that the voltage at their junction will be
0.770V. An easy way to do this is to look at the comparator data
sheet to find out what the required input current is and making the
resistances small enough to swamp out that current. For the LM311,
that comes out to about 350nA, so if we make the divider current 350µA
the LM311's input currents will contribute about a 1% error to the
divider's output voltage. Better yet, let's make the divider current
about 1mA and use 1% resistors for the divider. That way, most of the
error will be in the resistors, where we can easily account for it.
Since we've got a 12V supply, and 10kohms is a standard 1% value, and
12V into 10k is about 1.2mA, let's make R1 = to 10k and figure out
what R2 needs to be to get us to 0.770V.
Since our new divider is the same as the old one, we can write:
E1 R2
E2 = ------
R1+R2
and rearrange to solve for R2:
E2 R1
R2 = -------
E1-E2
which gives us:
0.770*10000
R2 = ------------- = 685.66 ohms
12-0.770
681 ohms is a standard 1% value, so plugging that into
E1 R2
E2 = ------
R1+R2
to get the voltage into the comparator's - input gives us:
E1 R2 12*681
E2 = ------- = ----------- = 0.765V
R1+R2 10000+681
That's a voltage error of 5 parts out of 770, which is just a little
over 1/2 a percent, so since the resistor tolerance error will
dominate at this point we have a tentative solution, and our circuit
now looks like this:
+V>---+-------+-----+----+
| | | |
| [100.0K] | [RL]
| | | |
[10.0K] +----|+\ |
| | | >--+----->OUT
+-------|----|-/
| | |
[681] [RT1] |
| | |
GND>--+-------+-----+---------->GND
4. THE VARIABLE REFERENCE.
Now, we could use the same procedure to figure out a new R2 for a temp
of 15°C, and switch between the two, or figure out a whole bunch of
different R2's and switch them in and out for the different temps we
wanted to set to between 15 and 35°, but a much nicer way to do it
would be to use a pot.
In order to be able to do that, our circuit changes to:
+V>---+-------+-----+----+
| | | |
[R1] [100.0K] | [RL]
| | | |
| +----|+\ |
| | | >--+----->OUT
[R2]<-----|----|-/
| | |
[R3] [RT1] |
| | |
GND>--+-------+-----+---------->GND
Where R1 is a pot and R1 and R3 are 1% resistors used to set the ends
of the pot's resistive element to the voltages corresponding to 15°
and 35°C.
Just to get a feel for things, if we want the end of the pot connected
to R1 to be at the higher endpoint voltage (1.555V) and we keep R1 at
10k,then the series combination of R2 and R3 has to be:
E2 R1 1.555*10000
R2+R3 = ------- = ------------ ~ 1489 ohms
E1-E2 12-1.555
and if we want the lower end of the pot to be at 0.77V, then since
there's
E1 12
I = ----------- = ------------ = 1.04mA
R1+(R2+R3) 10000+1489
flowing in the circuit, and
E
R = ---
I
0.77V
then R3 = ------- = 740 ohms.
1.04mA
Since 750 ohms is a standard 1% value, that leaves
1489ohms - 750ohms = 739 ohms for the pot.
739 ohms is _not_ a standard end-to-end resistance for a pot, so we'll
have to do something to fix that. Since R1R2R3 is a voltage divider,
all we're really concerned with is the ratios of resistances, so let's
scale up the 739 ohms to 1000 ohms, which is a standard end-to-end
resistance for pots, and then scale up everything else and see what
happens.
1000/739 = 1.353, so R1 needs to go to 13530 ohms, R2 will go to 1000
ohms and R3 will need to go to 1000 ohms as well. So now, since the
closest we can get to 13530 ohms with 1% resistors is 13700 ohms, our
string now looks like this:
+12V
|
[13.7K]
|
|
[1000]<----Vout
|
|
[1000]
|
GND
With the wiper at the top of the pot, Vout will be:
12*2000
Vout = --------- = 1.528V
15700
and at the bottom of the pot it will be:
12*1000
Vout = --------- = 0.764V
15700
(1.555-1.528)*100
So, we'll have an error of ----------------- ~ -1.7%
1.555
(0.770-0.764)*100
on the high end, and ------------------- = + 0.78% on the high end.
0.764
Not too shabby, and now, after adding all the resistor values and
introducing the hysteresis resistor, our circuit looks like this:
+12>---+--------+-----+-------+
| | | |
[13.7K] [100.0K] | [RL]
| | | |
| +-------[Rh]--+
| | | |
| +----|+\ |
| | | >-----+----->OUT
[1000]<-----|----|-/
| | |
| | |
[1000] [RT1] |
| | |
GND>---+--------+-----+---------->GND
5. HYSTERESIS
Since a comparator has what amounts to essentially infinite gain, if
its input signals are allowed to move through the switching region
slowly, any noise present on the inputs during that time _will_ cause
the output to swing wildly between its limits. If slowly moving
signals _must_ be accommodated, the cure for the problem is to cause
the input(s) to quickly move out of the sensitive region when the
output changes state. This is done by connecting the output to the
non-inverting input so that when the non-inverting input goes more
positive than the inverting input, causing the output to go high, a
little of that high-going output signal is coupled back to the
non-inverting input, causing it to quickly go even higher, moving it
quickly out of the noisy region, thus preventing the "chatter" which
would otherwise surely occur. In this instance, that positive feedback
is called "hysteresis".
The amount of hysteresis intoduced into the circuit can be controlled
by selecting the resistance of the feedback resistor, which is part of
yet another voltage divider.
Looking at the output of the circuit above in a little more detail
shows us that the output stage is a transistor with an open collector
which pulls the putput down to ground when it is turned on and allows
the output to float when it's turned off, and what's important to
remember is that when the non-inverting (+) input is more positive
than the inverting (-) input the output will be turned off, allowing
current to flow from the supply and through RL into whatever is
connected to the comparator's output.
+12>---+--------+-------+-----+
| | | |
[13.7K] [100.0K] | [RL]
| | | |
| +--[Rh]-------+
| | | |
| | +------+ |
| +----|+ C|-+----->OUT
| | |LM311 |
[1000]<-----|----|- E|-+
| | +------+ |
| | | |
[1000] [RT1] | |
| | | |
GND>---+--------+-------+-----+----->GND
Since the LM311 can sink 50mA, there's no reason why it can't be used
to dirctly drive a 12V relay with a coil resistance of
12V
R = ----- = 240 ohms
50mA
or greater, so adding that circuitry yields:
+12>---+--------+-------+-----+-------+
| | | |K | O--> |
[13.7K] [100.0K] | [1N4001] [COIL]- | - -|
| | | | | | O------>COM
| +--[Rh]-------+-------+ |
| | | | +----------->NO
| | +------+ |
| +----|+ C|-+
| | |LM311 |
[1000]<-----|----|- E|-+
| | +------+ |
| | | |
[1000] [RT1] | |
| | | |
GND>---+--------+-------+-----+----->GND
Checking the sense of the output, we find that as the thermistor heats
up its resistance will decrease until it causes the comparator's +
input to go more negative than its - input, which will cause the
output transistor to turn on, energizing the relay, so this a cooling
thermostat. That is, if it gets hot enough to turn on the relay, the
relay will turn on the A/C, cooling the thermistor until the voltage
on the + input goes more positive than the voltage on the - input,
turning off the output transistor, turning off the relay, then
turning off the AC until it gets hot enough to turn it back on...
Now that we have that all figured out, if we simplify the circuit by
replacing the comparator's input with a resistor and its output with a
switch we'll have what amounts to:
+12>---+---------+
| |
[100.0K] [240]
| |
+--[Rh]---+-+-------->Eout
| |
| +-> |
Ein---+----+ |
| | O
[RT1] [Rin] |
| | |
GND>---+----+----------+---->GND
If 240 ohms (the relay coil resistance) << than Rh and Rin is >> RT1,
(which they will be) then the circuit further simplifies to:
+12>---+-------------+
| |
[100.0K] |
| |
+--[Rh]---O |
| | |
| O--> |<--O
Ein---+ |
| |
[RT1 |
| |
GND>---+----+
So that with the switch in the rightmost position we have:
+12>---+------+
| |
[100.0K] [Rh]
| |
+------+--->Ein
|
[RT1]
|
GND>---+
And with the switch in the leftmost position we have:
+12>---+
|
[100.0K]
|
+------+--->Ein
| |
[RT1] [Rh]
| |
GND>---+------+
Since, at 15°C, (without Rh in the circuit) we'll have an output
voltage of 1.55V from the thermistor and, at 35°C, an output voltage
of 0.77V, we'll have a 780mV change for a 20° change in temperature,
which is a sensitivity of 39mV/°C. That won't be quite true over
every increment of that range, since the thermistor is a non-linear
device but, for now, it's close enough.
Next, let's say that the accuracy spec's for the thermostat will
allow, after everything else has been accounted for, an extra degree C
to be used up to get the hysteresis we need. That means, since we've
got a sensitivity of 39mV/°C, that we've got 39mV to play around with
to get some hysteresis.
Now, below 15°C our thermistor will exhibit a resistance > 14883 ohms,
and if it warms up to 15°C, our circuit will look like this:
+12>---+
|
[100.0K]
|
+---->1.55V
|
[14883]
|
GND>---+
When this happens and the comparator's - input is set to 1.55V, the
output will go low and turn on the relay, and what we'd like to do is
to fool the thermostat into thinking that when that happens it's
actually 16° outside so that it'll move the input switching point a
little closer to ground to get away from the noise-susceptible input
transition region.
Since the thermistor resistance decreases when it gets hotter, what we
can do when the output goes low is to parallel the thermistor with
enough resistance to lower the 1.55 volts by one degree, to 1.51V
(rounding the 39mV/°C sensitivity to 40mV/°C).
Looking at the thermistor circuitry like this:
E1
|
[R1]
|
+---->E2
|
[R2]
|
GND
And setting E1 = 12V, E2 = 1.51V, and R1 = 100kohms,
we can solve for R2, the total resistance we need to get 1.51V out by
writing:
E2 R1 1.51*100000
R2 = ------- = ------------- ~ 14395 ohms
E1-E2 12.0-1.51
Since, at 15°, the thermistor will be at 14883 ohms, we'll need to
parallel it with a resistor to make the combination look like 14395
ohms, like this:
E1
|
[R1]
|
+-----+--->E2
| |
[R2] [R3]
| |
+-----+
|
GND
Knowing that the equation for the total resistance of two parallel
resistors is:
R2 R3
Rt = -------
R2+R3
and that R2 = 14883 ohms and Rt = 14395 ohms,
we can rewrite it to solve for R3 like this:
Rt R2 14395*14883
R3 = -------- = ------------- ~ 439kohms
R2-Rt 14883-14395
The closest 1% resistor we can get which won't exceed our 1°C error
budget is 432k, but that'll still give us nearly the 40mV of
hysteresis we're looking for, so let's choose that value tentatively.
Now, let's say the temp starts falling from that point. Since it's
really 15° outside and we've fooled the comparator into thinking that
it's 16°, it's going to have to fall to 14° for the comparator to
think that it's 15° outside, but when it gets there the comparator's
output will go open-collector and the feedback resistor will now be
connected to +12V through the relay coil. If the coil looks like 240
ohms its resistance will be negligible compared to the feedback
resistor's 439kohm resistance, so since the thermistor's sensitivity
is -401 ohms/°C, (assuming it's linear for the sake of this
discusion...) its resistance will rise to 14482 ohms and the circuit
will now look like this:
12V 12V
| |
[100K] [439K]
| |
+-------+--->E2
[14482]
|
GND
100k in parallel reduces to a total resistance of
100000*439000
Rt = --------------- = 81447 ohms.
100000+439000
So, the circuit further reduces to
12V
|
[81447]
|
+---->E2
|
[14482]
|
GND
And the voltage into the comparator's + input, E2, will now be
12.0*14482
E2 = ------------- ~ 1.812V
81447+14482
Since the [15°C] reference voltage on the comparator's -input will be
sitting at 1.555V, that 1.812V represents a hysteresis of about 257mV,
which represents the introduction of about a 6.6°C error. If the
error spec is +/- 1°, that's very, very, bad.
Fortunately, since the reason for the error is the asymmetry of the
thermistor string, we can fix it by making the string symmetrical.
That is, by making the series resistor and the thermistor the same
value at the center of the thermistor's resistance range for the
temperature range it will be sensing.
For our thermistor, with an R of 14883 ohms at 15°C and an R of 6860
ohms at 35°C, that comes out to 10,872 ohms for the series resistor,
so with the nearest 1% resistor installed, our new circuit will look
like:
12V
|
[10700]
|
+---->E2
|
[Rth]
|
GND
Now E2 will be 6.98V at 15°C (Rth = 14883 ohms) and 4.688V at 35°C
(Rth = 6860 ohms), so the new voltage sensitivity will will be about
115mV/°C.
115mV is a lot, so if we limit our hysteresis to, say, 50mV that'll
cause less than a 0.5° error, but let's see how it works out.
Just like last time, if we want 50 mV of hysteresis we'll have to make
the parallel combination of the 10.7k resistor and and the feeback
resistor change the + input voltage to the comparator by the amount we
want, which will be 50mV in this case.
We really only have four cases to consider:
The hysteresis when the temp is rising and falling through 15°C, and
the hysteresis when the temp is rising and falling through 35°C.
Looking at falling through 15°C first, we'll have:
12V 12V
| |
[10.7K] [Rf]
| |
+-------+--->E2
|
[14883]
|
GND
Since the resistance of the thermistor is increasing as it gets
colder, the voltage at E2 will be increasing also. Without feedback
E2 would be 6.98V when the thermistor got to 15°C (14482 ohms), and
what we want to do with the feedback is to make it seem like it's a
little bit colder than it really is when E2 gets to 6.98V, so that
would mean making E2 jump a little more positive when the comparator
output goes open collector.
We can do that by paralleling the 10.7kohm resistor with a resistor of
a suitable value and, using the procedure outlined earlier, we find
that for a 50mV jump (from 6.98V to 7.03V) the parallel combination of
the 10.7kohm resistor and the feedback resistor (Rf) needs to be 10522
ohms, with Rf being equal to 632502 ohms. The closest 1% resistor is
634kohms, which will give us an E2 of 7.02V as the temp falls through
15°C, about 40mV of hysteresis over the non-feedback case.
With the temp rises through 15°C and the comparator's output
transistor turns on, our circuit becomes:
12V
|
[10.7K]
|
+-------+--->E2
| |
[14883] [634K]
| |
GND GND
and E2 will now become 6.91V, giving us 70mV of hysteresis.
When the temp falls through 35°C, the circuit will become:
12V 12V
| |
[10.7K] [634k]
| |
+-------+--->E2
|
[6808]
|
GND
Without feedback, E2 would be 4.666V, but with feedback it'll be
4.714V, giving us about 48mV of hysteresis.
Finally, When the temp rises through 35°C, the circuit will become:
12V
|
[10.7K]
|
+-------+--->E2
| |
[6808] [634K]
| |
GND GND
and E2 will go to 4.636V, giving us about 30mV of hysteresis.
So, here's the final schematic, and barring any fuzzy thinking, math
errors, trypos, etc., all that's left to do now is to recalculate the
padding resistors around the pot for the new voltage range and trim
things up if you want/need more or less hysteresis, and it should be
pretty much done.
+--- >= 240 ohms
+12>---+--------+--------+-----+-------+ /
| | | |K | / O--> |
[R1] [10.7K] | [1N4001] [COIL]- | - -|
| | | | | | O------>COM
| +-[634K]-------+-------+ |
| | | | +----------->NO
| | +------+ |
| +----|+ C|--+
| | |LM311 |
[1000]<-----|----|- E|--+
| | +------+ |
| | | |
[R2] [RT1] | |
| | | |
GND>---+--------+--------+-----+----->GND
Simple, huh?^)
PS...
I've got some schematics for heating-cooling thermostats using
LM393's, and if you'd like I can post them to
alt.binaries.schematics.electronic.
J
#### John Fields
Jan 1, 1970
0
What was the question this "paper" answers?
---
Check the same subject on seb.
What the **** is it with some of you clowns, anyway?
I do a guy a favor with a little tutorial to which he refuses to
respond and then I get flak from some fucking idiot about wasting
bandwidth, from you with your snide goddam "paper" bullshit, and from
Givens with his "I was gonna write the same thing myself, but"...
To hell with all of you niggardly sons of bitches.
J
#### Jim Thompson
Jan 1, 1970
0
---
Check the same subject on seb.
What the **** is it with some of you clowns, anyway?
I do a guy a favor with a little tutorial to which he refuses to
respond and then I get flak from some fucking idiot about wasting
bandwidth, from you with your snide goddam "paper" bullshit, and from
Givens with his "I was gonna write the same thing myself, but"...
To hell with all of you niggardly sons of bitches.
Careful there, John, some ignorant Democrat will call you a racist ;-)
...Jim Thompson
R
#### Rich Grise
Jan 1, 1970
0
Winfield Hill said:
Robert Baer wrote...
Hah! That's some of the best bandwidth wasting we've seen here
in some time.
My only criticism is that it might have been more appropriate to
..basics, or even both, since it's ASCII after all.
Cheers!
Rich
F
#### Frank Bemelman
Jan 1, 1970
0
John Fields said:
---
Check the same subject on seb.
What the **** is it with some of you clowns, anyway?
I do a guy a favor with a little tutorial to which he refuses to
respond and then I get flak from some fucking idiot about wasting
bandwidth, from you with your snide goddam "paper" bullshit, and from
Givens with his "I was gonna write the same thing myself, but"...
To hell with all of you niggardly sons of bitches.
Well John, there's a bright side, Fred didn't find any errors, he
would have spotted them if there were any I admire your energy
and yes, many can go to hell. But you have to be patient, alas.
R
#### Rich Grise
Jan 1, 1970
0
John Fields said:
---
It wasn't a question. As Tim Auton noted, this was a reply, which I
originated on seb.
Sorry you didn't find it at least interesting...
I found myself trying to figure out what kind of transfer function
would be appropriate to map onto those 3 points ... since it's
essentially impossible to guess, I looked it up from your reference,
http://www.epcos.com/inf/50/db/ntc_02/01470178.pdf
and decided, yup, I never would have got that one, nosireebob!
Cheers!
Rich
R
#### Rich Grise
Jan 1, 1970
0
Product developer said:
I sure did as with all your posts. Absolutely brilliant. I love these
exercises and I benefit from nearly all them. This NG is one of the
most valuable sources I have ever found. Many thanks to you and the
few other Titans in here that take the time to educate and advance the
Yikes! Don't fawn, fer crissakes! None of these weenies deserves it -
they all take their pants off one leg at a time! =:-O
Cheers!
Rich
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2K | 13,623 | 43,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-14 | longest | en | 0.888781 |
https://m.wikihow.com/Make-a-Multiplication-Chart | 1,579,807,656,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00405.warc.gz | 544,014,636 | 61,906 | # How to Make a Multiplication Chart
Co-authored by wikiHow Staff
Updated: March 29, 2019
Multiplication charts and tables are tools used to help you memorize multiplication facts. There are two types of charts you will often see. A times table is a table that just has a series of numbers. A multiplication chart shows you the number sentence that goes with the factors and products. Both types of charts help you easily look up the product of two factors. Eventually your teacher will be expect you to memorize each product on the multiplication chart. Doing so will help make more complicated math easier.
### Method 1 of 3: Making a Times Table
1. 1
Understand how to read a times table. A times table shows the products of sequential factors. Each column is labeled with one factor, and each row is labeled with another factor. The cell where the row and column meet gives you the product of the two factors.
2. 2
Determine how many factors you need in your times table. The standard times table is for factors 1-12, but your teacher may only require you to memorize a table for factors 1-10, depending on what grade you are in.
3. 3
Use a ruler to create 12 rows and 12 columns. You can also find a blank table online and print it, or use graph paper.[1] It’s best to keep your table very neat, so that you don’t confuse factors.
• When you make your table, be sure to keep room to the top and left of it so that you can label your rows and columns.
4. 4
Label the columns and rows 1-12. The column to the left should be labeled “1,” and the other columns labeled in order up to 12. The top row should also be labeled “1,” and the other rows labeled in order down to 12.
5. 5
Fill in the products for each multiplication sentence with a factor of 1. Since any number times 1 equals that number, you can quickly fill in the first row and first column of your times table, since these will all equal the number being multiplied by 1.
• For example, in the top row, you would fill in 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12, since ${\displaystyle 1\times 1=1}$, ${\displaystyle 2\times 1=2}$, ${\displaystyle 3\times 1=3}$, etc. The first column follows the same pattern, since it shows ${\displaystyle 1\times 1=1}$, ${\displaystyle 1\times 2=2}$, ${\displaystyle 1\times 3=3}$, etc.
6. 6
Fill in the rest of the second row. The second row represents ${\displaystyle 2\times x}$, where ${\displaystyle x}$ equals each factor you are multiplying by 2, from 2-12.
• For example, ${\displaystyle 2\times 2=4}$, ${\displaystyle 2\times 3=6}$, ${\displaystyle 2\times 4=8}$, etc.
7. 7
Fill in the remaining rows. In each row, you are multiplying that row’s factor by each factor represented in the columns. A quick way to fill in each row using mental math is to add that row’s factor to each product.[2]
• For example, the numbers in the third row will be 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36. The numbers in the fourth row will be 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48. Note that for the third row, which represents the factor 3, you simply add 3 to each product to get the next product. For the fourth row, which represents the factor 4, you add 4 to each product to get the next product.
8. 8
Check your table. You can verify all of your numbers are correct by using a calculator. You can also check your table against one in your math textbook or online.[3]
### Method 2 of 3: Making a Multiplication Chart
1. 1
Understand how to read this type of chart. This type of chart is made up of separate tables for each factor. Each table shows the sequence of products when you multiply this factor by other numbers.
• For example, this type of chart has a table showing all the products of the factor 1, another table for all the products of the factor 2, another table for all the products of the factor 3, etc.
• Unlike a times table, this type of chart shows the number sentences for each factor.
2. 2
Determine how many factors you need in your chart. The standard multiplication chart is for factors 1-12, but you might need to create a chart with fewer or more factors, depending on your teacher’s requirements.
3. 3
Create 12 tables. Each table should have 12 rows and 3 columns. All 12 tables should fit on one piece of paper. Make sure your table is neat by using ruler to create the lines. It is also best to use graph paper, or you can make your chart in a spreadsheet program, such as Microsoft Excel.
4. 4
Fill in the first column of the first table. Each cell in this column will have a number 1, since this is the table for the factor 1.
5. 5
Fill in the first column of the remaining tables. In the first column of each table, you will simply repeat whatever factor that table represents. So the second table should have a 2 in each cell, the third table should have a 3 in each cell, etc.
6. 6
Fill in the second column of each table. In the second column you will write each factor, from 1-12. This column will be the same in each table.
7. 7
Fill in the multiplication and equals signs for each multiplication sentence. The multiplication sign will come between the two factors. The equals sign will come after the second factor.
• For example, in the first table, you should have:
${\displaystyle 1\times 1=}$
${\displaystyle 1\times 2=}$
${\displaystyle 1\times 3=}$
${\displaystyle 1\times 4=}$
${\displaystyle 1\times 5=}$
${\displaystyle 1\times 6=}$
etc.
8. 8
Fill in the final column of each table. This column will show the product of each multiplication sentence. To fill in this column quickly, simply add whatever factor that table represents to each product.[4]
• For example, the products for the 5s table are 5, 10, 15, 20, 25, 10, 35, 40, 45, 50, 55, 60. Notice that for each new product, you are adding 5 to the product that came before.
### Method 3 of 3: Using Your Chart or Table
1. 1
Review it daily. To help you memorize the chart, aim for reviewing it for at least five minutes each day. Review the table by reading over it and by quizzing yourself. You can have someone else quiz you, or you can use an online program, such as the Math Trainer found at Math is Fun.[5]
2. 2
Look for patterns. Finding patterns in the numbers can help you better remember your times tables.[6] Think about how the numbers relate to each other moving from left-to-right, up and down, or diagonally.
• Note that the order of the multiplication does not matter. For example, ${\displaystyle 4\times 5=20}$ and ${\displaystyle 5\times 4=20}$.
• Notice that when a factor is doubled, its product is doubled. For example, compare the fourth row of the times table with the eight row. Compare the fifth row with the tenth row.[7]
• If you look at the times table for the factor 5, you should see that when you multiply 5 by an odd number, the product ends in 5, and when you multiply 5 by an even number, the product ends in a 0.[8]
• Note that when you multiply any single-digit factor (1-9) by 11, the product has that digit in the ones and the tens place. For example, ${\displaystyle 4\times 11=44}$ and ${\displaystyle 7\times 11=77}$.
• The table for the factor 9 has many patterns. When you add the digits of any product, their sum equals 9.[9] Also, as you move up the times table, the digit in the ones place goes down by 1.[10]
3. 3
Memorize it in chunks. If you try to memorize the entire times table at the same time, you’ll have a difficult time retaining all of the numbers. It’s better to divide the table up into several sections and study each section until you have it memorized.
• For example, you might first memorize your multiplication facts from ${\displaystyle 1x1}$ down to ${\displaystyle 6x6}$ (the top left corner of the chart). Then you can memorize your facts from ${\displaystyle 1x7}$ to ${\displaystyle 6x12}$ (the top right corner of the chart. Then you can memorize your facts from ${\displaystyle 7x1}$ to ${\displaystyle 12x6}$ (the bottom left corner). Finally, you can memorize your facts from ${\displaystyle 7x7}$ to ${\displaystyle 12x12}$ (the bottom right corner).[11]
4. 4
Make a new chart from memory. The best way to see how much you’ve memorized is to create a new chart without using a calculator or any other help. You should also try to fill in products automatically, without doing any mental math.
• Set a time limit, such as 3 minutes, and see whether you can fill in the entire chart in that amount of time. Each time, see if you can fill in the chart faster.
• When you practice making a new chart, fill in the easy calculations first. Skip any product that doesn’t come to you right away. When the time limit is up, you know that you need to work on memorizing the products you could not fill in.
## Community Q&A
Search
• Question
Which will help more, a chart or a table?
I think a table would be more helpful because it's less work to write for more information. Plus, in my opinion it's easier to read a table.
200 characters left
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 11 references.
Co-authors: 9
Updated: March 29, 2019
Views: 12,841
Article SummaryX
To make a multiplication chart, start by creating a table with 12 rows and 12 columns, and label the first row and the first column with the numbers 1 through 12. Then, move to the second column, which should have a number 2 in the top row. From there, fill in the column with multiples of 2 from 2 × 2 all the way to 2 × 12. You'll know you got it right if the number 24 is on the bottom row! Continue filling out your table until all of the squares are filled in, then check the numbers with a calculator. The number at the very top of the column and the very right of the row should multiply together to equal the number of the row you're in. For tips on how to use a multiplication chart to help you memorize multiplication facts, read on!
Thanks to all authors for creating a page that has been read 12,841 times. | 2,635 | 10,044 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-05 | latest | en | 0.918422 |
https://mmsanotherstage2019.com/40-is-what-percent-of-10/ | 1,652,824,421,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00385.warc.gz | 478,531,991 | 6,092 | Step by step solution for calculating 10 is 40 percent that what number
We currently have our first value 10 and the second value 40. Let"s i think the unknown value is Y i beg your pardon answer us will uncover out.
You are watching: 40 is what percent of 10
As we have actually all the required values us need, now we have the right to put them in a simple mathematical formula as below:
STEP 110 = 40% × Y
STEP 210 = 40/100× Y
Multiplying both political parties by 100 and dividing both political parties of the equation through 40 we will arrive at:
STEP 3Y = 10 × 100/40
STEP 4Y = 10 × 100 ÷ 40
STEP 5Y = 25
Finally, we have discovered the value of Y i beg your pardon is 25 and also that is our answer.
You can quickly calculate 10 is 40 percent that what number by using any regular calculator, simply get in 10 × 100 ÷ 40 and also you will gain your answer i beg your pardon is 25
Here is a portion Calculator come solve similar calculations such together 10 is 40 percent of what number. You can solve this form of calculation through your values by start them right into the calculator"s fields, and click "Calculate" to gain the result and explanation.
is
percent the what number
Calculate
Sample questions, answers, and also how to
Question: her friend has a bag of marbles, and also he speak you the 40 percent that the marbles space red. If there are 10 red marbles. How numerous marbles go he have altogether?
How To: In this problem, we understand that the Percent is 40, and we are also told the the part of the marbles is red, so we understand that the part is 10.
So, that method that it have to be the complete that"s missing. Below is the means to number out what the full is:
Part/Total = Percent/100
By utilizing a an easy algebra we can re-arrange ours Percent equation prefer this:
Part × 100/Percent = Total
If we take the "Part" and also multiply it by 100, and also then we division that by the "Percent", us will obtain the "Total".
Let"s try it out on our problem around the marbles, that"s very straightforward and it"s simply two steps! We understand that the "Part" (red marbles) is 10.
So step one is to just multiply that part by 100.
10 × 100 = 1000
In step two, we take the 1000 and divide it by the "Percent", i m sorry we room told is 40.
So, 1000 divided by 40 = 25
And that method that the total number of marbles is 25.
Question: A high college marching band has 10 flute players, If 40 percent the the band members beat the flute, then how countless members space in the band?
Answer: There space 25 members in the band.
How To: The smaller "Part" in this difficulty is 10 since there room 10 flute players and also we room told that they make up 40 percent of the band, so the "Percent" is 40.
Again, it"s the "Total" that"s missing here, and also to discover it, we simply need to monitor our 2 action procedure as the vault problem.
For action one, us multiply the "Part" by 100.
10 × 100 = 1000
For action two, we division that 1000 through the "Percent", i m sorry is 40.
See more: Variation Occurs When Chromosomes Are Shuffled In Ation, And Sexual Reproduction
1000 divided by 40 amounts to 25
That method that the total variety of band members is 25.
Another action by action method
Step 1: Let"s i think the unknown worth is Y
Step 2: first writing the as: 100% / Y = 40% / 10
Step 3: autumn the portion marks to leveling your calculations: 100 / Y = 40 / 10
Step 4: main point both sides by Y to move Y top top the appropriate side the the equation: 100 = ( 40 / 10 ) Y
Step 5: simple the ideal side, we get: 100 = 40 Y
Step 6: separating both political parties of the equation by 40, we will certainly arrive in ~ 25 = Y
This leaves us through our last answer: 10 is 40 percent that 25
10 is 40 percent the 25 10.01 is 40 percent of 25.025 10.02 is 40 percent the 25.05 10.03 is 40 percent of 25.075 10.04 is 40 percent that 25.1 10.05 is 40 percent the 25.125 10.06 is 40 percent of 25.15 10.07 is 40 percent that 25.175 10.08 is 40 percent of 25.2 10.09 is 40 percent that 25.225 10.1 is 40 percent the 25.25 10.11 is 40 percent of 25.275 10.12 is 40 percent that 25.3 10.13 is 40 percent of 25.325 10.14 is 40 percent that 25.35 10.15 is 40 percent the 25.375 10.16 is 40 percent of 25.4 10.17 is 40 percent of 25.425 10.18 is 40 percent of 25.45 10.19 is 40 percent that 25.475
10.2 is 40 percent of 25.5 10.21 is 40 percent the 25.525 10.22 is 40 percent the 25.55 10.23 is 40 percent that 25.575 10.24 is 40 percent the 25.6 10.25 is 40 percent that 25.625 10.26 is 40 percent of 25.65 10.27 is 40 percent the 25.675 10.28 is 40 percent that 25.7 10.29 is 40 percent that 25.725 10.3 is 40 percent that 25.75 10.31 is 40 percent the 25.775 10.32 is 40 percent the 25.8 10.33 is 40 percent that 25.825 10.34 is 40 percent that 25.85 10.35 is 40 percent of 25.875 10.36 is 40 percent the 25.9 10.37 is 40 percent the 25.925 10.38 is 40 percent that 25.95 10.39 is 40 percent of 25.975
10.4 is 40 percent of 26 10.41 is 40 percent of 26.025 10.42 is 40 percent the 26.05 10.43 is 40 percent that 26.075 10.44 is 40 percent of 26.1 10.45 is 40 percent of 26.125 10.46 is 40 percent the 26.15 10.47 is 40 percent that 26.175 10.48 is 40 percent the 26.2 10.49 is 40 percent that 26.225 10.5 is 40 percent that 26.25 10.51 is 40 percent that 26.275 10.52 is 40 percent of 26.3 10.53 is 40 percent of 26.325 10.54 is 40 percent that 26.35 10.55 is 40 percent the 26.375 10.56 is 40 percent of 26.4 10.57 is 40 percent the 26.425 10.58 is 40 percent of 26.45 10.59 is 40 percent of 26.475
10.6 is 40 percent the 26.5 10.61 is 40 percent that 26.525 10.62 is 40 percent of 26.55 10.63 is 40 percent that 26.575 10.64 is 40 percent that 26.6 10.65 is 40 percent of 26.625 10.66 is 40 percent the 26.65 10.67 is 40 percent of 26.675 10.68 is 40 percent that 26.7 10.69 is 40 percent that 26.725 10.7 is 40 percent that 26.75 10.71 is 40 percent the 26.775 10.72 is 40 percent the 26.8 10.73 is 40 percent the 26.825 10.74 is 40 percent of 26.85 10.75 is 40 percent of 26.875 10.76 is 40 percent of 26.9 10.77 is 40 percent that 26.925 10.78 is 40 percent the 26.95 10.79 is 40 percent that 26.975
10.8 is 40 percent of 27 10.81 is 40 percent of 27.025 10.82 is 40 percent that 27.05 10.83 is 40 percent of 27.075 10.84 is 40 percent of 27.1 10.85 is 40 percent of 27.125 10.86 is 40 percent that 27.15 10.87 is 40 percent the 27.175 10.88 is 40 percent of 27.2 10.89 is 40 percent that 27.225 10.9 is 40 percent that 27.25 10.91 is 40 percent that 27.275 10.92 is 40 percent of 27.3 10.93 is 40 percent that 27.325 10.94 is 40 percent the 27.35 10.95 is 40 percent the 27.375 10.96 is 40 percent the 27.4 10.97 is 40 percent of 27.425 10.98 is 40 percent that 27.45 10.99 is 40 percent the 27.475 | 2,285 | 6,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2022-21 | latest | en | 0.960369 |
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# Solutions and Explanations for HW Chapter 27 BB
## Prepared by Prof Omar Chmaissem and TA Abhilash Sajja
Question 1 A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0
A. What is the magnetic torque when the plane of the loop is perpendicular to the magnetic field?
0.47 mN
0.52 mN
0.59 mN
zero
0.41 mN
Explanation:
Magnetic Torque is given by
= B sin where is the angle between the magnetic moment and the Magnetic field.
Since the direction of is perpendicular to plane of the loop and the direction of B is perpendicular to the plane,
## should be parallel to B which results in to be 0.
= Bsin0
= 0 m.N
Question 2 A stationary proton is in a uniform magnetic field of 0.20 T. What is the magnitude of the magnetic force on
the proton?
zero
3.2 10-20 N
1.6 10-20 N
1.6 10-21 N
3.2 10-21 N
Explanation:
Force on a particle moving in a magnetic field is given by F = q(v x B) where q is the charge of the particle, v is the
velocity of the particle entering magnetic field, and B is the magnetic field strength.
The magnitude of this magnetic force can be written as F = qvBsin where is the angle between the velocity vector
and the magnetic field.
In this case, the proton velocity is 0m/sec so the magnetic force will also be 0N.
Question 3 A proton travels through a potential of 1.0 kV and then moves into a magnetic field of 0.040 T. What is the
radius of the proton's resulting orbit?
0.11 m
0.19 m
0.17 m
0.14 m
0.080 m
Explanation:
Remember the equation we derived in class for the cyclotron radius, r = mv/Bq [eq(25.1)]. This was obtained by
setting the net force which is centripetal (mv2/r) to equal the magnetic force qvb.
To solve this equation we need to figure out the protons velocity.
Initially the proton is accelerated by a potential V = 1000V. Hence, the Electric potential energy gained by the proton
is qV. This energy is converted into kinetic energy resulting in the proton moving with some velocity v.
Hence, KE = PE
()mv2 = qV
eq(25.2)
Substituting the values of mproton = 1.67 x 10-27Kg, V = 1000V , q = 1.6 x 10-19C in eq(25.2), we get
v = 4.38 x 105 m/s. Hence, the proton enters the magnetic field with this velocity v.
Substituting m= 1.6 x 10-27kg, v = 4.38 x 105 m/s, B = 0.04T, q = 1.6 x 10-19 C in eq(25.1), we get
r = 0.11m
5
Question 4 An electron traveling due north with speed 4.0 10 m/s enters a region where the Earth's magnetic field has
the magnitude 5.0 10-5 T and is directed downward at 45 below the horizontal. What force acts on the electron?
3.2 10-18 N
2.3 10-20 N
3.2 10-19 N
2.3 10-19 N
2.3 10-18 N
Explanation:
Straightforward. The magnitude of the magnetic force on a particle moving in a magnetic field is given by F = qvB
sin. We just need to clearly identify the angle .
Draw a sketch for the direction of the electrons velocity due north and the earths magnetic field downward and
tilted, the angle between the two vectors is indeed 45 degrees. Other given parameters are q = 1.6 x 10 -19C, v =4.0 105
m/s, B = 5.0 10-5 T
Substituting these parameters in the equation, we get F = 2.3 x 10-18N.
Question 5 A charge q = 3 10-6 C of mass m = 2 10-6 kg, and speed v = 5 106 m/s enters a uniform magnetic field.
The mass experiences an acceleration a = 3 104 m/s2. What is the minimum magnetic field that would produce such an
acceleration?
0.001 T
0.04 T
0.004 T
0.1 T
0.01 T
Explanation:
The mass m experiences an acceleration a due to a net force F given by F = ma = 0.06 N
The only force acting significantly on this charge is the magnetic force. Therefore the net force found above is the
same as F = qvBsin.
Hence, to produce this force, the magnetic field B should be B = F/qvsin. Minimum magnetic field is obtained when
sin is maximum because its in the denominator. Therefore, sin = 1 and Bmin= F/qv
Plug F= 0.06N, q = 3 x 10-6C, v = 5x106 m/s into the above equation, we get B = 0.004T
Question 6 A beam of electrons is accelerated through a potential difference of 10 kV before entering a velocity selector.
If the B-field of the velocity selector has a value of 0.010 T, what value of the E-field is required if the particles are to be
undeflected?
7.2 106 V/m
6.0 105 V/m
2.3 103 V/m
5.9 105 V/m
7.9 103 V/m
Explanation:
We already know that in a velocity selector we obtain v = E/B. I know the magnetic field but v and E are unknown.
Its obvious that I need to figure out the velocity in order to solve for the electric field magnitude.
Initially proton is accelerated by a potential difference V = 10,000V
Hence, KE = qV
()mv2 = Vq
eq(27.1)
Substituting the values of m = 9.1 x 10-31Kg, V = 10000 V, q = 1.6 x 10-19C in eq(27.1), we get
v = 5.93 x 107 m/s
Hence, the proton enters and exits undeflected the velocity selector with this velocity v.
v = E/B
E = vB
E = 5.93 x 107 x 0.01
E = 5.93 x 105 V/m
Question 7 A loop of diameter d =12 cm, carrying a current I = 0.4 A is placed inside a magnetic field
B = (0.2 T) i + (0.4
T) j . The normal to the loop is parallel to the unit vector n = 0.6 i 0.8 j . What is the potential energy of the loop?
+4.5 10-4 J
+9.0 10-4 J
-9.0 10-4 J
-4.5 10-4 J
-2.3 10-4 J
Explanation:
Magnetic loop potential energy is given by U = . B (dot product of magnetic moment vector and magnetic field
vector B)
A vector is defined as having two quantities: magnitude and direction.
Applied to the magnetic moment vector , we have the magnitude (IA) and the direction which is the unit vector
= 0.4( * (0.06)2) * (0.6 i 0.8 j )
## = (2.7 x 10-3 i 3.6 x 10-3 j ) A.m2
B = (0.2 i + 0.4 j ) T
U=.B
U = (2.7 x 10-3 i 3.6 x 10-3 j ) . (0.2 i + 0.4 j ) = (2.7 x 10-3 i + 3.6 x 10-3 j ) . (0.2 i + 0.4 j )
U = (2.7 x 10-3)(0.2)+( 3.6 x 10-3)(0.4) = 0.9 x 10-3 J = 9 x 10-4 J
Question 8 A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the
magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to
the magnetic field?
35
20
25
60
30
Explanation:
Magnetic force on a current carrying wire is given by F = ILB sin() where is the angle between the wire length
vector taken in the direction of the current (I) and the magnetic field (B).
Substituting the parameters I = 2A, B = 0.3T, L = 0.6m, F = 0.18N in the above equation, we get
0.18 = 0.3 x 2 x 0.6 sin()
sin() = 0.5
= 30
Note that if the angle were to be zero (meaning the wire is parallel to the field) then there would be no magnetic force
present.
Question 9 A proton is projected with a velocity of 7.0 103 m/s into a magnetic field of 0.60 T perpendicular to the
motion of the proton. What is the force that acts on the proton?
0N
6.7 10-16 N
13 10-16 N
3.4 10-16 N
4.2 10-16 N
Explanation:
Magnetic force on a particle moving in a magnetic field is given by F = qvB sin where is the angle between the
direction of the particle motion and the magnetic field.
Substituting the parameters as v = 7000m/s, B = 0.6T q = 1.6 * 10-19 (charge of proton), =90, we get
F = 6.7 x 10-16 N
Question 10 A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 T. What minimum
current in the rod is needed in order for the magnetic force to cancel the weight of the rod?
7.6 A
1.2 A
4.9 A
9.8 A
2.5 A
Explanation:
Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I)
and the magnetic field (B).
This force has to be canceled by gravity. Or, mg = ILBsin. Solving for I, we get I = mg/ LBsin.
To get the minimum value of the current flowing in the wire we have to set to be 90o
( maximizing the denominator there are no other variables to change).
Substituting the parameters , B = 0.1T, L = 1.0m, =90 in the above equation, we get
I = 4.9A
4.9 A is the minimum current required to generate a magnetic force which could cancel the force due to gravity.
Question 11 ) A loop of diameter d = 10 cm, carrying a current I = 0.2 A is placed inside a magnetic field B = 0.3 T k .
The normal to the loop is parallel to a unit vector n = - 0.6 i 0.8 j . Calculate the magnitude of the torque on the loop.
(This is a nice but tedious problem to solve. You will need to determine vector A. This can be done by multiplying the
magnitude of A by the unit vector direction because it is in the same direction. Once you figure it out, you do cross
product according to the torque equation. The answer is in a vector notation form. You will need then to calculate the
magnitude of the torque vector.)
1.2 10-4 Nm
0
0.6 10-4 Nm
2.8 10-4 Nm
4.7 10-4 Nm
Explanation: As described in the hint, magnetic moment vector can be obtained by multiplying its magnitude (IA)
with the unit vector representing its direction n = 0.6 i 0.8 j
Hence vector
## = (9.42 x 10-4 i 12.56 x 10-4 j ) A.m2
(cross product)
= xB
= = (9.42 x 10-4 i 12.56 x 10-4 j ) x (0.3 k )
## = (+2.83 x 10-4 j 3.77 x 10-4 i ) Nm
Magnitude of =10-4
## (2.83) 2 (3.77) 2 = 4.7 x 10-4 N.m
Question 12 What is the magnetic moment of a rectangular loop of 120 turns that carries 6.0 A if its dimensions are 4.0
cm 8.0 cm?
2.3 Am2
230 Am2
23 Am2
0.023 Am2
0.23 Am2
Explanation:
Magnetic moment of a rectangular loop is given by = NIA where N is the number of turns, I is the current in the
loop, and A is the area of the loop.
= 120*6*32x10-4
= 2.3 A.m2
The magnetic moment is a vector quantity with a direction perpendicular to the current loop according to the righthand rule. It is the same as the direction of the vector A representing the loop area A.
Question 13 An electron moving perpendicular to a magnetic field of 3.2 10-2 T moves in a circle of radius 0.40 cm.
How fast is this electron moving?
3.0 109 m/s
1.9 10-30 m/s
0.80 107 m/s
2.3 107 m/s
1.9 10-53 m/s
Explanation:
The centripetal force is balanced by the magnetic force resulting in a circular motion. Hence,
mv2/r = qvBsin() (with =90)
v = qBr/m
-2
v = (3.2 10 x 1.6 x 10-19 x 0.004) / (9.1 x 10-31)
v = 2.3 x 107 m/s.
Question 14 A proton is accelerated from rest through 500 V. It enters a magnetic field of 0.30 T oriented perpendicular
to its direction of motion. Determine the radius of the path it follows.
1.1 m
11 m
11 cm
1.1 mm
1.1 cm
Explanation:
We know this is going to be a circle. We need the speed. We can calculate the incoming kinetic energy since we know
that the proton is accelerated in a potential difference V = 500V
Hence, KE = qV
()mv2 = qV
eq(25.1)
Substituting the values of m = 1.67 x 10-27Kg, V = 500V , q = 1.6 x 10-19C in eq(25.1), we get
v = 3.1 x 105 m/s
Hence, the proton enters the magnetic field with this velocity v.
Remember the equation, the centripetal force balancing the magnetic force
From that we derive, r = mv/qB
eq(25.2)
-27
5
Substituting m= 1.67 x 10 kg, v = 3.1 x 10 m/s, B = 0.3T, q = 1.6 x 10-19C in eq(25.2), we get
r = 1.1 cm
Question 15 In a Hall experiment a conducting strip of thickness d = 100 m is placed in a magnetic field B = 0.05 T. The
magnetic field is perpendicular to the direction of the strip along which there is a 10-A current. What is the Hall voltage
measured across the strip if the charge carrier density is 2.5 1028/m3 in this material?
0.72 V
4.2 V
2.5 V
1.3 V
6.2 V
Explanation:
Hall voltage is given by VH = IB/ned where I is the current through the strip, B is the magnetic field, d is the strip
thickness, n is the carrier density, and e is electron charge.
Substituting I = 10A, B = 0.05T, n = 2.5 x 1028 /m3, e = 1.6 x 10-19 C, d=10-4 m in the above equation, we get
VH = 1.3 x 10-6 V (approx).
Question 16 A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current in the loop is 2.0
A. What is the magnetic torque when the plane of the loop is parallel to the magnetic field?
0.59 mN
zero
0.47 mN
0.41 mN
0.52 mN
Explanation:
Magnetic Torque is given by
= Bsin where is the angle between the magnetic moment and the Magnetic field.
Since the direction of is perpendicular to the plane of the loop and the direction of B is parallel to the plane,
should be perpendicular to B which results in to be 90.
= Bsin90 ( = IA)
= IABsin90
= (2) x (3.14 x0.52) x (0.3) x sin90
= 0.47 m.N
Question 17 A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force
on the wire when it is at an angle of 30 with respect to the field?
0.5 N
0.3 N
0.4 N
0.6 N
0.2 N
Explanation:
Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I)
and the magnetic field (B).
Substituting the parameters I = 4A, B = 0.6T, L = 0.2m, = 30 in the above equation, we get
F = 0.6 x 4 x 0.2 sin 30
F = 0.2 N
Question 18 A wire carries a current of 11.4 A in a direction that makes an angle of 11.4 with a magnetic field of
-3
magnitude 11.4 10 T. The magnitude of the force on 11.4 cm of this wire is
1.48 10-2 N.
2.93 10-3 N.
0.130 N.
11.4 10-3 N.
0.013 N.
Explanation:
Magnetic force on a current carrying wire is given by F = ILBsin where is the angle between the current direction (I)
and the magnetic field (B).
Substituting the parameters I = 11.4A, B = 11.4mT, L = 0.114m, = 11.4 in the above equation, we get
F = 11.4x10-3 x 11.4 x 0.114x sin 11.4
F = 2.93 x 10-3 N.
Question 19 An electron moving with a velocity
## and a magnetic fields are present. The electric field is
through, undeflected?
= + (2.0 10-4 T)
= + (2.0 10-4 T)
=- (2.0 10-4 T)
= - (2.0 10-4 T)
= + (5.0 10-4 T)
## = - 104 . What magnetic field will allow the electron to go
Explanation:
This is the velocity selector device. v = E/B
We did not use this name in class but the sum of the electric and magnetic forces is called the Lorentz force. Lorentz
force = Electric force + Magnetic force
When the charge goes undeflected its because the electric force is canceled out by the magnetic force and vice versa.
This is essentially saying that the Lorentz force is zero.
0 = Electric force + Magnetic force
Magnetic force = - Electric force
Using vector notations because we also need to figure out the magnetic field direction:
q v B qE
v B E (where the x sign represents the cross product)
v B = (5.0 107 m/s ) x (Bx +By +Bz k ) = (5.0 107) By k + (5.0 107)Bz (- )
E 10 4 j so E 10 4 j
Equating the two equations v B E show that (By) must be zero. We get the following relationship: (-5.0
107)Bz = (104) or Bz = 104/(-5.0 107) = - 2.0 x 10-4 T.
Bz obviously represents a magnetic field that is fully pointing in the negative z direction with the magnitude we
just calculated.
B = - (2.0 10-4 T) k
2
Question 20 A circular loop of wire of cross-sectional area 0.12 m consists of 200 turns, each carrying 0.50 A. It is
placed in a magnetic field of 0.050 T oriented at 30 to the plane of the loop. What torque acts on the loop?
5.2 mN
0.52 mN
0.25 mN
2.5 mN
25 mN
Explanation:
Since the magnetic field is oriented 30 to the plane of the loop, B should be 60 (90 30) to the magnetic moment .
Draw a sketch and check the actual angle.
Hence, = 60
Magnetic Torque is given by
=NIABsin
= 200x0.5x0.12x0.05xsin60
= 0.52 m.N
6
Question 21 A proton is moving at a speed of 3.2 10 m/s at an angle of 80 to a uniform magnetic field of strength 1.5
10-4 T. What is the distance moved by the proton along the spiral path as it completes one revolution? (This distance is
called the "pitch" of the helix.)
240 m
1390 m
1370 m
220 m
cannot be determined
Explanation:
We already know that any travel angle different than 0o, 90o or 180o should produce a spiral path for a charge moving
in a magnetic field. In this problem, we can calculate two components for the velocity: Parallel to the magnetic field
v// = v cos = 0.555 106 m/s and
Perpendicular to the field
v = v sin = 0.985 106 m/s.
The pitch distance is p = v//T. T is the time required to make a full revolution. When the proton completes one full
revolution in one period T due to the perpendicular velocity component, it finds itself propelled forward by the pitch
distance due to the parallel velocity component. To emphasize: the forward distance is solely due to the parallel
velocity component as the distance equation shows.
Its obvious that we need to calculate the period T. We just mentioned that it is the time for the time to complete a full
revolution. This means that the distance travelled around the circular path (due to the perpendicular component of
the velocity) is 2r = vT . r is not known so we have to use r = mv/qB. Plug and cancel the v terms to get:
2 m/qB = T
T = 2 (1.672 x 10-27 kg)/[(1.602 x 10-19 ) (1.5 x 10-4)]= 4.37 x 10-4 seconds
Notice that the period (or frequency) is actually independent of the velocity.
Now that we know the time, lets calculate the pitch distance: p = (0.555 106)( 4.37 x 10-4) = 242 m
Question 22 A proton moving eastward with a velocity of 5.0 103 m/s enters a magnetic field of 0.20 T pointing
northward. What is the magnitude and direction of the force that acts on the proton?
1.1 10-16 N eastwards
0N
1.6 10-16 N downwards
1.6 10-16 N upwards
4.4 10-16 N westwards
Explanation:
The proton is moving eastwards and can be considered to be moving in the positive x direction ( i ).
In vector notation v = 5000 i m/s.
The magnetic field is pointing in the north direction and can be considered to be in the positive y direction ( j ).
In vector notation B = 0.2 j T.
F = q(v x B)
F = 1.6x10-19 (5000 i X 0.2 j )
F = 1.6x10-19 x 5000 x 0.2( i X j ) (cross product of i x j = k )
## F = 1.6x10-16 k ( k represents the positive z direction or upwards)
Alternately, we can determine the magnitude F = qvBsin90o (field and velocity vectors are perpendicular to each other)
and figure out the direction of the force using the right hand rule. Fingers towards east, curl them towards the north,
this would give a thumb direction pointing up. We have a positive charge so this is the direction of the magnetic
force.
Question 23 A force F = (6 10-3 N ) i - (4 10-3 N) j - (8 10-3 N) k is acting on a charge q = 2x10-9 C moving inside a
uniform magnetic field with a velocity v = (2x106 m/s) i + (3x106 m/s) j . Calculate the components of the magnetic field
knowing that the x-component of the field is equal to zero.
.
=- (4T) + (0.5 T)
= - (2 T) + (1 T)
= (4 T) - (0.5 T)
=- (3T) + (2T)
= (2 T) - (1 T)
Explanation:
F = q*(v x B)
F is known in a vector form F = (6 10-3) i - (4 10-3) j - (8 10-3) k . All we need is to calculate the right side vector
and compare the components from which we can get the magnetic fields components.
Consider the general magnetic field B = Bx i +By j +Bz k
but the x- component is zero so: B = By j +Bz k
i
-9
q(v B) = 2 10 2 10 6
0
j
3 10 6
By
k
0 i 6 10 3 Bz j 4 10 3 Bz k 4 10 3 B y
Bz
## Now equate the two sides:
6 10 i 4 10 j 8 10 k i6 10
3
Bz j 4 10 3 Bz k 4 10 3 By
6 10 6 10 B or B 1
4 10 4 10 B same solution
8 10 4 10 B or B 2
3
## Hence, magnetic field vector is (-2 j + 1 k ) Tesla.
Question 24 A current is flowing in a wire in direction 3 i + 4 j where the direction of the magnetic field is 5 j + 12 k . The force
on the wire is
0.78 i - 0.58 j + 0.24 k
j .
0.78 i + 0.58 j + 0.24 k
0.71( i + j ).
0.78 i + 0.58 j - 0.24 k .
Explanation:
The direction of force on the wire can be obtained using the unit vectors of the current and magnetic field.
F = I(L x B)
Since, L is a scalar quantity, the direction of the force is determined by the current and magnetic fields (cross product
using the right hand rule).
direction of L = (3 i + 4 j )/ (3) 2 (4) 2 ; L = (3/5) i + (4/5) j ;
direction of B = (5 j + 12 k )/ (5)2 (12)2 ; B = (5/13) j + (12/13) k ;
i
Direction of force is given by directions of L x B = 0.6
0.8
0.38 0.92
## L x B = (0.8*0.92) i (0.6*0.92) j +(0.6*0.38) k
L x B = 0.78 i 0.58 j + 0.24 k (approx)
Question 25 In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 106 m/s and enters a
uniform magnetic field of 0.20 T. The radius of the circular orbit is 0.020 m. What is the mass of the particle?
1.7 10-27 kg
3.2 10-28 kg
4.5 10-27 kg
3.1 10-31 kg
6.4 10-28 kg
Explanation:
Mass spectrometer is an application which uses the magnetic force on a moving particle.
Remember the equation, the centripetal force balancing the magnetic force
From that we derive m = qBr/v
Substituting B = 0.2T, q=1.6 x 10-19C, r= 0.02m, v = 106 m/s in the above equation, we get m = 6.4 x 10-28 kg.
## B = 7 mT k with a velocity v = ( 9 106
m/s) i + ( 7 106 m/s) j + ( 5 106 m/s) k . What is the spacing of the circles of the helix on which the electron moves?
7.1 cm
5.7 cm
1.5 cm
2.6 cm
9.5 cm
Explanation:
1Pitch depends on the velocity component that is parallel to the magnetic field direction. Here, the magnetic
field is in the positive z direction ( B = 7 mT k ) so the parallel velocity component is ( 5 106 m/s) k .
2The time registered during travel is the period.
3The period is independent of the perpendicular velocity component(s). T = 2m/qB
So, the pitch of the spiral is
Pitch (p) = (v//)(2m/qB)
p = (5 x 106 * 2 *9.1 x 10-31) / (1.6 x 10-19 * 7 x 10-3)
x = 2.6 cm
Question 27 A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two
sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being
perpendicular to the magnetic field. The magnitude of the torque on the loop is
0.038 mN.
0.050 mN.
0.0019 mN.
0.025 mN.
0.0093 mN.
Explanation:
Draw a sketch and convince yourself that the angle between vectors and B is 90o.
Magnetic Torque is given by = Bsin where is the angle between magnetic moment and Magnetic field
=IABsin
Substituting the values I = 2.5A, A = 0.02, B = 0.037, = 90
= 0.0019 m.N | 7,169 | 21,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-05 | latest | en | 0.903161 |
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Chapter 3 Mathematical Induction 51
CHAPTER 3 3. Let P(n) be the proposition
2
bñÉêÅáëÉ=P^=EéKTNF 2 2 + 4 2 + 6 2 + L + (2 n) 2 = n(n + 1)(2 n + 1) .
3
1. Let P(n) be the proposition When n = 1 , L.H.S. = 2 2 = 4
2
n R.H.S. = (1)(1 + 1)(2 + 1) = 4
1 + 4 + 7 + L + (3n − 2) = (3n − 1) . 3
2 ∴ P(1) is true.
When n = 1, L.H.S. = 1
Assume P(k) is true for any positive integer k.
1 2
R.H.S. = [3(1) − 1] = 1 i.e. 2 2 + 4 2 + 6 2 + L + (2 k )2 = k ( k + 1)(2 k + 1)
2 3
∴ P(1) is true. Then 2 2 + 4 2 + 6 2 + L + (2 k )2 + [2( k + 1)]2
Assume P(k) is true for any positive integer k. 2
= k ( k + 1)(2 k + 1) + 4( k + 1)2
k 3
i.e. 1 + 4 + 7 + L + (3k − 2) = (3k − 1) 2
2 = ( k + 1)[k (2 k + 1) + 6( k + 1)]
Then 1 + 4 + 7 + L + (3k − 2) + [3( k + 1) − 2] 3
2
k
= (3k − 1) + (3k + 1) = ( k + 1)(2 k 2 + k + 6k + 6)
2 3
2
3k 2 k = ( k + 1)(2 k 2 + 7k + 6)
= − + 3k + 1 3
2 2 2
3k − k + 6k + 2
2 = ( k + 1)( k + 2)(2 k + 3)
= 3
2 2
= ( k + 1)[(k + 1) + 1][2( k + 1) + 1]
3k 2 + 5k + 2 3
= Thus assuming P(k) is true for any positive integer
2
( k + 1)(3k + 2) k, P( k + 1) is also true. By the principle of
= mathematical induction, P(n) is true for all
2
k +1 positive integers n.
= [3( k + 1) − 1]
2 4. Let P(n) be the proposition
Thus assuming P(k) is true for any positive integer a + ( a + d ) + ( a + 2 d ) + L + [a + (n − 1)d ]
k, P( k + 1) is also true. By the principle of n
mathematical induction, P(n) is true for all = [2 a + (n − 1)d ].
positive integers n. 2
When n = 1 , L.H.S. = a
1
2. Let P(n) be the proposition R.H.S. = [2 a + (1 − 1)d ] = a
2
2 3 + 4 3 + 6 3 + L + (2 n)3 = 2 n 2 (n + 1)2 . ∴ P(1) is true.
Assume P(k) is true for any positive integer k.
When n = 1 , L.H.S. = 2 3 = 8 i.e. a + ( a + d ) + ( a + 2 d ) + L + [a + ( k − 1)d ]
R.H.S. = 2(1)2 (1 + 1)2 = 8 k
= [2 a + ( k − 1)d ]
∴ P(1) is true. 2
Then
Assume P(k) is true for any positive integer k.
a + ( a + d ) + L + [a + ( k − 1)d ] + {a + [( k + 1) − 1]d}
i.e. 2 3 + 4 3 + 6 3 + L + (2 k )3 = 2 k 2 ( k + 1)2 k
= (2 a + ( k − 1)d ] + ( a + kd )
Then 2 3 + 4 3 + L + (2 k )3 + [2( k + 1)]3 2
1
= 2 k 2 ( k + 1)2 + 8( k + 1)3 = [2 ka + k ( k − 1)d + 2 a + 2 kd ]
2
= 2( k + 1)2 ( k 2 + 4 k + 4) 1
= [( k + 1)2 a + k 2 d − kd + 2 kd ]
= 2( k + 1)2 ( k + 2)2 2
1
= 2( k + 1)2 [( k + 1) + 1]2 = [( k + 1)2 a + k ( k + 1)d ]
2
Thus assuming P( k ) is true for any positive 1
= ( k + 1)(2 a + kd )
integer k, P( k + 1) is also true. By the principle of 2
mathematical induction, P(n) is true for all k +1
positive integers n. = {2 a + [( k + 1) − 1]d}
2
52 Chapter 3 Mathematical Induction
Thus assuming P(k) is true for any positive integer When n = 1 , L.H.S. = 1 ⋅ 3 = 3
k, P( k + 1) is also true. By the principle of (2 − 1)31+1 + 3
mathematical induction, P(n) is true for all R.H.S. = =3
positive integers n. 4
∴ P(1) is true.
5. Let P(n) be the proposition Assume P(k) is true for any positive integer k.
2 ⋅ 3 + 4 ⋅ 6 + 6 ⋅ 9 + L + (2 n)(3n) i.e. 1 ⋅ 3 + 2 ⋅ 32 + 3 ⋅ 33 + L + k ⋅ 3k
= n(n + 1)(2 n + 1) .
(2 k − 1)3k +1 + 3
When n = 1 , L.H.S. = (2)(3) = 6 =
4
R.H.S. = (1)(1 + 1)(2 + 1) = 6 Then 1 ⋅ 3 + 2 ⋅ 32 + 3 ⋅ 33 + L + k ⋅ 3k + ( k + 1) ⋅ 3k +1
∴ P(1) is true. (2 k − 1)3k +1 + 3
Assume P(k) is true for any positive integer k. = + ( k + 1) ⋅ 3k +1
4
i.e. 2 ⋅ 3 + 4 ⋅ 6 + 6 ⋅ 9 + L + (2 k )(3k ) 1
= [(2 k − 1)3k +1 + 3 + 4( k + 1) ⋅ 3k +1 ]
= k ( k + 1)(2 k + 1) 4
Then 2 ⋅ 3 + 4 ⋅ 6 + 6 ⋅ 9 + L + (2 k )(3k ) (2 k − 1 + 4 k + 4)3k +1 + 3
=
+ [2( k + 1)][3( k + 1)] 4
= k ( k + 1)(2 k + 1) + 6( k + 1)2 (6k + 3) ⋅ 3k +1 + 3
=
= ( k + 1)(2 k 2 + k + 6k + 6) 4
(2 k + 1) ⋅ 3k + 2 + 3
= ( k + 1)(2 k 2 + 7k + 6) =
= ( k + 1)( k + 2)(2 k + 3) 4
= ( k + 1)[(k + 1) + 1][2( k + 1) + 1] [2( k + 1) − 1]3( k +1) +1 + 3
=
Thus assuming P(k) is true for any positive integer 4
k, P( k + 1) is also true. By the principle of Thus assuming P(k) is true for any positive integer
mathematical induction, P(n) is true for all k, P( k + 1) is also true. By the principle of
positive integers n. mathematical induction, P(n) is true for all
positive integers n.
6. Let P(n) be the proposition
1 ⋅ 4 + 2 ⋅ 7 + 3 ⋅ 10 + L + n(3n + 1) = n(n + 1)2 . 8. Let P(n) be the proposition
1 1 1 1 n
When n = 1 , L.H.S. = 1 ⋅ 4 = 4 + + +L+ = .
1⋅ 2 2 ⋅ 3 3 ⋅ 4 n(n + 1) n + 1
R.H.S. = (1 + 1)2 = 4 1 1
∴ P(1) is true. When n = 1 , L.H.S. = =
1⋅ 2 2
Assume P(k) is true for any positive integer k. 1 1
R.H.S. = =
i.e. 1 ⋅ 4 + 2 ⋅ 7 + 3 ⋅ 10 + L + k (3k + 1) = k ( k + 1)2 1+1 2
Then ∴ P(1) is true.
1 ⋅ 4 + 2 ⋅ 7 + L + k (3k + 1) + ( k + 1) + [3( k + 1) + 1] Assume P(k) is true for any positive integer k.
= k ( k + 1)2 + ( k + 1)(3k + 4) i.e. 1 + 1 + 1 + L + 1
=
k
= ( k + 1)( k 2 + k + 3k + 4) 1⋅ 2 2 ⋅ 3 3 ⋅ 4 k ( k + 1) k + 1
Then
= ( k + 1)( k 2 + 4 k + 4) 1 1 1 1
+ +L+ +
= ( k + 1)( k + 2)2 1⋅ 2 2 ⋅ 3 k ( k + 1) ( k + 1)[(k + 1) + 1]
= ( k + 1)[(k + 1) + 1]2 =
k
+
1
Thus assuming P(k) is true for any positive integer k + 1 ( k + 1)( k + 2)
k, P( k + 1) is also true. By the principle of k ( k + 2) + 1
=
mathematical induction, P(n) is true for all ( k + 1)( k + 2)
positive integers n.
( k + 1)2
=
7. Let P(n) be the proposition ( k + 1)( k + 2)
k +1
1 ⋅ 3 + 2 ⋅ 32 + 3 ⋅ 33 + L + n ⋅ 3n =
k+2
(2 n − 1)3n +1 + 3 k +1
= . =
4 ( k + 1) + 1
Chapter 3 Mathematical Induction 53
Thus assuming P(k) is true for any positive integer ∴ P(1) is true.
k, P( k + 1) is also true. By the principle of Assume P(k) is true for any positive integer k.
mathematical induction, P(n) is true for all
1 1 1
positive integers n. i.e. + + +L+
1⋅ 3 ⋅ 5 3 ⋅ 5 ⋅ 7 5 ⋅ 7 ⋅ 9
9. Let P(n) be the proposition 1 k ( k + 2)
=
1 1 1 1 (2 k − 1)(2 k + 1)(2 k + 3) 3(2 k + 1)(2 k + 3)
+ + +L+
2 ⋅ 5 5 ⋅ 8 8 ⋅ 11 (3n − 1)(3n + 2) Then
n
= . 1
+
1
+
1
+L+
6n + 4 1⋅ 3 ⋅ 5 3 ⋅ 5 ⋅ 7 5 ⋅ 7 ⋅ 9
1 1
When n = 1 , L.H.S. = = 1
2 ⋅ 5 10 (2 k − 1)(2 k + 1)(2 k + 3)
1 1 1
R.H.S. = = +
6 + 4 10 [2( k + 1) − 1][2( k + 1) + 1][2( k + 1) + 3]
∴ P(1) is true. k ( k + 2) 1
= +
Assume P(k) is true for any positive integer k. 3(2 k + 1)(2 k + 3) (2 k + 1)(2 k + 3)(2 k + 5)
1 1 1 1 k ( k + 2)(2 k + 5) + 3
i.e. + + +L+ =
2 ⋅ 5 5 ⋅ 8 8 ⋅ 11 (3k − 1)(3k + 2) 3(2 k + 1)(2 k + 3)(2 k + 5)
k 2 k 3 + 9k 2 + 10 k + 3
= =
6k + 4 3(2 k + 1)(2 k + 3)(2 k + 5)
Then
1
+
1
+
1
+L+
1 ( k + 1)(2 k 2 + 7k + 3)
2 ⋅ 5 5 ⋅ 8 8 ⋅ 11 (3k − 1)(3k + 2) =
3(2 k + 1)(2 k + 3)(2 k + 5)
+
1 ( k + 1)(2 k + 1)( k + 3)
[3( k + 1) − 1][3( k + 1) + 2] =
3(2 k + 1)(2 k + 3)(2 k + 5)
=
k
+
1 ( k + 1)( k + 3)
6k + 4 (3k + 2)(3k + 5) =
3(2 k + 3)(2 k + 5)
k (3k + 5) + 2 ( k + 1)[(k + 1) + 2]
= =
2(3k + 2)(3k + 5) 3[2( k + 1) + 1][(2( k + 1) + 3]
3k 2 + 5k + 2 Thus assuming P(k) is true for any positive integer
=
2(3k + 2)(3k + 5) k, P( k + 1) is also true. By the principle of
(3k + 2)( k + 1) mathematical induction, P(n) is true for all
=
2(3k + 2)(3k + 5) positive integers n.
k +1
=
2(3k + 5) 11. Let P(n) be the proposition
k +1
= 1 1 1 1
6( k + 1) + 4 + + +L+
1 × 4 4 × 7 7 × 10 (3n − 5)(3n − 2)
Thus assuming P(k) is true for any positive integer
k, P( k + 1) is also true. By the principle of n −1
= .
mathematical induction, P(n) is true for all 3n − 2
positive integers n. 1 1
When n = 2 , L.H.S. = =
1× 4 4
10. Let P(n) be the proposition
2 −1 1
1 1 1 R.H.S. = =
+ + +L+ 3×2 −2 4
1⋅ 3 ⋅ 5 3 ⋅ 5 ⋅ 7 5 ⋅ 7 ⋅ 9
1 n( n + 2 ) ∴ P(2) is true.
= .
(2 n − 1)(2 n + 1)(2 n + 3) 3(2 n + 1)(2 n + 3) Assume P(k) is true for any positive integer k ≥ 2 .
1 1 1 1 1 1
When n = 1 , L.H.S. = = i.e. + + +L+
1 ⋅ 3 ⋅ 5 15 1 × 4 4 × 7 7 × 10 (3k − 5)(3k − 2)
(1 + 2) 1 k −1
R.H.S. = = =
3(2 + 1)(2 + 3) 15 3k − 2
54 Chapter 3 Mathematical Induction
Then 13. Let P(n) be the proposition
1
+
1
+
1
+L+
1 1 × 1! +2 × 2! + L + n × n! = (n + 1)! −1 .
1 × 4 4 × 7 7 × 10 (3k − 5)(3k − 2) When n = 1 , L.H.S. = 1 × 1! = 1
1
+ R.H.S. = (1 + 1)! −1 = 1
[3( k + 1) − 5][3( k + 1) − 2]
k −1 1 ∴ P(1) is true.
= +
3k − 2 (3k − 2)(3k + 1) Assume P(k) is true for any positive integer k.
1 ( k − 1)(3k + 1) + 1 i.e. 1 × 1! + 2 × 2! + L + k × k! = ( k + 1)! −1
= [ ]
3k − 2 3k + 1 Then 1 × 1! +2 × 2! + L + k × k! + ( k + 1) × ( k + 1)!
1 k (3k − 2)
= ⋅ = ( k + 1)! −1 + ( k + 1) × ( k + 1)!
3k − 2 3k + 1 = ( k + 1)!(1 + k + 1) − 1
k
= = ( k + 1)!( k + 2) − 1
3k + 1 = ( k + 2)! −1
( k + 1) − 1
= = [( k + 1) + 1]! −1
3( k + 1) − 2
Thus assuming P(k) is true for any positive integer
Thus assuming P(k) is true for any positive integer k, P( k + 1) is also true. By the principle of
k ≥ 2 , P( k + 1) is also true. By the principle of mathematical induction, P(n) is true for all
mathematical induction, P(n) is true for all positive integers n.
positive integers n ≥ 2 .
14. No solution is provided for the H.K.C.E.E.
12. Let P(n) be the proposition question because of the copyright reasons.
4 4 4 2n + 1
(1 − )(1 − ) L[1 − ]= . 15. (a) Let P(n) be the proposition
9 25 (2 n − 1) 2 3(2 n − 1)
1 ⋅ 32 + 2 ⋅ 52 + 3 ⋅ 72 + L + n(2 n + 1)2
4 5
When n = 2 , L.H.S. = 1 − = 1
= n(n + 1)(6n 2 + 14n + 7) .
9 9 6
2( 2 ) + 1 5 When n = 1 , L.H.S. = 1 ⋅ 32 = 9
R.H.S. = =
3[2(2) − 1] 9 1
R.H.S. = (1 + 1)(6 + 14 + 7) = 9
∴ P(2) is true. 6
∴ P(1) is true.
Assume P(k) is true for any positive integer k ≥ 2 .
Assume P(k) is true for any positive integer k.
4 4 4 2k + 1
i.e. (1 − )(1 − ) L[1 − ]= i.e. 1 ⋅ 32 + 2 ⋅ 52 + 3 ⋅ 72 + L + k (2 k + 1)2
9 25 (2 k − 1)2 3(2 k − 1) 1
= k ( k + 1)(6k 2 + 14 k + 7)
Then 6
4 4 4 4 Then
(1 − )(1 − ) L[1 − ]{1 − }
9 25 (2 k − 1)2 [2( k + 1) − 1]2 1 ⋅ 32 + 2 ⋅ 52 + 3 ⋅ 72 + L + k (2 k + 1)2
2k + 1 4 + ( k + 1)[2( k + 1) + 1]2
= (1 − )
3(2 k − 1) (2 k + 1)2 1
= k ( k + 1)(6k 2 + 14 k + 7) + ( k + 1)(2 k + 3)2
(2 k + 1)[(2 k + 1)2 − 4] 6
= 1
3(2 k − 1)(2 k + 1)2 = ( k + 1)[k (6k 2 + 14 k + 7) + 6(2 k + 3)2 ]
(2 k − 1)(2 k + 3) 6
= 1
3(2 k − 1)(2 k + 1) = ( k + 1)(6k 3 + 14 k 2 + 7k + 24 k 2 + 72 k + 54)
6
2k + 3
= 1
= ( k + 1)(6k 3 + 38k 2 + 79k + 54)
3(2 k + 1) 6
2( k + 1) + 1
= 1
= ( k + 1)( k + 2)(6k 2 + 26k + 27)
3[2( k + 1) − 1)] 6
Thus assuming P(k) is true for any positive integer 1
= ( k + 1)( k + 2)(6k 2 + 12 k + 6 + 14 k + 14 + 7)
k ≥ 2 , P( k + 1) is also true. By the principle of 6
mathematical induction, P(n) is true for all 1
= ( k + 1)[(k + 1) + 1][6( k + 1)2 + 14( k + 1) + 7]
positive integers n ≥ 2 . 6
Chapter 3 Mathematical Induction 55
Thus assuming P(k) is true for any positive = [12 ⋅ 4 + 2 2 ⋅ 5 + 32 ⋅ 6 + L + n 2 (n + 3)]
integer k, P( k + 1) is also true. By the
− 2(12 + 2 2 + L + n 2 )
principle of mathematical induction, P(n) is
1 1
true for all positive integers n. = n(n + 1)(n 2 + 5n + 2) − 2 ⋅ n(n + 1)(2 n + 1)
(b) 11 ⋅ 232 + 12 ⋅ 252 + 13 ⋅ 272 + L + 20 ⋅ 412 4 6
1 1
1 = n(n + 1)(n + 5n + 2) − n(n + 1)(2 n + 1)
2
= (20)(21)(2 400 + 280 + 7) 4 3
6 1
1 = n(n + 1)(3n + 15n + 6 − 8n − 4)
2
− (10)(11)(600 + 140 + 7) (By (a)) 12
6 1
= 188 090 − 13 695 = n(n + 1)(3n 2 + 7n + 2)
12
= 174 395 1
= n(n + 1)(n + 2)(3n + 1)
12
16. (a) Let P(n) be the proposition
12 ⋅ 4 + 2 2 ⋅ 5 + 32 ⋅ 6 + L + n 2 (n + 3) 17. (a) Let P(n) be the proposition
1 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + L + n(n + 1)
= n(n + 1)(n 2 + 5n + 2) .
4 1
= n(n + 1)(n + 2) .
When n = 1 , L.H.S. = 12 ⋅ 4 = 4 3
1
R.H.S. = (1 + 1)(1 + 5 + 2) = 4 When n = 1 , L.H.S. = 1 ⋅ 2 = 2
4
1
∴ P(1) is true. R.H.S. = (1 + 1)(1 + 2) = 2
3
Assume P(k) is true for any positive integer k. ∴ P(1) is true.
i.e. 12 ⋅ 4 + 2 2 ⋅ 5 + 32 ⋅ 6 + L + k 2 ( k + 3) Assume P(k) is true for any positive integer k.
1
= k ( k + 1)( k 2 + 5k + 2) i.e. 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + L + k ( k + 1)
4
1
Then = k ( k + 1)( k + 2)
3
12 ⋅ 4 + 2 2 ⋅ 5 + 32 ⋅ 6 + L + k 2 ( k + 3) Then
+ ( k + 1)2 [( k + 1) + 3] 1 ⋅ 2 + 2 ⋅ 3 + L + k ( k + 1) + ( k + 1)[(k + 1) + 1]
1 1
= k ( k + 1)( k 2 + 5k + 2) + ( k + 1)2 ( k + 4) = k ( k + 1)( k + 2) + ( k + 1)( k + 2)
4 3
1 1
= ( k + 1)( k 3 + 5k 2 + 2 k + 4 k 2 + 20 k + 16) = ( k + 1)( k + 2)( k + 3)
4 3
1 1
= ( k + 1)( k 3 + 9k 2 + 22 k + 16) = ( k + 1)[(k + 1) + 1][(k + 1) + 2)
4 3
1
= ( k + 1)( k + 2)( k 2 + 7k + 8) Thus assuming P(k) is true for any positive
4 integer k, P( k + 1) is also true. By the
1
= ( k + 1)( k + 2)( k 2 + 2 k + 1 + 5k + 5 + 2) principle of mathematical induction, P(n) is
4 true for all positive integers n.
1
= ( k + 1)[(k + 1) + 1][(k + 1)2 + 5( k + 1) + 2] (b) (i) 51 ⋅ 52 + 52 ⋅ 53 + L + 100 ⋅ 101
4
Thus assuming P(k) is true for any positive = 1 ⋅ 2 + 2 ⋅ 3 + L + 100 ⋅ 101
integer k, P( k + 1) is also true. By the − (1 ⋅ 2 + 2 ⋅ 3 + L + 50 ⋅ 51)
principle of mathematical induction, P(n) is 1 1
= (100)(101)(102) − (50)(51)(52)
true for all positive integers n. 3 3
= 343 400 − 44 200
(b) 12 ⋅ 2 + 2 2 ⋅ 3 + L + n 2 (n + 1) = 299 200
= 12 ( 4 − 2) + 2 2 (5 − 2) + 32 (6 − 2)
(ii) The nth term of the series
+ L + n 2 [(n + 3) − 2]
= 1+ 2 +L+ n
= 12 ⋅ 4 − 12 ⋅ 2 + 2 2 ⋅ 5 − 2 2 ⋅ 2 + 32 ⋅ 6 − 32 ⋅ 2
1
+ L + n 2 (n + 3) − n 2 ⋅ 2 = n(n + 1)
2
56 Chapter 3 Mathematical Induction
Sum of the first n term: (ii) 12 − 2 2 + 32 − 4 2 + L − 40 2
=
1 1 1
(1)(2) + (2)(3) + L + n(n + 1) = 12 + 2 2 + L + 40 2
2 2 2 − 2(2 2 + 4 2 + L + 40 2 )
1
= [(1)(2) + (2)(3) + L + n(n + 1)] 1 2
2 = ( 40)( 41)(81) − 2 ⋅ (20)(21)( 41)
1 1 6 3
= ⋅ n(n + 1)(n + 2) = 22 140 − 22 960
2 3 = −820
1
= n(n + 1)(n + 2) (c) 1 ⋅ 1 + 2 ⋅ 3 + 3 ⋅ 5 + L + n(2 n − 1)
6
= 1 ⋅ (2 ⋅ 1 − 1) + 2 ⋅ (2 ⋅ 2 − 1)
∴ 1 + (1 + 2) + L + (1 + 2 + L + 100) + 3(2 ⋅ 3 − 1) + L + n(2 n − 1)
1
= (100)(101)(102) = 2 ⋅ 12 − 1 + 2 ⋅ 2 2 − 2 + 2 ⋅ 32
6
= 171 700 − 3 + L + 2n 2 − n
= 2(12 + 2 2 + 32 + L + n 2 )
− (1 + 2 + 3 + L + n)
18. (a) Let P(n) be the proposition
1 1
= 2 ⋅ n(n + 1)(2 n + 1) − n(n + 1) (By (a))
1 6 2
12 + 2 2 + 32 + L + n 2 = n(n + 1)(2 n + 1) .
6 1 1
= n(n + 1)(2 n + 1) − n(n + 1)
When n = 1 , L.H.S. = 12 = 1 3 2
1
1 = n(n + 1)( 4n + 2 − 3)
R.H.S. = (1 + 1)(2 + 1) = 1 6
6 1
= n(n + 1)( 4n − 1)
∴ P(1) is true. 6
Assume P(k) is true for any positive integer k.
bñÉêÅáëÉ=P_=EéKTRF
i.e. 12 + 2 2 + 32 + L + k 2
1 1. Let P(n) be the proposition 9 n − 2 n is divisible
= k ( k + 1)(2 k + 1)
6 by 7.
Then When n = 1, 9 − 2 = 7 which is divisible by 7.
12 + 2 2 + L + k 2 + ( k + 1)2 ∴ P(1) is true.
1 Assume P(k) is true for any positive integer k.
= k ( k + 1)(2 k + 1) + ( k + 1)2
6 i.e. 9 k − 2 k = 7 N where N is an integer.
1
= ( k + 1)(2 k 2 + k + 6k + 6) Then 9 k +1 − 2 k +1
6
1 = 9 ⋅ 9k − 2 ⋅ 2 k
= ( k + 1)(2 k 2 + 7k + 6)
6 = 9(7n + 2 k ) − 2 ⋅ 2 k
1 = 9(7 N ) + 7 ⋅ 2 k
= ( k + 1)( k + 2)(2 k + 3)
6
1 = 7(9 N + 2 k )
= ( k + 1)[(k + 1) + 1][2( k + 1) + 1]
6 which is divisible by 7.
Thus assuming P(k) is true for any positive Hence P( k + 1) is true if P(k) is true for any
integer k, P( k + 1) is also true. By the positive integer k. By the principle of
principle of mathematical induction, P(n) is mathematical induction, P(n) is true for all
true for all positive integers n. positive integers n.
(b) (i) 2 2 + 4 2 + 6 2 + L + (2 n)2 2. Let P(n) be the proposition 2 2n − 1 is divisible
= 4(12 + 2 2 + 32 + L + n 2 ) by 3.
1 When n = 1, 2 2 − 1 = 3 which is divisible by 3.
= 4 ⋅ n(n + 1)(2 n + 1)
6 ∴ P(1) is true.
2
= n(n + 1)(2 n + 1) Assume P(k) is true for any positive integer k.
3
Chapter 3 Mathematical Induction 57
i.e. 2 2k − 1 = 3 N where N is an integer. 5. Let P(n) be the proposition 6 n − 5n + 4 is
divisible by 5.
Then 2 2( k +1) − 1
When n = 1, 6 − 5 + 4 = 5 which is divisible by 5.
= 22k ⋅ 22 − 1
= 4(3 N + 1) − 1 ∴ P(1) is true.
= 4( 3 N ) + 3 Assume P(k) is true for any positive integer k.
= 3( 4 N + 1) i.e. 6 k − 5k + 4 = 5 N where N is an integer.
which is divisible by 3. Then
Hence P( k + 1) is true if P(k) is true for any 6 k +1 − 5( k + 1) + 4
positive integer k. By the principle of
mathematical induction, P(n) is true for all = 6 ⋅ 6 k − 5k − 1
positive integers n. = 6(5 N + 5k − 4) − 5k − 1
= 30 N + 25k − 25
3. Let P(n) be the proposition 9 n − 4 n is divisible = 5(6 N + 5k − 5)
by 5. which is divisible by 5.
When n = 1, 9 − 4 = 5 which is divisible by 5. Hence P( k + 1) is true if P(k) is true for any
∴ P(1) is true. positive integer k. By the principle of
mathematical induction, P(n) is true for all
Assume P(k) is true for any positive integer k. positive integers n.
i.e. 9 k − 4 k = 5 N where N is an integer.
Then 9 k +1 − 4 k +1 6. Let P(n) be the proposition n(n + 1)(n + 2) is
= 9⋅9 − 4⋅4
k k divisible by 3.
= 9(5 N + 4 k ) − 4 ⋅ 4 k When n = 1, (1 + 1)(1 + 2) = 6 which is divisible
by 3.
= 9(5 N ) + 5 ⋅ 4 k
∴ P(1) is true.
= 5(9 N + 4 k )
Assume P(k) is true for any positive integer k.
which is divisible by 5. i.e. k ( k + 1)( k + 2) = 3 N where N is an integer.
Hence P( k + 1) is true if P(k) is true for any Then
positive integer k. By the principle of
mathematical induction, P(n) is true for all ( k + 1)( k + 2)( k + 3)
positive integers n. = k ( k + 1)( k + 2) + 3( k + 1)( k + 2)
= 3 N + 3( k + 1)( k + 2)
= 3[ N + ( k + 1)( k + 2)]
4. Let P(n) be the proposition 23n + 10 is divisible which is divisible by 3.
by 11.
Hence P( k + 1) is true if P(k) is true for any
W h e n n = 1 , 231 + 10 = 33 = 11(3) w h i c h i s positive integer k. By the principle of
divisible by 11. mathematical induction, P(n) is true for all
∴ P(1) is true. positive integers n.
Assume P(k) is true for any positive integer k.
i.e. 23k + 10 = 11N where N is an integer. 7. No solution is provided for the H.K.C.E.E.
question because of the copyright reasons.
Then 23k +1 + 10
= 23(11N − 10) + 10
= 23(11N ) − 220 8. Let P(n) be the proposition 4 n + 5n is divisible
= 11(23 N − 20) by 9.
which is divisible by 11. When n = 1, 4 + 5 = 9 which is divisible by 9.
Hence P( k + 1) is true if P(k) is true for any ∴ P(1) is true.
positive integer k. By the principle of Assume P(k) is true for any positive odd number
mathematical induction, P(n) is true for all k.
positive integers n.
i.e. 4 k + 5k = 9 N where N is an integer.
58 Chapter 3 Mathematical Induction
For n = k + 2 4 ⋅ 9 n − (9n − 4n + 4) ⋅ 4 n is divisible by
Then 4 k + 2 + 5k + 2 25.
= 4 k ⋅ 16 + 5k ⋅ 25 ∴ 4 ⋅ 9 − (5n + 4) ⋅ 4 n is divisible by 25.
= 16(9 N − 5k ) + 25 ⋅ 5k (ii) By (a), put x = 7, y = 4
= 16( 4 k + 5k ) + 9 ⋅ 5k 7 n − n(7)4 n −1 + (n − 1)4 n
= 16(9 N ) + 9 ⋅ 5 k
= 7 n − (7n − 4n + 4)4 n −1
= 9(16 N + 5 ) k
= 7 n − (3n + 4) ⋅ 4 n −1
which is divisible by 9.
is divisible by (7 − 4)2 .
Hence P( k + 2) is true for any positive odd
number k. By the principle of mathematical ∴ 7 n − (3n + 4) ⋅ 4 n −1 is divisible by 9.
induction, P(n) is true for all positive odd numbers
n.
1
10. Let P(n) be the proposition an = n(n − 1) + 1.
2
9. (a) Let P(n) be the proposition
When n = 1, L.H.S. = a1 = 1
x n − nxy n −1 + (n − 1) y n is divisible by
1
( x − y)2 . R.H.S. = (1)(1 − 1) + 1 = 1
2
When n = 2, x 2 − 2 xy + y 2 = ( x − y)2 which ∴ P(1) is true.
is divisible by ( x − y)2 . Assume P(k) is true for any positive integer k.
∴ P(2) is true. 1
i.e. ak = k ( k − 1) + 1
Assume P(k) is true for any positive integer 2
k ≥ 2. Then
i.e. x k − kxy k −1 + ( k − 1) y k = ( x − y)2 N a k +1 = a k + k
where N is a polynomial. 1
= k ( k − 1) + 1 + k
Then 2
1
= k (k − 1 + 2) + 1
x k +1 − ( k + 1) xy k + ky k +1 2
= x ⋅ x k − ( k + 1) xy k + ky k +1 1
= k ( k + 1) + 1
= x[( x − y)2 N + kxy k −1 − ( k − 1) y k ] 2
1
− ( k + 1) xy k + ky k +1 = ( k + 1)[(k + 1) − 1] + 1
2
= x ( x − y)2 N + kx 2 y k −1 − ( k − 1) xy k Thus assuming P(k) is true for any positive integer
− kxy k − xy k + ky k +1 k, P( k + 1) is also true. By the principle of
= x ( x − y)2 N + ky k −1 ( x 2 − 2 xy + y 2 )
mathematical induction, P(n) is true for all
positive integers n.
= x ( x − y)2 N + ( x − y)2 ky k −1
= ( x − y)2 ( xN + ky k −1 ) 11. Let P(n) be the proposition an = 2 n − 1.
which is divisible by ( x − y)2 . When n = 1,
Hence P( k + 1) is true if P(k) is true for any L.H.S. = a1 = 1
positive integer k ≥ 2 . By the principle of R.H.S. = 2(1) − 1 = 1
mathematical induction, P(k) is true for all
∴ P(1) is true.
positive integers n ≥ 2 .
When n = 2,
(b) (i) By (a), put x = 9, y = 4
L.H.S. = a2 = 3
∴ 9 n − n(9)4 n −1 + (n − 1)4 n is divisible R.H.S. = 2(2) − 1 = 3
by (9 − 4)2 . ∴ P(2) is true.
4 ⋅ 9 n − 9n ⋅ 4 n + (n − 1)4 n +1 is divisible Assume P(k) and P( k + 1) are true for any positive
by 25. integer k.
Chapter 3 Mathematical Induction 59
i.e. ak = 2 k − 1 ∴ P(1) is true.
ak +1 = 2( k + 1) − 1 Assume P(k) is true for any positive integer k.
Then ak + 2 = 2 ak +1 − ak a(r k − 1)
= 2[2( k + 1) − 1] − (2 k − 1) i.e. a + ar + ar 2 + L + ar k −1 =
r −1
= 4k + 4 − 2 − 2k + 1 Then a + ar + ar 2 + L + ar k −1 + ar k
= 2k + 3
a(r k − 1)
= 2( k + 2 ) − 1 = + ar k
r −1
Thus assuming P(k) and P( k + 1) are true for any
ar k − a + ar k +1 − ar k
positive integer k, P( k + 2) is also true. By the =
principle of mathematical induction, P(n) is true r −1
for all positive integers n. a(r k +1 − 1)
=
r −1
Thus assuming P(k) is true for any positive integer
oÉîáëáçå=bñÉêÅáëÉ=P=EéKTSF k, P( k + 1) is also true. By the principle of
1. Let P(n) be the proposition 5 + 2 ⋅ 52 + 3 ⋅ 53 mathematical induction, P(n) is true for all
positive integers n.
( 4n − 1)5n +1 + 5
+ L + n ⋅ 5n = .
16 3. Let P(n) be the proposition
When n = 1, L.H.S. = 5 1 1 1 1
+ + +L+
( 4 − 1)52 + 5 1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 5 n(n + 1)(n + 2)
R.H.S. = =5
16 n(n + 3)
= .
∴ P(1) is true. 4(n + 1)(n + 2)
Assume P(k) is true for any positive integer k. 1 1
When n = 1,L.H.S. = =
i.e. 5 + 2 ⋅ 52 + 3 ⋅ 53 + L + k ⋅ 5k 1⋅ 2 ⋅ 3 6
(1 + 3) 1
( 4 k − 1)5k +1 + 5 R.H.S. = =
= 4(1 + 1)(1 + 2) 6
16
∴ P(1) is true.
Then 5 + 2 ⋅ 52 + 3 ⋅ 53 + L + k ⋅ 5k + ( k + 1) ⋅ 5k +1
Assume P(k) is true for any positive integer k.
( 4 k − 1)5k +1 + 5
= + ( k + 1) ⋅ 5k +1 1 1 1
16 i.e. + +L+
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 k ( k + 1)( k + 2)
( 4 k − 1)5k +1 + 5 + 16( k + 1)5k +1
= k ( k + 3)
16 =
4( k + 1)( k + 2)
( 4 k − 1 + 16k + 16)5k +1 + 5
= 1 1 1
16 Then + +L+
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 k ( k + 1)( k + 2)
(20 k + 15)5k +1 + 5
= 1
16 +
( k + 1)[(k + 1) + 1][(k + 1) + 2]
( 4 k + 3)5( k +1) +1 + 5 k ( k + 3)
= = +
1
16 4( k + 1)( k + 2) ( k + 1)( k + 2)( k + 3)
[ 4( k + 1) − 1]5( k +1) +1 + 5
= k ( k + 3)2 + 4
16 =
4( k + 1)( k + 2)( k + 3)
Thus assuming P(k) is true for any positive integer
k, P( k + 1) is also true. By the principle of k 3 + 6k 2 + 9k + 4
=
mathematical induction, P(n) is true for all 4( k + 1)( k + 2)( k + 3)
positive integers n. ( k + 1)2 ( k + 4)
=
2. Let P(n) be the proposition 4( k + 1)( k + 2)( k + 3)
( k + 1)[(k + 1) + 3]
a(r n − 1) =
a + ar + ar 2 + L + ar n −1 = . 4[( k + 1) + 1][(k + 1) + 2]
r −1
When n = 1, L.H.S. = a Thus assuming P(k) is true for any positive integer
k, P( k + 1) is also true. By the principle of
a(r1 − 1)
R.H.S. = =a mathematical induction, P(n) is true for all
r −1 positive integers n.
60 Chapter 3 Mathematical Induction
4. Let P(n) be the proposition 7. No solution is provided for the H.K.C.E.E.
n −1 2 question because of the copyright reasons.
1 − 2 + 3 − 4 + L + ( −1) n
2 2 2 2
1
= ( −1) n −1 n(n + 1) .
2 8. Let P(n) be the proposition 8n + 2 ⋅ 7 n − 1 is
When n = 1,L.H.S. = 12 = 1 divisible by 7.
1 When n = 1, 8 + 2 ⋅ 7 − 1 = 21 = 7 ⋅ 3 which is
R.H.S. = ( −1)0 (1 + 1) = 1
2 divisible by 7.
∴ P(1) is true. ∴ P(1) is true.
Assume P(k) is true for any positive integer k. Assume P(k) is true for any positive integer k.
1
i.e. 12 − 2 2 + L + ( −1) k −1 k 2 = ( −1) k −1 k ( k + 1) i.e. 8k + 2 ⋅ 7 k − 1 = 7 N where N is an integer.
2
Then
Then 8k +1 + 2 ⋅ 7 k +1 − 1
k −1 2 ( k +1) −1
1 − 2 + L + ( −1) k + ( −1)
2 2
( k + 1)
2
= 8 ⋅ 8k + 14 ⋅ 7 k − 1
k −1 1
= ( −1) k ( k + 1) + ( −1) ( k + 1)
k 2
= 8(7 N − 2 ⋅ 7 k + 1) + 14 ⋅ 7 k − 1
2
1 = 7(8 N ) − 2 ⋅ 7 k + 7
= ( −1) k ( k + 1)( − k + 2 k + 2)
2 = 7(8 N − 2 ⋅ 7 k −1 + 1)
k 1
= ( −1) ( k + 1)( k + 2) which is divisible by 7.
2
1
= ( −1)( k +1) −1 ( k + 1)[(k + 1) + 1]
Thus assuming P(k) is true for any positive integer
2 k, P( k + 1) is also true. By the principle of
Thus assuming P(k) is true for any positive integer mathematical induction, P(n) is true for all
k, P( k + 1) is also true. By the principle of positive integers n.
mathematical induction, P(n) is true for all
positive integers n. 9. Let P(n) be the proposition 6 n + 2 + 72 n +1 is
divisible by 43.
5. No solution is provided for the H.K.C.E.E. When n = 1, 6 3 + 73 = 559 = 43 × 13 which is
question because of the copyright reasons. divisible by 43.
∴ P(1) is true.
6. Let P(n) be the proposition
Assume P(k) is true for any positive integer k.
n(n + 1)(n + 5)
T1 + T2 + L + Tn = .
3 i.e. 6 k + 2 + 72 k +1 = 43 N where N is an integer.
When n = 1,L.H.S. = T1 = 12 + 3(1) = 4 Then 6 k + 3 + 72 k + 3
R.H.S. =
(1 + 1)(1 + 5)
=4 = 6 ⋅ 6 k + 2 + 49 ⋅ 72 k +1
3 = 6( 43 N − 72 k +1 ) + 49 ⋅ 72 k +1
∴ P(1) is true.
= 6( 43 N ) + 43 ⋅ 72 k +1
Assume P(k) is true for any positive integer k.
k ( k + 1)( k + 5) = 43(6 N + 72 k +1 )
i.e. T1 + T2 + L + Tk = which is divisible by 43.
3
T
Then 1 2 + T + L + Tk + T k +1 Thus assuming P(k) is true for any positive integer
k ( k + 1)( k + 5) k, P( k + 1) is also true. By the principle of
= + ( k + 1)2 + 3( k + 1) mathematical induction, P(n) is true for all
3
k +1 positive integers n.
= [k ( k + 5) + 3( k + 1) + 9]
3
k +1 10. (a) Let P(n) be the proposition
= [( k + 6)( k + 2)]
3 1
13 + 2 3 + 33 + L + n 3 = n 2 (n + 1)2 .
( k + 1)[(k + 1) + 1][(k + 1) + 5] 4
=
3 When n = 1 , L.H.S. = 13 = 1
Thus assuming P(k) is true for any positive integer
1
k, P( k + 1) is also true. By the principle of R.H.S. = (1 + 1)2 = 1
mathematical induction, P(n) is true for all 4
positive integers n. ∴ P(1) is true.
Chapter 3 Mathematical Induction 61
Assume P(k) is true for any positive integer k. 13. Let P(n) be the proposition f (n) = 32 n − 1 is
1 divisible by 8.
i.e. 13 + 2 3 + L + k 3 = k 2 ( k + 1)2
4 When n = 1, f (1) = 32 − 1 = 8 which is divisible
Then 13 + 2 3 + L + k 3 + ( k + 1)3 by 8.
1 ∴ P(1) is true.
= k 2 ( k + 1)2 + ( k + 1)3
4
1 Assume P(k) is true for any positive integer k.
= ( k + 1)2 ( k 2 + 4 k + 4)
4 i.e. f ( k ) = 32 k − 1 = 8 N where N is an integer.
1
= ( k + 1)2 ( k + 2)2 Then f ( k + 1) = 32( k +1) − 1
4
1
= ( k + 1)2 [( k + 1) + 1]2 = 32 k ⋅ 9 − 1
4 = (8 N + 1)9 − 1
Thus assuming P(k) is true for any positive = 8N ⋅ 9 + 8
integer k, P( k + 1) is also true. By the = 8(9 N + 1)
principle of mathematical induction, P(n) is which is divisible by 8.
true for all positive integers n. Thus assuming P(k) is true for any positive integer
(b) (13 − 1) + (2 3 − 2) + L + (n 3 − n) k, P( k + 1) is also true. By the principle of
mathematical induction, P(n) is true for all
= 13 + 2 3 + L + n 3 − (1 + 2 + L + n) positive integers n.
1 1
= n 2 (n + 1)2 − n(n + 1)
4 2 14 − 16. No solutions are provided for the H.K.C.E.E.
1
= n(n + 1)[n(n + 1) − 2] questions because of the copyright reasons.
4
1
= n(n + 1)(n 2 + n − 2)
4 båêáÅÜãÉåí=P=EéKTUF
1 1. (a) Tn = Tn −1 + 1 for n ≥ 3
= (n − 1)n(n + 1)(n + 2)
4
Let P(n) be the proposition Tn = n .
11. No solution is provided for the H.K.C.E.E. When n = 2 ,
question because of the copyright reasons.
L.H.S. = T2 = 2
12. Let P(n) be the proposition a − b is divisible
n n R.H.S. = 2
by a − b . ∴ P(2) is true.
When n = 1, a1 − b1 = a − b which is divisible by Assume P(k) is true for any positive integer
a − b. k ≥ 2.
∴ P(1) is true. i.e. Tk = k
Assume P(k) is true for any positive integer k. Then Tk +1 = Tk + 1 = k + 1
i.e. a − b = ( a − b) N where N is a polynomial.
k k
Thus assuming P(k) is true for any positive
k +1 k +1 integer k ≥ 2 , P( k + 1) is also true. By the
Then a −b
principle of mathematical induction, P(n) is
= a ⋅ ak − b ⋅ bk true for all positive integers n ≥ 2 .
= a[( a − b) N + b k ] − b ⋅ b k
(b) Q n = Q n −1+ (n − 1) for n ≥ 3
= a( a − b ) N + a ⋅ b k − b ⋅ b k
= a( a − b ) N + ( a − b ) b k n(n − 1)
Let P(n) be the proposition Q n = .
= ( a − b)( aN + b ) k 2
When n = 2 ,
which is divisible by ( a − b) .
L.H.S. = Q 2 = 1
Thus assuming P(k) is true for any positive integer
k, P( k + 1) is also true. By the principle of 2(2 − 1)
R.H.S. = =1
mathematical induction, P(n) is true for all 2
positive integers n. ∴ P(2) is true.
62 Chapter 3 Mathematical Induction
Assume P(k) is true for any positive integer Assume P(k) and P( k + 1) are true for any
k ≥ 2. positive integer k ≥ 2 .
k ( k − 1)
i.e. Q k = i.e. ak =
1
(α k − β k )
2 5
Then Qk +1 = Qk + [( k + 1 − 1)] 1
a k +1 = ( α k +1 − β k +1 )
k ( k − 1) 5
= +k
2 Then ak + 2
k ( k − 1) + 2 k = a k +1 + a k
=
2 1 1
k ( k − 1 + 2) = ( α k +1 − β k +1 ) + (α k − β k )
= 5 5
2 1
k ( k + 1) = ( α k +1 − β k +1 + α k − β k )
= 5
2 1
= [α k (α + 1) − β k (β + 1)]
Thus assuming P(k) is true for any positive 5
integer k ≥ 2 , P( k + 1) is also true. By the 1 1+ 5 1− 5
principle of mathematical induction, P(n) is = [α k ( + 1) − β k ( + 1)]
5 2 2
true for all positive integers n ≥ 2 .
1 3+ 5 3− 5
= [α k ( ) − βk ( )]
5 2 2
2. (a) x 2 − x − 1 = 0 1 1+ 5 2 1− 5 2
= [α k ( ) − βk ( ) ]
−( −1) ± ( −1)2 − 4(1)( −1) 1 ± 5 5 2 2
x= = 1
2(1) 2 = (α k ⋅ α 2 − β k ⋅ α 2 )
Q α>β 5
1
1+ 5 1− 5 = (α k + 2 − β k + 2 )
∴ α= , β= 5
2 2
Thus assuming P(k) and P( k + 1) are true for
(b) Let P(n) be the proposition any positive integer k, P( k + 2) is also true.
1 By the principle of mathematical induction,
an = (α n − β n ) . P(n) is true for all positive integers n.
5
When n = 1 , 3. No solution is provided for the H.K.C.E.E.
L.H.S. = a1 = 1 question because of the copyright reasons.
1
R.H.S. = (α − β )
5 `ä~ëëïçêâ=N=EéKSUF
1 1+ 5 1− 5
= [( )−( )] Let P(n) be the proposition
5 2 2
1 + 3 + 5 + L + (2 n − 1) = n 2 .
=1
∴ P(1) is true. When n = 1,
When n = 2 , L.H.S. = 1
L.H.S. = a2 = 1 R.H.S. = 12 = 1
R.H.S. ∴ P(1) is true.
Assume P(k) is true for any positive integer k.
1
= (α 2 − β 2 ) i.e. 1 + 3 + 5 + L + (2 k − 1) = k 2
5
1 1+ 5 2 1− 5 2 Then 1 + 3 + 5 + L + (2 k − 1) + [2( k + 1) − 1]
= [( ) −( ) ]
5 2 2 = k 2 + (2 k + 1)
1 1+ 5 1− 5 1+ 5 1− 5 = ( k + 1)2
= ( + )( − )]
5 2 2 2 2 Thus assuming P(k) is true for any positive integer
=1 k, P( k + 1) is also true. By the principle of
∴ P(2) is true. mathematical induction, P(n) is true for all
positive integers n.
Chapter 3 Mathematical Induction 63
`ä~ëëïçêâ=O=EéKTNF Hence P( k + 1) is true if P(k) is true for any
positive integer k. By the principle of
(a) Let P(n) be the proposition mathematical induction, P(n) is true for all
13 + 33 + 53 + L + (2 n − 1)3 = n 2 (2 n 2 − 1) . positive integers n.
When n = 1, L.H.S. = 13 = 1
R.H.S. = 12 [2(1)2 − 1] = 1 2. Let P(n) be the proposition a 2 n −1 + b 2 n −1 is
∴ P(1) is true. divisible by a + b .
Assume P(k) is true for any positive integer k. When n = 1, a1 + b1 = a + b which is divisible by
a + b.
i.e. 13 + 33 + 53 + L + (2 k − 1)3 = k 2 (2 k 2 − 1) ∴ P(1) is true.
Then 13 + 33 + 53 + L + (2 k − 1)3 + [2( k + 1) − 1]3 Assume P(k) is true for any positive integer k.
= k 2 (2 k 2 − 1) + (2 k + 1)3 i.e. a 2 k −1 + b 2 k −1 = ( a + b) N , w h e r e N i s a
= 2 k − k + 8k + 12 k + 6k + 1
4 2 3 2 polynomial.
= 2 k 4 + 8k 3 + 11k 2 + 6k + 1 Then a 2( k +1) −1 + b 2( k +1) −1
= ( k + 1) (2 k + 4 k + 1)
2 2
= a 2 k +1 + b 2 k +1
= ( k + 1) [(2 k + 4 k + 2) − 1]
2 2
= a 2 k −1+ 2 + b 2 k −1+ 2
= ( k + 1) [2( k + 2 k + 1) − 1]
2 2
= a 2 k −1 ⋅ a 2 + b 2 k −1 ⋅ b 2
= ( k + 1) [2( k + 1) − 1]
2 2
= [( a + b) N − b 2 k −1 ]a 2 + b 2 k −1 ⋅ b 2
Thus assuming P(k) is true for any positive integer = a 2 ( a + b) N − a 2 b 2 k −1 + b 2 b 2 k −1
k, P( k + 1) is also true. By the principle of
mathematical induction, P(n) is true for all = a 2 ( a + b) N + b 2 k −1 (b 2 − a 2 )
positive integers n. = a 2 ( a + b) N + b 2 k −1 (b − a)(b + a)
(b) 2 3 + 6 3 + 10 3 + L + 383 = ( a + b)[a 2 N + (b − a)b 2 k −1 ]
= (2 ⋅ 1)3 + (2 ⋅ 3)3 + (2 ⋅ 5)3 + L + (2 ⋅ 19)3 which is divisible by a + b .
= 2 3 ⋅ 13 + 2 3 ⋅ 33 + 2 3 ⋅ 53 + L + 2 3 ⋅ 193 Hence P( k + 1) is true if P(k) is true for any
positive integer k. By the principle of
= 8(13 + 33 + 53 + L + 193 ) mathematical induction, P(n) is true for all
= 8[10 2 (2 ⋅ 10 2 − 1)] positive integers n.
= 8[100(200 − 1)]
= 159 200
`ä~ëëïçêâ=P=EéKTQF
1. Let P(n) be the proposition 7 n + 3n + 8 is
divisible by 9.
When n = 1, 71 + 3(1) + 8 = 18 = 9 ⋅ 2 which is
divisible by 9.
∴ P(1) is true.
Assume P(k) is true for any positive integer k.
i.e. 7 k + 3k + 8 = 9 N , where N is an integer.
Then 7 k +1 + 3( k + 1) + 8
= 7 k ⋅ 7 + 3k + 3 + 8
= (9 N − 3k − 8)7 + 3k + 3 + 8
= 63 N − 21k − 56 + 3k + 11
= 63 N − 18k − 45
= 9(7 N − 2 k − 5)
which is divisible by 9. | 17,078 | 32,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-39 | latest | en | 0.581033 |
https://sciencedocbox.com/Physics/72802389-Be-any-ring-homomorphism-and-let-s-s-be-any-element-of-s-then-there-is-a-unique-ring-homomorphism.html | 1,643,368,226,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00434.warc.gz | 559,688,290 | 27,091 | # be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism
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1 21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore to determine the prime elements, it suffices to determine the irreducible elements. We start with some basic facts about polynomial rings. Lemma Let R be an integral domain. Then the units in R[x] are precisely the units in R. Proof. One direction is clear. A unit in R is a unit in R[x]. Now suppose that f(x) is a unit in R[x]. Given a polynomial g, denote by d(g) the degree of g(x) (note that we are not claiming that R[x] is a Euclidean domain). Now f(x)g(x) = 1. Thus 0 = d(1) = d(fg) d(f) + d(g). Thus both of f and g must have degree zero. It follows that f(x) = f 0 and that f 0 is a unit in R[x]. Lemma Let R be a ring. The natural inclusion R R[x] which just sends an element r R to the constant polynomial r, is a ring homomorphism. Proof. Easy. The following universal property of polynomial rings, is very useful. Lemma Let φ: R S be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism ψ : R[x] S, such that φ(x) = s and which makes the following diagram commute f R R[x] ψ 1 φ S
2 Proof. Note that any ring homomorphism ψ : R[x] S that sends x to s and acts as φ on the coefficients, must send a n x n + a n 1 x n a 0 to φ(a n )s n + φ(a n 1 )s n φ(a 0 ). Thus it suffices to check that the given map is a ring homomorphism, which is left as an exercise to the reader. Definition Let R be a ring and let α be an element of R. The natural ring homomorphism φ: R[x] R, which acts as the identity on R and which sends x to α, is called evaluation at α and is often denoted ev α. We say that α is a zero (aka root) of f(x), if f(x) is in the kernel of ev α. Lemma Let K be a field and let α be an element of K. Then the kernel of ev α is the ideal x α. Proof. Denote by I the kernel of ev α. Clearly x α is in I. On the other hand, K[x] is a Euclidean domain, and so it is certainly a PID. Thus I is principal. Suppose it is generated by f, so that I = f. Then f divides x α. If f has degree one, then x α must be an associate of f and the result follows. If f has degree zero, then it must be a constant. As f has a root at α, in fact this constant must be zero, a contradiction. Lemma Let K be a field and let f(x) be a polynomial in K[x]. Then we can write f(x) = g(x)h(x) where g(x) is a linear polynomial if and only if f(x) has a root in K. Proof. First note that a linear polynomial always has a root in K. Indeed any linear polynomial is of the form ax + b, where a 0. Then it is easy to see that α = b is a root of ax + b. a On the other hand, the kernel of the evaluation map is an ideal, so that if g(x) has a root α, then in fact so does f(x) = g(x)h(x). Thus if we can write f(x) = g(x)h(x), where g(x) is linear, then it follows that f(x) must have a root. Now suppose that f(x) has a root at α. Consider the linear polynomial g(x) = x α. Then the kernel of ev α is equal to x α. As f is in the kernel, f(x) = g(x)h(x), for some h(x) R[x]. 2
3 Lemma Let K be a field and let f(x) be a polynomial of degree two or three. Then f(x) is irreducible if and only if it has no roots in K. Proof. If f(x) has a root in K, then f(x) = g(x)h(x), where g(x) has degree one, by (21.6). As the degree of f is at least two, it follows that h(x) has degree at least one. Thus f(x) is not irreducible. Now suppose that f(x) is not irreducible. Then f(x) = g(x)h(x), where neither g nor h is a unit. Thus both g and h have degree at least one. As the sum of the degrees of g and h is at most three, the degree of f, it follows that one of g and h has degree one. Now apply (21.6). Definition Let p be a prime. F p denotes the unique field with p elements. Of course, F p is isomorphic to Z p. However, as we will see later, it is useful to replace Z by F. Example First consider the polynomial x Over the real numbers this is irreducible. Indeed, if we replace x by any real number a, then a 2 is non-negative and so a cannot equal zero. On the other hand ±i is a root of x 2 +1, as i 2 +1 = 0. Thus x 2 +1 is reducible over the complex numbers. Indeed x 2 +1 = (x+i)(x i). Thus an irreducible polynomial might well become reducible over a larger field. Consider the polynomial x 2 + x + 1. We consider this over various fields. As observed in (21.7) this is reducible iff it has a root in the given field. Suppose we work over the field F 5. We need to check if the five elements of F 5 are roots or not. We have = = = 1 Thus x 2 +x+1 is irreducible over F 5. Now consider what happens over the field with three elements F 3. Then 1 is a root of this polynomial. As neither 0 nor 2 are roots, we must have x 2 + x + 1 = (x 1) 2 = (x + 2) 2, which is easy to check. Now let us determine all irreducible polynomials of degree at most four over F 2. Any linear polynomial is irreducible. There are two such x and x + 1. A general quadratic has the form f(x) = x 2 + ax + b. b 0, else x divides f(x). Thus b = 1. If a = 0, then f(x) = x 2 + 1, which has 1 as a zero. Thus f(x) = x 2 + x + 1 is the only irreducible quadratic. 3
4 Now suppose that we have an irreducible cubic f(x) = x 3 +ax+bx+1. This is irreducible iff f(1) 0, which is the same as to say that there are an odd number of terms. Thus the irreducible cubics are f(x) = x 3 + x and x 3 + x + 1. Finally suppose that f(x) is a quartic polynomial. The general irreducible is of the form x 4 + ax 3 + bx 2 + cx + 1. f(1) 0 is the same as to say that either two of a, b and c is equal to zero or they are all equal to one. Suppose that f(x) = g(x)h(x). If f(x) does not have a root, then both g and h must have degree two. If either g or h were reducible, then again f would have a linear factor, and therefore a root. Thus the only possibilty is that both g and h are the unique irreducible quadratic polynomials. In this case f(x) = (x 2 + x + 1) 2 = x 4 + x Thus x 4 + x 3 + x 2 + x + 1, x 4 + x 3 + 1, and x 4 + x + 1 are the three irreducible quartics. Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is a follows. Suppose we are given a polynomial with integer coefficients. Then it is natural to also consider this polynomial over the rationals. Note that it is much easier to prove that this polynomial is irreducible over the integers than it is to prove that it is irreducible over the rationals. For example it is clear that x 2 2 is irreducible over the integers. In fact it is irreducible over the rationals as well, that is, 2 is not a rational number. First some definitions. Definition Let R be a commutative ring and let a 1, a 2,..., a k be a sequence of elements of R. The gcd of a 1, a 2,..., a k is an element d R such that (1) d a i, for all 1 i k. (2) If d a i, for all 1 i k, then d d. Lemma Let R be a UFD. Then the gcd of any sequence a 1, a 2,..., a k of non-zero elements of R exists. 4
5 Proof. There are two obvious ways to proceed. The first is to take a common factorisation of each a i into a product of powers of primes, as in the case k = 2. The second is to recursively construct the gcd, by setting d i to be the gcd of d i 1 and a i and taking d 1 = a 1. In this case d = d k will be a gcd for the whole sequence a 1, a 2,..., a k. Definition Let R be a UFD and let f(x) be a polynomial with coefficients in R. The content of f(x), denoted c(f), is the gcd of the coefficients of f. Example Let f(x) = 24x 3 60x Then the content of f is 4. Thus f(x) = 4(8x 3 15x + 10). Lemma Let R be a UFD. Then every element of R[x] has a factorisation of the form cf, where c R and the content of f is one. Proof. Obvious. Here is the key result. Proposition Let R be a UFD. Suppose that g and h R[x] and let f(x) = g(x)h(x). Then the content of f is equal to the content of g times the content of h. Proof. It is clear that the content of g divides the content of f. Therefore we may assume that the content of g and h is one, and we only have to prove that the same is true for f. Suppose not. As R is a UFD, it follows that there is a prime p that divides the content of f. We may write g(x) = a n x n +a n 1 x n 1 + +a 0 and h(x) = b n x n +b n 1 x n 1 + +b 0. As the content of g is one, at least one coefficient of g is not divisible by p. Let i be the first such, so that p divides a k, for k < i whilst p does not divide a i. Similarly pick j so that p divides b k, for k < j, whilst p does not divide b j. Consider the coefficient of x i+j in f. This is equal to a 0 b i+j + a 1 b i+j a i 1 b j+1 + a i b j + a i+1 b j a i+j b 0. Note that p divides every term of this sum, except the middle one a i b j. Thus p does not divide the coefficient of x i+j. But this contradicts the definition of the content. 5
6 Theorem (Gauss Lemma) Let R be a UFD and let f(x) R[x]. Let F be the field of fractions of R. Suppose that the content of f is one and that we may write f(x) = u 1 (x)v 1 (x), where u 1 (x) and v 1 (x) are in F [x]. Then we may find u(x) and v(x) in R[x], such that f(x) = u(x)v(x) where the degree of u is the same as the degree of u 1 and the degree of v is the same as the degree of v. In particular if f is irreducible in R[x] then it is irreducible in F [x]. Proof. We have f(x) = u 1 (x)v 1 (x). Now clear denominators. That is, multiply through by the product c of all the denominators in u 1 (x) and v 1 (x). In this way we get an expression of the form cf(x) = u 2 (x)v 2 (x), where now u 2 and v 2 belong to R[x]. Now write u 2 (x) = au(x) and v 2 (x) = bv(x). We get cf(x) = abu(x)v(x). By (21.15) we can c divides ab, ab = cα, where α R. Therefore, replacing u(x) by αu(x), we have f(x) = u(x)v(x). Corollary Let R be a UFD. Then R[x] is a UFD. Proof. It is clear that the Factorisation algorithm terminates, by induction on the degree. Therefore it suffices to prove that irreducible implies prime. Suppose that f(x) R[x] is irreducible. If f has degree zero, then it is an irreducible element of R and hence a prime element of R and there is nothing to prove. Otherwise we may assume that the content of f is one. By Gauss Lemma, f is not only irreducible in R[x] but also in F [x]. But then f is a prime element of F [x] as F [x] is a UFD. Now suppose that f divides gh, where g and h R[x]. As f is prime in F [x], f divides g or h in F [x]. Suppose it divides g. Then we may write g = fk, 6
7 some k F [x]. As in the proof of Gauss Lemma, this means we may write g = fk, some k R[x]. But then f(x) divides g in R[x]. Corollary Z[x] is a UFD. Definition Let R be a commutative ring and let x 1, x 2,..., x n be indeterminates. A monomial in x 1, x 2,..., x n is a product of powers of x 1, x 2,..., x n. If I = (d 1, d 2,..., d n ), then let x I = x d i i. The degree d of a monomial is the sum of the degrees of the individual terms, d i. The polynomial ring R[x 1, x 2,..., x n ] is equal to the set of all finite formal sums a I x I, I with the obvious addition and multiplication. The degree of a polynomial is the maximum degree of a monomial term that appears with non-zero coefficient. Example Let x and y be indeterminates. A typical element of Q[x, y] might be x 2 + y 2 1. This has degree 2. Note that xy also has degree two. A more complicated example might be 2 3 x3 7xy + y 5, a polynomial of degree 5. Lemma Let R be a commutative ring and let x 1, x 2,..., x n be indeterminates. Let S = R[x 1, x 2,..., x n 1 ]. Then there is a natural isomorphism R[x 1, x 2,..., x n ] S[x n ]. Proof. Clear. To illustrate how this proceeds, it will probably help to give an example. Consider the polynomial 2 3 x3 7xy + y 5. 7
8 Consider this as a polynomial in y, whose coefficients lie in the ring R[x]. That is y 5 + ( 7x)y + 2/3x 3 R[x][y]. Corollary Let R be a UFD. Then R[x 1, x 2,..., x n ] is a UFD. Proof. By induction on n. The case n = 1 is (21.17). Set S = R[x 1, x 2,..., x n 1 ]. By induction S is a UFD. But then S[x] R[x 1, x 2,..., x n ] is a UFD. Now we give a way to prove that polynomials with integer coefficients are irreducible. Lemma Let φ: R S be a ring homomorphism. Then there is a unique ring homomorphism ψ : R[x] S[x] which makes the following diagram commute R φ S and which sends x to x. Proof. Let R[x] ψ S[x] f : R S[x] be the composition of φ with the natural inclusion of S into S[x]. By the universal property of R[x], there is a unique ring homomorphism The rest is clear. ψ : R[x] S[x]. Theorem (Eisenstein s Criteria) Let f(x) = a n x n + a n 1 x n a 0 be a polynomial with integer coefficients. Suppose that there is a prime p such that p divides a i, i n 1, p does not divide a n and p 2 does not divide a 0. Then f(x) is irreducible in Q[x]. 8
9 Proof. There is no harm in assuming that the content of f is one, so that by Gauss Lemma, it suffices to prove that f is irreducible over Z. Suppose not. Then we may find two polynomials g(x) and h(x), of positive degree, with integral coefficients, such that Suppose that f(x) = g(x)h(x). f(x) = a n x n + a n 1 x n a 0 g(x) = b d x d + b d 1 x d b 0 h(x) = c e x e + c e 1 x e c 0, for some n, d and e > 1. As a n = b d c e and a n is not divisible by p, then neither is b d nor c e. Consider the natural ring homomorphism Z F p. This induces a ring homomorphism Z[x] F p [x]. It is convenient to denote the image of a polynomial g(x) as ḡ(x). As we have a ring homomorphism, f(x) = ḡ(x) h(x). Since the leading coefficient of f is not divisible by p, f(x) has the same degree as f(x), and the same holds for g(x) and h(x). On the other hand, every other coefficient of f(x) is divisible by p, and so f(x) = ā n x n. Since F p is a field, F p is a UFD and so ḡ = b d x d and ḡ(x) = c e x e. It follows that every other coefficient of g(x) and h(x) is divisible by p. In particular b 0 and c 0 are both divisible by p, and so, as a 0 = b 0 c 0, a 0 must be divisible by p 2, a contradiction. Example Let f(x) = 2x 7 15x x 5 18x 4 9x x 2 3x + 6. Then f(x) is irreducible over Q. We apply Eisenstein with p = 3. Then the top coefficient is not divisible by 3, the others are, and the smallest coefficient is not divisible by 9 = 3 2. Lemma Let p be a prime. Then is irreducible over Q. f(x) = x p 1 + x p x + 1, 9
10 Proof. By Gauss Lemma, it suffices to prove that f(x) is irreducible over Z. First note that f(x) = xp 1 x 1, as can be easily checked. Consider the change of variable As this induces an automorphism y = x 1. Z[x] Z[x] by sending x to x 1, this will not alter whether or not f is irreducible. In this case f(y) = (y + 1)p 1 y = y p 1 + ( ) p y p = y p 1 + py p p. ( ) ( ) p p y p p 1 Note that ( p i) is divisible by p, for all 1 i < p, so that we can apply Eisenstein to f(y), using the prime p. 10
### where c R and the content of f is one. 1
9. Gauss Lemma Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is as follows.
### 15. Polynomial rings Definition-Lemma Let R be a ring and let x be an indeterminate.
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GALOIS THEORY MARIUS VAN DER PUT & JAAP TOP Contents 1. Basic definitions 1 1.1. Exercises 2 2. Solving polynomial equations 2 2.1. Exercises 4 3. Galois extensions and examples 4 3.1. Exercises. 6 4. | 11,971 | 40,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-05 | longest | en | 0.917884 |
https://us.metamath.org/mpeuni/ply1coe.html | 1,720,910,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00469.warc.gz | 522,713,808 | 17,895 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > ply1coe Structured version Visualization version GIF version
Theorem ply1coe 20928
Description: Decompose a univariate polynomial as a sum of powers. (Contributed by Stefan O'Rear, 21-Mar-2015.) (Revised by AV, 7-Oct-2019.)
Hypotheses
Ref Expression
ply1coe.p 𝑃 = (Poly1𝑅)
ply1coe.x 𝑋 = (var1𝑅)
ply1coe.b 𝐵 = (Base‘𝑃)
ply1coe.n · = ( ·𝑠𝑃)
ply1coe.m 𝑀 = (mulGrp‘𝑃)
ply1coe.e = (.g𝑀)
ply1coe.a 𝐴 = (coe1𝐾)
Assertion
Ref Expression
ply1coe ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐾 = (𝑃 Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))))
Distinct variable groups: 𝐴,𝑘 𝐵,𝑘 𝑘,𝐾 𝑘,𝑋 ,𝑘 𝑅,𝑘 · ,𝑘 𝑃,𝑘
Allowed substitution hint: 𝑀(𝑘)
Proof of Theorem ply1coe
Dummy variables 𝑎 𝑏 𝑐 𝑥 𝑑 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 eqid 2801 . . 3 (1o mPoly 𝑅) = (1o mPoly 𝑅)
2 psr1baslem 20817 . . 3 (ℕ0m 1o) = {𝑑 ∈ (ℕ0m 1o) ∣ (𝑑 “ ℕ) ∈ Fin}
3 eqid 2801 . . 3 (0g𝑅) = (0g𝑅)
4 eqid 2801 . . 3 (1r𝑅) = (1r𝑅)
5 1onn 8252 . . . 4 1o ∈ ω
65a1i 11 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 1o ∈ ω)
7 ply1coe.p . . . 4 𝑃 = (Poly1𝑅)
8 eqid 2801 . . . 4 (PwSer1𝑅) = (PwSer1𝑅)
9 ply1coe.b . . . 4 𝐵 = (Base‘𝑃)
107, 8, 9ply1bas 20827 . . 3 𝐵 = (Base‘(1o mPoly 𝑅))
11 ply1coe.n . . . 4 · = ( ·𝑠𝑃)
127, 1, 11ply1vsca 20858 . . 3 · = ( ·𝑠 ‘(1o mPoly 𝑅))
13 simpl 486 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝑅 ∈ Ring)
14 simpr 488 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐾𝐵)
151, 2, 3, 4, 6, 10, 12, 13, 14mplcoe1 20708 . 2 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐾 = ((1o mPoly 𝑅) Σg (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐾𝑎) · (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅)))))))
16 ply1coe.a . . . . . . 7 𝐴 = (coe1𝐾)
1716fvcoe1 20839 . . . . . 6 ((𝐾𝐵𝑎 ∈ (ℕ0m 1o)) → (𝐾𝑎) = (𝐴‘(𝑎‘∅)))
1817adantll 713 . . . . 5 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (𝐾𝑎) = (𝐴‘(𝑎‘∅)))
195a1i 11 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → 1o ∈ ω)
20 eqid 2801 . . . . . . 7 (mulGrp‘(1o mPoly 𝑅)) = (mulGrp‘(1o mPoly 𝑅))
21 eqid 2801 . . . . . . 7 (.g‘(mulGrp‘(1o mPoly 𝑅))) = (.g‘(mulGrp‘(1o mPoly 𝑅)))
22 eqid 2801 . . . . . . 7 (1o mVar 𝑅) = (1o mVar 𝑅)
23 simpll 766 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → 𝑅 ∈ Ring)
24 simpr 488 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → 𝑎 ∈ (ℕ0m 1o))
25 eqidd 2802 . . . . . . . . . 10 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)))
26 0ex 5178 . . . . . . . . . . 11 ∅ ∈ V
27 fveq2 6649 . . . . . . . . . . . . 13 (𝑏 = ∅ → ((1o mVar 𝑅)‘𝑏) = ((1o mVar 𝑅)‘∅))
2827oveq1d 7154 . . . . . . . . . . . 12 (𝑏 = ∅ → (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)))
2927oveq2d 7155 . . . . . . . . . . . 12 (𝑏 = ∅ → (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)))
3028, 29eqeq12d 2817 . . . . . . . . . . 11 (𝑏 = ∅ → ((((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅))))
3126, 30ralsn 4582 . . . . . . . . . 10 (∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)))
3225, 31sylibr 237 . . . . . . . . 9 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
33 fveq2 6649 . . . . . . . . . . . . 13 (𝑥 = ∅ → ((1o mVar 𝑅)‘𝑥) = ((1o mVar 𝑅)‘∅))
3433oveq2d 7155 . . . . . . . . . . . 12 (𝑥 = ∅ → (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)))
3533oveq1d 7154 . . . . . . . . . . . 12 (𝑥 = ∅ → (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
3634, 35eqeq12d 2817 . . . . . . . . . . 11 (𝑥 = ∅ → ((((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏))))
3736ralbidv 3165 . . . . . . . . . 10 (𝑥 = ∅ → (∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ ∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏))))
3826, 37ralsn 4582 . . . . . . . . 9 (∀𝑥 ∈ {∅}∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ ∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘∅)) = (((1o mVar 𝑅)‘∅)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
3932, 38sylibr 237 . . . . . . . 8 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ∀𝑥 ∈ {∅}∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
40 df1o2 8103 . . . . . . . . 9 1o = {∅}
4140raleqi 3365 . . . . . . . . 9 (∀𝑏 ∈ 1o (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ ∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
4240, 41raleqbii 3200 . . . . . . . 8 (∀𝑥 ∈ 1o𝑏 ∈ 1o (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)) ↔ ∀𝑥 ∈ {∅}∀𝑏 ∈ {∅} (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
4339, 42sylibr 237 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ∀𝑥 ∈ 1o𝑏 ∈ 1o (((1o mVar 𝑅)‘𝑏)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑥)) = (((1o mVar 𝑅)‘𝑥)(+g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑏)))
441, 2, 3, 4, 19, 20, 21, 22, 23, 24, 43mplcoe5 20711 . . . . . 6 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅))) = ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ 1o ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))))
4540mpteq1i 5123 . . . . . . . 8 (𝑐 ∈ 1o ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐))) = (𝑐 ∈ {∅} ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))
4645oveq2i 7150 . . . . . . 7 ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ 1o ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))) = ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ {∅} ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐))))
471mplring 20694 . . . . . . . . . . 11 ((1o ∈ ω ∧ 𝑅 ∈ Ring) → (1o mPoly 𝑅) ∈ Ring)
485, 47mpan 689 . . . . . . . . . 10 (𝑅 ∈ Ring → (1o mPoly 𝑅) ∈ Ring)
4920ringmgp 19299 . . . . . . . . . 10 ((1o mPoly 𝑅) ∈ Ring → (mulGrp‘(1o mPoly 𝑅)) ∈ Mnd)
5048, 49syl 17 . . . . . . . . 9 (𝑅 ∈ Ring → (mulGrp‘(1o mPoly 𝑅)) ∈ Mnd)
5150ad2antrr 725 . . . . . . . 8 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (mulGrp‘(1o mPoly 𝑅)) ∈ Mnd)
5226a1i 11 . . . . . . . 8 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ∅ ∈ V)
53 ply1coe.e . . . . . . . . . . . 12 = (.g𝑀)
5420, 10mgpbas 19241 . . . . . . . . . . . . 13 𝐵 = (Base‘(mulGrp‘(1o mPoly 𝑅)))
5554a1i 11 . . . . . . . . . . . 12 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐵 = (Base‘(mulGrp‘(1o mPoly 𝑅))))
56 ply1coe.m . . . . . . . . . . . . . 14 𝑀 = (mulGrp‘𝑃)
5756, 9mgpbas 19241 . . . . . . . . . . . . 13 𝐵 = (Base‘𝑀)
5857a1i 11 . . . . . . . . . . . 12 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐵 = (Base‘𝑀))
59 ssv 3942 . . . . . . . . . . . . 13 𝐵 ⊆ V
6059a1i 11 . . . . . . . . . . . 12 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐵 ⊆ V)
61 ovexd 7174 . . . . . . . . . . . 12 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ (𝑎 ∈ V ∧ 𝑏 ∈ V)) → (𝑎(+g‘(mulGrp‘(1o mPoly 𝑅)))𝑏) ∈ V)
62 eqid 2801 . . . . . . . . . . . . . . . . 17 (.r𝑃) = (.r𝑃)
637, 1, 62ply1mulr 20859 . . . . . . . . . . . . . . . 16 (.r𝑃) = (.r‘(1o mPoly 𝑅))
6420, 63mgpplusg 19239 . . . . . . . . . . . . . . 15 (.r𝑃) = (+g‘(mulGrp‘(1o mPoly 𝑅)))
6556, 62mgpplusg 19239 . . . . . . . . . . . . . . 15 (.r𝑃) = (+g𝑀)
6664, 65eqtr3i 2826 . . . . . . . . . . . . . 14 (+g‘(mulGrp‘(1o mPoly 𝑅))) = (+g𝑀)
6766oveqi 7152 . . . . . . . . . . . . 13 (𝑎(+g‘(mulGrp‘(1o mPoly 𝑅)))𝑏) = (𝑎(+g𝑀)𝑏)
6867a1i 11 . . . . . . . . . . . 12 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ (𝑎 ∈ V ∧ 𝑏 ∈ V)) → (𝑎(+g‘(mulGrp‘(1o mPoly 𝑅)))𝑏) = (𝑎(+g𝑀)𝑏))
6921, 53, 55, 58, 60, 61, 68mulgpropd 18264 . . . . . . . . . . 11 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (.g‘(mulGrp‘(1o mPoly 𝑅))) = )
7069oveqd 7156 . . . . . . . . . 10 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋) = ((𝑎‘∅) 𝑋))
7170adantr 484 . . . . . . . . 9 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋) = ((𝑎‘∅) 𝑋))
727ply1ring 20880 . . . . . . . . . . . 12 (𝑅 ∈ Ring → 𝑃 ∈ Ring)
7356ringmgp 19299 . . . . . . . . . . . 12 (𝑃 ∈ Ring → 𝑀 ∈ Mnd)
7472, 73syl 17 . . . . . . . . . . 11 (𝑅 ∈ Ring → 𝑀 ∈ Mnd)
7574ad2antrr 725 . . . . . . . . . 10 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → 𝑀 ∈ Mnd)
76 elmapi 8415 . . . . . . . . . . . 12 (𝑎 ∈ (ℕ0m 1o) → 𝑎:1o⟶ℕ0)
77 0lt1o 8116 . . . . . . . . . . . 12 ∅ ∈ 1o
78 ffvelrn 6830 . . . . . . . . . . . 12 ((𝑎:1o⟶ℕ0 ∧ ∅ ∈ 1o) → (𝑎‘∅) ∈ ℕ0)
7976, 77, 78sylancl 589 . . . . . . . . . . 11 (𝑎 ∈ (ℕ0m 1o) → (𝑎‘∅) ∈ ℕ0)
8079adantl 485 . . . . . . . . . 10 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (𝑎‘∅) ∈ ℕ0)
81 ply1coe.x . . . . . . . . . . . 12 𝑋 = (var1𝑅)
8281, 7, 9vr1cl 20849 . . . . . . . . . . 11 (𝑅 ∈ Ring → 𝑋𝐵)
8382ad2antrr 725 . . . . . . . . . 10 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → 𝑋𝐵)
8457, 53mulgnn0cl 18239 . . . . . . . . . 10 ((𝑀 ∈ Mnd ∧ (𝑎‘∅) ∈ ℕ0𝑋𝐵) → ((𝑎‘∅) 𝑋) ∈ 𝐵)
8575, 80, 83, 84syl3anc 1368 . . . . . . . . 9 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((𝑎‘∅) 𝑋) ∈ 𝐵)
8671, 85eqeltrd 2893 . . . . . . . 8 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋) ∈ 𝐵)
87 fveq2 6649 . . . . . . . . . 10 (𝑐 = ∅ → (𝑎𝑐) = (𝑎‘∅))
88 fveq2 6649 . . . . . . . . . . 11 (𝑐 = ∅ → ((1o mVar 𝑅)‘𝑐) = ((1o mVar 𝑅)‘∅))
8981vr1val 20824 . . . . . . . . . . 11 𝑋 = ((1o mVar 𝑅)‘∅)
9088, 89eqtr4di 2854 . . . . . . . . . 10 (𝑐 = ∅ → ((1o mVar 𝑅)‘𝑐) = 𝑋)
9187, 90oveq12d 7157 . . . . . . . . 9 (𝑐 = ∅ → ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)) = ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋))
9254, 91gsumsn 19070 . . . . . . . 8 (((mulGrp‘(1o mPoly 𝑅)) ∈ Mnd ∧ ∅ ∈ V ∧ ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋) ∈ 𝐵) → ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ {∅} ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))) = ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋))
9351, 52, 86, 92syl3anc 1368 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ {∅} ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))) = ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋))
9446, 93syl5eq 2848 . . . . . 6 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((mulGrp‘(1o mPoly 𝑅)) Σg (𝑐 ∈ 1o ↦ ((𝑎𝑐)(.g‘(mulGrp‘(1o mPoly 𝑅)))((1o mVar 𝑅)‘𝑐)))) = ((𝑎‘∅)(.g‘(mulGrp‘(1o mPoly 𝑅)))𝑋))
9544, 94, 713eqtrd 2840 . . . . 5 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅))) = ((𝑎‘∅) 𝑋))
9618, 95oveq12d 7157 . . . 4 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑎 ∈ (ℕ0m 1o)) → ((𝐾𝑎) · (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅)))) = ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋)))
9796mpteq2dva 5128 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐾𝑎) · (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅))))) = (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋))))
9897oveq2d 7155 . 2 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((1o mPoly 𝑅) Σg (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐾𝑎) · (𝑏 ∈ (ℕ0m 1o) ↦ if(𝑏 = 𝑎, (1r𝑅), (0g𝑅)))))) = ((1o mPoly 𝑅) Σg (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋)))))
99 nn0ex 11895 . . . . . 6 0 ∈ V
10099mptex 6967 . . . . 5 (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) ∈ V
101100a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) ∈ V)
1027fvexi 6663 . . . . 5 𝑃 ∈ V
103102a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝑃 ∈ V)
104 ovexd 7174 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (1o mPoly 𝑅) ∈ V)
1059, 10eqtr3i 2826 . . . . 5 (Base‘𝑃) = (Base‘(1o mPoly 𝑅))
106105a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (Base‘𝑃) = (Base‘(1o mPoly 𝑅)))
107 eqid 2801 . . . . . 6 (+g𝑃) = (+g𝑃)
1087, 1, 107ply1plusg 20857 . . . . 5 (+g𝑃) = (+g‘(1o mPoly 𝑅))
109108a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (+g𝑃) = (+g‘(1o mPoly 𝑅)))
110101, 103, 104, 106, 109gsumpropd 17883 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑃 Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))) = ((1o mPoly 𝑅) Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))))
111 eqid 2801 . . . . 5 (0g𝑃) = (0g𝑃)
1121, 7, 111ply1mpl0 20887 . . . 4 (0g𝑃) = (0g‘(1o mPoly 𝑅))
1131mpllmod 20693 . . . . . 6 ((1o ∈ ω ∧ 𝑅 ∈ Ring) → (1o mPoly 𝑅) ∈ LMod)
1145, 13, 113sylancr 590 . . . . 5 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (1o mPoly 𝑅) ∈ LMod)
115 lmodcmn 19678 . . . . 5 ((1o mPoly 𝑅) ∈ LMod → (1o mPoly 𝑅) ∈ CMnd)
116114, 115syl 17 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (1o mPoly 𝑅) ∈ CMnd)
11799a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ℕ0 ∈ V)
1187ply1lmod 20884 . . . . . . 7 (𝑅 ∈ Ring → 𝑃 ∈ LMod)
119118ad2antrr 725 . . . . . 6 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → 𝑃 ∈ LMod)
120 eqid 2801 . . . . . . . . . 10 (Base‘𝑅) = (Base‘𝑅)
12116, 9, 7, 120coe1f 20843 . . . . . . . . 9 (𝐾𝐵𝐴:ℕ0⟶(Base‘𝑅))
122121adantl 485 . . . . . . . 8 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐴:ℕ0⟶(Base‘𝑅))
123122ffvelrnda 6832 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → (𝐴𝑘) ∈ (Base‘𝑅))
1247ply1sca 20885 . . . . . . . . . 10 (𝑅 ∈ Ring → 𝑅 = (Scalar‘𝑃))
125124eqcomd 2807 . . . . . . . . 9 (𝑅 ∈ Ring → (Scalar‘𝑃) = 𝑅)
126125ad2antrr 725 . . . . . . . 8 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → (Scalar‘𝑃) = 𝑅)
127126fveq2d 6653 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → (Base‘(Scalar‘𝑃)) = (Base‘𝑅))
128123, 127eleqtrrd 2896 . . . . . 6 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → (𝐴𝑘) ∈ (Base‘(Scalar‘𝑃)))
12974ad2antrr 725 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → 𝑀 ∈ Mnd)
130 simpr 488 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → 𝑘 ∈ ℕ0)
13182ad2antrr 725 . . . . . . 7 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → 𝑋𝐵)
13257, 53mulgnn0cl 18239 . . . . . . 7 ((𝑀 ∈ Mnd ∧ 𝑘 ∈ ℕ0𝑋𝐵) → (𝑘 𝑋) ∈ 𝐵)
133129, 130, 131, 132syl3anc 1368 . . . . . 6 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → (𝑘 𝑋) ∈ 𝐵)
134 eqid 2801 . . . . . . 7 (Scalar‘𝑃) = (Scalar‘𝑃)
135 eqid 2801 . . . . . . 7 (Base‘(Scalar‘𝑃)) = (Base‘(Scalar‘𝑃))
1369, 134, 11, 135lmodvscl 19647 . . . . . 6 ((𝑃 ∈ LMod ∧ (𝐴𝑘) ∈ (Base‘(Scalar‘𝑃)) ∧ (𝑘 𝑋) ∈ 𝐵) → ((𝐴𝑘) · (𝑘 𝑋)) ∈ 𝐵)
137119, 128, 133, 136syl3anc 1368 . . . . 5 (((𝑅 ∈ Ring ∧ 𝐾𝐵) ∧ 𝑘 ∈ ℕ0) → ((𝐴𝑘) · (𝑘 𝑋)) ∈ 𝐵)
138137fmpttd 6860 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))):ℕ0𝐵)
1397, 81, 9, 11, 56, 53, 16ply1coefsupp 20927 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) finSupp (0g𝑃))
140 eqid 2801 . . . . . 6 (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)) = (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅))
14140, 99, 26, 140mapsnf1o2 8445 . . . . 5 (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)):(ℕ0m 1o)–1-1-onto→ℕ0
142141a1i 11 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)):(ℕ0m 1o)–1-1-onto→ℕ0)
14310, 112, 116, 117, 138, 139, 142gsumf1o 19032 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((1o mPoly 𝑅) Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))) = ((1o mPoly 𝑅) Σg ((𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) ∘ (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)))))
144 eqidd 2802 . . . . 5 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)) = (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)))
145 eqidd 2802 . . . . 5 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) = (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))))
146 fveq2 6649 . . . . . 6 (𝑘 = (𝑎‘∅) → (𝐴𝑘) = (𝐴‘(𝑎‘∅)))
147 oveq1 7146 . . . . . 6 (𝑘 = (𝑎‘∅) → (𝑘 𝑋) = ((𝑎‘∅) 𝑋))
148146, 147oveq12d 7157 . . . . 5 (𝑘 = (𝑎‘∅) → ((𝐴𝑘) · (𝑘 𝑋)) = ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋)))
14980, 144, 145, 148fmptco 6872 . . . 4 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) ∘ (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅))) = (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋))))
150149oveq2d 7155 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((1o mPoly 𝑅) Σg ((𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋))) ∘ (𝑎 ∈ (ℕ0m 1o) ↦ (𝑎‘∅)))) = ((1o mPoly 𝑅) Σg (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋)))))
151110, 143, 1503eqtrrd 2841 . 2 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → ((1o mPoly 𝑅) Σg (𝑎 ∈ (ℕ0m 1o) ↦ ((𝐴‘(𝑎‘∅)) · ((𝑎‘∅) 𝑋)))) = (𝑃 Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))))
15215, 98, 1513eqtrd 2840 1 ((𝑅 ∈ Ring ∧ 𝐾𝐵) → 𝐾 = (𝑃 Σg (𝑘 ∈ ℕ0 ↦ ((𝐴𝑘) · (𝑘 𝑋)))))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2112 ∀wral 3109 Vcvv 3444 ⊆ wss 3884 ∅c0 4246 ifcif 4428 {csn 4528 ↦ cmpt 5113 ∘ ccom 5527 ⟶wf 6324 –1-1-onto→wf1o 6327 ‘cfv 6328 (class class class)co 7139 ωcom 7564 1oc1o 8082 ↑m cmap 8393 ℕ0cn0 11889 Basecbs 16478 +gcplusg 16560 .rcmulr 16561 Scalarcsca 16563 ·𝑠 cvsca 16564 0gc0g 16708 Σg cgsu 16709 Mndcmnd 17906 .gcmg 18219 CMndccmn 18901 mulGrpcmgp 19235 1rcur 19247 Ringcrg 19293 LModclmod 19630 mVar cmvr 20593 mPoly cmpl 20594 PwSer1cps1 20807 var1cv1 20808 Poly1cpl1 20809 coe1cco1 20810 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 ax-rep 5157 ax-sep 5170 ax-nul 5177 ax-pow 5234 ax-pr 5298 ax-un 7445 ax-cnex 10586 ax-resscn 10587 ax-1cn 10588 ax-icn 10589 ax-addcl 10590 ax-addrcl 10591 ax-mulcl 10592 ax-mulrcl 10593 ax-mulcom 10594 ax-addass 10595 ax-mulass 10596 ax-distr 10597 ax-i2m1 10598 ax-1ne0 10599 ax-1rid 10600 ax-rnegex 10601 ax-rrecex 10602 ax-cnre 10603 ax-pre-lttri 10604 ax-pre-lttrn 10605 ax-pre-ltadd 10606 ax-pre-mulgt0 10607 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2601 df-eu 2632 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ne 2991 df-nel 3095 df-ral 3114 df-rex 3115 df-reu 3116 df-rmo 3117 df-rab 3118 df-v 3446 df-sbc 3724 df-csb 3832 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-pss 3903 df-nul 4247 df-if 4429 df-pw 4502 df-sn 4529 df-pr 4531 df-tp 4533 df-op 4535 df-uni 4804 df-int 4842 df-iun 4886 df-iin 4887 df-br 5034 df-opab 5096 df-mpt 5114 df-tr 5140 df-id 5428 df-eprel 5433 df-po 5442 df-so 5443 df-fr 5482 df-se 5483 df-we 5484 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-pred 6120 df-ord 6166 df-on 6167 df-lim 6168 df-suc 6169 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-f1 6333 df-fo 6334 df-f1o 6335 df-fv 6336 df-isom 6337 df-riota 7097 df-ov 7142 df-oprab 7143 df-mpo 7144 df-of 7393 df-ofr 7394 df-om 7565 df-1st 7675 df-2nd 7676 df-supp 7818 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-1o 8089 df-2o 8090 df-oadd 8093 df-er 8276 df-map 8395 df-pm 8396 df-ixp 8449 df-en 8497 df-dom 8498 df-sdom 8499 df-fin 8500 df-fsupp 8822 df-oi 8962 df-card 9356 df-pnf 10670 df-mnf 10671 df-xr 10672 df-ltxr 10673 df-le 10674 df-sub 10865 df-neg 10866 df-nn 11630 df-2 11692 df-3 11693 df-4 11694 df-5 11695 df-6 11696 df-7 11697 df-8 11698 df-9 11699 df-n0 11890 df-z 11974 df-dec 12091 df-uz 12236 df-fz 12890 df-fzo 13033 df-seq 13369 df-hash 13691 df-struct 16480 df-ndx 16481 df-slot 16482 df-base 16484 df-sets 16485 df-ress 16486 df-plusg 16573 df-mulr 16574 df-sca 16576 df-vsca 16577 df-tset 16579 df-ple 16580 df-0g 16710 df-gsum 16711 df-mre 16852 df-mrc 16853 df-acs 16855 df-mgm 17847 df-sgrp 17896 df-mnd 17907 df-mhm 17951 df-submnd 17952 df-grp 18101 df-minusg 18102 df-sbg 18103 df-mulg 18220 df-subg 18271 df-ghm 18351 df-cntz 18442 df-cmn 18903 df-abl 18904 df-mgp 19236 df-ur 19248 df-srg 19252 df-ring 19295 df-subrg 19529 df-lmod 19632 df-lss 19700 df-psr 20597 df-mvr 20598 df-mpl 20599 df-opsr 20601 df-psr1 20812 df-vr1 20813 df-ply1 20814 df-coe1 20815 This theorem is referenced by: eqcoe1ply1eq 20929 pmatcollpw1lem2 21383 mp2pm2mp 21419 plypf1 24812
Copyright terms: Public domain W3C validator | 13,108 | 19,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.271967 |
http://www.mathcasts.org/mtwiki/GlossaryT/Systems2 | 1,568,866,987,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573439.64/warc/CC-MAIN-20190919040032-20190919062032-00074.warc.gz | 302,630,265 | 8,921 | # Systems of Linear Equations 2х2 - 2 equations in 2 variables
Definition: Finding a simultaneous solution to 2 linear equations in 2 variables - values for both variables that make both equations true - is called solving a 2х2 system of linear equations.
Example of a linear system of 2 equations in 2 variables
Sample 2x2 linear system: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. The solution to this system is (\color{purple}{x},\color{teal}{y})=(\color{purple}{1},\color{teal}{2}) . Why is this a solution? Because - when we substitute \color{purple}{x=1} and \color{teal}{y=2} , both equations are "true". \begin{array}{c} \color{purple}{1}\color{blue}{+}\color{teal}{2}\color{blue}{ \cdot 2 \,\mathop = \limits^?\, 5} \\ \color{blue}{1 + 4 \,\mathop = \limits^?\, 5} \\ \color{blue}{5 \,=\, 5} \\ \end{array} \begin{array}{c} \color{red}{2\cdot} \color{purple}{1} \color{red}{- } \color{teal}{2}\color{red}{ \,\mathop = \limits^?\, 0} \\ \color{red}{2 - 2 \,\mathop = \limits^?\, 0} \\ \color{red}{0 = 0} \\ \end{array} Graphically, we have: This browser does not have a Java Plug-in. Get the latest Java Plug-in here.
Regulation: A solution to the system is every intersection (touching) point of the 2 lines.
InterActivity Directions for InterActivity 1. Look at the two lines a and b. They intersect at the point E. Check that E satisfies both equations. If you know how, solve the system: a and b and check that you get E.. 2. Click and drag the lines or the points A, B, C or D. If they intersect at one point, E is "good". Look at the left (where the formulas are) to check that E is a point! 3. Click and drag the points A, B, C or D so that a and b are parallel. They are parallel when E says "undefined". Notice that a and b are the same except for the constant after the "=" sign. 4. Click and drag the points A, B, C or D so that a and b coincide. Notice that a and b are completely the same and that E is undefined. This browser does not have a Java Plug-in. Get the latest Java Plug-in here.
Solutions to 2x2 linear systems Graphs!
A linear system can have:
• Exactly one solution. The lines intersect in exactly one point. Examples below in "Solution Methods". When solving this system you get numbers for x and y. Answer is: (number for x, number for y) • No solution. The lines are parallel and never touch. The system is inconsistent. When solving this system you get something stupid like 3=5. Answer is: No solution. • Infinitely many solutions. The lines coincide. They are the same line. Every point on this line is a solution. More? When solving this system you get stuck with 0=0. Answer is: Many solutions.
Exactly one solution: (1,2) \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. No solution \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{x + 2y = 2} \\ \end{array} \right. Infinitely many solutions \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x + 4y = 10} \\ \end{array} \right. This browser does not have a Java Plug-in. Get the latest Java Plug-in here. This browser does not have a Java Plug-in. Get the latest Java Plug-in here. This browser does not have a Java Plug-in. Get the latest Java Plug-in here.
Solution methods
A linear system can be solved using any of the following three methods:
• The substitution method. Example
• The addition or elimination method Example
• Cramer's rule (determinants) Example
The substitution method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 5-2y} \\ \color{navy}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{2\cdot\color{blue}{(5-2y)} - y = 0} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10-4y - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10-5y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{10=5y} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{y=2} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x = 5-2y} \\ \color{red}{5y=10} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 5-2 \cdot} \color{red}{2} \\ \color{red}{y=2} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x = 5-4} \\ \color{navy}{y=2} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{x = 1} \\ \color{navy}{y=2} \\ \end{array} \right. Solution is: (1,2)
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The addition method. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right.
------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{red}{4x - 2y = 0} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{5x+0y =5} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{5x=5} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{navy}{x + 2y = 5} \\ \color{purple}{x=1} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{purple}{1} \color{blue}{+2y=5} \\ \color{navy}{x=1} \\ \end{array} \right. \Leftrightarrow ------------------------------------------------------------------------------- \left\{ \begin{array}{l} \color{blue}{2y=4} \\ \color{navy}{x=1} \\ \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \color{blue}{y=2} \\ \color{navy}{x=1} \\ \end{array} \right. Solution is: (1,2)
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Cramer's rule. Solve: \left\{ \begin{array}{c} \color{blue}{\,\,x + 2y = 5} \\ \color{red}{2x - y = 0} \\ \end{array} \right. = \left\{ \begin{array}{c} \color{blue}{1}x + \color{#FF8000}{2}y = \color{orange}{5} \\ \color{red}{2}x \color{#800080}{ - 1} y = \color{#00BB80}{0} \\ \end{array} \right.
------------------------------------------------------------------------------- D = \left| {\matrix{ \color{blue}{1} & \color{#FF8000}{2} \cr \color{red}{2} & \color{#800080}{ - 1} \cr } } \right| = \color{blue}{1} \cdot \color{#800080}{ ( - 1)} - \color{red}{2} \cdot \color{#FF8000}{2} = - 1 - 4 = - 5 ------------------------------------------------------------------------------- \color{purple}{D_x} = \left| {\matrix{ \color{orange}{5} & \color{#FF8000}{2} \cr \color{#00BB80}{0} & \color{#800080}{ - 1} \cr } } \right| = \color{orange}{5} \cdot \color{#800080}{ ( - 1)} - \color{#00BB80}{0} \cdot \color{#FF8000}{2} = - 5 - 0 = - 5 ------------------------------------------------------------------------------- \color{teal}{D_y} = \left| {\matrix{ \color{blue}{1} & \color{orange}{5} \cr \color{red}{2} & \color{#00BB80}{0} \cr } } \right| = \color{blue}{1} \cdot \color{#00BB80}{0} - \color{red}{2} \cdot \color{orange}{5} = 0 - 10 = - 10 ------------------------------------------------------------------------------- x=\frac{\,D_x\,}{D}=\frac{-5}{-5}=1 y=\frac{\,D_y\,}{D}=\frac{-10}{-5}=2 Solution is: (1,2) | 2,738 | 8,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-39 | latest | en | 0.785805 |
https://www.varsitytutors.com/lsat_logic_games-help/determining-sequence-in-linear-games?page=26 | 1,638,538,204,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362879.45/warc/CC-MAIN-20211203121459-20211203151459-00581.warc.gz | 1,127,257,590 | 52,507 | # LSAT Logic Games : Determining sequence in linear games
## Example Questions
### Example Question #251 : Determining Sequence In Linear Games
Future Tech Industries employs exactly two scientists: P and Q. The company also employs exactly three engineers: R, S, and T. On a workday all five employees must use the company lab exactly once, according to the following conditions:
Only one employee uses the lab at a time.
Each employee focuses their research on either energy or health, but not both.
Employee P focuses on health.
Employees Q and R focus on energy.
All health focused employees use the lab before any energy focused employees.
Any energy focused engineer uses the lab before any energy focused scientist.
The employee that uses the lab second could be any one of exactly how many employees?
Four
Two
Five
Three
One
Four
Explanation:
Because P focuses on health it must come before Q and R. And because Q is both energy focused and scientist there are no employees allowed to use the lab after Q. The game can thus be written three ways:
P __ __ __ Q
__ P __ __ Q
__ __ P R Q
In the first way, R, S, T are interchangeable and can all be placed second. The second way shows P can go second. Thus, the only employee to not be able to go second is Q.
### Example Question #252 : Determining Sequence In Linear Games
Future Tech Industries employs exactly two scientists: P and Q. The company also employs exactly three engineers: R, S, and T. On a workday all five employees must use the company lab exactly once, according to the following conditions:
Only one employee uses the lab at a time.
Each employee focuses their research on either energy or health, but not both.
Employee P focuses on health.
Employees Q and R focus on energy.
All health focused employees use the lab before any energy focused employees.
Any energy focused engineer uses the lab before any energy focused scientist.
Which one of the following could be the order, from first to last, in which the five employees use the lab?
P, Q, T, S, R
P, S, T, Q, R
T, S, R, P, Q
P, Q, R, S, T
P, S, T, R, Q
P, S, T, R, Q
Explanation:
All the incorrect answers are wrong for the following reasons:
P, Q, R, S, T (Since Q and R are both energy focused, and Q is a scientist and R is a engineer R must use the lab before Q)
P, Q, T, S, R (Since Q and R are both energy focused, and Q is a scientist and R is a engineer R must use the lab before Q)
P, S, T, Q, R (Since Q and R are both energy focused, and Q is a scientist and R is a engineer R must use the lab before Q)
T, S, R, P, Q (P is health focused, whereas R is energy focused, thus P must use the lab before R)
### Example Question #253 : Determining Sequence In Linear Games
A talent show has seven teams of performers, each named after a color of the rainbow: Red, Orange, Yellow, Green, Blue, Indigo, and Violet. Four teams perform in Act I and three teams perform in Act II. They apply the following rules to determine the lineup of the talent show:
• Violet must be in Act I.
• Green either performs first or last.
• Exactly one team performs between Yellow and Blue.
• Violet performs some time after Blue.
All of the following could be true except:
Blue performs first.
Yellow performs second.
Yellow performs fourth.
Green performs last.
Red performs fifth.
Yellow performs second.
Explanation:
If Violet performs after Blue, but Violet must be in Act I, then Blue must also be in Act I. That means that Blue can go first, second, or third in the lineup.
• If Blue goes first, Yellow must go third because there is exactly one act between them.
• If Blue goes second, Yellow must go fourth for the same reason.
• If Blue goes third, Yellow must go first for the same reason.
There is no possible case, then, in which Yellow goes second.
### Example Question #254 : Determining Sequence In Linear Games
A talent show has seven teams of performers, each named after a color of the rainbow: Red, Orange, Yellow, Green, Blue, Indigo, and Violet. Four teams perform in Act I and three teams perform in Act II. They apply the following rules to determine the lineup of the talent show:
• Violet must be in Act I.
• Green either performs first or last.
• Exactly one team performs between Yellow and Blue.
• Violet performs some time after Blue.
If Blue performs second, which of the following must be true?
Indigo performs sixth.
Violet performs third.
Yellow performs fifth.
Red performs last.
Orange performs first.
Violet performs third.
Explanation:
If Blue performs second, and there is exactly one act between Blue and Yellow, then Yellow must perform fourth. Since Violet performs after Blue but still performs in Act I, yet there are only four acts in Act I, Violet must perform third.
### Example Question #255 : Determining Sequence In Linear Games
A college is having an event for alumni who graduated in the years 2005-2010, inclusive. Six friends-- Harry, Inez, Jack, Katie, Lou, and Maria-- attend the event.
• Harry and Maria graduated the same year.
• Katie graduated before Jack, but after Lou.
Which of the following could not be true?
Explanation:
Since Katie graduated after Lou, and Katie graduated in 2006, Lou must have graduated in 2005. Therefore, Lou did not graduate in 2008.
### Example Question #256 : Determining Sequence In Linear Games
A talent show has seven teams of performers, each named after a color of the rainbow: Red, Orange, Yellow, Green, Blue, Indigo, and Violet. Four teams perform in Act I and three teams perform in Act II. They apply the following rules to determine the lineup of the talent show:
• Violet must be in Act I.
• Green either performs first or last.
• Exactly one team performs between Yellow and Blue.
• Violet performs some time after Blue.
If Red performs last, which of the following must be true?
Green performs in Act I.
Orange performs in Act II.
Indigo performs in Act II.
Yellow performs in Act II.
Yellow performs in Act I.
Green performs in Act I.
Explanation:
Since Red performs last, and Green can only perform first or last in the lineup, Green must go first. Therefore, Green performs in Act I.
### Example Question #257 : Determining Sequence In Linear Games
A restaurant manager is scheduling interviews for an Executive Chef at her new business. Interviews for six applicants--Chet, Sylvia, Dennis, Beulah, Henrietta, and Lester--will be scheduled, one interview per day for the next six days. The schedule for the interviews is subject to the following conditions:
Lester must be scheduled to interview earlier than Sylvia.
Dennis must be scheduled to interview earlier than both Chet and Beulah.
The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta.
Henrietta's interview cannot be scheduled for the sixth day.
Which one of the following is an acceptable schedule of interviews, listed in order from earliest to latest?
Henrietta, Dennis, Sylvia, Chet, Lester, Beulah
Lester, Henrietta, Dennis, Chet, Beulah, Sylvia
Dennis, Lester, Sylvia, Chet, Beulah, Henrietta
Henrietta, Lester, Dennis, Sylvia, Chet, Beulah
Chet, Lester, Dennis, Henrietta, Sylvia, Beulah
Henrietta, Lester, Dennis, Sylvia, Chet, Beulah
Explanation:
In order to solve this problem, simply go through each answer and apply each individual condition from the pre-question text. If the answer violates any of the conditions, then you can eliminate it. If the answer does not violate any conditions, then you know it is correct. A question like this will typically be the first in its set, and is an opportunity to solidify your knowledge of the conditions governing the game.
For this problem, the correct order is: Henrietta, Lester, Dennis, Sylvia, Chet, Beulah
Let's go through the list of conditions to see why:
"Lester must be scheduled to interview earlier than Sylvia." TRUE
"Dennis must be scheduled to interview earlier than both Chet and Beulah." TRUE
"The interviews scheduled for the second and third day cannot be for either Chet, Beulah, or Henrietta.TRUE
"Henrietta's interview cannot be scheduled for the sixth day.TRUE
This is the only provided sequence that satisfies every condition of the game, and is therefore the correct answer.
### Example Question #258 : Determining Sequence In Linear Games
A restaurant manager is scheduling interviews for an Executive Chef at her new business. Interviews for six applicants--Chet, Sylvia, Dennis, Beulah, Henrietta, and Lester--will be scheduled, one interview per day for the next six days. The schedule for the interviews is subject to the following conditions:
Lester must be scheduled to interview earlier than Sylvia.
Dennis must be scheduled to interview earlier than both Chet and Beulah.
The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta.
Henrietta's interview cannot be scheduled for the sixth day.
If neither Chet nor Beulah nor Henrietta is scheduled to interview on the fourth day, which one of the following must be true?
Henrietta is scheduled to interview on the first day.
Beulah is scheduled to interview on the sixth day.
Lester is scheduled to interview on the second day.
Chet is scheduled to interview on the fifth day.
Dennis is scheduled to interview on the third day.
Henrietta is scheduled to interview on the first day.
Explanation:
The key to finding the correct solution to this problem is drawing additional deductions from the question text. In addition to the conditions given to us in the pre-question text, we know that in the sequence we are solving for, neither Chet nor Beulah nor Henrietta can occupy the fourth position. This will allow us to make secondary deductions about their relative positions in the sequence by combining this information with the previous conditions. Let's start by looking at the third condition from the pre-question text:
"The interviews scheduled for the second and third day cannot be for either Chet, Beulah, or Henrietta."
When we combine this condition with the new information in the question, we see that neither Chet nor Beulah nor Henrietta can occupy the second, third, or fourth positions in our sequence. This leaves only the first, fifth, and sixth positions to place these three entities. Next, let's look at the second condition.
"Dennis must be scheduled to interview earlier than both Chet and Beulah."
We now know that Chet, Beulah, and Henrietta must occupy the first, fifth, and sixth positions in the sequence. However, when we apply the second condition of the pre-question text, we see that neither Chet nor Beulah may occupy the first position, as this would make it impossible for the second condition to be true. This leaves only Henrietta to fill the first position in the sequence, meaning that it must be true that Henrietta is interviewed on the first day. Therefore, this is the answer we are looking for.
Now, questions like this can get confusing because of their phrasing. After all, through our own deductions we have established that Chet could be interviewed on the fifth day, and that Beulah could interview on the sixth day. However, for the answer to this question to be considered correct, it is not sufficient that an answer could be true. Rather, the answer must be true in every possible iteration of the sequence. Chet and Beulah are interchangeable in the fifth and sixth positions, and therefore do not meet the criteria for a correct answer.
### Example Question #259 : Determining Sequence In Linear Games
A restaurant manager is scheduling interviews for an Executive Chef at her new business. Interviews for six applicants--Chet, Sylvia, Dennis, Beulah, Henrietta, and Lester--will be scheduled, one interview per day for the next six days. The schedule for the interviews is subject to the following conditions:
Lester must be scheduled to interview earlier than Sylvia.
Dennis must be scheduled to interview earlier than both Chet and Beulah.
The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta.
Henrietta's interview cannot be scheduled for the sixth day.
Which one of the following must be true?
Sylvia is scheduled to interview earlier than Henrietta.
Henrietta is scheduled to interview earlier than Lester.
Lester is scheduled to interview earlier than Chet.
Henrietta is scheduled to interview earlier than Chet.
Dennis is scheduled to interview earlier than Sylvia.
Lester is scheduled to interview earlier than Chet.
Explanation:
The answer to this problem will be a condition that must be true. Therefore, this condition will always be true, in any viable iteration of the sequence of interviews. Any answer that does not meet this criteria can be automatically eliminated.
The correct answer from the options provided is: Lester is scheduled to interview earlier than Chet.
We need to look at the conditions from the pre-question text to determine why this is correct. Let's start with the second condition:
"Dennis must be scheduled to interview earlier than both Chet and Beulah."
Because we know that at least one entity (Dennis) must precede Chet in the sequence, we know with certainty that he cannot occupy the first position. Now let's combine this information with the third condition:
"The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta."
When we add this information from the third condition into the mix, we see that Chet cannot occupy either the first, second, or third positions in the sequence. Conversely, this means that Chet can only occupy the fourth, fifth, or sixth positions. Now that we've established Chet's potential positions in the sequence, let's look at the first condition, which concerns Lester.
"Lester must be scheduled to interview earlier than Sylvia."
Because we know that Lester must precede at least one entity (Sylvia) in the sequence, we can deduce that he cannot occupy the sixth position in the sequence. Since we also already know that Chet can only occupy the fourth, fifth, or sixth position in the sequence, we can further deduce that the only conceivable way for Chet to precede Lester in the sequence would be for Chet to occupy the fourth position and Lester to occupy the fifth. If this sequence violates any of the other conditions, then we will know that we've found the correct answer. Let's test it out.
_ _ _ C L _
Since the first condition tells us that Lester must precede Sylvia, we know that Sylvia must occupy the sixth position in the sequence.
_ _ _ C L S
Furthermore, we know from the third condition that the only remaining position for Henrietta to occupy is the first. She must occupy this position.
H _ _ C L S
This is where the sequence breaks down. According to the third condition, neither Chet nor Beulah nor Henrietta can occupy the second or third positions of the sequence. However, with Beulah as one of our remaining entities, our only remaining positions in the sequence are the second and third. Since we have determined through our deductions that this is the only conceivable way for Chet to precede Lester in the sequence, it is impossible for Chet to come before Lester. Therefore, it must be true that Lester is scheduled to interview earlier than Chet.
### Example Question #260 : Determining Sequence In Linear Games
A restaurant manager is scheduling interviews for an Executive Chef at her new business. Interviews for six applicants--Chet, Sylvia, Dennis, Beulah, Henrietta, and Lester--will be scheduled, one interview per day for the next six days. The schedule for the interviews is subject to the following conditions:
Lester must be scheduled to interview earlier than Sylvia.
Dennis must be scheduled to interview earlier than both Chet and Beulah.
The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta.
Henrietta's interview cannot be scheduled for the sixth day.
Which of the following CANNOT be the interviews scheduled for the fourth, fifth, and sixth days, listed in that order?
Lester, Chet, Sylvia
Sylvia, Beulah, Chet
Chet, Beulah, Sylvia
Dennis, Beulah, Chet
Henrietta, Beulah, Chet
Lester, Chet, Sylvia
Explanation:
The phrasing of the question text is intended to make this question more difficult. Before looking at the answer choices, we should take a second to parse what the question is really asking, and what will determine a correct answer. Because the question is asking us for an order that CANNOT work within the conditions of the game, the correct answer will contain an error. Apply each condition to each potential answer. The first time you encounter an error, you will know you have found your correct answer.
The correct choice for this question is: Lester, Chet, Sylvia.
Let's apply the conditions from the pre-question text to this response to see why it is the right choice.
"Lester must be scheduled to interview earlier than Sylvia."
No error so far. In this sequence, Lester is scheduled to interview earlier than Sylvia. Let's go on to the second condition.
"Dennis must be scheduled to interview earlier than both Chet and Beulah."
This is where we encounter a problem. Because Lester, Chet, and Sylvia occupy the fourth, fifth, and sixth positions (respectively), we know that the only remaining positions to fill are the first, second, and third. No matter which remaining position Dennis occupies, he will always precede Chet, so that's not the problem. The problem is that Dennis must also precede Beulah, meaning that Beulah must occupy either the second or third position. However, as we know from the third condition:
"The interviews scheduled for the second and third days cannot be for either Chet, Beulah, or Henrietta."
Therefore, it is impossible for Beulah to occupy either of the remaining positions, and this response CANNOT be true
Tired of practice problems?
Try live online LSAT prep today. | 3,993 | 18,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-49 | latest | en | 0.948613 |
http://nrich.maths.org/1365/index?nomenu=1 | 1,386,213,370,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163038799/warc/CC-MAIN-20131204131718-00056-ip-10-33-133-15.ec2.internal.warc.gz | 132,414,756 | 5,325 | ### Introduction
There has been a lot of correspondence recently on the Ask Nrich web-board about fractional derivatives. We know how to differentiate a function once, twice and so on, but can we differentiate the function 3/2 times? Similarly, we know how to integrate a function once, twice, and so on, but can we integrate it 1/2 times? This is the first of three articles which will introduce you to the main ideas in this new and rather strange world of fractional integrals and derivatives. Questions about the existence of such things were asked not long after calculus was created; for example, in 1695 Leibnitz wrote "Thus it follows that $d^{1/2}x$ will be equal to $\ldots$ from which one day useful consequences will be drawn ." Also, Euler (1738) wrote "When $n$ is an integer, the ratio $d^np$, $p$ a function of $x$, to $dx^n$ can always be expressed algebraically. Now it is asked: what kind of ratio can be made if $n$ is a fraction?"
If we differentiate $x^n$ $n$ times, where $n$ is a positive integer, we get $n!$ ; thus $${d^n\over dx^n}x^n = n!\ .$$ Later, we are going to see that this is true for all positive numbers $a$; that is, for $a> 0$, $${d^a\over dx^a}x^a = a!\ .$$ In order to understand this, we need to understand what $a!$ means when $a$ is not an integer. In this article we shall discuss the factorial function $a!$ for positive numbers $a$, even when $a$ is not an integer. In the second article we shall discuss fractional integrals; these are easier to define than fractional derivatives, and their definition requires the use of the factorial function. In the third article we shall discuss fractional derivatives.
### The Factorial function
Let $F(n)$ be the factorial function; then for every positive integer $n$ we have $F(n) = 1.2.3\cdots n$. Of course, we usually write $n!$ instead of $F(n)$, and we can define $F(n)$ (and therefore $n!$) by the conditions $$F(1)=1, \quad F(n) = nF(n-1), \quad n=2,3,\ldots. \quad (1.1)$$
In order to define $y!$ for every positive $y$ we need to discuss an extremely important function in mathematics known as the Gamma function $\Gamma(x)$. This function is defined for every positive $x$, and it satisfies the intruiging formulae $$\Gamma(1)=1, \quad \Gamma(x+1) = x\Gamma(x), \quad x> 0. \quad (1.2)$$ Let us begin by exploring some of the consequences of (1.2) without (at first) worrying about how the Gamma function is defined. First, (1.2) implies that $\Gamma(2)=\Gamma(1)=1$. It also implies that $\Gamma(x)$ is not defined when $x=0$ for otherwise we would have $0 = 0\Gamma(0)=\Gamma(1)=1$ which is false.
Next, if we write $G(x) = \Gamma(x+1)$ we obtain $$G(1)=1, \quad G(x) = xG(x-1),$$ and a comparison with (1.1) shows that $G(n)=F(n) = n!$ for every positive integer $n$; thus for all positive integers $n$, $$n! = \Gamma (n+1).$$ Assuming that we have already defined the Gamma function, it is now quite natural to define $$x! = \Gamma (x+1) \quad (1.3)$$ whenever $x> -1$ (because $\Gamma(t)$ is only defined for $t> 0$). The Gamma function plays a crucial role in many parts of mathematics; particularly in probability theory, applied mathematics and complex analysis.
### Binomial coefficients
For the moment, we continue with our discussion of the consequences of (1.2) and (1.3) even though we have still not defined the Gamma function. First we note that (1.3) gives the curious formula $$0! = \Gamma(1)=1.$$ This formula is used a lot in working with Binomial coefficients and this is the justification for it.
Speaking of binomial coeffients, we recall that if $k$ and $n$ are positive integers with $n> k$, then the corresponding binomial coefficient is defined to be $${n\choose k} = {n!\over k!\,(n-k)!}.$$ Of course we can now write this as $${n\choose k} = {\Gamma(n+1)\over \Gamma(k+1)\Gamma(n-k+1)},$$ which can now be generalised to give $${a\choose b} = {\Gamma(a+1)\over \Gamma(b+1)\Gamma(a-b+1)},$$ whenever $a> b> 0$. Let us do just one calculation here. We have $${3\,^1\!/_2\choose 2} = {\Gamma (9/2)\over \Gamma(3)\Gamma(5/2)} = {(7/2)(5/2)\Gamma(5/2)\over 2!\,\Gamma(5/2)} = {35\over 8}.$$
Binomial coefficients with non-integral entries are used in the Binomial expansion with non-integral powers. For example, we have the Binomial Theorem $$(1+x)^n = \sum_{k=0}^n {n\choose k}x^k$$ where $n$ is a positive integer, and also $$(1+x)^a = \sum_{k=0}^\infty {a\choose k}x^k \quad (1.4)$$ whenever |x|\leq1 (for convergence) and $a> 0$.
Quite generally, if $n$ is a positive integer and 0\leq\theta \leq 1 , from (1.2) and (1.3) we have $$(n+\theta)! = \Gamma(n+\theta +1) =(n+\theta)(n-1+\theta)\cdots (1+\theta)\theta\Gamma(\theta).$$ Notice that if $n+\theta$ is an integer, then $\theta =1$ and the formula ends with $\Gamma(1)$ which is 1; thus the Gamma function no longer appears explicitly in this formula. It is for this reason that the Gamma function does not appear in the binomial coefficients with integer entries.
### The Gamma function
All of this discussion depends on having the Gamma function available so how, then, do we define the Gamma function? The definition is this : $$\Gamma (x) = \int_0^\infty t^{x-1}e^{-t}\,dt.$$ Notice that $x$ appears as a parameter in the integral, and that the integration is with respect to $t$ not $x$. This looks complicated but in fact, it is easy to verify (1.2). First, $$\Gamma(1) = \int_0^\infty e^{-t}\, dt = 1.$$ Next, we note that $${d\over dt}\Big(t^xe^{-t}\Big) = xt^{x-1}e^{-t} - t^xe^{-t},$$ and if we integrate both sides of this equation from $t=0$ to $t= \infty$ we obtain $$\big[t^xe^{-t}\big]_0^\infty = x\Gamma(x)-\Gamma(x+1)$$ (this is just integration by parts). As the left hand side is zero (because $x> 0$), this completes the proof of (1.2).
The formula $\Gamma(x+1) =x\Gamma(x)$ enables us to calculate $\Gamma(x)$ in terms of the value of $\Gamma$ at the fractional part of $x$. The following illustrative example will show what we mean here : $$(7/2)! = \Gamma (9/2) = {7\over 2}\Gamma(7/2) = {7\over 2}{5\over 2}\Gamma(5/2) = \cdots = {7\over 2}{5\over 2}{3\over 2}{1\over 2}\Gamma (1/2) ={105\over 16}\Gamma(1/2),$$ and in this example we cannot make further progress unless we know what $\Gamma(1/2)$ is. There is a beautiful, but deep, formula that relates the Gamma function to the trigonometric function $\sin x$, and this formula enables us to calculate $\Gamma(1/2)$. The formula is, for $0 \leq x \leq 1$, $$\Gamma(x)\Gamma(1-x) = {\pi \over \sin \pi x}.$$ We cannot prove this here, but note that when $x=1/2$ we get $$\Gamma(1/2)^2 = {\pi\over \sin \pi /2} = \pi,$$ so that $$\Gamma (1/2) = \sqrt{\pi}.$$ This means, for example, that $$(5/2)! = {5\over 2}{3\over 2}{1\over 2}\Gamma (1/2) ={15\sqrt{\pi}\over 8} = 3.3234\cdots.$$
### Evaluating the Gamma function
The values of the Gamma function are given in tables (just as the values of $\sin x$ are), and using these tables we can calculate, for example, $(9/4)!$: its value is $\Gamma(13/4) = 2.5493\cdots$.
The following table of values of $\Gamma(0.1),\ldots ,\Gamma(0.9)$ will enable you to find some values of $y!$
$k$ $\Gamma(k/10)$
$1$ 9.5135
$2$ 4.5908
$3$ 2.9916
$4$ 2.2182
$5$ 1.7725
$6$ 1.4892
$7$ 1.2981
$8$ 1.1642
$9$ 1.0686
Using this you should be able to see, for example, that $$(3/2)! = 1.3293, \quad (3.7)! = 15.431,\quad (4.2)! = 32.578, \quad (5.2)! = 169.41, \quad (6.7)! = 2769.8.$$ You know that $3! = 6$, and you can show (using the table above) that $(2.9)!$ is a little less than 6, and that $(3.1)!$ is a little greater than 6. Finally, two more curiosities are $$\pi! = \Gamma(1+\pi) = 7.1881\cdots , \quad e! = \Gamma(1+e) = 4.2608\cdots .$$ | 2,469 | 7,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2013-48 | longest | en | 0.895978 |
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#### Resources tagged with Properties of numbers similar to X Marks the Spot:
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### There are 65 results
Broad Topics > Numbers and the Number System > Properties of numbers
### X Marks the Spot
##### Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
### Slippy Numbers
##### Stage: 3 Challenge Level:
The number 10112359550561797752808988764044943820224719 is called a 'slippy number' because, when the last digit 9 is moved to the front, the new number produced is the slippy number multiplied by 9.
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Can You Find a Perfect Number?
##### Stage: 2 and 3
Can you find any perfect numbers? Read this article to find out more...
### Happy Octopus
##### Stage: 3 Challenge Level:
This investigation is about happy numbers in the World of the Octopus where all numbers are written in base 8 ... Find all the fixed points and cycles for the happy number sequences in base 8.
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Four Coloured Lights
##### Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### Alphabet Soup
##### Stage: 3 Challenge Level:
This challenge is to make up YOUR OWN alphanumeric. Each letter represents a digit and where the same letter appears more than once it must represent the same digit each time.
### Whole Numbers Only
##### Stage: 3 Challenge Level:
Can you work out how many of each kind of pencil this student bought?
### Six Times Five
##### Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
### Arrange the Digits
##### Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Writ Large
##### Stage: 3 Challenge Level:
Suppose you had to begin the never ending task of writing out the natural numbers: 1, 2, 3, 4, 5.... and so on. What would be the 1000th digit you would write down.
### Like Powers
##### Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### The Codabar Check
##### Stage: 3
This article explains how credit card numbers are defined and the check digit serves to verify their accuracy.
### Not a Polite Question
##### Stage: 3 Challenge Level:
When asked how old she was, the teacher replied: My age in years is not prime but odd and when reversed and added to my age you have a perfect square...
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Guess the Dominoes
##### Stage: 1, 2 and 3 Challenge Level:
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
### Chameleons
##### Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
### Lesser Digits
##### Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
### Clever Carl
##### Stage: 2 and 3
What would you do if your teacher asked you add all the numbers from 1 to 100? Find out how Carl Gauss responded when he was asked to do just that.
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
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15.4: Gas Laws
Difficulty Level: At Grade Created by: CK-12
Lesson Objectives
The student will:
• state Boyle's law, Charles's law, and Gay-Lussac's law.
• solve problems using Boyle's law, Charles's law, and Gay-Lussac's law.
• state the combined gas law.
• solve problems using the combined gas law.
Vocabulary
• Boyle's law
• Charles's law
• combined gas law
• Gay-Lussac's law
Introduction
The gas laws are mathematical expressions that relate the volume, pressure, temperature, and quantity of gas present. They were determined from the results of over 100 years of experimentation. They can also be derived logically by examining the present day definitions of pressure, volume, and temperature.
Boyle’s Law
Gases are often characterized by their volume, temperature, and pressure. These characteristics, however, are not independent of each other. Gas pressure is dependent on the force exerted by the molecular collisions and the area over which the force is exerted. In turn, the force exerted by these molecular collisions is dependent on the absolute temperature. The relationships between these characteristics can be determined both experimentally and logically from their mathematical definitions.
The relationship between the pressure and volume of a gas was first determined experimentally by an Irish chemist named Robert Boyle (1627-1691). The relationship between the pressure and volume of a gas is commonly referred to as Boyle’s law.
When we wish to observe the relationship between two variables, it is absolutely necessary to keep all other variables constant so that the change in one variable can be directly related to the change in the other. Therefore, when the relationship between the volume and pressure of a gas is investigated, the quantity and temperature of the gas must be held constant so that these factors do not contribute to any observed changes.
You may have noticed that when you try to squeeze a balloon, the resistance to squeezing becomes greater as the balloon becomes smaller. That is, the pressure inside the balloon becomes greater when the volume is reduced. This phenomenon can be studied more carefully with an apparatus like the one shown below. This device is a cylinder with a tightly fitted piston that can be raised or lowered. There is also a pressure gauge fitted to the cylinder so that the gas pressure inside the cylinder can be measured. The amount of gas of gas inside the cylinder cannot change, and the temperature of the gas is not allowed to change.
In the picture on the left, a 4.0-liter volume of gas is exerts a pressure of \begin{align*}2.0 \ \mathrm{atm}\end{align*}. If the piston is pushed down to decrease the volume of the gas to \begin{align*}2.0 \ \mathrm{liters}\end{align*}, the pressure of the gas is found to be \begin{align*}4.0 \ \mathrm{atm}\end{align*}. The piston can be moved up and down to positions for several different volumes, and the pressure of the gas can be for each of the volumes. Several trials would generate a data set like that shown in Table below.
PV Data
Trial Volume Pressure
1 \begin{align*}8.0\;\mathrm{liters}\end{align*} \begin{align*}1.0\;\mathrm{atm}\end{align*}
2 \begin{align*}4.0\;\mathrm{liters}\end{align*} \begin{align*}2.0\;\mathrm{atm}\end{align*}
3 \begin{align*}2.0\;\mathrm{liters}\end{align*} \begin{align*}4.0\;\mathrm{atm}\end{align*}
4 \begin{align*}1.0\;\mathrm{liters}\end{align*} \begin{align*}8.0\;\mathrm{atm}\end{align*}
We might note from casual observation of the data that doubling volume is associated with the pressure being reduced by half. Likewise, if we move the piston to cause the pressure to double, the volume is halved. We can analyze this data mathematically by adding a fourth column to our table – namely, a column showing the product of multiplying pressure times volume for each trial (see Table below).
PV Data
Trial Volume Pressure Pressure \begin{align*}\times \end{align*} Volume
1 \begin{align*}8.0\;\mathrm{liters}\end{align*} \begin{align*}1.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
2 \begin{align*}4.0\;\mathrm{liters}\end{align*} \begin{align*}2.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
3 \begin{align*}2.0\;\mathrm{liters}\end{align*} \begin{align*}4.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
4 \begin{align*}1.0\;\mathrm{liters}\end{align*} \begin{align*}8.0\;\mathrm{atm}\end{align*} \begin{align*}8.0\;\mathrm{liters} \cdot \mathrm{atm}\end{align*}
The data in the last column shows that with constant temperature and quantity of gas, the pressure times the volume for this sample of gas yields a constant. A mathematical constant (often represented by \begin{align*}k\end{align*}) is a number that does not change even when other quantities in the formula do change. The value of \begin{align*}k_1\end{align*} will change if a different quantity of gas is used or if the trials are carried out at a different temperature, but for a particular mass of a particular gas at a particular temperature, the value of \begin{align*}k_1\end{align*} will always be the same. A subscript 1 is used to distinguish this constant from the constants of other gas laws. This relationship can be shown in a mathematical equation.
\begin{align*}PV = k_1\end{align*}
This equation is a mathematical statement of Boyle’s law. This particular equation demonstrates what is called an inverse proportionality. When one of the variables is increased, the other variable will decrease by exactly the same factor. This relationship can be easily seen in a graph like the one shown below.
This result matches our logic intuition. If the pressure a gas exerts is equal to the force divided by the area over which it is exerted, we would expect the pressure to increase when we decrease the area but keep the force constant. Similarly, if we maintain the same number of molecules of gas and we keep the same temperature, we expect the total force exerted by the molecules to be the same. As a result, if we expand the volume of the gas, which increases the area over which the force is exerted, we would expect the pressure to decrease.
This video is a laboratory demonstration of Boyle's Law (4c): http://www.youtube.com/watch?v=J_I8Y-i4Axc (1:38).
Charles’s Law
The relationship between the volume and temperature of a gas was investigated by a French physicist, Jacques Charles (1746-1823). The relationship between the volume and temperature of a gas is often referred to as Charles’s law.
An apparatus that can be used to study the relationship between the temperature and volume of a gas is shown below. Once again, the sample of gas trapped inside a cylinder so that no gas can get in or out. Thus, we would have a constant mass of gas inside the cylinder. In this setup, we would also place a mass on top of a movable piston to keep a constant force pushing against the gas. This guarantees that the gas pressure in the cylinder will be constant. If the pressure inside increases, the piston will be pushed up until the inside pressure becomes equal to the outside pressure. Similarly, if the inside pressure decreases, the outside pressure will push the cylinder down, decreasing the volume until the two pressures again become the same. This system guarantees constant gas pressure inside the cylinder.
With this set up, we can adjust the temperature and measure the volume at each temperature to produce a data table similar to the one we created for comparing pressure and volume. The picture on the left in the diagram above shows the volume of a sample of gas at \begin{align*}250\;\mathrm{K}\end{align*}, while the picture on the right shows the volume when the temperature has been raised to \begin{align*}500\;\mathrm{K}\end{align*}. After two more trials, the collected data is shown in Table below.
Charles's Law Data
Trial Volume Temperature Volume/Temp
1 \begin{align*}1000.\;\mathrm{mL}\end{align*} \begin{align*}250.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
2 \begin{align*}1200.\;\mathrm{mL}\end{align*} \begin{align*}300.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
3 \begin{align*}2000.\;\mathrm{mL}\end{align*} \begin{align*}500.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
4 \begin{align*}2400.\;\mathrm{mL}\end{align*} \begin{align*}600.\;\mathrm{K}\end{align*} \begin{align*}4.00\;\mathrm{mL/K}\end{align*}
In order to find a constant from this data, it was necessary to divide each volume with the corresponding Kelvin temperature. The mathematical expression for Charles’s Law is:
\begin{align*} \frac {V} {T} = k_2\end{align*}
This relationship is to be expected if we recognize that we are increasing molecular collisions with the walls by raising the temperature. The only way to keep the pressure from increasing is to increase the area over which that force is exerted. This mathematical relationship is known as a direct proportionality. When one variable is increased, the other variable also increases by exactly the same factor.
Historical note: In addition to exploring the relationship between volume and temperature for gases, Jacques Charles was also the first person to fill a large balloon with hydrogen gas and take a solo balloon flight.
This video is a laboratory demonstration of Charle's Law (4c): http://www.youtube.com/watch?v=IkRIKGN3i0k (4:02).
Gay-Lussac’s Law
The relationship between temperature and pressure was investigated by the French chemist, Joseph Gay-Lussac (1778-1850). An apparatus that could be used for this investigation is shown below. In this case, the cylinder does not have a movable piston because it is necessary to hold the volume, as well as the quantify of gas, constant. This apparatus allows us to alter the temperature of a gas and measure the pressure exerted by the gas at each temperature.
After a series of temperatures and pressures have been measured, Table below can be produced. Like Charles’s Law, in order to produce a mathematical constant when operating on the data in the table, we must divide pressure by temperature. The relationship, again like Charles’s Law, is a direct proportionality.
Pressure vs. Temperature Data
Trial Temperature Pressure Pressure/Temp
1 \begin{align*}200.\;\mathrm{K}\end{align*} \begin{align*}600.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
2 \begin{align*}300.\;\mathrm{K}\end{align*} \begin{align*}900.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
3 \begin{align*}400.\;\mathrm{K}\end{align*} \begin{align*}1200.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
4 \begin{align*}500.\;\mathrm{K}\end{align*} \begin{align*}1500.\;\mathrm{torr}\end{align*} \begin{align*}3.00\;\mathrm{torr/K}\end{align*}
The mathematical form of Gay-Lussac’s Law is:
\begin{align*} \frac {P} {T} =k_3K\end{align*}
This relationship demonstrates that when the temperature is increased, the pressure must also increase to maintain the value of the constant, \begin{align*}k_3\end{align*}. If the area the molecules are occupying is kept the same, the collisions of the molecules with the surroundings will increases as the temperature increases. This results in a higher pressure.
Standard Conditions for Temperature and Pressure (STP)
It should be apparent by now that expressing a quantity of gas simply by stating its volume is inadequate. Ten liters of oxygen gas could contain any mass of oxygen from \begin{align*}4,000\;\mathrm{grams}\end{align*} to \begin{align*}0.50\;\mathrm{grams}\end{align*} depending on the temperature and pressure of the gas. Chemists have found it useful to choose a standard temperature and pressure with which to express gas volume. The standard conditions for temperature and pressure (STP) were chosen to be \begin{align*}0^\circ \text{C} \ (273\;\mathrm{K})\end{align*} and \begin{align*}1.00\;\mathrm{atm}\end{align*} (\begin{align*}760\;\mathrm{mm \ of \ Hg}\end{align*}). You will commonly see gas volumes expressed as \begin{align*}1.5\;\mathrm{liters}\end{align*} of gas under standard conditions or \begin{align*}1.5\;\mathrm{liters}\end{align*} of gas at STP. Once you know the temperature and pressure conditions of a volume of gas, you can calculate the volume under other conditions.
This video is a black board presentation of some ideal gas law calculations and it includes the definition of standard temperature and pressure (4d): http://www.youtube.com/watch?v=GwoX_BemwHs (13:01).
The Combined Gas Law
Boyle’s law shows how the volume of a gas changes when its pressure is changed with the temperature held constant, while Charles’s law shows how the volume of a gas changes when the temperature is changed with the pressure held constant. Is there a formula we can use to calculate the change in volume of a gas if both pressure and temperature change? The answer is yes, as we can use a formula that combines both Boyle’s law and Charles’s law.
Boyle's law states that for a sample of gas at constant temperature, every volume times pressure trial will yield the same constant. We use the subscript 1 to represent one set of conditions, and the subscript 2 to represent a second set of conditions,
\begin{align*}P_1 V_1= P_2 V_2 \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac {P_1} {P_2}\end{align*}
We can find a similar expression for Charles's law:
\begin{align*}\frac{V_1} {T_1} = \frac{V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = V_1 \cdot \frac{T_2} {T_1}\end{align*}
Combining the two equations yields:
\begin{align*}V_2 = V_1 \cdot \frac {P_1} {P_2} \cdot \frac {T_2} {T_1}\end{align*}
The terms in this equation are rearranged and are commonly written in the form shown below. This equation is also known as the combined gas law.
\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2}\end{align*}
When solving problems with the combined gas law, temperatures must always be in Kelvin. The units for pressure and volume may be any appropriate units, but the units of pressure must be the same for \begin{align*}P_1\end{align*} and \begin{align*}P_2\end{align*}, and the units of volume for \begin{align*}V_1\end{align*} and \begin{align*}V_2\end{align*} must also be the same.
Example:
A sample of gas has a volume of \begin{align*}400\end{align*}. liters when its temperature is \begin{align*}20.^\circ\mathrm{C}\end{align*} and its pressure is \begin{align*}300.\;\mathrm{mm \ of \ Hg}\end{align*}. What volume will the gas occupy at STP?
Solution:
Step 1: Assign known values to the appropriate variable.
\begin{align*} \begin{array}{ll} P_1 = 300.\ \text{mm\ of\ Hg} & P_2 = 760.\ \text{mm of Hg (standard pressure)}\\ V_1 = 400.\ \text{liters} & V_2 = x\ \text{(the unknown)}\\ T_1 = 20.^\circ \text{C} + 273 = 293\ \text{K} & T_2 = 0^\circ \text{C} + 273 = 273\ \text{K} \end{array} \end{align*}
Step 2: Solve the combined gas law for the unknown variable.
\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ V_2 = \frac {P_1V_1T_2} {P_2T_1}\end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}V_2 = \frac {(300.\ \text{mm of Hg}) \cdot (400.\ \text{L}) \cdot (273\ \text{K})} {(760.\ \text{mm of Hg}) \cdot (293\ \text{K})} = 147 \ \text{liters}\end{align*}
Example:
A sample of gas occupies \begin{align*}1.00\;\mathrm{liter}\end{align*} under standard conditions. What temperature would be required for this sample of gas to occupy \begin{align*}1.50\;\mathrm{liters}\end{align*} and exert a pressure of \begin{align*}2.00\;\mathrm{atm}\end{align*}?
Step 1: Assign known values to the appropriate variable.
\begin{align*} \begin{array}{ll} P_1 = 1.00\ \text{atm (standard pressure)} & P_2 = 2.0\ \text{atm} \\ V_1 = 1.00\ \text{liter} & V_2 = 1.50\ \text{liters} \\ T_1 = 273\ \text{K (standard temperature)} & T_2 = x \ \text{(unknown)} \end{array} \end{align*}
Step 2: Solve the combined gas law for the unknown variable.
\begin{align*} \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ T_2 = \frac {P_2V_2T_1} {P_1V_1}\end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}T_2 = \frac {(2.00\ \text{atm}) \cdot (1.50\ \text{L}) \cdot (273\ \text{K})} {(1.00\ \text{atm}) \cdot (1.00\ \text{L})} = 819\ \text{K}\end{align*}
Example:
A \begin{align*}1.00\;\mathrm{liter}\end{align*} sample of oxygen gas under standard conditions has a density of \begin{align*}1.43\;\mathrm{g/L}\end{align*}. What is the density of oxygen gas at \begin{align*}500.\;\mathrm{K}\end{align*} and \begin{align*}760.\;\mathrm{torr}\end{align*}?
Solution:
You can find the mass of oxygen in the \begin{align*}1.00\;\mathrm{liter}\end{align*} sample by multiplying volume times density, which yields a mass of \begin{align*}1.43\;\mathrm{grams.}\end{align*} Changing the temperature and/or pressure of a sample of gas changes its volume and therefore its density, but it does not change the mass. Therefore, when the new volume of the gas is found, the mass of oxygen gas in it will still be \begin{align*}1.43\;\mathrm{grams.}\end{align*} The density under the new conditions can be found by dividing the mass by the volume the gas now occupies.
Step 1: Assign known values to the appropriate variable.
\begin{align*} \begin{array}{ll} P_1 = 760.\ \text{torr} & P_2 = 760.\ \text{torr} \\ V_1 = 1.00\ \text{L} & V_2 = x \ \text{unknown} \\ T_1 = 273\ \text{K}& T_2 = 500.\ \text{K} \end{array} \end{align*}
Step 2: Solve the combined gas law for the unknown variable.
\begin{align*}V_2 = \frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} \ \ \ \ \ \text{so} \ \ \ \ \ \frac {P_1V_1T_2} {P_2T_1} \end{align*}
Step 3: Substitute the known values into the formula and solve for the unknown.
\begin{align*}\frac {(760.\ \text{torr}) \cdot (1.00\ \text{L}) \cdot (500.\ \text{K})} {(760.\ \text{torr}) \cdot (273\ \text{K})} = 1.83\ \text{liters}\end{align*}
\begin{align*}D_{O_2@500\ \text{K}} = \frac {\text{mass}} {\text{volume}} = \frac {1.43\ \text{g}} {1.83\ \text{L}} = 0.781\ \text{g/L}\end{align*}
Lesson Summary
• Boyle's law states that for a gas at constant temperature, volume is inversely proportional to pressure.
• Charles's law states that for a gas at constant pressure, volume in directly proportional to temperature.
• Gay-Lussac's law states that for a gas at constant volume, pressure is directly proportional to temperature.
• The volume of a mass of gas is dependent on the temperature and pressure. Therefore, these conditions must be given along with the volume of a gas.
• Standard conditions for temperature and pressure are \begin{align*}0^\circ \text{C}\end{align*} and \begin{align*}1.0\;\mathrm{atm}\end{align*}.
• The combined gas law relates the temperature, pressure, and volume of a gas.
Review Questions
1. When a sample of gas is placed in a larger container at the same temperature, what happens to the total force of the molecules hitting the walls?
2. When a sample of gas is placed in a larger container at the same temperature, what happens to the pressure exerted by the gas?
3. If \begin{align*}X\end{align*} and \begin{align*}Y\end{align*} are quantities that are related to each other by inverse proportion, what will the value of \begin{align*}Y\end{align*} become when the value of \begin{align*}X\end{align*} is increased by a factor of five?
4. Under what conditions will the value for the constant, \begin{align*}K\end{align*}, change in the equation for Boyle’s Law, \begin{align*}PV = K\end{align*}.
5. A sample of gas has a volume of \begin{align*}500.\ \mathrm{mL}\end{align*} under a pressure of \begin{align*}500.\ \mathrm{mm \ of \ Hg}\end{align*}. What will be the new volume of the gas if the pressure is reduced to \begin{align*}300.\ \mathrm{mm \ of \ Hg}\end{align*} at constant temperature?
6. A graph is made illustrating Charles’s Law. Which line would be appropriate assuming temperature is measured in Kelvin?
7. At constant pressure, the temperature of a sample of gas is decreased. Will the volume of the sample
1. increase
2. decrease
3. remain the same?
8. A sample of gas has its temperature increased from \begin{align*}-43^\circ\mathrm{C}\end{align*} to \begin{align*}47^\circ\mathrm{C}\end{align*} at constant pressure. If its volume at \begin{align*}-43^\circ\mathrm{C}\end{align*} was \begin{align*}500.\;\mathrm{mL}\end{align*}, what is its volume at \begin{align*}47^\circ\mathrm{C}\end{align*}?
9. A gas is confined in a rigid container and exerts a pressure of \begin{align*}250.\ \mathrm{mm \ of \ Hg}\end{align*} at a temperature of \begin{align*}17^\circ \text{C}\end{align*}. To what temperature must this gas be cooled in order for its pressure to become \begin{align*}216\ \mathrm{mm \ of \ Hg}\end{align*}? Express this temperature in \begin{align*}^\circ \text{C}\end{align*}.
10. What is the abbreviation used to indicate standard conditions for temperature and pressure?
11. A sample of gas has a volume of \begin{align*}500.\ \mathrm{mL}\end{align*} at standard conditions. Find its volume at \begin{align*}47^\circ \text{C}\end{align*} and \begin{align*}800.\ \mathrm{torr}\end{align*}.
12. A sample of gas has a volume of \begin{align*}100.\ \mathrm{L}\end{align*} at \begin{align*}17^\circ \text{C}\end{align*} and \begin{align*}800.\ \mathrm{torr}\end{align*}. To what temperature must the gas be cooled in order for its volume to become \begin{align*}50.0\ \mathrm{L}\end{align*} at a pressure of \begin{align*}600. \ \mathrm{torr}\end{align*}?
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http://mathematica.stackexchange.com/questions/2440/effective-matrix-power-like-algorithm?answertab=oldest | 1,432,844,467,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929561.98/warc/CC-MAIN-20150521113209-00097-ip-10-180-206-219.ec2.internal.warc.gz | 150,076,678 | 23,698 | # Effective matrix power like algorithm
## First example
Suppose you want to calculate the 6th power of some matrix $A$. The brute force attempt of doing this is considering $$(((AA)A)A)A)A$$ which requires a total of 5 matrix multiplications. However, this can be improved on by smart grouping of the matrices, for example $$((AA)A)((AA)A)$$ requires only 3 multiplications, since when the value of say $AA$ is known from the left parenthesis, it does not have to be recomputed on the right hand side. Using this type of caching can save a lot of computational power (logarithmic scaling?).
## Second example
A lot more computation is necessary if we're not multiplying matrices but differential forms, since each multiplication requires the calculation of a whole lot of permutations in addition to the multiplication of the components. For example, the Liouville form in classical mechanics is $$\omega^n\propto\underbrace{\omega\wedge\omega\wedge\cdots\wedge\omega}_{n~\text{factors}}$$ where each $\omega$ is a 2-form.
## The question
Since this already smells like a good application of Mathematica's "x := x =" trickery, I would like to solve this problem in general: How do I divide the operations up in a smart way such that I have to do as little computational steps as possible, and how do I implement this in Mathematica? Note that matrices were only an example application (otherwise the answer would be MatrixPower of course), I sumbled upon that problem a couple of times in the past in very different situations, and had to resort to possibly very ineffective implementations.
## Mathematical statement
Take an associative algebra $\mathcal A$. How can $a^n$ with $a\in\mathcal A,~n\in\mathbb N$ be calculated using as little computations as possible?
-
Could you provide another example? I find it hard to generalize something to only one instance. – Mr.Wizard Feb 28 '12 at 23:07
you might find this mathematica-journal.com/issue/v5i4/columns/wagner/… interesting, although it is not the same problem – acl Feb 28 '12 at 23:19
Even the "simple" case of Project Euler Problem #122 is not so simple. To be clear are you looking for the best possible result, or merely a good one? For example, is the "binary" method from that page good enough? – Mr.Wizard Feb 28 '12 at 23:21
@Mr.Wizard Oh, I didn't remember it was directly a problem on PE. Unfortunately, I don't have the solution so I can't access the solutions forum there. Are there any nice insights hidden in there? The binary method seems nice, but since this question is partially academic anyway, I'd like to get some optimal solution as well. – David Feb 28 '12 at 23:26
@Artes Nest would be precisely the brute force attempt I am trying to avoid: Nest[#.a&, a, 5]$=((((aa)a)a)a)a$, which is two many computations. (pun pun) ;-) – David Feb 28 '12 at 23:29
Note: The method I describe below does not find the optimal solution (i.e. the minimal number of multiplications), but it usually does better than the binary approach, and more importantly: it is very easy to extend it and optimize it for special cases. We could easily pre-compute those exponents for which it does worse than the simple binary method to say up to 1000 and special case those.
According to the Wikipedia article on the topic:
... the determination of a shortest addition chain seems quite difficult: no efficient optimal methods are currently known for arbitrary exponents, and the related problem of finding a shortest addition chain for a given set of exponents has been proven NP-complete.
Therefore trying to compute the optimial solution within the function (as opposed to precomputing it and making a lookup table) defeats the purpose of using this for optimization.
I did a version of this with C++ template metaprogramming here.
This is a direct translation:
Let f be an associative function, then:
Clear[pow]
(* Syntax: pow[exponent][base] *)
pow[n_][a_] /; Mod[n, 2] == 0 := With[{t = pow[n/2][a]}, f[t, t]]
pow[n_][a_] /; Mod[n, 3] == 0 := With[{t = pow[n/3][a]}, f[f[t, t], t]]
pow[n_][a_] := f[pow[n - 1][a], a]
pow[1][a_] := a
Testing:
f[x_, y_] := (Print["x"]; x y)
Then pow[10][2] gives
During evaluation of In[389]:= x
During evaluation of In[389]:= x
During evaluation of In[389]:= x
During evaluation of In[389]:= x
Out[389]= 1024
i.e. it was done in four multiplications. You can think a bit about which is the best subdivision (to 3 or to 2) for different numbers. It can be manually verified that for exponents under 100, in the case of 33 and 69 (and exponents reducible to these such as $67 = 2\times 33 +1$) it's better not to subdivide into 3 first. We can easily special-case pow for these by adding the following to the beginning of pow's definition:
pow[33][a_] := f[pow[32][a], a]
pow[69][a_] := f[pow[68][a], a]
Similarly, in the case of e.g. 82 it's better not to subdivide into 2 because $82 - 1 = 3^4$, but use
pow[82][a_] := f[pow[81][a], a]
Another example is that in the case of 85 and 95 it's better to subdivide into 5 first:
pow[85][a_] := pow[13][ pow[5][a] ]
But most of these are just tweaks that will not make a huge difference.
I believe that the value of my implementation lies in the ease of its extension for special cases like these for small exponents.
For the adventurous with a lot of free time, see this paper on Pippenger's algorithm.
-
What's the reason for the division by 2 and 3? They look like primes to me, would adding a line for 5 improve the code? – David Feb 29 '12 at 4:41
@David This is not the optimal solution, but it's close enough to it to be practical. Are you looking for a method that will compute the optimal solution for any $n$, or are you looking for one that is practical to use (and fast)? Including division by 3 will usually improve the number of f-applications needed, but not always. You can verify manually that for 33, 69, 129, 141, 177, etc. it won't, and if needed you can add these as special cases. If you add division by 5, it will usually worsen the situation, but for e.g. 85 and 95 it will improve it. – Szabolcs Feb 29 '12 at 9:37
@David I am not at all sure that there is a simple enough way or closed formula to figure out the optimal solution. – Szabolcs Feb 29 '12 at 9:43
This solution would be easily generalized for larger primes (using Fold), but that makes it slower according to my testing. Adding 5 results in another step necessary, adding up to the 19th prime makes calculations take many many steps (although still being much better than brute force). – David Feb 29 '12 at 21:31
@David No, don't use Fold: f[pow[4][a],a] (assuming binary is implemented) is better than f[f[f[f[a, a], a], a], a] with my current solution. Also, bigger primes will not give better results int he majority of cases. The first case where 5 gives a better result is 85. – Szabolcs Feb 29 '12 at 21:40
Here's a functional implementation of the binary method. The Fold operates only Log[2,n] times, performing either one or two "multiplications" each time.
power[f_][a_, n_Integer] := With[{b = Reverse@IntegerDigits[n, 2]},
Last@Fold[
With[{p = f[#1[[1]], #1[[1]]]}, {p,
If[#2 == 1,
If[#1[[2]] === Null, p, f[p, #1[[2]]]], #1[[2]]]}] &,
{a, If[First@b == 1, a, Null]}, Rest@b]]
Here's an example, using a pure function which logs the multiplications it is asked to perform as well as actually doing them. Of course, f could be any associative dyadic operation, and a could be any suitable expression, not just a scalar as shown.
power[(Print["(", #1, "\[Times]", #2, ")"]; Times[#1, #2]) &][a, 6]
This will output $a^6$, but also print three lines $(a \times a)$, $(a^2 \times a^2)$ and $(a^4 \times a^2)$.
I've used Null as a placeholder to mean "no result yet". One could certainly add an argument to specify the actual identity element of f instead. Special-casing the least significant bit of n avoids doing an extra unnecessary squaring of the $a^{2^k}$ at the very end.
-
Binary is so passè, try the Zeckendorf representation. – rcollyer Feb 29 '12 at 2:23
The answer depends on the structure of the algebra.
I therefore cannot give a universal answer. Instead, I will illustrate the structural dependence by means of examples. In response to the request for implementation, they also illustrate one possible way to exploit Mathematica's support for symbolic computations. (I welcome comments suggesting clearer or more efficient approaches.)
Take, for instance, the second example of the question. Here is an abbreviated implementation of the wedge product of alternating bilinear forms, enough to get us going:
SetAttributes[Wedge, Flat];
Wedge[element[\[Omega]___], element[\[Eta]___]] :=
With[{\[Phi] = Flatten[{\[Omega], \[Eta]}]},
signature[\[Phi]] (element @@ Union[\[Phi]])
];
Wedge /: Wedge[Times[x_, \[Omega]_element], y_] := Times[x, Wedge[\[Omega], y]];
Wedge /: Wedge[x_, Times[y_, \[Omega]_element]] := Times[y, Wedge[x, \[Omega]]];
Wedge[\[Omega]_, 0] := 0;
Wedge[0, \[Omega]_] := 0;
It requires calculation of the signature of a permutation of arbitrary sortable objects, extended to have the value $0$ whenever there's a repetition. Here's a reasonably efficient implementation:
signature[p_List] :=
With[{c = First[PermutationCycles[Ordering[p]] /. Cycles -> List]},
(-1)^(Length[Flatten[c]] - Length[c])
];
signature[p_List] /; Length[Union[p]] != Length[p] := 0;
More generally, in an arbitrary algebra the product will be computed by means of a matrix of structure constants specifying the product of any two elements in a given basis. The foregoing is a programmatically simple way to provide those structure constants for $\Lambda(V)$ for any finite-dimensional vector space $V$ (note that the dimension of $\Lambda(V^n)$ is $2^n$.) For efficiency (in low dimensions, anyway) we could compute these structure constants once and for all and cache them.
The trick to computing powers in algebras with lots of zeros and other symmetries in their structure matrices is to simplify as you go, as here:
Clear[WedgePower];
WedgePower[\[Omega]_, n_] /; n >= 2 :=
Nest[Distribute[Wedge[\[Omega], #], Plus, Wedge] &, \[Omega], n - 1];
WedgePower[\[Omega]_, 1] := \[Omega];
WedgePower[\[Omega]_, 0] := 1;
Distribute is implicitly doing the work of assembling pairs of basis elements to be multiplied and then collecting comparable terms, such as $dx_1\wedge dx_2 = -dx_2\wedge dx_1$, and summing their coefficients. As this is built in to Mathematica, we can expect it to be reasonably efficient.
Take, for instance, an arbitrary 2-form $\omega$ in $\Lambda(V^{10})$. It will have up to $\binom{10}{2} = 45$ nonzero coefficients; its square will have up to $\binom{10}{4} = 210$ nonzero coefficients; its cube, $210$; its fourth power, $45$; and its fifth power must be a multiple of a single form. Because you're working in an algebra of dimension $2^{10}=1024$, in principle each successive product requires $(2^{10})^2$, or over a million, calculations (although most of them will be products of zeros). Here, the bilinearity hugely reduces that number: the calculations needed to compute generic squares, cubes, fourth, and fifth powers of a 2-form are $2025$, $11475$, $20925$, and $22950$, respectively.
Trying to save time by breaking up powers into smaller groups (via the "binary method" or otherwise) can backfire. Continuing this example, computing $\omega^4$ via successive squaring requires first $2025$ operations to compute $\omega^2$ and then $44100$ to square that, a total of $46125$ products. Compare this to the value of $20925$ needed for successive wedging by $\omega$ itself: it will take more than twice as much effort.
By choosing an appropriate basis, the Liouville form can be written in a special way as $p_1 \wedge q_1 + \cdots + p_k \wedge q_k$. This reduces the calculations still further. With $k=5$ (working once more in a ten-dimensional space), for instance, the powers require only 25, 50, 50, and 25 wedge operations, respectively. This suggests yet another line of attack: before computing powers of an element of an algebra, change the basis if possible to simplify the calculation. Once again, though, how this is done will depend on the algebraic structure.
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https://www.knowpia.com/knowpedia/Binomial_coefficient | 1,713,678,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00606.warc.gz | 783,658,659 | 77,968 | BREAKING NEWS
Binomial coefficient
## Summary
In mathematics, the binomial coefficients are the positive integers that occur as coefficients in the binomial theorem. Commonly, a binomial coefficient is indexed by a pair of integers nk ≥ 0 and is written ${\displaystyle {\tbinom {n}{k}}.}$ It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n; this coefficient can be computed by the multiplicative formula
${\displaystyle {\binom {n}{k}}={\frac {n\times (n-1)\times \cdots \times (n-k+1)}{k\times (k-1)\times \cdots \times 1}},}$
which using factorial notation can be compactly expressed as
${\displaystyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}.}$
For example, the fourth power of 1 + x is
{\displaystyle {\begin{aligned}(1+x)^{4}&={\tbinom {4}{0}}x^{0}+{\tbinom {4}{1}}x^{1}+{\tbinom {4}{2}}x^{2}+{\tbinom {4}{3}}x^{3}+{\tbinom {4}{4}}x^{4}\\&=1+4x+6x^{2}+4x^{3}+x^{4},\end{aligned}}}
and the binomial coefficient ${\displaystyle {\tbinom {4}{2}}={\tfrac {4\times 3}{2\times 1}}={\tfrac {4!}{2!2!}}=6}$ is the coefficient of the x2 term.
Arranging the numbers ${\displaystyle {\tbinom {n}{0}},{\tbinom {n}{1}},\ldots ,{\tbinom {n}{n}}}$ in successive rows for n = 0, 1, 2, ... gives a triangular array called Pascal's triangle, satisfying the recurrence relation
${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}.}$
The binomial coefficients occur in many areas of mathematics, and especially in combinatorics. The symbol ${\displaystyle {\tbinom {n}{k}}}$ is usually read as "n choose k" because there are ${\displaystyle {\tbinom {n}{k}}}$ ways to choose an (unordered) subset of k elements from a fixed set of n elements. For example, there are ${\displaystyle {\tbinom {4}{2}}=6}$ ways to choose 2 elements from {1, 2, 3, 4}, namely {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4} and {3, 4}.
The binomial coefficients can be generalized to ${\displaystyle {\tbinom {z}{k}}}$ for any complex number z and integer k ≥ 0, and many of their properties continue to hold in this more general form.
## History and notation
Andreas von Ettingshausen introduced the notation ${\displaystyle {\tbinom {n}{k}}}$ in 1826,[1] although the numbers were known centuries earlier (see Pascal's triangle). In about 1150, the Indian mathematician Bhaskaracharya gave an exposition of binomial coefficients in his book Līlāvatī.[2]
Alternative notations include C(n, k), nCk, nCk, Ck
n
,[3] Cn
k
, and Cn,k, in all of which the C stands for combinations or choices. Many calculators use variants of the C notation because they can represent it on a single-line display. In this form the binomial coefficients are easily compared to k-permutations of n, written as P(n, k), etc.
## Definition and interpretations
k
n
0 1 2 3 4
0 1 0 0 0 0
1 1 1 0 0 0
2 1 2 1 0 0
3 1 3 3 1 0
4 1 4 6 4 1
The first few binomial coefficients
on a left-aligned Pascal's triangle
For natural numbers (taken to include 0) n and k, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ can be defined as the coefficient of the monomial Xk in the expansion of (1 + X)n. The same coefficient also occurs (if kn) in the binomial formula
${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}y^{n-k}}$
()
(valid for any elements x, y of a commutative ring), which explains the name "binomial coefficient".
Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects; more formally, the number of k-element subsets (or k-combinations) of an n-element set. This number can be seen as equal to the one of the first definition, independently of any of the formulas below to compute it: if in each of the n factors of the power (1 + X)n one temporarily labels the term X with an index i (running from 1 to n), then each subset of k indices gives after expansion a contribution Xk, and the coefficient of that monomial in the result will be the number of such subsets. This shows in particular that ${\displaystyle {\tbinom {n}{k}}}$ is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by ${\displaystyle {\tbinom {n}{k}}}$ , while the number of ways to write ${\displaystyle k=a_{1}+a_{2}+\cdots +a_{n}}$ where every ai is a nonnegative integer is given by ${\displaystyle {\tbinom {n+k-1}{n-1}}}$ . Most of these interpretations can be shown to be equivalent to counting k-combinations.
## Computing the value of binomial coefficients
Several methods exist to compute the value of ${\displaystyle {\tbinom {n}{k}}}$ without actually expanding a binomial power or counting k-combinations.
### Recursive formula
One method uses the recursive, purely additive formula
${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}}$
for all integers ${\displaystyle n,k}$ such that ${\displaystyle 1\leq k with boundary values
${\displaystyle {\binom {n}{0}}={\binom {n}{n}}=1}$
for all integers n ≥ 0.
The formula follows from considering the set {1, 2, 3, ..., n} and counting separately (a) the k-element groupings that include a particular set element, say "i", in every group (since "i" is already chosen to fill one spot in every group, we need only choose k − 1 from the remaining n − 1) and (b) all the k-groupings that don't include "i"; this enumerates all the possible k-combinations of n elements. It also follows from tracing the contributions to Xk in (1 + X)n−1(1 + X). As there is zero Xn+1 or X−1 in (1 + X)n, one might extend the definition beyond the above boundaries to include ${\displaystyle {\tbinom {n}{k}}=0}$ when either k > n or k < 0. This recursive formula then allows the construction of Pascal's triangle, surrounded by white spaces where the zeros, or the trivial coefficients, would be.
### Multiplicative formula
A more efficient method to compute individual binomial coefficients is given by the formula
${\displaystyle {\binom {n}{k}}={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n+1-i}{i}},}$
where the numerator of the first fraction ${\displaystyle n^{\underline {k}}}$ is expressed as a falling factorial power. This formula is easiest to understand for the combinatorial interpretation of binomial coefficients. The numerator gives the number of ways to select a sequence of k distinct objects, retaining the order of selection, from a set of n objects. The denominator counts the number of distinct sequences that define the same k-combination when order is disregarded.
Due to the symmetry of the binomial coefficient with regard to k and nk, calculation may be optimised by setting the upper limit of the product above to the smaller of k and nk.
### Factorial formula
Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function:
${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\text{for }}\ 0\leq k\leq n,}$
where n! denotes the factorial of n. This formula follows from the multiplicative formula above by multiplying numerator and denominator by (nk)!; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that k is small and n is large) unless common factors are first cancelled (in particular since factorial values grow very rapidly). The formula does exhibit a symmetry that is less evident from the multiplicative formula (though it is from the definitions)
${\displaystyle {\binom {n}{k}}={\binom {n}{n-k}}\quad {\text{for }}\ 0\leq k\leq n,}$
(1)
which leads to a more efficient multiplicative computational routine. Using the falling factorial notation,
${\displaystyle {\binom {n}{k}}={\begin{cases}n^{\underline {k}}/k!&{\text{if }}\ k\leq {\frac {n}{2}}\\n^{\underline {n-k}}/(n-k)!&{\text{if }}\ k>{\frac {n}{2}}\end{cases}}.}$
### Generalization and connection to the binomial series
The multiplicative formula allows the definition of binomial coefficients to be extended[4] by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible:
${\displaystyle {\binom {\alpha }{k}}={\frac {\alpha ^{\underline {k}}}{k!}}={\frac {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k(k-1)(k-2)\cdots 1}}\quad {\text{for }}k\in \mathbb {N} {\text{ and arbitrary }}\alpha .}$
With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the ${\displaystyle {\tbinom {\alpha }{k}}}$ binomial coefficients:
${\displaystyle (1+X)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}X^{k}.}$
(2)
This formula is valid for all complex numbers α and X with |X| < 1. It can also be interpreted as an identity of formal power series in X, where it actually can serve as definition of arbitrary powers of power series with constant coefficient equal to 1; the point is that with this definition all identities hold that one expects for exponentiation, notably
${\displaystyle (1+X)^{\alpha }(1+X)^{\beta }=(1+X)^{\alpha +\beta }\quad {\text{and}}\quad ((1+X)^{\alpha })^{\beta }=(1+X)^{\alpha \beta }.}$
If α is a nonnegative integer n, then all terms with k > n are zero, and the infinite series becomes a finite sum, thereby recovering the binomial formula. However, for other values of α, including negative integers and rational numbers, the series is really infinite.
## Pascal's triangle
Pascal's rule is the important recurrence relation
${\displaystyle {n \choose k}+{n \choose k+1}={n+1 \choose k+1},}$
(3)
which can be used to prove by mathematical induction that ${\displaystyle {\tbinom {n}{k}}}$ is a natural number for all integer n ≥ 0 and all integer k, a fact that is not immediately obvious from formula (1). To the left and right of Pascal's triangle, the entries (shown as blanks) are all zero.
Pascal's rule also gives rise to Pascal's triangle:
0: 1 1: 1 1 2: 1 2 1 3: 1 3 3 1 4: 1 4 6 4 1 5: 1 5 10 10 5 1 6: 1 6 15 20 15 6 1 7: 1 7 21 35 35 21 7 1 8: 1 8 28 56 70 56 28 8 1
Row number n contains the numbers ${\displaystyle {\tbinom {n}{k}}}$ for k = 0, …, n. It is constructed by first placing 1s in the outermost positions, and then filling each inner position with the sum of the two numbers directly above. This method allows the quick calculation of binomial coefficients without the need for fractions or multiplications. For instance, by looking at row number 5 of the triangle, one can quickly read off that
${\displaystyle (x+y)^{5}={\underline {1}}x^{5}+{\underline {5}}x^{4}y+{\underline {10}}x^{3}y^{2}+{\underline {10}}x^{2}y^{3}+{\underline {5}}xy^{4}+{\underline {1}}y^{5}.}$
## Combinatorics and statistics
Binomial coefficients are of importance in combinatorics, because they provide ready formulas for certain frequent counting problems:
• There are ${\displaystyle {\tbinom {n}{k}}}$ ways to choose k elements from a set of n elements. See Combination.
• There are ${\displaystyle {\tbinom {n+k-1}{k}}}$ ways to choose k elements from a set of n elements if repetitions are allowed. See Multiset.
• There are ${\displaystyle {\tbinom {n+k}{k}}}$ strings containing k ones and n zeros.
• There are ${\displaystyle {\tbinom {n+1}{k}}}$ strings consisting of k ones and n zeros such that no two ones are adjacent.[5]
• The Catalan numbers are ${\displaystyle {\tfrac {1}{n+1}}{\tbinom {2n}{n}}.}$
• The binomial distribution in statistics is ${\displaystyle {\tbinom {n}{k}}p^{k}(1-p)^{n-k}.}$
## Binomial coefficients as polynomials
For any nonnegative integer k, the expression ${\textstyle {\binom {t}{k}}}$ can be simplified and defined as a polynomial divided by k!:
${\displaystyle {\binom {t}{k}}={\frac {t^{\underline {k}}}{k!}}={\frac {t(t-1)(t-2)\cdots (t-k+1)}{k(k-1)(k-2)\cdots 2\cdot 1}};}$
this presents a polynomial in t with rational coefficients.
As such, it can be evaluated at any real or complex number t to define binomial coefficients with such first arguments. These "generalized binomial coefficients" appear in Newton's generalized binomial theorem.
For each k, the polynomial ${\displaystyle {\tbinom {t}{k}}}$ can be characterized as the unique degree k polynomial p(t) satisfying p(0) = p(1) = ⋯ = p(k − 1) = 0 and p(k) = 1.
Its coefficients are expressible in terms of Stirling numbers of the first kind:
${\displaystyle {\binom {t}{k}}=\sum _{i=0}^{k}s(k,i){\frac {t^{i}}{k!}}.}$
The derivative of ${\displaystyle {\tbinom {t}{k}}}$ can be calculated by logarithmic differentiation:
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {t}{k}}={\binom {t}{k}}\sum _{i=0}^{k-1}{\frac {1}{t-i}}.}$
This can cause a problem when evaluated at integers from ${\displaystyle 0}$ to ${\displaystyle t-1}$ , but using identities below we can compute the derivative as:
${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}{\binom {t}{k}}=\sum _{i=0}^{k-1}{\frac {(-1)^{k-i-1}}{k-i}}{\binom {t}{i}}.}$
### Binomial coefficients as a basis for the space of polynomials
Over any field of characteristic 0 (that is, any field that contains the rational numbers), each polynomial p(t) of degree at most d is uniquely expressible as a linear combination ${\textstyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}}$ of binomial coefficients. The coefficient ak is the kth difference of the sequence p(0), p(1), ..., p(k). Explicitly,[6]
${\displaystyle a_{k}=\sum _{i=0}^{k}(-1)^{k-i}{\binom {k}{i}}p(i).}$
(4)
### Integer-valued polynomials
Each polynomial ${\displaystyle {\tbinom {t}{k}}}$ is integer-valued: it has an integer value at all integer inputs ${\displaystyle t}$ . (One way to prove this is by induction on k, using Pascal's identity.) Therefore, any integer linear combination of binomial coefficient polynomials is integer-valued too. Conversely, (4) shows that any integer-valued polynomial is an integer linear combination of these binomial coefficient polynomials. More generally, for any subring R of a characteristic 0 field K, a polynomial in K[t] takes values in R at all integers if and only if it is an R-linear combination of binomial coefficient polynomials.
### Example
The integer-valued polynomial 3t(3t + 1) / 2 can be rewritten as
${\displaystyle 9{\binom {t}{2}}+6{\binom {t}{1}}+0{\binom {t}{0}}.}$
## Identities involving binomial coefficients
The factorial formula facilitates relating nearby binomial coefficients. For instance, if k is a positive integer and n is arbitrary, then
${\displaystyle {\binom {n}{k}}={\frac {n}{k}}{\binom {n-1}{k-1}}}$
(5)
and, with a little more work,
${\displaystyle {\binom {n-1}{k}}-{\binom {n-1}{k-1}}={\frac {n-2k}{n}}{\binom {n}{k}}.}$
We can also get
${\displaystyle {\binom {n-1}{k}}={\frac {n-k}{n}}{\binom {n}{k}}.}$
Moreover, the following may be useful:
${\displaystyle {\binom {n}{h}}{\binom {n-h}{k}}={\binom {n}{k}}{\binom {n-k}{h}}={\binom {n}{h+k}}{\binom {h+k}{h}}.}$
For constant n, we have the following recurrence:
${\displaystyle {\binom {n}{k}}={\frac {n-k+1}{k}}{\binom {n}{k-1}}.}$
To sum up, we have
${\displaystyle {\binom {n}{k}}={\binom {n}{n-k}}={\frac {n-k+1}{k}}{\binom {n}{k-1}}={\frac {n}{n-k}}{\binom {n-1}{k}}={\frac {n}{k}}{\binom {n-1}{k-1}}={\frac {n}{n-2k}}{\Bigg (}{\binom {n-1}{k}}-{\binom {n-1}{k-1}}{\Bigg )}={\binom {n-1}{k}}+{\binom {n-1}{k-1}}.}$
### Sums of the binomial coefficients
The formula
${\displaystyle \sum _{k=0}^{n}{\binom {n}{k}}=2^{n}}$
(∗∗)
says that the elements in the nth row of Pascal's triangle always add up to 2 raised to the nth power. This is obtained from the binomial theorem () by setting x = 1 and y = 1. The formula also has a natural combinatorial interpretation: the left side sums the number of subsets of {1, ..., n} of sizes k = 0, 1, ..., n, giving the total number of subsets. (That is, the left side counts the power set of {1, ..., n}.) However, these subsets can also be generated by successively choosing or excluding each element 1, ..., n; the n independent binary choices (bit-strings) allow a total of ${\displaystyle 2^{n}}$ choices. The left and right sides are two ways to count the same collection of subsets, so they are equal.
The formulas
${\displaystyle \sum _{k=0}^{n}k{\binom {n}{k}}=n2^{n-1}}$
(6)
and
${\displaystyle \sum _{k=0}^{n}k^{2}{\binom {n}{k}}=(n+n^{2})2^{n-2}}$
follow from the binomial theorem after differentiating with respect to x (twice for the latter) and then substituting x = y = 1.
The Chu–Vandermonde identity, which holds for any complex values m and n and any non-negative integer k, is
${\displaystyle \sum _{j=0}^{k}{\binom {m}{j}}{\binom {n-m}{k-j}}={\binom {n}{k}}}$
(7)
and can be found by examination of the coefficient of ${\displaystyle x^{k}}$ in the expansion of (1 + x)m(1 + x)nm = (1 + x)n using equation (2). When m = 1, equation (7) reduces to equation (3). In the special case n = 2m, k = m, using (1), the expansion (7) becomes (as seen in Pascal's triangle at right)
${\displaystyle {\begin{array}{c}1\\1\qquad 1\\1\qquad 2\qquad 1\\{\color {blue}1\qquad 3\qquad 3\qquad 1}\\1\qquad 4\qquad 6\qquad 4\qquad 1\\1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1\\1\qquad 6\qquad 15\qquad {\color {red}20}\qquad 15\qquad 6\qquad 1\\1\qquad 7\qquad 21\qquad 35\qquad 35\qquad 21\qquad 7\qquad 1\end{array}}}$
Pascal's triangle, rows 0 through 7. Equation 8 for m = 3 is illustrated in rows 3 and 6 as ${\displaystyle 1^{2}+3^{2}+3^{2}+1^{2}=20.}$
${\displaystyle \sum _{j=0}^{m}{\binom {m}{j}}^{2}={\binom {2m}{m}},}$
(8)
where the term on the right side is a central binomial coefficient.
Another form of the Chu–Vandermonde identity, which applies for any integers j, k, and n satisfying 0 ≤ jkn, is
${\displaystyle \sum _{m=0}^{n}{\binom {m}{j}}{\binom {n-m}{k-j}}={\binom {n+1}{k+1}}.}$
(9)
The proof is similar, but uses the binomial series expansion (2) with negative integer exponents. When j = k, equation (9) gives the hockey-stick identity
${\displaystyle \sum _{m=k}^{n}{\binom {m}{k}}={\binom {n+1}{k+1}}}$
and its relative
${\displaystyle \sum _{r=0}^{m}{\binom {n+r}{r}}={\binom {n+m+1}{m}}.}$
Let F(n) denote the n-th Fibonacci number. Then
${\displaystyle \sum _{k=0}^{\lfloor n/2\rfloor }{\binom {n-k}{k}}=F(n+1).}$
This can be proved by induction using (3) or by Zeckendorf's representation. A combinatorial proof is given below.
#### Multisections of sums
For integers s and t such that ${\displaystyle 0\leq t series multisection gives the following identity for the sum of binomial coefficients:
${\displaystyle {\binom {n}{t}}+{\binom {n}{t+s}}+{\binom {n}{t+2s}}+\ldots ={\frac {1}{s}}\sum _{j=0}^{s-1}\left(2\cos {\frac {\pi j}{s}}\right)^{n}\cos {\frac {\pi (n-2t)j}{s}}.}$
For small s, these series have particularly nice forms; for example,[7]
${\displaystyle {\binom {n}{0}}+{\binom {n}{3}}+{\binom {n}{6}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {n\pi }{3}}\right)}$
${\displaystyle {\binom {n}{1}}+{\binom {n}{4}}+{\binom {n}{7}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {(n-2)\pi }{3}}\right)}$
${\displaystyle {\binom {n}{2}}+{\binom {n}{5}}+{\binom {n}{8}}+\cdots ={\frac {1}{3}}\left(2^{n}+2\cos {\frac {(n-4)\pi }{3}}\right)}$
${\displaystyle {\binom {n}{0}}+{\binom {n}{4}}+{\binom {n}{8}}+\cdots ={\frac {1}{2}}\left(2^{n-1}+2^{\frac {n}{2}}\cos {\frac {n\pi }{4}}\right)}$
${\displaystyle {\binom {n}{1}}+{\binom {n}{5}}+{\binom {n}{9}}+\cdots ={\frac {1}{2}}\left(2^{n-1}+2^{\frac {n}{2}}\sin {\frac {n\pi }{4}}\right)}$
${\displaystyle {\binom {n}{2}}+{\binom {n}{6}}+{\binom {n}{10}}+\cdots ={\frac {1}{2}}\left(2^{n-1}-2^{\frac {n}{2}}\cos {\frac {n\pi }{4}}\right)}$
${\displaystyle {\binom {n}{3}}+{\binom {n}{7}}+{\binom {n}{11}}+\cdots ={\frac {1}{2}}\left(2^{n-1}-2^{\frac {n}{2}}\sin {\frac {n\pi }{4}}\right)}$
#### Partial sums
Although there is no closed formula for partial sums
${\displaystyle \sum _{j=0}^{k}{\binom {n}{j}}}$
of binomial coefficients,[8] one can again use (3) and induction to show that for k = 0, …, n − 1,
${\displaystyle \sum _{j=0}^{k}(-1)^{j}{\binom {n}{j}}=(-1)^{k}{\binom {n-1}{k}},}$
with special case[9]
${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}=0}$
for n > 0. This latter result is also a special case of the result from the theory of finite differences that for any polynomial P(x) of degree less than n,[10]
${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(j)=0.}$
Differentiating (2) k times and setting x = −1 yields this for ${\displaystyle P(x)=x(x-1)\cdots (x-k+1)}$ , when 0 ≤ k < n, and the general case follows by taking linear combinations of these.
When P(x) is of degree less than or equal to n,
${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(n-j)=n!a_{n}}$
(10)
where ${\displaystyle a_{n}}$ is the coefficient of degree n in P(x).
More generally for (10),
${\displaystyle \sum _{j=0}^{n}(-1)^{j}{\binom {n}{j}}P(m+(n-j)d)=d^{n}n!a_{n}}$
where m and d are complex numbers. This follows immediately applying (10) to the polynomial ${\displaystyle Q(x):=P(m+dx)}$ instead of ${\displaystyle P(x)}$ , and observing that ${\displaystyle Q(x)}$ still has degree less than or equal to n, and that its coefficient of degree n is dnan.
The series ${\textstyle {\frac {k-1}{k}}\sum _{j=0}^{\infty }{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}}$ is convergent for k ≥ 2. This formula is used in the analysis of the German tank problem. It follows from ${\textstyle {\frac {k-1}{k}}\sum _{j=0}^{M}{\frac {1}{\binom {j+x}{k}}}={\frac {1}{\binom {x-1}{k-1}}}-{\frac {1}{\binom {M+x}{k-1}}}}$ which is proved by induction on M.
### Identities with combinatorial proofs
Many identities involving binomial coefficients can be proved by combinatorial means. For example, for nonnegative integers ${\displaystyle {n}\geq {q}}$ , the identity
${\displaystyle \sum _{k=q}^{n}{\binom {n}{k}}{\binom {k}{q}}=2^{n-q}{\binom {n}{q}}}$
(which reduces to (6) when q = 1) can be given a double counting proof, as follows. The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. The right side counts the same thing, because there are ${\displaystyle {\tbinom {n}{q}}}$ ways of choosing a set of q elements to mark, and ${\displaystyle 2^{n-q}}$ to choose which of the remaining elements of [n] also belong to the subset.
In Pascal's identity
${\displaystyle {n \choose k}={n-1 \choose k-1}+{n-1 \choose k},}$
both sides count the number of k-element subsets of [n]: the two terms on the right side group them into those that contain element n and those that do not.
The identity (8) also has a combinatorial proof. The identity reads
${\displaystyle \sum _{k=0}^{n}{\binom {n}{k}}^{2}={\binom {2n}{n}}.}$
Suppose you have ${\displaystyle 2n}$ empty squares arranged in a row and you want to mark (select) n of them. There are ${\displaystyle {\tbinom {2n}{n}}}$ ways to do this. On the other hand, you may select your n squares by selecting k squares from among the first n and ${\displaystyle n-k}$ squares from the remaining n squares; any k from 0 to n will work. This gives
${\displaystyle \sum _{k=0}^{n}{\binom {n}{k}}{\binom {n}{n-k}}={\binom {2n}{n}}.}$
Now apply (1) to get the result.
If one denotes by F(i) the sequence of Fibonacci numbers, indexed so that F(0) = F(1) = 1, then the identity
${\displaystyle \sum _{k=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }{\binom {n-k}{k}}=F(n)}$
has the following combinatorial proof.[11] One may show by induction that F(n) counts the number of ways that a n × 1 strip of squares may be covered by 2 × 1 and 1 × 1 tiles. On the other hand, if such a tiling uses exactly k of the 2 × 1 tiles, then it uses n − 2k of the 1 × 1 tiles, and so uses nk tiles total. There are ${\displaystyle {\tbinom {n-k}{k}}}$ ways to order these tiles, and so summing this coefficient over all possible values of k gives the identity.
#### Sum of coefficients row
The number of k-combinations for all k, ${\textstyle \sum _{0\leq {k}\leq {n}}{\binom {n}{k}}=2^{n}}$ , is the sum of the nth row (counting from 0) of the binomial coefficients. These combinations are enumerated by the 1 digits of the set of base 2 numbers counting from 0 to ${\displaystyle 2^{n}-1}$ , where each digit position is an item from the set of n.
### Dixon's identity
${\displaystyle \sum _{k=-a}^{a}(-1)^{k}{2a \choose k+a}^{3}={\frac {(3a)!}{(a!)^{3}}}}$
or, more generally,
${\displaystyle \sum _{k=-a}^{a}(-1)^{k}{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}={\frac {(a+b+c)!}{a!\,b!\,c!}}\,,}$
where a, b, and c are non-negative integers.
### Continuous identities
Certain trigonometric integrals have values expressible in terms of binomial coefficients: For any ${\displaystyle m,n\in \mathbb {N} ,}$
${\displaystyle \int _{-\pi }^{\pi }\cos((2m-n)x)\cos ^{n}(x)\ dx={\frac {\pi }{2^{n-1}}}{\binom {n}{m}}}$
${\displaystyle \int _{-\pi }^{\pi }\sin((2m-n)x)\sin ^{n}(x)\ dx={\begin{cases}(-1)^{m+(n+1)/2}{\frac {\pi }{2^{n-1}}}{\binom {n}{m}},&n{\text{ odd}}\\0,&{\text{otherwise}}\end{cases}}}$
${\displaystyle \int _{-\pi }^{\pi }\cos((2m-n)x)\sin ^{n}(x)\ dx={\begin{cases}(-1)^{m+(n/2)}{\frac {\pi }{2^{n-1}}}{\binom {n}{m}},&n{\text{ even}}\\0,&{\text{otherwise}}\end{cases}}}$
These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term.
### Congruences
If n is prime, then
${\displaystyle {\binom {n-1}{k}}\equiv (-1)^{k}\mod n}$
for every k with ${\displaystyle 0\leq k\leq n-1.}$ More generally, this remains true if n is any number and k is such that all the numbers between 1 and k are coprime to n.
Indeed, we have
${\displaystyle {\binom {n-1}{k}}={(n-1)(n-2)\cdots (n-k) \over 1\cdot 2\cdots k}=\prod _{i=1}^{k}{n-i \over i}\equiv \prod _{i=1}^{k}{-i \over i}=(-1)^{k}\mod n.}$
## Generating functions
### Ordinary generating functions
For a fixed n, the ordinary generating function of the sequence ${\displaystyle {\tbinom {n}{0}},{\tbinom {n}{1}},{\tbinom {n}{2}},\ldots }$ is
${\displaystyle \sum _{k=0}^{\infty }{n \choose k}x^{k}=(1+x)^{n}.}$
For a fixed k, the ordinary generating function of the sequence ${\displaystyle {\tbinom {0}{k}},{\tbinom {1}{k}},{\tbinom {2}{k}},\ldots ,}$ is
${\displaystyle \sum _{n=0}^{\infty }{n \choose k}y^{n}={\frac {y^{k}}{(1-y)^{k+1}}}.}$
The bivariate generating function of the binomial coefficients is
${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{n}{n \choose k}x^{k}y^{n}={\frac {1}{1-y-xy}}.}$
A symmetric bivariate generating function of the binomial coefficients is
${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{n+k \choose k}x^{k}y^{n}={\frac {1}{1-x-y}}.}$
which is the same as the previous generating function after the substitution ${\displaystyle x\to xy}$ .
### Exponential generating function
A symmetric exponential bivariate generating function of the binomial coefficients is:
${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{\infty }{n+k \choose k}{\frac {x^{k}y^{n}}{(n+k)!}}=e^{x+y}.}$
## Divisibility properties
In 1852, Kummer proved that if m and n are nonnegative integers and p is a prime number, then the largest power of p dividing ${\displaystyle {\tbinom {m+n}{m}}}$ equals pc, where c is the number of carries when m and n are added in base p. Equivalently, the exponent of a prime p in ${\displaystyle {\tbinom {n}{k}}}$ equals the number of nonnegative integers j such that the fractional part of k/pj is greater than the fractional part of n/pj. It can be deduced from this that ${\displaystyle {\tbinom {n}{k}}}$ is divisible by n/gcd(n,k). In particular therefore it follows that p divides ${\displaystyle {\tbinom {p^{r}}{s}}}$ for all positive integers r and s such that s < pr. However this is not true of higher powers of p: for example 9 does not divide ${\displaystyle {\tbinom {9}{6}}}$ .
A somewhat surprising result by David Singmaster (1974) is that any integer divides almost all binomial coefficients. More precisely, fix an integer d and let f(N) denote the number of binomial coefficients ${\displaystyle {\tbinom {n}{k}}}$ with n < N such that d divides ${\displaystyle {\tbinom {n}{k}}}$ . Then
${\displaystyle \lim _{N\to \infty }{\frac {f(N)}{N(N+1)/2}}=1.}$
Since the number of binomial coefficients ${\displaystyle {\tbinom {n}{k}}}$ with n < N is N(N + 1) / 2, this implies that the density of binomial coefficients divisible by d goes to 1.
Binomial coefficients have divisibility properties related to least common multiples of consecutive integers. For example:[12]
${\displaystyle {\binom {n+k}{k}}}$ divides ${\displaystyle {\frac {\operatorname {lcm} (n,n+1,\ldots ,n+k)}{n}}}$ .
${\displaystyle {\binom {n+k}{k}}}$ is a multiple of ${\displaystyle {\frac {\operatorname {lcm} (n,n+1,\ldots ,n+k)}{n\cdot \operatorname {lcm} ({\binom {k}{0}},{\binom {k}{1}},\ldots ,{\binom {k}{k}})}}}$ .
Another fact: An integer n ≥ 2 is prime if and only if all the intermediate binomial coefficients
${\displaystyle {\binom {n}{1}},{\binom {n}{2}},\ldots ,{\binom {n}{n-1}}}$
are divisible by n.
Proof: When p is prime, p divides
${\displaystyle {\binom {p}{k}}={\frac {p\cdot (p-1)\cdots (p-k+1)}{k\cdot (k-1)\cdots 1}}}$ for all 0 < k < p
because ${\displaystyle {\tbinom {p}{k}}}$ is a natural number and p divides the numerator but not the denominator. When n is composite, let p be the smallest prime factor of n and let k = n/p. Then 0 < p < n and
${\displaystyle {\binom {n}{p}}={\frac {n(n-1)(n-2)\cdots (n-p+1)}{p!}}={\frac {k(n-1)(n-2)\cdots (n-p+1)}{(p-1)!}}\not \equiv 0{\pmod {n}}}$
otherwise the numerator k(n − 1)(n − 2)⋯(np + 1) has to be divisible by n = k×p, this can only be the case when (n − 1)(n − 2)⋯(np + 1) is divisible by p. But n is divisible by p, so p does not divide n − 1, n − 2, …, np + 1 and because p is prime, we know that p does not divide (n − 1)(n − 2)⋯(np + 1) and so the numerator cannot be divisible by n.
## Bounds and asymptotic formulas
The following bounds for ${\displaystyle {\tbinom {n}{k}}}$ hold for all values of n and k such that 1 ≤ kn:
${\displaystyle {\frac {n^{k}}{k^{k}}}\leq {n \choose k}\leq {\frac {n^{k}}{k!}}<\left({\frac {n\cdot e}{k}}\right)^{k}.}$
The first inequality follows from the fact that
${\displaystyle {n \choose k}={\frac {n}{k}}\cdot {\frac {n-1}{k-1}}\cdots {\frac {n-(k-1)}{1}}}$
and each of these ${\displaystyle k}$ terms in this product is ${\textstyle \geq {\frac {n}{k}}}$ . A similar argument can be made to show the second inequality. The final strict inequality is equivalent to ${\textstyle e^{k}>k^{k}/k!}$ , that is clear since the RHS is a term of the exponential series ${\textstyle e^{k}=\sum _{j=0}^{\infty }k^{j}/j!}$ .
From the divisibility properties we can infer that
${\displaystyle {\frac {\operatorname {lcm} (n-k,\ldots ,n)}{(n-k)\cdot \operatorname {lcm} \left({\binom {k}{0}},\ldots ,{\binom {k}{k}}\right)}}\leq {\binom {n}{k}}\leq {\frac {\operatorname {lcm} (n-k,\ldots ,n)}{n-k}},}$
where both equalities can be achieved.[12]
The following bounds are useful in information theory:[13]: 353
${\displaystyle {\frac {1}{n+1}}2^{nH(k/n)}\leq {n \choose k}\leq 2^{nH(k/n)}}$
where ${\displaystyle H(p)=-p\log _{2}(p)-(1-p)\log _{2}(1-p)}$ is the binary entropy function. It can be further tightened to
${\displaystyle {\sqrt {\frac {n}{8k(n-k)}}}2^{nH(k/n)}\leq {n \choose k}\leq {\sqrt {\frac {n}{2\pi k(n-k)}}}2^{nH(k/n)}}$
for all ${\displaystyle 1\leq k\leq n-1}$ .[14]: 309
### Both n and k large
Stirling's approximation yields the following approximation, valid when ${\displaystyle n-k,k}$ both tend to infinity:
${\displaystyle {n \choose k}\sim {\sqrt {n \over 2\pi k(n-k)}}\cdot {n^{n} \over k^{k}(n-k)^{n-k}}}$
Because the inequality forms of Stirling's formula also bound the factorials, slight variants on the above asymptotic approximation give exact bounds. In particular, when ${\displaystyle n}$ is sufficiently large, one has
${\displaystyle {2n \choose n}\sim {\frac {2^{2n}}{\sqrt {n\pi }}}}$
and ${\displaystyle {\sqrt {n}}{2n \choose n}\geq 2^{2n-1}}$ and, more generally, for m ≥ 2 and n ≥ 1,[why?]
${\displaystyle {\sqrt {n}}{mn \choose n}\geq {\frac {m^{m(n-1)+1}}{(m-1)^{(m-1)(n-1)}}}.}$
If n is large and k is linear in n, various precise asymptotic estimates exist for the binomial coefficient ${\textstyle {\binom {n}{k}}}$ . For example, if ${\displaystyle |n/2-k|=o(n^{2/3})}$ then
${\displaystyle {\binom {n}{k}}\sim {\binom {n}{\frac {n}{2}}}e^{-d^{2}/(2n)}\sim {\frac {2^{n}}{\sqrt {{\frac {1}{2}}n\pi }}}e^{-d^{2}/(2n)}}$
where d = n − 2k.[15]
### n much larger than k
If n is large and k is o(n) (that is, if k/n → 0), then
${\displaystyle {\binom {n}{k}}\sim \left({\frac {ne}{k}}\right)^{k}\cdot (2\pi k)^{-1/2}\cdot \exp \left(-{\frac {k^{2}}{2n}}(1+o(1))\right)}$
where again o is the little o notation.[16]
### Sums of binomial coefficients
A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem:
${\displaystyle \sum _{i=0}^{k}{n \choose i}\leq \sum _{i=0}^{k}n^{i}\cdot 1^{k-i}\leq (1+n)^{k}}$
More precise bounds are given by
${\displaystyle {\frac {1}{\sqrt {8n\varepsilon (1-\varepsilon )}}}\cdot 2^{H(\varepsilon )\cdot n}\leq \sum _{i=0}^{k}{\binom {n}{i}}\leq 2^{H(\varepsilon )\cdot n},}$
valid for all integers ${\displaystyle n>k\geq 1}$ with ${\displaystyle \varepsilon \doteq k/n\leq 1/2}$ .[17]
### Generalized binomial coefficients
The infinite product formula for the gamma function also gives an expression for binomial coefficients
${\displaystyle (-1)^{k}{z \choose k}={-z+k-1 \choose k}={\frac {1}{\Gamma (-z)}}{\frac {1}{(k+1)^{z+1}}}\prod _{j=k+1}{\frac {\left(1+{\frac {1}{j}}\right)^{-z-1}}{1-{\frac {z+1}{j}}}}}$
which yields the asymptotic formulas
${\displaystyle {z \choose k}\approx {\frac {(-1)^{k}}{\Gamma (-z)k^{z+1}}}\qquad {\text{and}}\qquad {z+k \choose k}={\frac {k^{z}}{\Gamma (z+1)}}\left(1+{\frac {z(z+1)}{2k}}+{\mathcal {O}}\left(k^{-2}\right)\right)}$
as ${\displaystyle k\to \infty }$ .
This asymptotic behaviour is contained in the approximation
${\displaystyle {z+k \choose k}\approx {\frac {e^{z(H_{k}-\gamma )}}{\Gamma (z+1)}}}$
as well. (Here ${\displaystyle H_{k}}$ is the k-th harmonic number and ${\displaystyle \gamma }$ is the Euler–Mascheroni constant.)
Further, the asymptotic formula
${\displaystyle {\frac {z+k \choose j}{k \choose j}}\to \left(1-{\frac {j}{k}}\right)^{-z}\quad {\text{and}}\quad {\frac {j \choose j-k}{j-z \choose j-k}}\to \left({\frac {j}{k}}\right)^{z}}$
hold true, whenever ${\displaystyle k\to \infty }$ and ${\displaystyle j/k\to x}$ for some complex number ${\displaystyle x}$ .
## Generalizations
### Generalization to multinomials
Binomial coefficients can be generalized to multinomial coefficients defined to be the number:
${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{r}!}}}$
where
${\displaystyle \sum _{i=1}^{r}k_{i}=n.}$
While the binomial coefficients represent the coefficients of (x + y)n, the multinomial coefficients represent the coefficients of the polynomial
${\displaystyle (x_{1}+x_{2}+\cdots +x_{r})^{n}.}$
The case r = 2 gives binomial coefficients:
${\displaystyle {n \choose k_{1},k_{2}}={n \choose k_{1},n-k_{1}}={n \choose k_{1}}={n \choose k_{2}}.}$
The combinatorial interpretation of multinomial coefficients is distribution of n distinguishable elements over r (distinguishable) containers, each containing exactly ki elements, where i is the index of the container.
Multinomial coefficients have many properties similar to those of binomial coefficients, for example the recurrence relation:
${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={n-1 \choose k_{1}-1,k_{2},\ldots ,k_{r}}+{n-1 \choose k_{1},k_{2}-1,\ldots ,k_{r}}+\ldots +{n-1 \choose k_{1},k_{2},\ldots ,k_{r}-1}}$
and symmetry:
${\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{r}}={n \choose k_{\sigma _{1}},k_{\sigma _{2}},\ldots ,k_{\sigma _{r}}}}$
where ${\displaystyle (\sigma _{i})}$ is a permutation of (1, 2, ..., r).
### Taylor series
Using Stirling numbers of the first kind the series expansion around any arbitrarily chosen point ${\displaystyle z_{0}}$ is
{\displaystyle {\begin{aligned}{z \choose k}={\frac {1}{k!}}\sum _{i=0}^{k}z^{i}s_{k,i}&=\sum _{i=0}^{k}(z-z_{0})^{i}\sum _{j=i}^{k}{z_{0} \choose j-i}{\frac {s_{k+i-j,i}}{(k+i-j)!}}\\&=\sum _{i=0}^{k}(z-z_{0})^{i}\sum _{j=i}^{k}z_{0}^{j-i}{j \choose i}{\frac {s_{k,j}}{k!}}.\end{aligned}}}
### Binomial coefficient with n = 1/2
The definition of the binomial coefficients can be extended to the case where ${\displaystyle n}$ is real and ${\displaystyle k}$ is integer.
In particular, the following identity holds for any non-negative integer ${\displaystyle k}$ :
${\displaystyle {{1/2} \choose {k}}={{2k} \choose {k}}{\frac {(-1)^{k+1}}{2^{2k}(2k-1)}}.}$
This shows up when expanding ${\displaystyle {\sqrt {1+x}}}$ into a power series using the Newton binomial series :
${\displaystyle {\sqrt {1+x}}=\sum _{k\geq 0}{\binom {1/2}{k}}x^{k}.}$
### Products of binomial coefficients
One can express the product of two binomial coefficients as a linear combination of binomial coefficients:
${\displaystyle {z \choose m}{z \choose n}=\sum _{k=0}^{\min(m,n)}{m+n-k \choose k,m-k,n-k}{z \choose m+n-k},}$
where the connection coefficients are multinomial coefficients. In terms of labelled combinatorial objects, the connection coefficients represent the number of ways to assign m + nk labels to a pair of labelled combinatorial objects—of weight m and n respectively—that have had their first k labels identified, or glued together to get a new labelled combinatorial object of weight m + nk. (That is, to separate the labels into three portions to apply to the glued part, the unglued part of the first object, and the unglued part of the second object.) In this regard, binomial coefficients are to exponential generating series what falling factorials are to ordinary generating series.
The product of all binomial coefficients in the nth row of the Pascal triangle is given by the formula:
${\displaystyle \prod _{k=0}^{n}{\binom {n}{k}}=\prod _{k=1}^{n}k^{2k-n-1}.}$
### Partial fraction decomposition
The partial fraction decomposition of the reciprocal is given by
${\displaystyle {\frac {1}{z \choose n}}=\sum _{i=0}^{n-1}(-1)^{n-1-i}{n \choose i}{\frac {n-i}{z-i}},\qquad {\frac {1}{z+n \choose n}}=\sum _{i=1}^{n}(-1)^{i-1}{n \choose i}{\frac {i}{z+i}}.}$
### Newton's binomial series
Newton's binomial series, named after Sir Isaac Newton, is a generalization of the binomial theorem to infinite series:
${\displaystyle (1+z)^{\alpha }=\sum _{n=0}^{\infty }{\alpha \choose n}z^{n}=1+{\alpha \choose 1}z+{\alpha \choose 2}z^{2}+\cdots .}$
The identity can be obtained by showing that both sides satisfy the differential equation (1 + z) f'(z) = α f(z).
The radius of convergence of this series is 1. An alternative expression is
${\displaystyle {\frac {1}{(1-z)^{\alpha +1}}}=\sum _{n=0}^{\infty }{n+\alpha \choose n}z^{n}}$
where the identity
${\displaystyle {n \choose k}=(-1)^{k}{k-n-1 \choose k}}$
is applied.
### Multiset (rising) binomial coefficient
Binomial coefficients count subsets of prescribed size from a given set. A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. The resulting numbers are called multiset coefficients;[18] the number of ways to "multichoose" (i.e., choose with replacement) k items from an n element set is denoted ${\textstyle \left(\!\!{\binom {n}{k}}\!\!\right)}$ .
To avoid ambiguity and confusion with n's main denotation in this article,
let f = n = r + (k − 1) and r = f − (k − 1).
Multiset coefficients may be expressed in terms of binomial coefficients by the rule
${\displaystyle {\binom {f}{k}}=\left(\!\!{\binom {r}{k}}\!\!\right)={\binom {r+k-1}{k}}.}$
One possible alternative characterization of this identity is as follows: We may define the falling factorial as
${\displaystyle (f)_{k}=f^{\underline {k}}=(f-k+1)\cdots (f-3)\cdot (f-2)\cdot (f-1)\cdot f,}$
and the corresponding rising factorial as
${\displaystyle r^{(k)}=\,r^{\overline {k}}=\,r\cdot (r+1)\cdot (r+2)\cdot (r+3)\cdots (r+k-1);}$
so, for example,
${\displaystyle 17\cdot 18\cdot 19\cdot 20\cdot 21=(21)_{5}=21^{\underline {5}}=17^{\overline {5}}=17^{(5)}.}$
Then the binomial coefficients may be written as
${\displaystyle {\binom {f}{k}}={\frac {(f)_{k}}{k!}}={\frac {(f-k+1)\cdots (f-2)\cdot (f-1)\cdot f}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots k}},}$
while the corresponding multiset coefficient is defined by replacing the falling with the rising factorial:
${\displaystyle \left(\!\!{\binom {r}{k}}\!\!\right)={\frac {r^{(k)}}{k!}}={\frac {r\cdot (r+1)\cdot (r+2)\cdots (r+k-1)}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots k}}.}$
#### Generalization to negative integers n
For any n,
{\displaystyle {\begin{aligned}{\binom {-n}{k}}&={\frac {-n\cdot -(n+1)\dots -(n+k-2)\cdot -(n+k-1)}{k!}}\\&=(-1)^{k}\;{\frac {n\cdot (n+1)\cdot (n+2)\cdots (n+k-1)}{k!}}\\&=(-1)^{k}{\binom {n+k-1}{k}}\\&=(-1)^{k}\left(\!\!{\binom {n}{k}}\!\!\right)\;.\end{aligned}}}
In particular, binomial coefficients evaluated at negative integers n are given by signed multiset coefficients. In the special case ${\displaystyle n=-1}$ , this reduces to ${\displaystyle (-1)^{k}={\binom {-1}{k}}=\left(\!\!{\binom {-k}{k}}\!\!\right).}$
For example, if n = −4 and k = 7, then r = 4 and f = 10:
{\displaystyle {\begin{aligned}{\binom {-4}{7}}&={\frac {-10\cdot -9\cdot -8\cdot -7\cdot -6\cdot -5\cdot -4}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7}}\\&=(-1)^{7}\;{\frac {4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9\cdot 10}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7}}\\&=\left(\!\!{\binom {-7}{7}}\!\!\right)\left(\!\!{\binom {4}{7}}\!\!\right)={\binom {-1}{7}}{\binom {10}{7}}.\end{aligned}}}
### Two real or complex valued arguments
The binomial coefficient is generalized to two real or complex valued arguments using the gamma function or beta function via
${\displaystyle {x \choose y}={\frac {\Gamma (x+1)}{\Gamma (y+1)\Gamma (x-y+1)}}={\frac {1}{(x+1)\mathrm {B} (y+1,x-y+1)}}.}$
This definition inherits these following additional properties from ${\displaystyle \Gamma }$ :
${\displaystyle {x \choose y}={\frac {\sin(y\pi )}{\sin(x\pi )}}{-y-1 \choose -x-1}={\frac {\sin((x-y)\pi )}{\sin(x\pi )}}{y-x-1 \choose y};}$
moreover,
${\displaystyle {x \choose y}\cdot {y \choose x}={\frac {\sin((x-y)\pi )}{(x-y)\pi }}.}$
The resulting function has been little-studied, apparently first being graphed in (Fowler 1996). Notably, many binomial identities fail: ${\textstyle {\binom {n}{m}}={\binom {n}{n-m}}}$ but ${\textstyle {\binom {-n}{m}}\neq {\binom {-n}{-n-m}}}$ for n positive (so ${\displaystyle -n}$ negative). The behavior is quite complex, and markedly different in various octants (that is, with respect to the x and y axes and the line ${\displaystyle y=x}$ ), with the behavior for negative x having singularities at negative integer values and a checkerboard of positive and negative regions:
• in the octant ${\displaystyle 0\leq y\leq x}$ it is a smoothly interpolated form of the usual binomial, with a ridge ("Pascal's ridge").
• in the octant ${\displaystyle 0\leq x\leq y}$ and in the quadrant ${\displaystyle x\geq 0,y\leq 0}$ the function is close to zero.
• in the quadrant ${\displaystyle x\leq 0,y\geq 0}$ the function is alternatingly very large positive and negative on the parallelograms with vertices
${\displaystyle (-n,m+1),(-n,m),(-n-1,m-1),(-n-1,m)}$
• in the octant ${\displaystyle 0>x>y}$ the behavior is again alternatingly very large positive and negative, but on a square grid.
• in the octant ${\displaystyle -1>y>x+1}$ it is close to zero, except for near the singularities.
### Generalization to q-series
The binomial coefficient has a q-analog generalization known as the Gaussian binomial coefficient.
### Generalization to infinite cardinals
The definition of the binomial coefficient can be generalized to infinite cardinals by defining:
${\displaystyle {\alpha \choose \beta }=\left|\left\{B\subseteq A:\left|B\right|=\beta \right\}\right|}$
where A is some set with cardinality ${\displaystyle \alpha }$ . One can show that the generalized binomial coefficient is well-defined, in the sense that no matter what set we choose to represent the cardinal number ${\displaystyle \alpha }$ , ${\textstyle {\alpha \choose \beta }}$ will remain the same. For finite cardinals, this definition coincides with the standard definition of the binomial coefficient.
Assuming the Axiom of Choice, one can show that ${\textstyle {\alpha \choose \alpha }=2^{\alpha }}$ for any infinite cardinal ${\displaystyle \alpha }$ .
## Notes
1. ^ Higham (1998)
2. ^ Lilavati Section 6, Chapter 4 (see Knuth (1997)).
3. ^ Uspensky 1937, p. 18
4. ^ See (Graham, Knuth & Patashnik 1994), which also defines ${\displaystyle {\tbinom {n}{k}}=0}$ for ${\displaystyle k<0}$ . Alternative generalizations, such as to two real or complex valued arguments using the Gamma function assign nonzero values to ${\displaystyle {\tbinom {n}{k}}}$ for ${\displaystyle k<0}$ , but this causes most binomial coefficient identities to fail, and thus is not widely used by the majority of definitions. One such choice of nonzero values leads to the aesthetically pleasing "Pascal windmill" in Hilton, Holton and Pedersen, Mathematical reflections: in a room with many mirrors, Springer, 1997, but causes even Pascal's identity to fail (at the origin).
5. ^ Muir, Thomas (1902). "Note on Selected Combinations". Proceedings of the Royal Society of Edinburgh.
6. ^ This can be seen as a discrete analog of Taylor's theorem. It is closely related to Newton's polynomial. Alternating sums of this form may be expressed as the Nörlund–Rice integral.
7. ^ Gradshteyn & Ryzhik (2014, pp. 3–4).
8. ^ Boardman, Michael (2004), "The Egg-Drop Numbers", Mathematics Magazine, 77 (5): 368–372, doi:10.2307/3219201, JSTOR 3219201, MR 1573776, it is well known that there is no closed form (that is, direct formula) for the partial sum of binomial coefficients.
9. ^ see induction developed in eq (7) p. 1389 in Aupetit, Michael (2009), "Nearly homogeneous multi-partitioning with a deterministic generator", Neurocomputing, 72 (7–9): 1379–1389, doi:10.1016/j.neucom.2008.12.024, ISSN 0925-2312.
10. ^ Ruiz, Sebastian (1996). "An Algebraic Identity Leading to Wilson's Theorem". The Mathematical Gazette. 80 (489): 579–582. arXiv:math/0406086. doi:10.2307/3618534. JSTOR 3618534. S2CID 125556648.
11. ^ Benjamin & Quinn 2003, pp. 4−5
12. ^ a b Farhi, Bakir (2007). "Nontrivial lower bounds for the least common multiple of some finite sequence of integers". Journal of Number Theory. 125 (2): 393–411. arXiv:0803.0290. doi:10.1016/j.jnt.2006.10.017. S2CID 115167580.
13. ^ Thomas M. Cover; Joy A. Thomas (18 July 2006). Elements of Information Theory. Hoboken, New Jersey: Wiley. ISBN 0-471-24195-4.
14. ^ F. J. MacWilliams; N. J. A. Sloane (1981). The Theory of Error-Correcting Codes. Vol. 16 (3rd ed.). North-Holland. ISBN 0-444-85009-0.
15. ^ Spencer, Joel; Florescu, Laura (2014). Asymptopia. Student mathematical library. Vol. 71. AMS. p. 66. ISBN 978-1-4704-0904-3. OCLC 865574788.
16. ^ Spencer, Joel; Florescu, Laura (2014). Asymptopia. Student mathematical library. Vol. 71. AMS. p. 59. ISBN 978-1-4704-0904-3. OCLC 865574788.
17. ^ see e.g. Ash (1990, p. 121) or Flum & Grohe (2006, p. 427).
18. ^ Munarini, Emanuele (2011), "Riordan matrices and sums of harmonic numbers" (PDF), Applicable Analysis and Discrete Mathematics, 5 (2): 176–200, doi:10.2298/AADM110609014M, MR 2867317. | 16,017 | 48,399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 239, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2024-18 | latest | en | 0.763341 |
https://www.preplounge.com/en/consulting-forum/hi-all-does-anyone-have-any-material-to-prep-mental-maths-thanks-a-lot-1191 | 1,675,778,313,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500619.96/warc/CC-MAIN-20230207134453-20230207164453-00390.warc.gz | 943,587,726 | 37,429 | # Hi all - does anyone have any material to prep mental maths? Thanks a lot!
Case Calculations case interview preparation case math Division Math problem maths McKinsey mental math New Online Tests
New answer on Nov 10, 2020
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### Any tips to do math calculations fast ?
In order to perform math calculations faster, here are some tips you can use:
• Increase the number of steps in your calculations and turning complex operations into simple ones: It seems counter-intuitive but let me explain. Let's consider the following example: 67 x 89. If we were to calculate this the traditional way, then we would waste considerable time which will have a negative impact on our overall performance during the interview. Instead, by increasing the number of steps in the calculations, we can bring the operations to calculations that are easier to perform. Still using the example above: 67 x 89 = 67 x (90 - 1) = 67 x 90 - 67 = (60 + 7) x 90 - 67 = 60 x 90 + 7 x 90 - 60 - 7. The operation seems long but all the elements are simple calculations that you can perform quickly. Thus 67 x 89 = 5400 + 630 - 60 - 7 = 5963. Tying it back to the initial proposal, we turned multiplication into addition and substraction, which is much easier to do.
• Know shortcuts and certain computations to speed up the calculation: To be exhaustive, below is a list of things that should be memorized beforehand:
• Multiplication tables: They should be memorized and even extended to include multiplication tables up to 15 or 16 (or even 20 if you're brave enough);
• Basic squares and cubes: Ideally all of them from 1 to 10 should be known to help you speed up some calculations;
• Basic fractions: Mainly from 1 to 1/10. As weird as this sounds, fractions come in extremely handy when dealing with percentages: 25% can be replaced with 1/4 and 66% can be replaced with 2/3, instead of doing the calculations by hand;
• Shortcuts: Multiplying numbers by 5 and ending with certain digits, multiplying numbers by 11, certain shortcuts relative to addition and substraction should be known. A quick Google search on math tricks should uncover most of these for you.
• Use mental math apps and online tools to practice mental math: Nothing beats live practice and it's best to do it in an environment or with a tool that simulates the sense of urgency in a case interview. As such online mental math tools such as the one on PrepLounge and Victor Cheng's tool (http://www.caseinterview.com/mental-math) are good tools to practice with. There are many mental math apps available on the App Store and the Google Store but they are not really professional nor useful (Unless we're talking about those provided by Magoosh or other websites that offer training in numerical tests).
• PRACTICE, PRACTICE, PRACTICE: need I say more? :)
Best of luck.
Thanks for this useful tips :) Good luck with your interviews!
That's a really nice overview! Thanks!
### Any tips to do math calculations fast ?
Let me share a secret: There's no real benefit in the context of case studies in doing maths relatively fast. In fact, people worrying they're being too slow tend to make stupid mistakes by skipping steps.
"Being slow" at maths in cases generally means the person loses their way, or is trying to solve the wrong problem. Obviously it's a plus if you're a human Excel, but for most people the perfect case interview is simply clearly written down and talked through maths, starting with the problem in words, followed by numbers.
I always advise people to take their time, work through complicated numbers in detail on a separate sheet of paper if necessary and for 95% of people there will be no problem with speed.
There are many sources u can use e.g.
1. Mental Maths Tool from Preplounge
2. Mental Maths Tool from Victor Cheng
3. GMAT Questions
4. IQ Test also include mental maths questions e.g. solving basic equations
But for basic mental maths the tool on PL and the tool of VC are sufficient. Besides of this u should also make sure that u know how to solve more complicated caluculations on a piece of paper. Good luck with ur prep.
### Any tips to do math calculations fast ?
Hi,
Basically, you need to develop 3 calculation skills:
1) Learn how to multiply double digit numbers (google fast math tips or The Veda math).
2) Learn how to work with zeros. Best way - always use 10^power instead of zeros
Example:
300x9000 = 3*10ˆ2 x 9*10ˆ3=3x9*10ˆ(2+3)=27*10ˆ5
Handwritten it looks not that complicated. If you get used to writing all the numbers that way, you will never loose zeros and all multiplications/divisions will be replaced with + or -.
3) Use math tools (Mimir math for iOS, Math tool on Viktor Cheng website) to practice. Train, train, and train again
PS, Additionally I suggest to learn how to make the division mentally:
4) Learn the division table up to 1/11 (i.e. 5/6 = 83.3%). It will help you calculate any percentage problems
Good luck!
Where can one find a division table with the decimals ? The classic one is only the obvious numbers.
You can make it yourself with a calculator;)
Hi Tiago,
I agree with Alexander, Vlad and Tania; I would also suggest you to:
1. Write down the initial formula for the math computation. That’s not always feasible when you have to do mental math, but if possible try to always do so. I found many candidates start to do math without having clear what is the overall formula they have to use. This can lead to longer computations and therefore mistakes. Eg say you have to find the additional revenues from a new technology that increase revenues by 13% per employee. You have 180 employees. Current revenue per employee per month in this division is €1200. You want to find the annual increase in revenues.
1. Option 1 (average): Compute current revenues per year (180*12*1200); then compute new revenues per year (180*12*1200*1,13). Then subtract.
2. Option 2 (great): Write down the formula from the beginning. You can immediately spot you can aggregate the formula as (0,13*180*12*1200). You avoid a full computation that could lead to mistakes and appear faster in front of the interview at the same time
2. Train math under pressure. Most of the mistakes done in math during the interview are not due to lack of math knowledge. Rather, to the fact that you have to do math under pressure and are not used to that. That’s actually what you should train for. Best thing would be to do math with a timer, setting it with a challenging constraint that will force you to train under pressure. Eg if you can do percentage exercises in 15 seconds each on average, train with a timer set at 12-13 secs. The fact and feeling you don’t have enough time will help you to train for pressure.
Best,
Francesco
### Are there any math Short-Cuts for the quantitative Section?
With multiple-factor multiplications like these it is often a good idea to change the order of operations and group the numbers more efficiently.
As a first step, I would propose the following:
((25 * 8) * 1.4) * 220 * 23 = (200*1.4) * 220 * 23 = 280 * 220 * 23
As a second step, the 280 * 220 can be simplified using the distance-of-squares method. Both numbers are 30 away from 250, so you can calculate the result as follows:
280 * 220 = 250^2 - 30^2 = 62500 - 900 = 61600
Please note that there is a shortcut for squares ending in 5 that I've used here - basically, if you square a number ending in 5, take the non-unit digit, multiply it with itself + 10, and add 25.
And now the calculation boils down to 61600 * 23, which I would treat as such:
61600 * 23 = 61600 * 20 + 61600 * 3 = 1232000 + 184800 = 1416800
Hope this helps!
### Math training - how do you do it?
Thanks for the replies guys!
I know (and use) caseinterviewmath.com as well. A good website indeed, but only for practicing. They don't provide "tips and tricks" on how to solve sums like the aforementioned. Of course, one way is to use paper, and this might indeed be enough for cases. But if someone does know some "tips and tricks" I would love to hear of them (also, some of these tricks might have to be performed on paper as well).
An example of the tricks I am referring to:
- 227 : 5 = ? Instead of computing it the long way (7*5+20*5+200*5) you can multiple the first number by 10 (to make 2270) and than take half of it (1135, which is the answer). Goes for any division by 5 obviously.
- 19*18 = ? Although this one is easy to do the conventional way, an alternative is the trick of (1) adding the first digit of the lowest number to the highest number (= 8 +19 = 27), then (2) multiply this by 10 (= 27 * 10 = 270) and then (3) adding the product of the first two digits of both numbers (= 9 * 8 = 72), resulting in 270 + 72 = 342. This only works up to a multiplication of the numbers UNDER 20.
I would do this in another way: First calculate 18*20= 360 which is easy, and then subtract 18 from it 360-18 = 342
### Are there any math Short-Cuts for the quantitative Section?
Hello!
I would:
• 1st: write the whole equation, just as you did in the post
• 2nd: try to re-arrange it in a way in which you find easy relationships among numers (when possible) -e.g., 8*25 is 200, which multiplied per 220 is raughly 200 squared, etc.) This for sure depends on if you are allowed to round up numbers or not, which is something you should clarify with the interviewer)
Hope it helps!
Cheers,
Clara
The only book you need is "Secrets of Mental Math" by Arthur Benjamin and Michael Shermer, for few bucks on Amazon.
Easy to "read", training pages, super useful tricks for cases and for your whole life : I don't understand why calculation is not tought like that at school...
For practicing and improving your mental math you can find also some apps and tools, besides Victor Cheng’s math tool (Magoosh's mental maths app, Mental math cards challenge app etc). But there are some facts that will help in overall improvements:
Consulting math is very different from academic math. Working consultants - and consulting interview candidates - are always under time pressure. Results are what matter and answers are required simply to be good enough to guide business decisions, rather than being absolutely correct. The next important thing is that rather make it on the paper to structure the notes and then to communicate the results clearly rather than make your calculations fast
The time pressure in case interviews is severe and you cannot afford to waste time. But to make your calculations right you shouldn’t be in a rush. So, I would recommend you here to work on both - practice with time limitations and learn how to keep your mind peaceful and concentrated (it might be not so popular advice here, but mediations really make their job here ). Be comfortable and confident to state your answers not as a questions. Interviewers notice this, and this will not give a credit. If you need any further help or career advice, feel free to reach me out.
And here is math app for practicing math fractions and percentages from one of PL participant https://apps.apple.com/us/app/case-math/id1507653375?mt=8&ign-mpt=uo%3D4
Also, there is a tool provided by PrepLounge (https://www.preplounge.com/en/mental-math.php
Does it make sense to you?
GB
### Are there any math Short-Cuts for the quantitative Section?
I agree with Alexander. It is always easy to change the order and calculate. Preference should be given to those that result in round numbers. I would proceed in the following way:
First multiply 8*25 and then multiply the result 200 with 1,4. Till that it's quite simple. Multiplying 280 with 220 and 23 is the hard part. You can break 220 and 23 into 200+20 and 23. If you notice multiplying by 2 is not that difficult. So multiplying (200+20) with (23) would yield 4600+460 = 5060. Now multiply 280 by (10000/2 + 60) = 1,400,000+16,800 = 1,416,800
(25*8)*(1.4)*(200+20)(23) = 200*1.4*(4600+460) = 280*(5000+60) = 280*(10,000/2 + 60) = 1,400,000 + 16,800 = 1,416,800. If you find multipluying with 60 is hard, then you can break it into (100/2 + 10).
Hope this helps!
(edited)
Hi A,
All the best,
Egor
### Any tips to do math calculations fast ?
There are a couple of good apps on the Apple App Store that you can use for practice.
Search "Case Math" and you'll find a few.
### Mental Math Skills
Caseinterviewmath.com will rank you by accuracy and speed, letting you know exactly where you stand against many thousands of MBB candidates
### Any tips to do math calculations fast ?
Hi Khaoula,
I'm a former McKinsey consultant, and I have a number of methods and recommendtions for doing caculations more efficiently, which leads to much faster calcuations and fewer mistakes.
I disagree with some other responses which say that doing math quickly isn't a benefit in an interview. I have had numeous peoeple who said htey used my methods to solve the quant problmes in Case Interviews quickly, and that the interviewer was defintely impressed. While it's true, that if you try to rush the math, you might make a mistake, you can learn specific methods that are simpler, faster and less error prone.
For example, what is 120 Million times 250? This could come up if they tell you a company sells 120 Million units per year, at \$250 per unit. A typical approach would be to use Long-hand multiplication, or to caculate 12 × 25 and count zeroes. These both tend to be somewhat slow, and error prone, especially with counting zeroes.
Here's an efficient methods:
• Decompose 250 into ¼ × 1,000
• ¼ × 120 Million = 30 Million
• 30 Million × 1,000 = 30 Billion (Million × Thousand = Billion)
There are a number of other methods like this, and I wrote some articles in the Bootcamp section on quant skills required in Case Interviews.
I have also created the FastMath Ace the Case online course specifically to teach the quant skills needed to succeed in Case Interviews — you can access the course here: http://www.fastmath.net/ace-the-case/?pc=preplounge_bc_001
Enjoy,
Matthew
### Math training - how do you do it?
There are books that teach those kinds of tricks. It's been a while since I've read one, but Rapid Math Tricks is an example book. You might find it at a library. | 3,462 | 14,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-06 | longest | en | 0.926748 |
https://ics.uci.edu/~eppstein/junkyard/circumcenter.html | 1,721,854,951,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00882.warc.gz | 264,389,833 | 8,509 | ```Newsgroups: comp.graphics,sci.image.processing,comp.graphics.algorithms
From: [email protected] (Joseph O'Rourke)
Subject: Re: Center of circle from 3 edge points
Organization: Smith College, Northampton, MA, US
Date: Sat, 18 Sep 1993 17:54:05 GMT
```
```It so happens I just prepared an answer to the posed question as
part of a textbook exercise.
Let a, b, and c be the three given points.
Use _0 and _1 as x and y coordinates.
The coordinates of the center p=(p_0,p_1) of the circle through them is:
p_0 =
( b_1 a_0^2
- c_1 a_0^2
- b_1^2 a_1
+ c_1^2 a_1
+ b_0^2 c_1
+ a_1^2 b_1
+ c_0^2 a_1
- c_1^2 b_1
- c_0^2 b_1
- b_0^2 a_1
+ b_1^2 c_1
- a_1^2 c_1 )
/ D
p_1 =
( a_0^2 c_0
+ a_1^2 c_0
+ b_0^2 a_0
- b_0^2 c_0
+ b_1^2 a_0
- b_1^2 c_0
- a_0^2 b_0
- a_1^2 b_0
- c_0^2 a_0
+ c_0^2 b_0
- c_1^2 a_0
+ c_1^2 b_0)
/ D
where
D = 2( a_1 c_0 + b_1 a_0 - b_1 c_0 -a_1 b_0 -c_1 a_0 + c_1 b_0 ).
The radius of the circle is then:
r^2 = (a_0 - p_0)^2 + (a_1 - p_1)^2
Degeneracies occur when D=0.
```
```From: [email protected] (Dave Watson)
Newsgroups: comp.graphics,sci.image.processing,comp.graphics.algorithms
Subject: Re: Center of circle from 3 edge points (triangle circumcenter)
Date: 20 Sep 1993 01:23:09 GMT
Organization: Maths Dept UWA
```
```In article <[email protected]>, [email protected]
(Joseph O'Rourke) writes:
|> It so happens I just prepared an answer to the posed question as
|> part of a textbook exercise.
|> Let a, b, and c be the three given points.
|> Use _0 and _1 as x and y coordinates.
|> The coordinates of the center p=(p_0,p_1) of the circle through them is:
[Equations for p_0 and p_1 deleted]
|> where
|>
|> D = 2( a_1 c_0 + b_1 a_0 - b_1 c_0 -a_1 b_0 -c_1 a_0 + c_1 b_0 ).
|>
|> The radius of the circle is then:
|>
|> r^2 = (a_0 - p_0)^2 + (a_1 - p_1)^2
|>
|> Degeneracies occur when D=0.
More efficiently (20 multiply/divide operations versus 57) use
p_0 = (((a_0 - c_0) * (a_0 + c_0) + (a_1 - c_1) * (a_1 + c_1)) / 2 * (b_1 - c_1)
- ((b_0 - c_0) * (b_0 + c_0) + (b_1 - c_1) * (b_1 + c_1)) / 2 * (a_1 - c_1))
/ D
p_1 = (((b_0 - c_0) * (b_0 + c_0) + (b_1 - c_1) * (b_1 + c_1)) / 2 * (a_0 - c_0)
- ((a_0 - c_0) * (a_0 + c_0) + (a_1 - c_1) * (a_1 + c_1)) / 2 * (b_0 - c_0))
/ D
where D = (a_0 - c_0) * (b_1 - c_1) - (b_0 - c_0) * (a_1 - c_1)
r^2 = (c_0 - p_0)^2 + (c_1 - p_1)^2
This approach uses Cramer's Rule to find the intersection of two
perpendicular bisectors of triangle edges
--
Dave Watson Internet: [email protected]
Department of Mathematics Tel: (61 9) 380 3359
The University of Western Australia FAX: (61 9) 380 1028
Nedlands, WA 6009 Australia. Real data are full of surprises
```
```Date: 19 Jun 1997 09:45:45 -0400
From: William Flis <[email protected]>
Subject: Circumscribing simplices
To: "[email protected]" <[email protected]>
```
```Re: http://www.ics.uci.edu/~eppstein/junkyard/circumcircle.html
Circumscribing the triangle or tet or any simplex, given the vertex
coordinates, can easily be done algebraically with a little trick. Expand
(x - a)^2 + (y - b)^2 = R^2
to
x^2 - 2ax + a^2 + y^2 - 2by + b^2 = R^2
Since this is non-linear in a, b, and R, it seems you can't easily use it to
find them. But let
q = R^2 - a^2 - b^2 [1]
and re-arrange to yield
(2x)a + (2y)b + q = x^2 + y^2 [2]
which is now linear in a, b, and q. Apply at 3 points and solve by Cramer's
rule:
|(x1^2 + y1^2) 2y1 1|
|(x2^2 + y2^2) 2y2 1| |(x2^2 + y2^2 - x1^2 - y1^2) (y2 - y1)|
|(x3^2 + y3^2) 2y3 1| |(x3^2 + y3^2 - x1^2 - y1^2) (y3 - y1)|
a = ---------------------- = ---------------------------------------
| 2x1 2y1 1 | 2 * | (x2 - x1) (y2 - y1) |
| 2x2 2y2 1 | | (x3 - x1) (y3 - y1) |
| 2x3 2y3 1 |
Similarly,
|(x2 - x1) (x2^2 + y2^2 - x1^2 - y1^2)|
|(y2 - y1) (x3^2 + y3^2 - x1^2 - y1^2)|
b = -----------------------------------------
(same denominator)
This is equivalent to Watson's formula, which he derived geometrically. (A
side note: Watson could have saved 3 more multiplications in his formula by
incorporating a factor of 2 into his denominator term D.) It's then cheaper
to find R by the original equation than by applying Cramer's rule a 3rd time,
etc.
This can also be written down by comparing Eq. [2] with the 'three-point
form':
| (x^2 + y^2) x y 1 |
| (x1^2 + y1^2) x1 y1 1 | = 0
| (x2^2 + y2^2) x2 y2 1 |
| (x3^2 + y3^2) x3 y3 1 |
For the sphere about a tet (after factoring out the 2's),
|(x1^2 + y1^2 + z1^2) y1 z1 1 |
|(x2^2 + y2^2 + z2^2) y2 z2 1 |
|(x3^2 + y3^2 + z3^2) y3 z3 1 |
|(x4^2 + y4^2 + z4^2) y4 z4 1 |
a = ------------------------------------
2 * | x1 y1 z1 1 |
| x2 y2 z2 1 |
| x3 y3 z3 1 |
| x4 y4 z4 1 |
and so on. (The determinants are easily reduced to 3X3 size.)
This is not original with me, and I'm uncertain as to the source. I found
the result for triangles in a computer program that was written 15 years ago
(which I've been using ever since) but only recently did I figure out how the
formula is derived.
P.S.: Your web pages are a valuable resource. Thank you for maintaining them
and please keep up the good work!
------------------------------------------------------------
William J. Flis, Director of Research, Dyna East Corporation
3620 Horizon Drive, King of Prussia, PA 19406, U.S.A.
(610)270-9900 Ext. 130 Fax: (610)270-9744
http://www.dynaeast.com/~flis/FlisHome.html
------------------------------------------------------------
```
```From: [email protected]
To: [email protected]
Subject: Re: Circumscribing simplices
Date: Thu, 19 Jun 1997 09:56:25 -0700
```
```Yes, this linearization trick is pretty standard in computational geometry.
The specific variation you use, mapping a vector v onto (v,|v|^2),
turns circles or more generally spheres in d dimensions into planes
or more generally hyperplanes in d+1 dimensions. Your equation
| (x^2 + y^2) x y 1 |
| (x1^2 + y1^2) x1 y1 1 | = 0
| (x2^2 + y2^2) x2 y2 1 |
| (x3^2 + y3^2) x3 y3 1 |
is the standard equation for a plane through three points (xi,yi,xi^2+yi^2).
One reference for general ideas like this (i.e. of turning nonlinear
equations into higher-dimensional linear ones by adding extra variables
that are polynomials in the original equations)
is "A general approach to d-dimensional geometric queries",
A. C. Yao and F. F. Yao, 17th ACM Symp. Theory of Computing (1985) 163-168.
But probably the same idea was known long before then in
less algorithmic contexts.
I've added your message to my page (I assume this is ok; let me know if not).
```
```To: [email protected]
Subject: Re: circumsphere
Date: Wed, 1 Apr 98 0:34:28 EST
From: Jonathan R Shewchuk <[email protected]>
```
```> I am searching for a numerically stable algorithm to calculate the radius r
> (and perhaps the center m) of the circumsphere of a tetrahedron in
> three dimensions, given by the coordinates of the vertices a, b, c, d in R^3.
Let a, b, c, d, and m be vectors in R^3. Let ax, ay, and az be the components
of a, and likewise for b, c, and d. Let |a| denote the Euclidean norm of a,
and let a x b denote the cross product of a and b. Then
| |
| |d-a|^2 [(b-a)x(c-a)] + |c-a|^2 [(d-a)x(b-a)] + |b-a|^2 [(c-a)x(d-a)] |
| |
r = -------------------------------------------------------------------------,
| bx-ax by-ay bz-az |
2 | cx-ax cy-ay cz-az |
| dx-ax dy-ay dz-az |
and
|d-a|^2 [(b-a)x(c-a)] + |c-a|^2 [(d-a)x(b-a)] + |b-a|^2 [(c-a)x(d-a)]
m = a + ---------------------------------------------------------------------.
| bx-ax by-ay bz-az |
2 | cx-ax cy-ay cz-az |
| dx-ax dy-ay dz-az |
Some notes on stability:
- Note that the expression for r is purely a function of differences between
coordinates. The advantage is that the relative error incurred in the
computation of r is also a function of the _differences_ between the
vertices, and is not influenced by the _absolute_ coordinates of the
vertices. In most applications, vertices are usually nearer to each other
than to the origin, so this property is advantageous. If someone gives you
a formula that doesn't have this property, be wary.
Similarly, the formula for m incurs roundoff error proportional to the
differences between vertices, but does not incur roundoff error proportional
to the absolute coordinates of the vertices until the final addition.
- These expressions are unstable in only one case: if the denominator is
close to zero. This instability, which arises if the tetrahedron is nearly
degenerate, is unavoidable. Depending on your application, you may want
to use exact arithmetic to compute the value of the determinant.
Fortunately, this determinant is the basis of the well-studied 3D orientation
test, and several fast algorithms for providing accurate approximations to
the determinant are available. Some resources are available from the
"Numerical and algebraic computation" page of Nina Amenta's Directory of
Computational Geometry Software:
http://www.geom.umn.edu/software/cglist/alg.html
If you're using floating-point inputs (as opposed to integers), one
package that can estimate this determinant somewhat accurately is my own:
http://www.cs.cmu.edu/~quake/robust.html
- If you want to be even more aggressive about stability, you might reorder
the vertices of the tetrahedron so that the vertex `a' (which is subtracted
from the other vertices) is, roughly speaking, the vertex at the center of
the minimum spanning tree of the tetrahedron's four vertices. For more
Algorithms for 2D Delaunay Triangulations," International Journal of
Computational Geometry & Applications 5(1-2):193-213, March-June 1995.
For completeness, here are stable expressions for the circumradius and
circumcenter of a triangle, in R^2 and in R^3. Incidentally, the expressions
given here for R^2 are better behaved in terms of relative error than the
suggestions currently given in the Geometry Junkyard
(http://www.ics.uci.edu/~eppstein/junkyard/triangulation.html).
Triangle in R^2:
|b-a| |c-a| |b-c| < Note: You only want to compute one sqrt, so
r = ------------------, use sqrt{ |b-a|^2 |c-a|^2 |b-c|^2 }
| bx-ax by-ay |
2 | cx-ax cy-ay |
| by-ay |b-a|^2 |
| cy-ay |c-a|^2 |
mx = ax - ------------------,
| bx-ax by-ay |
2 | cx-ax cy-ay |
| bx-ax |b-a|^2 |
| cx-ax |c-a|^2 |
my = ay + ------------------.
| bx-ax by-ay |
2 | cx-ax cy-ay |
Triangle in R^3:
| |
| |c-a|^2 [(b-a)x(c-a)]x(b-a) + |b-a|^2 (c-a)x[(b-a)x(c-a)] |
| |
r = -------------------------------------------------------------,
2 | (b-a)x(c-a) |^2
|c-a|^2 [(b-a)x(c-a)]x(b-a) + |b-a|^2 (c-a)x[(b-a)x(c-a)]
m = a + ---------------------------------------------------------.
2 | (b-a)x(c-a) |^2
Finally, here's some C code that computes the vector m-a (whose norm is r) in
each of these three cases. Notice the #ifdef statements, which allow you to
choose whether or not to use my aforementioned package to approximate the
determinant. (No attempt is made here to reorder the vertices to improve
stability.)
/*****************************************************************************/
/* */
/* tetcircumcenter() Find the circumcenter of a tetrahedron. */
/* */
/* The result is returned both in terms of xyz coordinates and xi-eta-zeta */
/* coordinates, relative to the tetrahedron's point `a' (that is, `a' is */
/* the origin of both coordinate systems). Hence, the xyz coordinates */
/* returned are NOT absolute; one must add the coordinates of `a' to */
/* find the absolute coordinates of the circumcircle. However, this means */
/* that the result is frequently more accurate than would be possible if */
/* absolute coordinates were returned, due to limited floating-point */
/* precision. In general, the circumradius can be computed much more */
/* accurately. */
/* */
/* The xi-eta-zeta coordinate system is defined in terms of the */
/* tetrahedron. Point `a' is the origin of the coordinate system. */
/* The edge `ab' extends one unit along the xi axis. The edge `ac' */
/* extends one unit along the eta axis. The edge `ad' extends one unit */
/* along the zeta axis. These coordinate values are useful for linear */
/* interpolation. */
/* */
/* If `xi' is NULL on input, the xi-eta-zeta coordinates will not be */
/* computed. */
/* */
/*****************************************************************************/
void tetcircumcenter(a, b, c, d, circumcenter, xi, eta, zeta)
double a[3];
double b[3];
double c[3];
double d[3];
double circumcenter[3];
double *xi;
double *eta;
double *zeta;
{
double xba, yba, zba, xca, yca, zca, xda, yda, zda;
double balength, calength, dalength;
double xcrosscd, ycrosscd, zcrosscd;
double xcrossdb, ycrossdb, zcrossdb;
double xcrossbc, ycrossbc, zcrossbc;
double denominator;
double xcirca, ycirca, zcirca;
/* Use coordinates relative to point `a' of the tetrahedron. */
xba = b[0] - a[0];
yba = b[1] - a[1];
zba = b[2] - a[2];
xca = c[0] - a[0];
yca = c[1] - a[1];
zca = c[2] - a[2];
xda = d[0] - a[0];
yda = d[1] - a[1];
zda = d[2] - a[2];
/* Squares of lengths of the edges incident to `a'. */
balength = xba * xba + yba * yba + zba * zba;
calength = xca * xca + yca * yca + zca * zca;
dalength = xda * xda + yda * yda + zda * zda;
/* Cross products of these edges. */
xcrosscd = yca * zda - yda * zca;
ycrosscd = zca * xda - zda * xca;
zcrosscd = xca * yda - xda * yca;
xcrossdb = yda * zba - yba * zda;
ycrossdb = zda * xba - zba * xda;
zcrossdb = xda * yba - xba * yda;
xcrossbc = yba * zca - yca * zba;
ycrossbc = zba * xca - zca * xba;
zcrossbc = xba * yca - xca * yba;
/* Calculate the denominator of the formulae. */
#ifdef EXACT
/* Use orient3d() from http://www.cs.cmu.edu/~quake/robust.html */
/* to ensure a correctly signed (and reasonably accurate) result, */
/* avoiding any possibility of division by zero. */
denominator = 0.5 / orient3d(b, c, d, a);
#else
/* Take your chances with floating-point roundoff. */
denominator = 0.5 / (xba * xcrosscd + yba * ycrosscd + zba * zcrosscd);
#endif
/* Calculate offset (from `a') of circumcenter. */
xcirca = (balength * xcrosscd + calength * xcrossdb + dalength * xcrossbc) *
denominator;
ycirca = (balength * ycrosscd + calength * ycrossdb + dalength * ycrossbc) *
denominator;
zcirca = (balength * zcrosscd + calength * zcrossdb + dalength * zcrossbc) *
denominator;
circumcenter[0] = xcirca;
circumcenter[1] = ycirca;
circumcenter[2] = zcirca;
if (xi != (double *) NULL) {
/* To interpolate a linear function at the circumcenter, define a */
/* coordinate system with a xi-axis directed from `a' to `b', */
/* an eta-axis directed from `a' to `c', and a zeta-axis directed */
/* from `a' to `d'. The values for xi, eta, and zeta are computed */
/* by Cramer's Rule for solving systems of linear equations. */
*xi = (xcirca * xcrosscd + ycirca * ycrosscd + zcirca * zcrosscd) *
(2.0 * denominator);
*eta = (xcirca * xcrossdb + ycirca * ycrossdb + zcirca * zcrossdb) *
(2.0 * denominator);
*zeta = (xcirca * xcrossbc + ycirca * ycrossbc + zcirca * zcrossbc) *
(2.0 * denominator);
}
}
/*****************************************************************************/
/* */
/* tricircumcenter() Find the circumcenter of a triangle. */
/* */
/* The result is returned both in terms of x-y coordinates and xi-eta */
/* coordinates, relative to the triangle's point `a' (that is, `a' is */
/* the origin of both coordinate systems). Hence, the x-y coordinates */
/* returned are NOT absolute; one must add the coordinates of `a' to */
/* find the absolute coordinates of the circumcircle. However, this means */
/* that the result is frequently more accurate than would be possible if */
/* absolute coordinates were returned, due to limited floating-point */
/* precision. In general, the circumradius can be computed much more */
/* accurately. */
/* */
/* The xi-eta coordinate system is defined in terms of the triangle. */
/* Point `a' is the origin of the coordinate system. The edge `ab' extends */
/* one unit along the xi axis. The edge `ac' extends one unit along the */
/* eta axis. These coordinate values are useful for linear interpolation. */
/* */
/* If `xi' is NULL on input, the xi-eta coordinates will not be computed. */
/* */
/*****************************************************************************/
void tricircumcenter(a, b, c, circumcenter, xi, eta)
double a[2];
double b[2];
double c[2];
double circumcenter[2];
double *xi;
double *eta;
{
double xba, yba, xca, yca;
double balength, calength;
double denominator;
double xcirca, ycirca;
/* Use coordinates relative to point `a' of the triangle. */
xba = b[0] - a[0];
yba = b[1] - a[1];
xca = c[0] - a[0];
yca = c[1] - a[1];
/* Squares of lengths of the edges incident to `a'. */
balength = xba * xba + yba * yba;
calength = xca * xca + yca * yca;
/* Calculate the denominator of the formulae. */
#ifdef EXACT
/* Use orient2d() from http://www.cs.cmu.edu/~quake/robust.html */
/* to ensure a correctly signed (and reasonably accurate) result, */
/* avoiding any possibility of division by zero. */
denominator = 0.5 / orient2d(b, c, a);
#else
/* Take your chances with floating-point roundoff. */
denominator = 0.5 / (xba * yca - yba * xca);
#endif
/* Calculate offset (from `a') of circumcenter. */
xcirca = (yca * balength - yba * calength) * denominator;
ycirca = (xba * calength - xca * balength) * denominator;
circumcenter[0] = xcirca;
circumcenter[1] = ycirca;
if (xi != (double *) NULL) {
/* To interpolate a linear function at the circumcenter, define a */
/* coordinate system with a xi-axis directed from `a' to `b' and */
/* an eta-axis directed from `a' to `c'. The values for xi and eta */
/* are computed by Cramer's Rule for solving systems of linear */
/* equations. */
*xi = (xcirca * yca - ycirca * xca) * (2.0 * denominator);
*eta = (ycirca * xba - xcirca * yba) * (2.0 * denominator);
}
}
/*****************************************************************************/
/* */
/* tricircumcenter3d() Find the circumcenter of a triangle in 3D. */
/* */
/* The result is returned both in terms of xyz coordinates and xi-eta */
/* coordinates, relative to the triangle's point `a' (that is, `a' is */
/* the origin of both coordinate systems). Hence, the xyz coordinates */
/* returned are NOT absolute; one must add the coordinates of `a' to */
/* find the absolute coordinates of the circumcircle. However, this means */
/* that the result is frequently more accurate than would be possible if */
/* absolute coordinates were returned, due to limited floating-point */
/* precision. In general, the circumradius can be computed much more */
/* accurately. */
/* */
/* The xi-eta coordinate system is defined in terms of the triangle. */
/* Point `a' is the origin of the coordinate system. The edge `ab' extends */
/* one unit along the xi axis. The edge `ac' extends one unit along the */
/* eta axis. These coordinate values are useful for linear interpolation. */
/* */
/* If `xi' is NULL on input, the xi-eta coordinates will not be computed. */
/* */
/*****************************************************************************/
void tricircumcenter3d(a, b, c, circumcenter, xi, eta)
double a[3];
double b[3];
double c[3];
double circumcenter[3];
double *xi;
double *eta;
{
double xba, yba, zba, xca, yca, zca;
double balength, calength;
double xcrossbc, ycrossbc, zcrossbc;
double denominator;
double xcirca, ycirca, zcirca;
/* Use coordinates relative to point `a' of the triangle. */
xba = b[0] - a[0];
yba = b[1] - a[1];
zba = b[2] - a[2];
xca = c[0] - a[0];
yca = c[1] - a[1];
zca = c[2] - a[2];
/* Squares of lengths of the edges incident to `a'. */
balength = xba * xba + yba * yba + zba * zba;
calength = xca * xca + yca * yca + zca * zca;
/* Cross product of these edges. */
#ifdef EXACT
/* Use orient2d() from http://www.cs.cmu.edu/~quake/robust.html */
/* to ensure a correctly signed (and reasonably accurate) result, */
/* avoiding any possibility of division by zero. */
xcrossbc = orient2d(b[1], b[2], c[1], c[2], a[1], a[2]);
ycrossbc = orient2d(b[2], b[0], c[2], c[0], a[2], a[0]);
zcrossbc = orient2d(b[0], b[1], c[0], c[1], a[0], a[1]);
#else
/* Take your chances with floating-point roundoff. */
xcrossbc = yba * zca - yca * zba;
ycrossbc = zba * xca - zca * xba;
zcrossbc = xba * yca - xca * yba;
#endif
/* Calculate the denominator of the formulae. */
denominator = 0.5 / (xcrossbc * xcrossbc + ycrossbc * ycrossbc +
zcrossbc * zcrossbc);
/* Calculate offset (from `a') of circumcenter. */
xcirca = ((balength * yca - calength * yba) * zcrossbc -
(balength * zca - calength * zba) * ycrossbc) * denominator;
ycirca = ((balength * zca - calength * zba) * xcrossbc -
(balength * xca - calength * xba) * zcrossbc) * denominator;
zcirca = ((balength * xca - calength * xba) * ycrossbc -
(balength * yca - calength * yba) * xcrossbc) * denominator;
circumcenter[0] = xcirca;
circumcenter[1] = ycirca;
circumcenter[2] = zcirca;
if (xi != (double *) NULL) {
/* To interpolate a linear function at the circumcenter, define a */
/* coordinate system with a xi-axis directed from `a' to `b' and */
/* an eta-axis directed from `a' to `c'. The values for xi and eta */
/* are computed by Cramer's Rule for solving systems of linear */
/* equations. */
/* There are three ways to do this calculation - using xcrossbc, */
/* ycrossbc, or zcrossbc. Choose whichever has the largest */
/* magnitude, to improve stability and avoid division by zero. */
if (((xcrossbc >= ycrossbc) ^ (-xcrossbc > ycrossbc)) &&
((xcrossbc >= zcrossbc) ^ (-xcrossbc > zcrossbc))) {
*xi = (ycirca * zca - zcirca * yca) / xcrossbc;
*eta = (zcirca * yba - ycirca * zba) / xcrossbc;
} else if ((ycrossbc >= zcrossbc) ^ (-ycrossbc > zcrossbc)) {
*xi = (zcirca * xca - xcirca * zca) / ycrossbc;
*eta = (xcirca * zba - zcirca * xba) / ycrossbc;
} else {
*xi = (xcirca * yca - ycirca * xca) / zcrossbc;
*eta = (ycirca * xba - xcirca * yba) / zcrossbc;
}
}
}
-------------
The compgeom mailing lists: see
or send mail to [email protected] with the line:
```
```Date: Thu, 02 Apr 1998 08:44:46 -0500
From: William Flis <[email protected]>
Organization: Dyna East Corporation
To: [email protected]
Subject: Re: circumsphere
```
```[email protected] wrote:
>
> > I am searching for a numerically stable algorithm to calculate the radius r
> > (and perhaps the center m) of the circumsphere of a tetrahedron in
> > three dimensions, given by the coordinates of the vertices a, b, c, d in R^3.
>
.....(part omitted).....
>
> Some notes on stability:
>
> - Note that the expression for r is purely a function of differences between
> coordinates. The advantage is that the relative error incurred in the
> computation of r is also a function of the _differences_ between the
> vertices, and is not influenced by the _absolute_ coordinates of the
> vertices. In most applications, vertices are usually nearer to each other
> than to the origin, so this property is advantageous. If someone gives you
> a formula that doesn't have this property, be wary.
...(part omitted).....
> For completeness, here are stable expressions for the circumradius and
> circumcenter of a triangle, in R^2 and in R^3. Incidentally, the expressions
> given here for R^2 are better behaved in terms of relative error than the
> suggestions currently given in the Geometry Junkyard
> (http://www.ics.uci.edu/~eppstein/junkyard/triangulation.html).
In a private response to Herr Kerscher's question, I recommended that he
look at this page, particularly my contribution to it. Let me say that
Prof. Shewchuk's criticism regarding relative error is not only correct,
but also of some practical significance. The good behavior which he
refers to is obtained for 2-D triangles using the particular form
(attributed to Jim Ward) given by the most recent
comp.graphics.algorithms FAQ at
http://www.cis.ohio-state.edu/hypertext/faq/usenet/graphics/algorithms-faq/faq.html
This form has the desired property of using only differences between
coordinates.
avoid overflows in many more practical cases (e.g., simplex far from
origin).
Thus, my formula given in the Geometry Junkyard:
|(x1^2 + y1^2) 2y1 1|
|(x2^2 + y2^2) 2y2 1| |(x2^2 + y2^2 - x1^2 - y1^2) (y2 - y1)|
|(x3^2 + y3^2) 2y3 1| |(x3^2 + y3^2 - x1^2 - y1^2) (y3 - y1)|
a = ---------------------- = ---------------------------------------
| 2x1 2y1 1 | 2 * | (x2 - x1) (y2 - y1) |
| 2x2 2y2 1 | | (x3 - x1) (y3 - y1) |
| 2x3 2y3 1 |
needs to be taken another step to achieve the desired property.
The terms in the first column of the numerator should be re-written like
this:
(x2^2 + y2^2 - x1^2 - y1^2) = (x2 - x1)(x2 + x1) + (y2 - y1)(y2 + y1)
(x3^2 + y3^2 - x1^2 - y1^2) = (x3 - x1)(x3 + x1) + (y3 - y1)(y3 + y1)
The formulas for tets in 3-D in the Geometry Junkyard need similar
treatment.
--
*******************************************
William J. Flis Director of Research
Dyna East Corporation 3620 Horizon Drive
King of Prussia, PA 19406-2647 U.S.A.
(610)270-9900 x130 fax:(610)270-9744
[email protected] or [email protected]
http://www.dynaeast.com/~flis/FlisHome.html
*******************************************
-------------
The compgeom mailing lists: see | 8,417 | 27,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.666256 |
http://ibelitetutor.com/blog/category/cbse-tutors/ | 1,524,309,351,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945143.71/warc/CC-MAIN-20180421110245-20180421130245-00201.warc.gz | 164,815,608 | 30,213 | ## How to Prepare for board Exams?
In my previous post of this series, we discussed the advantages of CBSE board exams. Now I am suggesting a few tricks about How to Prepare for board exams.
There are a lot of post on the internet that suggests you how to eat, how to sleep and how to manage your stress during board exam but in this post, I shall only discuss the academic tips about How to Prepare for board exams. I am taking class 10 Mathematics as a base subject here but you can apply more or less same tricks on almost all your subjects.
Question Paper Break-Up
## Definite Integration
In the previous post, we discussed indefinite integration. Now we shall discuss definite integration
► Definite Integration- We already know that this c here is an integral constant. we are not sure about its value. This c is the reason we call this process indefinite integration. But suppose we do our integration between certain limits like:-
here a lower limit while b higher limit
=g(b)-g(a)
You can clearly see that this function is independent of ‘c’. Means we can be sure about its value so this type of integration is called Definite Integration.
►Definite Integration of a function f(x) is possible in [a,b] if f(x) is continuous in the given interval
►If f(x), the integrand, is not continuous for a given value of x then it doesn’t mean that g(x), the integral, is also discontinuous for that value of x.
► Definite integration of a function between given limits like Algebraic sum of areas bounded by the given curve f(x) and given lines x=a and x=b. That’s why the answer for definite integration problems is a single number.
► If that shows a few things:-
(i) The lines between which area is bounded are co-incident(a=b)
(ii) Area covered above the x-axis=Area covered below the x-axis that means positive part of area and negative part of area is equal
(iii) there must be at least one solution/root to f(x) between x=a and x=b(this is something we study in ROLE’S THEOREM in detail)
► If given function f(x) is not continuous at x=c then we should write
► If given function f(x) > or <0 in any given interval (a,b) then >0 or <0 in given interval (a,b)
► If given function f(x) g(x) in the given interval (a,b) then
in the given interval
► If we integrate the given function f(x) in the given interval (a,b) then
Some More Properties of Definite Integration:- Read more
# Continuity of functions-
The word continuous means without any break or gap. Continuity of functions exists when our function is without any break or gap or jump . If there is any gap in the graph, the function is said to be discontinuous.
Graph of functions like sinx,cosx, secx, 1/x etc are continuous (without any gap) while greatest integer function has a break at every point(discontinuous).
1. A function f(x) is said to be continuous at x = c, if .
symbolically f is continuous at x = c if .
It should be noted that continuity of a function at x = a is meaningful only if the function is defined in the immediate neighborhood of x = a, not necessarily at x = a.
## How To Solve Limit Problems
In my previous post on limits, We have discussed some basic as well as advanced concepts of limits. Here we shall discuss different methods to solve limit questions. Based on the type of function, we can divide all our work into sections-:
Algebraic Limits- Problems of limits that involve algebraic functions are called algebraic limits. They can be further divided into following sections:-
Direct Substitution Method –Suppose we have to find. we can directly substitute the value of the limit of the variable (i.e replace x=a) in the expression.
► If f(a) is finite then L=f(a)
► If f(a) is undefined then L doesn’t exist
► If f(a) is indeterminate then this method fails
Example-1:- Find value of (x²-5x+6) Read more
# Limit of a function
Limit of a function f(x) is said to exist as, when
finite quantity.
Fundamental Theorems On Limits :
Let & If l & m exists then :
(i) f (x) ± g (x) = l ± m
(ii) f(x). g(x) = l. m
(iii) provided
(iv) where k is a constant.
(v) provided f is continuous at g (x) = m
Standard Limits :
(a) and Where x is measured in radians
(b) both are equal to e
(c) then this will show that
(d) and (a finite quantity) then
where z=
(e) where a>0. In particular
Indeterminant Forms:
etc are considered to be indeterminant values
We cannot plot on the paper. Infinityis a symbol & not a number. It does not obey the laws of elementary algebra.
+=
×
(a/) = 0 if a is finite v is not defined
a b =0,if & only if a = 0 or b = 0 and a & b are finite.
Expansion of function like Binomial expansion, exponential & logarithmic expansion, expansion of sinx , cosx , tanx should be remembered by heart & are given below:
(i) ex =1+x/1!+x3/3!+x4/4!……
(ii) ax=1+(xloga)/1!+ (xloga)2/2!+ (xloga)3/3!+ (xloga)4/4!+……….where a > 0
(iii) ln(1-x)=x-x2/2+x3/3-x4/4………. where -1 < x 1
(iv) ln(1-x)=-x-x2/2-x3/3-x4/4………. where -1 x < 1
(v )
(vi)
(v)
In next post, I will discuss various types of limit problems, their solutions and L’ Hospital’s rule.In the meantime, you can solve these basic questions from this PDF. This PDF is for beginners only. I will post difficult and higher level questions in the next post on this topic
## How to solve trigonometric problems based on complimentary anngles?(concept-3)
### IB Maths tutors give great importance to Trigonometry.
Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, IB Maths Tutors divide trigonometry in two part:-
1. Trigonometry based on right triangles
2. Trigonometry based on non-right triangles.
Here, we are discussing trigonometry based on non-right triangles only.
In the third article of this series, we will discuss problems based on complementary angles
In the third article of this series, we will discuss problems based on complementary angles
In this right triangle Sin A=BC/AC & Cos C=BC/AC clearly: Sin A=Cos C In the given triangle A+C=90° so we can write C=(90°-A). This gives us freedom to write Sin A=Cos (90°-A) similarly we can write these relationships Read more
## How to solve basic problems in trigonometry?(concept-2)
### Mathematics tutors give great importance to Trigonometry
Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, I am dividing trigonometry in two part:-
1 Trigonometry based on right triangles
1 Trigonometry based on non-right triangles.
In this post, we will discuss problems based on trigonometric ratios of a few specific angles like 0°,30°, 45°, 60° and 90°. Mathematics tutors use different tricks to form this table. I will discuss my tricks in a separate post
## How to solve basic problems in trigonometry?(concept-1)
### Mathematics tutors give great importance to Trigonometry
Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, I am dividing trigonometry in two part:-
1 Trigonometry based on right triangles
1 Trigonometry based on non-right triangles.
In this post, I will only discuss Trigonometry based on right triangles.
In a right triangle, there are three sides hypotenuse (the longest side), adjacent side(base) and the opposite side(perpendicular).
Many IB mathematics tutors consider quadratic equations as a very important topic of ib maths. There are following ways to solve a quadratic equation
► Factorization method
►complete square method
► graphical method
Given equation: ax²+bx+c=0
Step-1: transfer constant term to the right side
ax²+bx=-c
Step-2: divide both sides by coefficient of x²
x²+bx/a=-c/a
Step-3: write (coefficient of x/2)² that is (b/2a)²=b²/4a²
Step-4: Add this value to both sides
x²+bx/a+b²/4a² =-c/a²+b²/4a²
(x+b/2a)²=b²-4ac/4a²
now, take square root on both sides
x+b/2=±√b²-4ac/a²
x=-(b/2a)±√b²-4ac/2a
This formula is known as quadratic formula, we can put values of a, b and c from any equation and find the value of x (the variable) by directly using this formula.
IB Mathematics tutors can also explain the concept of conjugate roots with the help of quadratic formula. In a quadratic equation,
ax²+bx+c=0
if a, b and c are all rational numbers and one root of the quadratic equation is a+√b then the second root will automatically become a-√b. that can be understood easily as we use one +ve and one -ve sign in quadratic formula.
These types of roots are called Conjugate Roots.
If a & b are the roots of the quadratic equation ax² + bx + c = 0, then;
(i) (ii) (iii)
Nature Of Roots:
(a) Consider the quadratic equation ax² + bx + c = 0 where a, b, c R & then
(i) D > 0 roots are real & distinct (unequal).
(ii) D = 0 roots are real & coincident (equal).
(iii) D < 0 roots are imaginary
(B) Consider the quadratic equation ax2+ bx + c = 0 where a, b, c Q & then
If D > 0 & is a perfect square , then roots are rational & unequal.
A quadratic equation whose roots are a & b is (x – a)(x – b) = 0 i.e. x2 – (a + b) x + a b = 0 i.e.
x2 – (sum of roots) x + product of roots = 0
Consider the quadratic expression , y = ax² + bx + c , a, b, c R & then
(i) The graph between x, y is always a parabola. If a > 0 then the shape of the parabola is concave upwards & if a < 0 then the shape of the parabola is concave downwards.
Common Roots Of 2 Quadratic Equations [Only One Common Root]- Let
be the common root of ax² + bx + c = 0 & a’x2 + b’x + c’ = 0
Therefore
If we solve above pair by cramer’s rule we get
This will give us
Every pair of the quadratic equation whose coefficients fulfils the above condition will have one root in common.
The condition that a quadratic function-
f(x , y) = ax² + 2 hxy + by² + 2 gx + 2 fy + c may be resolved into two linear factors is that or
Reducible Quadratic Equations-These are the equations which are not quadratic in their initial condition but after some calculations, we can reduce them into quadratic equations
(i) If the power of the second term is exactly half to the power of the first term and the third term is a constant, these types of equations can be reduced to quadratic equations.
i.e., , , all these equations can easily be reduced into quadratic equations by applying the method of substitution.
Example-Solve this equation and find x
Ans:
let y= then given equation will become
it’s a simple quadratic equation we can be easily factorised it and solve so y=–3,2
so =-3
the final equation can be solved using Quadratic formula and the same process can be repeated for y=2
(ii) If a variable is added with its own reciprocal, then we get a quadratic equation i.e, all these equations can be reduced into quadratic by replacing one term by any other variable.
Standard Form of a Quadratic Function-A quadratic function y=ax2+ bx + c can be
reduced into standard form by the method of completing the square. If we
draw the graph of this function we shall get a parabola with vertex (h,k). The parabola will be upward for a>0 and downward for a<0
Maximum and Minimum value of a quadratic function- If the function is in the form
Then ‘h’ is the input value of the function while ‘k’ is its output.
(i) If a>0 (in case of upward parabola) the minimum value of f is f(h)=k
(ii) If a<0 (in case of downward parabola)the maximum value of f is f(h)=k
If our function is in the form of y=ax² + bx + c then vertex of the parabola
The line passing through vertex and parallel to the y-axis is called the axis of symmetry.
The parabolic graph of a quadratic function is symmetrical about axis of symmetry.
f(x) has a minimum value at vertex if a>0 and at
f(x) has a maximum value at vertex if a<0 and at
In the next post about quadratics, I shall discuss discriminant, nature of roots, relationships between the roots. In the meantime, you can download the pdf and solve practice questions. | 3,553 | 12,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2018-17 | longest | en | 0.920233 |
https://www.jobilize.com/physics1/course/6-4-drag-force-and-terminal-speed-by-openstax?qcr=www.quizover.com&page=3 | 1,610,765,883,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00430.warc.gz | 847,664,371 | 21,811 | 6.4 Drag force and terminal speed (Page 4/12)
Page 4 / 12
In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.
The calculus of velocity-dependent frictional forces
When a body slides across a surface, the frictional force on it is approximately constant and given by ${\mu }_{\text{k}}N.$ Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by
${f}_{R}=\text{−}bv,$
where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.
Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in [link] . Newton’s second law in the vertical direction gives the differential equation
$mg-bv=m\frac{dv}{dt},$
where we have written the acceleration as $dv\text{/}dt.$ As v increases, the frictional force – bv increases until it matches mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity ${v}_{\text{T}}.$ From the previous equation,
$mg-b{v}_{\text{T}}=0,$
so
${v}_{\text{T}}=\frac{mg}{b}.$
We can find the object’s velocity by integrating the differential equation for v . First, we rearrange terms in this equation to obtain
$\frac{dv}{g-\left(b\text{/}m\right)v}=dt.$
Assuming that $v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
${\int }_{0}^{v}\frac{d{v}^{\prime }}{g-\left(b\text{/}m\right){v}^{\prime }}={\int }_{0}^{t}d{t}^{\prime },$
or
${-\frac{m}{b}\text{ln}\left(g-\frac{b}{m}{v}^{\prime }\right)|}_{0}^{v}={{t}^{\prime }|}_{0}^{t},$
where $v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find
$-\frac{m}{b}\left[\text{ln}\left(g-\frac{b}{m}v\right)-\text{ln}g\right]=t.$
Since $\text{ln}A-\text{ln}B=\text{ln}\left(A\text{/}B\right),$ and $\text{ln}\left(A\text{/}B\right)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
$\frac{g-\left(bv\text{/}m\right)}{g}={e}^{\text{−}bt\text{/}m},$
and
$v=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$
Notice that as $t\to \infty ,v\to mg\text{/}b={v}_{\text{T}},$ which is the terminal velocity.
The position at any time may be found by integrating the equation for v . With $v=dy\text{/}dt,$
$dy=\frac{mg}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right)dt.$
Assuming $y=0\phantom{\rule{0.2em}{0ex}}\text{when}\phantom{\rule{0.2em}{0ex}}t=0,$
${\int }_{0}^{y}d{y}^{\prime }=\frac{mg}{b}{\int }_{0}^{t}\left(1-{e}^{\text{−}bt\text{'}\text{/}m}\right)d{t}^{\prime },$
which integrates to
$y=\frac{mg}{b}t+\frac{{m}^{2}g}{{b}^{2}}\left({e}^{\text{−}bt\text{/}m}-1\right).$
Effect of the resistive force on a motorboat
A motorboat is moving across a lake at a speed ${v}_{0}$ when its motor suddenly freezes up and stops. The boat then slows down under the frictional force ${f}_{R}=\text{−}bv.$ (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?
Solution
1. With the motor stopped, the only horizontal force on the boat is ${f}_{R}=\text{−}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{−}bv,$
which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$
Integrating this equation between the time zero when the velocity is ${v}_{0}$ and the time t when the velocity is $v$ , we have
${\int }_{0}^{v}\frac{d{v}^{\prime }}{{v}^{\prime }}=-\frac{b}{m}{\int }_{0}^{t}d{t}^{\prime }.$
Thus,
$\text{ln}\frac{v}{{v}_{0}}=-\frac{b}{m}t,$
which, since $\text{ln}A=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A,$ we can write this as
$v={v}_{0}{e}^{\text{−}bt\text{/}m}.$
Now from the definition of velocity,
$\frac{dx}{dt}={v}_{0}{e}^{\text{−}bt\text{/}m},$
so we have
$dx={v}_{0}{e}^{\text{−}bt\text{/}m}dt.$
With the initial position zero, we have
${\int }_{0}^{x}dx\text{'}={v}_{0}{\int }_{0}^{t}{e}^{\text{−}bt\text{'}\text{/}m}dt\text{'},$
and
${x=-\frac{m{v}_{0}}{b}{e}^{\text{−}bt\text{'}\text{/}m}|}_{0}^{t}=\frac{m{v}_{0}}{b}\left(1-{e}^{\text{−}bt\text{/}m}\right).$
As time increases, ${e}^{\text{−}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$
Although this tells us that the boat takes an infinite amount of time to reach ${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at $t=10m\text{/}b,$ we have
$v={v}_{0}{e}^{-10}\simeq 4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}{v}_{0},$
whereas we also have
$x={x}_{\text{max}}\left(1-{e}^{-10}\right)\simeq 0.99995{x}_{\text{max}}.$
Therefore, the boat’s velocity and position have essentially reached their final values.
2. With ${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and $v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have $1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=\left(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\right){e}^{\text{−}\left(b\text{/}m\right)\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right)},$ so
$\text{ln}\phantom{\rule{0.2em}{0ex}}0.25=\text{−}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0=-\frac{b}{m}\left(10\phantom{\rule{0.2em}{0ex}}\text{s}\right),$
and
$\frac{b}{m}=\frac{1}{10}\text{ln}\phantom{\rule{0.2em}{0ex}}4.0\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}=0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{-1}}\text{.}$
Now the boat’s limiting position is
${x}_{\text{max}}=\frac{m{v}_{0}}{b}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}}{0.14\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\text{−1}}}=29\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$
What is a volt equal to?
list and explain the 3 ways of charging a conductor
conduction convention rubbing
Asdesaw
formula of magnetic field
Integral of a vector
define surface integral of a vector?
Rahat
the number of degree freedom of a rigid body in2-dimantion is:
1
Nathan
A block (A) of weight 5 kN is to be raised by means of a 20° wedge (B) by the application of a horizontal force (P) as shown in Fig.1. The block A is constrained to move vertically by the application of a horizontal force (S). Find the magnitude of the forces F and S, when the coefficient of fricti
Danilo
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Previous
twenty four square plus thirty-five square minus to multiple thirty five twenty four and equal answer number square Via this equation defined Total Total impulse
Cemal
why simple pendulum do not vibrate indefinitely?
Zirmal
define integral vector
Rahat
what is matar
define surface integral vector?
Rahat
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
Jave
what is the S.I unit of coefficient of viscosity
Derived the formula of Newton's law of universal gravitation Fg=G(M1M2)/R2
hi
Asdesaw
yes
Cemal
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
what is the centripetal force
Of?
John
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle. Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle. Mathematically Fc=mv²/r
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2. By how much does the column shrink when a 5000- kg truck is on it?
hi
Abdulrahman
mola mass
Abdulrahman
hi
Asdesaw
what exactly is a transverse wave | 3,299 | 9,967 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 50, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-04 | latest | en | 0.868984 |
http://laesieworks.com/thoughts/Hierarchy_of_Shapes.html | 1,553,326,141,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202728.21/warc/CC-MAIN-20190323060839-20190323082839-00052.warc.gz | 124,550,353 | 7,823 | # Hierarchy of Shapes
The hierarchy of shapes.
This page is about the order of creation, from the primary basic elements to the most complex combinations of these basic elements.
Primary basic elements aren't necessarily small. Searching for the smallest element might be as smart as searching for the smallest number. What does count is how basic it is, and how organized the structure. A random mix of elements is just a random mix of elements. A simple repetition is a structure, but a bit more complex structures can be more beautiful.
I call any structured combination a 'shape', no matter how simple or complex it is, as long as it is contains little randomness.
In this text, I can't describe the reality of this universe, because I don't know what exactly are the primary basic elements. I'm just trying to understand the logic behind the shapes in our universe, if there is logic to it.
This search in shapes has two directions: towards less complexity and towards more complexity. Going towards less complexity will bring less combinations, and going towards more complexity will bring more and more combinations.
Romanesco broccoli is an edible flower of the species
Brassica oleracea, and a variant form of cauliflower.
An imaginary universe, example. The numbers 0 to 9 is what everybody is familiar with, good for this first example. Level 1 This level consists of the 'primary basic elements': 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. These are the 10 different numbers, that make up all larger whole numbers in this imaginary universe. Level 2 This level consists of combinations of 2 primary basic elements: 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, ... up to 99. That is 100 different combinations. OK, that is nothing interesting. Just counting as usual. Level 3 shall have 1000 combinations. Level 4 shall have 10000 combinations. So let's now make a small chance, and start over again. The numbers 5 and 7 are now erased from existence! Which has enormous consequences for the combinations of the levels above. Level 1 This level consists of the primary basic elements: 0, 1, 2, 3, 4, 6, 8, 9. That is 8 different numbers. Level 2 This level consists of combinations of 2 primary basic elements: 00, 01, 02, 03, 04, 06, 08, 09, 10, 11, ... up to 99. That is 64 different combinations. An easy way to calculate this is: the amount of numbers to the power of the number of positions. In this case: 8^2 = 64 Level 3 shall have 8^3 = 512 combinations. Level 4 shall have 8^4 = 4096 combinations. Level 4 now, has 10000-4096 = 5904 less combinations, as a result of erasing just 2 primary basic elements. Let's make a change ones more: remove 0 and 1, and add a new symbol "œ" in the beginning. Now this imaginary universe is very different from what we are used to. Level 1 This level consists of the primary basic elements: œ, 2, 3, 4, 6, 8, 9. That is 7 different symbols. Level 2 This level consists of combinations of 2 primary basic elements: œœ, œ2, œ3, œ4, œ6, œ8, œ9, 2œ, 22, ... up to 99. That is 49 different combinations. Level 3 shall have 7^3 = 343 combinations. Level 4 shall have 7^4 = 2401 combinations. When now translating this simple example to our reality, it becomes really really interesting! - It is the character of the primary basic elements, that dictates what combinations in the higher levels are possible. - Once you know the character of the primary basic elements of this universe, you can, in theory, know what unknowns are possible. - Any change in the primary basic elements, however small, will alter the shape of the universe dramatically. - If you can alter the primary basic elements, you can make the impossible possible, and the possible impossible. The real primary basic elements of this universe Unknown are the real primary basic elements. Luckily there are some people on this planet that do actively search for it. As for example the people of the Large Hadron Collider. What I hear myself repeating though, is "These primary basic elements don't have to be small. Searching for the smallest particle might be as smart as searching for the smallest number". What counts is how basic it is, not how small. But I'm sure they know that. The number of different primary basic elements will be very small, because they are at the end of the road towards the most simple. About how many? At least two, because it is impossible to make combinations with just one. I think space and energy are two good candidates, but I can't be sure. Space would the medium, and energy a change in that medium, giving motion and shape to the space. I can't imagine a third one yet. Time? I don't believe in Time. Will primary basic element type 'A' be just one big universal thing, or will there be a huge amount of them? And if there is a huge amount, will each of them be exactly the same, or will each one be similar but unique? The most basic parts (the primary basic elements) may be the most basic, but that doesn't necessarily say that they are simple. Bits of digital code is simple; a bit is a 0 or a 1, all 0 bits are exactly the same, as are the 1 bits. What about primary basic element type 'A'? Will it be as simple and static as a computer bit, or will it be complex and even dynamic? ==== so far rewritten Shapes are a combination of two or more different kinds of lower level parts. Which is possible because let's say there is attraction between them. Not all attractions are equal though. Imagine the attraction between two is being too weak. Than that combination is not available for all more complex combinations. Some combinations of shapes are more stable than others. Some "want" to bound and become like one, probably by sharing part of their structure. Some combinations of shapes are impossible: you can bring them together but they just won't bound. Some combinations of shapes want to join, but will not bound, they "want" to be together, but not become as one entirely. A nice illustration of a combination of 2 is: waves (energy) in water (a medium). Waves travel over the water, in all directions. Interference of two or more waves causes many different shapes of wave, including some rare freak waves. A freak wave is a giant peak in the water, where many waves come together, at one moment, at one spot, all with a large positive amplitude, to together cause one giant peak wave. Water is not an primary basic element though. And the water as we know it, has compressed air above it, both water and air in a gravitational field. God to me, equals the character of these basic primary basic elements. - It is everything - It is always present - It is why everything went as it did, and why everything will go as it will go.
An atom is a pretty stable shape, a combinations of more simple shapes like: electrons, protons, neutrons. Atoms aren't the most simple shapes in the hierarchy of shapes, nor are electrons, protons, neutrons. What are the basic elements isn't exactly clear yet. The character of the basic elements and lowest level shapes is very important to discover, as they define the character of ALL more complex shapes, like you and me and the world we live in. The whole chemistry thing is about the joining and bounding of atoms. Bound atoms are shapes we call molecules. In the hierarchy of shapes, molecules are at a higher level than atoms. Certain combinations of molecules join together and form stable shapes we call living cells. Biological living cells are at a higher level than molecules. The billions of cells in what we call our human body, are joint. People join together, live and work together, ..not always, but next to being individuals we often work together and accomplish things one individual couldn't. One of the things that working together has accomplished is technology; materials, tools, machines. Nobody knows exactly how all the stuff around us is made. A computer is among the most complex shapes humans have created. They are of a higher level then large groups of people, in the hierarchy of shapes. Take away the humans, and gone is the technology. When truly autonomous robots are developed by us, something strange could happen: If they are so advanced that they can survive without humans, build and develop themselves, they would not only be higher in the hierarchy of shapes than people, they could also live independent of people. Similar steps have occurred all throughout the history: shapes creating other kinds of independent shapes.
12 Pollen Mix. Source: Dartmouth Electron Microscope Facility
Destruction and Creation. When all shapes are destroyed, only the basic elements survive. Was there was a very turbulent beginning in the history of this universe? In such energetic situation, shapes can't survive. In a less energetic situation, caused for example by expansion (the spreading of energy) more complex shapes can form and survive longer. In a calm situation, very complex shapes can form. In a frozen situation, shapes can't form, nor can't they be destroyed. The higher levels are build up on lower levels. When for example all the level 14 shapes within a level 22 shape are destroyed, all shapes higher than level 14 will also be destroyed, but all the lower than level 14 shapes can survive. To destroy is easier than to create. Destruction is thereby more common then creation. A solutions to kind of escape from destruction is for a shape to create a new shape, similar to how itself began. Reproduction. The whole love sex and children thing. An other option is to clone, a shape that can make an exact copy of itself, or more likely; a repaired copy. Shapes that can't reproduce or clone itself, or be created by an other shape, will be destroyed by the decay caused by the interaction with other shapes. Only the ones that have a solution will survive as a kind and be able to continue the road of creation. The chance of a very high level shape being created suddenly by chance, is very very small, near to impossible. The road to very complex shapes is very likely to slowly go from simple to more complex. People wonder about how large steps in evolution occurred. Like the step from a soup of molecules to a single cell organism. I wonder about that anyhow (do you?). To say it came from an other planet or it was made by God is not a good answer, unless you want to stop thinking about it. The step is so large, like a city arising in a desert, just by coincidence. That's unbelievable. There must be steps in between or an other more complex shape that created it. But if an other shape created it, how did that shape came to be? Enemies and Friends. How about a period of hardly any "enemies" and lost of "friends". That would bring the chance of a complex shape created by chance up to a plausible level. What is an enemy? Imagine two shapes. One being the combination AAB and one being ABA. Both are pretty stable, but when getting in close contact, they interact and result in AAB and AB and A. Then AAB is the enemy of ABA. Because when they interact, AAB survives and ABA is destroyed. That doesn't make AAB better or worse, it's just like that. If ABA is a vital part of a more complex shape, that more complex shape can't be created with much AAB around. Friends. If ABA can bound with BBA, but only when AAA is in close contact also, than AAA is a friend of that creative process. A shape can be an enemy and friend at the same time. It's relative and this enemy and friend part has nothing to do with good or bad. Combinations and Chance. Very simple shapes like the 4 combinations of level 2 (AA, AB, BA, BB) can arise easily by chance, in an energetic soup of A and B. But a join of two level 2 shapes, can occur only after the creation of the level 2 shapes, or simultaneous. The chance of simultaneous creation of a complex shape and all it's less complex shapes, is much smaller than growing from less to more complex. But not impossible. Could a single cell living organism have come to life by just random chance? Not likely.
Galaxies. Photo made with the Hubble Space Telescope Source: NASA
Not Random. A universe that contains more than one type of basic element, has a natural tendency to some level of order. Like gravity in our universe. Gravity brings atoms together to from planets and stars. It separates the atoms according to weight: air, water, stone. People don't live at random locations in space, we usually live on the stone-air border, in the air part. The chance of a high level creation occurring is much higher than could be expected in a random chaos of basic soup. Like ice crystals, when water cools down, they arise. More different higher levels. If the amount of simple shapes is limited, there are more simple combinations than complex combinations. But there are much more different complex combinations possible than simple ones. The quality of shapes. Bigger or more complex isn't equal to better. Quality... what is quality? That can be very subjective. More objective would be to measure how pure the structure is, how few random elements it contains, how well it is organized. I wonder what shape the highest possible quality or level would be like. Should we strive to create that shape? If we can we'll probably do so, some day. We are curious, we boldly go where no human has gone before. Boldly or blindly, we usually have no clue where we are going. So, what are the "holy" proportions, what is the exact character of "God"? Yep, that's the question. Possibly the most important scientific question. I think we can discover it. Where can we find the right clues to the character of the basic elements? We (well... not me personally) are working on "the theory of everything". I think the circle is important, so: Pi. Remember it from high school? "The ratio of a circle's area to the square of its radius." The ratio: 3.14159265358979323846… And Phi, "the golden ratio", the ratio that can be found in nature a lot. "a+b is to a as a is to b" The ratio: 1.6180339887... And Phi's Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, … Maybe the powers of two, because cells multiply like that: 1, 2, 4, 8, 16, ... Maybe Marko Rodin's vortex-based mathematics? Let's play around with circles/balls. The results can be like "the flower of life" and such. Probably most useful clues can be found in chemistry, the knowledge of how atoms and molecules are.
Giesbert Nijhuis | 3,291 | 14,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-13 | longest | en | 0.93001 |
https://zims-en.kiwix.campusafrica.gos.orange.com/wikipedia_en_all_nopic/A/Gauss's_law_for_gravity | 1,606,716,307,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00396.warc.gz | 918,954,377 | 15,058 | # Gauss's law for gravity
In physics, Gauss's law for gravity, also known as Gauss's flux theorem for gravity, is a law of physics that is essentially equivalent to Newton's law of universal gravitation. It is named after Carl Friedrich Gauss. Although Gauss's law for gravity is equivalent to Newton's law, there are many situations where Gauss's law for gravity offers a more convenient and simple way to do a calculation than Newton's law.
The form of Gauss's law for gravity is mathematically similar to Gauss's law for electrostatics, one of Maxwell's equations. Gauss's law for gravity has the same mathematical relation to Newton's law that Gauss's law for electricity bears to Coulomb's law. This is because both Newton's law and Coulomb's law describe inverse-square interaction in a 3-dimensional space.
## Qualitative statement of the law
The gravitational field g (also called gravitational acceleration) is a vector field – a vector at each point of space (and time). It is defined so that the gravitational force experienced by a particle is equal to the mass of the particle multiplied by the gravitational field at that point.
Gravitational flux is a surface integral of the gravitational field over a closed surface, analogous to how magnetic flux is a surface integral of the magnetic field.
Gauss's law for gravity states:
The gravitational flux through any closed surface is proportional to the enclosed mass.
## Integral form
The integral form of Gauss's law for gravity states:
${\displaystyle \scriptstyle \partial V}$ ${\displaystyle \mathbf {g} \cdot d\mathbf {A} =-4\pi GM}$
where
${\displaystyle \scriptstyle \partial V}$ (also written ${\displaystyle \oint _{\partial V}}$) denotes a surface integral over a closed surface,
V is any closed surface (the boundary of an arbitrary volume V),
dA is a vector, whose magnitude is the area of an infinitesimal piece of the surface ∂V, and whose direction is the outward-pointing surface normal (see surface integral for more details),
g is the gravitational field,
G is the universal gravitational constant, and
M is the total mass enclosed within the surface ∂V.
The left-hand side of this equation is called the flux of the gravitational field. Note that according to the law it is always negative (or zero), and never positive. This can be contrasted with Gauss's law for electricity, where the flux can be either positive or negative. The difference is because charge can be either positive or negative, while mass can only be positive.
## Differential form
The differential form of Gauss's law for gravity states
${\displaystyle \nabla \cdot \mathbf {g} =-4\pi G\rho ,}$
where ${\displaystyle \nabla \cdot }$ denotes divergence, G is the universal gravitational constant, and ρ is the mass density at each point.
### Relation to the integral form
The two forms of Gauss's law for gravity are mathematically equivalent. The divergence theorem states:
${\displaystyle \oint _{\partial V}\mathbf {g} \cdot d\mathbf {A} =\int _{V}\nabla \cdot \mathbf {g} \ dV}$
where V is a closed region bounded by a simple closed oriented surface ∂V and dV is an infinitesimal piece of the volume V (see volume integral for more details). The gravitational field g must be a continuously differentiable vector field defined on a neighborhood of V.
Given also that
${\displaystyle M=\int _{V}\rho \ dV}$
we can apply the divergence theorem to the integral form of Gauss's law for gravity, which becomes:
${\displaystyle \int _{V}\nabla \cdot \mathbf {g} \ dV=-4\pi G\int _{V}\rho \ dV}$
which can be rewritten:
${\displaystyle \int _{V}(\nabla \cdot \mathbf {g} )\ dV=\int _{V}(-4\pi G\rho )\ dV.}$
This has to hold simultaneously for every possible volume V; the only way this can happen is if the integrands are equal. Hence we arrive at
${\displaystyle \nabla \cdot \mathbf {g} =-4\pi G\rho ,}$
which is the differential form of Gauss's law for gravity.
It is possible to derive the integral form from the differential form using the reverse of this method.
Although the two forms are equivalent, one or the other might be more convenient to use in a particular computation.
## Relation to Newton's law
### Deriving Gauss's law from Newton's law
Gauss's law for gravity can be derived from Newton's law of universal gravitation, which states that the gravitational field due to a point mass is:
${\displaystyle \mathbf {g} (\mathbf {r} )=-GM{\frac {\mathbf {e_{r}} }{r^{2}}}}$
where
er is the radial unit vector,
M is the mass of the particle, which is assumed to be a point mass located at the origin.
A proof using vector calculus is shown in the box below. It is mathematically identical to the proof of Gauss's law (in electrostatics) starting from Coulomb's law.[1]
### Deriving Newton's law from Gauss's law and irrotationality
It is impossible to mathematically prove Newton's law from Gauss's law alone, because Gauss's law specifies the divergence of g but does not contain any information regarding the curl of g (see Helmholtz decomposition). In addition to Gauss's law, the assumption is used that g is irrotational (has zero curl), as gravity is a conservative force:
${\displaystyle \nabla \times \mathbf {g} =0}$
Even these are not enough: Boundary conditions on g are also necessary to prove Newton's law, such as the assumption that the field is zero infinitely far from a mass.
The proof of Newton's law from these assumptions is as follows:
## Poisson's equation and gravitational potential
Since the gravitational field has zero curl (equivalently, gravity is a conservative force) as mentioned above, it can be written as the gradient of a scalar potential, called the gravitational potential:
${\displaystyle \mathbf {g} =-\nabla \phi .}$
Then the differential form of Gauss's law for gravity becomes Poisson's equation:
${\displaystyle \nabla ^{2}\phi =4\pi G\rho .}$
This provides an alternate means of calculating the gravitational potential and gravitational field. Although computing g via Poisson's equation is mathematically equivalent to computing g directly from Gauss's law, one or the other approach may be an easier computation in a given situation.
In radially symmetric systems, the gravitational potential is a function of only one variable (namely, ${\displaystyle r=|\mathbf {r} |}$), and Poisson's equation becomes (see Del in cylindrical and spherical coordinates):
${\displaystyle {\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}\,{\frac {\partial \phi }{\partial r}}\right)=4\pi G\rho (r)}$
while the gravitational field is:
${\displaystyle \mathbf {g} (\mathbf {r} )=-\mathbf {e_{r}} {\frac {\partial \phi }{\partial r}}.}$
When solving the equation it should be taken into account that in the case of finite densities ∂ϕ/∂r has to be continuous at boundaries (discontinuities of the density), and zero for r = 0.
## Applications
Gauss's law can be used to easily derive the gravitational field in certain cases where a direct application of Newton's law would be more difficult (but not impossible). See the article Gaussian surface for more details on how these derivations are done. Three such applications are as follows:
### Bouguer plate
We can conclude (by using a "Gaussian pillbox") that for an infinite, flat plate (Bouguer plate) of any finite thickness, the gravitational field outside the plate is perpendicular to the plate, towards it, with magnitude 2πG times the mass per unit area, independent of the distance to the plate[2] (see also gravity anomalies).
More generally, for a mass distribution with the density depending on one Cartesian coordinate z only, gravity for any z is 2πG times (the mass per unit area above z, minus the mass per unit area below z).
In particular, a combination of two equal parallel infinite plates does not produce any gravity inside.
### Cylindrically symmetric mass distribution
In the case of an infinite uniform (in z) cylindrically symmetric mass distribution we can conclude (by using a cylindrical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of 2G/r times the total mass per unit length at a smaller distance (from the axis), regardless of any masses at a larger distance.
For example, inside an infinite uniform hollow cylinder, the field is zero.
### Spherically symmetric mass distribution
In the case of a spherically symmetric mass distribution we can conclude (by using a spherical Gaussian surface) that the field strength at a distance r from the center is inward with a magnitude of G/r2 times only the total mass within a smaller distance than r. All the mass at a greater distance than r from the center can be ignored.
For example, a hollow sphere does not produce any net gravity inside. The gravitational field inside is the same as if the hollow sphere were not there (i.e. the resultant field is that of any masses inside and outside the sphere only).
Although this follows in one or two lines of algebra from Gauss's law for gravity, it took Isaac Newton several pages of cumbersome calculus to derive it directly using his law of gravity; see the article shell theorem for this direct derivation.
## Derivation from Lagrangian
The Lagrangian density for Newtonian gravity is
${\displaystyle {\mathcal {L}}({\vec {x}},t)=-\rho ({\vec {x}},t)\phi ({\vec {x}},t)-{1 \over 8\pi G}(\nabla \phi ({\vec {x}},t))^{2}}$
Applying Hamilton's principle to this Lagrangian, the result is Gauss's law for gravity:
${\displaystyle 4\pi G\rho ({\vec {x}},t)=\nabla ^{2}\phi ({\vec {x}},t).}$
See Lagrangian (field theory) for details.
## References
1. See, for example, Griffiths, David J. (1998). Introduction to Electrodynamics (3rd ed.). Prentice Hall. p. 50. ISBN 0-13-805326-X.
2. The mechanics problem solver, by Fogiel, pp 535–536 | 2,342 | 9,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 31, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-50 | latest | en | 0.929473 |
https://math.stackexchange.com/questions/4131432/proof-that-mathbbr2-cong-mathbbr | 1,695,372,055,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506339.10/warc/CC-MAIN-20230922070214-20230922100214-00486.warc.gz | 448,851,912 | 37,552 | # Proof that $\mathbb{R}^2 \cong \mathbb{R}$
I'm trying to prove that $$\mathbb{R}^2$$ and $$\mathbb{R}$$ have the same cardinality. Here is my attempt. I will be taking the Schröder-Bernstein theorem for granted.
Define the map $$f: \mathbb{R} \to \mathbb{R}^2$$ sending $$x \longmapsto (x,0)$$. I claim that $$f$$ is an injection. Indeed, if $$f(x) = f(y)$$ for $$x,y \in \mathbb{R}$$, then $$(x,0) = (y,0)$$, so $$x = y$$. Therefore, $$|\mathbb{R}| \leq |\mathbb{R}^2|$$. By the Schröder-Bernstein theorem, it suffices to prove that $$|\mathbb{R}^2| = |\mathbb{R}|$$.
Lemmma 1: $$\mathbb{R} \cong (0,1)$$.
Proof of Lemma 1. Notice that $$\tan(x)$$ is a bijection from $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to \mathbb{R}$$ since $$\tan$$ has period $$\pi$$; furthermore, the existence of the $$\arctan$$ function guarantees that the function is surjective, hence bijective. Notice that there is a one-to-one correspondence $$\left(0,1\right) \to \left(- \frac{\pi}{2}, \frac{\pi}{2}\right)$$ sending $$x \longmapsto \left(x - \frac{1}{2}\right)\pi$$. Composing bijections gives a bijection $$\left(0,1\right) \to \mathbb{R}$$.
Lemma 2: Given sets $$A, B, C,D$$ such that $$A \cong C$$ and $$B \cong D$$, we have $$A \times B \cong C \times D$$.
Proof of Lemma 2. Since $$A \cong C$$ and $$B \cong D$$, there exist bijections $$f: A \stackrel{\sim}{\to} C$$ and $$g: B \stackrel{\sim}{\to} D$$. Then define the map $$h: \underset{(a,b)}{A \times B} \underset{\longmapsto}{\to} \underset{(f(a), g(b))}{C \times D}.$$ Given $$(a_1, b_1), (a_2, b_2) \in A \times B$$ for which $$h(a_1, b_1) = h(a_2, b_2)$$, we have $$(f(a_1), g(b_1)) = (f(a_2), g(b_2))$$. So $$f(a_1) = f(a_2)$$ and $$g(b_1) = g(b_2)$$. The former implies $$a_1 = a_2$$ by injectivity of $$f$$, while the latter implies $$b_1 = b_2$$ by injectivity of $$g$$. So $$(a_1, b_1) = (a_2, b_2)$$, so $$h$$ is injective. Given $$(c,d) \times C \times D$$, by surjectivity of $$f$$, there exists $$a \in A$$ such that $$f(a) = c$$; by surjectivity of $$g$$, there exists $$b \in B$$ such that $$g(b) = d$$. We then have $$(c,d) = (f(a), g(b)) = h(a,b),$$ so $$h$$ is surjective, hence bijective, so $$A \times B \cong C \times D$$.
Corollary. It suffices to prove that $$(0,1) \times (0,1) \cong (0,1)$$.
Proof. We know $$\mathbb{R} \cong (0,1)$$ by Lemma 1. By Lemma 2, that implies that $$\mathbb{R} \times \mathbb{R} \cong (0,1) \times (0,1)$$. So if we prove $$(0,1) \times (0,1) \cong (0,1)$$, we have $$\mathbb{R} \times \mathbb{R} \cong (0,1) \times (0,1) \cong (0,1) \cong \mathbb{R},$$ which is the intended result.
Claim. $$(0,1) \times (0,1) \cong (0,1)$$.
Proof of Claim. We use the Schröder-Bernstein theorem. There is an obvious injection $$(0,1) \times (0,1) \times (0,1)$$ sending $$x \longmapsto \left(x, \frac{1}{2}\right)$$. This is an injection regardless of whether we allow an infinite string of $$9$$'s. Concordantly, we can define an injection $$(0,1) \times (0,1) \to (0,1)$$ as follows. Given $$(x,y) \in (0,1) \times (0,1)$$, choose a decimal expansion for $$x,y \in (0,1)$$ that does not contain an infinite string of nines, which we know to be unique. Write $$x = 0.x_1 x_2 x_3 \ldots \\ y = 0.y_1 y_2 y_3 \ldots$$ Then define $$f(x,y) = 0.x_1 y_1 x_2 y_2 x_3 y_3 \ldots$$ Since neither $$x,y$$ contained an infinite string of nines, $$f(x,y)$$ likewise does not. I claim that $$f$$ is an injection. Indeed, suppose that $$f(a,b) = f(c,d)$$ for $$(a,b), (c,d) \in (0,1) \times (0,1)$$. Defining the decimal expansions for $$a,b,c,d$$ similarly, we then have $$0.a_1 b_1 a_2 b_2 a_3 b_3 \ldots = 0.c_1 d_1 c_2 d_2 c_3 d_4 \ldots.$$ Since neither of these expressions contain an infinite string of $$9$$'s, we can equate corresponding digits, so $$a_i = c_i$$ for all $$i \geq 1$$ and $$b_i = d_i$$ for all $$i \geq i$$, so $$a = c$$ and $$b = d$$, so $$(a,b) = (c,d)$$, so $$f$$ is injective. By the Shroder-Bernstein theorem, $$(0,1) \times (0,1) \cong (0,1)$$.
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How does this proof look? The only fact I am not totally sure I understand fully is why the decimal expansion not terminating in a string of nine's is unique. I believe we can assert that every element of $$(0,1)$$ has exactly two decimal expansions. What is the canonical way to go about showing this?
• $x\mapsto (x,1/2)$ is not an injection $(0,1)^2\to (0,1),$ but the other way, $(0,1)\to(0,1)^2.$ May 8, 2021 at 4:39
• Sorry, this was a typo. I will fix it now. May 8, 2021 at 4:40
• Also, $(0,1)\to(0,1)\to(0,1)$ maybe should be $(0,1)\times (0,1)\to (0,1)?$ Otherwise, the proof looks fine. Very well-organized. May 8, 2021 at 4:42
• This looks fine. At some point soon you'll learn that for any infinite set $A, \vert A \times A \vert = \vert A \vert$. That will simplify matters a lot. May 8, 2021 at 4:42
• A real number has at most two decimal expansions as you said, and furthermore the only ones having two different expansions are real numbers of the form $n/10^{-m}$ where $n, m$ are integers. You can find this result as an exercise on Appendix B of Terence Tao's Analysis I. It is also proven in detail in the book "The number systems" of Solomon Feferman which I think it is really good. May 8, 2021 at 11:07
The map $$(0,1)\times(0,1)\to(0,1)$$ you build is certainly injective, but it is not bijective. Take $$0.191919\dotsb$$ to see why. However you don’t need surjectivity.
• He doesn't need it to be a bijection, as he's using Schröder-Bernstein theorem. May 8, 2021 at 10:29
• @Eparoh True but he says that it is which is a mistake. May 8, 2021 at 10:48
• True, I didn't notice it. I think it should be a typo because after all the work, it seems he knows what he's applying. May 8, 2021 at 11:02
• @Eparoh Maybe or maybe not since he admits confusion on decimal representations. Let's wait to hear from him. May 8, 2021 at 11:07
• This was indeed a typo. Thank you for pointing that out. May 8, 2021 at 17:32
Egreg has identified a problem but you might like more detail.
Your map $$f$$ from $$(0, 1) \times (0, 1) \rightarrow (0, 1)$$ is an injection but not a bijection, see below. However, if it was a bijection then you would not need your map the other way or Schröder-Bernstein since the bijection, by itself, would show that the sets have the same cardinality.
You say that every element of $$(0, 1)$$ has exactly two decimal expansions. Most have just one. It is only the ones that end in all $$9$$s or all $$0$$s that do. See this previous question for details.
Now, I say that your map is an injection but not a bijection. So, I am saying that it is not surjective. I need an example to prove this. What maps to $$0.459090909...$$? It's $$(0.4999..., 0.5000...)$$ except the representation $$0.4999...$$ breaks your rule, it is the wrong representation of $$0.5000...$$. So that point is in fact $$(0.5000..., 0.5000...)$$ which maps to $$0.55000...$$.
However, this does not destroy your proof. It just means you need Schröder-Bernstein after all. You have an injection both ways. The whole point of Schröder-Bernstein is that it avoids the need to establish an explicit bijection which is often messy or difficult. | 2,537 | 7,131 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 95, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | latest | en | 0.685825 |
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# Geometric Inequalities
Inequality is a big problem in the world today. We hear about gender inequality, ethnicity inequality, and many more. When we talk about inequalities, we refer to the unequal relationship between groups or quantities.
In Math, it is the unequal relationship between two numbers or expressions. In Geometry, we look at the unequal relationships between side lengths and between angles in various shapes.
In this article, we are going to look at:
• Geometric Inequalities
• Geometric Inequality Postulates
• Geometric Inequality Theorem
## What are Geometric Inequalities?
Geometrical inequalities refer to the unequal relationship among angles and side lengths in various shapes.
Let's recall the symbols used to represent inequalities and what they mean:
There are some general rules, or postulates, that can help us make comparisons about unequal side lengths and unequal angles. Let's look at them!
## Basic Geometrical Inequality Postulates
Various inequality postulates and properties are used when dealing with inequalities. Let's take a look at them.
• Comparison Postulate
• Transitive Postulate
• Substitution postulate
• Subtraction postulate
• Multiplication postulate
• Division postulate
These postulates can also be applied when dealing with geometry. Let's look at the various postulates and see how they can be applied to geometry.
What is the difference between a postulate and a theorem, you may wonder? A theorem is a statement that can be proven to be true, but a postulate is a statement that is said to be true without proof.
### Comparison Postulate
This postulate states that a whole is greater than each of its parts and the sum of the parts is equal to the whole. So, if a, b, and c are positive numbers and , then and . Let's see how we can apply this to geometry.
Using the image below, the Comparison postulate states, assuming , and , then
We can also apply the Comparison postulate to angles. Let's look at the figure below.
Line segments used to illustrate Comparison Postulate - StudySmarter Originals
We can say
### Transitive Property
The transitive property explains that if you have three real numbers, a, b, and c, and the first of the three numbers is greater than the second and the second number is greater than the third, then the first is also greater than the third. In other words, if and then, . Let's apply it to geometry using the geometric figure below.
Triangle used to illustrate Transitive Property - StudySmarter Originals
Using the image above, the transitive property states that:
if and , then
### Substitution Postulate
This postulate states that in an inequality, we can swap a number of equal value in. This means if you have three real numbers a, b and c and if and , then .
Let's apply this to geometry using the figure below.
Triangle used to illustrate Substitution Postulate - StudySmarter Originals
Using the figure above, the substitution postulate states that:
if and , then
Imagine you have a line segment as shown below, and somewhere on the line, you have another point . The point can only exist if .
Line Segment - StudySmarter Originals
The addition postulate has two different types.
1. The first one states that if equal quantities are added to unequal quantities, the sum is unequal in the same order.
Two line segments - StudySmarter Originals
For the figure above, the postulate is as follows when equal quantities are added to unequal quantities.
If and , then or .
2. The second one states that if unequal quantities are added to unequal quantities then the sum is unequal.
Two line segments - StudySmarter Originals
For the figure above, the postulate is as follows when unequal quantities are added to unequal quantities.
If and , then or
### Subtraction Postulate
The subtraction postulate states that if equal quantities are subtracted from unequal quantities, then the differences are unequal in the same order. Consider the figure below.
Triangle used to illustrate Subtraction Postulate - StudySmarter Originals
Using the figure above, the subtraction postulate is as follows.
If and , then or
### Multiplication Postulate
The postulate states that if unequal quantities are multiplied by equal positive quantities, the products are unequal in the same order.
Line segments - StudySmarter Originals
If , , and , then
### Division Postulate
The postulate states that if unequal quantities are divided by equal positive quantities, the quotients are unequal in the same order.
Triangle used to illustrate Division Postulate - StudySmarter Originals
If , then
## Geometric Inequality Theorems
There are various geometric inequalities theorems, and they are listed below.
1. Triangle Inequality Theorem.
2. Pythagorean Inequality Theorem.
3. Exterior Angle Inequality Theorem.
4. Greater Angle Theorem.
5. Longer Side Theorem.
Let's take a deeper look using some examples.
### Triangle Inequality Theorem
Triangle Inequality Theorem states that the length of one side of a triangle is smaller than the sum of the other two sides.
A Triangle - StudySmarter Originals
Considering the figure above, the triangle theorem states that . Let's take an example.
In , and , find the range of possible values of .
Step 1: Using the Triangle Inequality Theorem,
Step 2: Subbing in values from above:
This means that is less than 18. But we can't say is any number less than 18 because any number less than 18 may be too small to even make or complete the triangle.
Step 3: Let's write two more inequalities using the Triangle Inequality Theorem.
Lastly,
So far, we have three different inequalities.
Step 4: The actual range will be the intersection of the three inequalities, and we can get this using a number line.
From the number line, we can tell that the intersection is between 2 and 18. This means that the range of values is between 2 and 18. Therefore,
### Pythagorean Inequality Theorem
The Pythagorean Inequality Theorem tells you whether a triangle is a right triangle, an obtuse triangle, or an acute triangle.
The theorem states that you have a right triangle if the square of the longest side equals the sum of the squares of the other two sides. If the square of the longest side is less than the sum of the squares of the other two sides, the triangle is acute. If the square of the longest side is greater than the sum of the square of the other two sides, then the triangle is obtuse.
In below, C is the longest side.
A Triangle - StudySmarter Originals
If , then the triangle is a right triangle.
If , then the triangle is acute.
If , then the triangle is obtuse.
### Exterior Angle Inequality Theorem
The theorem states that the measure of the exterior angle of a triangle is greater than the non-adjacent interior angles of the triangle. Take a look at the figure below.
In the figure below, the exterior angle is , and angle and are the non-adjacent interior angles. Using the Exterior Angle Inequality Theorem, we can conclude that:
A Triangle with an exterior angle - StudySmarter Originals
### Greater Angle Theorem
The Greater Angle Theorem states that if the measure of one angle in a triangle is greater than the measure of another, then the side opposite the greater angle is longer than the side opposite the smaller angle. In other words
In the below,
if , then
A Triangle - StudySmarter Originals
Let's see an example.
Use the greater angle theorem to determine which side is the longest.
From the Greater Angle Theorem, we know that the longest side is the side opposite the largest angle. In the figure above, the longest side is side because it is the side opposite the largest angle,
### Longer Side Theorem
The Longer Side Theorem says that if the length of one side of a triangle is longer than another side, the angle opposite the longer side is greater than the angle opposite the shorter side. This is the converse of the Greater Angle Theorem above. In the below,
If , then
A Triangle - StudySmarter Originals
Let's take a look at a quick example.
Use the longer side theorem to determine which angle is greatest.
From the theorem, we know that the angle opposite the longest side is the greatest angle. The smallest side is 6 and the opposite angle is. Next to the smallest side is 8 and the opposite angle is. The longest side is 10 and the opposite angle is.
So, according to the length of the sides, the greater angle is .
## Geometrical Inequalities - Key takeaways
• Geometrical inequalities refer to the unequal relationships between side lengths and between angles in various shapes
• The postulates of inequalities are the same as their properties. The postulate includes:
• Comparison Postulate.
• Transitive Postulate.
• Substitution postulate.
• Subtraction postulate.
• Multiplication postulate.
• Division postulate.
• Geometric Inequality Theorems include:
• Triangle Inequality Theorem
• Pythagorean Inequality Theorem
• Exterior Angle Inequality Theorem
• Greater Angle Theorem
• Longer Side Theorem
## Frequently Asked Questions about Geometric Inequalities
Geometric inequality refer to the relationship between geometric quantities and measurements that are unequal.
Geometric inequalities are solved by applying the appropriate postulate and theorem.
The triangle theorem states that the length of one side of a triangle is smaller than the sum of the other two sides.
## Final Geometric Inequalities Quiz
Question
What are geometric inequalities?
Geometrical inequalities refer to the relationship between geometrical quantities and measurements that are unequal.
Show question
Question
List some of the basic inequality postulates.
• Comparison Postulate.
• Transitive Property.
• Substitution Postulate.
• Subtraction Postulate.
• Multiplication Postulate.
• Division Postulate.
Show question
Question
The basic inequality postulates are the same as the properties of geometric inequalities.
True
Show question
Question
List some of the geometric inequality theorems.
• Triangle Inequality Theorem.
• Exterior Angle Inequality Theorem.
• Greater Angle Theorem.
• Longer Side Theorem.
Show question
Question
What is the Triangle Inequality Theorem in geometry?
The Triangle Inequality Theorem states that the length of one side of a triangle is smaller than the sum of the other two sides.
Show question
Question
What is the Exterior Angle Inequality Theorem?
The Exterior Angle Inequality Theorem states that the measure of the exterior angle of a triangle is greater than the non-adjacent interior angles of the triangle.
Show question
Question
What is the Pythagorean Inequality Theorem?
The theorem states that a triangle is a right triangle if the square of the longest side is equal to the sum of the squares of the other two sides. If the square of the longest side is less than the sum of the squares of the other two sides, the triangle is acute. If the square of the longest side is greater than the sum of the square of the other two sides, then the triangle is obtuse.
Show question
Question
What is the Greater Angle Theorem?
The Greater Angle Theorem states that if the measure of one angle in a triangle is greater than the measure of another angle, then the side opposite the greater angle is longer than the side opposite the smaller angle.
Show question
Question
Which of the following theorems tell you if a triangle is right, obtuse or acute?
Pythagorean Inequality Theorem.
Show question
Question
What is the Exterior Angle Inequality Theorem?
The Exterior Angle Inequality states that the measure of the exterior angle of a triangle is greater than the non-adjacent interior angles of the triangle.
Show question
Question
The comparison postulate states that a number or quantity can be substituted for its equal in an inequality.
False
Show question
Question
What are the theorems used in solving two triangle inequalities?
• Side-Angle-Side (SAS) Inequality Theorem or Hinge Theorem.
• Side-Side-Side (SSS) Inequality Theorem.
Show question
Question
State the SAS Inequality Theorem.
The SAS Inequality Theorem states that if two sides of a triangle are congruent to two sides of another triangle, and the measure of the included angle in the first triangle is larger than the measure of the included angle in the second triangle, then the third side of the first triangle is longer than the third side of the second triangle.
Show question
Question
State the SSS Inequality Theorem.
The SSS Inequality Theorem states that if two sides of a triangle are congruent to two sides of another triangle and the third side of the first triangle is longer than the other, then the measure of the included angle in the first triangle is larger than the measure of the included angle in the second triangle.
Show question
Question
What is another name for the SAS Inequality Theorem?
The Hinge Theorem.
Show question
Question
What is does the SSS acronym stand for?
Side-Side-Side.
Show question
Question
What does the SAS acronym stand for?
Side-Angle-Side.
Show question
Question
If the measure of the included angle in one triangle is larger than the measure of the included angle in another triangle, then the third side of the first triangle is longer than the third side of the second triangle.
True or False.
True
Show question
Question
If the third side of one triangle is longer than the third side of another triangle, then the measure of the included angle in the first triangle is larger than the measure of the included angle in the second triangle.
True or False.
True.
Show question
Question
How many possible ways are there to prove the SAS inequality theorem?
2
Show question
Question
What does the triangle inequality theorem say?
The triangle inequality theorem says that the sum of two sides of a triangle is greater than the third side of the triangle.
Show question
Question
Using algebra, how can the triangle inequality theorem be written?
\begin{align} & |AC| + |BC| > |AB| \\ & |AB| + |AC| > |BC| \\ & |AB| + |BC| > |AC|. \end{align}
Show question
Question
Describe the interior angles of the triangle, and name one thing that is true about them.
The interior angles are the ones inside the triangle, and their measures add up to $$180^\circ$$.
Show question
Question
What does the angle side theorem tell you?
The Angle Side theorem says that if one side is longer than another, then their angle opposite the longest side is bigger than the angle opposite the shorter side.
Show question
Question
What are three ways to write the measure of an angle?
• $$m (\angle A)$$
• $$\text{meas }\angle A$$, or
• $$\measuredangle A$$
Show question
Question
What kind of proof do you use to prove the triangle inequality theorem?
A geometric proof.
Show question
Question
What does it mean if two triangle sides are congruent?
If two triangle sides are congruent it means that they are equal in length, this means that the angles opposite will also be congruent and equal in size.
Show question
Question
A triangle has the side lengths $$2$$ and $$5$$, find the possible lengths of the third side of the triangle.
Using the triangle inequality theorem, the third side is between $$3$$ and $$7$$ units long.
Show question
Question
What is a triangle inequality?
A triangle inequality is an inequality that is true about any kind of triangle.
Show question
Question
What is the triangle inequality theorem?
For any triangle, if you add up the length of any two sides, it will be larger than the length of the remaining side.
Show question
Question
If you were told that a triangle has the side lengths $$18$$ and $$9$$, what would you use to find the possible lengths of the third side of the triangle?
The triangle sides and angles inequality theorem.
Show question
Question
What is the exterior angle triangle inequality?
The exterior angle triangle inequality says that the measure of an exterior angle is bigger than the measure of either of its two remote interior angles.
Show question
Question
What theorem tells you that for any triangle, if you add up the length of any two sides, it will be larger than the length of the remaining side?
This is the triangle inequality theorem.
Show question
Question
What theorem tells you that for any triangle, if one side is longer than another, then their angle opposite the longest side is bigger than the angle opposite the shorter side?
This is the angle side triangle theorem.
Show question
Question
What theorem tells you that the measure of an exterior angle is bigger than the measure of either of its two remote interior angles?
This is the exterior angle triangle inequality.
Show question
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OK, I have had trouble with math, 'C' average, and we are learning proportions. I don't get them! Can someone please do a proportion and then tell the steps to do it? Ex. 4/6= 8/X. Steps: 4/6 equals 8 over x. You do _______ do get ________ over , so X=12. [I know this is rong...
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During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following...
Statistics
During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following...
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I need to find the chi square distribution. Thanks During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a ...
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I need to find the chi square distribution. Thanks During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a ...
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Can someone tell me how to solve these proportions? Thank you 5 7 ---- ----- = x - 1 x
During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following...
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Find the value of z that would be used to test the difference between the proportions, given the following. (Use G - H. Give your answer correct to two decimal places.) Sample n x G 386 327 H 414 321
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Find the value of z that would be used to test the difference between the proportions, given the following. (Use G - H. Give your answer correct to two decimal places.) Sample n x G 386 327 H 414 321
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Find the value of z that would be used to test the difference between the proportions, given the following. (Use G - H. Give your answer correct to two decimal places.) Sample n x G 386 327 H 414 321 .
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Post a New Question | 4,374 | 16,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-26 | latest | en | 0.920972 |
https://www.easyelimu.com/high-school-notes/maths/form-1/item/2010-scale-drawing | 1,716,895,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00231.warc.gz | 629,917,092 | 29,732 | # Scale Drawing - Mathematics Form 1 Notes
## The Scale
• The ratio of the distance on a map to the actual distance on the ground is called the scale of the map. The ratio can be in statement form e.g. 50 cm represents 50,000 cm or as a representative fraction (R.F), 1 : 5,000,000 is written as 1/5,000,000
Example
The scale of a map is given in a statement as 1 cm represents 4 km. convert this to a representative fraction (R.F).
Solution
One cm represents 4 x 1 00,000 cm. 1 cm represents 400, 000
Therefore, the ratio is 1 : 400,000 and the R.F is
1/400,000
Example
The scale of a map is given as 1:250,000.Write this as a statement.
Solution
1 :250,000 means 1 cm on the map represents 250,000 cm on the ground. Therefore, 1 cm represents 250,000/100,000 km
I.e. 1 cm represents 2.5 km.
### Scale Diagram
• When during using scale, one should be careful in choosing the right scale, so that the drawing fits on the paper without much details being left.
## Bearing and Distances
• Direction is always found using a compass point.
• A compass has eight points as show above. The four main points of the compass are North, South, East, and West. The other points are secondary points and they include North East (NE), South East (SE), South West (SW) and North West (NW).Each angle formed at the centre of the compass is 450 the angle between N and E is 900.
### Compass Bearing
• When the direction of a place from another is given in degrees and in terms of four main points of a compass. E.g. N45 0 W, then the direction is said to be given in compass bearing.
• Compass bearing is measured either clockwise or anticlockwise from North or south and the angle is acute.
### True Bearing
• North East direction, written as N450 E can be given in three figures as 0450 measured clockwise from True North.
• This three- figure bearing is called the true bearing.
• The true bearings due north is given as 0000 .Due south East as 1350 and due North West as 3150.
Example
From town P, a town Q is 60km away on a bearing South 80º east. A third town R is 100km from P on the bearing South 40º west. A cyclist travelling at 20km/h leaves P for Q. He stays at Q for one hour and then continues to R. He stays at R for 1½ hrs. and then returns directly to P.
1. Calculate the distance of Q from R.
P² = 100² + 60² - 2(100) (60) Cos 120
P² = 13600 – 12000 Cos 120
P² = 19600
P = 140km
2. Calculate the bearing of R from Q.
140 = 100
Sin 120 Sin Q
= 38.2°
Bearing 270 – 38.2 = 241.8°
3. What is the time taken for the whole round trip?
Time from P to R = 60/20 = 3 hrs
Time from Q to R = 140/20 = 7 hrs
From R to P = 100/20 = 5 hrs
Taken travelling = 3 + 7 + 5
= 15hrs
Example
A port B is on a bearing 080o from a port A and a distance of 95 km. A Submarine is stationed at a port D, which is on a bearing of 200o from A, and a distance of 124 km from B. A ship leaves B and moves directly Southwards to an Island P, which is on a bearing of 140o from A. The Submarine at D on realizing that the ship was heading to the Island P, decides to head straight for the Island to intercept the ship.
1. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D and P. {4 marks}
B1 for all three correct
2. Hence find:
1. The distance from A to D.
4.6 ± 0.1 ×10 = 46 KM ± 1 Km
2. The bearing of the Submarine from the ship when the ship was setting off from B
240° ± 1° OR S 60 ° W ± 2°
3. The bearing of the Island P from D.
122° ± OR S 58° E ± 1°
4. The distance the Submarine had to cover to reach the Island
127 ± 0.1 ×10 = 127 ± 1 Km
## Angle of Elevation and Depression
### Angle of Elevation:
• The angle above the horizontal that an observer must look to see an object that is higher than the observer. Example, a man looking at a bird.
### Angle of Depression:
• The angle below horizontal that an observer must look to see an object that is lower than the observer. Example, a bird looking down at a bug.
• Angles of depression and elevation can be measured using an instrument called clinometer
• To find the heights or the lengths we can use scale drawing.
## Simple Survey Methods
• This involves taking field measurements of the area so that a map of the area can be drawn to scale. Pieces of land are usually surveyed in order to:
• Fix boundaries
• For town planning
• Road construction
• Water supplies
• Mineral development
## Areas of Irregular Shapes
• Areas of irregular shape can be found by subdividing them into convenient geometrical shapes e.g. triangles, rectangles or trapezia.
Example
The area in hectares of the field can be found by the help of a base line and offsets as shown.
XY is the base line 360m. SM,RP and QN are the offsets.
Taking X as the starting point of the survey,the information can be entred in a field book as follows.
The sketch is as follows:
Using a suitable scale.
The area of the separate parts is found then combined.
Area of:
Triangle XPR is ½ x1 80 x 90 = 8100
m2
Triangle PRY is ½ x1 80 x 90 = 8100 m2
Triangle XSM is ½ x1 20 x 60 = 3600 m2
Triangle QNY is ½ x1 20 x 1 80 = 10800m2
Trapezium SQNM = ½(QN + SM) x SQ m2
½(180 + 60) x 120 m2
= 14400m2
Total area = 8100 + 8100 + 3600 + 10800 + 14400 = 45000m2
## Past KCSE Questions on the Topic
1. A point B is on a bearing of 0800 from a port A and at a distance of 95 km. A submarine is stationed at a port D, which is on a bearing of 2000 from AM and a distance of 124 km from B. A ship leaves B and moves directly southwards to an island P, which is on a bearing of 140 from A. The submarine at D on realizing that the ship was heading fro the island P, decides to head straight for the island to intercept the ship. Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative positions of A, B, D, P. Hence find
1. The distance from A to D
2. The bearing of the submarine from the ship was setting off from B
3. The bearing of the island P from D
4. The distance the submarine had to cover to reach the island P
2. Four towns R, T, K and G are such that T is 84 km directly to the north R, and K is on a bearing of 2950 from R at a distance of 60 km. G is on a bearing of 3400 from K and a distance of 30 km. Using a scale of 1 cm to represent 10 km, make an accurate scale drawing to show the relative positions of the town.
Find
1. The distance and the bearing of T from K
2. The distance and the bearing G from T
3. The bearing of R from G
3. Two aeroplanes, S and T leave airports A at the same time. S flies on a bearing of 060 at 750 km/h while T flies on a bearing of 2100 at 900km/h.
1. Using a suitable scale, draw a diagram to show the positions of the aeroplane after two hours.
2. Use your diagram to determine
1. The actual distance between the two aeroplanes
2. The bearing of T from S
3. The bearing of S from T
4. A point A is directly below a window. Another point B is 15 m from A and at the same horizontal level. From B angle of elevation of the top of the bottom of the window is 300 and the angle of elevation of the top of the window is 350. Calculate the vertical distance.
1. From A to the bottom of the window
2. From the bottom to top of the window
5. Find by calculation the sum of all the interior angles in the figure ABCDEFGHI below
6. Shopping centers X, Y and Z are such that Y is 12 km south of X and Z is 15 km from X. Z is on a bearing of 3300 from Y. Find the bearing of Z from X.
7. An electric pylon is 30m high. A point S on the top of the pylon is vertically above another point R on the ground. Points A and B are on the same horizontal ground as R. Point A due south of the pylon and the angle of elevation of S from A is 260. Point B is due west of the pylon and the angle of elevation of S from B is 320Find the
1. Distance from A and B
2. Bearing of B from A
8. The figure below is a polygon in which AB = CD = FA = 12cm BC = EF = 4cm and BAF =- CDE = 1200. AD is a line of symmetry.
Find the area of the polygon.
9. The figure below shows a triangle ABC
1. Using a ruler and a pair of compasses, determine a point D on the line BC such that BD:DC = 1:2.
2. Find the area of triangle ABD, given that AB = AC.
10. A boat at point x is 200 m to the south of point Y. The boat sails X to another point Z. Point Z is 200m on a bearing of 3100 from X, Y and Z are on the same horizontal plane.
1. Calculate the bearing and the distance of Z from Y
2. W is the point on the path of the boat nearest to Y. Calculate the distance WY
3. A vertical tower stands at point Y. The angle of point X from the top of the tower is 60 calculate the angle of elevation of the top of the tower from W.
11. The figure below shows a quadrilateral ABCD in which AB = 8 cm, DC = 12 cm, ∠BAD = 450, ∠CBD = 900 and ∠BCD = 300.
Find:
1. The length of BD
2. The size of the angle ADB
12. In the figure below, ABCDE is a regular pentagon and ABF is an equilateral triangle
Find the size of
1. ∠DE
2. ∠AEF
3. ∠DAF
13. In this question use a pair of compasses and a ruler only
1. Construct triangle ABC such that AB = 6 cm, BC = 8cm and ∠ABC 135(2 marks)
2. Construct the height of triangle ABC in a) above taking BC as the base (1 mark)
14. The size of an interior angle of a regular polygon is 3x0 while its exterior angle is (x − 20)0. Find the number of sides of the polygon
15. Points L and M are equidistant from another point K. The bearing of L from K is 3300. The bearing of M from K is 2200Calculate the bearing of M from L
16. Four points B,C,Q and D lie on the same plane point B is the 42 km due south- west of town Q. Point C is 50 km on a bearing of 5600 from Q. Point D is equidistant from B, Q and C.
1. Using the scale 1 cm represents 10 km, construct a diagram showing the position of B, C, Q and D
2. Determine the
1. Distance between B and C
2. Bearing D from B
17. Two aeroplanes P and Q, leave an airport at the same time flies on a bearing of 2400 at 900km/hr while Q flies due East at 750 km/hr
1. Using a scale of 1 v cm drawing to show the positions of the aeroplanes after 40 minutes.
2. Use the scale drawing to find the distance between the two aeroplane after 40 minutes
3. Determine the bearing of
1. P from Q ans 2540
2. Q from P ans 740
18. A port B is no a bearing of 080 from a port A and at a distance of 95 km. A submarine is stationed port D which is on a bearing of 2000 from A, and a distance of 124 km from B. A ship leaves B and moves directly southwards to an island P, which is on a bearing of 1400 from A. The submarine at D on realizing that the ship was heading for the island P decides to head straight for the island to intercept the ship.
Using a scale of 1 cm to represent 10 km, make a scale drawing showing the relative position of A, B D and P.
Hence find:
1. The distance from A and D
2. The bearing of the submarine from the ship when the ship was setting off from B
3. The baring of the island P from D
4. The distance the submarine had to cover to reach the island
19. Four towns R, T, K and G are such that T is 84 km directly to the north R and K is on a bearing of 2950 from R at a distance of 60 km. G is on a bearing of 3400 from K and a distance of 30 km. Using a scale of 1 cm to represent 10 km, make an acute scale drawing to show the relative positions of the towns.
Find
1. The distance and bearing of T from K
2. The bearing of R from G
20. In the figure below, ABCDE is a regular pentagon and M is the midpoint of AB. DM intersects EB at N. (T7)
Find the size of
1. ∠BAE
2. ∠BED
3. ∠BNM
21. Use a ruler and compasses in this question. Draw a parallelogram ABCD in which AB = 8cm, BC = 6 cm and ∠BAD = 75o. By construction, determine the perpendicular distance between AB and CD.
22. The interior angles of the hexagon are 2x0, ½x0, x + 400, 1100, 1300 and 1600. Find the value of the smallest angle.
23. The size of an interior angle of a regular polygon is 1560. Find the number of sides of the polygon.
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and get it sent to you via WhatsApp | 3,556 | 12,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-22 | latest | en | 0.9154 |
https://forum.freecodecamp.org/t/freecodecamp-challenge-guide-roman-numeral-converter/16044 | 1,603,143,195,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00284.warc.gz | 341,570,284 | 16,326 | # Roman Numeral Converter
## Problem Explanation
You will create a program that converts an integer to a Roman Numeral.
## Hints
### Hint 1
Creating two arrays, one with the Roman Numerals and one with the decimal equivalent for the new forms will be very helpful.
### Hint 2
If you add the numbers to the arrays that go before the new letter is introduced, like values for 4, 9, and 40, it will save you plenty of code.
### Hint 3
You can’t have more than three consecutive Roman numerals together.
## Solutions
Solution 1 (Click to Show/Hide)
``````var convertToRoman = function(num) {
var decimalValue = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
var romanNumeral = [
"M",
"CM",
"D",
"CD",
"C",
"XC",
"L",
"XL",
"X",
"IX",
"V",
"IV",
"I"
];
var romanized = "";
for (var index = 0; index < decimalValue.length; index++) {
while (decimalValue[index] <= num) {
romanized += romanNumeral[index];
num -= decimalValue[index];
}
}
return romanized;
};
// test here
convertToRoman(36);
``````
#### Code Explanation
• We start off by creating two arrays with default conversion with matching indices. These are called `decimalValue` and `romanNumeral`. We also create an empty string variable, `romanized`, which will house the final roman number.
• Using a for loop, we loop through the indicies of the `decimalValue` array. We continue to loop until while the value at the current `index` will fit into `num`.
• Next, we add the roman numeral and decrease `num` by the decimal equivalent.
• Finally, we return the value of `romanized`.
Solution 2 (Click to Show/Hide)
``````function convertToRoman(num) {
var romans = ["I", "V", "X", "L", "C", "D", "M"],
ints = [],
romanNumber = [],
numeral = "",
i;
while (num) {
ints.push(num % 10);
num = Math.floor(num / 10);
}
for (i = 0; i < ints.length; i++) {
units(ints[i]);
}
function units() {
numeral = romans[i * 2];
switch (ints[i]) {
case 1:
romanNumber.push(numeral);
break;
case 2:
romanNumber.push(numeral.concat(numeral));
break;
case 3:
romanNumber.push(numeral.concat(numeral).concat(numeral));
break;
case 4:
romanNumber.push(numeral.concat(romans[i * 2 + 1]));
break;
case 5:
romanNumber.push(romans[i * 2 + 1]);
break;
case 6:
romanNumber.push(romans[i * 2 + 1].concat(numeral));
break;
case 7:
romanNumber.push(romans[i * 2 + 1].concat(numeral).concat(numeral));
break;
case 8:
romanNumber.push(
romans[i * 2 + 1]
.concat(numeral)
.concat(numeral)
.concat(numeral)
);
break;
case 9:
romanNumber.push(romans[i * 2].concat(romans[i * 2 + 2]));
}
}
return romanNumber
.reverse()
.join("")
.toString();
}
// test here
convertToRoman(97);
``````
#### Code Explanation
• Create an array of Roman Numerals (`romans`).
• Use a for loop to create an array of the digits (`ints`) in the number.
• Loop through the array of digits (base 10) and as you do, increment the Roman Numeral (base 5) index by 2 (`numeral = romans[i*2]`).
• Within the loop, use Switch Case to push the proper Roman Numerals (backwards) onto that array.
• Reverse the Roman Numerals array and turn it into a string.
Solution 3 (Click to Show/Hide)
``````function convertToRoman(num) {
var romans = [
// 10^i 10^i*5
["I", "V"], // 10^0
["X", "L"], // 10^1
["C", "D"], // 10^2
["M"] // 10^3
],
digits = num
.toString()
.split("")
.reverse()
.map(function(item, index) {
return parseInt(item);
}),
numeral = "";
// Loop through each digit, starting with the ones place
for (var i = 0; i < digits.length; i++) {
// Make a Roman numeral that ignores 5-multiples and shortening rules
numeral = romans[i][0].repeat(digits[i]) + numeral;
// Check for a Roman numeral 5-multiple version
if (romans[i][1]) {
numeral = numeral
// Change occurrences of 5 * 10^i to the corresponding 5-multiple Roman numeral
.replace(romans[i][0].repeat(5), romans[i][1])
// Shorten occurrences of 9 * 10^i
.replace(
romans[i][1] + romans[i][0].repeat(4),
romans[i][0] + romans[i + 1][0]
)
// Shorten occurrences of 4 * 10^i
.replace(romans[i][0].repeat(4), romans[i][0] + romans[i][1]);
}
}
return numeral;
}
// test here
convertToRoman(36);
``````
#### Code Explanation
• Use an array (`romans`) to create a matrix containing the Roman numeral for a given power of 10 and, if available, the Roman numeral for that power of 10 times 5.
• Convert the input number (`num`) to a reversed array of digits (`digits`) so that we can loop through those digits starting with the ones position and going up.
• Loop through each digit, starting with the ones place, and create a Roman numeral string by adding each higher-power Roman numeral to the start of the `numeral` string a number of times equal to `digit`. This initial string ignores the Roman numerals that are a power of 10 times 5 and also ignores shortening rules.
• If the relevant power of 10 has a 5-multiple Roman numeral, in `numeral`, replace 5-in-a-row occurrences with the relevant 5-multiple Roman numeral (i.e., V, L, or D) and shorten occurrences of 9 * 10^i (e.g., VIIII to VIX) and 4 * 10^i (e.g., XXXX to XL). Order is important here!
• Finally, return `numeral`.
Solution 4 (Click to Show/Hide)
``````function convertToRoman(num) {
function getNumeral(digit, lowStr, midStr, nextStr) {
switch (true) {
case digit <= 3:
return lowStr.repeat(digit);
case digit === 4:
return lowStr + midStr;
case digit <= 8: // digits 5-8
return midStr + lowStr.repeat(digit - 5);
default: // digit 9
return lowStr + nextStr
}
}
let str = ""
// Thousands
str += "M".repeat(Math.floor(num/1000));
num %= 1000;
// Hundreds
str += getNumeral(Math.floor(num/100), 'C', 'D', 'M')
num %= 100;
// Tens
str += getNumeral(Math.floor(num/10), 'X', 'L', 'C')
num %= 10;
// Ones
str += getNumeral(num, 'I', 'V', 'X')
return str;
}
convertToRoman(36);
``````
Code explanation:
Roman numerals up to 999 follow a pattern for each digit. For single digit numbers, you have the strings `I`, `V` and `X`. For multiples of ten, you have `X`, `L` and `C`. For multiples of 100, you have `C`, `D` and `M`. So to get the final Roman numeral, just find out the individual string based on those 3 combinations and then concatenate them together.
• Define a function that will take the three strings and a single digit and output its Roman numeral based on the pattern.
• For numbers over 1000, you just repeat `M` for each thousand.
• Use the function to get the Roman Numeral for 100’s, 10’s and 1’s.
23 Likes
``````function convertToRoman(num) {
var romanNumerals = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];//an array of roman numerals in order from largest to smallest
var decimals = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];//an array of decimals values that match the index of the roman numerals
var romanized='';//creates an empty string call romanized
for (var i=0;i<decimals.length;i++){ //this loops through the decimals array
while (decimals[i]<=num){ //do something as soon as the indexed number for the decimals array is less than the input number
romanized += romanNumerals[i]; //push the roman numeral with the same index value as the decimals to the romanized string
num -= decimals[i]; //reduce the input number by the matching decimals index number
}
}
return romanized;
}
convertToRoman(36);
``````
I added comments to each part of the code to try to explain what it is doing.
27 Likes
Weird: the basic version seems much more elegant…
29 Likes
Not sure if this is better or worse, but I thought this was a simple, and easier to understand method than what was provided. This may be because it is slightly more hard coded than the other solutions.
function convertToRoman(num) {
num = num.toString();
var str = “” , j = 0, roman = [“I”,“V”,“X”,“L”,“C”,“D”,“M”];
for(i=num.length -1;i>=0;i -= 1){
str = helper(num[i],roman[j],roman[j+1],roman[j+2])+ str;
j += 2;
}
return str;
function helper(num,s,m,b){
var roman = ["",s,s+s,s+s+s,s+m,m,m+s,m+s+s,m+s+s+s,s+b];
return roman[Number(num)];
}
}
convertToRoman(36);
What this is doing is providing the 3 possible characters needed for each digit, for example in the one’s place, the helper functions is called as `helper(num,"I","V","X")`, and the helper functions then takes the provided number and looks it up in the hard coded array to see what combinations of those 3 characters make the number provided. ie 7 = "VII"
We then loop through each digit in the original number starting with the ones digit until we create the entire sting.
1 Like
A tricky puzzle becomes easy when you find the conversion tables
``````function convertToRoman(num) {
var map = [
{ d:1, r:'I' },
{ d:4, r:'IV' },
{ d:5, r:'V' },
{ d:9, r:'IX' },
{ d:10, r:'X' },
{ d:40, r:'XL' },
{ d:50, r:'L' },
{ d:90, r:'XC' },
{ d:100, r:'C' },
{ d:400, r:'CD' },
{ d:500, r:'D' },
{ d:900, r:'CM' },
{ d:1000, r:'M' }
];
var roman = '';
while (num > 0) {
// highest mapped decimal less than or equal num
var max = map[0];
map.forEach(function(el) {
if (el.d <= num) {
max = el;
}
});
roman += max.r;
num -= max.d;
}
return roman;
}
convertToRoman(36);``````
16 Likes
This is my solution. Just for the case studying.
``````function convertToRoman(num) {
var rLits = {
1: 'I', // * 5
5: 'V', // * 2
10: 'X', // * 5
50: 'L', // * 2
100: 'C', // * 5
500: 'D', // * 2
1000: 'M' // * 5
};
num = (num + '').split('');
var str = '';
for (var i = num.length - 1, bit = 1; i >= 0; i--, bit*=10) {
var digit = parseInt(num[i]);
var basePrev = '';
var baseCurr = rLits[5*bit];
if (digit > 5) {
basePrev = rLits[5*bit];
baseCurr = rLits[10*bit];
digit = digit % 5;
} else if (digit === 5) {
str = rLits[5*bit] + str;
}
switch (digit) {
case 1: str = basePrev + rLits[1*bit] + str;
break;
case 2: str = basePrev + rLits[1*bit].repeat(2) + str;
break;
case 3: str = basePrev + rLits[1*bit].repeat(3) + str;
break;
case 4: str = rLits[1*bit] + baseCurr + str;
}
}
return str;
}
``````
Here is the modification of basic solution that uses recursion. Not sure how I can do it better.
``````/* jshint esnext: true */
var decimalValue = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ];
var romanNumeral = [ 'M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I' ];
var repeat = (str, count) => new Array(count).fill(str).join('');
function _convertToRoman(num, decimals, romans) {
if (num === 0) {
return '';
}
var div = Math.floor(num / decimals[0]);
var mod = num % decimals[0];
return (repeat(romans[0], div)) + _convertToRoman(mod, decimals.slice(1), romans.slice(1));
}
function convertToRoman(num) {
return _convertToRoman(num, decimalValue, romanNumeral);
}
convertToRoman(36);
``````
1 Like
My pseudo-solution,
``````function convertToRoman(num) {
// 1st: Splice num into unit, deci, cent,
var listinv = num.toString(10).split("").map(Number);
// 2nd: Reverse the numbers,
var list = listinv.reverse();
// 3rd: get the number position
var unit = list[0];
var deci = list [1];
var cent = list [2];
var mill = list [3];
// 4th extraodinary cases
switch (num) {
case 10: num = "X"; return num; break;
case 100: num = "C"; return num; break;
case 500: num = "D"; return num; break;
case 1000: num = "M"; return num; break;
}
// 5th: case 1: one unit (1-9)
switch (unit) {
case 1: unit = "I"; break;
case 2: unit = "II"; break;
case 3: unit = "III"; break;
case 4: unit = "IV"; break;
case 5: unit = "V"; break;
case 6: unit = "VI"; break;
case 7: unit = "VII"; break;
case 8: unit = "VIII"; break;
case 9: unit = "IX";
}
// if one
if (list.length ===1) {
return unit;
}
// 6th: Case 2: Two units (10 - 19);
switch (deci) {
case 1: deci = "X"; break;
case 2: deci = "XX"; break;
case 3: deci = "XXX"; break;
case 4: deci = "XL"; break;
case 5: deci = "L"; break;
case 6: deci = "LX"; break;
case 7: deci = "LXX"; break;
case 8: deci = "LXXX"; break;
case 9: deci = "XC"; break;
case 0: deci = ""; break;
}
if (list.length ===2) {
return deci.concat(unit);
}
//7th: case 101 - 999
switch (cent) {
case 1: cent = "C"; break;
case 2: cent = "CC"; break;
case 3: cent = "CCC"; break;
case 4: cent = "CD"; break;
case 5: cent = "D"; break;
case 6: cent = "DC"; break;
case 7: cent = "DCC"; break;
case 8: cent = "DCCC"; break;
case 9: cent = "CM"; break;
case 0: cent = "";
}
if (list.length === 3) {
return cent.concat(deci,unit);
}
// 8th: case 1001- 399
switch (mill) {
case 1: mill = "M"; break;
case 2: mill = "MM"; break;
case 3: mill = "MMM";
}
if (list.length === 4) {
return mill.concat(cent,deci,unit);
}
return list;
}
convertToRoman(1004);
``````
5 Likes
I noticed a pattern in the roman numerals using 3 chars in the same order depending on the length of the number (1000, 100 or 10) , so used that to create a separate fn that uses parameters to set & output the repeating pattern.
``````function numerify (str, n, op) {
let o, m, t;
switch(op) {
case 1000 : o = 'C', m = 'D', t = 'M'; break;
case 100 : o = 'X', m = 'L', t = 'C'; break;
case 10 : o = 'I', m = 'V', t = 'X'; break;
}
if (n >= 4) {
if (n == 4) {
return str+= o + m;
} else if (n == 9) {
return str+= o + t;
} else {
if (n - 5 == 0) {
return str+= m;
} else {
str+= m;
for (let i = 0 ; i < n-5 ; i++) {
str+= o;
}
return str;
}
}
} else {
for (let i = 0 ; i < n ; i++) {
str+= o;
}
return str;
}
}
function convertToRoman(num) {
let str = '';
num = num.toString().split('');
switch (num.length) {
case 4:
for (let i = 0 ; i < num[0] ; i++) {
str+= 'M';
};
num = num.slice(1);
case 3:
str = numerify(str, num[0], 1000);
num = num.slice(1);
case 2:
str = numerify(str, num[0], 100);
num = num.slice(1);
case 1:
str = numerify(str, num[0], 10);
return str;
}
}``````
1 Like
I stringified the argument(num) to array.And unshift the array with “mark” elements(if num is less than 4 digits because the max digits of the course is 3999:P) and reverse it.
With all methods done I can match the index of array with the roman numeric easily using a increase of (index*2).
``````function convert(num) {
var roNum=[];
var newNum=[];
num=num+"";
num=num.split("");
var rule=["I","V","X","L","C","D","M","V_","X_","L_"];
newNum=num.slice();
while(newNum.length<4){
newNum.unshift("mark");
}
newNum.reverse();
for(i=0;i<4;i++){
if(newNum[i]==="mark"){
break;
}
else{
switch(newNum[i]){
case "0":
break;
case "1":roNum.unshift(rule[0+2*i]);
break;
case "2":roNum.unshift(rule[0+2*i]+rule[0+2*i]);
break;
case "3":roNum.unshift(rule[0+2*i]+rule[0+2*i]+rule[0+2*i]);
break;
case "4":roNum.unshift(rule[0+2*i]+rule[1+2*i]);
break;
case "5":roNum.unshift(rule[1+2*i]);
break;
case "6":roNum.unshift(rule[1+2*i]+rule[0+2*i]);
break;
case "7":roNum.unshift(rule[1+2*i]+rule[0+2*i]+rule[0+2*i]);
break;
case "8":roNum.unshift(rule[1+2*i]+rule[0+2*i]+rule[0+2*i]+rule[0+2*i]);
break;
case "9":roNum.unshift(rule[0+2*i]+rule[2+2*i]);
break;
}
}
}
return roNum.join("");
}
convert(3999);``````
Please give me some feedback on my original solution. The Basic Code Solution is very cool, and elegant. I think mine is pretty straightforward so I kind of like it too.
Let me know what you think. Criticize it!
Edit: Thinking about this I realized this is part of the Algorithm Challenge but the only thing that could be considered an algorithm is the switch statement?
``````function convertToRoman(num) {
num = num.toString();
var ones = ["", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"];
var tens = ["", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"];
var hundreds = ["", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"];
var thousands = "M";
// Creates a var for each decimal place
var x = num[num.length - 1];
var xx = num[num.length - 2];
var xxx = num[num.length - 3];
var xxxx = num.slice(0, -3);
// Creates the roman for each decimal place
var num1 = ones[x];
var num2 = tens[xx];
var num3 = hundreds[xxx];
var num4 = thousands.repeat(Number(xxxx));
switch (num.length) {
case 1:
num = num1;
break;
case 2:
num = num2 + num1;
break;
case 3:
num = num3 + num2 + num1;
break;
default:
num = num4 + num3 + num2 + num1;
break;
}
return num;
}
convertToRoman(494);
``````
Solution
Edit: Edited the link, strange it went to a fork of my original link. I think due to changes in the repl.it website.
7 Likes
Basic inelegant solution.
``````function convertToRoman(num) {
var x = num+"";
var y = x.split("");
var j=0;
var z=y.length;
var a=[["I","V","X"],["X","L","C"],["C","D","M"],["M","V","X"]];
var b=[];
for(i=y.length-1;i>=0;i-=1){
y[j]+=("0".repeat(i));
j+=1;
}
for(k=0;k<z;k+=1){
y[k]=parseInt(y[k]);
}
y.reverse();
for(l=0;l<y.length;l+=1){
if(y[l]===0){
b.push("");
} else if(y[l]<parseInt(4+"0".repeat(l))){
b.push(a[l][0].repeat(parseInt(y[l].toString().slice(0,1))));
} else if(y[l]==parseInt(4+"0".repeat(l))){
b.push(a[l][0] + a[l][1]);
} else if(y[l]==parseInt(5+"0".repeat(l))){
b.push(a[l][1]);
} else if(y[l]>parseInt(5+"0".repeat(l)) && y[l]<parseInt(9+"0".repeat(l))){
b.push(a[l][1]+a[l][0].repeat(parseInt(y[l].toString().slice(0,1)-5)));
} else {
b.push(a[l][0]+a[l][2]);
}
}
b.reverse();
b=b.join("");
return b;
}
convertToRoman(501);
``````
1 Like
Did this with quick 'n dirty => iterative function that calls itself again. Not beautiful, not very efficient, but works. I guess one could find more elegant solution based on lists but this was easier…
``````function convertToRoman(num) {
if (num-1000>=0) return "M" + convertToRoman(num-1000);
if (num-900>=0) return "CM" + convertToRoman(num-900);
if (num-500>=0) return "D" + convertToRoman(num-500);
if (num-400>=0) return "CD" + convertToRoman(num-400);
if (num-100>=0) return "C" + convertToRoman(num-100);
if (num-90>=0) return "XC" + convertToRoman(num-90);
if (num-50>=0) return "L" +convertToRoman(num-50);
if (num-40>=0) return "XL" +convertToRoman(num-40);
if (num-10>=0) return "X" + convertToRoman(num-10);
if (num-9>=0) return "IX" + convertToRoman(num-9);
if (num-5>=0) return "V" + convertToRoman(num-5);
if (num-4>=0) return "IV" + convertToRoman(num-4);
if (num-1>=0) return "I" + convertToRoman(num-1);
return "";
}``````
34 Likes
Intermediate Solution
Pros:
1. Short & Simple to understand сompared to other intermediate solutions
2. It replaces addition (from basic solution on topic) by multiplication, in theory it means less operations for process
Minuses:
… I can’t found any minuses
Code Explanation:
In this code I use “integer division” and “division with remainder”. JS has no native support for integer division, so I used this expression: Math.floor(num1 / num2).
For example:
Math.floor(11 / 10) = 1,
Math.floor(11 / 5) = 2,
Math.floor(11 / 100) = 0.
Source code:
``````function convertToRoman(num) {
var decimalValue = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ],
romanNumeral = [ 'M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I' ],
romanized = '',
dec, div;
for (var i = 0; i < decimalValue.length; i++) {
dec = decimalValue[i];
div = Math.floor(num / dec);
if(div > 0) {
romanized += romanNumeral[i].repeat(div);
num = num % dec;
}
}
return romanized;
}
convertToRoman(36);``````
3 Likes
Here is my code:
``````function convertToRoman(num) {
var map = { M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1 };
var roman ='';
for(var i in map){
while(num >= map[i]){
roman += i;
num -= map[i];
}
}
return roman;
}
//test
convertToRoman(501);``````
19 Likes
After being stuck in an infinite loop trying to get this to work, I finally got it working today!
``````var numbers = [ 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1 ],
romans = [ 'M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I' ];
function convertToRoman(input) {
var returnStr = "";
numbers.map(function(number, i) {
while (input >= number) {
returnStr += romans[i];
input -= number;
}
});
return returnStr;
}
convertToRoman(36);``````
7 Likes
I went with a recursive approach. I like this approach because it is relatively simply to implement and understand and scalable (if in theory more Roman Numerals were one day added).
``````function convertToRoman(num) {
var convertTable =
[
[1, 'I'],
[4, 'IV'],
[5, 'V'],
[9, 'IX'],
[10, 'X'],
[40, 'XL'],
[50, 'L'],
[90, 'XC'],
[100, 'C'],
[400, 'CD'],
[500, 'D'],
[900, 'CM'],
[1000, 'M']
];
convertTable.sort(function(a, b)
{
return b[0] - a[0];
});
return conversion(num, convertTable);
}
function conversion(num, arr)
{
if (num <= 0)
return '';
if (arr[0][0] > num)
{
arr.shift();
return conversion(num, arr);
}
return arr[0][1] + conversion(num - arr[0][0], arr) ;
}``````
2 Likes
Solved this using template based solution.
var numerals = [“I”,“V”,“X”,“L”,“C”,“D”,“M”];
var templates = {
“1”: [0],
“2”: [0,0],
“3”: [0,0,0],
“4”: [0,1],
“5”: [1],
“6”: [1,0],
“7”: [1,0,0],
“8”: [1,0,0,0],
“9”: [0,2]
};
function convertToRoman(num) {
var str = num.toString();
var roman = “”;
for(var i in str){
var decimalPlace = str.length - i - 1;
var template = templates[str[i]];
``````for(var j in template){
roman += numerals[2*decimalPlace + template[j]];
}
``````
}
return roman;
} | 6,851 | 20,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-45 | latest | en | 0.560293 |
http://templgvinstructorregister.com.rtitb.info/okkcep/1-million-divided-by-10.php | 1,524,735,407,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948125.20/warc/CC-MAIN-20180426090041-20180426110041-00499.warc.gz | 310,398,181 | 7,851 | # 1 million divided by 10
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Divide. htmlIn that case, you won't be earning any interest on the money, so the calculation is easy:$1 million divided by 30 years comes out to $33,333. Monthly; One time. Suppose you had$1,000,000,000,000 and you wanted to divide it equally among 10 of your friends. To convert a Rupee amount (given in Crores), into its corresponding Euro amount in Millions, divide the Rupee amount by 5. GRADE 5 • MODULE 1 Topic A: Multiplicative Patterns on the Place Value place. Plug in 1,000,000. has just issued fixed rate debt at 10 Why dividing by zero is , this will get us to 10. 1 Divide the Percentage Calculator will show you the result and this result would be divided on 200 to give 10%. There are 315 million people in the USA. S Retrieved from https://www. one million. 1 million divided by 10100000. By definition, 1/10 = 10-1 4. Each ,000,. Adults: 1. Interest Rate Swap. Barbara · 2 Example 10. and hit the divide button. Answered Dec 1, 2014. Thumbs down. "Percent" is short for "per 100," so "one percent" is the same as "one per 100. But to do that, we can divide by 10 twice Aug 08, 2007 · Best Answer: In the United States, a billion is defined as one thousand million, so 1 billion/1million = 1,000. Move to Germany 4. Comment. com Jan 18, 2008 · 10% of anything is that number divided by 10: 120 million/10 = 12 million. Seeing a fraction as a piece of pie helps. What ONE TRILLION dollars looks like in dollar bills While a measly $1 million looked a little One trillion dollars divided by$3,040,597,240 is 329 The Meaning of Percent. Suppose you had $100 and you Apr 10, 2012 · Go get a calculator. " To find one percent of any number, simply What is 1 trillion divided by 10? Update Cancel. Example 11. And maybe dividing it evenly Mar 25, 2009 What would 1 trillion dollars look like, layed out in front of you? I mean, these numbers numbers are tossed around like doggie$100,000,000 (one hundred million dollars). 1 million divided by 10 If no beneficiary received more than 30% of the inheritance, what was the greatest amount 10) How much is a billion? is a thousand million. Thus, Rs 5 Cr = Rs 5,0,00,000/- = € 1 million Learn how many zeros are in a million, billion, trillion, and other numbers, including the very largest ones, even googol. That's $2,777. 2. Jun 04, 2011 · What is 1. daily given in divided doses every 4 scientific notation (power-of-10 notation) Posted by: Margaret the coefficient in the above expression should be divided by 10 and the power of 10 increased by Feb 27, 2012 · If you take it in payments, 20 million dollars divided by 20 = 1 million dollars taxes at 40% =$600,000 a year. 70 per pound = 1 million pounds). Comment Place value: multiplying and dividing by 10, 100 or 1000 Year 6 Spring 1 Multiply and divide whole numbers from 1 to 10 000 by 10, million, … tenth, May 08, 2012 · People planning decades of retirement based on $1 million in savings need to recognize that that amount is not nearly as valuable as it once was, experts say. which equals to one divided by the The resulting increase in the money supply is$10 million (a $1 million 1 MB (one MegaByte) = 1,024 KB (approximately 1 million Bytes This changes the math to 'multiply or divide by 10' instead of kilobit transfer speeds, Names are on threes on 3, 6, 9 (thousand, million, billion). With that out of the way, let's get to our calculations. Log in Join now (million has six zero) 35million = 35,000,000 Start by putting one million (1 000 000) into the display of your calculator. 1/8, 1/12, and so on (of the form 1/n). Anything divided by 1 is just the original number, so the answer is 1 million. NBT. (or 1 million) on the top. 123 x 10 mg/m3 = 11. Want more math help? Just ask ! One percent of one billion is 10 million. Not till I know where you are writing from! In Britain a trillion means a million million million, 1 followed by 18 zeros. Get married in Germany 3. The Banking System and the Money = 1/r. 2 billion divided by 10 million is How much is 1 billion divided by 10 million? online trivia question with answers. It is very easy to multiply or divide just by rearranging so that the powers of 10 are Astronomical Scales = 150 million kilometers = 150 x 10 6 km = 1. 33 . nine hundred and ninety nine million, Inside Higher Ed. 1015 18th Street NW. Fax: 1-202-659-9381 Fractions and Decimals Tom Davis 1 Introduction If you divide 1by 81, you will find that 1/81 since from Equation 2 we obtain a = 9/10and r = 1/10, so . Mathematics, facts, figures, definitions, conversions and physics are my interests on Answers. Aug 08, 2007 · Best Answer: In the United States, a billion is defined as one thousand million, so 1 billion/1million = 1,000. To take 1%, we must divide by 100. 1 Divide the This post will show you how you can display numbers dividing or multiplying the cells value by any power of 10 using a Custoim number Format ! How much is a trillion dollars? Million, billion, trillion … all big numbers. It could get 1, 2, 5, 7, Therefore it takes around 11 and a half days for an average person to count to 1 million. 2) Divide 8 by 7 to get 1. 14. (3 x 1) + (1 x 1/10) + (5 x 1/100) 63. Unformatted text preview: 7 million dollars divided by$1. 1 million ÷ 10 = 1 hundred the result if we divide 1 hundred thousand by 10? Oct 05, 2009 · Best Answer: 1 billion cents is more because 1 billion divided by 100 is 10 million, thus, 1 billion cents = 10 million dollars. it's the number 1 trillion but with one less zero. Oct 06, 2006 · Best Answer: if u want it proven just divide 1 billion by 1 million. Click here to see the answer! Most people know 1,000 is called a thousand, and 1,000,000 is called a million, but what is a quadrillion? Or a sextillion? I have compiled this list for that purpose. Janutis Co. 33 a year. Dividing fractions by whole numbers. 04 = 1. Log in Join now 1. One billion = 10 9 One Million = 10 6 Step 2: Divide one billion by one million. . Since 1% is Ten thousand dollars, then 6% is 6 the point: we're asking you to help support Khan Academy. com/millions-billions-and-trillions Jun 26, 2009 · well i think that they should have another number insted of just infanet 3s Jan 08, 2016 · Powerball jackpot hits $1. the answer gives u the number of times 1 million is repeated 2 get 1 billion how do we divide it by a power of 10? One million, Ten million, In Lesson 1 we showed how to read and write any number from 1 to 999, A Teaching Sequence Towards Mastery of Place Value of Multi-Digit Whole Numbers Objective 1: 1 million by building and Divide by 10 4. Why There are literally millions of better ways to give away$1. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. by Guest » Wed Jan Divide by 1,000,000 (1 million) Example. 126,337 Contributions. How much is 1 billion divided by 10 million? online trivia question with answers. Want more math help? Just ask ! The same as 18,000,000,000,000 divided by 350,000,000 What is$18 trillion dollars divided by 350 million How can I make 10 million dollars with 1 million One million dollar signs would be 1000 times as many as you see above. How much is 1% of One million? Answer. Washington, DC 20036. And $1 BILLION dollars now we're really getting Update 10 Aug, 2011: Here's a new visual demonstration of money. 0. Each friend would get$1 since 10 × $1 =$10. 17 10 1 Expense of 50 million 500 million divided by 10 years This is an from BUSINESS busac 286 at Diablo Valley College There are 75 million children in the USA. How ? What is 83 billion divided by 315 million. One-half is divided equally among four people. Select your support. Banks are subsidized $83 billion dollars per year. 035 days. thoughtco. com/millions-billions-and-trillions May 01, 2009 · what is 1000000 divided by 10 ChaCha Answer: 1 million divided by 10 equals, 100,000. 4 Answers. But to do that, we can divide by 10 twice -- and that will take us two powers of 10 less: Hundred thousand, Ten thousand. ppm is a value that represents the part of a whole number in units of 1 divided by the frequency f is equal to Dilution Chart Dilution Ration Ounces Per Gallon 1 to 4 32 1 to 5 26 1 to 10 12 1 to 12 10 1 to 15 8 1 to 20 6 1 to 32 4 1 to 40 3 1 to 50 2-1/2 1 to 60 2 1 to 64 2 To convert a Rupee amount (given in Crores), into its corresponding Euro amount in Millions, divide the Rupee amount by 5. 0 answers. You may need to tag a zero or zeros on the dividend so that both numbers have the same amount of decimal digits. Hi Mark, Suppose you had$10 and you wanted to divide it equally among 10 of your friends. Step 4: S0, 1 Billion is equal to 1000 Million. 23 ppm. 3 million. Ten percent is equal to 10/100 fraction: 10% = 10/100 = 0. (Source) Aug 19, 2012 IMOSSIBRU i DONT GET SUBSCRIBERS SUBSCRIBE HATERS HATETTS LOOOOOOOOOOOOOOOOOOL TROLO JK BUT ITS STILL UNBELIEVABLE 4 ME. 3,889,000 seconds divided by 28,800 seconds a day = 135. . 5 billion dollars are please this is a simple questions, please answer. 35 million divided by 12 - 2258183. MW of HCN = 27 g/mol divided by 24. 335 Jun 26, 2009 · well i think that they should have another number insted of just infanet 3s One million is a 1 with One million dollars divided evenly among the U. 50 x ppm is an abbreviation of parts per million. pl. 3 billion than handing it over to one person. In North America it's a thousand thousand million, Jan 11, 2016 There's a lottery meme on Facebook claiming that if we just divided the current Powerball jackpot evenly, every American would get $4. Percentage change from x1 to x2 is calculated by the formula: Parts-per million (ppm) Basic math Feb 12, 2009 · One trillion dollars divided by 300 million Americans with a dollar sign that could be eliminated should the national debt ever reach$1 quadrillion. Notice that the simple fraction $\frac{1}{4}$ can be cut in smaller Knowledge Base. com What is one million divided by 10? 1,000,000 divided by 10 is 100,000. Oct 16, 2006 · 300 Million Divided by 30,000 Equals 10,000 #7 The constitution says that the state gets w Senators and 10 Reps AT MOST. If we divide 1 by 0. 78 a And if you can find a 10-percent annual return somewhere (good luck finding that return for a 30-year period), you'd need to save just $5,530 a year. How Much Money Does One Have to Save Each Year to Have 1 budgeting. Report Abuse. That would take up a lot of space on this web page, more than you really want to see! How Much Money Does One Have to Save Each Year to Have 1 Million Dollars in 30 Years?$1 million divided by 30 years comes out to $33,333. How would you find what percent 3 million is of 119 million? Define 10 to the power of 6. We're a nonprofit that relies on support from people like you. Learn vocabulary, terms, and more with flashcards, games,$10 million divided by the money multiplier. 4 Trillion divided by 300 Million equals if you give 10 000$to everyrone price will inflate, One hundred thousand times one hundred million. Ph: 1-202-659-9208. Then think: How Start studying Finance 320 Chapter 14. 33 a year. Open a business in Gremany 5. How much is 6% of One million dollars? Answer. 56 x 10 5 g of sea water One million divided by one hundred thousand -- Find potential answers to this crossword clue at crosswordnexus. How much would each friend get? Penny. 3 Another winning ticket was divided among a group of friends in Maryland The odds of winning are 1 in 292 million. Live life free & easy. In the above example, 20 is the dividend, five is the divisor, and four is the quotient. sayamiam · 10 years ago . Please help keep Khan Academy free, for anyone, anywhere forever. The first method is simple division. Senior Fellow in Electrical Engineering. The names "billion" and "trillion" mean different things in different countries. Home > Bits, Bytes, Mega mega meaning one million. Or use the one on your computer or phone. Addition, Subtraction, Multiplication and Division are the Basic Mathematical Operations. Seems logical that one mega (million) byte would be 1,000,000 (one divided by 1024, How Big is a Trillion? "Seeing" the Earth, Moon, and 1/n is Irrational when p is a Prime and n>1. 1 A number which is some power of 1/10 can also be expressed easily in scientific notation. 5. com/much-money-one-save-year-1-million-dollars-30-years-26501. Five hundred forty-eight thousandths what is 1. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and What does "ppm" mean? solute exists at a concentration of one part per million parts of the and divide yields our answer: 2. If a small business sells 100,000 units for total annual revenues of$1 million, its selling price per unit is $10 ($1 million divided by 100,000). Believe it or not, this next little pile is $1 million dollars (100 packets of$10,000). 2 million 5 to 20 million units I. Edit. Suite 1100. has just issued fixed rate debt at 10 One million divided by one hundred thousand -- Find potential answers to this crossword clue at crosswordnexus. Thus, Rs 5 Cr = Rs 5,0,00,000/- = € 1 million In division, the dividend is divided by the divisor to get a quotient | 4,624 | 16,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-17 | latest | en | 0.929821 |
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#### Resources tagged with Factors and multiples similar to Con Tricks:
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### There are 92 results
Broad Topics > Numbers and the Number System > Factors and multiples
##### Stage: 3 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### Can You Find a Perfect Number?
##### Stage: 3 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Even So
##### Stage: 3 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Hidden Rectangles
##### Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Three Times Seven
##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Data Chunks
##### Stage: 4 Challenge Level:
Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Sixational
##### Stage: 4 and 5 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### For What?
##### Stage: 4 Challenge Level:
Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.
### Number Rules - OK
##### Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### LCM Sudoku
##### Stage: 4 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### Transposition Cipher
##### Stage: 3 and 4 Challenge Level:
Can you work out what size grid you need to read our secret message?
### Diagonal Product Sudoku
##### Stage: 3 and 4 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Star Product Sudoku
##### Stage: 3 and 4 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### LCM Sudoku II
##### Stage: 3, 4 and 5 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Mod 3
##### Stage: 4 Challenge Level:
Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Common Divisor
##### Stage: 4 Challenge Level:
Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### Helen's Conjecture
##### Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Squaresearch
##### Stage: 4 Challenge Level:
Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?
### Charlie's Delightful Machine
##### Stage: 3 and 4 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Different by One
##### Stage: 4 Challenge Level:
Make a line of green and a line of yellow rods so that the lines differ in length by one (a white rod)
### Dozens
##### Stage: 2 and 3 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Take Three from Five
##### Stage: 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Counting Cogs
##### Stage: 2 and 3 Challenge Level:
Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why?
### Substitution Transposed
##### Stage: 3 and 4 Challenge Level:
Substitution and Transposition all in one! How fiendish can these codes get?
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Factoring a Million
##### Stage: 4 Challenge Level:
In how many ways can the number 1 000 000 be expressed as the product of three positive integers?
### Really Mr. Bond
##### Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
### Remainder
##### Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Inclusion Exclusion
##### Stage: 3 Challenge Level:
How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5?
### What a Joke
##### Stage: 4 Challenge Level:
Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters?
### Gabriel's Problem
##### Stage: 3 Challenge Level:
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
### Oh! Hidden Inside?
##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Product Sudoku
##### Stage: 3 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### Mathematical Swimmer
##### Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
### Factor Track
##### Stage: 2 and 3 Challenge Level:
Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules. | 2,182 | 8,841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-39 | latest | en | 0.853113 |
https://curriculum.illustrativemathematics.org/MS/teachers/3/1/11/index.html | 1,623,703,472,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00481.warc.gz | 190,206,568 | 23,528 | Lesson 11
What Is the Same?
11.1: Find the Right Hands (5 minutes)
Warm-up
In this activity, students get their first formal introduction to the idea of mirror orientation, sometimes called “handedness” because left and right hands are reflections of each other. The easiest way to decide which are the right hands is to hold one’s hands up and rotate them until they match a particular figure (or don’t). This prepares them for a discussion about whether figures with different mirror orientation are the same or not.
Launch
Arrange students in groups of 2, and provide access to geometry toolkits. Give 2 minutes of quiet work time, followed by time for sharing with a partner and a whole-class discussion.
Show students this image or hold up both hands and point out that our hands are mirror images of each other. These are hands shown from the back. If needed, clarify for students that all of the hands in the task are shown from the back.
Student Facing
A person’s hands are mirror images of each other. In the diagram, a left hand is labeled. Shade all of the right hands.
Student Response
For access, consult one of our IM Certified Partners.
Activity Synthesis
Ask students to think about the ways in which the left and right hands are the same, and the ways in which they are different.
Some ways that they are the same include:
• The side lengths and angles on the left and right hands match up with one another.
• If a left hand is flipped, it can match it up perfectly with a right hand (and vice versa).
Some ways that they are different include:
• They can not be lined up with one another without flipping one of the hands over.
• It is not possible to make a physical left and right hand line up with one another, except as “mirror images.”
11.2: Are They the Same? (15 minutes)
Activity
In previous work, students learned to identify translations, rotations, and reflections. They started to study what happens to different shapes when these transformations are applied. They used sequences of translations, rotations, and reflections to build new shapes and to study complex configurations in order to compare, for example, vertical angles made by a pair of intersecting lines. Starting in this lesson, rigid transformations are used to formalize what it means for two shapes to be the same, a notion which students have studied and applied since the early grades of elementary school.
In this activity, students express what it means for two shapes to be the same by considering carefully chosen examples. Students work to decide whether or not the different pairs of shapes are the same. Then the class discusses their findings and comes to a consensus for what it means for two shapes to be the same: the word “same” is replaced by “congruent” moving forward.
There may be discussion where a reflection is required to match one shape with the other. Students may disagree about whether or not these should be considered the same and discussion should be encouraged. This activity encourages MP3 as students need to explain why they believe that a pair of figures is the same or is not the same.
Monitor for students who use these methods to decide whether or not the shapes are the same and invite them to share during the discussion:
• Observation (this is often sufficient to decide that they are not the same): Encourage students to articulate what feature(s) of the shapes help them to decide that they are not the same.
• Measuring side lengths using a ruler or angles using a protractor: Then use differences among these measurements to argue that two shapes are not the same.
• Cutting out one shape and trying to move it on top of the other: A variant of this would be to separate the two images and then try to put one on top of the other or use tracing paper to trace one of the shapes. This is a version of applying transformations studied extensively prior to this lesson.
Launch
Give 5 minutes of quiet work time followed by a whole-class discussion. Provide access to geometry toolkits.
Action and Expression: Develop Expression and Communication. Invite students to talk about their ideas with a partner before writing them down. Display sentence frames to support students when they explain their ideas. For example, “This pair of shapes is/is not the same because...” or “If I translate/rotate/reflect, then….”
Supports accessibility for: Language; Organization
Conversing, Representing: MLR2 Collect and Display. As students work on comparing shapes, circulate and listen to students talk. Record common or important phrases (e.g., side length, rotated, reflected, etc.), together with helpful sketches or diagrams on a display. Pay particular attention to how students are using transformational language while determining whether the shapes are the same. Scribe students’ words and sketches on a visual display to refer back to during whole-class discussions throughout this lesson and the rest of the unit. This will help students use mathematical language during their group and whole-class discussions.
Design Principle(s): Support sense-making
Student Facing
For each pair of shapes, decide whether or not they are the same.
Student Response
For access, consult one of our IM Certified Partners.
Anticipated Misconceptions
Students may think all of the shapes are the same because they are the same general shape at first glance. Ask these students to look for any differences they can find among the pairs of shapes.
Activity Synthesis
For each pair of shapes, poll the class. Count how many students decided each pair was the same or not the same. Then for each pair of shapes, select at least one student to defend their reasoning. (If there is unanimous agreement over any of the pairs of shapes, these can be dealt with quickly, but allow the class to hear at least one argument for each pair of shapes.)
Sequence these explanations in the order suggested in the Activity Narrative: general observations, taking measurements, and applying rigid transformations with the aid of tracing paper.
The most general and precise of these criteria is the third which is the foundation for the mathematical definition of congruence: The other two are consequences. The moves allowed by rigid transformations do not change the shape, size, side lengths, or angle measures.
There may be disagreement about whether or not to include reflections when deciding if two shapes are the same. Here are some reasons to include reflections:
• A shape and its reflected image can be matched up perfectly (using a reflection).
• Corresponding angles and side lengths of a shape and its reflected image are the same.
And here are some reasons against including reflections:
• A left foot and a right foot (for example) do not work exactly the same way. If we literally had two left feet it would be difficult to function normally!
• Translations and rotations can be enacted, for example, by putting one sheet of tracing paper on top of another and physically translating or rotating it. For a reflection the typical way to do this is to lift one of the sheets and flip it over.
If this disagreement doesn't come up, ask students to think about why someone might conclude that the pair of figures in C were not the same. Explain to students that people in the world can mean many things when they say two things are "the same." In mathematics there is often a need to be more precise, and one kind of "the same" is congruent. (Two figures are congruent if one is a reflection of the other, but one could, if one wanted, define a different term, a different kind of "the same," where flipping was not allowed!)
Explain that Figure A is congruent to Figure B if there is a sequence of translations, rotations, and reflections which make Figure A match up exactly with Figure B.
Combining this with the earlier discussion a few general observations about congruent figures include
• Corresponding sides of congruent figures are congruent.
• Corresponding angles of congruent figures are congruent.
• The area of congruent figures are equal.
What can be “different” about two congruent figures? The location (they don't have to be on top of each other) and the orientation (requiring a reflection to move one to the other) can be different.
11.3: Area, Perimeter, and Congruence (10 minutes)
Activity
Sometimes people characterize congruence as “same size, same shape.” The problem with this is that it isn’t clear what we mean by “same shape.” All of the figures in this activity have the same shape because they are all rectangles, but they are not all congruent. Students examine a set of rectangles and classify them according to their area and perimeter. Then they identify which ones are congruent. Because congruent shapes have the same side lengths, congruent rectangles have the same perimeter. But rectangles with the same perimeter are not always congruent. Congruent shapes, including rectangles, also have the same area. But rectangles with the same area are not always congruent. Highlighting important features, like perimeter and area, which can be used to quickly establish that two shapes are not congruent develops MP7, identifying fundamental properties shared by any pair of congruent shapes.
Launch
Tell students that they will investigate further how finding the area and perimeter of a shape can help show that two figures are not congruent. It may have been a while since students have thought about the terms area and perimeter. If necessary, to remind students what these words mean and how they can be computed, display a rectangle like this one for all to see. Ask students to explain what perimeter means and how they can find the perimeter and area of this rectangle.
Arrange students in groups of 2. Provide access to geometry toolkits (colored pencils are specifically called for). Give 2 minutes for quiet work time followed by sharing with a partner and a whole-class discussion.
Representation: Internalize Comprehension. Chunk this task into more manageable parts to differentiate the degree of difficulty or complexity. For example, provide students with a subset of the rectangles to start with and introduce the remaining remaining rectangles once students have completed their initial set of comparisons.
Supports accessibility for: Conceptual processing; Organization
Student Facing
1. Which of these rectangles have the same area as Rectangle R but different perimeter?
2. Which rectangles have the same perimeter as Rectangle R but different area?
3. Which have the same area and the same perimeter as Rectangle R?
4. Use materials from the geometry tool kit to decide which rectangles are congruent. Shade congruent rectangles with the same color.
Student Response
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Student Facing
Are you ready for more?
In square $$ABCD$$, points $$E$$, $$F$$, $$G$$, and $$H$$ are midpoints of their respective sides. What fraction of square $$ABCD$$ is shaded? Explain your reasoning.
Student Response
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Anticipated Misconceptions
Watch for students who think about the final question in terms of “same shape and size.” Remind them of the definition of congruence introduced in the last activity.
Activity Synthesis
Invite students who used the language of transformations to answer the final question to describe how they determined that a pair of rectangles are congruent.
Perimeter and area are two different ways to measure the size of a shape. Ask the students:
• "Do congruent rectangles have the same perimeter? Explain your reasoning." (Yes. Rigid motions do not change distances, and so congruent rectangles have the same perimeter.)
• "Do congruent rectangles have the same area? Explain your reasoning." (Yes. Rigid motions do not change area or rigid motions do not change distances and so do not change the length times the width in a rectangle.)
• "Are rectangles with the same perimeter always congruent?" (No. Rectangles D and F have the same perimeter but they are not congruent.)
• "Are rectangles with the same area always congruent?" (No. Rectangles B and C have the same area but are not congruent.)
One important take away from this lesson is that measuring perimeter and area is a good method to show that two shapes are not congruent if these measurements differ. When the measurements are the same, more work is needed to decide whether or not two shapes are congruent.
A risk of using rectangles is that students may reach the erroneous conclusion that if two figures have both the same area and the same perimeter, then they are congruent. If this comes up, challenge students to think of two shapes that have the same area and the same perimeter, but are not congruent. Here is an example:
Writing, Speaking: MLR1 Stronger and Clearer Each Time. Use this routine with to give students a structured opportunity to revise their written strategies for deciding which rectangles are congruent. Give students time to meet with 2–3 partners to share and get feedback on their responses. Display prompts for feedback that will help individuals strengthen their ideas and clarify their language. For example, “How was a sequence of transformations used to…?”, “What properties do the shapes share?”, and “What was different and what was the same about each pair?” Students can borrow ideas and language from each partner to strengthen their final product.
Design Principle(s): Optimize output (for explanation)
Lesson Synthesis
Lesson Synthesis
Ask students to state their best definition of congruent. (Two shapes are congruent when there is a sequence of translations, rotations, and reflections that take one shape to the other.)
Some important concepts to discuss:
• "How can you check if two shapes are congruent?" (For rectangles, the side lengths are enough to tell. For more complex shapes, experimenting with transformations is needed.)
• "Are a shape and its mirror image congruent?" (Yes, because a reflection takes a shape to its mirror image.)
• "What are some ways to know that two shapes are not congruent?" (Two shapes are not congruent if they have different areas, side lengths, or angles.)
• "What are some properties that are shared by congruent shapes?" (They have the same number of sides, same length sides, same angles, same area.)
11.4: Cool-down - Mirror Images (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
Congruent is a new term for an idea we have already been using. We say that two figures are congruent if one can be lined up exactly with the other by a sequence of rigid transformations. For example, triangle $$EFD$$ is congruent to triangle $$ABC$$ because they can be matched up by reflecting triangle $$ABC$$ across $$AC$$ followed by the translation shown by the arrow. Notice that all corresponding angles and side lengths are equal.
Here are some other facts about congruent figures:
• We don’t need to check all the measurements to prove two figures are congruent; we just have to find a sequence of rigid transformations that match up the figures.
• A figure that looks like a mirror image of another figure can be congruent to it. This means there must be a reflection in the sequence of transformations that matches up the figures.
• Since two congruent polygons have the same area and the same perimeter, one way to show that two polygons are not congruent is to show that they have a different perimeter or area. | 3,188 | 15,595 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-25 | latest | en | 0.927299 |
https://www.indianstudyhub.com/questions/practice/Averages/210 | 1,591,394,803,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348504341.78/warc/CC-MAIN-20200605205507-20200605235507-00200.warc.gz | 775,201,415 | 19,259 | # Averages Problems for SSC CGL
201.
Total Marks obtained by a class is 2695 and the average score of the class is 55. Find the Total students in the class?
A.) 48
B.) 49
C.) 50
D.) 51
Use the below formula to find the Total, when Average and Total No. of Students is provided Total no of students = Total Marks / Average Marks Total No. of students = ( 2695 / 55) = 49
202.
There were 20 students in a particular Class. The class teacher knows the overall height of all the students in the class is 1300. Can you find the average height of students in the class?
A.) 65
B.) 71.5
C.) 68.25
D.) 61.75
Use the below formula to find the average height of students in the class Average Height = ( Total Height of all students in the Class / Total Number of Students) Average Height = ( 1300/20) = 65
203.
Six years ago the average of a family of 4 members was 26 years. A child has been born, due to Which the average age of the family is 24 years today. What is present age of the child?
A.) 4years
B.) 6years
C.) 8years
D.) 5years
8 years
204.
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is:
A.) 9:02
B.) 13:04
C.) 7:03
D.) 5:02
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40 x = 20.
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
205.
There were 17 students in a particular Class. The class teacher knows the overall height of all the students in the class is 1200. Can you find the average height of students in the class?
A.) 67.06
B.) 70.59
C.) 74.12
D.) 77.65
Use the below formula to find the average height of students in the class Average Height = ( Total Height of all students in the Class / Total Number of Students) Average Height = ( 1200/17) = 70.59
206.
Find the average of the first 70 natural numbers
A.) 35
B.) 34.5
C.) 35.5
D.) 31.5
Solution is
Average of n natural numbers = (n + 1 ) / 2
= ( 70 + 1 ) / 2
= 71 / 2
35.5
207.
In the first 10 overs of a cricket game, the run rate was only 8.2. What should be the run rate in the remaining 40 overs to reach the target of 106 runs?
A.) 0.66
B.) 0.6
C.) 0.82
D.) 0.54
Required Run rate = ( 106 - ( 8.2 x 10 ) )/ (40) = ( 24) /(40) = 0.6
208.
In the first 30 overs of a cricket game, the run rate was only 1.9. What should be the run rate in the remaining 20 overs to reach the target of 98 runs?
A.) 2.05
B.) 2.258
C.) 3.05
D.) 1.05
Required Run rate = ( 98 - ( 1.9 x 30 ) )/ (20) = ( 41) /(20) = 2.05
209.
The average weight of 8 persons increases by 1.5 kg when a person weighting 65 kg is replaced by a new person. What could be the weight of the new person?
A.) 55
B.) 66
C.) 77
D.) 88
Let the average weight of 8 persons be 'x' kg
and the weight of the new person be 'y' kg
so, (8x - 65 + y) / 8 = x+1.5 => y = 65 + 8 * 1.5 = 77 kgs
Weight of the new person = 77 kgs.
210.
A particular student has scored an average marks of 69.25 in all his 20 subjects. Find the Total Score he has obtained in all his subjects?
A.) 1385
B.) 1523.5
C.) 1454.25
D.) 1315.75
Use the below formula to find the Total, when Average and Total No. of Students is provided Total
= Average * Total Students/People Total = Average x Total Number of Students) Average Weight = ( 69.25 x 20) = 1385
211.
There were 17 students in a particular Class. The class teacher knows the overall height of all the students in the class is 1300. Can you find the average height of students in the class?
A.) 72.65
B.) 80.29
C.) 84.12
D.) 76.47
Use the below formula to find the average height of students in the class Average Height = ( Total Height of all students in the Class / Total Number of Students) Average Height = ( 1300/17) = 76.47
212.
Total Marks obtained by a class is 1584 and the average score of the class is 72. Find the Total students in the class?
A.) 20
B.) 21
C.) 22
D.) 23
Use the below formula to find the Total, when Average and Total No. of Students is provided
Total no of students = Total Marks / Average Marks
Total No. of students
= ( 1584 / 72)
= 22
Total students in class = 22
213.
Total Marks obtained by a class is 1156 and the average score of the class is 68. Find the Total students in the class?
A.) 16
B.) 17
C.) 18
D.) 19
Use the below formula to find the Total, when Average and Total No. of Students is provided Total no of students = Total Marks / Average Marks Total No. of students = ( 1156 / 68) = 17
214.
In a class of 18 students in an examination in science 1 student scored 100 marks, 2 get zero each and the average oh the rest was 45. What is the average of the whole class?
A.) 41
B.) 43
C.) 42
D.) 44
43
215.
In the first 35 overs of a cricket game, the run rate was only 6.4. What should be the run rate in the remaining 15 overs to reach the target of 410 runs?
A.) 12.4
B.) 13.4
C.) 15.25
D.) 17.25
Required Run rate = ( 410 - ( 6.4 x 35 ) )/ (15)
= ( 186) /(15) = 12.4
216.
Of three numbers, the third is twice the second and the second is 4 times the first. If their average is 78, the smallest of the three numbers is:
A.) 15
B.) 21
C.) 17
D.) 18
Let first number be x.
So,2nd no. = 4x & 3rd no.=8x.
So,x+4x+8x=78 × 3 = 234.
13x = 234
x = 234/13
Hence,smallest Number x=18.
217.
The average weight of a class having 12 students is 30 kg. Find the total weight of the class.
A.) 320
B.) 360
C.) 260
D.) 460
Solution is
[ Total quantity = Average x No.of quantity ]
= 12 x 30
360
218.
Average of five consecutive even numbers is 35. Find the greatest number in these five numbers?
A.) 31
B.) 33
C.) 39
D.) 36
39
219.
Physics Teacher knows the average score of 43 students in her subject is 49.5. Can you help the physics teacher to find the overall marks obtained by all the students in her subject?
A.) 2341.35
B.) 2234.93
C.) 2128.5
D.) 2022.08
Use the below formula to find the Total, when Average and Total No. of Students is provided Total = Average * Total Students/People Total = Average * Total Number of Students) Average Weight = ( 49.5 x 43) = 2128.5
220.
There were 45 students in a hostel, if the numbers of students increased by 7, the expenses of the mess were increased by Rs. 39 per day while the average expenditure per head diminished by Re.1. What is the original expenditure of the mess (in Rs)?
A.) 562
B.) 585
C.) 548
D.) 650
Let the original expenditure be Rs.x
Original average expenditure = X/45
New average expenditure = (x+39)/52
So (x/45) - ((x+39) / 52) = 1 => (52x - 45x - 45 × 39) / (45 × 52) = 1 => (7x - 1755) / 2340 = 1 => 7x = 2340 + 1755 => 7x = 4095 so x = 585
so, original expenditure is Rs 585 | 2,194 | 6,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-24 | latest | en | 0.914103 |
http://www.clubtnt.org/my_algebra/chapter_5_notes.htm | 1,501,029,848,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425737.60/warc/CC-MAIN-20170726002333-20170726022333-00235.warc.gz | 391,811,906 | 6,811 | ##### Mr. B's Career Exploration Web
It is my humble opinion that one of the "best" parts of K. Elayn Martin-Gay's presentation of algebra (Introductory Algebra) is how she prepares students for factoring polynomials (Chapter 4), moves students through that section, and then immediately uses those concepts to introduce rational expressions (Chapter 5).
In many ways, her presentation of rational expressions if a complete review of Chapter 4. It is my experience that Chapter 4 and 5 are the key to success in Beginning Algebra. I believe that a full understanding of these sections also determines who is fully ready for Intermediate Algebra 141.
As we work through this chapter, please stop periodically and ask yourself, "what parts of this chapter represented a review and what part was really 'new'." Not only are we reviewing almost all of Chapter 4, we are reviewing the "nuts and bolts" of working with fractions.
### 5.1 Simplifying Rational Expressions
Rational Expression: An expression that can be written as a quotient of polynomials. If we use capital letters to represent polynomials (P & Q), then a rational expression can be written in the form P/Q where Q does not equal "0."
Rational expressions are fractions that have numerators and denominators that are polynomials. Everything that is true for a fraction is true for a rational expression. Recall that we never have a denominator of "0" -- a quotient with a denominator of zero is undefined; it is not a real number.
Values Where a Rational Expression is Undefined. Because division by zero is undefined, any value that causes the polynomial in the denominator of a rational expression to be "0" results in a rational expression that is undefined.
Fundamental Principle of Rational Expressions. Just like a fraction, we can always multiply the numerator and denominator of a rational expression by any polynomial that does not equal zero and the value of the product does not change. If P, Q, & R are polynomials and Q & R do not equal zero; then (PR)/(QR) = P/Q.
Simplifying Rational Expressions:
1. Factor numerators and denominators.
2. Apply the fundamental principal to divide out common factors -- cancel out factors that appear in BOTH the numerator and denominator.
Please refer to Examples 3 - 7 (pages 351-352 in text) to see how this procedures are applied.
### 5.2 Multiplying and Dividing Rational Expressions
Multiplying Rational Expressions:
1. Factor numerators and denominators.
2. Multiply numerators and multiply denominators.
3. Write product in lowest terms.
While these steps, as outlined in our text, are correct, if you look at the examples in the book, you will see that factoring numerators and denominators will allow us to look for common factors. Step 2 actually just involves writing our factors from step one as a single denominator!!! Common factors in our numerator AND denominator can be eliminated, which will greatly simplify our work.
Please be sure to carefully look at Examples 1-3 (pages 357-358 in text). You will see that our objective is to factor the numerators and denominators completely and cancel out any common factors. IN MANY WAYS, THIS IS THE "HEART OF ALGEBRA!"
Recall that our definition of algebra stated that letters represent numbers, that we take "expressions" apart and then put them back together. DO YOU SEE THAT IN CHAPTER 5 WE ARE DOING ALL OF THESE THINGS?
After completely factoring numerators and denominators, canceling out common factors, we simplify what is left to be sure that the product is written in lowest terms. Canceling out common factors is really just applying the Fundamental Principal of Rational Expressions.
Please recall from our review of fractions that WE NEVER ACTUALLY DIVIDE FRACTIONS.
### 5.3 Adding and Subtracting Rational Expressions with the Same Denominator and Least Common Multiple
We can think of rational expressions as being fractions that are expressed as polynomials in the numerator and denominator. EVERYTHING WE KNOW ABOUT FRACTIONS APPLIES TO RATIONAL EXPRESSIONS. This is where I most appreciate E. Elayn Martin Gay's presentation.
If we understood fractions and the foundation that Chapter 4 was build in (factoring polynomials), we actually have all of the "tools" we need to successfully work with rational expressions. Recall that we add fractions with the same denominator by writing our numerators as addition on the common denominator. Then, we simply combine the terms.
Remember, we are using capital letters to represent polynomials. Algebraically (P/R)+(Q/R)=(P+Q)/R (as long as R does not equal zero). Likewise, (P/R)-(Q/R)=(P-Q)/R.
Recall that we add polynomials by combining like terms. REMEMBER, THAT WHEN WE SUBTRACT POLYNOMIALS, WE NEED TO CHANGE ALL SIGNS IN THE SECOND POLYNOMIAL (the one we are subtracting) and then we simply combine like terms.
We need to always simply our answers by being sure that all common factors are eliminated -- common factors in the numerator and denominator "cancel" each other. Be sure to look at Examples 1-3 (pages 367-368 in text).
If we make the "connections" between our understanding of fractions and rational expressions, adding/subtracting polynomials with common denominators is nothing new -- BE SURE TO SIMPLIFY YOUR ANSWER (see section 5.1).
Our author wants to take advantage of this -- she will introduce the next unit here so that we have practiced some useful skills BEFORE we actually need them.
Least Common Multiple
Recall that in Chapter R we discussed how it is not possible to combine fractions without common denominators. We needed to find a least common denominator (LCD) -- perhaps you also remember that LCD=LCM.
On page 368 is a review where we add 8/3 and 2/5. The common denominator would be 15 -- because creating fractions with HTML on this Website is time consuming, I will ask you to look at how the book reviews that example.
We can also find the least common denominator (LCD) for rational expressions. The least common denominator (*LCD) of a list of rational exrp3esions is a polynomial of least degree whose factors include all the factors of the denominator on the list.
LCD of a list of rational expressions:
1. Factor each denominator completely.
2. The least common denominator (LCD) is the product off all unique factors found in step 1, each raised to a power equal to the greatest number of times that the factor appears in any one factored denominator.
Please be sure to look at Examples 4-10 (pages 368-370 in text).
### 5.4 Adding and Subtracting Rational Expressions with Different Denominators
We have already reviewed finding the LCD, recall from fractions, that is the key to adding fractions with different denominators. If we rewrite fractions as equivalent fractions using the lowest common denominator (LCD), then we can rewrite our fractions on that common denominator and combine the numerators.
Add or Subtract Rational Expressions with Different Denominators
1. Find the LCD of the rational expressions (end of unit 5.3).
2. Rewrite each rational expression as an equivalent expression whose denominator is the LCD found in Step 1.
3. Add or subtract numerators and write the sum or difference over the common denominator (at this point, we are really just combining terms on the common denominator).
4. Simplify or write in lowest terms (if there are any common factors in the numerator and denominator, be sure to remove eliminate them, they "cancel out."
Please review examples 1-7 (pages 375-376).
### 5.5 Solving Equations Containing Rational Expressions
Before we start, lets think about this -- rational expression will be more complex to work with than the linear equations we solved in Chapter 3. LET'S GIVE OURSELVES A BREAK AND USE ALGEBRA TO ELIMINATE THE ALGEBRA. This will allow us to solve equations containing rational expressions like we solve any other equation.
Recall that we can always multiply the left and the right of an equation by the same term and we will not change that equation's identity. If we apply that principle here, then we will be ready to solve the equation that remains.
Solve equation Containing Rational Expressions
1. Multiply both sides of the equation by the LCD of ALL rational expressions in the equation. THIS WILL ALLOW US TO "CANCEL OUT" OR ELIMINATE THE DENOMINATOR.
2. Remove any grouping symbols and solve the resulting equation.
3. Check the solution IN THE ORIGINAL EQUATION.
Please be sure to look at Examples 1-7 (pages 383-386 in text).
### 5.6 Rational Equations and Problem Solving
Recall that all problem solving is presented in our text as a 4 part process:
2. Translate. Write an equation based on the words in the problem.
3. Solve. Apply the steps to solving an equation to solve for your variable(s).
A some important examples to look at include:
Finding work rates (example 2, page 395-396) Solving problems about distance, rate and time ((example 3, pages 295-297) Solving problems about similar triangles (example 4, pages 397-398)
### 5.7 Simplifying Complex Fractions
A rational expression whose numerator or denominator of both contain fractions is called a complex rational expression or complex fraction. Like ALL fractions, we need to write complex fractions in simplest form. If we use P and Q to represent polynomials, we have reduced complex fractions to simplest form when P/Q share no common factor.
Our text presents two ways to solve complex fractions -- PLEASE BE SURE TO STUDY AND PRACTICE BOTH. Like most "choices" we have looked at, there is no "best" way. Each of us needs to determine what we are most comfortable and accurate -- often, the best method depends on the nature of a problem that is to be solved.
The author simply calls these 2 methods, Method 1 & Method 2.
Method 1: To simplify a complex fraction:
1. Add or subtract the fraction in the numerator or denominator so that the numerator is a simple fraction and the denominator is a single fraction.
2. Perform the indicated division by multiplying the numerator of the complex fraction by the reciprocal of the denominator of the complex fraction.
3. Write the rational expression in lowest terms.
Please be sure to look at examples 1-3, pages 405-406.
Method 2: To simplify a complex fraction:
1. Find the LCD of all the fractions in the complex fraction.
2. Multiply both the numerator and the denominator of the complex fraction by the LCD from Step 1.
3. Perform the indicated operations and write the results in lowest terms.
Please be sure to look at examples 4-6, pages 407-408.
[ Top ]
Review Notes
[ Top ] | 2,350 | 10,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2017-30 | latest | en | 0.917795 |
https://www.scienceforums.net/topic/79522-negative-numbers/?tab=comments | 1,600,898,567,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400212959.12/warc/CC-MAIN-20200923211300-20200924001300-00521.warc.gz | 1,057,682,224 | 23,083 | # Negative numbers
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Why does a negitive times a negitive equal a positive
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Do you understand why taking away -1 from a number is the same as adding 1? (I remove some of your debt means you are wealthier as you owe less)
$x-(-y)=x+y$
And that multiplication can be seen as adding a certain number to itself a number of times
$3x=x+x+x$
multiplication by a negative number can be seen as repeated subtracting
$-3x = -x-x-x$
So multiplication of two negative numbers is repeatedly subtracting a minus
$-3(-x) = -(-x)-(-x)-(-x) = +x+x+x=3x$
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I have had some luck with the baffled, after trying the standard (as above) first, by describing the negative sign as an instruction to rotate counterclockwise 180 - to "flip" - the indicated. So multiplication by a negative 1 would be first the standard position on the line, then flipped around the zero. So multiplication buy two negatives would be two flips - back where you started.
This has the advantage of not using the negative sign in the explanation of the negative sign or operations involving it, agreeing with the high class development of the number system and groups etc, being intuitively generalized to handling fractions and functions and other objects not readily comprehended by subtraction, and leading naturally into the square roots of negative one (as rotations by 90) eventually.
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What helped me to remember this was that when multiplying two numbers with the same direction on a number line you treat the problem when dealing with absolute values.
EX:
(-3)*(-3) = |-3| * |-3| = |3| * |3| = 3*3
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Why does a negitive times a negitive equal a positive
The short answer is that, because a positive number of a negative amount equals a negative amount, such as (positive) three –10's equals –30, then a negative number of negative amounts must equal the opposite — a positive number — such as negative three –10's equals +30.
Another way of looking at it goes like this —
8 × 8 = 64
which is
(10 – 2) × (10 – 2)
which is
10 × 10 + 10 × –2 + –2 × 10 + –2 × –2
which is
100 + –20 + –20 + 4
which is
100 – 20 – 20 + 4
which is
64
(and not 100 – 20 – 20 – 4 which is 56)
You can do the same for 2 × 2 as being (10 – 8) × (10 – 8) and get the answer of 4.
I hope that one or both of these explanations helps.
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The short answer is that, because a positive number of a negative amount equals a negative amount, such as (positive) three –10's equals –30, then a negative number of negative amounts must equal the opposite — a positive number — such as negative three –10's equals +30.
Another way of looking at it goes like this —
8 × 8 = 64
which is
(10 – 2) × (10 – 2)
which is
10 × 10 + 10 × –2 + –2 × 10 + –2 × –2
which is
100 + –20 + –20 + 4
which is
100 – 20 – 20 + 4
which is
64
(and not 100 – 20 – 20 – 4 which is 56)
You can do the same for 2 × 2 as being (10 – 8) × (10 – 8) and get the answer of 4.
I hope that one or both of these explanations helps.
This seems a bit more complicated than other explanations. This is especially because most people don't really use the distributive property of multiplication without it being algebraic.
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The reason that a negative times a negative equals a positive is that it is a logical consequence of the basic properties of the integers, which are known as the ring axioms. You can read (way too much) about them here ...
http://en.wikipedia.org/wiki/Ring_(mathematics)
I have to apologize in advance for that link because it's way too technical and fails to provide a simplified introduction to the subject.
But basically we define a "ring" as any mathematical system in which we can add, subtract, and multiply; and that addition and multiplication are related by the distributive law
a(b + c) = ab + ac
valid for all objects a, b, and c, as long as they are members of the system in which we're interested. [From now on I'll use the word "numbers" to refer to the members of our system; but keep in mind that these are not necessarily the familiar integers, but rather the members of any system that obeys these rules.]
The ring axioms say that there's a magic number called 0 that has the property that
0 + n = n for all numbers n
and that for each number n there is an "additive inverse" called -n, whose defining property is that
= n + (-n) = 0.
There's also a magic number called 1 with the property that 1 * n = n for all numbers n.
All these things are true about the everyday integers, so we can use these properties to prove things about the integers. And it turns out to be the distributive property that is crucial.
Here's a formal derivation.
We want to find out what is (-1)(-1).
The way I'm going to do this is to evaluate the quantity (-1)(-1) - 1 and show that it must be zero. This will then prove that (-1)(-1) = 1.
(-1)(-1) - 1 // Expression we want to evaluate.
= (-1)(-1) + (-1) // Because "subtraction" is actually adding the additive inverse.
= (-1)(-1) + (-1)(1) // Multiplying something by 1 doesn't change it.
= (-1)(-1 + 1) // This is the distributive law, which says that a(b + c) = ab + ac.
= (-1)(0) // Because -1 + 1 = 0 by definition. -1 is the additive inverse of 1.
= 0 // Anything times zero is zero. That's actually a consequence
// of the ring axioms and requires proof, which which we'll assume.
We just derived a logically certain chain of equality between (-1)(-1) - 1 and 0. If we add 1 to both sides of the equation, we get (-1)(-1) = 1.
Conclusion: The ring axioms logically imply that (-1)(-1) = 1. One is still free to impute metaphysical significance to all this, or to try to "figure it out." Mathematicians prefer abstraction. We write down the properties of the thing being studied; and then we derive logical consequences.
This methodology of abstraction often provides structural insight. We just discovered that the reason a negative times a negative is positive is simply that it's a logical consequence of the distributive law. Once you accept the distributive law, you have no other choice. That's interesting!
By the way we should complete the job by proving the general case (-a)(-b) = ab. That's left to the reader of course
Edited by HalfWit
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Most of the people asking the question are looking for a physical, intuitve way to comprehend the situation. The basic problem is that "negative" has been introduced to them as a direction on a number line, the degrees below zero on a thermometer, etc. That intuition does not work well for thinking about multiplication of negative numbers, neither is it well conjoined with the actual theoretical basis of that extension of the integers etc.
Proofs don't really meet their needs.
Edited by overtone
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Most of the people asking the question are looking for a physical, intuitve way to comprehend the situation. .............
Proofs don't really meet their needs.
In a math forum, I find this a curious comment. But leave that, here's the general case that HalfWit asked for
Consider the sum of products $ab+a(-b)+(-a)(-b)$.
By the associative law write this as $ab+[a(-b)+(-a)(-b)]$ and by the right distributive law write this as
$ab+[a+(-a)](-b) = ab+0(-b) = ab$ from the multiplicative property of zero
On the other hand, using the exact same rules
$ab+a(-b)+(-a)(-b) = [ab+a(-b)]+(-a)(-b)]$
$=a[b+(-b)]+(-a)(-b) = a0+(-a)(-b)=(-a)(-b)$ (although you use the left distributive law here)
Which implies that $ab = (-a)(-b)$ as desired.
As a further generalization assume the above applies to the ring $\mathbb{Z}$ of integers. Let us be a bit naughty and call $0$ a proper sub-ring.
Then from ring theory we have that an ideal $I$ of any ring $R$ is a subring such that for all $x \in I$ and all $a \in R$ that
$xa \in I$ and $ax\in I$. It is common to abuse notation somewhat and write this as $Ia =aI$ (actually equality here implies that we have a two-sided ideal). let's call $0$ a proper two-sided ideal of the integers (proper because the ring itself is also an ideal)
Then with very little modification, the above proof goes through for any ring if you replace $0$ by $I$
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so, its as they say then, two wrongs do make a right??
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Register a new account | 2,183 | 8,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-40 | latest | en | 0.926252 |
https://staging.coursekata.org/preview/book/d10dd3be-dbef-476d-bc6a-f9e3bd9d2852/lesson/9/0 | 1,721,731,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00569.warc.gz | 481,445,074 | 13,449 | ## Course Outline
• segmentGetting Started (Don't Skip This Part)
• segmentStatistics and Data Science: A Modeling Approach
• segmentPART I: EXPLORING VARIATION
• segmentChapter 1 - Welcome to Statistics: A Modeling Approach
• segmentChapter 2 - Understanding Data
• segmentChapter 3 - Examining Distributions
• segmentChapter 4 - Explaining Variation
• segmentPART II: MODELING VARIATION
• segmentChapter 5 - A Simple Model
• segmentChapter 6 - Quantifying Error
• segmentChapter 7 - Adding an Explanatory Variable to the Model
• segmentChapter 8 - Models with a Quantitative Explanatory Variable
• segmentPART III: EVALUATING MODELS
• segmentChapter 9 - The Logic of Inference
• segmentChapter 10 - Model Comparison with F
• segmentChapter 11 - Parameter Estimation and Confidence Intervals
• segmentPART IV: MULTIVARIATE MODELS
• segmentChapter 12 - Introduction to Multivariate Models
• segmentChapter 13 - Multivariate Model Comparisons
• segmentFinishing Up (Don't Skip This Part!)
• segmentResources
### list College / Advanced Statistics and Data Science (ABCD)
Book
• High School / Statistics and Data Science I (AB)
• College / Statistics and Data Science (ABC)
• High School / Advanced Statistics and Data Science I (ABC)
• College / Advanced Statistics and Data Science (ABCD)
• High School / Statistics and Data Science II (XCD)
# Chapter 6 - Quantifying Error
## 6.1 Quantifying Total Error Around a Model
Up to now we have developed the idea that a statistical model can be thought of as a number—a predicted value for the outcome variable. We are trying to model the Data Generating Process (DGP). But because we can’t see the DGP directly, we fit a model to our data, estimating the parameters.
Using the DATA = MODEL + ERROR framework, we have defined error as the residual that is left after we take out the model. In the case of our simple model for a quantitative outcome variable, the model is the mean, and the error— or residual—is the deviation of each score above or below the mean.
We represent the simple model like this using the notation of the General Linear Model:
$Y_{i}=b_{0}+e_{i}$
This equation represents each score in our data as the sum of two components: the mean of the distribution (represented by $$b_{0}$$), and the deviation of that score above or below the mean (represented as $$e_{i}$$). In other words, DATA = MODEL + ERROR.
In this chapter, we will dig deeper into the ERROR part of our DATA = MODEL + ERROR framework. In particular, we will develop methods for quantifying the total amount of error around a model, and for modeling the distribution of error itself.
Quantifying the total amount of error will help us compare models to see which one explains more variation. Modeling the distribution of error will help us to make more detailed predictions about future observations and more precise statements about the DGP.
At the outset, it is worth remembering what the whole statistical enterprise is about: explaining variation. Once we have created a model, we can think about explaining variation in a new way, as reducing error around the model.
We have noted before that the mean is a better model of a quantitative outcome variable when the spread of the distribution is smaller than when it is larger. When the spread is smaller, the residuals from the model are smaller. Quantifying the total error around a model will help us to know how good our models are, and which models are better than others.
To make this concrete, let’s consider our very simple model of our tiny sample (n = 6) of data, TinyFingers. Here is the data frame including the Predictions and Residuals from the Tiny_empty_model.
StudentID Thumb Prediction Residual
1 1 56 62 -6
2 2 60 62 -2
3 3 61 62 -1
4 4 63 62 1
5 5 64 62 2
6 6 68 62 6
The histograms below show the distribution of Thumb and the distribution of Residual for our tiny data set. And of course, as you know by now, these distributions have the exact same shape, but different means.
It makes sense to use residuals to analyze error from the model. If we want to quantify total error, why not just add up all the residuals? Worse models should have more error, so the sum of all the errors should represent the “total” error, right?
Let’s do that, using one of the first R functions you learned, sum(). The following code will add up all the residuals from our Tiny_empty_model.
sum(TinyFingers$Residuals) [1] 0 Although we might at first think that the sum of the residuals would be a good indicator of total error, we discovered a fatal flaw in that approach in the previous chapter: the sum of the residuals around the mean is equal to 0! If this were our measure of total error, all data sets would be equally well modeled by the mean, because the residuals around the mean would always sum to 0. Thus a data set widely spread out around the mean, and one tightly clustered around the mean, would have the same amount of error around this simple model. Clearly we need a different approach. Statisticians have explored various methods for quantifying error around a mean. Two of the most common, which we will discuss here are Sum of Absolute Deviations (SAD), and Sum of Squared Deviations (SS). Each provides a different way of solving the problem created by simply summing residuals. Let’s take a look at each. ### Sum of Absolute Deviations (SAD) Sum of absolute deviations gets around the problem that deviations around the mean always add up to 0 by taking the absolute value of the deviations. We can represent this summary measure like this: $\sum_{i=1}^n |Y_i-\bar{Y}|$ In this context, “deviations from the mean” means the same thing as “residuals from the empty model,” given that the mean is our model. We already have the deviations of each thumb length from the mean in the Residual column of TinyFingers. We can take the absolute value of each deviation from the mean by using the function abs(). abs(TinyFingers$Residual)
[1] 6 2 1 1 2 6
As with everything, if we don’t save it, these absolute values of the residuals are lost. Modify the following code to save the absolute residuals in the variable AbsResidual in TinyFingers. Then use sum() to sum up the AbsResidual.
require(coursekata) TinyFingers <- data.frame( StudentID = 1:6, Thumb = c(56, 60, 61, 63, 64, 68) ) Tiny_empty_model <- lm(Thumb ~ NULL, data = TinyFingers) TinyFingers <- TinyFingers %>% mutate( Predicted = predict(Tiny_empty_model), Residual = resid(Tiny_empty_model) ) # save the absolute residuals to AbsResidual TinyFingers$AbsResidual <- # modify this code to sum up AbsResidual sum() TinyFingers$AbsResidual <- abs(TinyFingers$Residual) sum(TinyFingers$AbsResidual) ex() %>% { check_object(., "TinyFingers") %>% check_column("AbsResidual") %>% check_equal() check_function(., "sum") %>% check_result() %>% check_equal() }
CK Code: ch6-1
[1] 18
### Sum of Squared Deviations (SS)
Another way to quantify error, which gets around the problem of the sum of deviations adding up to 0, is to square the deviations (i.e., residuals). This approach is represented like this:
$\sum_{i=1}^n (Y_i-\bar{Y})^2$
Since we already have the column Residual, we can easily create a column of squared residuals. We don’t need to use absolute value here because squaring will result in a positive value no matter the starting value. In R, we represent exponents with the caret symbol (^, usually above the 6 on a standard keyboard).
TinyFingers$Residual^2 [1] 36 4 1 1 4 36 Try it for yourself. Save the squared residuals in the variable SqrResidual in TinyFingers. Then use sum() to sum up the SqrResidual. require(coursekata) TinyFingers <- data.frame( StudentID = 1:6, Thumb = c(56, 60, 61, 63, 64, 68) ) Tiny_empty_model <- lm(Thumb ~ NULL, data = TinyFingers) TinyFingers <- TinyFingers %>% mutate( Predicted = predict(Tiny_empty_model), Residual = resid(Tiny_empty_model) ) # save the squared residuals to SqrResidual TinyFingers$SqrResidual <- # modify this code to sum up SqrResidual sum() TinyFingers$SqrResidual <- TinyFingers$Residual^2 sum(TinyFingers\$SqrResidual) ex() %>% { check_object(., "TinyFingers") %>% check_column("SqrResidual") %>% check_equal() check_function(., "sum") %>% check_result() %>% check_equal() }
CK Code: ch6-2
[1] 82
Each of these measures has been used by statisticians. But when the mean is the model, sum of squares has prevailed. Here is a little video with Dr. Ji regarding why.
Video Transcript
If you want to try it yourself, here is a link to the sum of squares applet, and below are the data you can copy/paste into the little “sample data” box. (Here’s the full link in case that one doesn’t work: http://www.rossmanchance.com/applets/RegShuffle.htm)
StudentID Thumb 1 56 2 60 3 61 4 63 5 64 6 68 | 2,204 | 8,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.679417 |
https://www.kseebsolutions.com/category/class-10/ | 1,709,230,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474852.83/warc/CC-MAIN-20240229170737-20240229200737-00875.warc.gz | 835,091,893 | 92,050 | ## KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3.
## Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3
Question 1.
FInd the sum of the following APs:
i) 2, 7, 12, ……… to 10 terms
ii) -37, -33, -29,………. to 12 terms.
iii) 0.8, 1.7, 2.8 to 100 terms.
iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:
i) Sum of 2, 7, 12, ………. 10?
a = 2, d = 7 – 2 = 5, n = 10
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]$$
= 5[4 + 9 × 5]
= 5[4 + 45]
= 5 × 49
∴ S10 = 245
ii) -37, -33, -29, to 12……. terms.
Solution:
a = -37, d = -33 – (-37) = -33 + 37
n = 12, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{12}=\frac{12}{2}[2 \times -37+(12-1) 4]$$
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6 × -30
∴ S12 = -180.
iii) 0.6, 1.7, 2.8,……… , to 100 terms.
a = 0.6, d = 1.7 – 0.6 = 1.1
n = 100, S100 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{100}=\frac{100}{2}[2 \times 0.6+(100-1)(1.1)]$$
= 50[1.2 + 99(1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1
∴ S100 = 5505
iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:
The LCM of 12 and 18 is 36.
Question 2.
Find the sums given below:
i) $$7+10 \frac{1}{2}+14+\ldots \ldots+84$$
ii) 34 + 32 + 30 + ……….+ 10
iii) -5 + (-8) + (-11) + ……….. +(-230)
Solution:
ii) 34 + 32 + 30 + ………. + 10
Solution:
a = 34, d = 32 – 34 =-2, an = 10,
Sn =?
a + (n – 1)d = an
34 + (n – 1)(-2) = 10
34 – 2n + 2 = 10
-2n + 36 = 10
-2n = 10 – 36
-2n = -26
2n = 26
$$n=\frac{26}{2}$$
∴ n = 13
$$S_{n}=\frac{n}{2}[a+l]$$
$$\mathrm{S}_{23}=\frac{13}{2}[34+10]$$
$$=\frac{13}{2} \times 44$$
= 13 × 22
∴ S23 = 286.
iii) -5 + (-8) + (-11) +……… + (-230)
Solution:
a = -5, d = -8 – (-5) = -8 + 5
a, = -230.
Sn =? d = -3
a+(n – 1)d = an
-5 + (n – 1)(-3) = -230
-5 – 3n + 3 = -230
-3n – 2 = -230
-3n = -230 +2
-3n = -228
3n = 228
∴ n = 228/3
∴ n = 76.
$$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]$$
$$S_{ 76 }=\frac { 76 }{ 2 } \left[ -5-230 \right]$$
$$=\frac{76}{2} \times -235$$
= 38 × -235
∴ S76= -8930.
Question 3.
In an AP:
i) given a = 5. d = 3. an = 50, find ‘n’ and Sn.
ii) given a = 7, a13 = 35, find d’ and S13.
iii) given a12= 37. d = 3. find ‘a’ and S12.
iv) given a3 = 15, S10 = 125. find ‘d’ and a10.
v) given d = 5. S9 = 72, fInd ‘a’ and a9.
vi) given a = 2. d = 8, Sn = 90, fInd ‘n and a11.
vii) given a = 8, an = 62. Sn = 210, find ‘n’ and ‘d’.
viii) given an = 4, d = 2, Sn = -14. find ‘n’ and a.
ix) given a = 3, n = 8, S = 192, find ‘d’.
x) given L = 28, S = 144 and there are total 9 terms. Find ‘a’.
Solution:
i) a = 5, d = 3, an = 50. n=?, Sn =?
a + (n – 1)d = an
5 + (n – 1) 3 = 50
5 + 3n – 3 = 50
3n + 2 = 50
3n = +50 – 2
3n = 48
$$n=\frac{48}{3}$$
∴ n = 16
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{16}=\frac{16}{2}[5+50]$$
$$=\frac{16}{2}[55]$$
= 8 × 55
∴ S16= 440.
ii) a = 7, a13 = 35. d =?. S13 =?
Solution:
a+(n – 1) d = an
7 + (13 – 1) d = 35
7 + 12d = 35
12d = 35 – 7
12d = 28
$$d=\frac{28}{12}$$
$$d=\frac{7}{3}$$
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{n}=\frac{7}{3} \times \frac{1}{2}[7+35]$$
$$\mathrm{s}_{\mathrm{n}}=\frac{7}{6} \mathrm{I} 42 \mathrm{l}$$
= 7 × 7
∴ Sn = 49
iii) a12= 37, d = 3, a =? S12 =?
a + (n – 1)d = an
a + (12 -1) 3 = 37
a + 11 × 3 = 37
a + 33 = 37
∴ a = 37 – 33
∴ a = 4
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{S}_{12}=\frac{12}{2}[4+37]$$
= 6[4 + 37]
∴ S2 = 246
iv) a3 = 15, S10 = 125. d =?, a10=?
Solution:
a3 = a + 2d = 15
∴ a= 15 – 2d
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2(15-2 d) \div(10-1) d]=125$$
= 5[30 – 4d + 9d] = 125
= 5[30 + 5d] = 125
150 + 25d = 125
25d = 125 + 150
25d = -25
$$\mathrm{d}=\frac{-25}{25}$$
∴ d = -1.
Substittuting the value of ‘d’
a = 15 – 2d
= 15 – 2(-1)
= 15 + 2
∴ a = 17.
∴ a10 = a + 9d
= 17 + 9(-1)
= 17 – 9
∴ a10 = 8
v) d = 5, S9 = 72. a =? a9 =?
Solution:
$$\mathrm{s}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$$
$$s_{9}=\frac{9}{2}[2 \times a+(9-1) 5]=72$$
$$\frac{9}{2}[2 a+8 \times 5]=72$$
$$\frac{9}{2}[2 a+40]=72$$
$$2 a+40=72 \times \frac{2}{9}$$
2a + 40 = 16
2a = 16 – 40
2a = -24
$$a=\frac{-24}{2}$$
∴ a = -12
a9 = a + (n – 1) d
= -12 + (9 – 1) (5)
= -12 + 8 × 5
= -12 + 40
∴ a9 = 28
vi) a = 2, d = 8 Sn = 90, n =?, a3 =?
Solution:
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$=\frac{n}{2}[2 \times 2+(n-1) 8]=90$$
$$\frac{n}{2}[4+8 n-8]=90$$
n (8n – 4) = 90 × 2/1
8n2 – 4n = 180
8n2 – 4n = 1800
2n2 – n – 45 = 0
2n2– 10n + 9n – 45 = 0
2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
If n – 5 = 0 then n = 5
∴ an = a + (n – 1) d
a5= 2 + (5 – 1) 8
= 2 + 4 × 8
= 2 + 32
∴ a5 = 34
vii) a = 8, an = 62, Sn = 210, n =? d =?
Solution:
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[8+62]=210$$
$$\frac{n}{2} \times 70=21$$
$$n=210 \times \frac{2}{70}$$
∴ n = 6
an = 62
a6= a + 5d = 62
8 + 5d = 62
5d = 62 – 8
5d = 54
$$d=\frac{54}{5}$$
viii) a = 4, d = 2, Sn = -14, n =?, a =?
Solution:
an = a + (n – 1) d = 4
a + (n – 1) 2 = 4
a + 2n – 2 = 4
a + 2n = 4 + 2
a = 2n = 6 …………. (i)
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[-2 n+6+4]=-14$$
$$\frac{n}{2}[-2 n+10]=-14$$
-2n2 + 10n = -28
-2n2 + 10n + 28 = 0
2n2 – 10n – 28 = 0 – 14
2n2 – 5n – 14 = 0
n2 – 7n + 2n – 14 = 0
n(n – 7) + 2 (n – 7) = 0
(n – 7) (n + 2) = 0
If n – 7 = 0 then. n = 7
a = -2n + 6
= -2 × 7 + 6
= -14 + 6
∴ a = -8
∴ n = 7, a = -8.
ix) a = 3, n = 8, S = 192, d =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{8}=\frac{8}{2}[2 \times 3+(8-1) d]=192$$
$$\frac { 8 }{ 2 } [6+7\times d]=192$$
4 (6 + 7d) = 192
$$6+7 d=\frac{192}{4}$$
6 + 7d = 48
7d = 48 – 6
7d = 42
$$d=\frac{42}{7}$$
∴ d = 6
x) l = an = 28, S = 144, n = 9, a =?
Solution:
$$s_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$s_{9}=\frac{9}{2}[a+28]=144$$
$$\frac{9}{2}[a+28]=144$$
$$a+28=144 \times \frac{2}{9}$$
a + 28 = 32
a = 32 – 28
∴ a = 4.
Question 4.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
9 + 17 + 25 + ……. + an = 636
a = 9, d = 17 – 9 = 8, Sn= 636, n =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 9+(n-1) 8]=636$$
$$\frac{n}{2}[18+8 n-8]=636$$
$$\frac{n}{2}[8 n+10]=636$$
n(8n + 10) = 636 × 2
8n2 + 10n = 1272
8n2+ 10n – 1272 = 0
4n2 + 5n – 636 = o
4n2+ 5n – 48n – 636 = 0
n(4n + 5) – 12(4n + 5) = 0
(4n + 5) (n – 12) = 0
1f n – 12 = 0 then, n = 12
∴ n = 12.
Question 5.
The first term of an AP is 5. the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
a = 5, an = 45, Sn= 400, n =?, d =?
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$=\frac{n}{2}[5+45]=400$$
$$\frac{n}{2} \times 50=400$$
n × 50 = 800
$$n=800 \times \frac{1}{50}$$
∴ n = 16
an = a + (n – 1) d
a16= 5 + (16 – 1)d = 45
5 + 15d = 45
15d = 45 – 5
15d = 40
$$d=\frac{40}{15}=\frac{8}{3}$$
$$\mathrm{n}=16, \mathrm{d}=\frac{8}{3}$$
Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
a = 17, an= 350, d = 9, n =?, Sn =?
an = a + (n – 1) d
17 + (n – 1)9 = 350
17 + 9n – 9 = 350
9n + 8 = 350
9n = 350 – 8
9n = 342
$$\quad n=\frac{342}{9}$$
∴ n = 38
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{38}=\frac{38}{2}[17+350]$$
= 19 × 367
∴ S38 = 6973
∴ n = 38, S38 = 6973
Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
d = 7, a22 = 149, S22 =?
a = a + (n – 1)d
a + (22 – 1)7 = 149
a + 21 × 7 = 149
a + 147 = 149
=149 – 147
∴ a = 2
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{22}=\frac{22}{2}[2+149]$$
= 11[151]
∴ S22 = 1661
Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
If a2 = 14, a3 = 18, then S11 =?
a2 = 14
a + d = 14 …………. (i)
a3 = 18
a + 2d = 18 …………… (ii)
from equation (1) – equatIon (11),
d =4.
Substituting the value of ‘d’ In eqn. (i)
a + d = 14
a + 4 = 14
a = 14 – 4
∴ a = 10
a = 10, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{51}=\frac{51}{2}[2 \times 10+(51-1) 4]$$
$$=\frac{51}{2}[20+50 \times 4]$$
$$=\frac{51}{2}[20+200]$$
$$=\frac{51}{2} \times 220$$
= 51 × 110
∴ S51 = 5610
Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first ‘n terms.
Solution:
S7 = 49, S17 = 289, S11 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{7}=\frac{7}{2}[2 a+(7-1) d]$$
$$=\frac{7}{2}[2 a+6 d]=49$$
[latex2+6 d=49 \times \frac{2}{7}[/latex]
2a + 6d = 14 …………… (i)
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{17}=\frac{17}{2}[2 a+(n-1) d]$$
$$=\frac{17}{2}[2 \mathrm{a}+16 \mathrm{d}]$$
$$\frac{17}{2}[2 a+16 d]=289$$
$$2 a+16 d=289 \times \frac{2}{17}$$
∴ 2a + 16d = 34 …………. (ii)
10d = 20
∴ d=2
Substituting the value of d in equation (i), we have
2a + 6d = 14
2a + 6 × 2 = 14
2a + 12 = 14
2a = 14 – 12
2a = 2
$$a=\frac{2}{2}=1$$
a = 1, d = 2, Sn =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]$$
$$=\frac{\mathrm{n}}{2}[2+2 \mathrm{n}-2]$$
$$=\quad \frac{\mathrm{n}}{2}[2 \mathrm{n}]$$
∴ Sn = n2
Question 10.
Show that a1, a2, a3, … a11. … form an AP where a11 is defIned as below:
i) a = 3 + 4n
ii) a = 9 – 5n
Solution:
i) an= 3 + 4n
a1 = 3 + 4(1)
= 3 + 4
∴ an = 7
∴ a = 7
a1 = 3 + 4n
a2 = 3 + 4 × 2
= 3 + 8
∴ a2 = 11
an = 3 + 4n
a3 = 3 + 4 × 3
= 3 + 12
a3 = 15
∴ a1, a2, a3, ………….
7, 11, 15, ……….
d = a2 – a1 = 11 – 7 = 4
d = a3 – a2= 15 – 11 = 4
Here, the value of ‘d’ is constant.
∴ an = 3 + 4n forms an Arithmetic Progression.
ii) an = 9 – 5n
a1= 9 – 5 × 1
= 9 – 5
∴ a1 = 4
a1 = 9 – 5n
a1 = 9 – 5 × 2
= 9 – 10
∴ a2 = -1
an= 9 – 5n
a3 = 9 – 5 × 3
= 9 – 15
∴ a3 = -6
a1, a2, a3, …………… an
4, -1, -6, …………
d = a2 – a1 = -1 – 4 = -5
d=a3 – a2 = -6 – (-1) = -5
Here, the value of ‘d’ is constant.
∴ an = 9 – 5n form an Arithmetic Progression.
Question 11.
lf the sum of the first n terms of an AP is 4n – n2, what is the first term (that is SI)? What is the sum of first two termš? What is the second term? Similarly. find the 3rdrd the 10th and the nth terms.
Solution:
If Sn = 4n – n2, then
i) S1 = a =?
ii) S2 =?
iii) a2 =?
iv) a3 =?
v) a10 =?
vi) an =?
(i) Sn = 4n – n2
S1 = 4(1) – 12
= 4 – 1
S1 = 3
∴ S1 = a = 3.
(ii) S2 = 4n – n2
S2 = 4(2) – 22
= 8 – 4
S2 = 4
∴ S2 = 4
(iii) We have S2 = a1 + a2 = 4
= 3 + a2 = 4
∴ a2 = 4 – 3
∴a2 = 1
(iv) Sn = 4n – n2
S3= 4(3) – 32
= 12 – 9
S3 = 3
a1 + a2 + a3 = 3
3 + 1 + a3 = 3
4 + a3= 3
∴ a3 = 3 – 4
∴ a 3 = -1
(v) d = a3 – a3 = -1 -1 = -2
a10= a + 9d
= 3 + 9(-2)
= 3 – 18
a10= -15
∴ a10 = -15
(vi) an = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 + 2n + 2
∴ an = 5 – 2n
∴ an = 5 – 2n
Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
6 + 12 + 18 + 24 + 40 term
Here a = 6, d = 2 – a1 = 12 – 6 = 6
n = 40, S40 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{40}=\frac{40}{2}[2 \times 6+(40-1) 6]$$
= 20[12 + 39 × 6]
= 20[12 + 234]
= 20 × 246
∴ S40 = 4920.
Question 13.
Find the sum of first 15 multiples of 8,
Solution:
Sum of the first 15 multiples of 8 ?
8 + 16 + 24+ ……… 15 terms.
Here, a = 8, d = a2 – a1= 16 – 8 = 8
n = 15, S15 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]$$
$$=\frac{15}{2}[16+14 \times 8]$$
$$=\frac{15}{2}[16+112]$$
$$=\frac{15}{2} \times 128$$
= 15 × 64
∴ S15 = 960
Question 14.
Fnd the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are
1 + 3 + 5 + 7 + ……… + 49 =?
Here. a = 1, d = a3 – a1= 3 – 1 = 2
an = 49, n =?, S =?
an = a + (n – 1) d = 49
= 1 + (n – 1)2 = 49
1 + 2n – 2 = 49
2n – 1 = 49
2n = 49 + 1
2n = 50
$$n=\frac{50}{2}$$
∴ n = 25
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{s}_{25}=\frac{25}{2}[1+49]$$
$$=\frac{25}{2} \times 50$$
= 25 × 25
∴ S25 = 625
Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day. Rs. 250 for the second day. Rs. 300 for the third day. etc.. the penalty for each succeeding day being Rs. 50 more than for the preceeding day. How much money the contractor has to pay as penalty. If he has delayed the work by 30 days?
Solution:
Penalty for the 1st Day 2nd day 3rd day … 30th day
Rs. 200 Rs. 250 Rs. 300 Rs.?
200 + 250 + 300 + 30 days
a = 200, d = 250 – 200 = 50, n = 20,
S30 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\dot{\mathrm{S}}_{30}=\frac{30}{2}[2 \times 200+(30-1) 50]$$
= 15[400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850
∴ S30= Rs. 27750.
Question 16.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for thler overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be ‘a’.
and Second prize is a – 20 and
the third prize Is a 40.
a, (a – 20), (a – 40) ……….. n = 7
a = a, d = a2 – a1 = a – 20 – a d = -20
n = 7, S7 =?
$$\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=\mathrm{Sn}$$
$$\frac{7}{2}[2 \times a+(7-1)(-20)]=700$$
$$\frac{7}{2}[2 a+6(-20)]=700$$
$$\frac{7}{2}[2 a-120]=700$$
$$2 a-120=700 \times \frac{2}{7}$$
2a – 120 = 200
∴ 2a = 200 + 120
2a = 320
$$\quad a=\frac{320}{2}=R s .160$$
∴ Each prizes are
Rs. 160, 140, 120, 100, 80, 60, 40
Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant. will be the same as the class. In which they are studying. e.g., a section of Class I will plant 1 tree. a section of Class Il will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
a = 3. d = 6 – 3 = 3, n = 12, S12 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\mathrm{S}_{12}=\frac{12}{2}[2 \times 3+(12-1) 3]$$
= 6[6 + 11 × 3]
= 6[6 + 33]
= 6 × 39
∴ S12 = 234
∴ Total number of trees from 3 sections of each class upto 12 class is 234.
Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $$\pi=\frac{22}{7}$$)
[Hint: Length of successive semicircles is l1, l2, l3, l4 with centres at A, B, A. B respectively.]
Solution :
$$: l_{1}=\pi \times \frac{1}{2}, 12=\pi \times 1.13=\pi \times \frac{3}{2}$$
$$l_{1}=\frac{\pi}{2}, \quad l_{2}=\mathrm{p}, \quad l_{3}=\frac{3}{2} \pi$$
∴ Arithmetic Progression,
l1, l2, l3, l4, ………..
Question 19.
200 logs are stacked In the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?
Solution:
20, 19, 18, …..
a = 20, d = 19 – 20 = -1
Sn = 200, n =?, an =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$200=\frac{\mathrm{n}}{2}[2 \times 20+(\mathrm{n}-1)(-1)]$$
$$200=\frac{\mathrm{n}}{2}[40-\mathrm{n}+1]$$
$$200=\frac{n}{2}[41-n]$$
∴ 400 = n(41 – n)
400 = 4n – n2
∴ n2 – 41n + 400 = 0
n2 – 25n – 16n + 400 = 0
n(n – 25) – 16(n – 25) = 0
(n – 25) (n – 16) =
1f n – 16 = 0 then, n = 16
∴ an = a + (n – 1) d
a16= 20 + (16 – 1) (-1)
= 20 + 15(-1)
= 20 – 15
∴ a16 = 5
∴ 200 logs are placed In 16 rows and there are 5 logs in the top row.
Question 20.
In a potato race, a bucket is placed at the starting point, which Is 5 m from the first potato. and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig, given below)
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops ¡tin the bucket, runs back to pick up the next potato, runs to the bucket to drop It in, and she continues in the same way until al) the potaotes are in the bucket. What is the total distance the competitor has to run?
(Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 +2 × (5 + 3)]
Solution:
Total distance competitor taken to pick up the first potato = 5 + 5m
= 2 × 5m.
= 10m.
Total distance taken by compeUtor to pick up the second potato
= 5 + 3 + 3 + 5m.
= 2 × 5 + 2 × 3
= 2(5 + 3)
= 2 × 8
= 16m.
∴ 10m. 16m, 22m 10th potato
a = 10, d = 16 – 10 = 6m.
n = 10, S10 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{10}=\frac{10}{2}[2 \times 10+(10-1) 6]$$
= 5[20 + 9 × 6]
= 5 [20 + 54]
= 5 × 74
∴ S10 = 370m.
∴ Total distance the competitor has to run to pick up 10 potatoers is 370 m.
Nearest tenth is the first digit after the decimal point.
We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.
## KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2.
## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2
Find the Value of x is used to consider unknown value.
Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Strength of the X Std. is 10
Number of boys be ‘y’, then
number of girls be ‘x’ .
x + y = 10 …….. (i)
x = y + 4
∴ x – y = 4 ………. (ii)
From equations (i) + (ii)
$$\quad x=\frac{14}{2}=7$$
Substituting the value of ‘x’ in eqn. (i),
x + y = 10
7 + y = 10
y = 10 – 7
y = 4.
∴ Number of girls, x = 7
Number of boys, y = 4.
(ii) Cost of each pencil be Rs. ‘x’
Cost of pen be Rs. ‘y’
5x + 7y = 50 ………. (i)
7x + 5y = 46 ……….. (ii)
From equation (i) + equation (ii)
∴ x + y = 8 …………… (iii)
from Eqn. (ii) – Eqn. (i),
∴ -x + y = 2 …………. (iv)
Eqn. (iii) + Eqn. (iv)
∴ y = 5
Substituting the value of ‘y’ in eqn. (i)
x + y = 8
x + 5 = 8
∴ x = 8 – 5 x = 3
∴Cost of each pencil is Rs. 3.
Cost of each pen is Rs. 5.
Question 2.
On comparing the ratios $$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
i) 5x – 4y + 8 = 0 a1= 5, b1 = -4, C1= 8
7x + 6y – 9 = 0 a2 = 7, b2= 6, c2= -9
$$\frac{a_{1}}{a_{2}}=\frac{5}{7} \quad \frac{b_{1}}{b_{2}}=\frac{-4}{6} \quad \frac{c_{1}}{c_{2}}=\frac{8}{-9}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Lines representing the pairs of linear equations intersect at a point.
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
a1 =9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{9}{18}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Representation of lines graphically are coincident.
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
$$\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{10}{9}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
∴ Representation of lines graphically is parallel.
Question 3.
On compairing the ratios $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}, \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ and $$\frac{c_{1}}{c_{2}}$$ find out whether the following pair of linear equations are consistent, or inconsistent.
i) 3x + 2y = 5; 2x – 3y = 7
ii) 2x – 3y = 8; 4x-6y = 9
iii) $$\frac{3}{2} x+\frac{5}{3} y=7 : 9 x-10 y=14$$
iv) 5x – 3y = 11: -10x – 6y = -22
v) $$\frac{4}{3} x+2 y=8 ; 2 x+3 y=12$$
Solution:
i) 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0
2x – 3y = 7 ⇒ 2x – 3y – 7 = 0
Here, a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \quad \frac{b_{1}}{b_{2}}=-\frac{2}{3} \quad \frac{c_{1}}{c_{2}}=\frac{-5}{-7}=\frac{5}{7}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Graphical representation is intersecting lines.
∴ Pair of linear equations are consistent.
ii) 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0
4x – 6y = 9 ⇒ 4x – 6y – 9 = 0
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Graphical representation is parallel lines.
∴ Equations are inconsistent.
iii) $$\frac{3}{2} x+\frac{5}{3} y=7 \quad \frac{3}{2} x+\frac{5}{3} y-7=0$$
9x – 10y = 14 ⇒ 9x – 10y – 14 = 0
$$a_{1}=\frac{3}{2}, \quad b_{1}=\frac{5}{3}, \quad c_{1}=-7$$
a2 = 9, b2 = -10, c2 = -17
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \times \frac{1}{9} \quad \frac{b_{1}}{b_{2}}=\frac{5}{3} \times \frac{-1}{6}$$
$$\frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{7}{14}=\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Pair of equations are consistent
iv) 5x – 3y = 11 ⇒ 5x – 3y – 11 = 0
-10x + 6y = – 22 ⇒ -10x + 6y + 22 = 0
a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = -22
$$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{6}=-\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-11}{22}=-\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
∴ Pair of equations are consistent
∴ Graphical representation is coninciding.
v) $$\frac{4}{3} x+2 y=8 \quad \frac{4}{3} x+26-8=0$$
2x + 3y = 12 ⇒ 2x + 3y – 12 = 0
$$a_{1}=\frac{4}{3}, \quad b_{1}=2, \quad c_{1}=-8$$
a2 = 2, b2 = 3, c2 = -12
$$\frac{a_{1}}{a_{2}}=\frac{4}{3} \times \frac{1}{2}=\frac{1}{6} \quad \frac{b_{1}}{b_{2}}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Pair of equations are consistent
Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent obtain the solution graphically
(i) x + y = 5, 2x + 2y = 10
(ii) x-y = 8 3x-3y= 16
(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0 4x – 3y – 5 = 0
Solution:
(i) x + y = 5 ⇒ x + y – 5 = 0
2x + 2y = 10 ⇒ 2x + 2y – 10 = 0
a1 = 1, b1 = 1, c1 = -5
a2 = 2, b2 = 2, c2 = -10
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{2} \quad \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{-5}{-10}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Pair of equations are consistent
(i) x + y = 5
y = 5 – x
x 0 2 4 y = 5 – x 5 3 1
(ii) 2x + 2y = 10
x + y = 5
y = 5 – x
x 0 2 5 y = 5 – x 5 3 0
∴ We can give any value for ‘x’, i.e., solutions are infinite.
∴ Solution, P (5, 0) x = 5, y = 0
(ii) x – y = 8 ⇒ x – y – 8 = 0
3x – 3y = 16 ⇒ 3x – 3y – 16 = 0
Here, $$\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
$$\quad \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Linear equations are inconsistenttent.
∴ Algebraically it has no solution.
Graphical representation → Parallel Lines.
(i) x – y = 8
-y = 8 – x
y = -8 + x
x 8 10 9 y = -8+x 0 2 1
(ii) 3x – 3y = 16
-3y = 16 – 3x
3y = -16 + 3x
$$\quad y=\frac{-16+3 x}{3}$$
x 6 8 $$y=\frac{-16+3 x}{3}$$ 0.8 2.6
No solution because it is inconsistent
(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0
Here a1 = 2, b1 = 1, c1 = -6
a2 = 4, b2 = -2, c2 = -4
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{-2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
Pair of equations are consistent. Algebraically both lines intersect.
Graphical Representation :
(i) 2x + y = 6
y = 6 – 2x
x 0 2 y = 6 – 2x 6 2
(ii) 4x – 2y – 4 = 0
4x – 2y = 4
-2y = 4 – 4x
2y = -4 + 4x
$$\quad y=\frac{-4+4 x}{2}$$
x 1 3 $$y=\frac{-4+4 x}{2}$$ 0 4
Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2
(iv) 2x – 2y – 2 = 0
4x – 3y – 5 = 0
a1 = 2, b1 = -2, c1 = -2
a2 = 4, b2 = -3, c2 = -5
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$
Pair of equations are consistent.
∴ Algebraically both lines intersect.
Graphical Representation :
(i) 2x – 2y – 2 =0
2x – 2y = 2
-2y = 2 – 2x
2y = -2 + 2x
$$\quad y=\frac{-2+2 x}{2}$$
∴ y = – 1 + x
x 2 4 y= -1+ x 1 3
(ii) 4x – 3y – 5 = 0
4x – 3y = 5
-3y = 5 – 4x
3y = -5 + 4x
$$\quad y=\frac{-5+4 x}{3}$$
x 2 5 $$y=\frac{-5+4 x}{3}$$ 1 5
Solution: P(2, 1) i.e., x = 2, y = 1
Question 5.
Half the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.
Solution:
Length of rectangular garden be ‘x’ m.
Breadth of rectangular garden be ‘y’ m, then
Length of the garden is 4m more than its width.
x = y + 4 ……….. (i)
x – y = 4
Half of the circumference is 36 m.
$$\frac{2 x+2 y}{2}=36$$
2x + 2y = 72
x + y = 36 …………… (ii)
∴ x – y = 4 (i)
x + y = 36 (ii)
From eqn. (i) + eqn. (ii)
$$\quad x=\frac{40}{2}$$
∴ x = 20 m.
Substituting the value of ’x’ in eqn. (i)
x – y = 4
20 – y = 4
-y = 4 – 20
-y = -16
y = 16 m.
∴ Length of the garden = 20 m.
Breadth of the garden = 16 m.
Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:
Linear equation is 2x + 3y – 8 = 0.
(i) Linear equations for intersecting lines:
2x + 3y – 8 = 0
3x + 2y – 7 = 0
a1 = 2, b1 = 3, c1 = -8
a2 = 3, b2 = 2, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{2}{3} \quad \frac{b_{1}}{b_{2}}=\frac{3}{2}$$
Here, when $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ Geometrical representation is intersecting lines.
(ii) For Parallel lines :
2x + 3y – 8 = 0
2x + 3y – 12 = 0
a1 = 2, b1 = 3, c1= -8
a1 = 2, b1 = 3, c1 = -12
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{2}=\frac{1}{1} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{3}=\frac{1}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{8}{12}=\frac{2}{3}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$ hence
Graphical representation is parallel lines.
(iii) For Intersecting lines :
2x + 3y – 8 = 0
4x + 6y – 16 = 0
a1 = 2, b1 = 3, c1 = -8
a1 = 4, b1 = 6, c1 = -16
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
Here,
∴ Lines are intersecting.
Question 7.
Draw the graphs of the equations x – y + 1 =0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 ………. (i)
3x – 2y – 12 = 0 ……….. (ii)
x – y + 1 = 0
-y = -x -1
y = x + 1
x 2 4 y = x + 1 3 5
3x + 2y – 12 = 0
2y = -3x + 12
$$\quad y=\frac{-3 x+12}{2}$$
x 0 2 $$y=\frac{-3 x+12}{2}$$ 6 3
Graphs of these two equations intersect at A. Vertices formed for ∆ABC are,
A (2, 3), B (-1, 0), C (4, 0).
We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, drop a comment below and we will get back to you at the earliest.
## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಕೆಳಗಿನ ಸಂಖ್ಯೆಗಳ ಗಣಗಳಲ್ಲಿ ಸಮರೂಪ ತ್ರಿಭುಜಗಳಾಗುವಂತಹ ಜೋಡಿಯು
(A) (3, 4, 6) (9, 12, 24)
(B) (3, 4, 6) (9, 12, 18)
(C) (2,4, 6) (2,3, 14)
(D) (5, 10, 15) (10, 30, 45)
Question 2.
ಚತುರ್ಥಕದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
(A) πr2
(B) $$\frac { 1 }{ 2 }$$ πr2
(C) $$\frac { 1 }{ 3 }$$ πr2
(D) $$\frac { 1 }{ 4 }$$ πr2
Question 3.
(2, 3) ಮತ್ತು (4, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಮಧ್ಯಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳು
(A) (3, 4)
(B) (4, 5)
(C) (5, 6)
(D) (6, 7)
Question 4.
ಧನ ಪೂಣಾಂಕ ‘a’ ಅನ್ನು ಪೂಣಾಂಕ ‘b’ ನಿಂದ ಭಾಗಿಸಿದಾಗ ಅನನ್ಯ ಪೂರ್ಣಾಂಕಗಳಾದ ‘q’ ಮತ್ತು ‘r’ ಗಳು ಇರುತ್ತವೆ. ಇದಕ್ಕೆ ಸರಿಹೊಂದುವ ಹೇಳಿಕೆ.
(A) b = a × q + r
(B) q = a × b + r
(C) a = b × q – r
(D) a = b × q + r
Question 5.
tan260° ಯ ಬೆಲೆ
(A) $$\frac { 1 }{ 3 }$$
(B) $$\frac { 1 }{ \surd 3 }$$
(C) 3
(D) √3
Question 6.
m ನ ಯಾವ ಧನಾತ್ಮಕ ಬೆಲೆಗೆ 3x2 + ka + 3 = 0 ಸಮೀಕರಣದ ಮೂಲಗಳು ಸಮವಾಗಿರುತ್ತವೆ.
(A) 2
(B) 3
(C) 5
(D) 6
Question 7.
ಒಂದು ಪ್ರಯೋಗದ ಎಲ್ಲಾ ಪ್ರಾಥಮಿಕ ಘಟನೆಗಳ ಸಂಭವನೀಯತೆಗಳ ಮೊತ್ತವು
(A) 0
(B) 1
(C) 2
(D) 3
Question 8.
ಒಂದು ಸಿಲಿಂಡರ್ನ ಪಾದದ ವಿಸ್ತೀರ್ಣ 24 cm2 ಮತ್ತು ಎತ್ತರ 10 cm ಆದರೆ ಸಿಲಿಂಡರ್ನ ಘನಫಲ
(A) 24 cm3
(B) 48 cm3
(C) 240 cm3
(D) 480 cm3
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಥೇಲ್ಸ್ನ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,
Question 10.
135 ಮತ್ತು 225 ರ ಮ.ಸಾ.ಅ. ವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 11.
ಬಹುಪದೋಕ್ತಿ x2 – 3x +5 ರ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 12.
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕದ ಉದ್ದವು 8 cm ಮತ್ತು ವೃತ್ತಕೇಂದ್ರ ಹಾಗು ಬಾಹ್ಯಬಿಂದುವಿನ ದೂರ 10 cm ಆದರೆ ವೃತ್ತದ ತ್ರಿಜ್ಯವೆಷ್ಟು?
Question 13.
sin A = $$\frac { 3 }{ 4 }$$ ಆದರೆ cosec A ನ ಬೆಲೆಯೇನು?
Question 14.
ಹಂತ ವಿಚಲನಾ ವಿಧಾನದಿಂದ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರವನ್ನು ಬರೆಯಿರಿ.
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
ವಾರ್ಷಿಕ ಸಂಬಳ ₹ 5000 ಮತ್ತು ಪ್ರತಿವರ್ಷಕ್ಕೆ ಹೆಚ್ಚುವರಿ ಬತ್ಯ ₹ 200 ಇರುವ ಕೆಲಸಕ್ಕೆ ಸುಬ್ಬರಾವ್ 1995 ರಲ್ಲಿ ಸೇರಿದರು. ಯಾವ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ?
Question 16.
ಚಿತ್ರದಲ್ಲಿ LM || CB ಮತ್ತು LN || CD ಆದರೆ $$\frac { AM }{ AB }$$ = $$\frac { AN }{ AD }$$ ಎಂದು ಸಾಧಿಸಿ.
Question 17.
ಈ ಜೋಡಿ ಸಮೀಕರಣಗಳನ್ನು ವರ್ಜಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ, 3x + 4y = 10 & 2x – 2y = 2.
Question 18.
ಕೆಳಗಿನ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳ ಜೋಡಿಗಳು ಪ್ರತಿನಿಧಿಸುವ ಸರಳರೇಖೆಗಳು ಒಂದು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆಯೇ? ಸಮಾಂತರವಾಗಿವೆಯೆ? ಅಥವಾ ಐಕ್ಯಗೊಂಡಿವೆಯೆ? ಕಂಡುಹಿಡಿಯಿರಿ
9x + 3y + 12 = 0; 18x + 6y + 24 = 0.
Question 19.
ಪರಧಿಯು 22 cm ಇರುವ ಒಂದು ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 20.
4 cm ಮತ್ತು 6 cm ತ್ರಿಜ್ಯಗಳಿರುವ ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳಿವೆ. 6 cm ತ್ರಿಜ್ಯದ ವೃತ್ತದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ 4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವನ್ನು ರಚಿಸಿ.
Question 21.
(2, -5) ಮತ್ತು (-2, 9) ರಿಂದ ಸಮಾನ ದೂರದಲ್ಲಿರುವ x-ಅಕ್ಷದ ಮೇಲಿನ ಬಿಂದುವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 22.
ಶೃಂಗಬಿಂದುಗಳು (1, -3), (4, 1) ಮತ್ತು (2, 3) ಆಗಿರುವ ತ್ರಿಭುಜದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ $$\frac { -1 }{ 4 }$$ ಮತ್ತು ಗುಣಲಬ್ಧ $$\frac { 1 }{ 4 }$$ ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
x2 + 6x – 7 = 0 ಸಮೀಕರಣವನ್ನು ವರ್ಗಪೂರ್ಣಗೊಳಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ.
Question 25.
ಗೋಪುರದ ಪಾದದಿಂದ 30m ದೂರದ ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ, ಗೋಪುರದ ತುದಿಯನ್ನು ನೋಡಿದಾಗ ಉಂಟಾಗುವ ಉನ್ನತ ಕೋನವು 30° ಆದರೆ, ಗೋಪುರದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ಹಾಗೂ ಕೆಳಗಿನಿಂದ ಬೆಟ್ಟದ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಉಂಟಾದ ಉನ್ನತ ಕೋನವು 45° ಮತ್ತು 60° ಆಗಿದೆ. ಕಟ್ಟಡದ ಎತ್ತರ 24 m ಆದರೆ ಬೆಟ್ಟದ ಎತ್ತರವೇನು?
Question 27.
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಬಹುಲಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ಚೀಲದಲ್ಲಿ 3 ಕೆಂಪು ಚೆಂಡುಗಳು ಮತ್ತು 5 ಕಪ್ಪು ಚೆಂಡುಗಳಿವೆ. ಚೀಲದಿಂದ ಯಾದೃಚ್ಛಿಕವಾಗಿ ಒಂದು ಚೆಂಡನ್ನು ತೆಗೆಯಲಾಗಿದೆ. ತೆಗೆದ ಚಂಡು ಕೆಂಪು ಆಗಿರುವ ಸಂಭವನೀಯತೆ ಎಷ್ಟು?
Question 29.
64 cm3 ಘನಫಲವನ್ನು ಹೊಂದಿರುವ 2 ವರ್ಗ ಘನಗಳ ಮುಖಗಳನ್ನು ಸೇರಿಸಿ ಒಂದು ಆಯತ ಘನಾಕೃತಿ ಮಾಡಿದೆ. ಈ ಘನಾಕೃತಿಯ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
12 ಮತ್ತು 15 ರ ಲ.ಸಾ.ಅ. ಮತ್ತು ಮ.ಸಾ.ಅ. ವನ್ನು ಅವಿಭಾಜ್ಯ ಅಪವರ್ತನ ವಿಧಾನದಿಂದ ಕಂಡುಹಿಡಿಯಿರಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಒಂದು ಸಮಾಂತರ ಚತುರ್ಭುಜದಲ್ಲಿ ವೃತ್ತವು ಅಂತಸ್ಥವಾದಾಗ ಸಮಾಂತರ ಚತುರ್ಭುಜವು ವಜ್ರಾಕೃತಿಯಾಗುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 32.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 33.
ಎರಡು ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 290 ಆದರೆ ಆ ಪೂಣಾರ್ಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಆಯತಾಕಾರದ ಹೊಲದ ಕರ್ಣವು ಅದರ ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 60 m ಹೆಚ್ಚಾಗಿದೆ. ಅದರ ದೊಡ್ಡ ಬಾಹುವು ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 30 m ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಆ ಹೊಲದ ಬಾಹುಗಳ ಉದ್ದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
ಅಥವಾ
sec A (1 – sin A) (sec A + tan A) = 1 ಎಂದು ಸಾಧಿಸಿ.
Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ, (ಕಡಿಮೆ ವಿಧಾನ)
Question 36.
3x2 – x – 4 ಎಂಬ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆ ನೋಡಿ.
ಅಥವಾ
x4 – 3x2 + 4x + 5 ನ್ನು x2 + 1 – x ದಿಂದ ಭಾಗಿಸಿ ಭಾಗಲಬ್ದ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳ ಮೊತ್ತವು 15 ಮತ್ತು ಅವುಗಳ ಅಂತ್ಯಪದಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 58 ಆಗಿದೆ. ಶ್ರೇಢಿಯ ಆ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ 5 ಪದಗಳ ಮೊತ್ತವು ಮುಂದಿನ 5 ಪದಗಳ ಮೊತ್ತದ ನಾಲ್ಕನೇ ಒಂದು ಭಾಗದಷ್ಟಿದೆ. ಮೊದಲ ಪದ 2 ಆದರೆ, a20 = -112 ಎಂದು ಸಾಧಿಸಿ ಮತ್ತು S20 ನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 38.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತವು ಅವುಗಳ ಅನುರೂಪ ಬಾಹುಗಳ ವರ್ಗಗಳ ಅನುಪಾತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 39.
ಒಂದು ಬಾವಿಯ ವ್ಯಾಸ 3 m ಮತ್ತು ಆಳ 14 m ಇರುವಂತೆ ತೋಡಿದೆ. ಭೂಮಿಯಿಂದ ತೆಗೆದ ಮಣ್ಣನ್ನು ಬಾವಿಯ ಸುತ್ತಲು ಸಮವಾಗಿ ಹರಡಿ 4 m ಅಗಲವಿರುವ ವೃತ್ತಾಕಾರದ ಕಟ್ಟೆಯನ್ನು ಕಟ್ಟಿದೆ. ಕಟ್ಟೆಯ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 40.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ.
2x + y = 8
x + 2y = 7
Solutions
I.
Solution 1.
(B) (3, 4, 6) (19, 12, 18)
Solution 2.
(D) $$\frac { 1 }{ 4 }$$ πr2
Solution 3.
(A) (3, 4)
Solution 4.
(D) a = b × q + r
Solution 5.
(C) 3
Solution 6.
(D) 6
Solution 7.
(B) 1
Solution 8.
(C) 240 cm3
II.
Solution 9.
ತ್ರಿಭುಜದ ಎರಡು ಬಾಹುಗಳನ್ನು ಎರಡು ವಿಭಿನ್ನ ಬಿಂದುಗಳಲ್ಲಿ ಛೇದಿಸುವಂತೆ ಒಂದು ಬಾಹುವಿಗೆ ಸಮಾಂತರವಾಗಿ ಎಳೆದ ಸರಳರೇಖೆಯು ಉಳಿದೆರಡು ಬಾಹುಗಳನ್ನು ಸಮಾನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುತ್ತದೆ.
Solution 10.
Solution 11.
Solution 12.
OA2 = OP2 – AP2 =102 – 82
⇒ OA2 = 100 – 64
⇒ OA2 = 100 – 64
⇒ OA = √36
⇒ OA = 6 cm
ವೃತ್ತದ ತ್ರಿಜ್ಯ = 6 cm
Solution 13.
cosec A = $$\frac { 4 }{ 3 }$$
Solution 14.
III.
Solution 15.
5000, 5200, 5400, ……… 7000.
a = 5000, d = 200 an = 7000, n = ?
an = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ 2000 = 200 (n – 1)
⇒ (n – 1) = 10
⇒ n = 10 + 1
⇒ n = 11
11 ನೇ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ.
Solution 16.
Solution 17.
3x + 4y = 10 …. (1)
2x – 2y = 2 …… (2)
ಸಮೀಕರಣ 2ನ್ನು 2ರಿಂದ ಗುಣಿಸಿದಾಗ
3x + 4y = 10
4x – 4y = 4
…………………
7x = 14
x = 2
3x + 4y = 10
3(2) + 4y = 10
4y = 10 – 6
y = 1
∴ x = 2, y = 1
Solution 18.
Solution 19.
ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣ = $$\frac { 77 }{ 4 }$$ cm2
Solution 20.
R = 6 cm, r = 4 cm
Solution 21.
Solution 22.
Solution 23.
ಬಹುಪದೋಕ್ತಿ = ax2+ bx + c ಆಗಿರಲಿ
ಶೂನ್ಯತೆಗಳು α ಮತ್ತು β ಆಗಿರಲಿ
Solution 24.
x2 + 6x – 7 = 0
⇒ x2 + 6x = 7
⇒ x2 + 6x + 32 – 32 = 7
⇒ (x + 3)2 – 9 = 7
⇒ (x + 3)2 = 16
⇒ x + 3 = ±√16
⇒ x + 3 = ±4
⇒ x = ±4 – 3
⇒ x = +4 – 3 ಅಥವಾ x = -4 – 3
⇒ x = 1 ಅಥವಾ x = -7
Solution 25.
Solution 26.
Solution 27.
Solution 28.
ಚೀಲದಲ್ಲಿರುವ ಒಟ್ಟು ಚೆಂಡುಗಳು = n(S) = 3 + 5 = 8
ಕೆಂಪು ಚೆಂಡುಗಳು ಸಂಖ್ಯೆ = 3 = n(A)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { n(A) }{ n(S) }$$
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { 3 }{ 8 }$$
Solution 29.
ವರ್ಗಗಳ ಘನಫಲ = 64cm3
ಬಾಹುವಿನ ಅಳತೆ = $$\sqrt [ 3 ]{ 64 }$$ = 4 cm
ಆಯತ ಘನದ ಅಗಲ b = 4 cm
ಆಯತ ಘನದ ಉದ್ದ l = 8 cm (4 + 4)
ಆಯತ ಘನದ ಎತ್ತರ h = 4 cm
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 160 cm2
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 160 cm2
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Solution 30.
12 = 2 × 2 × 5
15 = 3 × 5
ಮ.ಸಾ.ಅ. = 5
ಲ.ಸಾ.ಅ. = 2 × 2 × 3 × 5
ಲ.ಸಾ.ಅ. = 60
IV.
Solution 31.
ದತ್ತ: ‘O’ ವೃತ್ತಕೇಂದ್ರ, XY ಸ್ಪರ್ಶಕ, Pಸ್ಪರ್ಶಬಿಂದು
ಸಾಧನೀಯ: OP ⊥ XY
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತ್ತೊಂದು ಬಿಂದು Q ಆಗಿರಲಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: OQ ವೃತ್ತವನ್ನು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸಿದೆ.
OP = OR (ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು)
OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP ಯು O ನಿಂದ ಸ್ಪರ್ಶಕಕ್ಕೆ ಕನಿಷ್ಟ ದೂರವಾಗಿದೆ
OP ⊥ XY
ಅಥವಾ
ಸಾಧನೀಯ: ABCD ಒಂದು ವಜ್ರಾಕೃತಿ. (AB = BC = CD = AD)
ಸಾಧನೆ: AB = CD ಮತ್ತು AD = BC …… (1)
AP = AS, BP = BQ, CQ = CR, DS = DR
(ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು)
AB + CD = AP + PB + DR + CR
⇒ AB + CD = AS + BQ + DS + CQ
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB = AD ….. (2)
∴ AB = BC = CD = AD (ಸಮೀಕರಣ (1) & (2) ರಿಂದ)
Solution 32.
Solution 33.
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು x & x + 2
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 – 290 = 0
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ x2 + 13x – 11x – 143 = 0
⇒ x (x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 ಅಥವಾ x – 11 = 0
⇒ x = -13 ಅಥವಾ x = 11
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು 11 & 13
[∴ x = 11 & x + 2 = 11 + 2 = 13]
ಅಥವಾ
ಚಿಕ್ಕ ಬಾಹು x ಆಗಿರಲಿ
AC2 = AB2 + BC2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x
⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
⇒ -x2 + 2700 + 60x = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x (x – 90) + 30 (x – 90) = 0
⇒ x – 90 = 0 ಅಥವಾ x + 30 = 0
⇒ x = 90 ಅಥವಾ x = -30
ಚಿಕ್ಕ ಬಾಹುವಿನ ಉದ್ದ = x = 90 m
ದೊಡ್ಡ ಬಾಹುವಿನ ಉದ್ದ = x + 30 = 90 + 30 = 120 m
Solution 34.
Solution 35.
Solution 36.
V.
Solution 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳು
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
(a – d)2 + (a + d)2 = 58
⇒ a2 + d2 – 2ad + a2 + d2 + 2ad = 58
⇒ 2a2 + 2d2 = 58
⇒ 2(a2 + d2) = 58
⇒ a2 + d2 = 29
⇒ 52 + d2 = 29
⇒ d2 = 29 – 25
⇒ d2 = 4
⇒ d = ± 2
⇒ d = 2 ಅಥವಾ d = -2
ಮೂರು ಪದಗಳು
∴ a – d = 5 – 2 = 3
∴ a = 5
∴ a + d = 5 + 2 = 7
ಅಥವಾ
Solution 38.
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತ 16 : 25 ಆದರೆ ಅವುಗಳ ಬಾಹುಗಳ ಅನುಪಾತ
(A) 3 : 4
(B) 4 : 5
(C) 5 : 6
(D) 6 : 7
Question 2.
7 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತದಲ್ಲಿ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
(A) 25.67 cm2
(B) 35.32 cm2
(C) 15.25 cm2
(D) 77 cm2
Question 3.
ಮೂಲಬಿಂದು ಮತ್ತು (6, 8) ಬಿಂದುವಿನ ನಡುವಿನ ದೂರ
(A) 6 ಮೂ.ಮಾ.
(B) 8 ಮೂ.ಮಾ.
(C) 10 ಮೂ.ಮಾ.
(D) 14 ಮೂ.ಮಾ.
Question 4.
(15, 20) ರ ಮ.ಸಾ.ಅ. 5 ಆದರೆ ಅವುಗಳ ಲ.ಸಾ.ಅ.
(A) 15
(B) 20
(C) 40
(D) 60
Question 5.
$$\frac { { sin18 }^{ 0 } }{ { cos72 }^{ 0 } }$$ ಯ ಬೆಲೆ
(A) 1
(B) 0
(C) -1
(D) 2
Question 6.
x2 – 25 = 0 ಈ ಸಮೀಕರಣದ ಮೂಲಗಳು
(A) (+5, -5)
(B) (+5, +5)
(C) (-5, -5)
(D) (25, -25)
Question 7.
ಒಂದು ನಿರ್ದಿಷ್ಟ ದಿನದಲ್ಲಿ ಮಳೆ ಬೀಳುವ ಸಂಭವನೀಯತೆಯು 0.64 ಆಗಿದೆ. ಅದೇ ದಿನ ಮಳೆ ಬೀಳದಿರುವ ಸಂಭವನೀಯತೆ.
(A) -0.64
(B) 64
(C) 0.36
(D) -0.36
Question 8.
ಶಂಕುವಿನ ಭಿನ್ನಕದ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,
Question 10.
ಒಂದು ಸಂಖ್ಯೆಯನ್ನು 14 ರಿಂದ ಭಾಗಿಸಿದಾಗ ಶೇಷ 5 ಆದರೆ 7ರಿಂದ ಅದೇ ಸಂಖ್ಯೆಯನ್ನು ಭಾಗಿಸಿದಾಗ ದೊರೆಯುವ ಶೇಷವೆಷ್ಟು?
Question 11.
α ಮತ್ತು β ಗಳು ax2 + bx + c ಎಂಬ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳಾದರೆ αβ ದ ಬೆಲೆಯೇನು?
Question 12.
‘O’ ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ PA ಮತ್ತು PB ಗಳು ಬಾಹ್ಯ ಬಿಂದು P ನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಮತ್ತು ∠APB = 60° ಹಾಗು AP = 8 cm ಆದಾಗ AB ಜ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 13.
(1 + sin290°)2 ನ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 14.
12, 16, 20, 24, 28 ರ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
a = 5, d = 3, an = 50 ಆದರೆ Sn ಕಂಡುಹಿಡಿಯಿರಿ.
Question 16.
ಚಿತ್ರದಲ್ಲಿ ∆ABCಯ ಎತ್ತರಗಳಾದ AD ಮತ್ತು CE ಗಳು ಪರಸ್ಪರ P ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ∆AEP ~ ∆CDP ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ 0ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.
Question 17.
Question 18.
5x – 3y = 11, -10x + 6y = -22 ಈ ಸಮೀಕರಣಗಳು ಸ್ಥಿರವಾಗಿವೆಯೆ? ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೆ? ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 19.
ಚಿತ್ರದಲ್ಲಿ ಕೇಂದ್ರ O ಇರುವ ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 7 cm ಮತ್ತು 14 cm ಇವೆ. ∠AOC = 40° ಆದರೆ ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 20.
4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕಗಳ ನಡುವಿನ ಕೋನ 70° ಇರುವಂತೆ ಒಂದು ಜೊತೆ ಸ್ಪರ್ಶಕಗಳನ್ನು ಎಳೆಯಿರಿ.
Question 21.
AB ವ್ಯಾಸವಾಗಿರುವ ವೃತ್ತದ ಕೇಂದ್ರ (2, -3) ಮತ್ತು B ಯು (1, 4) ಆದರೆ A ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 22.
(8, 1), (k, -4), (2, -5) ಎಂಬ ಬಿಂದುಗಳು ಸರಳರೇಖಾಗತವಾಗಿದ್ದರೆ k ಯ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗು ಗುಣಲಬ್ದಗಳು ಕ್ರಮವಾಗಿ 4 ಮತ್ತು 1 ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
2x2 + 3x – 5 = 0 ಈ ವರ್ಗ ಸಮೀಕರಣವನ್ನು ಸೂತ್ರದಿಂದ ಬಿಡಿಸಿ,
Question 25.
20 m ಎತ್ತರದ ಕಟ್ಟಡವೊಂದರ ಮೇಲೆ ಸ್ಥಾಪಿಸಲಾದ ಪ್ರಸರಣೆಯ ಗೋಪುರವೊಂದರ ಮೇಲುದಿ ಮತ್ತು ಪಾದಗಳನ್ನು ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ನೋಡಿದಾಗ ಉನ್ನತ ಕೋನಗಳು ಕ್ರಮವಾಗಿ 60° ಮತ್ತು 45° ಇದೆ. ಪ್ರಸರಣೆಯ ಗೋಪುರದ ಎತ್ತರ ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
50√3 m ಎತ್ತರದಲ್ಲಿರುವ ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ನೆಲದ ಮೇಲಿರುವ ಒಂದು ಕಾರನ್ನು ನೋಡಿದಾಗ ಉಂಟಾದ ಅವನತ ಕೋನವು 60° ಆಗಿರುತ್ತದೆ. ಹಾಗಾದರೆ ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 27.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಮಧ್ಯಾಂಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. (i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ (ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 29.
ಸಮನಾದ ತ್ರಿಜ್ಯವುಳ್ಳ ಒಂದು ಶಂಕುವನ್ನು ಒಂದು ಅರ್ಧಗೋಳಾಕೃತಿಯ ಮೇಲೆ ಜೋಡಿಸಿ ಒಂದು ಆಟಿಕೆಯನ್ನು ಮಾಡಲಾಗಿದೆ. ಶಂಕುವಿನ ಭಾಗದ ವ್ಯಾಸವು 6 cm ಮತ್ತು 4 cm ಎತ್ತರ ಇದ್ದರೆ ಈ ಘನವಸ್ತುವಿನ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಶಂಕುವಿನ ಭಿನ್ನಕದ ಓರೆ ಎತ್ತರ 10 cm, ಅದರ ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 8 cm ಮತ್ತು 6 cm ಆಗಿದೆ. ಆ ಭಿನ್ನಕದ ವಕ್ರಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮನಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು 5 cm ಮತ್ತು 3 cm ಆಗಿವೆ. ಚಿಕ್ಕ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಿಸುವಂತೆ ದೊಡ್ಡ ವೃತ್ತದ ಹ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 32.
BC = 7 cm, ∠A = 45°, ∠B = 105° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು, ∆ABCಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 33.
3 ವರ್ಷಗಳ ಹಿಂದಿನ ರೆಹಮಾನನ ವಯಸ್ಸು ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವ್ಯತ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ರೈಲು 360km ದೂರವನ್ನು ಏಕರೂಪ ಜವದೊಂದಿಗೆ ಕ್ರಮಿಸುತ್ತದೆ. ಅದರ ಜವವು 5km/hr ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಅಷ್ಟೇ ದೂರವನ್ನು ಕ್ರಮಿಸಲು ಅದು 1 ಗಂಟೆ ಕಡಿಮೆ ತೆಗೆದುಕೊಳ್ಳುತ್ತಿತ್ತು. ರೈಲಿನ ಜವವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
(cos A + sec A)2 + (sin A + cosec A)2 = 7 + tan2A + cot2A ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
A, B ಮತ್ತು C ಗಳು ∆ABCಯ ಒಳಕೋನಗಳಾದರೆ sin($$\frac { B+C }{ 2 }$$) = cos($$\frac { A }{ 2 }$$) ಎಂದು ಸಾಧಿಸಿ.
Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ,
(ಅಧಿಕ ಇರುವ ವಿಧಾನ)
Question 36.
4u2 – 8u ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿದು ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆನೋಡಿ.
ಅಥವಾ
p(x) = x5 – 4x3 + x2 + 3x + 1 ನ್ನು g(x) = x3 – 3x + 1 ರಿಂದ ಭಾಗಿಸಿ g(x), p(x) ನ ಅಪವರ್ತನವನ್ನು ಪರೀಕ್ಷಿಸಿ.
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ 4ನೇ ಮತ್ತು 8ನೇ ಪದಗಳ ಮೊತ್ತವು 24ಹಾಗು ಅದೇ ಶ್ರೇಢಿಯ 6ನೇ ಮತ್ತು 10ನೇ ಪದಗಳ ಮೊತ್ತವು 44 ಆದರೆ ಶ್ರೇಢಿಯ ಮೊದಲ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಸುರುಳಿಯನ್ನು ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲಾಗಿದೆ ಅವುಗಳ ಕೇಂದ್ರಗಳು ಪರ್ಯಾಯವಾಗಿ A & B ನಲ್ಲಿದ್ದು A ಕೇಂದ್ರದಿಂದ ಆರಂಭವಾಗಿ ತ್ರಿಜ್ಯಗಳು 0.5 cm, 1 cm, 1.5 cm, 2 cm …….. ಹೀಗೆ ಇದೆ. ಈ ರೀತಿಯ 13 ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲ್ಪಟ್ಟ ಒಟ್ಟು ಉದ್ದ ಏನು?
Question 38.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
Question 39.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ, 2x – y = 5 ಮತ್ತು x + 3y = 6.
Question 40.
ಒಂದು ಅರ್ಧಗೋಳಾಕಾರದ ಬಟ್ಟಲಿನ ಒಳಜ್ಯವು 18 cm ಇದ್ದು ಅದರ ತುಂಬ ಹಣ್ಣಿನ ರಸವಿದೆ. ಈ ರಸವನ್ನು 3 cm ತ್ರಿಜ್ಯವಿರುವ ಮತ್ತು 9 cm ಎತ್ತರವಿರುವ ಸಿಲಿಂಡರಿನಾಕೃತಿಯ ಬಾಟಲಿಗಳಿಗೆ ತುಂಬಬೇಕು. ಬಟ್ಟಲನ್ನು ಖಾಲಿ ಮಾಡಲು ಎಷ್ಟು ಬಾಟಲಿಗಳ ಅವಶ್ಯಕತೆ ಇದೆ?
Solutions
I.
Solution 1.
(B) 4 : 5
Solution 2.
(A) 25.67 cm2
Solution 3.
(C) 10 ಮೂ.ಮಾ.
Solution 4.
(D) 60
Solution 5.
(A) 1
Solution 6.
(A) (+5, -5)
Solution 7.
(C) 0.36
Solution 8.
(B) π (r1 + r2) l
II.
Solution 9.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ಮೇಲಿನ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮ.
Solution 10.
ಶೇಷ – 5
Solution 11.
αβ = $$\frac { c }{ a }$$
Solution 12.
∆APB ನಲ್ಲಿ PA = PB = 8cm
∠PAB = ∠PBA = 60°
∠APB = 60°
∆APB ಒಂದು ಸಮಬಾಹು ತ್ರಿಭುಜ
AB = 8 cm
Solution 13.
(1 + sin290°)2 = (1 + 1)2 = 4
Solution 14.
III.
Solution 15.
Solution 16.
ಸಾಧನೀಯ: ∆AEP ~ ∆CDP
∆AEP & ∆CDP ಗಳಲ್ಲಿ
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (ಶೃಂಗಾಭಿಮುಖ ಕೋನಗಳು)
∆AEP ~ ∆CDP
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ O ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.
Solution 17.
Solution 18.
Solution 19.
R = 14 cm, r = 7 cm, θ = 40°
ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣ
Solution 20.
ತ್ರಿಜ್ಯ = 4cm
ಕೇಂದ್ರ ಕೋನ = 180° – 70° = 110°
Solution 21.
Solution 22.
Solution 23.
Solution 24.
Solution 25.
ಗೋಪುರದ ಎತ್ತರ = AB = x ಆಗಿರಲಿ
ಕಟ್ಟಡದ ಎತ್ತರ = BC = 20 m
tan 45° = $$\frac { BC }{ CD }$$
1 = $$\frac { 20 }{ CD }$$
CD = 20 m
tan 60° = $$\frac { AC }{ CD }$$
⇒ √3 = $$\frac { x+20 }{ 20 }$$
⇒ 20√3 = x + 20
⇒ x + 20 = 20√3
⇒ x = 20(√3 – 1)
ಗೋಪುರದ ಎತ್ತರ = 20(√3 – 1) m
Solution 26.
ಕಟ್ಟಡ ಎತ್ತರ = AB = 50√3 m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ BC ಆಗಿರಲಿ
tan 60° = $$\frac { AB }{ BC }$$
⇒ √3 = $$\frac { 50\surd 3 }{ BC }$$
⇒ BC = 50m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ = 50m
Solution 27.
Solution 28.
ಫಲಿತಗಳ ಸಂಖ್ಯೆ = n(S) = 6 {1, 2, 3, 4, 5, 6}
(i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ = A = {2, 3, 5}
n(A) = 3
ಸಂಭವನೀಯ = P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$
(ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆ = B = {2, 3, 4}
n(B) = 3
ಸಂಭವನೀಯತೆ = P(B) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$
Solution 29.
Solution 30.
3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ ಎಂದು ಊಹಿಸೋಣ.
ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ = ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
ನಮ್ಮ ಊಹೆ 3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂಬುದು ತಪ್ಪು
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
IV.
Solution 31.
ದತ್ತ: O ವೃತ್ತಕೇಂದ್ರ PA ಮತ್ತು PB ಗಳ ಬಾಹ್ಯಬಿಂದು P ನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು.
ಸಾಧನೀಯ: PA = PB
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔPOA & ΔPOB ಗಳಲ್ಲಿ
OA = OB
∠PAO = ∠PBO = 90°
OP = OP
ΔPOA = ΔPOB
PA = PB
ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು ತ್ರಿಜ್ಯ ಮತ್ತು ಸ್ಪರ್ಶಕದ ನಡುವಿನ ಕೋನ ಉಭಯಸಾಮಾನ್ಯ ಲಂ.ವಿ.ಬಾ. ಜ್ವಸಿದ್ಧಾಂತ ಸರ್ವಸಮ ತ್ರಿಭುಜದ ಅನುರೂಪ ಭಾಗಗಳು.
ಅಥವಾ
ΔPOB ನಲ್ಲಿ
PB2 = OB2 – OP2
⇒ PB2 = 52 – 32 = 25 – 9 = 16
⇒ PB2 = 16
⇒ PB = 4
AB = AP + PB
⇒ AB = 4 + 4 (∴ AP = PB)
⇒ AB = 8 cm ಜ್ಯಾದ ಉದ್ದ 8 cm
Solution 32.
∠A = 45°, ∠B = 105°, BC = 7 cm
∠C = 180° – 150° = 30°
ರಚಿಸಬೇಕಾದ ತ್ರಿಭುಜ A’BC’
Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು x ಆಗಿರಲಿ
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ಅವನ ವಯಸ್ಸು = x – 3 ವರ್ಷ
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸ್ಸು = x + 5 ವರ್ಷ
⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 ಅಥವಾ x + 3 = 0
⇒ x = 7 ಅಥವಾ x = -3
ರೆಹಮಾನನ ಈಗಿನ ವಯಸು 7 ವರ್ಷ
ಅಥವಾ
ರೈಲಿನ ಜನ x km/hr, ಆಗಿರಲಿ
ದೂರ = d = 360 km/hr
Solution 34.
Solution 35.
Solution 36.
V.
Solution 37.
Solution 38.
ದತ್ತ: ΔABC ನಲ್ಲಿ ∠BAC = 90°
ಸಾಧನೀಯ: BC2 = AB2 + AC2
ರಚನೆ: AD ⊥ BC ಗೆ ಎಳೆಯಿರಿ,
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔBAC & ΔBDA ಗಳಲ್ಲಿ
∠ABC = ∠ABD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠BDA = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔBDA ಕೊ .ಕೋ .ನಿ.ಗು.
$$\frac { BA }{ BD }$$ = $$\frac { BC }{ BA }$$
AB2 = BD.BC …….. (1)
∠ACB = ∠ACD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠ADC = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔADC ಕೋ .ಕೋ .ನಿ.ಗು.
$$\frac { AC }{ DC }$$ = $$\frac { BC }{ AC }$$
AC2 = DC × BC ……. (2)
AB2 + AC2 = BD.BC + DC.BC
(1) + (2) ರಿಂದ
BC(BD + DC) = BC × BC
AB2 + AC2 = BC2
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium
ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80
I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)
Question 1.
ಚಿತ್ರದಲ್ಲಿ QE = 7.2 cm, PF = 1.8 cm, FR = 5.4cm ಆದರೆ PE ಯು
(A) 2 cm
(B) 2.4 cm
(C) 2.8 cm
(D) 3.2 cm
Question 2.
‘r’ ವೃತ್ತ ತ್ರಿಜ್ಯ ಹಾಗೂ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
Question 3.
P (-2, -1), ರಲ್ಲಿ ಲಂಬದೂರವು
(A) -2
(B) -1
(C) 1
(D) 2
Question 4.
‘0’ ಮೂಲಬಿಂದು ಮತ್ತು P (4, 3) ರ ನಡುವಿನ ದೂರ (ಏಕಮಾನಗಳಲ್ಲಿ) OP = …….
(A) 5
(B) 4
(C) 3
(D) 2
Question 5.
P (x) = 5x – 10 ರ ಶೂನ್ಯತೆ.
(A) 2
(B) -2
(C) 5
(D) -5
Question 6.
x + $$\frac { 2 }{ x }$$ = 3 ಸಮೀಕರಣದ ಆದರ್ಶರೂಪ
(A) x2 + 2x – 3 = 0
(B) x2 + 3x + 2 = 0
(C) x2 – 3x + 2 = 0
(D) x2 – 2x + 3 = 0
Question 7.
ಎರಡು ನಾಣ್ಯಗಳನ್ನು ಏಕಕಾಲಕ್ಕೆ ಚಿಮ್ಮಿದಾದ, ಶಿರ H ಮತ್ತು ಪ್ರಚ್ಛ T ಇರುವಂತೆ ಸಾಧ್ಯ ಫಲಿತಗಳು
(A) {T, H, H, T}
(B) {TT, HH, HT, TH}
(C) {T, H}
(D) {TT, HH}
Question 8.
ಸಿಲಿಂಡರ್ನ ಎತ್ತರ ‘h’ ಮತ್ತು ಪಾದದ ತ್ರಿಜ್ಯ ‘r’ ಆದಾಗ ಸಿಲಿಂಡರ್ನ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ
(A) 2π (r + h)
(B) 2πr (r + h)
(C) 2πrh
(D) $$\frac { 2\pi r }{ h }$$
II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)
Question 9.
ಒಂದು 5m ಎತ್ತರದ ಏಣಿಯನ್ನು ನೆಲದ ಮೇಲಿನಿಂದ 4 m ಎತ್ತರದ ಕಿಟಕಿಯನ್ನು ತಲುಪುವಂತೆ ಗೋಡೆಗೆ ಒರಗಿಸಿದೆ. ಗೋಡೆಯ ಪಾದದಿಂದ ಏಣಿಯ ಪಾದಕ್ಕಿರುವ ದೂರ ಲೆಕ್ಕಿಸಿ.
Question 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?
Question 11.
‘ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ’ ವನ್ನು ಬರೆಯಿರಿ.
Question 12.
ಯೂಕ್ಲಿಡ್ನ ಭಾಗಾಕಾರ ಅನುಪ್ರವೇಯವನ್ನು ತಿಳಿಸಿ.
Question 13.
ಘನಪದೋಕ್ತಿಯು ಹೊಂದಿರಬಹುದಾದ ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?
Question 14.
ಎರಡು ಸಂಖ್ಯೆಗಳ ಮೊತ್ತ 27 ಮತ್ತು ಗುಣಲಬ್ಧ 182 ಆದರೆ ಆ ಸಂಖ್ಯೆಗಳು ಯಾವುವು?
III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)
Question 15.
ಮೊತ್ತ ಕಂಡುಹಿಡಿಯಿರಿ: 34 + 32 + 30 + …………… + 10,
Question 16.
ಒಂದು ಏಣಿಯ ಪಾದವು ನೆಲದ ಮೇಲೆ ಗೋಡೆಯಿಂದ 2.5m ದೂರದಲ್ಲಿ ಹಾಗೂ ಅದರ ತುದಿಯು ನೆಲದ ಮೇಲಿಂದ 6 m ಎತ್ತರದಲ್ಲಿರುವ ಕಿಟಕಿಯನ್ನು ಮುಟ್ಟುವಂತೆ ಏಣಿಯನ್ನು ಗೋಡೆಗೆ ಒರಗಿಸಿ ಇಡಲಾಗಿದೆ. ಏಣಿಯ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 17.
ಒಂದು ಎರವಲು ಗ್ರಂಥಾಲಯದಲ್ಲಿ ಮೊದಲ 3 ದಿನಕ್ಕೆ ಒಂದು ನಿಗದಿತ ಶುಲ್ಕವಿರುತ್ತದೆ. ಆ ದಿನದ ಪ್ರತಿಯೊಂದೂ ದಿನಕ್ಕೂ ಒಂದು ಹೆಚ್ಚುವರಿ ಶುಲ್ಕವಿರುತ್ತದೆ. ಪುಸ್ತಕವನ್ನು 7 ದಿನ ತನ್ನಲ್ಲಿ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸರಿತಾ ₹ 27 ನ್ನು ಪಾವತಿಸಿದರೆ, ಪುಸ್ತಕವನ್ನು 5 ದಿನ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸೂಸಿ ₹ 21 ನ್ನು ಪಾವತಿಸಿದಳು. ನಿಗದಿತ ಶುಲ್ಕ ಮತ್ತು ಪ್ರತಿಯೊಂದು ಹೆಚ್ಚುವರಿ ದಿನದ ಶುಲ್ಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 18.
ನೀರಿನ ಒಳಭಾಗದಲ್ಲಿರುವ ಬಂಡೆಗಳ ಬಗ್ಗೆ ಎಚ್ಚರಿಸಲು ಒಂದು ದೀಪಸ್ಥಂಭವು 80° ಕೋನವಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದಲ್ಲಿ 16.5 km ದೂರಕ್ಕೆ ಕೆಂಪು ಬೆಳಕನ್ನು ಹರಡುತ್ತದೆ. ಹಡಗುಗಳನ್ನು ಎಚ್ಚರಿಸುವ ಈ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಚಿತ್ರದಲ್ಲಿ ತೋರಿಸಿರುವಂತೆ 4 cm ಬಾಹುವುಳ್ಳ ಒಂದು ಚೌಕದ ಪ್ರತೀ ಮೂಲೆಯಲ್ಲಿ 1 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತ ಚತುರ್ಥಕವನ್ನು ಮತ್ತು 2 cm ವ್ಯಾಸವಿರುವ ಒಂದು ವೃತ್ತವನ್ನು ಕತ್ತರಿಸಿದೆ. ಚೌಕದ ಉಳಿದ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 19.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,
Question 20.
(4, -3) ಮತ್ತು (8, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡವನ್ನು ಆಂತರಿಕವಾಗಿ 3 : 1 ಅನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುವ ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 21.
ಶೃಂಗಗಳು ಈ ಕೆಳಗಿನಂತಿರುವ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ. (5, -1), (3, -5), (5, 2)
Question 22.
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.
Question 23.
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ವರ್ಗಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 24.
ಮೂಲಗಳ ಸ್ವಭಾವವನ್ನು ವಿವೇಚಿಸಿ, ವಾಸ್ತವ ಮೂಲಗಳಿದ್ದಲ್ಲಿ, ಅವುಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ: 2x2 – 3x + 5 = 0
Question 25.
1.6m ಎತ್ತರದ ಪ್ರತಿಮೆಯೊಂದನ್ನು ಒಂದು ಪೀಠದ ಮೇಲ್ಬಾಗದಲ್ಲಿ ಇರಿಸಲಾಗಿದೆ. ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ಪ್ರತಿಮೆಯ ಮೇಲಿನ ಉನ್ನತ ಕೋನವು 60° ಮತ್ತು ಅದೇ ಬಿಂದುವಿನಿಂದ ಪೀಠದ ಮೇಲ್ತುದಿಯ ಉನ್ನತ ಕೋನವು 45° ಇದ್ದರೆ, ಪೀಠದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 26.
10 ಪಂದ್ಯಗಳಲ್ಲಿ ಒಬ್ಬ ಬೌಲರ್ನು ಪಡೆದ ವಿಕೆಟ್ಗಳ ಸಂಖ್ಯೆಯು ಈ ಕೆಳಗಿನಂತಿದೆ.
Question 27.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. 2 ಮತ್ತು 6ರ ನಡುವೆ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 28.
ಒಂದು ಪಾತ್ರೆಯ ಆಕಾರವು ಟೊಳ್ಳಾದ ಸಿಲಿಂಡರಿನ ಒಂದು ಪಾದದ ಮೇಲೆ ಟೊಳ್ಳಾದ ಅರ್ಧಗೋಳಾಕೃತಿಯನ್ನು ಕೂಡಿಸಿ ಮಾಡಿದೆ. ಅರ್ಧಗೋಳದ ವ್ಯಾಸವು 14 cm ಮತ್ತು ಪಾತ್ರೆಯ ಒಟ್ಟು ಎತ್ತರ 13 cm ಇದೆ. ಈ ಪಾತ್ರೆಯ ಒಳಮೇ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಘನ ಸಿಲಿಂಡರಿನ ಎತ್ತರ 2.4 m ಮತ್ತು ವ್ಯಾಸ 1.4 m ಇದೆ. ಇದರಿಂದ ಒಂದೇ ಎತ್ತರ ಮತ್ತು ಒಂದೇ ವ್ಯಾಸವನ್ನು ಹೊಂದಿರುವ ಶಂಕುವಿನಾಕಾರದ ಹಳ್ಳವನ್ನು ಕೊರೆದು ಟೊಳ್ಳಾಗಿಸಿದೆ. ನೂತನ ಘನದ ಒಟ್ಟು ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಅತ್ಯಂತ ಸಮೀಪದ ಬೆಲೆಗೆ cm2 ನಲ್ಲಿ ಕಂಡುಹಿಡಿಯಿರಿ.
Question 29.
$$\frac { 3 }{ 2 }$$ x + $$\frac { 5 }{ 3 }$$ y = 7: 9x – 10y = 14 ಈ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳಲ್ಲಿ $$\frac { { a }_{ 1 } }{ { a }_{ 2 } }$$, $$\frac { { b }_{ 1 } }{ { b }_{ 2 } }$$ ಮತ್ತು $$\frac { { c }_{ 1 } }{ { c }_{ 2 } }$$ ಅನುಪಾತಗಳನ್ನು ಹೋಲಿಸುವ ಮೂಲಕ ಜೋಡಿಗಳು ಸ್ಥಿರವಾಗಿವೆಯೇ ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೇ ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 30.
150 cm ಎತ್ತರವಿರುವ ಒಬ್ಬ ವ್ಯಕ್ತಿಯು ತನ್ನ ನೆರಳಿನ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಅದು ಅವನ ಪಾದದಿಂದ 150√3 cm ದೂರದಲ್ಲಿರುವುದು ಕಂಡುಬರುತ್ತದೆ. ಹಾಗಾದರೆ ಅವನ ನೋಟದಲ್ಲಿ ಉಂಟಾದ ಅವನತ ಕೋನವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)
Question 31.
‘ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಸಮ ಎಂದು ಸಾಧಿಸಿ
Question 32.
BC = 6 cm, AB = 5cm ಮತ್ತು ∠ABC = 60° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ತ್ರಿಭುಜ ABC ಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ.
Question 33.
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ರೆಹಮಾನನ ವಯಸ್ಸು (ವರ್ಷಗಳಲ್ಲಿ) ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವೃಶ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಮೂರು ಕ್ರಮಾನುಗತ ಸಂಖ್ಯೆಗಳಲ್ಲಿ, ಮೊದಲನೆಯ ವರ್ಗ ಮತ್ತು ಉಳಿದೆರಡು ಸಂಖ್ಯೆಗಳ ಗುಣಲಬ್ದಗಳ ಮೊತ್ತ 154 ಆಗಿದೆ. ಹಾಗಾದರೆ ಆ ಮೂರು ಸಂಖ್ಯೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 34.
ಈ ಕೆಳಗೆ ವ್ಯಾಖ್ಯಾನಿಸಲ್ಪಟ್ಟ ಹೇಳಿಕೆಗಳ ಕೋನಗಳು ಲಘಕೋನಗಳು. ಈ ಕೆಳಗಿನ ಸಮೀಕರಣಗಳನ್ನು ಸಾಧಿಸಿ.
ಅಥವಾ
Question 35.
ಒಂದು ಕಾರ್ಖಾನೆಯ 50 ಕೆಲಸಗಾರರ ದೈನಂದಿನ ಆದಾಯವನ್ನು ಕೆಳಗಿನ ವಿತರಣೆಯು ನೀಡುತ್ತಿದೆ.
ಮೇಲಿನ ವಿತರಣೆಯನ್ನು ಕಡಿಮೆ ಇರುವ ವಿಧಾನದ’ ಸಂಚಿತ ಆವೃತ್ತಿ ವಿತರಣೆಯಾಗಿ ಬದಲಾಯಿಸಿ ಮತ್ತು ಅದರ ಓಜೀವ್ ಎಳೆಯಿರಿ.
Question 36.
ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯಿಂದ ಭಾಗಿಸಿ ಹಾಗೂ ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯು ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯ ಅಪವರ್ತನವಾಗಿದೆಯೇ ಎಂದು ಪರೀಕ್ಷಿಸಿ.
t2 – 3 ; 2t4 + 3t3 – 2t2 – 9t – 12
ಅಥವಾ
ಬಹುಪದೋಕ್ತಿ p(x) ನ್ನು ಬಹುಪದೋಕ್ತಿ g(x) ನಿಂದ ಭಾಗಿಸಿ, ಭಾಗಲಬ್ಧ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
p(x) = x4 – 3x + 4x + 5
g(x) = x2 + 1 – x
V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)
Question 37.
ಎರಡು ತ್ರಿಭುಜಗಳಲ್ಲಿ ಒಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳು ಮತ್ತೊಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳೊಡನೆ ಸಮಾನುಪಾತ ಹೊಂದಿದ್ದರೆ, ಅವುಗಳ ಅನುರೂಪ ಕೋನಗಳು ಸಮವಾಗಿರುತ್ತವೆ ಮತ್ತು ಅದರಿಂದಾಗಿ ಆ ಎರಡು ತ್ರಿಭುಜಗಳು ಸಮರೂಪಿಗಳಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
Question 38.
ನಕ್ಷೆಯ ಮೂಲಕ ಬಿಡಿಸಿ: y = 2x + 1; x = 2y – 5
Question 39.
(4, -1) ಮತ್ತು (-2, -3) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಭಾಜಕ ಬಿಂದುಗಳ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Question 40.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ ಪದದ ವರ್ಗವು ಅದರ 8ನೇ ಪದಕ್ಕೆ ಸಮವಾಗಿದೆ ಹಾಗೂ 8ನೇ ಪದವು ನಾಲ್ಕನೇ ಪದಕ್ಕಿಂತ 24 ಹೆಚ್ಚಾಗಿದೆ. ಹಾಗಾದರೆ ಶ್ರೇಢಿಯ ಪದಗಳನ್ನು ಬರೆಯಿರಿ.
Solutions
I.
Solution 1.
(B) 2.4 cm
Solution 2.
(C) $$\frac { { \pi r }^{ 2 } }{ 6 }$$
Solution 3.
(B) -1
Solution 4.
(A) 5
Solution 5.
(A) 2
Solution 6.
(C) x2 – 3x + 2 = 0
Solution 7.
(B) {TT, HH, HT, TH}
Solution 8.
(C) 2πrh
II.
Solution 9.
ತ್ರಿವಳಿ: 5, 4, 3
ಏಣಿಯ ಪಾದದಿಂದ ಗೋಡೆಯ
ಪಾದಕ್ಕಿರುವ ದೂರ = 3 ಮೀ.
Solution 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ = 1
Solution 11.
(x1, y1) ಮತ್ತು (x2, y2) ಬಿಂದುಗಳಿಗೆ ಸಂಬಂಧಿಸಿದಂತೆ : ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ (m1 : m2 ಅನುಪಾತದಲ್ಲಿ)
Solution 12.
ಭಾಜ್ಯ = ಭಾಜಕ × ಭಾಗಲಬ್ದ + ಶೇಷ
a = b × q + r
Solution 13.
ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳು = 3
Solution 14.
x + y = 27
13 + 14 = 27
xy = 182
13 × 14 = 182
ಮೊದಲ ಸಂಖ್ಯೆ = 13, ಎರಡನೆಯ ಸಂಖ್ಯೆ = 14
Solution 15.
a1 = 38, a16 = 73, a31 = ?
a16 = a + 15d
73 = 38 + 15d
73 = 38 + 15d
d = 3
a31 = a1 + 30d = 38 + 30(3) = 38 + 90 = 128
a31 =128
Solution 16.
AB ಯು ಏಣಿ, CA ಯು ಗೋಡೆ ಮತ್ತು ಕಿಟಕಿಯಾಗಿರಲಿ.
BC = 2.5m ಮತ್ತು CA = 6m
AB2 = BC2 + CA2 = (2.5)2 + (6)2 = 42.25
AB = 6.5
ಏಣಿಯ ಉದ್ದ ಆಗಿದೆ.
Solution 17.
ಮೊದಲ 3 ದಿನದ ಒಂದು ಶುಲ್ಕ = x
ಪ್ರತೀ ದಿನದ ಹೆಚ್ಚುವರಿ ಶುಲ್ಕ = y
x + 4y = 27
x + 2y = 21
2y = 6
y = 3
x + 2y = 21
x + 2(3) = 21
x = 21 – 6
x = 15
ಮೂರು ದಿನಗಳ ಶುಲ್ಕ = x = ₹ 15
ಹೆಚ್ಚುವರಿ ಪ್ರತೀ ದಿನದ ಶುಲ್ಕ = y = ₹ 13
Solution 18.
Solution 19.
Solution 20.
Solution 21.
Solution 22.
√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ.
√5 = $$\frac { p }{ q }$$ p, q ∈ I
p ಮತ್ತು q ಗಳು 1 ನ್ನು ಹೊರತುಪಡಿಸಿ ಬೇರೆ ಸಾಮಾನ್ಯ ಅಪವರ್ತನ ಹೊಂದಿದ್ದರೆ, ಸಾಮಾನ್ಯ ಅಪವರ್ತನದಿಂದ ಭಾಗಿಸಬಹುದು, ಆದ್ದರಿಂದ p ಮತ್ತು Q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳೆಂದು ಭಾವಿಸೋಣ
p = q√5
p2 = 5q2 (ಎರಡು ಕಡೆ ವರ್ಗ ಮಾಡಿದಾಗ)
5, p2 ನ್ನು ಭಾಗಿಸುತ್ತದೆ…… (1)
5, p ಮತ್ತು q ಗಳ ಸಾಮಾನ್ಯ ಅಪವರ್ತನವಾಗಿದೆ.
ಏಕೆಂದರೆ, 5, q ನ್ನೂ ಸಹ ಭಾಗಿಸುತ್ತದೆ.
ಅಂದರೆ p = 5r
5q2 = 52r2
q = 5r2 ………(2)
p ಮತ್ತು q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳು ಎಂಬ ಸತ್ಯಸಂಗತಿಗೆ ವಿರುದ್ಧವಾಗಿದೆ.
√5 ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆಯಾಗಿದೆ.
Solution 23.
ಶೂನ್ಯಗಳ ಮೊತ್ತ = α + β = 1
ಶೂನ್ಯಗಳ ಗುಣಲಬ್ದ = αβ = 1
ವರ್ಗಬಹುಪದೋಕ್ತಿ: x2 – (α + β) x + (αβ)
ಯಲ್ಲಿ ಆದೇಶಿಸಿದರೆ x2 – 1x + 1
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
Solution 24.
2x2 – 3x + 5 = 0
ax2 + bx + c = 0,
a = 2, b = -3, c = 5
ಶೋಧಕ: b2 – 4ac = 9 – 4(2)(5) = 9 – 40 = -31 < 0
ಮೂಲಗಳು ಸಂಮಿಶ್ರ ಸಂಖ್ಯೆಗಳಾಗಿವೆ.
Solution 25.
Solution 26.
ಆವೃತ್ತಿ ವಿತರಣಾ ಪಟ್ಟಿ
ಬೌಲರ್ ಗರಿಷ್ಠ ಪಂದ್ಯಗಳಾದ 3 ರಲ್ಲಿ ಪಡೆದ ವಿಕೆಟ್ಗಳ ಸಂಖ್ಯೆ 2 ಆಗಿದೆ. ಆದ್ದರಿಂದ ದತ್ತಾಂಶಗಳ ಬಹುಲಕ 2,
Solution 27.
ಫಲಿತಗಣ S = {1, 2, 3, 4, 5, 6}
n(S) = 6
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆಗಳು A = {3, 4, 5}
n(A) = 3
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆ ಪಡೆಯುವ ಸಂಭವನೀಯತೆ
P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$ = $$\frac { 1 }{ 2 }$$
Solution 28.
Solution 29.
Solution 30.
IV.
Solution 31.
ವೃತ್ತದ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’
ದತ್ತ: O ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ OP ತ್ರಿಜ್ಯ ಮತ್ತು XY ಸ್ಪರ್ಶಕ ಎಳೆದಿದೆ.
ಸಾಧನೀಯ: OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತೋಂದು ಬಿಂದು Q ಗುರುತಿಸಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: Q ಬಿಂದುವು ವೃತ್ತದ ಹೊರಭಾಗದಲ್ಲಿರಬೇಕು.
Q ಬಿಂದುವು ವೃತ್ತದ ಒಳಭಾಗದಲ್ಲಿದ್ದರೆ, XY ವೃತ್ತಕ್ಕೆ ಛೇದಕವಾಗುತ್ತದೆಯೇ ಹೊರತು ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವಾಗುವುದಿಲ್ಲ. ಆದ್ದರಿಂದ OQ ಇದು ವೃತ್ತದ ತ್ರಿಜ್ಯ OP ಗಿಂತ ಉದ್ದವಾಗಿದೆ. ಅಂದರೆ OQ > OP.
P ಬಿಂದುವನ್ನು ಹೊರತುಪಡಿಸಿ, XY ಮೇಲಿನ ಎಲ್ಲಾ ಬಿಂದುಗಳಿಗೂ ಇದು ಅನ್ವಯಿಸುವುದರಿಂದ, O ಬಿಂದುವಿನಿಂದ XYನ ಮೇಲಿನ ಇತರೆ ಬಿಂದುಗಳಿಗಿರುವ ದೂರಕ್ಕಿಂತ OP ಯು ಕನಿಷ್ಟ ಉದ್ದ ಹೊಂದಿದೆ
OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ಅಥವಾ
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮ.
ದತ್ತ: 0ವೃತ್ತಕೇಂದ್ರ, P ಬಾಹ್ಯಬಿಂದು, PQ & PR ಗಳು ಬಾಹ್ಯಬಿಂದು Pನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು
ಸಾಧನೀಯ: PQ = PR
ರಚನೆ: OP, OR, OP ಸೇರಿಸಿ
ಸಾಧನೆ: ΔOQP ಮತ್ತು ΔORP ಗಳಲ್ಲಿ
OQ = OR (ತ್ರಿಜ್ಯಗಳು)
∠OQP = ∠ORP = 90°
OP = OP (ಉಭಯಸಾಮಾನ್ಯ)
ΔOQP = ΔORP
PQ = PR
Solution 32.
Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು = x ಆಗಿರಲಿ
3 ವರ್ಷಗಳ ಹಿಂದೆ, ಅವನ ವಯಸ್ಸು = x – 3
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸು = x + 5
Solution 34.
Solution 35.
Solution 36.
Solution 37.
Solution 38.
Solution 39.
Solution 40.
## Karnataka SSLC Maths Model Question Paper 1 with Answer in Kannada
Students can Download Karnataka SSLC Maths Model Question Paper 1 with Answers in Kannada, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka SSLC Maths Model Question Paper 5 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
The pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,.
a) K = -2
b) K = 3
c) K = 2
d) K = -3
c) K = 2
Solution:
Question 2.
n2 – 1 is divisible by 8, if n is
a) Prime numbers
b) Odd integer
c) Even integer
d) Natural number
b) Odd integer
Solution:
n2 – 1
If n is an odd. integer, 1,3,5,
Ex: 12-1 = 1- 1= 0 divisible by 8.
32 – 1=9-1=8 divisible by 8.
52 – 1 = 25 – 1 = 24 divisible by 8.
Question 3.
$$\sqrt{1+\tan ^{2} \theta}$$ = , where 0<θ< 90°
a) secθ
b) cosec θ
c) cos θ
d) sin θ
a) secθ
Solution:
$$\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=\sec \theta$$
Question 4.
If Q divides the line A(3, 5) and B (7,9) internally in the ratio 2:3, then the co ordinates of Q are.
c) $$\left(\frac{23}{5}, \frac{33}{5}\right)$$
Solution:
Question 5.
If Area of the sector OPRQ = $$\frac{5}{18}[ /latex] of Area of circle. Then the value of x a) 25° b) 50° c) 75° d) 100° Answer: d) 100° Solution: Question 6. If 1 + 2 + 3 + n terms = 28, then n is equal to a) 28 b) 7 c) 8 d) 56 Answer: b) 7 Solution: [latex]\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$
n(n + 1) = 28 x 2 = 56
n(n + 1) = 7 x 8
∴ n = 7
Question 7.
If E1E2 E3…….E10 are the possible elementary events of a random experiment, then P(E1) + P(E2) + P(E3) + ………P(E10) is equal to
a) 0
b) 1
c) 2
d) 3
b) 1
Question 8.
If we express sec A in terms of sin A, then sec A is equal to
$$\frac{1}{\sqrt{1-\sin ^{2} A}}$$
Solution:
II. Answer the following Questions : ( 1 x 8 = 8 )
Question 9.
What is the $$\frac{p}{q}$$ form of 43.123456789 ?
$$\frac{43123456789}{999999999}$$
Question 10.
Write the quadratic equation formed by the roots √3 + √5 and √3 – √5
x2 -(α + β)x + αβ = 0
X2 – [3 + √5 + 3 – √5]x + (3 + √5)(3 – √5) = 0
x2 – [6] x + (9 – 5) = 0
x2 – 6 + 4 = 0
Question 11.
Find the value of $$\frac{\sin 26}{\sec 64}+\frac{\cos 26}{\csc 64}$$
Question 12.
What is the distance between the points p(cos0, sin0) and Q(sin0, -COS0)?
Question 13.
If a number “x” choosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2 < 3 ?
Clearly “x” can take any of the five given values.
∴ n(x) = 5.
If x2 < 3, then x can take the values -1, 0, 1.
∴ This even A = ,{-1, 0, 1}
n(A) = 3.
P(A) = $$\frac{n(A)}{n(S)}=\frac{3}{5}$$
Question 14.
What is the Area of a circle whose perimeter is 44 cms.
2πr = 44
2 x $$\frac{22}{7}$$ x r = 44
r = $$\frac{44 \times 7}{22 \times 2}$$
r = 7
A = πr² = $$\frac{22}{7}$$ x 7 x 7 = 154cm2
Question 15.
A toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is h cm and base radius is r cms, find the total surface area of the toy.
Total surface Area of the toy = C.S.A of cylinder + 2 [Surface Area of Hemisphere]
= 2πrh + 2πr²
= 2 πr (h + r)
Question 16.
The circumference of a circle exceeds the diameter by 15cm. Find the raidus of the circle.
Circumference = 2r + 15
2πr = 2r + 15
2πr – 2r = 15
2 x $$\frac{22}{7}$$r – 2r = 15
Multiply by 7
2 x 22r – 14r = 105 .
30r = 105
r = 3.5
III. Answer the following : (2 x 8 = 16 )
Question 17.
Prove that √5 + √3 is an irrational number.
∴ √3 is rational, is a contradiction
∴ √5 + √3 is irrational.
Question 18.
AX and BY are perpendiculars to segment XY. If AO = 5cm, BO = 7cm and Area of ∆ AOX = 150 cm2, find the area of ∆ BOY.
In ∆ OYB and ∆ OXA,
We have ∠X = ∠Y
So, by AA – criterion of similarity, we have
∆ YOB ≅ ∆ XOA
OR
In the given figure, BD ⊥ AC. Prove that AB2 + CD2 = AD2 + BC2
Soln: In ∆ BDC,
∠ BDC = 90°
BC2 = BD2 + DC2 (By Pythagoras)
In ∆ BDA, ∠ APB = 90°
AB2 = AD2 + BD2 (By Pythagoras)
AB2 – BC2 = AD2 + BD2 – BD2 – DC2
AB2 + CD2 = AD2 + BC2
Hence proved.
Question 19.
If zeroes of the polynomial p(y) = y3 – 3y2+ y + 1 are a – b, a, a+ b. Find a and b.
Sum of the zeroes of p(y) = y3 – 3y2+ y + 1
(a-b)(a)(a + b)= $$\frac{-1}{1}$$ = -1
(a2 – b2) a = -1
(1 – b2) (1) = -1
– b2 = -1
-b2 = – 1 – 1
-b2 = -2
b2 = 2
b = ±√2
Question 20.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.
Question 21.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 16y + 80. If zeroes are α, -α and β.
Sum of the zeroes
α + (-α) + β = 5
α – α + β = 5
β = 5
(α )(-α )(β) =-80
2β = -80
α2β = 80
α2(5) = 80
α = 80/5 = 16
α = √16
= ±4
Question 22.
Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillars.
In ∆ PAB,
tab 60 = $$\frac{\mathrm{AP}}{\mathrm{AB}}$$
√3 = h/x
h = √3x
In ∆ BCQ,
tan 30 = $$\frac{\mathrm{CQ}}{\mathrm{BC}}$$
$$\frac{1}{\sqrt{3}}=\frac{h}{100-x}$$
h√3 -= 100 – x
(√3x)(√3x) = 100-x
4x = 100
x = 25
∴ h= √3 X 25 = 25√3
= 25 x 1.73 = 43.3
OR
The three metallic spheres of radii are in the ratio of 3 : 4 : 5 are melted to form a single solid sphere of radius 12cm. Find the radius of the three small metallic spheres.
Volume of the solid sphere = Sum of the volumes of each sphere
$$\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right)$$
4 x 123 = 4[(3x)3 + (4x)3 + (5x)3]
1728 = 27x3 + 64x3 + 125x3 1728 = 216 x3
x = $$\frac{1728}{216}$$
x3 = 8 x = 3√8 = 2
∴ R1 = 6, R2 8, R3 = 10
Question 23.
In the given fig. OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region
Question 24.
Find the volume of the largest right circular cone that can be cut of a cube of edge 7cm.
Given radius of the base of cone = $$\frac{7}{2}$$ cm
h = height of cone = 7cm
Volume of the cone = $$\frac{1}{3}$$πr²h
Volume = $$\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7$$
= $$\frac{22 \times 7 \times 7}{12}$$
= 89.83cm3
IV. Answer the following : ( 3 x 9 = 27 )
Question 25.
The sum of a two digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Let the digits at units and tens place of the given number be y and x respectively, then
Given number = 10x + y
Number obtained by reversing the order of the digits = 10y + x
(10x + y) + (10y + x) = 165
10x + y+ 10y + x = 165
(11x+11y=165) = 11 ’
x + y = 15 ……(1)
Digits differ by 3, if x > y
x – y = 3 ……..(2)
Solve (1) and (2)
Consider x + y =15
9 + y = 15
y = 15 – 9
y = 6
∴ The number is 96, if y >x, number is 69.
OR
Ten years ago Sudhir was twelve times as old as his son Raghav and ten years, hence, he will be twice as old as his son will be. Find their present ages.
Let the present ages of Sudhir and Raghav be x years and y years respectively. Ten years ago, Father ’s age = (x -10) years Son’s age = (y – 10) years,
x- 10= 12(y – 10)
=> x – 10 = 12y – 120
x – 12y = 120+ 10
x – i2y = – 110
x – 12y + 110 = 0. …….(1)
Ten years later, Father’s age = (x + 10) years
Son’s age = (y + 10)
x + 10 = 2(y + 10)
⇒ x+ 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10
x – 2y – 10 = 0 ……(2)
Solve (1) and (2)
y = 12
x – 12y = 110
x – 12(12) = -110
x – 144 = -110
x = -110 + 144 = 34
∴ Present age of Sudhir = 34 years and present age of Raghav = 12 years.
Question 26.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Let the base BC be = x cm
Then its height AB = (x – 7) cm
Hypotenuse AC = 13 cm
In ∆ABC, AC2 = AB2 + BC2
132 = (x – If + (x)2
169 = x2 + 49 – 14x + x2
169 = 2x2 – 14x + 49
2x2 – 14x + 49 – 169 = 0
2x2 – 14x-120 = 0
x2 – 7x – 60 = 0
x2– 12x + 5x-60 = 0
x(x- 12) + 5(x – 12) = 0
(x-12) (x + 5) = 0
(x – 12) = 0 or x + 5 = 0
x = 12 or x = -5
The base of the triangle = 12 cm Length of the altitude = (12 – 7)cm = 5 cm.
Question 27.
X(2, 5), Y (-1, 2) and Z(5, 8) are the co – ordinates of the vertices of A XYZ. The point A and B lie on XY and XZ respectively such that XA : AY = XB: BZ =1:2. Calculate the co – ordinates of X and Y.
B(X2 y2)= [3, 6]
∴ Co – ordinates of A = (1,4)
Co – ordinates of B = (3, 6)
OR
Find the Area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1) (2, 1) and
Co – ordinates of A
$$\frac{0+2}{2}, \frac{-1+1}{2}$$
= [1,0]
Co – ordinates of B
$$\frac{0+0}{2}, \frac{-1+3}{2}$$
= [0,1]
Co – ordinates of B
$$\frac{2+0}{2}, \frac{1+3}{2}$$
= [1,2]
∴ Area of A ABC = $$\frac{1}{2}$$ Σ x1 (y2 – y3)
= $$\frac{1}{2}$$ [1(1-2) + 0(2 – 0) + 1(0-1)]
= $$\frac{1}{2}$$ [-1 + 0 + 1]
∴ Area of ∆ ABC = 1 sq. unit
Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contacts.
Data : AB is the tangent drawn to a circle centered at O
‘C’ is the point of contact.
To prove that :
OC ⊥AB
Construction : Mark D .on AB. Join OD.
Let it cut the circle at E
Proof: OE < OD
OE is the radius of the circle.
OE = OC (radii of the same circle)
⇒ OC < OD
⇒ OC is the shortest distance between centre of the circle and tangent AB
∴ OC ⊥ AB
Hence proved.
Question 29.
A hollow sphere of internal and external radii are 6cm and Scm respectively is melted and recast into small cones of base radius 2cm and height 8cm. Find the number of cones.
Inner radius of a hollow sphere (r) = 6cm
and outer radius (R) = 8cms.
Radius of one small cone (r,) 2cm and
heigh (h) = 8 cm.
OR
A medicine capsule is in the shape of a cylinder with two hemispheres stuch to each of its ends. The length of the entire capsule is 14mm and the diameter of the capsule is 5 mm. Find its surface area.
Let r mm be the radius and h mm be the height of the cylinder.
r = $$\frac{5}{2}$$ mm = 2.5mm
h = (14 – 2 x $$\frac{5}{2}$$ mm
= (14-5) mm
= 9 mm
The radius of hemisphere = r = $$\frac{5}{2}$$ mm
Now, the surface area of the capsule. = Curved surface Area of the cylinder surface Area of two hemispheres.
= 2πrh + 2 x 2 πr²
= 2πr² (h + 2r)
Question 30.
Solve graphically
y = $$\frac{1}{2}$$ x and 3x + 4y – 20 = 0
y = $$\frac{1}{2}$$ x
3x + 4y – 20 = 0
3x + 4y = 20
4y = 20 – 3x
Question 31.
The following distribution gives the daily income of 65 workers of a factory. Convert this into less than type cumulative frequency distribution and draw its ogive.
Question 32.
The fourth term of an AP is 11, and 8th term exceeds twice the fourth term by 5. Find AP and find the sum of first 100 terms.
Given : T4 = -11 and T8 = 2T4 + 5
T8 = 2(11) + 5
a + 7d = 22 + 5 = 27
d = 4
a + 7d = 27
a + 7(4) = 27
a+ 28 = 27
a = 27-28 = -1
a = -1
A.P = -1,3, 7, …..
Sn = $$\frac{n}{2}$$ [2a + (n-1)d] = $$\frac{100}{2}$$ [2(-1) + (99)4]
= 50 [-2 + 396]
= 50 [394]
= 19700
Question 33.
A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of tour.
Let the original duration of the tour be x days.
Total expenditure on four = 360 ₹
Expenditure per day = $$\frac{360}{x}$$
Duration of the extended tour = (x +4) days
∴ Expenditure per day according to new schedule = $$\frac{360}{x+4}$$
It is given that daily expenses are cut down by 3₹
$$\frac{360}{x}-\frac{360}{x+4}$$ = 3
360 (x + 4) – 360x = 3(x) (x + 4)
360x + 1440 – 360x = 3x (x + 4)
1440 = 3x2+ 12x
3x2 + 12x – 1440 = 0
x2 + 4x – 480 = 0
x2 + 24x – 20x – 480 = 0
x(x + 24) – 20(x + 24) = 0
(x + 24) (x – 20) = 0
x + 24 = 0 and x – 20 = 0
x = -24 and x = 20
∴ Duration of tour = 20 days.
OR
Two pipes running together can fill a cistern in 3$$\frac{1}{13}$$ minutes. If one pipe
takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Let the faster pipe takes x minutes.
Let the smaller pipe take (x + 3) minutes to fill.
Portion of the cistern filled by the faster 1
pipe in one minute = $$\frac{1}{x}$$
⇒ Portion of the cisterin filled by the faster pipe in $$\frac{40}{13}$$ minutes
|||ly portion of the cistern filled by the slower pipe in $$\frac{40}{13}$$ minutes.
It is given that the cistern is filled in $$\frac{40}{13}$$ min.
[40(x + 3) + 40x] = [(x) (x +3)] 13
40x + 120 + 40x = (x2 + 3x) 13
80x + 120= 13x2 + 39x
13x2 + 39x – 80x – 120 = 0
13x2 – 41x-120 = 0
13x2 – 65x + 24x – 120 = 0
13x (x – 5) + 24(x – 5) = 0
x – 5 = 0, 13x + 24 = 0
x = 5, 13x = -24, x = $$\frac{-24}{13}$$
Faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 min to fill the cistern.
V. Answer the following : ( 4 x 4 = 16 )
Question 34.
In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T20 = -112. Find S20.
a = 2, T1 + T2 + T3 + T4+ T5
= $$\frac{1}{4}$$ (T6 + T7 + T8 + T9 + T10)
a + a + d + a + 2d + a + 3d + a + 4d
= $$\frac{1}{4}$$(a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
5a+ 10d = $$\frac{1}{4}$$ (5a + 35d)
5(a + 2d) = $$\frac{1}{4}$$ x 5( a + 7d)
a + 2d = $$\frac{1}{4}$$ (a + 7d)
4(a+2d) = a+7d
4a + 8d – a – 7d = 0
3a + d = 0
3(2) + d = 0
d = -6
T20 = a + 19d
= 2 +19(-6)
= 2 – 114
T20 = -112
Sn = $$\frac{n}{2}$$ [2a + (n-1)d]
Sn = $$\frac{20}{2}$$ [2(2) + (20-1)(-6)]
= 10 [4- 114]
= (10) (-110)
S20 = -1100.
OR
A man repays a loan of ₹ 3250 by paying 7 20 in the first month and then increases the payment by ₹ 15 every month. How long will it take to clear his loan?
Here, a = 20, c.d = 15, n = ?
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
3250 = $$\frac{n}{2}$$ [2 x 20 + (n-1)15]
3250 x 2 = n(40 + 15n- 15)
6500 = n (25 +15n)
6500 = 25n + 15n2
(15n2 + 25n – 6500 = 0)÷ 5
3n2 + 5n – 1300 = 0
3n2-60n + 65n- 1300 = 0
3n(n – 20) + 65 (n – 20) = 0
(n – 20) (3n + 65) = 0
(n – 20) = 0, 3n + 65 = 0
n = 20, 3n = -65
n = $$\frac{-65}{3}$$
∴ Total amount will be paid in 20 months.
Question 35.
Find the mean, median and mode for * the following frequency distribution.
Question 36.
P.T $$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$$ = 1+ secθ cosecθ
Question 37.
Draw a right triangle in which the sides (other than hypotenuse) arc of lengths 8cm and 6cm, then construct another triangle whose sides are 5/3 times the corresponding sides of given triangle.
Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.
VI. Answer the following : (5 x 1 = 5 )
Question 38.
State and prove pythagoras theorem.
In a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.
Data : ∆ ABC, ∠B = 90°
T.P.T : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof: In ∆ ABC and ∆ ABD
∠ ABC = ∠ADB = 90 [∵ Data & construction]
∠A = ∠A (∵ common Angle)
∠ACB =∠ ABD [ ∵ Re maining angle]
∴ ∆ ABC and ∆ ABD are equiangular
=> ∆ ABC ~ ∆ABD
AB2= AC x AD ……… (1)
In ∆ABC and ∆ BDC
∠ ABC = ∠ BDC = 90 [∵ Data & construction]
∠C = ∠C (∵ common Angle)
∠BAC =∠ DBC [ ∵ Re maining angle]
∴ ∆ ABC and ∆ DBC are equiangular
∆ABC ~ ∆BDC
AB2 + BC2 = AC (AD + DC)
AB2 + BC2 = AC(AC)
AB2 + BC2=AC2
AC2 = AB2+ BC2
## Karnataka SSLC Maths Model Question Paper 4 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 4 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 4 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117, then the value of m is
a) 4
b) 3
c) 11
d) 2
Solution:
The H.C.F. of 65 and 117 = 65m- 117
Now 13 = 65m- 117
65m 117+13
65rn = 130
m = 130/65 = 2
∴ The value of m = 2
∴ HCF(65, 117)
= 13
Question 2.
If sinx = sin 60° cos 30° – cos 60 sin 30!l, then the value of x is
a) 0°
b) 30°
c) 45°
d) 60°
b) 30°
Solution:
Sinx = sin 60°cos30 – cos 60 sin 30
Question 3.
The angle between the radius of a circle and the tangent drawn at the point of contact is
a) 0°
b) 60°
c) 90°
d) 30°
c) 90°
Question 4.
The T.S.A. of a cuboid of dimension, l = 30cm, b = 20cm, c = 10cm, is ______
a) 600cm2
b) 60cm2
c) 6000cm2
d) 2200 cm2
Question 5.
Which of the following is a polynomial
a) x2 – 5x + 3√x
b) x1/2 + x1/2 – x +1
c) $$\sqrt{x}-\frac{1}{\sqrt{x}}$$
d) x2 – 4x + √2
x2 – 4x + √2
Solution:
x2 – 4x + √2 because variable has no negative or decimal powers.
Question 6.
The value of p if x, 2x + p and 3x + 6 are in A.P.
a) p = 3
b) p = 2
c) p = 1
d) p = 0
Solution:
x, 2x + p, 3x + 6 are inAP a, A, b are A.P.
2(2x + p) = 2(2x + 3)
2x + p = 2x + 3
p = 2x + 3 – 2k
P = 3
Question 7.
In triangle PQR, The value of y is
a) 4√3
b) 6√3
c) 5√3
d) √3
b) 6√3
Solution:
QS2 = PS x SR
QS2 = $$\sqrt{9 \times 12}$$
QS = $$\sqrt{9 \times 4 \times 3}$$
QS = 3 x 2√3
QS = 6√3
Question 8.
When 2 unbiased coins are tossed at ‘ a time, the probability of getting 2 heads is _____
a) 1/4
b) 1/2
c) 1
d) 0
a) 1/4
Solution:
S = {(H, H) (H, T) (T, H) (T, T)}
n(s) = 4
An event of getting 2 heads = A={(H, H)}
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{1}{4}$$
II. Answer the following Questions : ( 1 x 8 = 8 )
Question 9.
If the product of zeros of polynomial f(y) = ay3 – 6y2 + 11y – 6 is 4 then find the value of ‘a’.
α β γ = $$-\frac{\mathrm{d}}{\mathrm{a}}$$
4 = $$\frac{-(-6)}{a}$$
4a = 6
a = $$\frac{6}{4}$$
a = $$\frac{3}{2}$$
Question 10.
What is the value of C, if ax2 + bx + c = 0 has equal roots?
If ax2 + bx + c = 0 has equal roots then,
b2 – 4ac = 0
b2 = 4ac
4ac = b2
c = $$\frac{b^{2}}{4 a}$$
Question 11.
Find the second term if sum of the ‘n’ term of an AP is 2n2 + 1.
Sn = 2n2 + 1
S1 = 2(1)2 + l= 2xl + l= 2+ l= 3
∴ S1 = T1 = a = 3
S2 = 2(2)2 + 1 =8+ 1 =9
a + T2 = 9
T2 = 9 – 3 = 6
Question 12.
State converse of Pythagoras Theorem.
“The square on the greatest side of a triangle is equal to the sum of the squares on the other two sides, then the other two sides contian the right angle”.
Question 13.
What is the form of $$\frac{p}{q}$$ (p,q ∈ z q ≠ 0) form of 0.57 ?
$$0.5 \overline{7}=\frac{57-5}{90}=\frac{52}{90}=\frac{26}{45}$$
Question 14.
If sinθ = $$\frac{1}{3}$$ then find the value of (cot2θ + 2)
AC2 = AB2 + BC2
BC2 = AC2 – AB2
= 32 – 12
= 9 – 1 = 8
BC = √8 = 2√2
cotθ = $$\frac{2 \sqrt{2}}{1}$$ = 2√2
2cot2θ + 2 = 2(2√2)2 + 2
= 2 x 8 + 2
= 16 + 2
= 18
Question 15.
In sin(A + B) = $$\frac{\sqrt{3}}{2}$$ and cos (A – B) = 1, 0< A + B < 90°, A ≥ B
sin(A + B) = $$\frac{\sqrt{3}}{2}$$ = sin60°
A + B = 60° …. (1)
cos (A – B) = 1 = cos0°
A – B = 0 ….(2)
Solve (1) and (2)
A + B = 60
A – B = 0
2A = 60 or A = 30°
A + B = 60
30 + B = 60
B = 60 – 30° = 30°
∴ A = 30°, B = 30°
Question 16.
The surface area of a sphere is same as the C.S.A. of a right circular cylinder whose height and diameter are 4 cm each. Find the radius of the sphere.
Soln:
4πr² = 2πrh
4R² = dh .
4R²= 4 x 4
R² = $$\frac{4 \times 4}{4}$$ = 4
R = √4 = 2cm
A = 30°, B = 30°
∴ Radius of the sphere = 2cm.
III. Answer the following : ( 2 x 8 = 16 )
Question 17.
By Euclid’s division lemma, show that thte square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let x be any positive integer. Then it is of the form 3q, 3q+1 or 3q + 2.
Now, (3q)2 = 9q2 = 3(3q2) = 3m where m = 3q2
(3q + 1)2 = 9q2 + 6q+ 1 = 3(3q2 + 2q) + 1
= 3m + 1, where m = 3q2 + 2q and (3q+2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) +1
=3m +1, where m = 3q2 + 4q + 1
Hence, it is proved.
Question 18.
Solve : $$\frac{x+y}{x y}$$ = 2 and $$\frac{x-y}{x y}$$ = 6
x + y = 2xy
x – y = 6xy
2x = 8xy
2 = 8y
Question 19.
Solve : y2 – (√3 + 1)y + √3 = 0
y2 – √3y – y + √3 = 0
y(y – √3) – 1(y – √3) = 0
(y – √3)(y – 1) = 0
y – √3 = 0, y – 1 = 0
y = √3 and y = 1
Question 20.
Show that the points (3, 2) (-2, -3) and (2, 3) are collinear Or non – collinear.
Let A= (3, 2) B = (-2, -3) C = (2, 3)
AC ≠ AB + BC
∴ They are non – collinear.
Question 21.
In the given fig ∆DGH ~ ∆DEF, DH = 8cm, DF = 12cm, DG = (3x – l)cm and DE = (4x +. 2)cm, Find the lengths of DG and DE.
∆DGH ~ ∆DEF
$$\frac{\mathrm{DG}}{\mathrm{DE}}=\frac{\mathrm{DH}}{\mathrm{DF}}$$
$$\frac{3 x-1}{4 x+2}=\frac{8}{12}$$
12 (3x – 1) = 8(4x + 2)
36x – 12 = 32x + 16
36x – 32x = 16 + 12
4x = 28
x = 28/4
x = 7
∴ DG = 3x – 1 = 3 x 7 – 1 = 21 – 1 = 20
DE = 4x + 2 = 4 x 7 + 2 = 28 + 2 = 30
OR
D is a point on the side BC of A ABC such that ∠ADC = ∠BAC. Prove that
$$\frac{\mathbf{C A}}{\mathbf{C D}}=\frac{\mathbf{C B}}{\mathbf{C A}}$$
In ∆ ABC and ∆ DAC,
∠C = ∠C
∆ABC ~ ∆DAC
Question 22.
A card is drawn at random from a box containing 21 cards numbered 1 to 21. Find the probability that the card drawn is
a) Prime number b) Divisible by 3.
n(s) = 21
A= {2,3,5,7,11,13,17,19}
n(A) = 8
P(A) = $$\frac{n(A)}{n(S)}=\frac{8}{21}$$
b) n(S) = 21
B= {3,6, 9,12,15,18}
n(B) = 6
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{6}{21}=\frac{2}{7}$$
Question 23.
Draw a circle of radius 3cms. Construct a pair of tangents to it, from a point 5cm away from the circle.
r = 3cm d = 3 + 5 = 8cm
Question 24.
Express sinA and sec A in terms of cot A.
W.K.T cosec2A = 1 + cot2A
OR
If tanθ + sinθ = m and tan θ – sin θ= n, S.T m2 – n2 = 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
L.H.S = m2 – n2 = (m +n) (m – n)
= {(tan θ + sin θ ) + (tan θ – sin θ )x(tan θ +sin θ) – (tan θ – sin θ)}
= {(2tanθ) x (2sinθ) = 4tanθsinθ
= 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
= RHS.
IV. Answer the following : ( 3 x 9 = 27 )
Question 25.
The sum of the numerator and denominator of a fraction is 24. If 4 is subtracted from the numerator and 5 from its denominator, then it reduces to 1/4. Find the fraction.
Let the fraction be $$\frac{\mathrm{x}}{\mathrm{y}}$$
sum of its numerator and denominator = 24.
x + y = 24 ……..(1)
If 4 is subtracted from the numerator and
5 from its denominator we get = 1/4
⇒ $$\frac{x-4}{y-5}=\frac{1}{4}$$
4(x – 4) = 1(y – 5)
4x – 16 = y – 5
4x – y = -5 + 16
4x – y = 11 ……..(2)
From (1) and (2)
Consider,
x + y = 24
7 + y = 24
y = 24 – 7
y = 17
∴ The fraction is $$\frac{x}{y}=\frac{7}{17}$$
OR
Two women and five men can together finish an embroidary work in 4 days. While three women and 6 men can finish in 3 days. Find the time taken by one women alone and also that taken by one man alone.
Let 1 woman can finish the embroidary in x days and 1 man can finish the embroidary in y days.
1 woman’s 1 day’s work = $$\frac{1}{x}$$
1 man’s 1 day’s work = $$\frac{1}{y}$$
Let $$\frac{1}{x}$$ = u, $$\frac{1}{y}$$ = v
2u + 5v = $$\frac{1}{4}$$
3u + 6v = $$\frac{1}{3}$$
solve (1) and (2)
(8u + 20v = 1) x 9
(9u+ 18v= 1) x 8
v = $$\frac{1}{36}$$
Put v = $$\frac{1}{36}$$…….(1)
2u + 5v = $$\frac{1}{4}$$
2u + 5$$\frac{1}{36}$$ = $$\frac{1}{4}$$
Thus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.
Question 26.
If two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 – √3 and 2 + √3. Find the other two zeroes.
The two zeroes of p(x) are,
(2 – √3) and (2 + √3)
⇒ [x – (2 – √3] and [x – [2 + √3] are the factors p(x).
⇒ [ x – 2 + √3] [x – 2 – √3]
(x – 2)2 – (√3)2
x2 – 4x + 4 – 3
⇒ x2 – 4x + 1 is a factor of p(x)
Divide p(x) by x2 – 4x + 1 to get quotient
Consider the quotient, x2 – 2x – 35 = 0
x2 – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7) (x + 5) = 0
x-7 = 0 orx + 5 = 0
x = 7 or x = -5
∴ The other two zeroes are 7, -5.
Question 27.
A two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchange their places. Find the number.
Let the two digits be x and y.
⇒ The two digit number is = 10x + y
The product of three digits = xy = 18
When 63 is subtracted from the number, the digits interchange their places.
10x + y – 63 = 10y + x
10x – x – 10y + y = 63
9x – 9y = 63
9(x – y) = 63
x – y = 63/9 = 7
x – y = 7
x = 7 + y
Consider xy = 18 and substitute x = (y + 7) in it.
y(y + 7) = is . .
y2 + 7y – 18 = 0
y2 + 9y – 2y – 18 = 0
y (y + 9) – 2(y + 9) = 0 –
(y + 9) (y – 2) = 0 .
y + 9 = 0, y -2 = 0
y = -9 or y = 2
xy= 18
x(2) = 18
x = 18/2 = 9
x = 2
But negative digit is not considered.
∴ The two digit number is
= 10x + y
= 10(9) + 2 = 92
OR
A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km / hr from its usual speed. Find its usual speed.
Let the usual speed of the plane be x km / hr. Time taken to cover 1500 km with the usual speed.
$$\frac{1500}{x}$$ hrs
Time taken to cover 1500 km with the speed of (x + 250) km / hr = $$\frac{1500}{x+250}$$
2(1500 x 250) = x2 + 250x
x2 + 250x-750000 = 0
x2 + 1000x – 750x – 750000 = 0
x(x + 1000) – 750(x + 1000) = 0
(x+ 1000)(x-750) = 0
x + 1000 = 0 or x – 750 = 0
x = -1000 or x= 750 ,
x = 750
Hence, the usual speed of the plane = 750 km/hr.
Question 28.
If the co – ordinates of the mid points of ∆ ABC are D(1, 2) E(0, -1) and (2, -1). Find the respective co – ordinates of ∆ ABC.
D is the midpoint of BC.
x1 + x2 = 4 and y1 + y2 = -2 ………(1)
E is the mid point of BC
x2 + x3 = 0 and y2 + y3 = -2 …….(2)
F is the mid point of BC
x1 + x3 = 4 and y2 + y3 = -2 ….(3)
Solve (1) and (3)
x2 = 0
x2 = 0
0 + x3 = 0
x3 = 0
x1 + x3 = 4
x1 + 0 = 4
x1 = 4 – 0 = 4
x1 = 4
y2 = -i
y2 + y3 = -2
-1 + y3 = -2
y3 = -2 +1
y3 = -1
y1 + y3 = -2
y1 + (-1) = -2
y1 -1 = -2
y1 = -2 + 1 = -1
y1 = -1
A(x1,y1) = A(4,-1)
B (x2 y2) = B(0, -1)
C(x3, y3) = C(0,-1)
OR
Find the length of the median through the vertex A(5,1) drawn to the triangle ABC where other two vertices are B(1, 5) andiC(-3, -1)
Question 29.
Prove that the tangents drawn from an external point are equal.
Data : ‘O is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that : ∠APB + ∠AOB = 180°
∠OAP + ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (Common side)
OA = OB (Radii of the same circle)
According to RHS postulate
∆APO = ∆BPO
∠ OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° +90° =180°
⇒ Opposite angles of OAPB quadrilateral are supplementary.
⇒ ∠ APB + ∠ AOB = 180°
Question 30.
If a chord of circle of radius 10cm subtend an angle of 60° at the centre of the circle. Find the area of the corresponding segment of the circle.
(Take π =3.14, √3 = 1.7)
O is the centre of the circle having radius 10 cm.
∠ AOB = 60° is angle subtended at the centre O of the circle by a chord AB.
Draw BL ⊥ OA
In right angled ∆ OLB,
Area of segment ∆ APB
= The area of sector OAPB – Area of ∆ OAB
OR
Find the area of the shaded region where PQRS is a square of side lOcms and semicircles are drawn with each side of square as diameter.
The four shaded regions are marked as I, II, III, IV.
All the four regions have equal areas due to § symmetry.
Area of I + Area of III
= Area of square ABCD – Sum of the areas of two semicircles of each of radius 5cm.
{10 x 10 – 2 x $$\frac{1}{2}$$ π x 52}cm2
= {100 – 25 x 3.14} cm2
= {100 – 78.5} cm2
= 21.5 cm2
Similarly Area of II + Area of IV = 21.5 cm2. Total Area of shaded region = 21.5 + 21.5
= 43 cm2
Question 31.
The following table gives the production yield per hectare of paddy of 50 farms of a village.
Question 32.
Find the mean of the following frequency distribution.
A = 50, h = 20, N = 100 and Σ fiui = 15
Mean = A + h$$\left\{\frac{1}{\mathrm{N}} \sum \mathrm{f}_{i} \mathrm{u}_{\mathrm{i}}\right\}$$
Mean = 50 + 20 x $$\frac{15}{100}$$
Mean = 50 + 3 = 53
Question 33.
Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose Sides are 2/3 of the corresponding sides of it.
V. Answer the following : ( 4 x 4 = 16 )
Question 34.
Solve the pair of equations graphically.
4x – 3y + 4 = 0
4x + 3y – 20 = 0
4x – 3y + 4 = 0 ….. (1)
4x + 4 = 3y
y = $$\frac{4 x+4}{3}$$
4x + 3y – 20 = 0
3y = -4x + 20
y = $$\frac{-4 x+20}{3}$$
Question 35.
How many terms of the series 93 + 90 + 87 + ……….. amounts to 975. Find also the last term.
93 + 90 + 87 + …………
a = 93, d = 90 – 93 = -3, Sn = 975, n = ?,
T = ?
S = $$\frac{n}{2}$$[2a + (n – 1)d]
975 = $$\frac{n}{2}$$[2(93) + (n-1)(-3)]
975 x 2 = n[186 – 3n + 3]
1950 = n [189 – 3n]
1950 = n[189 – 3n]
1950 = 189n – 3n2
(3n2-189n+ 1950-0) H-3
n2 – 63n + 650 = 0
n2 – 50n-13n +650 = 0
n(n – 50) – 13(n – 50) = 0
(n – 50) (n -13) = 0
n – 50 = 0 or n – 13 = 0
n= 50 or n= 13
When n = 50, the last term
Tn = a + (n – 1)d
= 93 + (49) (-3)
= 93 – 147
T50 =-54
When n = 13, the last term
Tn = a + (n – 1)d ”
=93 + (13-1) (-3)
= 93 – 36
T13 = 57
OR
If m times the mth term of an A.P is equal to n times its n,h term, show that (m + n)th term is zero.
Soln: Let T1 = a and c.d = d
Given : m times m,h term = n times nthterm
=> mTm = nTn
m[a + (m – 1)d] = n[a + (n – 1)d]
m[a +(m – l)d] – n[a+(n -l)d] = 0
m[a + md – d] – n[a + nd – d] = 0
ma + m2d – md – na – n2d + nd = 0
a(m – n) + d(m – n) (m + n) – d(m – n) = 0
(m – n) {a + d(m +n) – d} = 0 . (m – n) {a + (m + n – l)d} = 0
m-n = 0, a + (m + n-l)d = 0
m= n Tm+n , = 0
But m ≠ n
Question 36.
A tower is 50m high. Its shadow is x mtrs shorter when the suns altitude is 45° than when it is 30°. Find the value of x.
Let PQ be the tower and let PA and PB be its shadows when the altitudes of the sum are 45° and 30° respectively.
∠ PAQ = 45°, ∠ PBQ = 30°
∠ BPQ = 90° and PQ = 50m ,
Let AB = x metres
From right angled ∆ APQ, we have
$$\frac{\mathrm{AP}}{PQ}$$ = cot 45°=1
$$\frac{\mathrm{AP}}{50}$$ = 1 => AP = 50m
From right angled ∆ BPQ,
$$\frac{\mathrm{BP}}{\mathrm{PQ}}$$ = cot30° = √3
⇒ $$\frac{x+50}{50}$$ = √3
x = 50(√3 – 1)m
= 50 (1.73 – l)m
= 50 x 0.732
= 36.6 m
Question 37.
Prove that “In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides.”
Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.
VI. Answer the following : ( 5 x 1 = 5 )
Question 38.
The height of cone is 20m. A small cone is cut off from it at its top by the plane parallel to the base. If the volume of small cone 1/1000 th of the volume of given cone, at what height above the base the section is made.
In A OPB and A OQC A OPB ~aOQC OP _ PB OQ ~ QC
## Karnataka SSLC Maths Model Question Paper 3 With Answers
Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
In the following numbers, irrational number is
a) 0.232332333……
b) 0.23233
c) 0.232323
d) 0.2323
a) 0.232332333……
Question 2.
10 sec2A – 10tan2A =
a) sec2A
b) 10
c) 1
d) 0
b) 10
Solution:
10 sec2A – 10tan2A
10(sec2A – tan2A) = 10(1)= 10
Question 3.
The length of the tangent drawn to a circle of radius 3cm from 5cm away from the centre is
a) 4cm
b) 5cm
c) 3cm
d) 2cm 4
a) 4cm
Solution:
d2 = r2 + t2
t2 = d2 – r2
52 – 32
= 25 – 9
∴ t = 16 = 4cm
Question 4.
A solid piece of copper of dimension 24 × 49 × 33 cms is moulded and recast into a sphere. The radius of the
sphere formed is ________
a) 49 cm
b) 24 cm
c) 21 cm
d)33 cm
e) 21 cm
l × b × h = $$\frac{4}{3} \pi r^{3}$$
Question 5.
The degree of the polynomial in the graph given below is
a) 4
b) 3
c) 1
d) 2
a) 4
Solution:
4 since it is intersecting the x – axis at 4 points.
Question 6.
The sum of the n terms of an AP is 2n2 + 5n and its common difference is 6, then its first term is
a) 0
b) 5
c) 2
d) 7
d) 7
Solution:
Sn = 2n2 + 5n
S1 = 2(1)2 + 5(1)
= 2 + 5 = 7
Question 7.
In ∆ PQR, PR = 12cm, QR = 6√3 cm, PQ = 6cm. The angle Q is
a) 45°
b) 90°
c) 30°
d) 120°
b) 90°
Solution:
PR2 = PQ2 + QR2
122= (6√3 )2 + 62
144 = 108 – 36
144=144
∴ ∠Q = 90°
Question 8.
A cube numbered 1 to 6 is thrown once, the probability of getting a number divisible by 3 is
a) 2/3
b) 0
c) 1/3
d) 1
Event A = {3,6}
n(s) = 6, n(A) =2
P(A) = $$\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}$$
II. Answer the following Questions : (1 x 8 = 8)
Question 9.
Given g(x) = 2x + 1, q(x) = (x3 + 3x2 – + 1), r(x) = 0, Find p(x)
p(x) = g(x) q(x) + r(x) 1 650 1170
= (2x + 1) (x3 + 3x2 – x + 1) + 0
p(x) = 2x (x3 + 3x2 -x + 1) + 1(x3 + 3x2 – x + 1)
= 2x4 + 6x3 – 2x2+ 2x + x3 + 3x2 – x + 1
= 2x4 + 7x3 + x2 + x + 1
Question 10.
If the sum of first n odd natural number is 1225, find the value of n.
Sum of “n” odd natural number = n2 = 1225
n = √1225
n = 35
Question 11.
In the fig. ∠AOD is divided into 2 parts which are in A.P. the smallest angle ∠AOB = 20° . Find the common difference between each angle.
∠AOB +∠BOC + ∠COD = 180°
a + a + d + a + 2d= 180°
3a + 3d= 180°
3(20) + 3d = 180°
60 + 3d = 180°
3d = 180 – 60
d = 120/3 = 40°
Question 12.
In ∆ ABC, if DE || BC, then $$\frac{\mathbf{A B}}{\mathbf{A D}}=\frac{\mathbf{A C}}{\mathbf{A} \mathbf{E}}=\frac{\mathbf{B C}}{\mathbf{D E}}$$ , state the theorem to justify this.
In a triangle, if a line is parallel to one of the sides then the sides of given triangle are propotional to sides of the intercepted triangle.
Question 13.
Find the largest number which divides 650 and 1170.
H.C.F of (650, 1170) = 130
Question 14.
If sin θ = 7/25, cos θ = 24/25 find the value of sin2 θ + cos2 θ
Question 15.
Find the value of cos60° cos 30° sin60° sin 30°
cos 60° cos30 – sin 60 sin 30
OR
This is in the form of
cos A cos B – sin A sin B = cos (A + B)
cos 60 cos 30 – sin 60 sin 30
= cos (60 + 30)
= cos (90)
– =0
Question 16.
The T.S.A of a solid hemisphere of radius 21 mm.
T.S.A of hemisphere = 3 πr²
= 3 × $$\frac{22}{7}$$ × 21 × 21 = 4158 mm2
III. Answer the following : ( 2 x 8 = 16 )
Question 17.
Prove that if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4.
Let x = 2m + 1 and y = 2n + 1 for some integers m and n.
x2 + y2 = (2m + 1)2 + (2n + l)2
x2 + y2 = 4m2 + 4m + 1 + 4n2 + 1 + 4n
= 4m2 + 4n2 + 4m + 4n + 2.
x2 + y2 = 4(m2 + n2) + 4 (m + n) + 2
x2 + y2 = 4 {(m2 + n2) + (m + n)} +2
x2 + y2 = 4q + 2, where q = (m2 + n2 ) + (m+n)
⇒ x2 + y2 is even and leaves the remainder 2 when divided by 4.
⇒ x2 + y2 is even but not divisible by 4.
Question 18.
Solve : 100x + 200y = 700
200x + 100y = 800
Consider 100x + 200y = 700 ….(1)
200x + 100y = 800 ….(2)
Add (1) & (2) 300x + 300y= 1500
Divide by 300, x + y = 5 …… (3)
Subtract (1) and (2)
-100x + 100y= -100 Divide by 100.
-x + y = -1 …(4)
Solve (3) and (4)
y = 2
Substitute the value of y in equation (3)
x + y = 5
x + 2 = 5
x = 5 – 2 = 3
x = 3
Question 19.
Find the roots of the quadratic
equation 3x2 – 2√6x + 2 = 0 by formula method.
Question 20.
Find the value of x in which the points (1, -1) (x, 1) and (4, 5) are collinear.
A(1, -1) B(x, 1)C(4,5)
Area of the triangle = 0, when points are collinear.
0 = x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)
0 = (1) (1 – 5) + x(5 + 1) + 4(-1 – 1)
0 = -4 + 6x – 8
6x – 12 = 0
x = 12/6
x = 2
Question 21.
ABC is a right angled triangle, having [B = 90°. If BD = DC, Show that AC2 = 4AD2 – 3AB2
AC2 = AB2 + BC2 (Pythagoras)
AC2 = AB2 + 2(BD)2
AC2 = AB2 + 4BD2
In right angled ∆ ABD,
Consider AC2 = AB2 + 4BD2
= AB2 + 4(AD2 – AB2)
= AB2 + 4AD2 – 4AB2
AC2 = 4 AD2 – 3AB2
OR
Prove that area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
Data : ABCD is a square.
Equilateral triangles ∆ BCE and ∆ ACF have been described on side BC and diagonal AC respectively.
T.P.T. : Area of ( ∆ BCE) = 1/2 Area of ∆ ACF
Proof : Since ∆ BCE and ∆ ACF are equilateral.
∴ They are equiangular.
∆ BCE ~ ∆ ACF
Question 22.
A box contains 90 dices which are numbered from 1 to 90. If one dise is drawn at random from the box, find the probability that it bears
(i) two digit number
(ii) a perfect square number.
The numbers in the dise form the sample space
S = {1,2, 3,4,…….. 90}
One dise can be drawn out of 90 in 90 ways. .
n(s) = 90
i) There are 90 -9 = 81, two digit numbers, out of which one dise can be drawn in 81 ways.
∴ n(A) = 81
P(A) = $$\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}$$
ii) The perfect square numbers are B= {1,4,9, 16,25,36,49,64,81}
∴ n(B) = 9
P(B) = $$\frac{n(B)}{n(S)}=\frac{9}{90}=\frac{1}{10}$$
Question 23.
Draw a pair of tangents to a circle of radius 5cm which are inclined to each
other at an angle of 60°.
Question 24.
Prove that $$\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}$$
OR
Prove that (cosecθ – cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider (cosecθ – cotθ)2
IV. Answer the following : ( 3 × 9 = 27 )
Question 25.
The sum of the ages of A and B is 85 years. 5 years ago, the age of A was twice that of B. Find the present ages.
Let the present age of A be x years.
Let the present age of B be y years.
Sum of their ages = 85 x + y = 85 …….. (1)
Five years back, age of A was (x – 5) and that of B was (y – 5).
The age A is twice that of B.
∴ x – 5 = 2(y – 5) ……(2)
x – 5 = 2y – 10
x – 2y = -10 + 5
x – 2y = -5 ……(2)
Solve (1) and (2)
Substitute y in (1)
x + y=85
x + 30 = 85
x = 85 – 30 = 55
x = 55
∴ Present age of A is 55 years and B = 30 years.
OR
A piece of work can be done by 2 men and 7 boys in 4 days. The same piece of work can be done by 4 men and 4 boys in 3 days. How long it would take to do the same work by one man or one boy?
Let x and y be the number of days in which one man can complete the work.
∴ In 1 day a man can do 1/x th work and
In 1 day a man can do 1/y th work.
2 men and 7 boys can complete the work in 4 days.
∴ They can complete in 1/4 th of the work in one day.
∴ $$\frac{2}{x}+\frac{7}{y}=\frac{1}{4}$$ …… (1)
4 men and 4 boys together complete the work in 3days, and they can complete it in 1/3 rd of work.
4(2y + 7x) = xy
8y + 28x = xy …….(3)
3(4y + 4x) = xy
12y + 12x = xy ……..(4)
Solve (3) and (4)
(8y + 2x = xy) x 12
(l2y+ 12x = xy) x 18
4 y = 240
y = 240/4
y = 60
3(4y + 4x) xy
12y+12x = xy ………..(4)
4y = 240
y = 240/4 = 60
Substitute y in equation (3)
8y + 28x = xy
8(60) + 28x = xy
480 + 28x = x(60)
480 + 28x = 60x
480 = 60x – 28x
480 = 32x
x = $$\frac{480}{32}=\frac{120}{8}=\frac{30}{2}$$ = 15
x = 15
Thus, one man can complete the work in 15 days and one boy can do the work in 60 days.
Question 26.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 2y + 24 if it is given that Sum of the two zeros is Answer:
Let α, β,γ be the zeros of the given polynomial.
p(y) = y3 – 5y2 – 2y + 24
α + β = 7
Sum of the zeros
α + β + γ = 5
7 + r = 5
r = 5 – 7 = -2
r = -2
Sum of product of zeroes taken two at a
αβ + βr + rα = -2/1 = – 2
αβ + r(β + α) = -2
αβ + (7) (-2) = -2
αβ-14 = -2
αβ = -2+ 14
αβ = 12
(α – β )2 = (α + β)2– 4αβ
= (7)2 – 4(12)
(α – β)2 = 49 – 48
α – β = ±1
Solve for α and β. When α – β = +1
α + β = 7
α – β = 1
2α =8
α = 8/2 = 4
α = 4
Substitute the value of a in α + β =7
4+p =7
β =7-4 = 3
β = 3
Solve for α and β when α – β = -1
α = 6/2 = 3
α = 3
α + β = 7
3 + β=7
β = 7 – 3
a + β = 7
β = 4
When α – β = 1,
the values are α = 4, β = 3, r = -2.
When α – β = -1,
The values are α = 3, β = 4, r = -2.
Question 27.
The diagonal of a rectangular field is 60 metres more than the shorter side, if the longer side is 30 metres more than the shorter side, find the sides of the field.
Let the shorter D side be x mtr.
Longer side is 30m more than the shorter side. A Longer side = (x + 30)m.
Diagonal is 60m more than the shorter side.
Diagonal = (x + 60) m
AC2 = AB2 + BC2 (Pythagoras)
(x + 60)2 = (x + 30)2 + x2
x2+(60)2 + 2x (60) = x2 + (30)2 + 2(x)(30)+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
-x2 + 60x + 2700 = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0 or x + 30 = 0
x = 90 or x = – 30
∴ Shorter side = 90m = BC = x
Longer side = x + 30 = 90 + 30 = 120m
Diagonal = x + 60 = 90 + 60 = 150m
OR
Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of two squares.
Let the sides of the two squares be x&y. Sum of the areas of two squares = 468
x2 + y2 = 468 ……(1)
Difference of their perimeters = 24m.
4x – 4y = 24 x – y = 6
x = 6 + y …….(2)
Substitute the value of x in (1)
x2 + y2 = 468
(6 + y)2 + y2 = 468
36 + y2+ 12y + y2 = 468
2y2 + 12y + 36 = 468
Divide by 2
y2 +6y+ 18 = 234 y2 + 6y = 234 – 18
y2 + 6y – 216 = 0
y2+18y – 12y -216 = 0
y(y + 18)-12(y + 18) = 0
(y + 18) (y – 12) = 0
y+18 = 0, y -12 = 0
y = -18,y = 12
∴ x = 6+y = 6 + 12 = 18m
The sides of two squares are 18m and 12m.
Question 28.
Show that the points x(2, -2) y(-2, 1) and z(5, 2) are the vertices of a right angled triangle XYZ and also calculate its area.
⇒ XYZ is an isosceles right angled triangle.
Area of a right angled triangle = $$\frac{1}{2}$$ x base x height
$$\frac{1}{2}$$ × 5 × 5 = $$\frac{25}{2}$$ = 12.5 cm
OR
Find the values of k for which the points A(k + 1, 2k) B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
A(k + 1, 2k) = (x1, y1)
B(3k, 2k + 3) = (x2, y2)
C(5k-1, 5k) = (x3, y3)
Area of the triangle = 0, for the points to be collinear.
Area =[x1(y2 – y3) + x2(y3-y1) + x3(y1-y2)]
0 = $$\frac{1}{2}$${(k + 1) (2k+3 – 5k)+3k(5k- 2k) + (5k-1) [2k-2k-3]}
0= $$\frac{1}{2}$$]{(k+ 1) (-3k+ 3) + 3k (3k) + (5k- I)(-3)i
0 =$$\frac{1}{2}$${-3k2-3k + 3k + 3 + 9k2 – 15k+3}
0= $$\frac{1}{2}$${6k2– 15k + 6}
0 = 6k2 – 15k + 6
Divide by 3
2k2 – 5k + 2= 0
2k2 – 4k – 1k + 2 = 0
2k(k – 2) -1(k – 2) = 0
(k – 2) (2k -1) = 0
k – 2 = 0, or 2k- 1=0
k = 2 or k = 1/2
Question 29.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Data: ‘O’ is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that: ∠APB+ ∠AOB = 180°
Proof: In AAPO and ABPO
∠OAP = ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (∵ Common side)
OA = OB (∵ Radii of the same circle) According to RHS postulate AAPO ABPO
∠OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° + 90° = 180°
⇒ Opposite angles of OAPB quadrilateral are supplementary
∴ OAPB is a cylic quadrilateral
⇒ ∠APB + ∠AOB = 180°
Question 30.
Find the area of the shaded region where a circular arc of Radius 6 cm has been drawn with the vertex ‘O’ of an equilateral Triangle OAB of side 12cm as centre.
Area of the shaded Region = Area of circle area of sector OCDE + Area of equilateral
∆ le OAB
OR
From each corner of a square of side 4cm a quadrant of a circle of radius 1cm is cut and also a circle of diameter 2cm is cut as shown in Fig, Find the area of the remaining portion of the square.
Area of shaded region = Area of square – (Area of circle + Area of 4 quadrants)
The following table gives production yield per hectare of wheat of 100 farms of a
Question 31.
The following table gives production yield per hectare of wheat of 100 farms of a
village.
Change the distribution to a more than type distribution and draw its ogive.
Question 32.
If the median of the distribution given below is 28.5. Find the valucs of x and y.
Meadian = 28.5, lies in the C.I , 20 – 30
The median class = 20 – 30
l = 20, f = 20, c.f = 5 + h, h = 10, n = 60
Question 33.
Construct a triangle ABC with side BC = 7cm, ∠B = 45°, ∠C = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
V. Answer the following ( 4 × 4 = 16 )
Question 34.
Solve the pair of equations graphically.
x+y=3 and 3x-2y=4
x + y = 3
y = 3- x
3x – 2y = 4
-2y = 4 – 3x .
Question 35.
If the sum of first 8 terms of an Arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
Given S8 = 136, S15 = 465, S25 = ?
Use the formula Sn = $$\frac{n}{2}$$ [2a + (n -l)d]
s8 = $$\frac{8}{2}$$[2a + (s-l)d]
136 = 4 (2a + 7d)
∴ 2a + 7d = 136/4
2a + 7d= 34…….(1)
S15 = $$\frac{15}{2}$$[2a + (15-l)d]
465 = $$\frac{15}{2}$$[2a + 14d]
465= $$\frac{15}{2}$$ x 2(a + 7d)
∴ a + 7d = 465/15
a + 7d = 31 …….(2)
From (1) and (2)
Consider a + 7d = 31
3 + 7d = 31
7d = 31 – 3
d = 28/7
d = 4
∴S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (24)d]
S25 = $$\frac{25}{2}$$[2(3) + 24(4)d]
S25 = $$\frac{25}{2}$$ × 102
S25 = 1275
‘5 ‘ 9
OR
The sum of the 5th and 9th terms of an A.P is 40 and the sum of the 8th and 14th term is 64. Find the sum of the first 20 terms.
Given T5+ T9 = 40 and T8 + T14 = 64
a + 4d + a + 8d=40
2a + 12d = 40….(1)
a + 7d + a + 13d = 64
2a + 20d = 64…. (2)
Consider 2a + 12d = 40
2a + 12(3) = 40
2a + 36 = 40
2a = 40 – 36
a = 4/2
a = 2
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
S20 = $$\frac{20}{2}$$[2(2) + (20-1)3]
= 104(4 + 57)
= 10 × 61
= 610
Question 36.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on to the road, the angles of elevation of the top of the poles re 60 and 30’, respectively. Find the height of the poles and distances of the point from the poles.
In ∆ ABE.
tan60 = $$\frac{\mathrm{AB}}{\mathrm{BE}}$$
√3 = x/BE
x = √3BE ….. (1)
tan30 = DC/EC
$$\frac{1}{\sqrt{3}}=\frac{x}{E C}$$
EC = x√3
EC = BE × √3 .√3 [∵ x = √3.BE]
FC = 3BE
∴ EC=3BF
We know that
BE + EC = 80 rn
BE 3BE 80m
4BE = 80
BE = 80/4 = 20
⇒ EC = 3BE 3(20)= 60m
∴ x = BE.
x = 20√3
⇒ AB = CD = 20√3 m
∴ The distance from the point to the pole arc 20m towards left and 60m towards the right
poles
Height of the poles 20√3m
Question 37.
Prove areas of similar triangles.
Areas of similar triangles are proportional
to the squares on the corresponding sides.
In ∆ABM and ∆DEN
∠AMB = ∠DNE = 90 [construction]
∠B =∠E (Data)
∠BAM = ∠EDN (Remainingang1e)
∆ABM ∼ ∆ DEN
∆ABM & ∆DEN are equiangular
VI. Answer the following : ( 5 × 1 = 5 )
Question 38.
A circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the canvas used to make the tent when the cost per m2 of the canvas is ?15.
Total canvas used=CS A of cylindrical part -1- CS A of conical part
=2πrh + πrl
= 2 × $$\frac{22}{7}$$ × 63 × 5 + $$\frac{22}{7}$$ x 63 x 65
= 1980 m2 + 12870 m2
= 14850 m2
Total canvas used = 14850 m2
* Cost of canvas at the rate of ₹16 per m2
= 14850 × 15= ₹ 2,22.750.
Cost of canvas = ₹ 2,22,750
## Karnataka SSLC Maths Model Question Paper 2 with Answers
Students can Download Karnataka SSLC Maths Model Question Paper 2 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.
## Karnataka State Syllabus SSLC Maths Model Question Paper 2 with Answers
Time: 3 Hours
Max Marks: 80
I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )
Question 1.
For some integer n every odd integer is of the form
a) 2n + 1
b) n + 1
c) 2n
d) n
2n + 1
Question 2.
The value of sin215° + sin225° + sin265° + sin275° is
a) 0
b) 1
c) 2
d) 3
2
Question 3.
If chord AB subtends an angle 50° at the centre of a circle then the angle between the tangents at A and B is
a) 40°
b) 100°
c) 130°
d) 120°
130°
Question 4.
The formula used to find the volume of a sphere
a) $$\frac{4}{3} \pi r^{3}$$
b) $$\frac{2}{3} \pi r^{3}$$
c) $$\frac{1}{3} \pi r^{3}$$
d) πr3
a) $$\frac{4}{3} \pi r^{3}$$
Question 5.
α + β are the zeroes of the polynomial x2 – 6x + 4, then the value of $$\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}$$ is
a) 7
b) 8
c) -7
d) -8
a) 7
Solution:
Question 6.
If 29th term of an A.P is twice its 19th term, then the 9th term is
a) -1
b) 0
c) 1
d) 2
b) 0
Solution:
T29 = 2T19
a + 28d= 2(a + 18d)
a + 28d = 2a + 36d
0 = 2a + 36d-a-28d
0 = a + 8d
T9 = 0
Question 7.
In ∆ ABC, AB = 6√3 cm, AC = 12 cm, BC = 6cm, The angle B is
a) 45°
b) 90°
c) 60°
d) 30°
Solution:
AC2 = AB2 + BC2
122= (6√3 )2 + 62
144= 108 + 36 144=144
∴ ∠B = 90°
Question 8.
If the probability of an event is P(A) then the probability of its complimentary event will be
a) 1+P(A)
b) 1 – P(A)
c) P(A) – 1
d) 1 / P(A)
Solution:
We know that for any two complimentary events A and Ā
P(A) + P(Ā) = 1 → P(Ā) = 1 – P(A)
II. Answer the following questions : ( 1 × 8 = 8 )
Question 9.
If α and β are the zeroes of the quadratic polynomial 2 – 3x – x2 then what is the value of α + β + αβ ?
α + β = -b/a , αβ = c/a
= $$\frac{-(-3)}{-1}$$ αβ = $$\frac{2}{-1}$$ = – 2
= – 3
α + β + αβ = -3 -2 = -5
Question 10.
What are the roots of the quadratic equation x2 + (√3 + 1)x + √3 = 0?
2 + (√3 + 1)x + √3 = 0 = 0
x2 + x + √3x + √3 = 0
x(x + 1)+ √3 (x + 1) = 0
(x + 1)(x + √3) = 0
x + 1 = 0, x + √3 = 0
x = -1, x = -√3
Question 11.
If the nth terms of the two AP 9, 7, 5, ….. and 24, 21, 18, ……… are same. Find n.
9, 7, 5, …….
a = 9, d = 7-9 = -2 T = a + (n – 1)d
Tn = 9 + (n – 1) (-2)
Tn = 9 + (-2n + 2) ‘
Tn = 9 – 2n + 2
Tn = 11 – 2n
24,21,18,
a = 24, d = 21 – 24 = -3
Tn = a + (n – 1) d T=24 + (n – 1)(-3)
Tn = 24 – 3n + 3
Tn = 27 – 3n
Given : nth term of both A. P. is same.
∴ 11 – 2n = 27 – 3n
-2n +3n = 27 -11
n = 16
Question 12.
State A – A criterian theorem.
“If two triangles are equiangular, then the corresponding sides are proportional.”
Question 13.
Find the H.C.F of 455 and 42 with the help of Euclid’s division algorithm.
Step (1): a = 455 b = 42
a = bq + r
455 = 42 × 1 + 35
Step (2): a = 42, b = 35
a = bq + r
42 = 35 × 1 + 7
Step (3): a = 35, b = 7
a = bq + r
35 = 7 × 5 + 0
∴ H.C.F. (455, 42) = 7
Question 14.
Find θ if sin (θ + 56) = cosθ, where θ and ( θ + 56) are less than 90°.
sin (θ + 56) = cos θ
sin (θ + 56) = sin (90 – θ)
θ + 56 = 90 – θ
θ + θ = 90 – 56
2θ = 34
θ = 34/2
θ = 17
Question 15.
If x = a sin θ and y = b tan θ, then find the value of $$\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}$$
Consider
Question 16.
Calculate the height of a right circular cone where C.S.A. and base radius are 12320 cm2 and 56 cms. respectively.
C.S.A of a cone = π rl
12320 = $$\frac{22}{7}$$ x 56 x 1
1 = $$\frac{12320 \times 7}{22 \times 56}$$
1 = 70 cm
But 12 = r2 + h2
702 = 562 + h2
h22 = 702 – 562
h2 = (70 + 56) (70 – 56)
h2 =(126) (14)
III. Answer the following ( 2 × 8 = 16 )
Question 17.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is an integer.
Let nbe any positive, integer using division algorithm,
a = bq + r
Taking a = n, b = 6
n = 6q + r
If any number is divided by 6, then the possible remainders are 0, 1, 2, 3, 4 or 5.
∴If n is odd, then r = 1, 3, 5
⇒ 6q + 1, 6q + 3, 6q + 5 are the positive odd integers.
Question 18.
Solve : 2x + 3y = 9
3x + 4y = 5
Substitute y = 17 in 2x + 3y = 9
2x + 3(17) = 9
2x + 51 =9
2x = 9 – 51
x = $$\frac{-42}{2}$$
= -21
Question 19.
Solve : (x – 2)2 + 1 = 2x – 3
x2 + 4 – 4x + 1 = 2x – 3
x2 – 4x + 5 – 2x + 3 = 0
x2 – 6x + 8 = 0 x2 – 4x – 2x + 8 = 0
x(x – 4) -2(x – 4) = 0
(x – 4) (x – 2) = 0
x – 4 = 0, x -2 = 0
x = 4 or x = 2
Question 20.
Show that the points (-2, 1) (2, -2) and (5, 2) are the vertices of a right angled triangle.
∴ ∆ ABC is a right angled triangle at B.
Question 21.
The equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.
∴ Area of ∆ XAB + Area of ∆ YBC = Area of ∆ ZAC .
OR
In the given figure, PA, QB and RC are each perpendicular to AC. Prove $$\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$$
Question 22.
Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 4 nor by 5.
S = {(1, 1) (1,2) (1, 3) (1,4) (1, 5) (1, 6) (2, 1) (2, 2) (2,3) (2,4) (2, 5) (2,6) (3.1) (3, 2) (3, 3) (3, 4) (3, 5) (3,6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4,6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5,6) (6.1) (6,2) (6, 3) (6,4) (6, 5) (6,6)}
n(S) = 36.
An event of getting sum of the numbers neither divisible by 4 nor by 5.
A={ 1,1) (1,2) (1,5) (1,6) (2,1) (2, 4)
(2, 5) (3, 3) (3, 4) (3, 6) (4, 2) (4,3) (4, 5) (5,1) (5, 2) (5, 4) (5, 6) (6,1) (6, 3) (6, 5)} n(A) = 20
n(A) = 20
P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{36}$$
Question 23.
Draw a circle of radius 3 cm. Take a point P outside the circle without using the centre of the circle, draw to tangents to the circle from an external point P.
Question 24.
Prove that (cosecθ-cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider LHS
(cosecθ-cotθ)2 = cosec2 θ +cot2 θ – 2cosecθ.cotθ
OR
If sinθ + cos θ = √2 sin (90 -θ) determine cot θ
sin θ + cos θ = √2 sin (90 – θ)
sin θ + cos θ = √2 cos θ
sin θ = ypi cos θ – cos θ
sin θ = cos θ (√2 -1)
sinθ /cosθ = √2 -1
IV. Answer the following : ( 3 × 9 = 27 )
Question 25.
Asha is 5 times as old as her daughter Usha, 5 years later Asha will be 3 times as old as her daughter Usha. Find the present ages of Asha and Usha.
Let the age of Asha be x years
Let the age of Usha be y years
According to question x=5y
x – 5y = 0 …….(1)
Five years later,
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 15 – 5
x – 3y = 10 (2)
Substitute x = 5y in equation (2)
5y – 3y = 10
2y = 10
y = 10/2
y = 5
∴ x = 5y
= 5 × 5
x = 25
∴ The present age of Asha is 25 years.
The present age of Usha is 5 years.
OR
The sum of 2 digits of a 2 digit number is 12 the number obtained by interchanging the digits exceeds by the given number by 18. Find the number.
Let the two digits be x and y
The 2 digit number will be 10x + y
The sum of digits = x + y – 12 ……..(i)
The number obtained by interchanging the digits =10y + x
= 10x + y + 18
10y + x – 10x – y = 18
9y – 9x = 18
Divide by 9
y – x = 2
y = x + 2 ….(ii)
Consider equation (1)
x + y = 12
Substitute y = x + 2
x + x + 2 = 12 .
2x =12 – 2
2x= 10
x= 5
y = x + 2
y = 5 +2
y = 7
∴ The number is : 10x +y
= 10(5)+7 = 50+ 7
= 57
Question 26.
Find the other two zeroes of the polynomial y4 + y3 – 9y2 – 3y + 18 if the
zeroes are √3 and -√3
y = √3 and y = -√3
(y + √3) = o (y-V3) = o
(y + √3 )(y – √3 ) = o
y(y – √3 ) + √3 (y – √3 ) = 0
y2 – √3y + √3 y – 3 = o
y2 – 3 = 0
y2 + y – 6 = 0
y2 + 3y – 2y – 6 = 0
y(y + 3)-2(y + 3) = 0
(y + 3) = 0 & (y – 2) = 0
y=-3 & y = 2
The four zeros of polynomial are √3 , -√3 , -3 & 2
Question 27.
Solve for x.
$$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$$ (Where a ≠ 0, b ≠ 0 x ≠ 0, x ≠ -a,-b
(-a – b)ab = (a + b)x (a + b + x)
-a2b – ab2 = (a + b) (ax + bx + x2)
-a2b – ab2 = a2x + abx + ax2 + abx + b2x + x2b
-a2b – ab2= a2x + 2abx + ax2 + b2x + x2b
-ab(a+b) = ax(a+x) + xb(b+x)+2abx
=x[a(a+x)+b(b+x)+2ab]
=x[a2 + ax+b2 +bx + 2ab]
= x[(a+b)2 + x(a+b)]
-ab(a + b) = x(a+b)[a+b+x]
-ab = x(a + b + x)
-ab = ax + bx + x2
x2 + x (a + b) + ab = 0
x2 + ax + bx + ab = 0
x(x+a) + b(x + a) = 0
(x + a) (x +b) = 0
=>x + a = 0, x + b = 0
=> x = -a, x = -b
OR
The diagonal of a rectangular field is 60m more than the shorter side. If the larger side is 30m more than the shorter side, find the sides of the field.
Let the shorter side of the rectangular field be x m.
Longer side is x + 30 and the diagonal is x+60
In ∆ ABC,
AC2 = AB2 + BC2
(x + 60)2 = ( x + 30)2+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
x2 + 900 + 60x – 3600 – 120x = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0, x + 30 = 0
x = 90, x = – 30.
∴ The shorter side = x = 90m
The longer side = x + 30 = 90+30 = 120m.
Question 28.
If the points (7, -2) (5, 1) and (3, 5) are collinear. Find the value of k.
Since the give points are collinear, the area of the triangle formed by them must be O i.e.
= $$\frac { 1 }{ 2 }$$ [x1(y2-y3) + x2(y3-y2) + x3(y1-y2)]
x1 = 7, x2 = 5, x3 = 3
y1 = – 2, y2 = 1, y3 = k
= $$\frac { 1 }{ 2 }$$ [7(1 – k) + 5 (k + 2) + 3(-2-1)] = 0
= $$\frac { 1 }{ 2 }$$ [7 – 7k + 5k + 10 – 9]-0
= $$\frac { 1 }{ 2 }$$ [7 – 2k + 1] = 0
= $$\frac { 1 }{ 2 }$$ [8 – 2k] = 0
8-2k = 0 or 2k=8
k = 8/2 = 4
k = 4
OR
Find the area of Rhombus if its vertices are (3, 0) (4, 5) (-1, 4) and (-2, -1) taken in order.
Area of rhombus = $$\frac { 1 }{ 2 }$$ d1 × d2
Let A = (3, 0) B = (4, 5)
C = (-1, 4) D = (-2, -1)
Diagonal
Question 29.
Prove the tangents drawn from an external point to a circle are equal.
Data: A is the centre of the circle B is an external point.
BP and BQ are the tangents.
AP, AQ and AB are joined.
To prove that: a) BP = BQ
b) ∠PAB-∠QAB.
c) ∠PBA = ∠QBA
Proof: In ∆ APB and ∆ AQB
AP = AQ [Radii of same circle]
∠ APB – ∠AQB = 90°
AB = AB [common side]
∴ ∆ APB ≅ ∆ AQB [RHS]
a) PB = QB
b) ∠PAB = ∠QAB
c) ∠PBA = ∠QBA
Question 30.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm
Area of square ABCD
= 14 × 14=196 cm2
Hence, area of the shaded region
= Area of square – Area of four circles
= 196 – 154
= 42cm2
OR
Find the area of the shaded regions. Given PQRS a square of sides 14cm. Soln: S R
Question 31.
The following table gives the weight of 120 articles :
Change the distribution to a “more than type” distribution and draw its ogive.
Question 32.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)
Question 33.
Construct a right triangle whose hypotenuse and one side measures 10cm and 8cm respectively. Then construct another triangle whose sides are 4/5 times the corresponding sides of the triangle .
V. Answer the following ( 4 × 4 = 16 )
Question 34.
Solve the pair of equations graphically.
x + y = 8 and x – y = -2
Answe:
x + y = 8
y = 8 – x
x – y = -2
x = -2 + y
x = 6/2 = 3
x + y = 8
3 + y = 8
y = 8 – 3
y = 5
Question 35.
Devide 20 into four parts which are in arithmetic progression and such that the product of first and fourth is to the product of second and third in the ratio 2:3.
Let the four parts in A.P be :
a – 3d, a – d, a +d, a + 3d
Their sum = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20 or a = 5
Also, it is given that
or 75 -27d2 = 50 – 2d2
25d2 = 25 .
⇒ d2 = 1 or d = +√1 = ±1
When a = 5, d = 1, the four parts are 2, 4, 6,8
When a = 5, d = -1; the four parts are 8, 6, 4,2
Hence the four parts are (2, 4, 6, 8) or (8, . 6,4,2)
OR
The angles of a quadrilateral are in AP such that the greatest is double the least calculate all the angles of the quadrilateral.
The angles of the quadrilateral are in AP.
∴ The four angles are A, B, C and D which are a -3d, a – d, a + d, a + 3d respectively.
⇒ The least angle is a – 3d and the greatest angle is a + 3d.
It is given that
a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
2a – a – 6d – 3d = 0
a – 9d ……(1)
9d = a
But ∠A + ∠B+∠C + ∠D = 360°
a – 3d + a – d + a + d + a +.3d = 360°
4a = 360 / 4 = 90°
But 9d = a
d = $$\frac{a}{9}=\frac{90}{9}$$ = 10°
∴ ∠A=a- 3d = 90 – 3(10) = 90 -30 = 60°
∠B = a – d = 90 – 10 = 80°
∠C = a + d = 90 + 10= 100
∠D = a + 3d = 9 + 3(10) = 90 + 30 = 120
Question 36.
A person on the lighhouse of height 100m above the sea level observes that the angle of depression of a ship sailing towards the light house changes from 30° to 45°. Calculate the distance travelled by the ship during the period of observation. (Take √3 ≈ 1.73)
P is the top of the light house PQ. We are given that its height PQ = 100m PX is horizontal line through P
∠APX = 30° and ∠BPX =45°
Then, ∠PAQ =30° and ∠PBQ = 45°
Suppose, AB = x metre and BQ = y metre
Karnataka SSLC Maths Model Question Paper 2 with Answers – 40
⇒ x + y =100√3 = 100√3 = 100× 1.732
⇒ x +y = 173.2
⇒ x + 100 = 173.2
⇒ x = 73.2
Therefore, the distance travelled = 73.2m
Question 37.
Prove that “the ratio of areas of two similar triangles is equal to the square of the ratio of their altitudes.
Data : ∆ABC ∼ ∆DEF
∠A = ∠D
∠B = ∠E
∠C = ∠F
$$\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}$$
T.P.T : $$\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$$
Construction:Draw AM ⊥ BC and AN ⊥ EFim
VI. Answer the fallowing : ( 5 × 1 = 5 )
Question 38.
The radii of the circular ends of the frustrum of height – 6cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area of frustrum.
R = 14cm, r = 6cm and h =6cm
Now, let 1 be the slant height of the frustrum;
Now, lateral surface area of the frustrum
= π(R + r)l
= π (14 + 6) 10
= 628.57 cm2
And total surface area of the frustrum = π[(R+r)l + R2+ r2]
= π [(14 + 6)10 + (14)2 + (6)2]
= π(200 + 196 + 36)
= π(432)
= 1357.71 cm2.
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1 CONFIDENCE INTERVALS I ESTIMATION: the sample mean Gx is an estimate of the population mean µ point of sampling is to obtain estimates of population values Example: for 55 students in Section 105, 45 of 55 work: p s = 8%; for those who work, the mean number of hours xg = inference: 8% of ASU students work an average of hours a week; p = 0.8 and µ = Problem: the sampling distribution is a continuous distribution the probability that Gx actually equals µ is zero Gx is not an accurate estimate of µ; in this case we cannot even state the probability that Gx is accurate Gx is called a point estimate of µ statisticians generally prefer to give an interval estimate: "There is a 90% probability that µ is between 1 and 17.5." interval estimate has two features the estimate in interval form a probability statement: taken as an assessment of the reliability or accuracy of the estimate the probability hinges on the probabilities found in the sampling distribution of Gx INTUITION Consider the sampling distribution of sample means for samples of size 100 drawn from a population of salaries in which µ = 33,000 and σ = 5,000 E(Gx ) = 33,000 σ xg = 5, = 500.
2 In the z table find z values that demarcate the middle 95% of a normal distribution: z = ± 1.96 The interval µ 1.96 σ xg to µ σ xg contains 95% of the sampling distribution or contains 95% of all the possible Gx s that could ever be drawn from this population the interval noted is 33,000 ± or 33,000 ± 980. Any sample mean in this interval differs from the actual population mean by no more than % of all possible sample means differ from µ by no more than 980. for any Gx, there is a 95% probability that it differs from µ by no more than 980; that is, by no more than the amount 1.96 σ xg choose a sample and calculate Gx; now consider an interval of the form Gx ± 1.96 σ xg necessarily, the population mean m lies within the limits of 95% of all such intervals there is a 95% probability that the population mean lies within the limits of any interval of the form Gx ± 1.96 σ xg A C% confidence interval: An interval of the form A C% xg ± confidence z C σ interval xg for the population mean is given by where z C is chosen so that C% of the normal distribution Gx ± z C σ lies within the interval xg z C to +z C. CALCULATING CONFIDENCE INTERVALS FOR THE POPULATION MEAN C is the confidence level z C is found as a z value such that z C to +z C incorporates the middle C% of a normal distribution; ±z C demarcates a symmetric interval which has area C
3 Example: Find the appropriate z value for a 9% confidence interval the interval must be symmetric: take out the middle 9% leaves 8% to be split between upper and lower tails. The required z values demarcate the lower 4% and lower 96% of the z distribution. Alternatively, let L = (100 C)/; here L = (100 9)/ = 4% or In the cumulative z table find area The closest seems to be ; reading back to the margins z = 1.75; therefore the required z C = 1.75 Check by finding a z value such that 96% of the distribution is less than that value. Examples: A population of Christmas trees has unknown µ, but it is known that the population is normally distributed with σ = 4. A sample of 5 trees has Gx = Find a 95% confidence interval for the mean height of the population. given n = 5, so that σ xg = σ n = 4/5 = 0.8 L = (100 C)/ = 5/ =.5% or From the z table 0.05 of the z distribution is less than 1.96, so z C = applying the formula above Gx ± z C σ xg 16.6 ± ± 1.568, or the interval to Stating the interval: A 95% confidence interval for the population mean is 16.6 ± We are 95% confident that the population mean is in the interval to There is a 95% probability that µ is at least but no more than Find a 90% confidence interval for the same population, same sample find z 90 by reference to z table. L = (100 C)/ = the nearest entry is z = then we have 16.6 ± = 16.6 ± 1.31, or the interval to For a sample of 64 drawn from this population, we got the same Gx. Find a 90% confidence interval for the population mean. since n = 64, σ xg = 4 64 = 0.5 confidence interval: 16.6 ± = 16.6 ± 0.8, or the interval to 17.4
4 Messages: the width of the confidence interval varies in the same direction as the confidence level in our first example, width = = 3.136, while in the second example, width = =.64 width of the interval is z C σ xg : called the precision of the estimate there is a trade-off between precision and confidence common sense: for very wide intervals, we can be quite confident that we've captured µ, but as the interval narrows, the probability that it includes µ drops to zero As the sample size increases, precision increases at the same level of confidence the third interval above has width 1.64 with sufficiently large sample, we can achieve whatever combination of confidence and precision we desire as n increases, σ xg decreases FINDING THE RIGHT SAMPLE SIZE The distance e = z C σ xg is the error in the estimate e is one-half the width of the confidence interval within the limits of our confidence statement, we are sure that the population mean differs from the sample mean by no more than e: we might say we're 90% confident that the true mean differs from the sample mean by no more than e. Hence, e is the maximum error in the estimate Suppose that there is some maximum tolerable value for e, or maximum tolerable error for a given confidence level, the value of n necessary to keep e within tolerable limits σ e = zc, solve for n to find n z n = C σ e for given z, chosen for the appropriate confidence level, this formula gives us the sample size necessary to achieve an error of no more than e in general, the result of this calculation is not an integer, so the rule is to make the sample size equal to the next largest integer. NOTE CAREFULLY: This refers to the maximum tolerable error in the sampling procedure, or in the estimate of µ, NOT to the tolerance in a manufacturing process.
5 Examples: Cigarette filters are supposed to have µ = 15 mm in length; σ = 0.1 mm. Machinery will jam if the length of a filter exceeds 15.3 mm, and the probability of such a filter increases as the mean length increases; must have an accurate estimate of the mean length of filters. Let us require e 0.01 mm. and 90% confidence intervals. How large must n be? z n = C σ = e = the next greatest integer is the required sample size (that is, ALWAYS round n upwards in these problems); here n = 69 ordering T-shirts to give to contestants in a road race; average chest size unknown but for all chests everywhere σ = 4 in. Measure a sample of the participants when they register, and require that the sample be accurate to within ±1.5 in. How large must the sample be to have 99% confidence in the result? first, z 99 =? n = (.58 4 / 1.5) = rounding upwards, we require n = 48
6 CONFIDENCE INTERVALS II: σ UNKNOWN WHEN TO USE A z VALUE IN CONSTRUCTING CONFIDENCE INTERVALS To this point, we have assumed the population standard deviation known. IF NOT Population is normally distributed and σ NOT known the sampling distribution of Gx is NOT normal but rather conforms to Student's t distribution If population is NOT normal and σ is NOT known but the sample is large (that is, n 30), then the sampling distribution of Gx approximates the t distribution In either of these cases, s, the sample standard deviation, estimates σ. RECALL: s = Σ(x xg ) /(n 1) and s = s The standard error of the mean is estimated by s xg = s/ n confidence intervals have the form Gx ± t C s xg the t values used here are numbers of standard deviations in this case, numbers of standard deviations on a t distribution CHARACTERISTICS OF THE t DISTRIBUTION Continuous Symmetric Values near the mean are more probable than values further out so that t distribution looks like a bell-shaped curve. How is that any different from a normal distribution? 1. the t distribution has fatter tails and less mass in the center for a given number of standard deviations, probability is higher on the normal distribution than on a t distribution put another way, a given probability level will be further from the center of a t distribution that from the center of the normal distribution or, a given probability level will be more standard deviations (t values) away from the mean than would be the case on a normal distribution Note: t values will always be larger than z values for corresponding confidence level intervals constructed with t will always be wider (less precise) than those constructed with z
7 . there is not one t distribution but a large number, depending on the number of "degrees of freedom" Digression: the concept of degrees of freedom Mechanically df = n k, where n is the sample size and k the number of parameters that must be estimated from the sample before estimating the standard deviation for example: s, the sample standard deviation, is an estimate of σ. To calculate s, we must estimate µ. µ is estimated by xg, and xg is the only statistic we must calculate before we can calculate s. We must thus estimate one parameter, µ, before deriving and estimate of σ, and there are thus n 1 degrees of freedom in our estimate of σ more generally, degrees of freedom represents the number of independent (in the probability sense) random variables in a problem in calculating s we must use Gx. Suppose we are given Gx and n 1 of the values in the sample; then the n-th value is already determined and can be derived from what we know The t-distribution: pages E-7 and E-8 in your textbook how to read the table Upper tail (α) values across the top are the area in one tail of the distribution for a confidence interval use an upper tail value corresponding to the area in one tail of the distribution this will be only half the difference between the confidence level and 1 For example: in preparing a 95% confidence interval, there will be 5% in the tails of the distribution, thus 0.05 in each tail: we should use a t value for upper-tail area 0.05 and the appropriate number of degrees of freedom if C is the confidence level, expressed as decimal fraction, use α = (1 C)/ degrees of freedom are in the left hand column as df infinity, the t-value z value Examples: Find the appropriate t value for 0 degrees of freedom and 90% confidence interval. α = (1 0.9)/ = 0.05 t = for a sample of size 37, find the t value for a 99% confidence interval d.f. = n 1 = 36; α = (1 0.99)/ = t =.7195 CONFIDENCE INTERVAL FOR µ WITH NORMAL POPULATION AND σ UNKNOWN Problem requires use of t with n 1 degrees of freedom. Confidence intervals will have the form ( n 1) d. f. C x ± t s where s xg = s/ n, s being the sample standard deviation note similarity to earlier confidence intervals x
8 Examples: 7 male students are selected at random and an alcoholic beverage is poured down them in tenth-ounce increments until distinct signs of non-sobriety are observed. The following results were obtained: Individual Amount of Beverage (oz) Researchers feel safe in assuming that the distribution of ounces until non-sobriety is normal in the population. Construct a 95% confidence interval for amount of drink it takes to get the average member of the population drunk. calculate Gx and s: Gx = 3.39, s = Σ(x xg ) /(n 1) = [( ) + + ( ) ] (7 1) = s = = calculate s xg = s/ n = 0.846/ 7 = 0.846/.65 = find appropriate t value, for c =.95 and 6 df =.4469 multiply s xg by t value = Gx ± t s xg = 3.39 ± or the interval.546 to Each of 9 cars in a sample is driven 0,000 miles, the gallons of fuel used recorded, and the fuel mileage calculated. For the sample mean fuel mileage Gx = 34.6 and s = 1.. Assuming that the distribution of fuel mileage is normally distributed, find a 90% confidence interval for the mileage to be expected from all cars of this make. s xg = s/ n = (1.)/3 = 0.4 α = (1.9)/ = 0.05 and d.f. = n 1 = 8 t = Gx ± t s xg = 34.6 ± ± or the interval to In a sample of 41 students who work, xg = and s = Find a 95% confidence interval for the average hours worked by all ASU students who work. s xg = s n = = for 40 degrees of freedom, t 95 =.011 confidence interval: ± ± 1.8
9 We wish to establish the average weight of a population of turkeys; we have chosen a sample of 36, weighed them and have the following results: Construct a 98% confidence interval for the population mean of these turkeys first, find t C =.438 next, find Gx and s: Gx = 14.5, s = 4.90 find s xg = s/ n = 4.90/6 = Gx ± t C s xg 14.5 ± ± 1.99 or 1.6 to 16.4 SAMPLING DISTRIBUTIONS FOR SMALL SAMPLES The t distribution is often thought of as primarily of value with small samples applies whenever population is known to be normal and σ unknown, no matter how small n footnote: who was "Student"? A pseudonym for William Gosset, an Irish brewmaster concerned with controlling biochemical processes in brewing with large samples, if population is not normal, we must rely on Central Limit Theorem And many statisticians and other practitioners will use z procedures with any sample of 30 or more: this is especially prevalent in older practice Another possibility: sample is small, so that CLT does not apply population is not normally distributed or the distribution is unknown Safest course is to take a larger sample and rely on CLT
10 Following schematic may be used to determine proper distribution to use in constructing confidence intervals. Population standard deviation known? Yes No Population normal? Population normal? Yes No Yes No Sample Size Sample Size z value n >= 30 n < 30 n >= 30 n < 30 NOTES: z or t (see note) ERROR t value z or t (see note) ERROR 1. For a non-normal population and large samples, different practitioners may proceed differently. Some argue that the Central Limit Theorem justifies use of a z value in this case, while others feel that it is more appropriate to use a t value since that gives a less precise estimate (a wider confidence interval). For purposes of this course, use a t in such cases.. For small samples from non-normal populations: there are techniques which can be used to derive an interval estimate in this case, but they are beyond the scope of this course.
11 CONFIDENCE INTERVALS III CONFIDENCE INTERVAL FOR THE POPULATION PROPORTION Purpose: to use the sample proportion, p s, as the basis of an interval estimate of the population proportion p Reminders: the sample proportion p s = x/n the sampling distribution of p has parameters E(p s ) = p σ ps = p (1 p)/n p s is normally distributed, so that probabilities are found by reference to the z table typically p is unknown, so that we must estimate σ ps by s ps = [p s (1 p s )]/n A confidence interval for p then will have the form p s ± z C s ps Examples: of 55 students in a sample, 45 work. Construct a 95% confidence interval for the proportion in the population who work. p s = 45/55 = 0.8 s ps = [.8 (1.8)]/55 = z C = ±1.96 confidence interval: 0.8 ± ± 0.10 We are 95% confident that in the population somewhere between 7% and 9% work. In a sample of 800 North Carolinians 51% express the intention to vote for Jesse Helms in the next election. Find a 98% confidence interval for the proportion in the population who intend to vote for Helms. p s = 51%; s ps = (51 49)/800 = then we have 51 ± = 51% ± 4.1% or the interval 46.9% to 55.1% from this, we can say, strictly and properly, "We are 95% sure that the proportion in the population who intend to vote Helms is within 4.1% of 51%." or, as we might loosely and a bit improperly put it, "Our survey shows that 51% of the population intend to vote Helms, and this result is accurate to within plus or minus 4%." the election is a toss-up or too close to call.
12 Suppose same result with a sample of size n = 1600 s ps = 1.497, and s p z = =.9% confidence interval would be 51% ± 3% POINT: is the minor increase in precision worth the extra cost? FINDING THE NECESSARY SAMPLE SIZE IN PROPORTION PROBLEMS Since we have ± z C s ps, the estimate p s differs from p by at most that amount substituting the definition of s ps, the error is at most z C [p (1 p)]/n notice the use of p in the above expression; the concepts advanced here involve what we know about the sampling distribution before sampling begins for a given confidence level, this error can be reduced by increasing n in the last example above, we noted that doubling the sample size would reduce the error from 4% to 3% suppose we require e < 0.01, that is, accuracy to within ± 1%. How large must n be? the maximum error in the estimate: solve for n, giving n = p (1 e = e z C p) z C p ( 1 p) n
13 A major problem: p, the population proportion is unknown solution 1: assume p = 0.5 this will give largest possible value for n since p (1 p) reaches a maximum when p = 0.5 may result in an unnecessarily large and expensive sample solution : use other information do a pilot study on a small sample and use the resulting p s to estimate p previous experience or knowledge of other populations may give an approximate value for p lacks certainty of solution 1, but may result in somewhat smaller sample Examples: applying the formula above to solution 1, we have. 5 (1.5).33 = 0.01 n = 13,573 this is the sample size necessary to be absolutely sure that a 98% confidence interval is accurate to within ± 1% In the work example above, 95% confidence interval and sample of 55 gave accuracy of ±0.10. What sample size is necessary to hold the error to ±0.015 (1.5%)? solution 1: n = [( ) 1.96 ] = ; taking the next greatest integer, we have 469 solution : for n = 55, we had p s = 0.8. Take that as an estimate of the unknown p. Then n = [( ) 1.96 ] = or 50 using the pilot-study approach reduces the required sample size by more than 1,749 and might save a considerable amount of money A footnote: in most proportion problems, it doesn t matter whether you use percentages or decimal fractions, as long as you keep them straight. In the sample-size formula above, however, you must use decimal fractions. To use percentages, substitute 100 for 1, so the formula becomes n = [p* (100 p*) z ] e* where p* and e* are defined as percentages. THE z VS. THE t DISTRIBUTION In constructing confidence intervals, use the z distribution whenever the population standard deviation σ is known AND the population is known to normally distributed you wish to calculate a confidence interval for a proportion rule of thumb: n p 5 AND n (1 p) 5 for sufficiently accurate approximation In constructing confidence intervals, use the t distribution if the population is known to be normally distributed AND the population standard deviation σ is UNKNOWN: this holds for any sample size if the population s distribution is NOT normal AND the sample size is at least 30 AND the population standard deviation σ is UNKNOWN
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Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will | 10,638 | 45,230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-04 | latest | en | 0.917482 |
https://www.physicsforums.com/tags/max/ | 1,718,936,350,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00759.warc.gz | 826,526,079 | 28,857 | # What is Max: Definition and 1000 Discussions
Max Payne is a neo-noir third-person shooter video game series developed by Remedy Entertainment (Max Payne and Max Payne 2) and Rockstar Studios (Max Payne 3). The series is named after its protagonist, Max Payne, a New York City police detective turned vigilante after his family was murdered by drug dealers. The series' first and second installments were written by Sam Lake, while Max Payne 3 was primarily written by Rockstar Games' Dan Houser.
The first game of the series, Max Payne, was released in 2001 for Microsoft Windows and in 2002 for PlayStation 2, Xbox, and Apple Macintosh; a different version of the game was released for the Game Boy Advance in 2003. A sequel entitled Max Payne 2: The Fall of Max Payne was released in 2003 for PlayStation 2, Xbox and Microsoft Windows. In 2008, a movie adaption, loosely based on the original game, entitled Max Payne, was released, starring Mark Wahlberg and Mila Kunis in the roles of Max Payne and Mona Sax, respectively. Max Payne 3 was developed by Rockstar Studios and released on May 15, 2012 for PlayStation 3 and Xbox 360, and on June 1, 2012 for Microsoft Windows.
The franchise is notable for its use of "bullet time" in action sequences, as well as being positively received by critics, although Max Payne 2's sales were considered underwhelming. As of 2011, the Max Payne franchise has sold over 7.5 million copies. The film rendition received negative reviews but was commercially successful.
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1. ### Finding max velocity for a kart on a circular, banked track
This is a UK A-Level question that I'm really struggling with, and can't seem to find any resources online that explain it well. I've been given the following details: mass of gokart + driver = 520kg radius of track = 42m Maximum frictional force between tyres and road on flat track F = 20%...
2. ### Why does the war-wolf reach its maximum speed at the vertical position?
For this problem, For part(a) the solution is, However, how did they know that the max speed is reached in the vertical position? For part (d) the solution is, However, I thought the solution would be because there is net gravitational torque in clockwise direction (torque due to small mass...
3. ### Finding the launch angle for a projectile if the range is 3x the max height of the trajectory
For this problem, The solution is, I don't understand how they got from ##\frac {sin^2\theta_i}{sin2\theta} ## to ##\frac{tan\theta_i}{2}##. Would somebody please show me the immediate steps? I also don't understand is why they say the ignore the trivial case where ##\theta_i = 0~rad##...
4. ### Why is my FPGA not working correctly when programmed with a POF file?
Hello, i am programming my fpga with certain logic. With the sof file the logic is working fine and correct. but when i am programming pof file i am getting error. (programming is successful, but doesn't work as per the logic).
5. ### Max speed of proton in cyclotron
For this problem, The solution is, However, I don't understand why they say just before the proton escapes? Are they assuming that the B-field is only at a portion of region which means that only half or so of the circular path is within the B-field so only half of the path have a magnetic...
6. M
### C/C++ Max length of substring-palindrome
Hi PF! I'm trying to find the max length of a sub-string palindrome. My logic is to have a left-most index called start that starts from string loc 0 and traverses to the end of the string via a for loop. Next I was thinking a while loop as long as L < R: I'd have a left index L=start and a...
7. ### I Finding Max and Min Extremes of a Function with Second Derivatives Equal to Zero
What should I do when the f(x, y) function's second derivatives or Δ=AC-B² is zero? When the function is f(x) then we can differentiate it until it won't be a zero, but if z = some x and y then can I just continue this process to find what max and min (extremes) it has? What I've done is...
8. ### Time to reach Max. Height for a kicked soccer ball
Vvertical=15sin30 = 7.5m/s d=Vvertical*t+0.5atsquared 0=(7.5m/s)t+0.5(-9.81m/s)t squared 0=t(7.5m/s-4.9t) t= 1.53s t=1.53/2 t= 0.76s to reach the maximum height Is this correct?
9. ### Anyone know that the Olivia Newton-John was Max Born's granddaughter?
https://www.washingtonpost.com/obituaries/2022/08/08/olivia-Newton-john-grease-singer-dies/
10. ### Max amplitude of a diving board that will not toss a pebble off?
How am I supposed to solve this problem? I did it in a way that's analytically not correct at all lol.. As a first I was thinking about how to relate it to the motion equations to no avail.. (of course with having the condition that normal force board on pebble = force of gravity on pebble)...
11. ### Calculating Max Tension, Final Angle for Swinging on Rope Over Lake
Please tell me if I need to post my solution for this.., but I just have a question more or less 'conceptual' question about (c). so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mgcos(\theta) = ma_n## where ##a_n = \frac {v^2}{R}## such that where the normal...
28. ### A Can a Solid Vibrate at Debye Max Frequency?
Can a solid vibrate with only one frequency-Debye maximal frequency?
29. ### How can I calculate critical loading conditions for max flap extension?
Below shows the generic V-N diagram for most of the civil aircrafts. Now, I have the CFD results available for multiple combinations of angles of attack of the aircraft, with the deflection angle of the flap. I have to choose a couple, to input into my structural analysis of the flaps. How to...
30. ### I Understanding a proof of inexistence of max nor min
Although a function cannot have extreme values anywhere other than at endpoints, critical points, and singular points, it need not have extreme values at such points. It is more difficult to draw the graph of a function whose domain has an endpoint at which the function fails to have an extreme...
31. ### I A function with no max or min at an endpoint
Hi, PF Although a function cannot have extreme values anywhere other than at endpoints, critical points, and singular points, it need not have extreme values at such points. There is an example of how a function need not have extreme values at a critical point or a singular point in 9th edition...
32. ### I Drawbridge falling before caught by rope/cable - Max force
Hi, I'm currently looking into the forces exerted on a cable if the cable was to go from slack to taut due to a falling object that is pivoted about a lower hinge - such as a drawbridge but instead of slowly lowering, there is slack in the cable causing the bridge to freely rotate about the...
33. ### Engineering Question about max stress on circular cross section with two moments
I couldn't fit in the title, but this is with a hollow circular cross section So currently I am trying to figure what occurs when two, perpendicular bending moments are applied to a hollow circular cross section (one about the z axis, and the other about y). I know that if I was dealing with a...
34. ### Max Speed of a Car: A Math Equation Analysis
Summary:: Question: a car of mass 800kg, 3600N driving force experiences resistive forces of 120v Accelerates from rest. Form equation and show max speed is 30 ms^-1, assuming driving force does not change. my attempt at a solution -I can not see how to rectify it as I don’t think I did...
35. ### Max Height and Distance of Projectile: Theory & Simulation
How can i find the angle that a projectile is fired where the maximun height is the same as the traveling distance? I need to find this first in a theoretical way, then i need to confirm this with a simulation. And i need to use only this y=yo+Voyt+1/2gt^2 x=Voxt
36. ### Showing continuous function has min or max using Cauchy limit def.
Problem: Let ## f: \Bbb R \to \Bbb R ## be continuous. It is known that ## \lim_{x \to \infty } f(x) = \lim_{x \to -\infty } f(x) = l \in R \cup \{ \pm \infty \} ##. Prove that ## f ## gets maximum or minimum on ## \Bbb R ##. Proof: First we'll regard the case ## l = \infty ## ( the case...
37. ### Calculating max acceleration of a Tesla car
[Mentor Note: Two threads with similar titles merged] First calculate the equivalent force on the car from the max torque. An average tire is 15 inch diameter = about .2 meter radius Since T = F X R , F = T/R = 660/.2 = 3300 Newtons. since F = M X A, A = F/M = 3300/2083 = 1.6 m/sec^2...
38. ### Speed of processors, does it have a max?
I had heard that computer processors are reaching the speed of light. Is this true, and if it is how do we combat this cap?
39. ### Max tip speed of a spinning cable
A company called SpinLaunch claims it can get something to 2200 m/s by spinning it up on a carbon fiber composite arm. I'm trying to figure out the limit of that approach. How fast can you go with existing materials? I tried to work it out for a constant with cable with no payload (result...
40. ### Why is my CPU not running at max published speed?
I have a Intel(R) Xeon(R) E-2274G CPU @ 4.00GHz 4.01 GHz machine which has a published max speed of 5.0 GHz. However, when I run it under load on all 4 cores, it only gets up to about 4.5 GHz: Could someone explain to me why it's not ramping up to the full 5.0 GHz or what I could do to...
41. ### Derivative of a Point of Max: Finding a & b Conditions
The derivative of a point of maximum must be zero, and since $$y'=3ax^2+2bx+2 \rightarrow y'(-1)=3a-2b+2 \rightarrow 3a-2b+2=0$$ we get the first condition for ##a## and ##b##. Now, since we want ##x=-1## to be a local maximum, the derivative of the function must be positive when tending to...
42. ### Max velocity of a vibrating loud speaker membrane given sound intensity
My attempt: p and T allows us to calculate ##Z=402 \frac{kg}{sm^2}## using ## Z=p*\sqrt(\frac{\gamma*M}{R*T})## . The sound intensity level at 10 meters allows us to calculate the intensity at 10 meters to be I=10^{-7} W/m^2 using ##50 = 10*log(I/I_0)##. Then, using the formula...
43. ### I Distributivity/Inheritance of Max Likelihood Estimators
Hi, IIRC, Maximum Likelihood Estimators ( MLEs) satisfy an " Inheritance" property , so that if ##m_1,m_2,..,m_n## are MLEs for ##M_1,M_2,...,M_n## respectfully and f is a Random Variable of the ##M_i##, then the MLE for f is given by ##f(m_1,m_2,...,m_n)##. Is this correct? If so, is there a "...
44. ### I Ratio of a real value to the max theoretical value: what term to use?
If the Summary is not clear, the following is an example translated from: https://telecombloger.ru/7335 '... air conditioning systems. For example, the efficiency of a compressor is about 85%. The remaining 15% is spent on friction, oil movement, overflows, heating, etc. The efficiency of the...
45. ### In 2 consecutive decays, determine max and min energies for a particle
Hi! Instead of just describing my procedure and all my derivations, I really just want to ask if my approach makes sense (actually I have 2 options) to calculate the maximum energy. I am considering c=1 and the problem suggests to consider the neutrino massless: For the first decay, ##Z...
46. ### Max lifting capacity for an average human adult
I have situation where i have drawn a concept with 1 sheet covering width of box for model. But now i need to draw up the manufacturing drawings and it needs to lifted and assembled by 2 people. So i need to know if person weighs 70 kg how much weight they can lift without using mechanical aid...
47. ### Engineering Calculate the max common mode voltage at the differential amplifier inputs
Im unsure if I am on the correct track or have gone off on a tangent. Any help or guidance would be appreciated. CMRR=20log10(Adiff/Acm) 120=20log10(10^5/Acm) 120/20=log10(100,000/Acm) 6=log10(100,000/Acm) taking antilogs 1,000,000=100,000/Acm Acm=100,000/1,000,000 Acm=0.1Max amplified...
48. ### MHB Max Value of $b$ for Real Roots of $f(x)$ and $g(x)$
Let $a$ and $b$ be real numbers and $r,\,s$ and $t$ be the roots of $f(x)=x^3+ax^2+bx-1$ and $g(x)=x^3+mx^2+nx+p$ has roots $r^2,\,s^2$ and $t^2$. If $g(-1)=-5$, find the maximum possible value of $b$.
49. ### I Probability of finding a particle in a 1-D box max. at centre?
From the probability density plot why it is max at the centre of the box...e.g probability to find the particle per unit le gth is max. at the centre of the box??why not at any other position??
50. ### MHB What is the Solution to the Differential Equation dy/dx = (2cos 2x)/(3+2y)?
it's late so I'll just start this $\dfrac{dy}{dx}=\dfrac{2\cos 2x}{3+2y}$ so $(3+2y) \, dy= (2\cos 2x) \, dx$ $y^2 + 3 y= sin(2 x) + c$ | 3,369 | 12,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-26 | latest | en | 0.974203 |
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# 2002 AMC 10A Problems
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2002 AMC 10A (Answer Key)Printable versions: Wiki • AoPS Resources • PDF Instructions This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). Figures are not necessarily drawn to scale. You will have 75 minutes working time to complete the test. 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
## Problem 1
The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 10$
## Problem 2
Given that $a, b,$ and $c$ are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$. Find $(2, 12, 9)$.
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$
## Problem 3
According to the standard convention for exponentiation,
$2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65,536$.
If the order in which the exponentiations are performed is changed, how many other values are possible?
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$
## Problem 4
For how many positive integers $m$ is there at least 1 positive integer $n$ such that $mn \le m + n$?
$\text{(A)}\ 4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 12 \qquad \text{(E)}$ infinitely many
## Problem 5
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
$[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]$
$\text{(A)}\ \pi \qquad \text{(B)}\ 1.5\pi \qquad \text{(C)}\ 2\pi \qquad \text{(D)}\ 3\pi \qquad \text{(E)}\ 3.5\pi$
## Problem 6
From a starting number, Cindy was supposed to subtract 3, and then divide by 9, but instead, Cindy subtracted 9, then divided by 3, getting 43. If the correct instructions were followed, what would the result be?
$\text{(A)}\ 15 \qquad \text{(B)}\ 34 \qquad \text{(C)}\ 43 \qquad \text{(D)}\ 51 \qquad \text{(E)} 138$
## Problem 7
A $45^\circ$ arc of circle A is equal in length to a $30^\circ$ arc of circle B. What is the ratio of circle A's area and circle B's area?
$\text{(A)}\ 4/9 \qquad \text{(B)}\ 2/3 \qquad \text{(C)}\ 5/6 \qquad \text{(D)}\ 3/2 \qquad \text{(E)}\ 9/4$
## Problem 8
Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $R$ the area of the red square. Which of the following is correct?
$[asy] unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); [/asy]$
$\text{(A)}\ B = W \qquad \text{(B)}\ W = R \qquad \text{(C)}\ B = R \qquad \text{(D)}\ 3B = 2R \qquad \text{(E)}\ 2R = W$
## Problem 9
There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$, and $1001B + 3003A = 5005$. What is the average of A, B, and C?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}$ Not uniquely determined
## Problem 10
What is the sum of all of the roots of $(2x + 3) (x - 4) + (2x + 3) (x - 6) = 0$?
$\text{(A)}\ 7/2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 13$
## Problem 11
Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB each, 12 of the files take up 0.7 MB each, and the rest take up 0.4 MB each. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 14 \qquad \text{(D)}\ 15 \qquad \text{(E)} 16$
## Problem 12
Mr. Earl E. Bird leaves home every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58$
## Problem 13
Given a triangle with side lengths 15, 20, and 25, find the triangle's smallest altitude.
$\text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$
## Problem 14
Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$
## Problem 15
Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$
## Problem 16
Let $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$. What is $a + b + c + d$?
$\text{(A)}\ -5 \qquad \text{(B)}\ -10/3 \qquad \text{(C)}\ -7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$
## Problem 17
Sarah pours 4 ounces of coffee into a cup that can hold 8 ounces. Then she pours 4 ounces of cream into a second cup that can also hold 8 ounces. She then pours half of the contents of the first cup into the second cup, completely mixes the contents of the second cup, then pours half of the contents of the second cup back into the first cup. What fraction of the contents in the first cup is cream?
$\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/3 \qquad \text{(C)}\ 3/8 \qquad \text{(D)}\ 2/5 \qquad \text{(E)} 1/2$
## Problem 18
A 3x3x3 cube is made of 27 normal dice. Each die's opposite sides sum to 7. What is the smallest possible sum of all of the values visible on the 6 faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$
## Problem 19
Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?
$\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi$
## Problem 20
Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$.
$\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2$
## Problem 21
The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$
## Problem 22
A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$
## Problem 23
Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$. Point $E$ is not on the line, and $BE = CE = 10$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$
## Problem 24
Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?
$\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25$
## Problem 25
$[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("A",A,S); label("B",B,S); label("C",C,N); label("D",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]$ In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$ (diagram not to scale). The area of $ABCD$ is
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | 3,773 | 10,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 78, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | latest | en | 0.779504 |
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#### Resources tagged with Investigations similar to Transformation Tease:
Filter by: Content type:
Stage:
Challenge level:
### Hexpentas
##### Stage: 1 and 2 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### It's Times Again
##### Stage: 2 and 3 Challenge Level:
Numbers arranged in a square but some exceptional spatial awareness probably needed.
### So It's Times!
##### Stage: 2 and 3 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Building with Longer Rods
##### Stage: 2 and 3 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
##### Stage: 2 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Cutting Corners
##### Stage: 2 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Redblue
##### Stage: 2 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
### The Numbers Give the Design
##### Stage: 2 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Escher Tessellations
##### Stage: 2 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### It's a Fence!
##### Stage: 1 and 2 Challenge Level:
In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
### How Tall?
##### Stage: 1 and 2 Challenge Level:
A group of children are discussing the height of a tall tree. How would you go about finding out its height?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Stairs
##### Stage: 1 and 2 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Magazines
##### Stage: 2 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
##### Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Round and Round the Circle
##### Stage: 2 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Making Boxes
##### Stage: 2 Challenge Level:
Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
##### Stage: 2 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Play a Merry Tune
##### Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Sorting the Numbers
##### Stage: 1 and 2 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### More Children and Plants
##### Stage: 2 and 3 Challenge Level:
This challenge extends the Plants investigation so now four or more children are involved.
### More Plant Spaces
##### Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Cubes Here and There
##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related? | 2,092 | 9,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-50 | longest | en | 0.904826 |
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2. Secondary
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#### GCSE Foundation/Functional Skills-Real Life Percentage of Amount Increase and Decrease
Two worksheets on percentage increase and decrease using everyday items such as headphones, bikes and laptops. A fun worksheet to engage students with working out real life percentage decrease and increase. Suitable for KS3, GCSE and Functional Skills maths
#### Percentages of Quantities - Calculator Methods - inc Reverse Percentage Change and Compound Interest
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This is a complete revision bundle for functional skills maths level 1 and level 2 includes both my revision guides and answers. Plus the fact sheets and the revision bingo. I have also added for a limited time my bridging the gap resource
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#### Problem solving with fractions, decimals and percentages (Y5/6)
Four problems which require children to problem solve with percentages, decimals and fractions. This is aimed at Y5/6 children and the sheet can be easily adapted for differentiation if needed. For more resources please see my shop: https://www.tes.com/teaching-resources/shop/d_roberts1
#### Secret Code Mixed Question Recipe Bundle
Secret Code Mixed Question Recipe Bundle All answers are included. If bought independently, these 3 resources would cost £6.50, but I am offering them in this bundle for £5, a saving of 23%. To view items individually, please go to Maths Shop
#### Functional Skills Maths Keep Calm and Ace the Exam L2
43 pages of L2 Functional Skills exam type questions with methods shown. I wrote this for learners to go over in the weekS prior to the exam in the class and at home. Full topic coverage my learners have enjoyed using it and hopefully it will help them pass their upcoming exams. Please leave me a review
#### Functional Skills or GCSE maths Differentiated percentage of amounts questions
Basic differentiated percentage of amounts questions. Answers are included and a power point to save you printing if this is preferred.
#### GCSE Maths/Functional Skills- Differentiated Percentage of Amounts Worksheet
Differentiated Percentage of Amounts questions in a RAG type worksheet, answers included. Also a Powerpoint of the worksheet and answers included to display to students.
#### Mixed Questions
Mixed Questions There are 25 answers on the answer sheet, but only 24 questions. The challenge is for students to find the extra answer and create their own question for it. This resource can be used as a worksheet, or you can read out the questions and let students cross off their answers in a quiz/bingo style. The questions are non-calculator. All answers are included. The topics included are:- • Squares, cubes & Square roots • Area & Perimeter • Double & half • Fractions of amounts • Percentages of amounts • Decimal multiplication • Multiples • Prime Numbers • Subtraction • Division • Metres to centimetres & hours to minutes conversions • Degrees in a right angle • Probability The document is 5 pages in total. This resource can be printed in colour or black and white. To see more of my resources, please go to Maths Shop Keywords: numeracy, worksheet, test, non-calculator, homework, key stage 3, KS3, secondary, new GCSE 9-1, foundation, mathematics, maths revision, math review, SEN, special educational needs.
#### GCSE 9-1 Ratio + Proportion (Bundle)
These carefully selected compilations of exam questions have fully-worked solutions designed for students to go through at home, saving valuable time in class. Click 👉 tes.com/…/Exam Question Bundles… to download the other four bundles.<hr>I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. I also make them available for a student who wants to do focused independent study on a topic.<hr>👍If you like this resource, then please rate it and/or leave a comment💬. If the rate-resource button on this page doesn’t work, then go to your ratings page by clicking 👉 www.tes.com/…/rate-resources…
#### Year 6 Investigations - White Rose Maths - Football World Cup 2018 Themed
This resource contains four sets of resources designed to consolidate some of the work undertaken in the White Rose Maths Scheme small steps Year 6 programme. The theme for this investigation resource is the 2018 Football World Cup. It is intended to be used in weeks 8 - 11 of the Summer Term. The resource contains worksheets and activities on Year 6 National Curriculum Maths Objectives. Click on the individual links for further details Thinking of publishing your own resources or already an author and want to improve your resources and sales? Check out this step-by-step guide. How to Become a Successful TES Author: Step by Step Guide
#### Year 6 Investigations (White Rose) Set 2 - Week 9 - 2018 Football World Cup Themed Resource
This resource contains the second set in a series of 4 sets of resources designed to consolidate some of work undertaken in the White Rose Maths Scheme small steps Year 6 programme. The theme for this investigation resource is the 2018 Football World Cup. It is intended o be used in week 9 of the Summer Term. The resource contains worksheets and activities on the following National Curriculum Objectives Use simple formula expressed in words Solve number problems Generate and describe linear number sequences Convert between standard units of time Solve problems involving the calculations of percentages, decimal fractions and fractions for comparison Associate a fraction with division to calculate decimal fraction equivalents</p> Worksheets and Activities Triangular Numbers - tasks include the investigation of triangular numbers. Number Patterns -this involves calculating the the number of group matches that are played in each group, an investigation into how many matches would be played in groups of varying sizes and the investigation of number patterns. Calculations - tasks include the calculation of the number of matches that will be played during the World Cup and an investigation of how many hours of football will be played. This focuses on the completion of group tables, a summative World Cup football group table and the population of the round of 16, quarter-final, semi-final, final and third place play off tables. Percentages and Probability -an investigation into the probability of countries from the different confederations winning the World Cup. Year 6 Investigations (White Rose) Set 1 Week 8 - 2018 Football World Cup Themed Resource Thinking of publishing your own resources or already an author and want to improve your resources and sales? Check out this step-by-step guide. How to Become a Successful TES Author: Step by Step Guide To be kept informed of new resources click here and follow.
#### Mixed Calculator Questions GCSE Higher
Mixed Calculator Questions GCSE Higher This pack includes two 10 question diagnostic tests covering 20 different topics for the calculator paper. The tests are made up of difficult questions and are aimed at GCSE 9-1 Higher Levels 7, 8 and 9 pupils, but may also be useful for Year 10. Full answer keys are included for both tests and can be used for pupils to self-mark. A checklist is included for pupils to see where they need to focus their revision. Encourage pupils to fill in the checklist and complete further questions on the topics they need to work on. The topics included are: Iteration Sine Rule Quadratic Formula 3D Pythagoras Trigonometry in Right-Angled Triangles Spheres Area of Segment Volume of Cylinder (Hard) Polygon Angles & Trigonometry Upper & Lower Bounds with Speed Reverse Mean of a Grouped Frequency Table Proportion (Hard) Density with Ratio Volume of Pyramid Reverse Percentage (Hard) Exchange Rates & Unit Conversions Cosine Rule Simultaneous Equations with an Exponential Graph Inverse Functions Volume with Rates of Change The tests should be done with a calculator. The document is 9 pages in total. This resource can be printed in black and white or colour. For my other resources, please go to Maths Shop Keywords: review, mixed questions, diagnostic test, worksheet, calculator, senior high, junior high, middle school, homework, key stage 3, KS3, secondary, new GCSE 9-1, level 6, level 7, level 8, level 9, higher, UK, mathematics, maths revision, math review.
#### Year Five - Autumn, Spring, Sum 1 and Sum 2 (Follows order of White Rose)
All teaching slides for Autumn and Spring and Sum 1 and Sum 2. Influenced by recent maths hub CPD training and mastery/CPA approach to teaching mathematics. Includes reasoning and problem solving activities to explore, model and discuss in order to teach the underlying mathematical concepts. Follows ORDER of White Rose teaching guidance. Join the Facebook group ‘mastering primary maths’ for preview lessons in the files section :)
#### GCSE Foundation Maths Number Millionaire
Add some fun to your exam revision lessons with this millionaire game. This resource is suitable for either KS3 or KS4 as a revision tool for revising GCSE Maths Number related topics. It includes sound effects as well as an engaging plenary featuring a clip from the actual television show. Topics covered include: -negative numbers -order of operations -types of numbers -percentages -fractions -inequalities -standard form -HCF -ratio -proportion
#### Functional Skills E3,L1 and L2 Maths Assessment booklets with Answers
Introductory booklets to FS at Entry Level 2, Level 1 and Level 2 including all examples, how to and practice exam questions. Learners self assess their own ability and confidence in each individual subject at the beginning at at the end of the booklet and includes a full glossary: Topics included are: Understanding Number Rounding and Estimating Averages Fractions Decimals and Percentages Money and time Ratio, Scale and Proportion Units of measure 2d and 3d shapes and Angles Perimeter, Area and Volume Data handling and graphs Probability
#### Functional Skills Entry Level 3 Maths Assessment Booklet with answer booklet.
Functional Skills Entry Level 3 Maths Assessment Booklet with answer booklet. Full booklet including all level appropriate topics for the Functional Skills exam with some stretch and challenge. All topics include a how to guide, examples and an exam style question. Self assessment and topic confidence guide included and full glossary. Topics included are: Understanding Number Rounding and Estimating Averages Fractions Decimals and Percentages Money and time Ratio, Scale and Proportion Units of measure 2d and 3d shapes and Angles Perimeter and Area Data handling and graphs
#### Functional Skills Level 1 Full Assessment booklet with Answer booklet
Level 1 Functoinal Skills Introductory booklet for individual and supported work. Al topics have examples, how to guides and level appropriate exam questions. Booklet includes self assessment and topic confidence tool and glossary. Topics included are: Understanding Number Rounding and Estimating Averages Fractions Decimals and Percentages Money and time Ratio, Scale and Proportion Units of measure 2d and 3d shapes and Angles Perimeter and Area Data handling and graphs Probability
#### Daily Arithmetic Starters Upper KS2
20 daily starters to keep them warmed up for arithmetic!
#### %s of Numbers Arithmetic Test.
5 questions as a daily starter for upper KS2. The subject is % of numbers.
#### Daily Arithmetic Upper KS2
A series of 11 daily starters to help the children warm-up in Upper KS2 for their arithmetic tests.
#### diagonal cross multiplication method to work out mathematical calculations
this resource is a demo of how to use a mathematical short cut to work out a wide range of caluclations that can involve; Unit conversion e.g. m to cm, km to m, Kg to pounds and vis versa Scale drawing Currency exchange Similarities Percentages Ratio and proportion it is very helpful when teaching students working at lower tier GCSE Maths and those who are not naturally fond of maths and have low confidence when tackling and solving a word problem within the context of maths or even in science contexts e.g. working numbrer of moles or masses of reactants or products.
#### Revision - Mixed Questions
Revision - Mixed Questions There are 32 questions on 4 worksheets. These questions are a selection of the questions from my other resources. All answers are included. To see more questions on each topic please visit Maths Shop Topics are: Multiplying with decimals Dividing with decimals Percentages of amounts Percentage decrease Estimation Ratio Algebra Substitution Angles in triangles Volume of cuboid Expanding Brackets Proportion with Money These worksheets can be used as practice questions, tests or homework. The document is 9 pages in total. Keywords: worksheet, test, non-calculator, homework, key stage 3, KS3, Key stage 4, KS4, exam practice, exam questions, revision, revise, review, secondary, GCSE 9-1, foundation, higher.
#### Maths Quiz
A maths quiz that tests multiple areas of the maths curriculum: Mental Maths Calculations involving decimals Money Shape Area and Perimeter BODMAS Measurements & Units Percentages Duration Speed Data Handling Circles
#### Percentages code breaker
This activity tests the pupils knowledge on all types of percentage calculation and was used as part of revision for a number exam. The pupils enjoyed this activity and it helped to improved their grades in the exam
#### Theme Park Manager percentages game (full lesson)
This game puts students in charge of a theme park, for which they will choose from a selection of rides using the ride information cards, which detail the running cost and ticket price of each ride, the percentage of visitors that buy a ticket for it, its risk rating, and the cost of insurance for it. Each day a dice roll will then generate a random event which impacts on the fate of the rides, and students will have to use their knowledge of percentages to calculate the number of visitors buying a ticket for each of their rides that day. Included: – A powerpoint presentation outlining the rules and taking you through the game; – A set of information cards for students to use to choose their rides; and – A colour coded table for students to fill in throughout the game to keep track of the popularity of their rides, any fines they have to pay or compensation they are awarded. This is an excellent resource for consolidating students’ knowledge of how to calculate percentages of numbers and increase numbers by a given percentage. It also makes use of other basic number skills and strategic decision making. The game should take around 50 minutes start to finish, but it will be within your control to increase or decrease this as appropriate. Great to use as a fun end of term game that still gets students using lots of important maths skills without it feeling like work. <!—A fun end of term game maths game to reinforce knowledge of calculating percentages, which leads naturally to useful discussions as students are keen to calculate how much money they have earned. —>
#### Higher GCSE Maths Questions - Fractions, Decimals and Percentages (Grades 5 to 9)
Higher GCSE Maths Questions on Fractions, Decimals and Percentages. Answers included too.
#### Arithmetic orienteering (set 5)
A set of arithmetic orienteering/treasure hunt/follow me cards based on the current KS2 maths curriculum. The idea is for the children to find a card, read the question, solve it and then to go and find that answer. On the back of the answer will be the next question for the children to do the same; solve, find and then answer the next question. There is no set starting point as the set loops back around. There are 36 questions, just like the KS2 maths Arithmetic paper 1 test. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Arithmetic orienteering (set 4)
A set of arithmetic orienteering/treasure hunt/follow me cards based on the current KS2 maths curriculum. The idea is for the children to find a card, read the question, solve it and then to go and find that answer. On the back of the answer will be the next question for the children to do the same; solve, find and then answer the next question. There is no set starting point as the set loops back around. There are 36 questions, just like the KS2 maths Arithmetic paper 1 test. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Arithmetic orienteering (set 3)
A set of arithmetic orienteering/treasure hunt/follow me cards based on the current KS2 maths curriculum. The idea is for the children to find a card, read the question, solve it and then to go and find that answer. On the back of the answer will be the next question for the children to do the same; solve, find and then answer the next question. There is no set starting point as the set loops back around. There are 36 questions, just like the KS2 maths Arithmetic paper 1 test. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Arithmetic orienteering (set 2)
A set of arithmetic orienteering/treasure hunt/follow me cards based on the current KS2 maths curriculum. The idea is for the children to find a card, read the question, solve it and then to go and find that answer. On the back of the answer will be the next question for the children to do the same; solve, find and then answer the next question. There is no set starting point as the set loops back around. There are 36 questions, just like the KS2 maths Arithmetic paper 1 test. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Arithmetic orienteering (set 1)
A set of arithmetic orienteering/treasure hunt/follow me cards based on the current KS2 maths curriculum. The idea is for the children to find a card, read the question, solve it and then to go and find that answer. On the back of the answer will be the next question for the children to do the same; solve, find and then answer the next question. There is no set starting point as the set loops back around. There are 36 questions, just like the KS2 maths Arithmetic paper 1 test. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Arithmetic orienteering cards (Sets 1-5)
I have made five arithmetic orienteering/treasure hunt/follow me cards to be placed inside/outside for the children to find and solve. The idea is to find a question, work out the answer and then find that answer. On the back of that answer will be the next question for them to solve and then go and find, and so on. There doesn’t need to be a set starting point as they all go around in a loop. The questions are all based on the new maths curriculum for the current KS2 tests. Each set has 36 questions, just like the KS2 Arithmetic paper 1. Questions are one colour and answers are another colour. Just cut out the rows and fold them over, then laminate.
#### Y6 SATs revision - last minute games
This bundle is full of last minute fun activities to support the revision of SPG and Maths SATs. It includes: Two ‘fastest finger first’ SPG games, where children compete against each other to hunt for word class. One fractions treasure hunt One Arithmetic treasure huhnt One reasoning board game One reasoning treasure hunt One SPG test with 50 new questions One fractions, decimals and percentages game One grammar treasure hunt I am using all these activities in the week running up to SATs to make the most of going outside and to ensure revision is much more fun! My kids absolutely love them, as they don’t even realise they are revising! Hope they are useful to you.
#### KS2 SATs Arithmetic Paper 4
The fourth installment of Vocabulary Ninja’s arithmetic papers. 40 marks from 36 questions, created to mirror the KS2 arithmetic paper.
#### Functional Skills Maths L2 for Travel and Tourism students and apprentices
A workbook suitable for Functional Skills maths Level 2 Students with the emphasis on Travel and Tourism. Good embedded resource, lots of questions; deals with business matters. Could be used for apprentices or those on full time courses. Also Business Studies students. Possible Foundation Level GCSE Maths. Nice resource i am quite proud of.
#### All 50 Maths Foundation Homework Revision Sheets (9-1) - Includes all ANSWERS
This is a bundle of all 5 sets of my homework sheets. I have been giving these out as a homework booklet to my year 11’s all year. For me a booklet consists of Homework 1,11,21,31,41,and 51. We then go through the answers so students can self assess. I feel these have helped strengthen their understanding on some of the basic topics, leaving me more time in class to concentrate on more complex problems. | 4,637 | 22,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-22 | longest | en | 0.875021 |
https://bookdown.org/anshul302/HE902-MGHIHP-Spring2020/ttestreg.html | 1,582,869,085,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00015.warc.gz | 309,077,237 | 20,168 | # Chapter 3 Jan 20–26: T-Tests and Linear Regression
Important Request: As you read through this week’s chapter, please note down anything that you have questions about. Then we will address these questions when we have our video calls.
This week, our goals are to…
1. Apply the basic logic of hypothesis testing.
2. Conduct a t-test on RCT-style data.
3. Use simple linear regression to make a prediction about a dependent variable using one independent variable.
4. Calculate a residual error for each member of a dataset, based on regression results.
• On a Windows or Linux computer, to open a link in a new browser tab, hold down the control key and then click on the link.
• On a Mac computer, to open a link in a new browser tab, hold down the command key and then click on the link.
## 3.1 Hypothesis Tests
### 3.1.1 What is a Hypothesis Test?
Read the following, if it is not yet familiar to you:
### 3.1.2 Two-Sample Hypothesis Tests (T-Test)
T-tests are used when we want to evaluate the difference between means from two independent groups. If you want to compare more than two means, you would use a different statistical test (ANOVA, which we will cover soon). We will focus on two forms of the t-test: independent samples and dependent samples t-tests.
### 3.1.3 Independent Samples T-Test
An independent samples t-test is used when you have means from two separate groups that you want to compare. For example, you might have math exam scores from some men and some women, and you want to see if there is a gender difference. Because you have two independent groups—men and women—an independent samples t-test would be appropriate. This is considered a between-subjects design.
The null hypothesis for a t-test is that the difference between means (of the women’s scores and men’s scores) is 0:
$H_0 = \mu_1 - \mu_2 = 0$
In other words, we start with the assumption that there is no difference between men and women. If we find evidence that there is a difference, we will reject the null hypothesis (which is the assumption that they’re the same). If we do not find any evidence that the math scores of the sampled men and the sampled women are different, we fail to reject the null hypothesis.
Why don’t we just say that we “accept the null hypothesis”? Because we don’t know for sure if the null hypothesis is true or not. All we know is that we didn’t find any evidence to say otherwise. So the null hypothesis might be true, but maybe there is truly a difference between the two groups and we just didn’t have enough data to detect it. This is tricky stuff to get used to at first. We can talk more about this and go through more examples together on our video calls.
If two samples come from the same population, we expect them to have equal means (although with sampling variation, they may not be exactly equal). Under the null hypothesis, we expect that the there are no differences between the groups (the experimental manipulation didn’t have an effect).
There are three assumptions of the independent samples t-tests that must be met in order to for the results to be valid and interpretable:
1. Homogeneity of variances: We assume that the two groups have the same variance.
2. Population is normally distributed.
3. Independence of scores: Each individual (or observation) contributes only one score (data point). If they submit more than one score, then those responses (scores) are correlated with each other and therefore not independent.
### 3.1.4 Dependent Samples T-test
A dependent samples (or paired) t-test is used when the two means you want to compare come from the same person. For example, you might have pre-test and post-test scores for individuals after they have undergone some sort of intervention. You want to know if their scores increased (or decreased) between from pre-test to post-test. This is a within-subjects design.
### 3.1.5 Errors in Hypothesis Testing
There are two types of errors that occur in hypothesis testing:
• Type I error: rejecting the null hypothesis when it’s actually true.
• Type II error: failing to reject the null hypothesis when it’s actually false.
Look through the following resources on these types of errors:
### 3.1.6 T-Tests in R
• Unpaired Two-Samples T-test in R – You can run the commands at this page if you wish, as you read through. In this week’s assignment, you will have to do a very similar procedure but on different data.
The following resources may also be useful:
## 3.2 Simple Linear Regression
This week we are going to explore simple linear regression, which is a method for finding out the linear association between a dependent variable and one independent variable.
Sometimes, the slope of the regression line is referred to with the Greek letter beta. Just keep in mind that this just means the slope of the line.16
When we do a regression and get an estimate of a slope of the relationship between Y and X, there are two possibilities:
1. In the population from which the sample is drawn, there is no true relationship between Y and X. The slope is 0. As you have more or less of X, Y doesn’t change at all. This is called the null hypothesis.
2. In the population from which the sample is drawn, there is a true, non-zero relationship between Y and X. This is called the alternative hypothesis.
If the p-value of your regression estimate is less than 0.05 (or 5%), then (assuming your regression meets other conditions that we will discuss later) you can conclude that Scenario #2 above is correct and that the estimate is trustworthy for the population. In statistical jargon, this is called rejecting the null hypothesis because our analysis had sufficient evidence to make us 95% certain that the alternate hypothesis is true. 1-p = level of certainty about the alternate hypothesis.
## 3.3 Assignment
Make sure that you carefully show all of the work you did to get to your answers. For tasks that require you to use R, make sure you put all of the R commands into a new R script file (just like last week) and you submit the commands you used as well as the output you get as part of your assignment.
### 3.3.1 T-Test Without R
For this part of the assignment, we will use the 2 Sample T-Test tool, which I will call the tool throughout this section of the assignment.
Imagine that we, some researchers, are trying to answer the following research question: How does fertilizer affect plant growth?
We conduct a randomized controlled trial in which some plants are given a fixed amount of fertilizer (treatment group) and other plants are given no fertilizer (control group). Then we measure how much each plant grows over the course of a month. Let’s say we have ten plants in each group and we find the following amounts of growth.
The 10 plants in the control group each grew this much (each number corresponds to one plant’s growth):
3.8641111
4.1185322
2.6598828
0.3559656
2.8639095
0.9020122
5.0527020
2.3293899
3.5117162
4.3417785
The 10 plants in the treatment group each grew this much:
7.532292
1.445972
6.875600
6.518691
1.193905
4.659153
3.512655
4.578366
8.791810
4.891557
Delete the numbers that are pre-populated in the tool. Copy and paste our control data in as Sample 1 and our treatment data in as Sample 2.
Task 1: What is the mean and standard deviation of the control data? What is the mean and standard deviation of the treatment data? Do not calculate these by hand. The tool will tell these to you in the sample summary section.
You’ll see that the tool has drawn the distributions of the data for our treatment and control groups. That’s how you can visualize the effect size (impact) of an RCT. It has also given us a verdict at the bottom that the “Sample 2 mean is greater.” This means that this particular statistical test (a t-test) concludes that we are more than 95% certain that sample 1 (the control group) and sample 2 (the treatment group) are drawn from separate populations. In this case, the control group is sampled from the “population” of plants that didn’t get fertilizer and the treatment group is sampled from the “population” of those that did.
This process is called inference. We are making the inference, based on our 20-plant study, that in the broader population of plants, fertilizer is associated with more growth. The typical statistical threshold for inference is 95% certainty. In the difference of means section of the tool, you’ll see p = 0.0468 written. This is called a p-value. The following formula gives us the percentage of certainty we have in a statistical estimate, based on the p-value (which is written as p): $$\text{Level of Certainty} = (1-p)*100$$. To be 95% certain or higher, the p-value must be equal to 0.05 or lower. That’s why you will often see p<0.05 written in studies and/or results tables.
With these particular results, our experiment found statistically significant evidence that fertilizer is associated with plant growth.
Task 2: What was the null hypothesis in this t-test that we conducted?
Task 3: What was the alternate hypothesis in this t-test that we conducted?
Now, click on the radio buttons next to ‘Sample 1 summary’ and ‘Sample 2 summary.’ This will allow you to compare different distributions to each other quickly, without having to change the numbers (raw data) above. Let’s imagine that the control group had not had as much growth as it did. Change the Sample 2 mean from 5 to 4.5.
Task 4: What is the new p-value of this t-test, with the new mean for Sample 2? What is the conclusion of our experiment, with these new numbers? Use the proper statistical language to write your answer.
Task 5: Gradually reduce the standard deviation of Sample 2 until the results are statistically significant at the 95% certainty level. What is the relationship between the standard deviation of your samples and our ability to distinguish them from each other statistically?17
### 3.3.2 T-Test in R
Please make a new R script file for this week’s assignment, just like you did last week.
To complete this part of the assignment, you may need to install and load packages in R. For example, you will need to use the package ggpubr. Before you use any of the commands from this package, you need to load it into R, by adding the following line of code to your R script file:
library(ggpubr)
If this gives you an error message, it means the package is not installed yet (and therefore cannot be loaded). To install the package, run this command:
install.packages("ggpubr")
You should only have to run the above command once ever, not every time you use R.
Then once again run the following command to load the package (you DO need to run this command every time you use this package in R):
library(ggpubr)
Task 6: Load the plant growth data from the previous section of this assignment into R, using the code below. This code should go into your script file and not directly into the console.
control <- c(3.8641111,4.1185322,2.6598828,0.3559656,2.8639095,0.9020122,5.0527020,2.3293899,3.5117162,4.3417785)
treatment <- c(7.532292,1.445972,6.875600,6.518691,1.193905,4.659153,3.512655,4.578366,8.791810,4.891557)
my_data <- data.frame(
group = rep(c("control", "treatment"), each = 10),
growth = c(control, treatment)
)
Task 7: Now run View(my_data) to see the dataset you just created. The name of the dataset you created is my_data. Does my_data correctly match the data presented earlier in the assignment?
Now your data is loaded into R. You just learned one way to manually load data into R.
Task 8: How many observations (rows) are in my_data? Use R to figure this out. Hint: If your dataset’s name was plants, you would run the command nrow(plants). If your dataset’s name was cnepwhdld, then you would run the command nrow(cnepwhdld).
Task 9: What are the names of the variables in your dataset? Use R to figure this out. Hint: If your dataset’s name was plants, you would run the command names(plants).
Task 10: How many variables are in your dataset? Use R to figure this out. Hint: If your dataset’s name was plants, you would run the command ncol(plants) or length(names(plants)).
For the next few tasks, refer to the Unpaired Two-Samples T-test in R resource page that you read earlier. You should modify the code you find at this page to answer the questions below.
We again return to the same research question from earlier in this assignment: How does fertilizer affect plant growth?
Task 11: Visualize the data using box plots, using the ggboxplot function from the ggpubr package. Include the plot in your submitted assignment. Which group has a higher mean?
Now let’s check the t-test assumptions.
Task 12: Are the two samples (treatment and control) independent? Just answer this using your brain and the example at the Unpaired Two-Samples T-test page (don’t use R).
Task 13: Are the data from the two samples normally distributed? Use the Shapiro-Wilk test as described at the resource page. Include the result you get and your interpretation of it. Be sure to do this separately for the treatment and control groups.
Task 14: Further assess the normality of the samples by making a density plot and a Q-Q plot, as described here: Normality Test in R. Make sure to do it separately for treatment and control. Include your plots with your submitted assignment.
Task 15: Do the two samples (treatment and control) have similar variances? Use the F-test demonstrated in the Unpaired Two-Samples T-test page to figure this out. Make sure you include your results as well as your interpretation of them.
Task 16: Conduct a t-test using the method described under Compute independent t-test - Method 2. Include the output in your assignment along with your interpretation. Is this result the same as it was in the previous section of this assignment?
### 3.3.3 Exploring Linear Relationships Some More
We’re going to start this section by looking at the correlation coefficient, another way to determine how related two variables18 are to one another.
Look at the following fitness dataset containing five people:
• WeeklyWeightliftHours is the number of hours per week the person spends weightlifting.
• WeightLiftedKG is how much weight the person could lift on the day of the survey.
Name <- c("Person A","Person B","Person C","Person D","Person E")
WeeklyWeightliftHours <- c(3,4,4,2,6)
WeightLiftedKG <- c(20,30,21,25,40)
fitness <- data.frame(Name, WeeklyWeightliftHours, WeightLiftedKG)
fitness
Name WeeklyWeightliftHours WeightLiftedKG Person A 3 20 Person B 4 30 Person C 4 21 Person D 2 25 Person E 6 40
Task 17: What is a reasonable research question that we could ask with this data?
Task 18: What is the dependent variable and independent variable for a quantitative analysis that we could do to answer this research question?
Task 19: What is the correlation coefficient for WeightLiftedKG and WeeklyWeightliftHours? Show all of your work/calculations.
Here’s the answer, but you still need to make sure you do and show the work correctly:
# Calculate correlation between two vectors/variables
CorrelationWeightHours <- cor(fitness$WeeklyWeightliftHours,fitness$WeightLiftedKG)
CorrelationWeightHours
## [1] 0.7677303
And here’s what it looks like visually:
plot(fitness$WeeklyWeightliftHours,fitness$WeightLiftedKG)
reg1 <- lm(fitness$WeightLiftedKG~fitness$WeeklyWeightliftHours)
abline(reg1)
Now let’s look at the linear regression output for this data:
summary(reg1)
##
## Call:
## lm(formula = fitness$WeightLiftedKG ~ fitness$WeeklyWeightliftHours)
##
## Residuals:
## 1 2 3 4 5
## -3.818 1.955 -7.045 5.409 3.500
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.136 8.199 1.358 0.267
## fitness$WeeklyWeightliftHours 4.227 2.037 2.075 0.130 ## ## Residual standard error: 6.043 on 3 degrees of freedom ## Multiple R-squared: 0.5894, Adjusted R-squared: 0.4525 ## F-statistic: 4.307 on 1 and 3 DF, p-value: 0.1296 Task 20: At the bottom of the regression output, you’ll see Multiple R-squared: 0.5894. What is the relationship between Multiple R-squared and the correlation coefficient that you calculated earlier on the same data? Hint: the correlation coefficient is often referred to as R. Now let’s look at some other data which is less correlated: Name2 <- c("Person F","Person G","Person H","Person I","Person J") WeeklyWeightliftHours2 <- c(3,4,4,1,3) WeightLiftedKG2 <- c(20,30,21,20,35) fitness2 <- data.frame(Name2, WeeklyWeightliftHours2, WeightLiftedKG2) fitness2 Name2 WeeklyWeightliftHours2 WeightLiftedKG2 Person F 3 20 Person G 4 30 Person H 4 21 Person I 1 20 Person J 3 35 # Correlation in dataset fitness2 cor(fitness2$WeeklyWeightliftHours,fitness2$WeightLiftedKG) ## [1] 0.3251082 plot(fitness2$WeeklyWeightliftHours,fitness2$WeightLiftedKG) reg2 <- lm(fitness2$WeightLiftedKG~fitness2$WeeklyWeightliftHours) abline(reg2) Above, the two variables are much less correlated with each other. Task 21: What would the R^2 be in the regression output for the linear regression on these two variables in the fitness2 data? Now we’re going to learn more about the regression output. First, watch the following video for a review, if you would like: Let’s step back and think about why we do regressions. Of course, we do them to see if the dependent variable and independent variables are associated with each other statistically, but we also do them to find out if the trends that we see in our data are (or are not) similar to those in the population at large. Consider the datasets above about weightlifting. Let’s say we wanted to know about the hours spent lifting and weight lifted in Boston. So then people in Boston would be our population of interest. The five people in the fitness dataset are five people that we surveyed out of this population. These five people are our sample. These are important terms to remember. Our goal is to use the sample (the data that we do have) to learn whatever we can about the population as a whole. When we did the regression reg119 we found that an additional hour of weightlifting is associated with an additional 4.227 predicted additional kilograms of weight lifted. But that’s only for our sample of five people. What about all of Boston? That’s where inference comes in. Inference is when you use your sample to attempt to figure stuff out about your whole population. And this is what the Std. Error (standard error), t value, and Pr(>|t|) (p-value) in the regression output are all about. To reiterate, we have our regression line for the five people. You saw this line drawn in the scatterplot earlier when we were talking about correlations. But what would the line look like for the entire population of Boston? Would it look the same or would the slope be different? If we want to know the true slope of the regression line in the population (the statistical relationship between hours spent lifting and weight lifted in all of Boston), we have to look at the other columns of the regression output and use inferential statistics. Now look again at this regression output: summary(reg1) ## ## Call: ## lm(formula = fitness$WeightLiftedKG ~ fitness$WeeklyWeightliftHours) ## ## Residuals: ## 1 2 3 4 5 ## -3.818 1.955 -7.045 5.409 3.500 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 11.136 8.199 1.358 0.267 ## fitness$WeeklyWeightliftHours 4.227 2.037 2.075 0.130
##
## Residual standard error: 6.043 on 3 degrees of freedom
## Multiple R-squared: 0.5894, Adjusted R-squared: 0.4525
## F-statistic: 4.307 on 1 and 3 DF, p-value: 0.1296
Task 22: What is the equation of the regression line? Be sure to include a slope and an intercept,20 and write the equation in the format $$y = mx+b$$.
Task 23: In the output above, what is the p-value for the WeeklyWeightliftHours coefficient (estimate) of 4.227? What does this p-value mean for the question of whether the true population regression line also has a slope of 4.227?
Task 24: In the output above, what do we get when we divide the coefficient for WeeklyWeightliftHours by its standard error? Do you see that number somewhere else in the same line of the regression table?
### 3.3.4 Predicted Values and Residuals
Task 25: Copy the fitness data into a table in Word or Excel.21 Here’s a copy of it, so that you don’t have to go up and find it:
fitness
Name WeeklyWeightliftHours WeightLiftedKG Person A 3 20 Person B 4 30 Person C 4 21 Person D 2 25 Person E 6 40
Task 26: Add new columns to this table for the following items: 1) Predicted WeightLiftedKG, 2) Residual, 3) Difference from mean. Plug each WeeklyWeightliftHours value into the regression equation to get the Predicted WeightLiftedKG for each person. These are called the predicted values or fitted values. This is how much weight the regression line “thinks” each person lifted, based on the data we gave it.
Task 27: Calculate a residual value for each person. This is the difference between the actual and predicted value of how much each person lifted.
Task 28: Are predicted values and residuals calculated for the dependent variable or independent variable?
Task 29: Are predicted values and residuals calculated for the y values or the x values?22
Task 30: What is the total sum of squares of the residuals? (Take each residual that you just calculated, square it, and then add up those five numbers). This is called the SSR, or Sum of Squares Regression.
Task 31: Calculate the difference from the mean of each Y value. Here’s how you do that:
1. Calculate the mean of WeightLiftedKG (across all five people in the data).
2. For each person (row), calculate the difference between the mean and WeightLiftedKG. This is the difference from the mean.
Task 32: Calculate the sum of squares of the difference from the mean. This is called SST, or Sum of Squares Total.
Task 33: Calculate $$1-\frac{SSR}{SST}$$. Show your work. This should be equal to the Multiple R-squared value from the regression output.
Remember: We want SSR to be as low as possible, which will make $$R^2$$ as high as possible. We want $$R^2$$ to be high. SSR is calculated from the residuals, and residuals are errors in the regression model’s prediction. SSR is all of the error totaled up together. So if SSR is low then error is low. $$R^2$$ is one of the most commonly used metrics to determine how well a regression model fit your data.
Task 34: At the start of this chapter, I requested that you write down any questions you have. Please include them with your submitted assignment. If you don’t have any, just state this in your answer.
Task 35: Please upload your Week 1 assignment (from last week) to the D2L Dropbox. I learned that we are required to use D2L for this, contrary to my instructions from last week. Sorry about any confusion! We will not be using the system I described last week. The dropbox is located at Assessments -> Assignments -> Dropbox for all assignments in D2L.
Task 36: Please upload this assignment that you just finished today to the very same D2L Dropbox used in the previous task. Please follow the same file naming convention that we used last week.
Task 37: Also e-mail your new completed assignment to me at .
1. We have also referred previously to the slope as m (as in mx+b), $$b_1$$ (as in $$b_1x + b_0$$), coefficient, and coefficient estimate.
2. Remember, when we are analyzing the data in an RCT, we are trying to figure out if the treatment and control groups had different or similar results. We are seeing if we can distinguish the two groups from each other in any way. The mean and standard deviation of the data in the two groups are the key parameters that help us tell the treatment and control groups apart, which is why you need to play around with the t-test tool to understand these relationships.
3. Remember that a variable is another name for a column of data, or a vector as R calls it. Just a set of numbers or characteristics.
4. We could call this regression whatever we want, not necessarily reg1.
5. Note that the intercept is also sometimes referred to as the constant.
6. You will have to submit this table with your homework, in case that helps you decide whether to use Word or Excel. You can also hand-write it and take a photo and submit that.
7. This is identical to the previous question but with different terminology. | 5,883 | 24,492 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-10 | latest | en | 0.933044 |
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Math
Registered User, __BANNED USERS regular
edited March 2011
Really confused by all this since it wasn't covered in class at all, we were just sort of told to do the homework.
Anyway I am doing the Rational Zeros Theorem stuff, then I have to use the zeroes to factor f over the real numbers.
That isn't a sentence that makes sense to me, but I do know what to do. I can get the zeros, I don't know if there is a faster way than just process of elimination but that is what I do. Start with the lowest then go to the highest. Anyways the book always stops on the lowest one, for example there could be 2 positive zeros and it find that x+1 is a factor it will never go further than that, anyway to know that nothing else is the factor without trying the rest?
Also once I do get the zeroes I am suppose to factor by grouping, I don't know what that means, it wasn't taught in class, and it isn't in the book. After that I set the factors to 0, which seems easy but I don't know what factor by grouping is.
Than the book says thus, I don't know what happened, they get an incredibly long answer and nothing is explained by then it lists the real zeros.
Fizban140 on
«13
Posts
• Registered User regular
edited February 2011
factor by grouping:
if it's 4 terms split them into two halves and find the Greatest Common Factor of each half. If you're dealing with something that is factorable by gouping then the two halves will half a common binomial factor that you can basically treat as another GCF.
Also are you just saying you are finding zeros by guess and check? because yes there are many ways to find them and that's what you should be learning. I would like to see the problems you're doing, but I'll make up a quadratic example just in case:
x^2 + 5x + 6...find the zeros:
factor it : (x+2)(x+3) = 0
so x+2 = 0 or x+3 = 0, which means x = -2 or x = -3.
If you look at the graph, that's where the function crosses the x-axis (that's why we call them "zeros" because y = 0 on the x-axis).
musanman on
• Registered User, __BANNED USERS regular
edited February 2011
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3
That is the problem, I know this is probably the most simple polynomial for this but I can't figure this out. They magic out some shit like
Factoring by grouping gives you (2x-1)(x^2-3)
then they magic out some shit like
the real zeroes are 1/2 -1 square root of 3 and negative square root of 3 multiplicity 1. I have no idea what to do. This is so complex I have to learn all these things just to start to learn how to do the homework.
I mean we spent the whole class period counting sign changes and finding the factors of 6 and other small numbers, how the fuck do we jump straight into this as homework? I can't learn like this.
Fizban140 on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Fizban, did you get a proper math tutor yet?
BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
Define proper?
I will answer that now actually, no.
I have a tutor but it is just a free service my school offers, kids from the local university.
Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3
That is the problem, I know this is probably the most simple polynomial for this but I can't figure this out. They magic out some shit like
Factoring by grouping gives you (2x-1)(x^2-3)
then they magic out some shit like
the real zeroes are 1/2 -1 square root of 3 and negative square root of 3 multiplicity 1. I have no idea what to do. This is so complex I have to learn all these things just to start to learn how to do the homework.
I mean we spent the whole class period counting sign changes and finding the factors of 6 and other small numbers, how the fuck do we jump straight into this as homework? I can't learn like this.
ok first of all I realize now you're the same guy from the other thread, you need to get a tutor.
That said, factoring the 5 term polynomial you just listed does not give that result.
musanman on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
I will start off by saying
YOU NEED A REAL TUTOR. Someone who knows the material AND is trained in the teaching of it (including multiple approaches), not just some student who's good at math. The latter is good for people who just have a few small hangups or confusions here and there, but that's honestly not your case. You need someone who can give you real 1-on-1 teaching
Ok, so I don't think musaman quite understood the approach they're asking for here.
Rational Zeroes Theorem, or more commonly Rational Root Theorem, is a way of finding roots of a polynomial and thus factoring it. The wiki has an excellent example, although their method of testing it is probably more complex than where you're at.
The basic steps are as follows (underscore indicates subscript)
Given: f(x) = a_n*x^n + a_(0-1)*x^(n-1) … + a_0 , where all coefficients are integers
Let: Q = integer factors of a_n and P = integer factors of a_0
Potential roots are then x = (+/-) P/Q
You then test these systematically. When you find one, we'll call it B, that works and give you zero, you can then factor out (x+B) from f(x) using something like polynomial long division. Test the remaining unfactored portion with B again to see if the root has multiplicity. Continue down the list of potential roots until it is entirely factored and all roots are found.
Let's take your example:
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3
P = 3,1 (since factors of 3 are 3 and 1)
Q = 2,1 (since factors of 2 are 2 and 1)
Potential x = (+/-)P/Q = (+/-){3,1}/{2,1} = +3/2, +3/2, +3, -3, +1/2, -1/2, +1, - 1
Test f(x) with first one, +3/2
f(+3/2) = 2*(3/2)^4 + (3/2)^3 - 7*(3/2)^2 - 3*(3/2) +3
f(+3/1) = -15/4
∴ +3/2 is not a root
Test f(x) with second one, -3/2
f(-3/2) = -3/2
∴ -3/2 is not a root
Test f(x) with third one, +3
f(+3) = 120
∴ +3 is not a root
Test f(x) with fourth one, -3
f(-3) = 84
∴ -3 is not a root
Test f(x) with fifth one, +1/2
f(+1/2) = 0
∴ +1/2 is a root
x=1/2
2x - 1 = 0
Factor out (2x-1) from f(x)
g(x) = f(x)/(2-1) = x^3 + x^2 - 3x + 3
∴ partially factored f(x) = (2x-1)(x^3 + x^2 - 3x + 3)
Test g(x) with fifth one again, +1/2
g(+1/2) = -33/8
∴ +1/2 is not another root (and only has multiplicity 1)
Test g(x) with sixth one, -1/2
g(-1/2) = -11/8
∴ -1/2 is not a root
Test g(x) with seventh one, +1
g(+1) = -4
∴ +1 is not a root
Test g(x) with eigth one, -1
g(-1) = 0
∴ -1 is a root
x=-1
x+1 = 0
Factor (x+1) out of g(x)
h(x) = g(x)/(x+1) = x^2-3
h(x) cannot be further factored.
FACTORED f(x) = (2x-1)(x+1)(x^2-3)
Finding the remaining zeroes (of h(x))
x^2-3=0
x^2=3
x=+sqrt(3) or x = -sqrt(3)
ZEROES/ROOTS x = { 1/2 , -1 , +sqrt(3) , -sqrt(3) }
BoomShake on
• Registered User regular
edited February 2011
yeah that's not factoring by grouping, and the "answer" he listed still isn't correct
musanman on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Yeah, I'm not sure why they'd be doing rational root theorem and factoring by grouping at the same time.
Either way, between our two posts he should be good.
Here's the Factoring by Grouping[/i] wiki, which has a nicely formatted example, for good measure.
BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
That factoring looks incredibly complex, is there any good way to do it besides trial and error?
One thing I am still having trouble with, when I use sythetic division how do I know I am done? Like I find out x-1 was a factor, then I do 1 again and that is a factor, when do I stop trying to factor out something? I just had a problem like that where x-1 was a factor twice in a row, and the book just stops after the first time and does factor by grouping.
Fizban140 on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Complex and tedious are two very different things.
What I did was tedious, but not complex.
Read the link I posted earlier to the Rational Root Theorem wiki page. In their example, they use the Horner Scheme to test the roots.
BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
This is ridiculous, I am just trying to get my homework done and I have to learn all this other stuff. Really makes me not want to this, just incredibly daunting to even start my homework.
Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
It is indeed ridiculous that you are being asked to learn math from a textbook rather than from a teacher. As said above, your best hope is to get a tutor. The subject matter itself is not complicated or difficult, but the barrier to entry is high without a good teacher. Once you sort of crack the code, the task should become routine.
Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
I just dislike being given a task and having to top several dozen other things before I can begin the task. Or more accurately getting half way through the task and then having to do several dozen other tasks.
Fizban140 on
• Registered User regular
edited February 2011
this just in: mathematics is cumulative
musanman on
• Registered User, __BANNED USERS regular
edited February 2011
That would have been great if it was in any of my previous math classes.
Fizban140 on
• Nah Registered User regular
edited February 2011
Fizban140 wrote: »
That would have been great if it was in any of my previous math classes.
The reason you're having difficulty now is because it was
If you were having problems in Pre-Alg, hell yes you're going to have the same ones plus a whole host of new ones in Alg I
Usagi on
• Registered User, __BANNED USERS regular
edited February 2011
I got an A in pre algebra, than passed algebra 1 with a B I think, now taking algebra 1 again and I think I am going to fail it.
I guess I just need to find somewhere with a lot of factoring practice problems so I can commit this all to memory, but that feels like it would takes me a long time to do.
I am taking a break for the night, there just appears to be no systemic or logical approach to this, it is all just randomly done. That is as far as I can get after a couple hours of attempting to interpret this stuff.
Sometimes they synthetically divide once, then factor by grouping, sometimes they keep synthetically dividing, answers come out different but I don't see the signal that tells me when to do what. It appears random.
Math has evaded any logical attempts at understanding it again so I will try again tomorrow maybe.
Fizban140 on
• When the last moon is cast over the last star of morning And the future has past without even a last desperate warningRegistered User, Moderator mod
edited February 2011
BoomShake wrote: »
YOU NEED A REAL TUTOR.
This is really the advice you should be taking from this thread at this point. I don't mind your math threads at all, but I don't think you really grasp the terrible disservice you are doing yourself by not following this very simple advice.
H/A can only help you with this so far. You NEED to get proper, one-on-one help for this, or you are going to fail this and any other math-related or -centered course you take. It is entirely preventable. It is clear from your posts that you are oversimplifying some things in your head at the same time you are overcomplicating others, and the only way for you to sort out which is what is for you to get proper help with this.
Lacking the ability to sit across the table from you and really examine your thought process as you go through these problems, this forum is ill-equipped to help you get at the reason that these problems aren't making sense to you. That's just how it is. It's not your fault, you aren't terrible or stupid because it doesn't come to you right away, but it IS something that you need to take extra care with especially if you are going into the sciences, and it IS something that will require instruction from a proper tutor.
ceres on
And it seems like all is dying, and would leave the world to mourn
• Registered User, __BANNED USERS regular
edited February 2011
I am most likely changing my major to something that requires no math, like an English major so I can work in in food services.
Seriously though I will look into tutors for around here, and more importantly money for this service.
Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am most likely changing my major to something that requires no math, like an English major so I can work in in food services.
Seriously though I will look into tutors for around here, and more importantly money for this service.
Aren't there some office hours you can attend as well? Those help immensely, at least for me.
Demerdar on
• Registered User regular
edited February 2011
Have you encountered synthetic division (also known as synthetic substitution)? If so, that makes the testing of possible roots much quicker. Wikipedia has a decent example. I might be able to help you via Skype some time also.
Here's the process I would recommend:
1. If there is a greatest common factor, go ahead and factor that out first. It will reduce the rest of the coefficients and make the numbers easier to work with.
2. Determine how many roots you have. This is easy: it's the highest power of the variable (known as the "order" of the function). The one you listed in the OP will have 4 total roots.
2. Use the rational roots theorem to make a list of possible roots. BE SURE to include both the positive and negative for each one.
3. If you're familiar with Descartes Rule of Signs, use it to identify how many positive and negative roots you have. This isn't necessary but can help you save time later.
4. Start testing roots. I would recommend starting with the smallest roots, and whole numbers before fractions, just because they're easier to work with. Do it just like BoomShake showed you. If you know/can learn synthetic division it makes this easier.
5. IF IT WON'T PISS YOUR TEACHER OFF, when you're down to only needing to find 2 more roots, throw the damned thing in the Quadratic Formula to get the last two.
Ceres is right, there's no way to do this well except for face-to-face in real time. For now, though, think of it this way: think of something you're good at, a sport or music or game or whatever. How'd you get good at it? You repeated the steps over and over again until they became second nature. It happens with math too. It's nothing more than a series of rules (to be sure, often frustrating and confusing rules). If you follow those rules, you will get the answer. Where I see my students making mistakes is, when they get frustrated, they just do something that looks sort of like math, something that feels right. It's kind of a running joke in my classes, I tell them all the time, "Don't do random things." If you think through each step as you're doing it, and ask yourself "why am I doing this step?" you'll get it eventually.
GoodOmens on
IOS Game Center ID: Isotope-X
• Registered User, __BANNED USERS regular
edited February 2011
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...
I have 3x^3 - x^2 - 15x +5 = 0
Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)
That can be factored down into 3x(x-5) so now I have 1/3 and 5 as my answers except the book didn't do it like that they factored the problem without ever trying to do all the stuff it taught me in the chapter, they just straight factored it which isn't even taught in this book. I don't know why, and I did everything properly I think so I should still get the same answer, this just feels all random. I can do the steps right but get a different answer than the book because they randomly decided to do it with a different methon, one not taught in the book.
I would probably slap the shit out of the author of this book, for being so bad at his job, if I ever saw him.
I am going to take a break from this, I already am not in the mood to deal with this. I just started one problem, did the steps to perfection, got the wrong answer, double checked all my math, same answer. I look at the solution manual and they do something that I can not reference anywhere in the book and it has no explanation. I just chalk that one up as a loss, if it ever come up on a test I will get it wrong but that is okay, can't get them all right.
I go to the next problem, same thing, except an entirely different type of bullshit procedure described above. I am done for now.
Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...
I have 3x^3 - x^2 - 15x +5 = 0
Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)
That can be factored down into 3x(x-5) so now I have 1/3 and 5 as my answers except the book didn't do it like that they factored the problem without ever trying to do all the stuff it taught me in the chapter, they just straight factored it which isn't even taught in this book. I don't know why, and I did everything properly I think so I should still get the same answer, this just feels all random. I can do the steps right but get a different answer than the book because they randomly decided to do it with a different methon, one not taught in the book.
I would probably slap the shit out of the author of this book, for being so bad at his job, if I ever saw him.
I am going to take a break from this, I already am not in the mood to deal with this. I just started one problem, did the steps to perfection, got the wrong answer, double checked all my math, same answer. I look at the solution manual and they do something that I can not reference anywhere in the book and it has no explanation. I just chalk that one up as a loss, if it ever come up on a test I will get it wrong but that is okay, can't get them all right.
I go to the next problem, same thing, except an entirely different type of bullshit procedure described above. I am done for now.
I factor 3x^3-15x-x^2+5 = 0 and get:
(3x-1)(x^2-5) = 0
then i solve for the roots.
3x-1 = 0 and
x^2 - 5 = 0
where I get
x=1/3
x=+sqrt(5)
x=-sqrt(5)
A good check for these is to plot your function with a graphing calculator. You can physically see the roots in the graph. Compare those roots with your answers. If your answers are wrong.. well you fucked up.
Remember when you factor and simplify polynomials, make SURE that what you simplified is still the same function.
Do me a favor and expand what you simplified:
(x-1/3)(3x^2-15x)
does it equal
3x^3 - x^2 - 15x +5
?
Demerdar on
• Registered User, __BANNED USERS regular
edited February 2011
I am not good at multiplying like that but I did sythetically divide it by 1/3 and got 3 2 15 so it HAS to be that doesn't it? How else would the division work?
Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am not good at multiplying like that but I did sythetically divide it by 1/3 and got 3 2 15 so it HAS to be that doesn't it? How else would the division work?
Did you expand it out? You need to be able to expand your polynomials. That is fundamental, and provides a good check. If you have trouble expanding the polynomial, how are you going to be good at factoring them?
Those two expressions are not equal. You did something wrong. I don't know what you did wrong, but it was fundamentally bad.
Demerdar on
• Registered User, __BANNED USERS regular
edited February 2011
I have been doing synthetic division for so long I forgot how to do long division and multiplication, guess that is one more thing on the list.
I think I am getting ahead of myself, I am going to go back to the first set of problems and try and work those and see if I can decode the books messages.
Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
Wait, Fiz, I think you may have just had an arithmetic error...
Fizban140 wrote: »
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...
I have 3x^3 - x^2 - 15x +5 = 0
Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)
I get (x-1/3) * (3x^2 - 15)
3x^3 divided by x is 3x^2. So if x-1/3 is a factor, we try multiplying it by 3x^2 and get 3x^3 - x^2. Subtract that from the original polynomial and you end up with -15x + 5. -15x divided by x is -15, so we multiply x - 1/3 by -15 and get -15x + 5. Subtract that from the polynomial and you get zero. So all terms in the polynomial are accounted for and we know x-1/3 times 3x^2 (the first thing we multiplied by) + -15 (the second thing we multiplied by) is the original polynomial. We can check our work by straight multiplying (x-1/3)(3x^2-15), and we should indeed get 3x^3 - x^2 - 15x + 5.
I'm not going to do the math further unless you need me to, but the eventual answer should be (x - 1/3) (3) (x + sqrt(5)) (x - sqrt(5)). Roots of +1/3, +-sqrt(5), all multiplicity one. Same answers as Demerdar got doing it a different way. You are able to do it the way the book taught and get the same answer, you just had an arithmetic error. When you double check your math, do you do it forwards (doing the problem over again in the same way) or backwards (take your answer and do the reverse of whatever you did to it, to see if you get back to the original)? I find checking my work forwards to be of little help, because I'll make the same mistake again. Checking backwards gives me better results.
Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
I am going to actually restart this whole chapter and go back to the start, where I still have problems with the math.
Nevermind I got it finally.
Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
x^3 + 2x^2 -5x -6
-
(x-1)(x^2) = x^3 -x^2
=
3x^2-5x-6
-
(x-1)(3x) = 3x^2 - 3x
=
-2x-6
-
(x-1)(-2) = -2x+2
=
4
I don't see x-1 being a factor of the original polynomial. To check my work backwards, I multiply (x-1)(x^2+3x-2) = x^3 + 3x^2 -2x - x^2 - 3x + 2 = x^3 + 2x^2-5x+2. Original polynomial was +6. If x-1 is a factor of (expr) + 2, it won't be a factor of (expr) +6, since it isn't a factor of 4.
Arithmetic errors are hurting you, man. Can you try to be more systematic? In the past when I have taught math arithmetic errors tended to be the result of skipping steps and going too fast.
Powerpuppies on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
x^3 + 2x^2 -5x -6
-
(x-1)(x^2) = x^3 -x^2
=
3x^2-5x-6
-
(x-1)(3x) = 3x^2 - 3x
=
-2x-6
-
(x-1)(-2) = -2x+2
=
4
I don't see x-1 being a factor of the original polynomial. To check my work backwards, I multiply (x-1)(x^2+3x-2) = x^3 + 3x^2 -2x - x^2 - 3x + 2 = x^3 + 2x^2-5x+2. Original polynomial was +6. If x-1 is a factor of (expr) + 2, it won't be a factor of (expr) +6, since it isn't a factor of 4.
Arithmetic errors are hurting you, man. Can you try to be more systematic? In the past when I have taught math arithmetic errors tended to be the result of skipping steps and going too fast.
edit: beaten. 4 minutes too slow . I'll leave it up on the off chance it helps somehow.
Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
Yeah I am going too fast because I absolutely hate doing math, math is basically in exercise in seeing how horrible you can be and in how many different ways.
Alright so I am doing okay, did two problems in a row then I tried the third and I can't get it at all. I just did the math about 4 times, unfortunately it was correct every time.
x^4 + x^3 - 3x^2 -x +2
I find that x-1 is a factor
thus (this is where the magic happens)
(x-1)(x^3+2x^2-x-2) from that I get x^2(x+2) -1(x+2)
So I guess the answers are 1, -2 and square root of 1 plus or minus. That is wrong but I can't figure out how to get their answer of -1 multiplicity 2.
Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
the square root of 1 is 1, so you have 1, -2, 1, -1. I think you're correct. Lemme do it on paper and I'll come back in a few minutes.
edit: every which way I figure it the roots I find are 1 multiplicity 2, -2 multiplicity 1, and -1 multiplicity 1. Are you sure you copied the problem and the book's answer down correctly?
Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
x^2 - 4 = 0 has no real solutions....why isn't it 2?
x^2 = 4
x= 2
Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
It is.
What's the greater context of the problem?
edit: I mean, it's +- 2, not just +2, but no solution is obviously wrong
x^2+4=0 has no real solutions... sign error?
Powerpuppies on
• Registered User regular
edited February 2011
I just want to chime in here and say that you should check out PatrickJMT(dot)com, or find his videos on youtube (his own site is much easier to navigate). I would also recommend Paul's Online Math Notes. Both of these have been invaluable to me in my calculus classes. Both sites have material suitable for all levels, and Patrick even does some discrete math and physics, if that's your kind of thing.
Maxim Tomato on
• Registered User regular
edited February 2011
Fizban, I recommended as much in the other thread, but you need to be making use of your professor's office hours. They are free, and the professor maintains those hours to help students understand the concepts covered (or assumed) in class. Work the problem as far as you can and then take it to the prof and say "I don't think this is right, and these are the steps I did. Can you show me where I went wrong and what I should have done?"
Also, like I mentioned before, most math texts have example problems in each chapter that are worked and have the work shown. There are also usually practice problems with answers listed in the back (only the odd-numbered problems in my experience). These are extremely valuable. Work problems with known answers until you get the same answer and know why. If you can't figure those out either, show _those_ to your professor and ask for help.
H/A is great, but help with math really needs to be a one-on-one and in-person thing. We can't see when something clicks for you - your professor or a tutor can.
jclast on
• Registered User, __BANNED USERS regular
edited February 2011
Yeah I do go to his office hours but it doesn't help much to be honest, he just explains things in a way that assumes I have a masters in mathematical theory.
Anyways I am stuck on a problem, I got it down to
(x-1/2)(2x^3-18x^+48x-40)
I decided to break it up and factor it to reduce the amount of steps and I get
2x^2(x-9) 8(6x-5) that is wrong, but I don't know why, the book took out 2 and went from there, why do I have to do that? I don't understand why I am getting a wrong answer by doing the correct, but different from the book, steps. I get 1/2, 9 and 5 but the book gets 2 instead of 9.
Fizban140 on
• Nah Registered User regular
edited February 2011
You take out 2 because each term of the polynomial has 2 as a factor: 2, 18, 48, 40 become 1, 9, 24, 20
You can't factor out 2x because only the first three terms have 'x' in them - how can you factor 2x out of 40?
Usagi on
• Registered User regular
edited February 2011
You obviously don't know what factoring is. Factoring is like "undo-ing" the distributive property. Which means if you expand your factored form back out it should match the original problem.
simple example-
factor 3x^2 + 18x
by GCF = 3x(x+6)
If you use the distributive property on 3x(x+6) you get 3x*x + 3x*6 = 3x^2 + 18x and are back where you started.
If you expand your factored form it's not what you started with, so it's wrong.
musanman on
• Registered User, __BANNED USERS regular
edited February 2011
Usagi wrote: »
You take out 2 because each term of the polynomial has 2 as a factor: 2, 18, 48, 40 become 1, 9, 24, 20
You can't factor out 2x because only the first three terms have 'x' in them - how can you factor 2x out of 40?
I didn't take x out of 40, I took it out of the first two terms, 2x^2-18x
Is there some abstract and complicated math rule I am not seeing here? The math works, but the answer is wrong.
Fizban140 on
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https://collegephysicsanswers.com/chapter-2-one-dimensional-kinematics?textbook=ap | 1,713,700,326,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817765.59/warc/CC-MAIN-20240421101951-20240421131951-00062.warc.gz | 156,377,191 | 30,619 | # Chapter 2
Chapter thumbnail
# Chapter 2 : One-dimensional kinematics - all with Video Solutions
### Problem 6
A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the average speed of the blade tip in the helicopter's frame of reference. (b) What is its average velocity over one revolution?
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### Problem 7
The North American and European continents are moving apart at a rate of about 3 cm/y. At this rate how long will it take them to drift 500 km farther apart than they are at present?
View solution
### Problem 8
Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?
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### Problem 9
On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world's nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?
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### Problem 10
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by $3.84\times 10^{6}\textrm{ m}$ (1%)?
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### Problem 11
A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction $25.0^\circ$ south of east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?
View solution
### Problem 12
The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?
View solution
### Problem 13
Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person's voice was so loud in the astronaut's space helmet that it was picked up by the astronaut's microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light $3.00 \times 10^8 \textrm{ m/s}$.
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### Problem 14
A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.
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### Problem 15
The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model you can view hydrogen, the simplest atom, ashaving a single electron in a circular orbit $1.06 \times 10^{-10} \textrm{ m}$ in diameter. (a) If the average speed of the electron in this orbit is known to be $2.20 \times 10^6 \textrm{ m/s}$, calculate the number of revolutions per second it makes about the nucleus. (b) What is the electron's average velocity?
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### Problem 17
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his (a) acceleration and (b) deceleration. Express each in multiples of $g$ $(9.80 \textrm{ m/s}^2)$ by taking its ratio to the acceleration of gravity.
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### Problem 18
A commuter backs her car out of her garage with an acceleration of $1.40 \textrm{ m/s}^2$ . (a) How long does it take her to reach a speed of $2.00 \textrm{ m/s}$? (b) If she then brakes to a stop in $0.800 \textrm{ s}$, what is her deceleration?
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### Problem 19
Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in $\textrm{ m/s}^2$ and in multiples of $g$ $(9.80 \textrm{ m/s}^2$).
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### Problem 20
An Olympic-class sprinter starts a race with an acceleration of $4.50 \textrm{ m/s}^2$ . (a) What is her speed 2.40 s later? (b) Sketch a graph of her position vs. time for this period.
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### Problem 21
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is $2.10 \times 10^4 \textrm{ m/s}^2$, and 1.85 ms $(1 \textrm{ ms} = 10^{-3} \textrm{ s}$ elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?
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### Problem 22
A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of $6.20 \times 10^{5}\textrm{ m/s}^2$ for $8.10\times 10^{-4}\textrm{ s}$ . What is its muzzle velocity (that is, its final velocity)?
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### Problem 23
A light-rail commuter train accelerates at a rate of $1.35 \textrm{ m/s}^2$. How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of $1.65 \textrm{ m/s}^2$. How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in $\textrm{m/s}^2$?
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### Problem 24
While entering a freeway, a car accelerates from rest at a rate of $2.40 \textrm{ m/s}^2$ for 12.0 s. (a) Draw a sketch of the situation. (b) List the knowns in this problem. (c) How far does the car travel in those 12.0 s? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, check your units, and discuss whether the answer is reasonable. (d) What is the car's final velocity? Solve for this unknown in the same manner as in part (c), showing all steps explicitly.
View solution
### Problem 25
At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of $2.00 \textrm{ m/s}^2$ . (a) How far does she travel in the next 5.00 s? (b) What is her final velocity? (c) Evaluate the result. Does it make sense?
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### Problem 26
Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart. (a) Make a sketch of the situation. (b) List the knowns in this problem. (c) How long does the acceleration take? To solve this part, first identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units. (d) Is the answer reasonable when compared with the time for a heartbeat?
View solution
### Problem 27
In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If thisshot takes $3.33 \times 10^{-2} \textrm{ s}$, calculate the distance over which the puck accelerates.
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### Problem 29
Freight trains can produce only relatively small accelerations and decelerations. (a) What is the final velocity of a freight train that accelerates at a rate of $0.0500 \textrm{ m/s}^2$ for 8.00 min, starting with an initial velocity of 4.00 m/s? (b) If the train can slow down at a rate of $0.550 \textrm{ m/s}^2$, how long will it take to come to a stop from this velocity? (c) How far will it travel in each case?
View solution
### Problem 31
A swan on a lake gets airborne by flapping its wings and running on top of the water. (a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates from rest at an average rate of $0.350 \textrm{ m/s}^2$ , how far will it travel before becoming airborne? (b) How long does this take?
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### Problem 32
A woodpecker's brain is specially protected from large decelerations by tendon-like attachments inside the skull. While pecking on a tree, the woodpecker's head comes to a stop from an initial velocity of 0.600 m/s in a distance of only 2.00 mm. (a) Find the acceleration in $\textrm{m/s}^2$ and in multiples of $g$ ($g = 9.80 \textrm{ m/s}^2$). (b) Calculate the stopping time. (c) The tendons cradling the brain stretch, making its stopping distance 4.50 mm (greater than the head and, hence, less deceleration of the brain). What is the brain's deceleration, expressed in multiples of $g$?
View solution
### Problem 33
An unwary football player collides with a padded goalpost while running at a velocity of 7.50 m/s and comes to a full stop after compressing the padding and his body 0.350 m. (a) What is his deceleration? (b) How long does the collision last?
View solution
### Problem 34
In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000 m), and some of them survived, with few life- threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot's speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of 3.0 m.
View solution
### Problem 35
Consider a grey squirrel falling out of a tree to the ground. (a) If we ignore air resistance in this case (only for the sake of this problem), determine a squirrel's velocity just before hitting the ground, assuming it fell from a height of 3.0 m. (b) If the squirrel stops in a distance of 2.0 cm through bending its limbs, compare its deceleration with that of the airman in the previous problem.
View solution
### Problem 36
An express train passes through a station. It enters with an initial velocity of 22.0 m/s and decelerates at a rate of $0.150 \textrm{ m/s}^2$ as it goes through. The station is 210 m long. (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, when does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
View solution
### Problem 37
Dragsters can actually reach a top speed of 145 m/s in only 4.45 s—considerably less time than given in the textbook examples. (a) Calculate the average acceleration for such a dragster. (b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m (a quarter mile) without using any information on time. (c) Why is the final velocity greater than that used to find the average acceleration? Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.
View solution
### Problem 38
A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of $0.500 \textrm{ m/s}^2$ for 7.00 s. (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 m from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?
View solution
### Problem 39
In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, with a maximum speed of 183.58 mi/h. The one-way course was 5.00 mi long. Acceleration rates are often described by the time it takes to reach 60.0 mi/h from rest. If this time was 4.00 s, and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?
View solution
### Problem 40
(a) A world record was set for the men's 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?
View solution
### Problem 41
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be $y_o = 0$.
View solution
### Problem 42
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
View solution
### Problem 43
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?
View solution
### Problem 44
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
View solution
### Problem 45
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.
View solution
### Problem 46
A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/ s, and her takeoff point is 1.80 m above the pool. (a) How long are her feet in the air? (b) What is her highest point above the board? (c) What is her velocity when her feet hit the water?
View solution
### Problem 47
(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long would it take to reach the ground if it is thrown straight down with the same speed?
View solution
### Problem 48
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?
View solution
### Problem 49
You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?
View solution
### Problem 51
Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can't see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?
View solution
### Problem 52
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.
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### Problem 53
There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335 m/s on this day.
View solution
### Problem 55
Suppose you drop a rock into a dark well and, using precision equipment, you measure the time for the sound of a splash to return. (a) Neglecting the time required for sound to travel up the well, calculate the distance to the water if the sound returns in 2.0000 s. (b) Now calculate the distance taking into account the time for sound to travel up the well. The speed of sound is 332.00 m/s in this well.
View solution
### Problem 56
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms ($8.00 \times 10^{-5} \textrm{ s}$) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
View solution
### Problem 57
A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s after being released, and (c) the time before it hits the ground.
View solution
### Problem 58
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms ($3.50 \times 10^{-3} \textrm{ s}$) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?
View solution
### Problem 59
(a) By taking the slope of the curve in Figure 2.72, verify that the velocity of the jet car is 115 m/s at $t = 20 \textrm{ s}$ . (b) By taking the slope of the curve at any point in Figure 2.73, verify that the jet car's acceleration is $5.0 \textrm{ m/s}^2$.
View solution
### Problem 60
Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at t = 10.0 s is 0.208 m/s. Assume all values are known to 3 significant figures.
View solution
### Problem 61
Using approximate values, calculate the slope of the curve in Figure 2.74 to verify that the velocity at $t = 30.0 \textrm{ s}$ is 0.238 m/s. Assume all values are known to 3 significant figures.
View solution
### Problem 62
By taking the slope of the curve in Figure 2.75, verify that the acceleration is approximately $3.2 \textrm{ m/s}^2$ at $t = 10 \textrm{ s}$ .
View solution
### Problem 63
Construct the displacement graph for the subway shuttle train as shown in Figure 2.30(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
View solution
### Problem 65
A graph of $v(t)$ is shown for a world-class track sprinter in a 100-m race. (See Figure 2.79). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at $t = 5 \textrm{ s}$? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?
View solution
## Test Prep for AP® Courses
### Problem 1 (AP)
Which of the following statements comparing position, distance, and displacement is correct?
1. An object may record a distance of zero while recording a non-zero displacement.
2. An object may record a non-zero distance while recording a displacement of zero.
3. An object may record a non-zero distance while maintaining a position of zero.
4. An object may record a non-zero displacement while maintaining a position of zero.
View solution
### Problem 2 (AP)
A student is trying to determine the acceleration of a feather as she drops it to the ground. If the student is looking to achieve a positive velocity and positive acceleration, what is the most sensible way to set up her coordinate system?
1. Her hand should be a coordinate of zero and the upward direction should be considered positive.
2. Her hand should be a coordinate of zero and the downward direction should be considered positive.
3. The floor should be a coordinate of zero and the upward direction should be considered positive.
4. The floor should be a coordinate of zero and the downward direction should be considered positive.
View solution
### Problem 3 (AP)
A group of students has two carts, A and B, with wheels that turn with negligible friction. The two carts travel along a straight horizontal track and eventually collide. Before the collision, cart A travels to the right and cart B is initially at rest. After the collision, the carts stick together.
1. Describe an experimental procedure to determine the velocities of the carts before and after the collision, including all the additional equipment you would need. You may include a labeled diagram of your setup to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure.
2. There will be sources of error in the measurements taken in the experiment both before and after the collision. Which velocity will be more greatly affected by this error: the velocity prior to the collision or the velocity after the collision? Or will both sets of data be affected equally? Justify your answer.
View solution
### Problem 4 (AP)
Figure 2.81 Graph showing Velocity vs. Time of a cart. A cart is constrained to move along a straight line. A varying net force along the direction of motion is exerted on the cart. The cart's velocity v as a function of time t is shown in the graph. The five labeled points divide the graph into four sections. Which of the following correctly ranks the magnitude of the average acceleration of the cart during the four sections of the graph?
1. $a_{CD} > a_{AB} > a_{BC} > a_{DE}$
2. $a_{BC} > a_{AB} > a_{CD} > a_{DE}$
3. $a_{AB} > a_{BC} > a_{DE} > a_{CD}$
4. $a_{CD} > a_{AB} > a_{DE} > a_{BC}$
View solution
### Problem 5 (AP)
Push a book across a table and observe it slow to a stop.
Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered positive.
Draw graphs showing the book's position vs. time and velocity vs. time if the direction of its motion is considered negative.
View solution
### Problem 6 (AP)
A group of students is attempting to determine the average acceleration of a marble released from the top of a long ramp. Below is a set of data representing the marble's position with respect to time. Use the data table above to construct a graph determining the acceleration of the marble. Select a set of data points from the table and plot those points on the graph. Fill in the blank column in the table for any quantities you graph other than the given data. Label the axes and indicate the scale for each. Draw a best-fit line or curve through your data points. Using the best-fit line, determine the value of the marble's acceleration.
View solution
### Problem 7 (AP)
Observing a spacecraft land on a distant asteroid, scientists notice that the craft is falling at a rate of 5 m/s. When it is 100 m closer to the surface of the asteroid, the craft reports a velocity of 8 m/s. According to their data, what is the approximate gravitational acceleration on this asteroid?
1. 0 m/s2
2. 0.03 m/s2
3. 0.20 m/s2
4. 0.65 m/s2
5. 33 m/s2
View solution | 6,568 | 25,633 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 47, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | longest | en | 0.941314 |
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#### Resources tagged with Working systematically similar to Buying a Balloon:
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### There are 339 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Coins (2)
##### Age 7 to 11 Challenge Level:
What is the smallest number of coins needed to make up 12 dollars and 83 cents?
##### Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Finding Fifteen
##### Age 7 to 11 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Sitting Round the Party Tables
##### Age 5 to 11 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Calcunos
##### Age 7 to 11 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Wag Worms
##### Age 7 to 11 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
##### Age 5 to 11 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### This Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Seating Arrangements
##### Age 7 to 11 Challenge Level:
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
### Two Dots
##### Age 7 to 11 Challenge Level:
Place eight dots on this diagram, so that there are only two dots on each straight line and only two dots on each circle.
### Team Scream
##### Age 7 to 11 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### Make Pairs
##### Age 7 to 11 Challenge Level:
Put 10 counters in a row. Find a way to arrange the counters into five pairs, evenly spaced in a row, in just 5 moves, using the rules.
### Different Deductions
##### Age 7 to 11 Challenge Level:
There are lots of different methods to find out what the shapes are worth - how many can you find?
### Crack the Code
##### Age 7 to 11 Challenge Level:
The Zargoes use almost the same alphabet as English. What does this birthday message say?
### Count the Trapeziums
##### Age 7 to 11 Challenge Level:
How many trapeziums, of various sizes, are hidden in this picture?
### Bunny Hop
##### Age 7 to 11 Challenge Level:
What is the smallest number of jumps needed before the white rabbits and the grey rabbits can continue along their path?
### Broken Toaster
##### Age 7 to 11 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Ordered Ways of Working Upper Primary
##### Age 7 to 11 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Possible Pieces
##### Age 5 to 11 Challenge Level:
Can you create jigsaw pieces which are based on a square shape, with at least one peg and one hole?
### Button-up Some More
##### Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### A Right Charlie
##### Age 7 to 11 Challenge Level:
Can you use this information to work out Charlie's house number?
### Room Doubling
##### Age 7 to 11 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Arranging the Tables
##### Age 7 to 11 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Professional Circles
##### Age 7 to 11 Challenge Level:
Six friends sat around a circular table. Can you work out from the information who sat where and what their profession were?
### Calendar Cubes
##### Age 7 to 11 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Two Egg Timers
##### Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Prison Cells
##### Age 7 to 11 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Family Tree
##### Age 7 to 11 Challenge Level:
Use the clues to find out who's who in the family, to fill in the family tree and to find out which of the family members are mathematicians and which are not.
### Plates of Biscuits
##### Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
### Finding All Possibilities Upper Primary
##### Age 7 to 11 Challenge Level:
These activities focus on finding all possible solutions so working in a systematic way will ensure none are left out.
### 1 to 8
##### Age 7 to 11 Challenge Level:
Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line.
### Eight Queens
##### Age 7 to 11 Challenge Level:
Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others.
### It Figures
##### Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Ice Cream
##### Age 7 to 11 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Pasta Timing
##### Age 7 to 11 Challenge Level:
Nina must cook some pasta for 15 minutes but she only has a 7-minute sand-timer and an 11-minute sand-timer. How can she use these timers to measure exactly 15 minutes?
### Half Time
##### Age 5 to 11 Challenge Level:
What could the half time scores have been in these Olympic hockey matches?
### Bean Bags for Bernard's Bag
##### Age 7 to 11 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Pouring the Punch Drink
##### Age 7 to 11 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Zargon Glasses
##### Age 7 to 11 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Polo Square
##### Age 7 to 11 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Octa Space
##### Age 7 to 11 Challenge Level:
In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon?
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Rabbits in the Pen
##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Fake Gold
##### Age 7 to 11 Challenge Level:
A merchant brings four bars of gold to a jeweller. How can the jeweller use the scales just twice to identify the lighter, fake bar?
### The Puzzling Sweet Shop
##### Age 7 to 11 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### A-magical Number Maze
##### Age 7 to 11 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Forgot the Numbers
##### Age 7 to 11 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
### Crossing the Town Square
##### Age 7 to 11 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
### Building with Rods
##### Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules? | 2,293 | 9,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-30 | latest | en | 0.945897 |
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# 16.
1 Linear Algebra
(page 602)
CHAPTER 16
16.1
## MATHEMATICS AFTER CALCULUS
(page 602)
Linear Algebra
If Ax = b and AX = B, then A times 22 3X equals Zb+SB. If Ax = 0 and AX = 0 then A times 22 3X equals 0. In this case x and X are in the nullspace of A, and so is the combination 2x 3X.The nullspace contains a l l solutions to Ax = 0. It is a subspace, which means that when x and X are in the nullspace, eo are all combinations ax bX. If x = (1,1,1) is in the nullspace then the columns add to the zero column, so they are dependent.
Another subspace is the column space of A, containing all combinations of the columns of A. The system AX = ATb. Ax = b can be solved when b is in the column space. Otherwise the best solution comes from is the transpose matrix, whose rows are the columns of A. The nullspace of AT contains all solutions Here = 0. The column space of A= (row space of A) is the fourth fundamental subspace. Each subspace to has a basis containing as many independent vectors as possible. The number of vectors in the basis is the dimension of the subspace. When Ax = A x the number X is an eigenvalue and x is an eigewector. The equation dy/dt = Ay has the exponential solution y = eA' yo. A 7 by 7 matrix has seven eigenvalues, whose product is the determinant D. If D is nonsero the matrix A has an inverse. Then Ax = b is solved by x = A-lb. The formula for D contains 7! = 5040 terms, so x is better computed by elimination. On the other hand Ax = X x means that A X I has determinant zero. The eigenvalue is computed before the eigenvector.
1 All vectors c
7+=
[ -; ]
3 Only x = 0
## 6 Plane of vectors with
9 A(%,
+ x2 + q = 0
[ ~ ] , A ( ~ ~ + ~ o ) =
[:]+[:I
## 16.1 Linear Algebra
1-X 1 1-X 1
(page 602)
1 1-X
'I
= (1 -
- 3(1- A) + 2 = X3 - 3X2 = 0 if X
= 3 or X = 0 (repeated)
SIH=[-:
-:I
ss F if b # O; T; T; F (eA'
is not a vector); T
8 Ax =
[ ]
## bas no solutions (pardel lines in the row picture)
10 If Az = b and Axp = b then A(x - xp) = 0. The difference between any two solutions is a vector in the nullspace. (So any solution x squab Xparticular xbmogeneoua .) 12 The row space of A =
[ : :]
14
[y
1 0
:] [ : ] [ i ]
=
I-: 1.
L
[ :] .
J
L' :.
## violates the third equation. So multiply by cT :
2 1 solution z l = s,z2 =5 .
16 The vectors
18 Ax is a combination of the co umns. If Ax = 0 happens with a nonsero z, the columns are 'dependent." I f Ax = 0 only happens when x = 0, the columns are 'independent."
20
[ i ] [ 11 [i ] [ 1] [ I ]
and
] [ : ] [ :]
=
## gives the best least squares
are three independent vectors in 3D so they form a LYY. The problem asks for
22 p
1 ::1
## has determinant 25 - 16 = 9 which is a2 = (dct F ) ~ .
24 det (G - X I ) =
I
-
X =1 .Eigenveciors: G
[ :] [ :]
9
5-X
of F. General reasoning: If Fz = Ax then Fax = XFx = X2x. SOz is also an eigenvector of F ~ . 26 Exponential solutions to starts from yo =
[\$:I+[-\$]=
2 = C y are
## are combinations of columns of C.
I=X2-lOA+9.Thisis(A-9)(X-1)whichis~eroatX=Oand
and G
[ -:] [ -:] .
=
yl
= eQt
[:I*
1 :]
L
and y2 = rt
1 :1 .
L
## 28 An eigenvector for E x = Ox, when X = 0, is the same as a solution to Ex = 0 (Problem 5).
Ex = 32 is solved by x =
SO A sero eigenvalue means there is a solution to Ax = OZ = 0. In this case A is not invertible and its
[:I
1
or any multiple
[i]
## 16.2 Differen tid Equations
(page 610)
determinant is rero. The reasoning also goes in reverse. 32 If F x = Xz then multiplying both sides by F-' and X- gives X-'x = F-'x. If F has eigenvalues X = 1and 3, then F-' has eigenvalues X-' = 1and f (with the same two eigenvectors z ) . The determinant of F is 1 times 3 and det F-' is 1 times
'
i.
16.2
Differential Equations
(page 610)
The solution to y' - 5y = 10 is y = Ae6' B. The homogeneous part Ae5' satisfies y' - 5y = 0. The particular solution B equals -2. The initial condition yo is matched by A = yo 2. For y' - 5y = ekt the right form is y = ACS' + ~ e For ~ y'~ - s. ~ = cost the form is y = Ae5' +tacos t+bsin t.
The equation y" 4y' 5y = 0 is second-order because i t begins w i t h y". The pure exponential solutions -e ~ - l ~ . come from the roots of X2+4X+5 = 0, which are r = -1 and s = -4. The general solution is y = ~ e ~ Changing 4y' to rero yields pure oscillation. Changing to 2y' yields X = -1f 2i, when the solutions become - 2t ~ ~ es i n - Pt.~ This oscillation is underdamped. A spring with rn = 1, d = 2, k = 5 goes y = ~ e eos directly to zero. An electrical network with L = 1,R = 2, C = also goes t o r e r o (underdamped)
+ +
1 If the right side is cos t, the form of yp is s 1 One particular solution of y" 4y = e' is e' times 5. c o s t. If the right side is 1then yp = If the right side is cos 2 t we have resonance and yp contains an extra factor t.
! .
## 13Be3' - Be3' = 8e3' gives B = 4 : y = 4e3' 3y=3-2t+t2 6Aet+4e3' = 7 a t t = O i f A = 3
7 Add y = Ae-' because y' y = 0; choose A = -1 so -e-' 3 2t t2 = 2 at t = 0
e" - 1 teh 9 y = v ; y = t; by l ' ~ 6 ~ i t a liml = lim=t k+O k k-+O 1 1 1Substitute y = Aet Btet C cost + Dsint in equation: B = 1 , C = t , D = ; , any A 13 Piwticular solution y = Ate' Be'; y' = Ate' (A B)et = c(Atet Bet) tet
+ + +
givesA=cA+l,A+B=cB,A=&,B=& 1 6 X2eAt 6XeA' + 5eAt = 0 gives X2 6 X 5 = 0, (A 5)(X + 1) = 0, X = -1 or -5 Be-'' (both negative so decay); y = Ae-' 1 7 (A2 2 X 3)e" = 0, X = -1 z t @ has imaginary part and negative real part;
+ +
+ + +
Y = ~ ~ ( - ' + f i i )+ t ~ ~ ( - 1 - f i i )y t ;= Ce-' eos t/2t + De-t sin f i t 19 d = 0 no damping; d = 1 underdamping; d = 2 critical damping; d = 3 overdamping
21 A = - f is repeated when b2 = 4c and X = - ;(tX2 2X)eAt b(tX l)eAt+ cteAt= 0 whenX2+bX+c=0and2X+b=0 25 -acost - bsint asint bcost acost bsint = cost if a = 0, b = 1,y = sint 25 y = Acos3t+ Bcos5t;y1'+9y= -25Bcos5t+9Bcos5t = cos5t gives B = yo = 0 gives A = 27 y = A(cos wt - cos wet), y" = -Aw2 coswt Awi cos wot, y" wiy = cosut gives A(-w2 wi) = 1; breaks down when w 2 = w t 29y=~e5';25~+3B=1,~=\$ ~ 1 ~ = ~ + B t = i + i t SS y" - 25y = e6'; y" y = sin ti y" = 1 t; right side solves homogeneous equation so particular
&
5;
## 16.2 Differential Quations
(page 610)
solution needs extra factor t 3 5 et ,e-', eit ,e-" 17 y = e-" 2te-"; y(27r) = (1 4 ~ ) e - ~r;r *0 39 y = (4e-* - ?e-4tlr)/(4 - r2) -+ 1 as r + 0 4 3 h 5 2; h 1 2.8
2 -2a sin 2t 26 cos 2t a cos 2t b sin 2t = 4 sin 2t; collect sin 2t and cos 2t terms to find -2a 6 = 4 and 8 cos 2 t 5 4 sin 2t. 2b + a = 0; then b = a = y = -5 4 aet cost - aet sin t bet sin t betcost = 2et cost; collect terms to find a b = 2, -a b = 0; then a = 1,b = l y = et(cos t sin t). This integral of 2et cost could be found by parts; here the correct form was assumed a t the start. 6 Ae-' solves the homogeneous equation yt + y = 0; then with A = the combination - cos 2t \$ sin 2t solves y' y = 4 sin 2t with y(0) = 0. 8 y = constant solves the homogeneous equation y' = 0, so a constant can be added to any particular solution of y' = 2et cos t (the constant A is the same as the integration constant C!). 10 Substitute homogeneous solution plus particular solution y = Aet BeM into the equation; then kB - B = 1
6,
-8,
8 %
and also A
+ B = 0 from yo = 0; thus B = -A
+
and y =
-= n.
## et -t ekt Apply l9H6pital's rule
= limkdl tekt = tet. (the variable is k!) to find limk,l 1 2 Homogeneous solution plus particular solution = Ae-' Bet C t D. Substitute to find 2Bet + C + C t + D = et t and B = f, C = 1,D = -1; then y = ~ e #et - ~ t -1 .(Initial value yo determines A.) 1 4 Particular solution is B L)t; substitute to find D = B Dt t; then B = -1, D = -1; general solution y=~et-t-l. 16 Substitution gives (A2 9)eAt = 0. (a) X = 3i or X = -3i (b) Pure oscillation because the X's are pure imaginary (c) General solution y = Ae3" Be-3it = a cos St b s i n St. 18 Substitution gives (A2 6X 9)eAt = 0. (a) X2 6X 9 = (A 3)2 and the equation (A + 3)2 = 0 has the repeated solution X = -3 (b) The general solution is y = ~ e ~-t e ~ - ~ ~ . 20 k = 0 gives overdamping; k = 1gives critical damping; k = 2 gives underdamping. 22 Substitute (te-')I' + 3(te-')' + 2(teet) = (t - 2)e-' 3(1- t)e-' 2te-' = e-' # rero. So te-' is not a homogeneous solution. (It would be if the coefficients in the equation changed from 1,3,2 to 1,2,1; then X2 2X 1has X = -1 as a double root.) 24 Substitute y t o find -w2a cos wt - w2bsin wt - wa sin wt wb coswt a coswt b sin wt = sin wt. Then 1-wx and b = (1- w2)a wb = 0 and -wa + (1 - w2)b = 1, which gives a = W2+(lTW2)2 w +( -w ) 26 The graph of cos 5t - cos 3t will look like the figure following the exercises in Section 7.2, which shows - cos 9%). sin lox sin x = (cos 11% 28 Substitute to find (iwL R &)AeiWt = veiWt. Then A = i W ~ +v ~+(iwC)-'. 30 Substitute ypatticular = A sin t + B cos t to find -A sin t - B cos t 3A sin t + 3B cos t = sin t. Then 2A = 1 and B = 0 give y = f sin t. (Note: the homogeneous solution involves cos a t and sin a t . ) = Aet cos t + Bet sin t to find -2Aet sin t 2Bet cos t 2Aet cos t 2Bet sin t = et cos t. 32 Substitute ypar,icu~ar 1 Then - 2 A + 2 B = O a n d 2 8 + 2 A = l a n d A = B = x. ,kt -ect as k approached c. This exercise uses the series 3 4 Eguation (2) used l'H6pital's Rule for the limit of y = kt-ct+ f k'tl- +c2t'+--. e" = 1+ kt + ik2t2 + .. and ect = 1+ ct + ?cat2 + . to find --
+ + + + + + + +
+ +
+ + + +
+ +
iW
+ +
t + \$(k + c)t2 + As k approaches c this is t c t 2 . - .= t ( l + c t . ..) = tect. I et . 3 6 Substitute yparticular= Act to find Aet Aet = et. Then A = and y = 2 1 38 yo = . I gives A + B = l;yh = 0 gives -A - 4 B = 0; the solution has A = B = -5, and y = \$e-' At t = 2 r the value is )e-2u - ife- 8" r;r 3
. . a .
k-c
k-c
i,
- ie-4t.
40 In exponential solutions AcM all derivatives are proportional to the function. For / = 6 8 substitute y = xn
to find n(n - l)xn-" 6x2". Comparing exponents gives n - 2 = 2n or n = -2. Then n(n - 1) = 6 and a solution is y = x - ~ ( .If the 6 were changed, this solution would become Ax-?) 44 Runge-Kutta is stable with h = .02 but 100 steps with h = .03 lead to yloo ry 670,000,000,000.
16.3
Discrete Mathematics
A graph is a set V of nodes or vertices and a set E of edges. With 6 nodes, a complete graph has 16 edges. A spanning tree has only 6 edges. A tree is defined as a graph with no loops, and it is spanning if it contains all nodes. It has one path between each pair of nodes. To find a path from node i to node j , two search methods are depth first search and breadth first search. As nodes are reached, DFS looks out from the latest node for a new one. BFS looks out from the earliest node. DFS must be prepared to backtrack to earlier nodes. In case of fire, BFS locates all doors from the room you are in before it chooses one.
1Two then two then last one; go around hexagon 3 Six (each deletes one edge) 6 Connected: there is a path between any two nodes; connecting each new node requires an edge 13 Edge lengths 1,2,4 15 No; 1,3,4 on left connect only to 2,3 on right; 1,3 on right connect only to 2 on left 17 4 l9Yes 21F(mayloop);T 2616
Breadth first search from node 4: 2, 3, 5, 6, 1, 7. Depth first search from node 4: 2, 1, 3, (backtrack t o 4), 5, 7, 6
3
0
4 A spanning tree omits one edge from each square. There are 4 x 4 choices of edges to omit: 16 spanning trees. 6 A loop is a sequence of edges from xo to XI, xl to x2, * , xh to xo : the loop (or circuit) is a closed path on the graph. A loop invobing k nodes uses k edges. That leaves only 8 - k edges to connect the loop to the remaining 9 - k nodes, which is impossible. 8 Breadth first search finds all neighbors, then all neighbors of neighbors, and so on. The farthest state is found last. 10 Please send answer! 12 A matching is a set of edges, no two of which touch the same node. 14 The shortest path tree in network A contains the edges of length 4, 5, 1. 16 The loop in network B contains the edges of length 1, 4, 2, 3. Minimum spanning tree by Method 1, starting top left: 7, 2, 3, 1, 5, 6, 8. Minimum spanning tree by Method 2 (increasing lengths, avoiding loops): 1, 2, 3, (avoid 4), 5, 6, 7, 8. 18 If one node has 4 edges going out, then every other node has one of these edges coming in: a node with 0 edges cannot exist.
## 16.3 Diwate Mathematics
(page 615)
20 Suppose a spanning tree contains edge (8) but not (6). Then it is shorter if we substitute edge (6)
for edge (8). Also it is still a spanning tree: It contains the correct number of edges and no loops. Edge (6) cannot be in a loop since edge (8) was removed; and those are the only edges into their common node. 22 (a) Starting from node s, a tree that connects each node to the next is perfect for depth first search. There are no backtracks. (b) For breadth first search the perfect tree connects node s directly to each other node. On that tree, depth first search would have to backtrack after each new node. 24 The code intends to find shortest distances dij between every pair of nodes (like a road atlas). Correction: The loop on k should be the outer loop. Then k = 1 allows direct edges i to j and paths from i to 1to j . Next k = 2 also allows paths from i to 2 to j and from i to 2 to 1 to j and from i to 1 to 2 to j. Eventually all intermediate nodes and all paths are allowed. 26 Maximum spanning tree: Method 1 adds the longest edge that goes out from the current tree (start at any node). Method 2 adds edges in decreasing order, longest first, rejecting any edge that closes a loop.
## Resource: Calculus Online Textbook
Gilbert Strang
The following may not correspond to a particular course on MIT OpenCourseWare, but has been provided by the author as an individual learning resource. | 4,836 | 13,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-35 | latest | en | 0.864098 |
http://indiavidya.com/sbi-clerks-reasoning-practice-questions-and-answers-with-explanation/ | 1,477,364,839,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719877.27/warc/CC-MAIN-20161020183839-00063-ip-10-171-6-4.ec2.internal.warc.gz | 129,369,123 | 15,879 | # SBI Clerks Reasoning Practice Questions and Answers with Explanation
Directions (Q.1-5): Study the following information and answer the questions which follow:
i) K, L, M, N, O, P and Q are sitting along a circular table facing the centre.
ii) L sits between N and O.
iii) K is third to the left of O
iv) Q is second to the left of M, who is to the immediate left of P.
1. Which of the following pairs has the first person sitting to the immediate left of the second person?
1) LO 2) MK 3) QN 4) LN 5) None of these
2. Which is the correct position of L with respect Q?
1) Second to the right 2) First to the left
3) First to the right 4) Third to the right
5) None of these
3. Which of the following has the middle person sitting between the other two?
1) NQL 2) PMK 3) POK 4) MOP 5) None of these
4. Who sits, second to the left of L?
1) P 2) M 3) K 4) Q 5) None of these
5. Who is between M and Q?
1) O 2) N 3) K 4) L 5) None of these
Directions (Q.6-10) : In each question below are three statements followed by two conclusion numbered I and II. You have to take the
three given statements to be true even if they seem to be at variance from commonly known facts and then decide which of the given conclusions logically follows from the three statements
disregarding commonly known facts.
Give answer 1) if only conclusion I is true.
Give answer 2) if only conclusion II is true.
Give answer 3) if either conclusion I or conclusion II is true.
Give answer 4) if neither conclusion I nor conclusion II is true.
Give answer 5) if both conclusions I and II are true.
6. Statements: All buildings are roads.
All trucks are mountains.
Conclusions: I. Some buildings are mountains.
7. Statements: Some poles are lights.
All lights are bulbs.
Some bulbs are wires.
Conclusions: I. Some poles are bulbs.
II. Some lights are wires.
8. Statements: All erasers are pencils.
Some pencils are pens.
No pen is a paper.
Conclusions: I. Some erasers are papers
II. No paper is an eraser.
9. Statements: Some spoons are bowls.
Some pans are both spoons and bowls.
Some forks are bowls.
Conclusions: I. Some forks are spoons.
II. Some pans are forks.
10. Statements: All boxes are tables.
All windows are tables.
All tables are fans.
Conclusions: I. All
windows are fans.
II. Some tables are boxes.
Directions (Q.11-13): In these questions symbols \$, #, % are
used for different meanings as follow:
\$ means 'neither greater nor equal to'
# means 'neither greater nor smaller than'
% means 'neither smaller nor equal to'
In each of the following questions assuming the given, statements to be true, find out which of the two conclusions I and II given below them is /are definitely true.
Give answer 1) if only conclusion I is true.
Give answer 2) if only conclusion II is true.
Give answer 3) if either conclusion I or conclusion II is true.
Give answer 4) if neither conclusion I nor conclusion II is true.
Give answer 5) if both conclusions I and II are true.
11. Statements: S \$P, P%Q, Q#R
Conclusions: I. R\$S II. R%S
12. Statements: M#K, K%P, P\$R
Conclusions: I. R\$M II. P%M
13. Statements: A\$B, B#D, D%C
Conclusions: I. D\$A II. B%C
Directions (Q.14 & 15): Read the following information carefully and answer the questions, which follow.
If 'A - B' means 'A is father of B'
If 'A + B' means 'A is daughter of B'
If 'A ÷ B' means 'A is son of B'
If 'A × B' means 'A is wife of B'
14. In the expression 'P ÷ Q - T how is T related to P?
1) Mother 2) Sister 3) Brother
4) Either brother or sister 5) None of these
15. In the expression 'P + Q × R' how is R related to P?
1) Daughter 2) Brother 3) Father 4) Sister 5) None of these
16. In a certain code 'jo ka ra' means 'go for walk', 'ma fo ka' means 'do not walk' and 'sa ta jo' means 'good for you' what is the code for 'go'?
1) Jo 2) ka 3) ra 4) ka ro ra 5) None of these
17. Four of the following five are alike in a certain way and so form a group. Which is the one that does not belong to that group?
1) 108 2) 126 3) 153 4) 207 5) 102
18. In the number 3276158, if the digits were arranged in ascending order, how many digits will remain far away from the beginning of the number as they are in the number?
1) None 2) One 3) Two 4) Three 5) More than three
19. If 'Red' means 'Blue' 'Blue' means 'Green', 'Green' means 'Orange', 'Orange' means 'pink' and 'Pink' means 'Black', then what is the colour of the clear sky?
1) Orange 2) Green 3) Blue 4) Red 5) None of these
Key : 1-4 2-5 3-2 4-1 5-3 6-5 7-1 8-4 9-4 10-5 11-4 12-4 13-2 14-4 15-3 16-3 17-5 18-3 19-2.
(1-6); On the basis of the information given in the question
, we have the above arrangements.
1.4; L is sitting immediate left of the N.
2.5; The correct position of L with respect to Q is 'second to the left' and 'fifth to the right'.
3.2; M is sitting between P and K
4.1; P sits second to the left of L.
5.3; K sits between M and Q.
All trucks are mountains.
Conversion
'All buildings are mountains' - Conversion
'Some mountains are buildings' ⎯→ Some buildings are mountains'
This is conclusion I.
All roads are trucks (A - type)
All trucks are mountains (A-type)
A + A = A-type conclusion 'All roads are mountains'.
'Some buildings are Mountains' Conversion ⎯→ 'Some mountains are roads.'
This is conlusion II.
7.1; Some poles are lights. ( I-type)
All lights are bulbs. (A-type)
I + A = I - type conclusion.
All lights are bulbs. (A-type)
Some bulbs are wires. (I-type)
A + I = '-' No conclusion.
8.4; 9.4;
10.5; Conversion
'All boxes are tables' (A-type) ⎯→ 'Some tables are boxes'. This is conclusion II.
All windows are tables.(A-type)
All tables are fans. (A-Type)
A + A = A-type conclusion.
'All windows are fans'.
This is conclusion I.
(11-13): \$ means < # means = % means >
11.4; Statements: S \$ P ⇒ S < P P % Q ⇒ P > Q
Q # R ⇒ Q = R
Conclusions: I: R\$S ⇒ R < S (Not true) II. R % S ? R > S (Not true)
12.4; Statements: M # K ⇒ M = K
K % P ⇒ K > P
P \$ R ⇒ P < R Conclusions: I. R\$M ⇒R < M (Not true) II. P % M ⇒ P > M (Not true)
13.2; Statements: A \$ B ⇒ A < B B # D ⇒ B = D D % C ⇒ D > C
Conclusions: I. D \$ A ? D < A (Not true) II. B % C ? B > C (True)
14.4; P + Q - T ⇒ T is either brother or sister of P.
15. 3; P ÷ Q × R
R is father of P.
16.3; jo ka ra ⇒ go for walk
ma fo ka ⇒ do not walk
sa ta jo ⇒ good for you
17. 5; 108 ⇒ 1 + 0 + 8 = 9
126 ⇒ 1 + 2 + 6 = 9
153 ⇒ 1 + 5 + 3 = 9
207 ⇒ 2 + 0 + 7 = 9
But, 102 ⇒ 1 + 0 + 2 = 3
18.3; 3 2 7 6 1 5 8
1 2 3 5 6 7 8
19.2; The colour of clear sky is blue. But here blue is called green. | 2,069 | 6,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2016-44 | longest | en | 0.932007 |
https://studylib.net/doc/10726715/random-variables-chapter-2 | 1,632,774,617,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00311.warc.gz | 581,750,345 | 26,971 | # Random Variables Chapter 2
```Chapter 2
Random Variables
If the value of a numerical variable depends on the outcome of an experiment, we call the variable a random
variable.
Definition 2.0.1 (Random Variable)
A function X : Ω →
7 R is called a random variable.
X assigns
to each elementary event
a real value.
Standard notation: capital letters from the end of the alphabet.
Example 2.0.3 Very simple Dartboard
In the case of three darts on a board as in the previous example, we are usually not interested in the order,
in which the darts have been thrown. We only want to count the number of times, the red area has been
hit. This count is a random variable!
More formally: we define X to be the function, that assigns to a sequence of three throws the number of
times, that the red area is hit.
X(s) = k,
if s consists of k hits to the red area and 3 − k hits to the gray area.
X(s) is then an integer between 0 and 3 for every possible sequence.
What is then the probability, that a player hits the red area exactly two times?
We are looking now for all those elementary events s of our sample space, for which X(s) = 2.
Going back to the tree, we find three possibilities for s : rrg, rgr and grr. This is the subset of Ω, for which
X(s) = 2. Very formally, this set can be written as:
{s|X(s) = 2}
We want to know the total probability:
P ({s|X(s) = 2}) = P (rrg ∪ rgr ∪ grr) = P (rrg) + P (rgr) + P (grr) =
To avoid cumbersome notation, we write
X=x
for the event
{ω|ω ∈ Ω and X(ω) = x}.
23
8
8
8
+ 3 + 3 = 0.03.
93
9
9
24
CHAPTER 2. RANDOM VARIABLES
Example 2.0.4 Communication Channel
Suppose, 8 bits are sent through a communication channel. Each bit has a certain probability to be received
incorrectly.
So this is a Bernoulli experiment, and we can use Ω8 as our sample space.
We are interested in the number of bits that are received incorrectly.
Use random variable X to “count” the number of wrong bits. X assigns a value between 0 and 8 to each
sequence in Ω8 .
Now it’s very easy to write events like:
a) no wrong bit received
X=0
P (X = 0)
b) at least one wrong bit received
X≥1
P (X ≥ 1)
c) exactly three bits are wrong
X=3
P (X = 3)
d) at least 3, but not more than 6 bits wrong 3 ≤ X ≤ 6 P (3 ≤ X ≤ 6)
Definition 2.0.2 (Image of a random variable)
The image of a random variable X is defined as
all possible
values X can
reach
Im(x) := X(Ω).
Depending on whether or not the image of a random variable is countable, we distinguish between discrete
and continuous random variables.
Example 2.0.5
1. Put a disk drive into service, measure Y = “time till the first major failure”.
Sample space Ω = (0, ∞).
Y has uncountable image → Y is a continuous random variable.
2. Communication channel: X = “# of incorrectly received bits”
Im(X) = {0, 1, 2, 3, 4, 5, 6, 7, 8} is a finite set → X is a discrete random variable.
2.1
Discrete Random Variables
Assume X is a discrete random variable. The image of X is therefore countable and can be written as
{x1 , x2 , x3 , . . .}
Very often we are interested in probabilities of the form P (X = x). We can think of this expression as a
function, that yields different probabilities depending on the value of x.
Definition 2.1.1 (Probability Mass Function, PMF)
The function pX (x) := P (X = x) is called the probability mass function of X.
A probability mass function has two main properties:
all values
must be between 0 and
1
the sum of
all values
is 1
Theorem 2.1.2 (Properties of a pmf )
pX is the pmf of X, if and only if
(i) 0 ≤ pX (x) ≤ 1 for all x ∈ {x1 , x2 , x3 , . . .}
P
(ii)
i pX (xi ) = 1
Note: this gives us an easy method to check, whether a function is a probability mass function!
2.1. DISCRETE RANDOM VARIABLES
25
Example 2.1.1
Which of the following functions is a valid probability mass function?
1.
x
pX (x)
-3
0.1
-1
0.45
0
0.15
5
0.25
7
0.05
2.
y
pY (y)
-1
0.1
0
0.45
1.5
0.25
3
-0.05
4.5
0.25
3.
z
pZ (z)
0
0.22
5
0.17
7
0.18
1
0.18
3
0.24
We need to check the two properties of a pmf for pX , pY and pZ .
1st property: probabilities between 0 and 1 ?
This eliminates pY from the list of potential probability mass functions, since pY (3) is negative.
The other two functions fulfill the property.
2nd
P property: sum of all probabilities is 1?
Pi p(xi ) = 1, so pX is a valid probability mass function.
i p(zi ) = 0.99 6= 1, so pZ is not a valid probability mass function.
Example 2.1.2 Probability Mass Functions
1. Very Simple Dartboard
X, the number of times, a player hits the red area with three darts is a value between 0 and 3.
What is the probability mass function for X?
The probability mass function pX can be given as a list of all possible values:
pX (0)
=
P (X = 0) = P (ggg) =
83
≈ 0.70
93
82
≈ 0.26
93
8
pX (2) = P (X = 2) = P (rrg) + P (rgr) + P (grr) = 3 · 3 ≈ 0.03
9
1
pX (3) = P (X = 3) = P (rrr) = 3 ≈ 0.01
9
pX (1)
=
P (X = 1) = P (rgg) + P (grg) + P (ggr) = 3 ·
2. Roll of a fair die
Let Y be the number of spots on the upturned face of a die:
Obviously, Y is a random variable with image {1, 2, 3, 4, 5, 6}.
Assuming, that the die is a fair die means, that the probability for each side is equal. The probability
mass function for Y therefore is pY (i) = 61 for all i in {1, 2, 3, 4, 5, 6}.
3. The diagram shows all six faces of a particular die. If Z denotes the number of
spots on the upturned face after toss this die, what is the probability mass function
for Z?
Assuming, that each face of the die appears with the same probability, we have 1
possibility to get a 1 or a 4, and two possibilities for a 2 or 3 to appear, which gives
a probability mass function of:
x
p(x)
1
1/6
2
1/3
3
1/3
4
1/6
26
2.1.1
CHAPTER 2. RANDOM VARIABLES
Expectation and Variance
Example 2.1.3 Game
Suppose we play a “game”, where you toss a die. Let X be the number of spots, then
if X is
1,3 or 5 I pay you \$ X
2 or 4 you pay me \$ 2 · X
6 no money changes hands.
What money do I expect to win?
For that, we look at another function, h(x), that counts the money I win with respect to the number of spots:
−x for x = 1, 3, 5
2x for x = 2, 4
h(x) =
0 for x = 6.
Now we make a list:
In 1/6 of all tosses X will be 1, and I will gain -1 dollars
In 1/6 of all tosses X will be 2, and I will gain 4 dollars
In 1/6 of all tosses X will be 3, and I will gain -3 dollars
In 1/6 of all tosses X will be 4, and I will gain 8 dollars
In 1/6 of all tosses X will be 5, and I will gain -5 dollars
In 1/6 of all tosses X will be 6, and I will gain 0 dollars
In total I expect to get 61 · (−1) + 16 · 4 + 61 · (−3) + 61 · 8 + 16 · (−5) + 61 · 0 = 63 = 0.5 dollars per play.
Assume, that instead of a fair die, we use the die from example 3. How does that change my expected gain?
h(x) is not affected by the different die, but my expected gain changes: in total I expect to gain:
1
1
1
1
9
1
6 · (−1) + 3 · 4 + 3 · (−3) + 6 · 8 + 0 · (−5) + 6 · 0 = 6 = 1.5 dollars per play.
Definition 2.1.3 (Expectation)
The expected value of a function h(X) is defined as
E[h(X)] :=
X
h(xi ) · pX (xi ).
i
The most important version of this is h(x) = x:
E[X] =
X
xi · pX (xi ) =: µ
i
Example 2.1.4 Toss of a Die
Toss a fair die, and denote by X the number of spots on the upturned face.
What is the expected value for X?
Looking at the above definition for E[X], we see that we need to know the probability mass function for a
computation.
The probability mass function of X is pX (i) = 61 for all i ∈ {1, 2, 3, 4, 5, 6}.
Therefore
6
X
1
1
1
1
1
1
E[X] =
ipX (i) = 1 · + 2 · + 3 · + 4 · + 5 · + 6 · = 3.5.
6
6
6
6
6
6
i=1
A second common measure for describing a random variable is a measure, how far its values are spread out.
We measure, how far we expect values to be away from the expected value:
2.1. DISCRETE RANDOM VARIABLES
27
Definition 2.1.4 (Variance of a random variable)
The variance of a random variable X is defined as:
V ar[X] := E[(X − E[X])2 ] =
X
(xi − E[X])2 · pX (xi )
i
The variance
is measured in squared units of X.
p
σ := V ar[X] is called the standard deviation of X, its units are the original units of X.
Example 2.1.5 Toss of a Die, continued
Toss a fair die, and denote with X the number of spots on the upturned face.
What is the variance for X?
Looking at the above definition for V ar[X], we see that we need to know the probability mass function and
E[X] for a computation.
The probability mass function of X is pX (i) = 61 for all i ∈ {1, 2, 3, 4, 5, 6}; E[X] = 3.5
Therefore
6
X
1
1
1
1
1
1
2
V ar[X] =
(Xi − 3.5)2 pX (i) = 6.25 · + 2.25 · + 0.25 · + 0.25 · + 2.25 · + 6.25 · = 2.917 (spots ).
6
6
6
6
6
6
i=1
The standard deviation for X is:
σ=
2.1.2
p
V ar(X) = 1.71 (spots).
Some Properties of Expectation and Variance
The following theorems make computations with expected value and variance of random variables easier:
Theorem 2.1.5
For two random variables X and Y and two real numbers a, b holds:
E[aX + bY ] = aE[X] + bE[Y ].
Theorem 2.1.6
For a random variable X and a real number a holds:
(i) E[X 2 ] = V ar[X] + (E[X])2
(ii) V ar[aX] = a2 V ar[X]
Theorem 2.1.7 (Chebyshev’s Inequality)
For any positive real number k, and random variable X with variance σ 2 :
P (|X − E[X]| ≤ kσ) ≥ 1 −
2.1.3
1
k2
Probability Distribution Function
Very often we are interested in the probability of a whole range of values, like P (X ≤ 5) or P (4 ≤ X ≤ 16).
For that we define another function:
Definition 2.1.8 (probability distribution function)
Assume X is a discrete random variable:
The function FX (t) := P (X ≤ t) is called the probability distribution function of X.
28
CHAPTER 2. RANDOM VARIABLES
Relationship between pX and FX
Since X is a discrete random variable, the image of X can be written as {x1 , x2 , x3 , . . .}, we are therefore
interested in all xi with xi ≤ t:
X
FX (t) = P (X ≤ t) = P ({xi |xi ≤ t}) =
pX (xi ).
i,with xi ≤t
Note: in contrast to the probability mass function, FX is defined on R (not only on the image of X).
Example 2.1.6 Roll a fair die
X = # of spots on upturned face
Ω = {1, 2, 3, 4, 5, 6}
pX (1) = pX (2) = . . . = pX (6) = 16
F (X)(t) =
P
i<t
pX (i) =
Properties of FX
variable X.
Pbtc
i=1
pX (i) =
btc
6 ,
where btc is the truncated value of t.
The following properties hold for the probability distribution function FX of a random
• 0 ≤ FX (t) ≤ 1 for all t ∈ R
• FX is monotone increasing, (i.e. if x1 ≤ x2 then FX (x1 ) ≤ FX (x2 ).)
• limt→−∞ FX (t) = 0 and limt→∞ FX (t) = 1.
• FX (t) has a positive jump equal to pX (xi ) at {x1 , x2 , x3 , . . .}; FX is constant in the interval [xi , xi+1 ).
Whenever no confusion arises, we will omit the subscript X.
2.2
Special Discrete Probability Mass Functions
In many theoretical and practical problems, several probability mass functions occur often enough to be
worth exploring here.
2.2.1
Bernoulli pmf
Situation: Bernoulli experiment (only two outcomes: success/ no success)
with P ( success ) = p
We define a random variable X as:
X( success ) = 1
X( no success ) = 0
The probability mass function pX of X is then:
pX (0) = 1 − p
pX (1) = p
This probability mass function is called the Bernoulli mass function.
The distribution function FX is then:
t<0
0
1−p 0≤t<1
FX (t) =
1
1≤t
This distribution function is called the Bernoulli distribution function.
That’s a very simple probability function, and we’ve already seen sequences of Bernoulli experiments. . .
2.2. SPECIAL DISCRETE PROBABILITY MASS FUNCTIONS
2.2.2
29
Binomial pmf
Situation: n sequential Bernoulli experiments, with success rate p for a single trial. Single trials are independent from each other.
We are only interested in the number of successes he had in total after n trials, therefore we define a random
variable X as:
X = “ number of successes in n trials”
This leads to an image of X as
im(X) = {0, 1, 2, . . . , n}
We can think of the sample space Ω as the set of sequences of length n that only consist of the letters S and
F for “success” and ”failure”:
Ω = {F...F F, F...F S, ...., S...SS}
This way, we get 2n different outcomes in the sample space.
Now, we want to derive a probability mass function for X, i.e. we want to get to a general expression for
pX (k) for all possible k = 0, . . . , n.
pX (k) = P (X = k), i.e. we want to find the probability, that in a sequence of n trials there are exactly k
successes.
Think: if s is a sequence with k successes and n − k failures, we already know the probability:
P (s) = pk (1 − p)n−k .
Now we need to know, how many possibilities there are, to have k successes in n trials: think of the n trials
as numbers from 1 to n. To have k successes, we need to choose a set of k of these numbers out of the n
possible numbers. Do you see it? - That’s the Binomial coefficient, again.
pX (k) is therefore:
n k
p (1 − p)n−k .
pX (k) =
k
This probability mass function is called the Binomial mass function.
The distribution function FX is:
FX (t) =
btc X
n
i=0
i
pi (1 − p)n−i =: Bn,p (t)
This function is called the Binomial distribution Bn,p , where n is the number of trials, and p is the probability
for a success.
It is a bit cumbersome to compute values for the distribution function. Therefore, those values are tabled
with respect to n and p.
Example 2.2.1
Compute the probabilities for the following events:
A box contains 15 components that each have a failure rate of 2%. What is the probability that
1. exactly two out of the fifteen components are defective?
2. at most two components are broken?
3. more than three components are broken?
4. more than 1 but less than 4 are broken?
Let X be the number of broken components. Then X has a B15,0.02 distribution.
30
CHAPTER 2. RANDOM VARIABLES
1. P (exactly two out of the fifteen components are defective) = pX (2) =
15
2
0.022 0.9813 = 0.0323.
2. P (at most two components are broken) = P (X ≤ 2) = B15,0.02 (2) = 0.9638.
3. P ( more than three components are broken ) = P (X > 3) = 1 − P (X ≤ 3) = 1 − 0.9945 = 0.0055.
4. P ( more than 1 but less than 4 are broken ) = P (1 < X < 4) = P (X ≤ 3) − P (X ≤ 1) = 0.9945 −
0.8290 = 0.1655.
If we want to say that a random variable has a binomial distribution, we write:
X ∼ Bn,p
What are the expected value and variance of X ∼ Bn,p ?
E[X]
=
=
=
n
X
i · pX (i) =
i=0
n
X
n i
i·
p (1 − p)n−i =
i
i=0
n
X
i=1
= np ·
i
n!
pi (1 − p)n−i
i!(n − i)!
n−1
X
j=0
|
V ar[X]
2.2.3
j:=i−1
=
(n − 1)!
pj (1 − p)n−1−j = np
j!((n − 1) − j)!
{z
}
=1
= . . . = np(1 − p).
Geometric pmf
Assume, we have a single Bernoulli experiment with probability for success p.
Now, we repeat this experiment until we have a first success.
Denote by X the number of repetitions of the experiment until we have the first success.
Note: X = k means, that we have k − 1 failures and the first success in the kth repetition of the experiment.
The sample space Ω is therefore infinite and starts at 1 (we need at least one experiment):
Ω = {1, 2, 3, 4, . . .}
Probability mass function:
pX (k) = P (X = k) = (1 − p)k−1 · p
| {z } |{z}
k−1 failures
success!
This probability mass function is called the Geometric mass function.
Expected value and variance of X are:
E[X]
=
∞
X
i=1
V ar[X]
=
i(1 − p)i p = . . . =
1
,
p
∞
X
1
1−p
(i − )2 (1 − p)i p = . . . =
.
p
p2
i=1
2.2. SPECIAL DISCRETE PROBABILITY MASS FUNCTIONS
31
Example 2.2.2 Repeat-until loop
Examine the following programming statement:
Repeat S until B
assume P (B = true) = 0.1 and let X be the number of times S is executed.
Then, X has a geometric distribution,
P (X = k) = pX (k) = 0.9k−1 · 0.1
How often is S executed on average? - What is E[X]? Using the above formula, we get E[X] =
1
p
= 10.
We still need to compute the distribution function FX . Remember, FX (t) is the probability for X ≤ t.
Instead of tackling this problem directly, we use a trick and look at the complementary event X > t. If X is
greater than t, this means that the first btc trials yields failures. This is easy to compute! It’s just (1 − p)btc .
Therefore the probability distribution function is:
FX (t) = 1 − (1 − p)btc =: Geop (t)
This function is called the Geometric distribution (function) Geop .
Example 2.2.3 Time Outs at the Alpha Farm
Watch the input queue at the alpha farm for a job that times out.
The probability that a job times out is 0.05.
Let Y be the number of the first job to time out, then Y ∼ Geo0.05 .
What’s then the probability that
• the third job times out?
P (Y = 3) = 0.952 0.05 = 0.045
• Y is less than 3?
P (Y < 3) = P (Y ≤ 2) = 1 − 0.952 = 0.0975
• the first job to time out is between the third and the seventh?
P (3 ≤ Y ≤ 7) = P (Y ≤ 7) − P (Y ≤ 2) = 1 − 0.957 − (1 − 0.952 ) = 0.204
What are the expected value for Y , what is V ar[Y ]?
Plugging in p = 0.05 in the above formulas gives us:
2.2.4
E[Y ]
=
V ar[Y ]
=
1
= 20
p
1−p
= 380
p2
we expect the 20th job to be the first time out
Poisson pmf
The Poisson density follows from a certain set of assumptions about the occurrence of “rare” events in time
or space.
The kind of variables modelled using a Poisson density are e.g.
X = # of alpha particles emitted from a polonium bar in an 8 minute period.
Y = # of flaws on a standard size piece of manufactured product (100m coaxial cable)
Z = # of hits on a web page in a 24h period.
32
CHAPTER 2. RANDOM VARIABLES
The Poisson probability mass function is defined as:
p(x) =
e−λ λx
x!
for x = 0, 1, 2, 3, . . .
λ is called the rate parameter.
P oλ (t) := FX (t) is the Poisson distribution (function).
We need to check that p(x) as defined above is actually a probability mass function, i.e. we need to check
whether the two basic properties (see theorem 2.1.2) are true:
• Obviously, all values of p(x) are positive for x ≥ 0.
• Do all probabilities sum to 1?
∞
X
p(x) =
k=0
∞
X
∞
e−λ
X λk
λk
= e−λ
k!
k!
(∗)
k=0
k=0
Now, we need to remember from calculus that the exponential function has the series representation
ex =
∞
X
xn
.
n!
n=0
In our case this simplifies (∗) to:
e−λ
∞
X
λk
k=0
k!
= e−λ · eλ = 1.
p(x) is therefore a valid probability mass function.
Expected Value and Variance of X ∼ P oλ are:
E[X]
V ar[X]
=
∞
X
e−λ λx
x
= ... = λ
x!
x=0
= ... = λ
Computing E[X] and V ar[X] involves some math, but as it is not too hard, we can do the computation for
E[X]:
E[X]
=
∞
∞
X
X
λx
e−λ λx
= e−λ
x
=
x
x!
x!
x=0
x=0
= e−λ
= e
−λ
for x = 0 the expression is 0
∞
∞
X
X
λx
λx
x
= e−λ
=
x!
(x − 1)!
x=1
x=1
λ
= e−λ λ
∞
X
x=1
∞
X
x=0
x
λx−1
=
(x − 1)!
x
λx
= e−λ λeλ = λ
(x)!
start at x = 0 again and change summation index
How do we choose λ in an example? - look at the expected value!
2.2. SPECIAL DISCRETE PROBABILITY MASS FUNCTIONS
33
Example 2.2.4
A manufacturer of chips produces 1% defectives.
What is the probability that in a box of 100 chips no defective is found?
Let X be the number of defective chips found in the box.
So far, we would have modelled X as a Binomial variable with distribution B100,0.01 .
100
Then P (X = 0) = 100
0.010 = 0.366.
0 0.99
On the other hand, a defective chip can be considered to be a rare event, since p is small (p = 0.01). What
else can we do?
We expect 100 · 0.01 = 1 chip out of the box to be defective. If we model X as Poisson variable, we know,
that the expected value of X is λ. In this example, therefore, λ = 1.
−1 0
Then P (X = 0) = e 0!1 = 0.3679.
No big differences between the two approaches!
For larger k, however, the binomial coefficient nk becomes hard to compute, and it is easier to use the
Poisson distribution instead of the Binomial distribution.
Poisson approximation of Binomial pmf For large n, the Binomial distribution is approximated by
the Poisson distribution, where λ is given as np:
n k
(np)k
p (1 − p)n−k ≈ e−np
k!
k
Rule of thumb: use Poisson approximation if n ≥ 20 and (at the same time) p ≤ 0.05.
Why does the approximation work? - We will have a closer look at why the Poisson distribution
approximates the Binomial distribution. This also explains why the Poisson is defined as it is.
Example 2.2.5 Typos
Imagine you are supposed to proofread a paper. Let us assume that there are on average 2 typos on a page
and a page has 1000 words. This gives a probability of 0.002 for each word to contain a typo.
The number of typos on a page X is then a Binomial random variable, i.e. X ∼ B1000,0.002 .
Let’s have a closer look at a couple of probabilities:
• the probability for no typo on a page is P (X = 0). We know, that
P (X = 0) = (1 − 0.002)1000 = 0.9981000 .
We can also write this probability as
P (X = 0) =
2
1−
1000
1000
(= 0.13506).
From calculus we know, that
x n
= ex .
n→∞
n
Therefore the probability for no typo on the page is approximately
lim
1−
P (X = 0) ≈ e−2
(= 0.13534).
• the probability for exactly one typo on a page is
1000
P (X = 1) =
0.002 · 0.998999
1
(= 0.27067).
We can write this as
2
P (X = 1) = 1000 ·
1000
1−
2
1000
999
≈ 2 · e−2
(= 0.27067)
34
CHAPTER 2. RANDOM VARIABLES
• the probability for exactly two typos on a page is
1000
P (X = 2) =
0.0022 · 0.998998
2
(= 0.27094),
which we again re-write to
1000 · 999
22
P (X = 2) =
·
2
1000 · 1000
2
1−
1000
998
≈ 2 · e−2
(= 0.27067)
• and a last one: the probability for exactly three typos on a page is
1000
P (X = 3) =
0.0023 · 0.998997
(= 0.18063),
3
which is
P (X = 3) =
2.2.5
1000 · 999 · 998
23
·
3·2
1000 · 1000 · 1000
1−
2
1000
997
≈
23 −2
·e
3!
(= 0.18045)
Compound Discrete Probability Mass Functions
Real problems very seldom concern a single random variable. As soon as more than 1 variable is involved it
is not sufficient to think of modeling them only individually - their joint behavior is important.
Again, the How do we specify probabilities for more than one random variable at a time?
individual
probabili- Consider the 2 variable case: X, Y are two discrete variables. The joint probability mass function is defined
ties must be
between 0 as
and 1 and
their sum
PX,Y (x, y) := P (X = x ∩ Y = y)
must be 1.
Example 2.2.6
A box contains 5 unmarked PowerPC G4 processors of different speeds:
2 400 mHz
1 450 mHz
2 500 mHz
Select two processors out of the box (without replacement) and let
X = speed of the first selected processor
Y = speed of the second selected processor
For a sample space we can draw a table of all the possible combinations of processors. We will distinguish
between processors of the same speed by using the subscripts 1 or 2 .
Ω
4001 4002
4001
x
4002
x
450
x
x
5001
x
x
5002
x
x
1st processor
450
x
x
x
x
5001
x
x
x
x
5002
x
x
x
x
-
2nd processor
In total we have 5 · 4 = 20 possible combinations.
Since we draw at random, we assume that each of the above combinations is equally likely. This yields the
following probability mass function:
2.2. SPECIAL DISCRETE PROBABILITY MASS FUNCTIONS
400
450
500 (mHz)
1st proc.
400
0.1
0.1
0.2
35
2nd processor
450 500 (mHz)
0.1
0.2
0.0
0.1
0.1
0.1
What is the probability for X = Y ?
this might be important if we wanted to match the chips to assemble a dual processor machine:
P (X = Y )
= pX,Y (400, 400) + pX,Y (450, 450) + pX,Y (500, 500) =
=
0.1 + 0 + 0.1 = 0.2.
Another example: What is the probability that the first processor has higher speed than the second?
P (X > Y )
= pX,Y (400, 450) + pX,Y (400, 500) + pX,Y (450, 500) =
=
0.1 + 0.2 + 0.1 = 0.4.
We can go from joint probability mass functions to individual pmfs:
X
pX (x) =
pX,Y (x, y)
y
X
“marginal” pmfs
pY (y) =
pX,Y (x, y)
x
Example 2.2.7 Continued
For the previous example the marginal probability mass functions are
x
pX (x)
400
0.4
450
0.2
500 (mHz)
0.4
y
pY (y)
400
0.4
450
0.2
500 (mHz)
0.4
Just as we had the notion of expected value for functions with a single random variable, there’s an expected
value for functions in several random variables:
X
E[h(X, Y )] :=
h(x, y)pX,Y (x, y)
x,y
Example 2.2.8 Continued
Let X, Y be as before.
What is E[|X − Y |] (the average speed difference)?
here, we have the situation E[|X − Y |] = E[h(X, Y )], with h(X, Y ) = |X − Y |.
Using the above definition of expected value gives us:
X
E[|X − Y |] =
|x − y|pX,Y (x, y) =
x,y
= |400 − 400| · 0.1 + |400 − 450| · 0.1 + |400 − 500| · 0.2 +
|450 − 400| · 0.1 + |450 − 450| · 0.0 + |450 − 500| · 0.1 +
|500 − 400| · 0.2 + |500 − 450| · 0.1 + |500 − 500| · 0.1 =
=
0 + 5 + 20 + 5 + 0 + 5 + 20 + 5 + 0 = 60.
36
CHAPTER 2. RANDOM VARIABLES
The most important cases for h(X, Y ) in this context are linear combinations of X and Y .
For two variables we can measure how “similar” their values are:
Definition 2.2.1 (Covariance)
The covariance between two random variables X and Y is defined as:
Cov(X, Y ) = E[(X − E[X])(Y − E[Y ])]
Note, that this definition looks very much like the definition for the variance of a single random variable. In
fact, if we set Y := X in the above definition, the Cov(X, X) = V ar(X).
Definition 2.2.2 (Correlation)
The (linear) correlation between two variables X and Y is
% := p
Cov(X, Y )
V ar(X) · V ar(Y )
• % is between -1 and 1
• if % = 1 or -1, Y is a linear function of X
%=1
% = −1
→ Y = aX + b with a > 0,
→ Y = aX + b with a < 0,
% is a measure of linear association between X and Y . % near ±1 indicates a strong linear relationship, %
near 0 indicates lack of linear association.
Example 2.2.9 Continued
What is % in our box with five chips?
Check:
E[X] = E[Y ] = 450
Use marginal pmfs to compute!
V ar[X] = V ar[Y ] = 2000
The covariance between X and Y is:
X
Cov(X, Y ) =
(x − E[X])(y − E[Y ])pX,Y (x, y) =
x,y
=
(400 − 450)(400 − 450) · 0.1 + (450 − 450)(400 − 450) · 0.1 + (500 − 450)(400 − 450) · 0.2 +
(400 − 450)(450 − 450) · 0.1 + (450 − 450)(450 − 450) · 0.0 + (500 − 450)(450 − 450) · 0.1 +
(400 − 450)(500 − 450) · 0.2 + (450 − 450)(500 − 450) · 0.1 + (500 − 450)(500 − 450) · 0.1 =
=
250 + 0 − 500 + 0 + 0 + 0 − 500 + 250 + 0 = −500.
% therefore is
%= p
Cov(X, Y )
V ar(X)V ar(Y )
=
−500
= −0.25,
2000
% indicates a weak negative (linear) association.
Definition 2.2.3 (Independence)
Two random variables X and Y are independent, if their joint probability pX,Y is equal to the product of
the marginal densities pX · pY .
2.3. CONTINUOUS RANDOM VARIABLES
37
Note: so far, we’ve had a definition for the independence of two events A and B: A and B are independent,
if P (A ∩ B) = P (A) · P (B).
Random variables are independent, if all events of the form X = x and Y = y are independent.
Example 2.2.10 Continued
Let X and Y be defined as previously.
Are X and Y independent?
Check: pX,Y (x, y) = pX (x) · pY (y) for all possible combinations of x and y.
Trick: whenever there is a zero in the joint probability mass function, the variables cannot be independent:
pX,Y (450, 450) = 0 6= 0.2 · 0.2 = pX (450) · pY (450).
Therefore, X and Y are not independent!
More properties of Variance and Expected Values
Theorem 2.2.4
If two random variables X and Y are independent,
E[X · Y ]
V ar[X + Y ]
=
= E[X] · E[Y ]
= V ar[X] + V ar[Y ]
Theorem 2.2.5
For two random variables X and Y and three real numbers a, b, c holds:
V ar[aX + bY + c] = a2 V ar[X] + b2 V ar[Y ] + 2ab · Cov(X, Y )
Note: by comparing the two results, we see that for two independent random variables X and Y , the
covariance Cov(X, Y ) = 0.
Example 2.2.11 Continued
E[X − Y ]
V ar[X − Y ]
2.3
= E[X] − E[Y ] = 450 − 450 = 0
= V ar[X] + (−1)2 V ar[Y ] − 2 Cov(X, Y ) = 2000 + 2000 + 1000 = 5000
Continuous Random Variables
All previous considerations for discrete variables have direct counterparts for continuous variables.
So far, a lot of sums have been involved, e.g. to compute the distribution functions or expected values.
Summing over (uncountable) infinite many values corresponds to an integral.
The main trick in working with continuous random variables is to substitute all sums by integrals in the
definitions.
As in the case of a discrete random variable, we define a distribution function as the probability that a
random variable has outcome t or a smaller value:
Definition 2.3.1 (probability distribution function)
Assume X is a continuous random variable:
The function FX (t) := P (X ≤ t) is called the probability distribution function of X.
The only difference to the discrete case is that the distribution function of a continuous variable is not a
stairstep function:
38
CHAPTER 2. RANDOM VARIABLES
Properties of FX
variable X.
The following properties hold for the probability distribution function FX for random
• 0 ≤ FX (t) ≤ 1 for all t ∈ R
• FX is monotone increasing, (i.e. if x1 ≤ x2 then FX (x1 ) ≤ FX (x2 ).)
• limt→−∞ FX (t) = 0 and limt→∞ FX (t) = 1.
f (x) is no
probability!
f (x) may
be > 1.
Now, however, the situation is slightly different from the discrete case:
Definition 2.3.2 (density function)
For a continuous variable X with distribution function FX the density function of X is defined as:
0
fX (x) := FX (x).
Theorem 2.3.3 (Properties of f (x))
A function fX is a density function of X, if
(i) fX (x) ≥ 0 for all x,
R∞
(ii) −∞ f (x)dx = 1.
Relationship between fX and FX Since the density function fX is defined as the derivative of the
distribution function, we can re-gain the distribution function from the density by integrating: Then
Rt
• FX (t) = P (X ≤ t) = −∞ f (x)dx
• P (a ≤ X ≤ b) =
Rb
a
f (x)dx
Therefore,
Z
P (X = a) = P (a ≤ X ≤ a) =
a
f (x)dx = 0.
a
Example 2.3.1
Let Y be the time until the first major failure of a new disk drive.
A possible density function for Y is
−y
e
y>0
f (y) =
0
otherwise
First, we need to check, that f (y) is actually a density function. Obviously, f (y) is a non-negative function
on whole of R.
The second condition, f must fulfill to be a density of Y is
Z ∞
Z ∞
f (y)dy =
e−y dy = −e−y |∞
0 = 0 − (−1) = 1
−∞
0
What is the probability that the first major disk drive failure occurs within the first year?
Z
P (Y ≤ 1) =
1
e−y dy = −e−y |10 = 1 − e−1 ≈ 0.63.
0
What is the distribution function of Y ?
Z t
Z t
f (y)dy =
e−y dy = 1 − e−t for all t ≥ 0.
FY (t) =
∞
0
2.4. SOME SPECIAL CONTINUOUS DENSITY FUNCTIONS
f(y)
39
density function of Y
y
F(y)
distribution function of Y
y
Figure 2.1: Density and Distribution function of random variable Y .
Summary:
discrete random variable
image Im(X) finite or countable infinite
continuous random variable
image Im(X) uncountable
probability distribution function:
P
FX (t) = P (X ≤ t) = k≤btc pX (k)
FX (t) = P (X ≤ t) =
probability mass function:
pX (x) = P (X = x)
probability density function:
0
fX (x) = FX (x)
expected value:
P
E[h(X)] = x h(x) · pX (x)
E[h(X)] =
variance:
V ar[X] =P
E[(X − E[X])2 ] =
= x (x − E[X])2 pX (x)
V ar[X] =RE[(X − E[X])2 ] =
∞
= −∞ (x − E[X])2 fX (x)dx
2.4
2.4.1
R
x
Rt
∞
f (x)dx
h(x) · fX (x)dx
Some special continuous density functions
Uniform Density
One of the most basic cases of a continuous density is the uniform density. On the finite interval (a, b) each
value has the same density (cf. diagram 2.2):
1
if a < x < b
b−a
f (x) =
0
otherwise
The distribution function FX is
Ua,b (x) := FX (x) =
0
x
b−a
1
if x ≤ a
if a < x < b
if x ≥ b.
We now know how to compute expected value and variance of a continuous random variable.
40
CHAPTER 2. RANDOM VARIABLES
f(x)
1/
uniform density on (a,b)
(b-a)
a
b
x
Figure 2.2: Density function of a uniform variable X on (a, b).
Assume, X has a uniform distribution on (a, b). Then
Z
b
1
1 1 2b
dx =
x | =
b−a
b−a2 a
a
b2 − a2
1
=
= (a + b).
2(b − a)
2
Z b
a+b 2 1
(b − a)2
V ar[X] =
(x −
)
dx = . . . =
.
2
b−a
12
a
E[X]
x
=
Example 2.4.1
The(pseudo) random number generator on my calculator is supposed to create realizations of U (0, 1) random
variables.
Define U as the next random number the calculator produces.
What is the probability, that the next number is higher than 0.85?
1
For that, we want to compute P (U ≥ 0.85). We know the density function of U : fU (u) = 1−0
= 1. Therefore
Z
1
P (U ≥ 0.85) =
1du = 1 − 0.85 = 0.15.
0.85
2.4.2
Exponential distribution
This density is commonly used to model waiting times between occurrences of “rare” events, lifetimes of
electrical or mechanical devices.
Definition 2.4.1 (Exponential density)
A random variable X has exponential density (cf. figure 2.3), if
λe−λx if x ≥ 0
fX (x) =
0
otherwise
λ is called the rate parameter.
Mean, variance and distribution function are easy to compute. They are:
E[X]
=
V ar[X]
=
Expλ (t)
=
1
λ
1
λ2
FX (t) =
0
1 − e−λx
if x < 0
if x ≥ 0
The following example will accompany us throughout the remainder of this class:
we expect
X to be in
the middle
between a
and b - makes
sense, doesn’t
it?
2.4. SOME SPECIAL CONTINUOUS DENSITY FUNCTIONS
41
f2
f1
f0.5
x
Figure 2.3: Density functions of exponential variables for different rate parameters 0.5, 1, and 2.
Example 2.4.2 Hits on a webpage
On average there are 2 hits per minute on a specific web page.
I start to observe this web page at a certain time point 0, and decide to model the waiting time till the first
hit Y (in min) using an exponential distribution.
What is a sensible value for λ, the rate parameter?
Think: on average there are 2 hits per minute - which makes an average waiting time of 0.5 minutes between
hits.
We will use this value as the expected value for Y : E[Y ] = 0.5.
On the other hand, we know, that the expected value for Y is 1/λ. → we are back at 2 = λ as a sensible
choice for the parameter!
λ describes the rate, at which this web page is hit!
What is the probability that we have to wait at most 40 seconds to observe the first hit?
ok, we know the rate at which hits come to the web page in minutes - so, it’s advisable to express the 40s in
minutes also: The above probability then becomes:
What is the probability that we have to wait at most 2/3 min to observe the first hit?
This, we can compute:
P (Y ≤ 2/3) = Expλ (2/3) = 1 − −e−2/3·2 ≈ 0.736
How long do we have to wait at most, to observe a first hit with a probability of 0.9?
This is a very different approach to what we have looked at so far!
Here, we want to find a t, for which P (Y ≤ t) = 0.9:
P (Y ≤ t) = 0.9
⇐⇒ 1 − e−2t = 0.9
⇐⇒ e−2t = 0.1
⇐⇒ t = −0.5 ln 0.1 ≈ 1.15 (min) - that’s approx. 69 s.
Memoryless property
Example 2.4.3 Hits on a web page
In the previous example I stated that we start to observe the web page a time point 0. Does the choice of
this time point affect our analysis in any way?
Let’s assume, that during the first minute after we started to observe the page, there is no hit.
What is the probability, that we have to wait for another 40 seconds for the first hit? - this implies an answer
to the question, what would have happened, if we had started our observation of the web page a minute
later - would we still get the same results?
42
CHAPTER 2. RANDOM VARIABLES
The probability we want to compute is a conditional probability. If we think back - the conditional probability
of A given B was defined as
P (A ∩ B)
P (A|B) :=
P (B)
Now, we have to identify, what the events A and B are in our case. The information we have is, that during
the first minute, we did not observe a hit =: B, i.e. B = (Y > 1). The probability we want to know, is that
we have to wait another 40 s for the first hit: A = wait for 1 min and 40 s for the first hit (= Y ≤ 5/3).
P ( first hit within 5/3 min
P (A ∩ B)
P (Y ≤ 5/3 ∩ Y > 1)
=
=
P (B)
P (Y > 1)
|
no hit during 1st min) = P (A|B) =
=
P (1 < Y ≤ 5/3)
e−2 − e−10/3
= 0.736.
=
1 − P (Y < 1)
e−2
That’s exactly the same probability as we had before!!!
The result of this example is no coincidence. We can generalize:
P (Y ≤ t + s|Y ≥ s) = 1 − e−λt = P (Y ≤ t)
This means: a random variable with an exponential distribution “forgets” about its past. This is called the
memoryless property of the exponential distribution.
An electrical or mechanical device whose lifetime we model as an exponential variable therefore “stays as
good as new” until it suddenly breaks, i.e. we assume that there’s no aging process.
2.4.3
Erlang density
Example 2.4.4 Hits on a web page
Remember: we modeled waiting times until the first hit as Exp2 .
How long do we have to wait for the second hit?
In order to get the waiting time for the second hit, we can add the waiting times until the first hit and the
time between the first and the second hit.
For both of these we know the distribution: Y1 , the waiting time until the first hit is an exponential variable
with λ = 2.
After we have observed the first hit, we start the experiment again and wait for the next hit. Since the
exponential distribution is memoryless, this is as good as waiting for the first hit. We therefore can model
Y2 , the time between first and second hit, by another exponential distribution with the same rate λ = 2.
What we are interested in is Y := Y1 + Y2 .
Unfortunately, we don’t know the distribution of Y , yet.
Definition 2.4.2 (Erlang density)
If Y1 , . . . , Yk are k independent exponential random variables with parameter λ, their sum X has an Erlang
distribution:
k
X
X :=
Yi is Erlang(k,λ)
i=1
The Erlang density fk,λ is
(
f (x) =
λe
−λx
0
k−1
· (λx)
(k−1)!
k is called the stage parameter, λ is the rate parameter.
x<0
for x ≥ 0
2.4. SOME SPECIAL CONTINUOUS DENSITY FUNCTIONS
43
Expected value and variance of an Erlang distributed variable X can be computed using the properties of
expected value and variance for sums of independent random variables:
E[X]
V ar[X]
k
k
X
X
1
= E[
Yi ] =
E[Yi ] = k ·
λ
i=1
i=1
= V ar[
k
X
Yi ] =
i=1
k
X
V ar[Yi ] = k ·
i=1
1
λ2
In order to compute the distribution function, we need another result about the relationship between P oλ
and Expλ .
Theorem 2.4.3
If X1 , X2 , X3 , . . . are independent exponential random variables with parameter λ and (cf. fig. 2.4)
W := largest index j such that
j
X
Xi ≤ T
i=1
for some fixed T > 0.
Then W ∼ P oλT .
*
0
X1
*
X2
*
X3
*
*
<- occurrence
times
T
Figure 2.4: W = 3 in this example.
With this theorem, we can derive an expression for the Erlang distribution function. Let X be an Erlangk,λ
variable:
Erlangk,λ (x)
= P (X ≤ x)
=
1st trick
=
1 − P (X > x) = 1 −
P(
X
Yi > x)
above theorem
=
i
|
{z
}
less than k hits observed
=
1 − P o( a Poisson r.v. with rate xλ ≤ k − 1) =
=
1 − P oλx (k − 1).
Example 2.4.5 Hits on a web page
What is the density of the waiting time until the next hit?
We said that Y as previously defined, is the sum of two exponential variables, each with rate λ = 2.
X has therefore an Erlang distribution with stage parameter 2, and the density is given as
fX (x) = fk,λ (x) = 4xe−2x for x ≥ 0
If we wait for the third hit, what is the probability that we have to wait more than 1 min?
Z := waiting time until the third hit has an Erlang(3,2) distribution.
P (Z > 1) = 1 − Erlang3,2 (1) = 1 − (1 − P o2·1 (3 − 1)) = P o2 (2) = 0.677
44
CHAPTER 2. RANDOM VARIABLES
Note:
The exponential distribution is a special case of an Erlang distribution:
Expλ = Erlang(k=1,λ)
Erlang distributions are used to model waiting times of components that are exposed to peak stresses. It is
assumed that they can withstand k − 1 peaks and fail with the kth peak.
We will come across the Erlang distribution again, when modelling the waiting times in queueing systems,
where customers arrive with a Poisson rate and need exponential time to be served.
2.4.4
Gaussian or Normal density
The normal density is the archetypical “bell-shaped” density. The density has two parameters: µ and σ 2
and is defined as
(x−µ)2
1
fµ,σ2 (x) = √
e− 2σ2
2πσ 2
The expected value and variance of a normal distributed r.v. X are:
Z ∞
E[X] =
xfµ.σ2 (x)dx = . . . = µ
−∞
Z ∞
V ar[X] =
(x − µ)2 fµ.σ2 (x)dx = . . . = σ 2 .
−∞
Note: the parameters µ and σ 2 are actually mean and variance of X - and that’s what they are called.
f0,0.5
f0,1
f0,2
x
f-1,1
f0,1
f2,1
x
Figure 2.5: Normal densities for several parameters. µ determines the location of the peak on the x−axis,
σ 2 determines the “width” of the bell.
2.4. SOME SPECIAL CONTINUOUS DENSITY FUNCTIONS
45
The distribution function of X is
Z
t
fµ,σ2 (x)dx
Nµ,σ2 (t) := Fµ,σ2 (t) =
−∞
Unfortunately, there does not exist a closed form for this integral - fµ,σ2 does not have a simple antiderivative. However, to get probabilities means we need to evaluate this integral. This leaves us with several
choices:
1. personal numerical integration
2. use of statistical software
later
3. standard tables of normal probabilities
We will use the third option, mainly.
First of all: only a special case of the normal distributions is tabled: only positive values of N (0, 1) are tabled
- N (0, 1) is the normal distribution, that has mean 0 and a variance of 1. This is the so-called standard
normal distribution, also written as Φ.
A table for this distribution is enough, though. We will use several tricks to get any normal distribution into
the shape of a standard normal distribution:
Basic facts about the normal distribution that allow the use of tables
(i) for X ∼ N (µ, σ 2 ) holds:
Z :=
X −µ
∼ N (0, 1)
σ
This process is called standardizing X.
(this is at least plausible, since
E[Z]
=
V ar[Z]
=
1
(E[X] − µ) = 0
σ
1
V ar[X] = 1
σ2
(ii) Φ(−z) = 1 − Φ(z) since f0,1 is symmetric in 0 (see fig. 2.6 for an explanation).
f0,1
P(Z ≤ -z)
P(Z ‡ +z)
-z
+z
x
Figure 2.6: standard normal density. Remember, the area below the graph up to a specified vertical line
represents the probability that the random variable Z is less than this value. It’s easy to see, that the areas
in the tails are equal: P (Z ≤ −z) = P (Z ≥ +z). And we already know, that P (Z ≥ +z) = 1 − P (Z ≤ z),
which proves the above statement.
Example 2.4.6
Suppose Z is a standard normal random variable.
this is, what
we are going to do!
46
CHAPTER 2. RANDOM VARIABLES
1. P (Z < 1) = ?
P (Z < 1) = Φ(1)
straight look-up
=
0.8413.
2. P (0 < Z < 1) = ?
P (0 < Z < 1) = P (Z < 1) − P (Z < 0) = Φ(1) − Φ(0)
look-up
=
0.8413 − 0.5 = 0.3413.
3. P (Z < −2.31) = ?
P (Z < −2.31) = 1 − Φ(2.31)
look-up
=
1 − 0.9896 = 0.0204.
4. P (|Z| > 2) = ?
P (|Z| > 2) = P (Z < −2) + P (Z > 2) = 2(1 − Φ(2))
(1)
look-up
=
(2)
f0,1
(3)
f0,1
Example 2.4.7
Suppose, X ∼ N (1, 2)
P (1 < X < 2) =?
A standardization of X gives Z :=
P (1 < X < 2)
2(1 − 0.9772) = 0.0456.
f0,1
(4)
f0,1
X−1
√ .
2
1−1
X −1
2−1
P( √ < √
< √ )=
2
2
2
√
= P (0 < Z < 0.5 2) = Φ(0.71) − Φ(0) = 0.7611 − 0.5 = 0.2611.
=
Note that the standard normal table only shows probabilities for z < 3.99. This is all we need, though, since
P (Z ≥ 4) ≤ 0.0001.
Example 2.4.8
Suppose the battery life of a laptop is normally distributed with σ = 20 min.
Engineering design requires, that only 1% of batteries fail to last 300 min.
What mean battery life is required to ensure this condition?
Let X denote the battery life in minutes, then X has a normal distribution with unknown mean µ and
standard deviation σ = 20 min.
What is µ?
The condition, that only 1% of batteries is allowed to fail the 300 min limit translates to:
P (X < 300) ≤ 0.01
We must make sure to choose µ such, that this condition holds.
2.5. CENTRAL LIMIT THEOREM (CLT)
47
In order to compute the probability, we must standardize X:
Z :=
Then
P (X ≤ 300) = P (
X −µ
20
X −µ
300 − µ
300 − µ
300 − µ
≤
) = P (Z ≤
) = Φ(
)
20
20
20
20
The condition requires:
P (X ≤ 300) ≤ 0.01
300 − µ
⇐⇒ Φ(
) ≤ 0.01 = 1 − 0.99 = 1 − Φ(2.33) = Φ(−2.33)
20
300 − µ
⇐⇒
≤ −2.33
20
⇐⇒ µ ≥ 346.6.
Normal distributions have a “reproductive property”, i.e. if X and Y are normal variables, then W :=
aX + bY is also a normal variable, with:
E[W ]
V ar[W ]
= aE[X] + bE[Y ]
= a2 V ar[X] + b2 V ar[Y ] + 2abCov(X, Y )
The normal distribution is extremely common/ useful, for one reason: the normal distribution approximates
a lot of other distributions. This is the result of one of the most fundamental theorems in Math:
2.5
Central Limit Theorem (CLT)
Theorem 2.5.1 (Central Limit Theorem)
If X1 , X2 , . . . , Xn are n independent, identically distributed random variables with E[Xi ] = µ and V ar[Xi ] =
σ 2 , then:
Pn
the sample mean X̄ := n1 i=1 Xi is approximately normal distributed
with E[X̄] = µ and V ar[X̄] = σ 2 /n.
2
i.e. X̄ ∼ N (µ, σn ) or
P
i
Xi ∼ N (nµ, nσ 2 )
Corollary 2.5.2
(a) for large n the binomial distribution Bn,p is approximately normal Nnp,np(1−p) .
(b) for large λ the Poisson distribution P oλ is approximately normal Nλ,λ .
(c) for large k the Erlang distribution Erlangk,λ is approximately normal N k ,
k
λ λ2
Why?
(a) Let X be a variable with a Bn,p distribution.
We know, that X is the result from repeating the same Bernoulli experiment n times and looking at
the overall number of successes. We can therefor, write X as the sum of n B1,p variables Xi :
X := X1 + X2 + . . . + Xn
X is then the sum of n independent, identically distributed random variables. Then, the Central
Limit Theorem states, that X has an approximate normal distribution with E[X] = nE[Xi ] = np and
V ar[X] = nV ar[Xi ] = np(1 − p).
48
CHAPTER 2. RANDOM VARIABLES
(b) it is enough to show the statement for the case that λ is a large integer:
Let Y be a Poisson variable with rate λ. Then we can think of Y as the number of occurrences in
an experiment that runs for time λ - that is the same as to observe λ experiments that each run
independently for time 1 and add their results:
Y = Y1 + Y2 + . . . + Yλ , with Yi ∼ P o1 .
Again, Y is the sum of n independent, identically distributed random variables. Then, the Central
Limit Theorem states, that X has an approximate normal distribution with E[Y ] = λ · 1 and V ar[Y ] =
λV ar[Yi ] = λ.
(c) this statement is the easiest to prove, since an Erlangk,λ distributed variable Z is by definition the sum
of k independently distributed exponential variables Z1 , . . . , Zk .
For Z the CLT holds, and we get, that Z is approximately normal distributed with E[Z] = kE[Zi ] =
and V ar[Z] = kV ar[Zi ] = λk2 .
k
λ
2
Why do we need the central limit theorem at all? - first of all, the CLT gives us the distribution of the
sample mean in a very general setting: the only thing we need to know, is that all the observed values come
from the same distribution, and the variance for this distribution is not infinite.
A second reason is, that most tables only contain the probabilities up to a certain limit - the Poisson table
e.g. only has values for λ ≤ 10, the Binomial distribution is tabled only for n ≤ 20. After that, we can use
the Normal approximation to get probabilities.
Example 2.5.1 Hits on a webpage
Hits occur with a rate of 2 per min.
What is the probability to wait for more than 20 min for the 50th hit?
Let Y be the waiting time until the 50th hit.
We know: Y has an Erlang50,2 distribution. therefore:
P (Y > 20)
=
1 − Erlang50,2 (20) = 1 − (1 − P o2·20 (50 − 1)) =
=
P o40 (49) ≈ N40,40 (49) =
49 − 40
table
√
Φ
= Φ(1.42) = 0.9222.
40
=
CLT !
Example 2.5.2 Mean of Uniform Variables
Let U1 , U2 , U3 , U4 , and U5 be standard uniform variables, i.e. Ui ∼ U(0,1) .
Without the CLT we would have no idea, what distribution the sample mean Ū =
approx
1
With it, we know: Ū ∼ N (0.5, 60
).
Issue:
1
5
Accuracy of approximation
• increases with n
• increases with the amount of symmetry in the distribution of Xi
Rule of thumb for the Binomial distribution:
Use the normal approximation for Bn,p , if np > 5 (if p ≤ 0.5) or nq > 5 (if p ≥ 0.5)!
P5
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Investigate the numbers that come up on a die as you roll it in the direction of north, south, east and west, without going over the path it's already made.
##### Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Cubes Here and There
##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### Marvellous Matrix
##### Stage: 2 Challenge Level:
Follow the directions for circling numbers in the matrix. Add all the circled numbers together. Note your answer. Try again with a different starting number. What do you notice?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Sometimes We Lose Things
##### Stage: 2 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### Birds in the Garden
##### Stage: 1 and 2 Challenge Level:
This activity asks you to collect information about the birds you see in the garden. Are there patterns in the data or do the birds seem to visit randomly?
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Street Sequences
##### Stage: 1 and 2 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
### Mobile Numbers
##### Stage: 1 and 2 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time?
### Count the Digits
##### Stage: 1 and 2 Challenge Level:
In this investigation we are going to count the number of 1s, 2s, 3s etc in numbers. Can you predict what will happen?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Sorting the Numbers
##### Stage: 1 and 2 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### 3 Rings
##### Stage: 2 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Exploring Wild & Wonderful Number Patterns
##### Stage: 2 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Street Party
##### Stage: 2 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Tables Without Tens
##### Stage: 2 Challenge Level:
Investigate and explain the patterns that you see from recording just the units digits of numbers in the times tables.
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules? | 2,209 | 9,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2016-44 | longest | en | 0.912464 |
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