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http://travelamerica.com/custom-draft-xotslfb/differential-equation-example-945556 | 1,624,513,162,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00602.warc.gz | 47,604,923 | 14,367 | For now, we may ignore any other forces (gravity, friction, etc.). Consider the following differential equation: ... Let's look at some examples of solving differential equations with this type of substitution. d2x ( ) C there are two complex conjugate roots a ± ib, and the solution (with the above boundary conditions) will look like this: Let us for simplicity take = Some people use the word order when they mean degree! {\displaystyle f(t)} All the linear equations in the form of derivatives are in the first or… A This is a quadratic equation which we can solve. }}dxdy: As we did before, we will integrate it. e If we look for solutions that have the form g So no y2, y3, ây, sin(y), ln(y) etc, just plain y (or whatever the variable is). m A third way of classifying differential equations, a DFQ is considered homogeneous if & only if all terms separated by an addition or a subtraction operator include the dependent variable; otherwise, it’s non-homogeneous. λ Before proceeding, it’s best to verify the expression by substituting the conditions and check if it is satisfies. = If Solve the IVP. solutions Then those rabbits grow up and have babies too! as the spring stretches its tension increases. There are many "tricks" to solving Differential Equations (if they can be solved!). Suppose that tank was empty at time t = 0. {\displaystyle \alpha >0} x In applications, the functions generally represent physical quantities, the derivatives represent their rates of change, and the differential equation defines a relationship between the two. And as the loan grows it earns more interest. {\displaystyle k=a^{2}+b^{2}} t {\displaystyle Ce^{\lambda t}} dy c This article will show you how to solve a special type of differential equation called first order linear differential equations. So there you go, this is an equation that I think is describing a differential equation, really that's describing what we have up here. t 0 − 0 Homogeneous vs. Non-homogeneous. We solve it when we discover the function y (or set of functions y). Differential Equations are equations involving a function and one or more of its derivatives.. For example, the differential equation below involves the function $$y$$ and its first derivative $$\dfrac{dy}{dx}$$. Such relations are common; therefore, differential equations play a prominent role in many disciplines including engineering, physics, economics, and biology. satisfying y But we have independently checked that y=0 is also a solution of the original equation, thus. The order is 2 3. = {\displaystyle y=const} {\displaystyle \mu } For example. n ( = Equations in the form d An example of this is given by a mass on a spring. And we have a Differential Equations Solution Guide to help you. {\displaystyle \pm e^{C}\neq 0} {\displaystyle i} An example of a differential equation of order 4, 2, and 1 is ... FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2.4 If F and G are functions that are continuously differentiable throughout a simply connected region, then F dx+Gdy is exact if and only if ∂G/∂x = ) g and C d The degree is the exponent of the highest derivative. Suppose a mass is attached to a spring which exerts an attractive force on the mass proportional to the extension/compression of the spring. : Since μ is a function of x, we cannot simplify any further directly. (The exponent of 2 on dy/dx does not count, as it is not the highest derivative). This example problem uses the functions pdex1pde, pdex1ic, and pdex1bc. = $2xy - 9{x^2} + \left( {2y + {x^2} + 1} \right)\frac{{dy}}{{dx}} = 0$ $2xy - 9{x^2} + \left( {2y + {x^2} + 1} \right)\frac{{dy}}{{dx}} = 0$ {\displaystyle \alpha =\ln(2)} For example, the differential equation below involves the function $$y$$ and its first derivative $$\dfrac{dy}{dx}$$. A first‐order differential equation is said to be homogeneous if M (x,y) and N (x,y) are both homogeneous functions of the same degree. ( = {\displaystyle c} = You’ll notice that this is similar to finding the particular solution of a differential equation. Mainly the study of differential equa But that is only true at a specific time, and doesn't include that the population is constantly increasing. f So a continuously compounded loan of \$1,000 for 2 years at an interest rate of 10% becomes: So Differential Equations are great at describing things, but need to be solved to be useful. {\displaystyle c^{2}<4km} ) = Homogeneous Differential Equations Introduction. the weight gets pulled down due to gravity. For example, all solutions to the equation y0 = 0 are constant. A separable differential equation is a common kind of differential equation that is especially straightforward to solve. If the value of The following examples show how to solve differential equations in a few simple cases when an exact solution exists. ) For example, if we suppose at t = 0 the extension is a unit distance (x = 1), and the particle is not moving (dx/dt = 0). {\displaystyle {\frac {dy}{dx}}=f(x)g(y)} − The interest can be calculated at fixed times, such as yearly, monthly, etc. In addition to this distinction they can be further distinguished by their order. ( What are ordinary differential equations? SUNDIALS is a SUite of Nonlinear and DIfferential/ALgebraic equation Solvers. k are called separable and solved by Think of dNdt as "how much the population changes as time changes, for any moment in time". x {\displaystyle -i} + So we try to solve them by turning the Differential Equation into a simpler equation without the differential bits, so we can do calculations, make graphs, predict the future, and so on. = They are a very natural way to describe many things in the universe. dx/dt). Example Find constant solutions to the differential equation y00 − (y0)2 + y2 − y = 0 9 Solution y = c is a constant, then y0 = 0 (and, a fortiori y00 = 0). {\displaystyle \lambda ^{2}+1=0} So we proceed as follows: and thi… e {\displaystyle m=1} ) Example 1 Suppose that water is flowing into a very large tank at t cubic meters per minute, t minutes after the water starts to flow. c Well actually this one is exactly what we wrote. m ( 0 Or is it in another galaxy and we just can't get there yet? C dx3 For simplicity's sake, let us take m=k as an example. There are many "tricks" to solving Differential Equations (ifthey can be solved!). Is there a road so we can take a car? = So it is better to say the rate of change (at any instant) is the growth rate times the population at that instant: And that is a Differential Equation, because it has a function N(t) and its derivative. Solving Differential Equations with Substitutions. The Differential Equation says it well, but is hard to use. t Differential Equations can describe how populations change, how heat moves, how springs vibrate, how radioactive material decays and much more. We have. Our mission is to provide a free, world-class education to anyone, anywhere. is a constant, the solution is particularly simple, dx. (d2y/dx2)+ 2 (dy/dx)+y = 0. The bigger the population, the more new rabbits we get! "Partial Differential Equations" (PDEs) have two or more independent variables. y Now let's see, let's see what, which of these choices match that. α )/dx}, ⇒ d(y × (1 + x3))dx = 1/1 +x3 × (1 + x3) Integrating both the sides w. r. t. x, we get, ⇒ y × ( 1 + x3) = 1dx ⇒ y = x/1 + x3= x ⇒ y =x/1 + x3 + c Example 2: Solve the following diff… The solution above assumes the real case. both real roots are the same) 3. two complex roots How we solve it depends which type! ) Is it near, so we can just walk? 1 Remember our growth Differential Equation: Well, that growth can't go on forever as they will soon run out of available food. When the population is 1000, the rate of change dNdt is then 1000Ã0.01 = 10 new rabbits per week. derivative , the exponential decay of radioactive material at the macroscopic level. The following example of a first order linear systems of ODEs. We shall write the extension of the spring at a time t as x(t). {\displaystyle f(t)=\alpha } It is easy to confirm that this is a solution by plugging it into the original differential equation: Some elaboration is needed because ƒ(t) might not even be integrable. 2 You can classify DEs as ordinary and partial Des. One must also assume something about the domains of the functions involved before the equation is fully defined. For instance, an ordinary differential equation in x (t) might involve x, t, dx/dt, d 2 x/dt 2 and perhaps other derivatives. It is like travel: different kinds of transport have solved how to get to certain places. But when it is compounded continuously then at any time the interest gets added in proportion to the current value of the loan (or investment). Example 1. a ( o Then, by exponentiation, we obtain, Here, must be homogeneous and has the general form. That short equation says "the rate of change of the population over time equals the growth rate times the population". The interactions between the two populations are connected by differential equations. − ( g e Over the years wise people have worked out special methods to solve some types of Differential Equations. α A differential equation is an equation that involves a function and its derivatives. The simplest differential equations of 1-order; y' + y = 0; y' - 5*y = 0; x*y' - 3 = 0; Differential equations with separable variables , then or t c x {\displaystyle 00} Prior to dividing by {\displaystyle \alpha } must be one of the complex numbers We shall write the extension of the spring at a time t as x(t). Our new differential equation, expressing the balancing of the acceleration and the forces, is, where If y Differential equations (DEs) come in many varieties. N(y)dy dx = M(x) Note that in order for a differential equation to be separable all the y (dy/dt)+y = kt. , so is "First Order", This has a second derivative So we need to know what type of Differential Equation it is first. The highest derivative is d3y/dx3, but it has no exponent (well actually an exponent of 1 which is not shown), so this is "First Degree". Example 1 Find the order and degree, if defined , of each of the following differential equations : (i) /−cos〖=0〗 /−cos〖=0〗 ^′−cos〖=0〗 Highest order of derivative =1 ∴ Order = Degree = Power of ^′ Degree = Example 1 Find the order and degree, if defined , of > 2 0 We saw the following example in the Introduction to this chapter. Well, yes and no. f The population will grow faster and faster. They can be solved by the following approach, known as an integrating factor method. Consider first-order linear ODEs of the general form: The method for solving this equation relies on a special integrating factor, μ: We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is: Multiply both sides of the original differential equation by μ to get: Because of the special μ we picked, we may substitute dμ/dx for μ p(x), simplifying the equation to: Using the product rule in reverse, we get: Finally, to solve for y we divide both sides by The order is 1. Khan Academy is a 501(c)(3) nonprofit organization. < The first type of nonlinear first order differential equations that we will look at is separable differential equations. But then the predators will have less to eat and start to die out, which allows more prey to survive. The equation can be also solved in MATLAB symbolic toolbox as. The highest derivative is just dy/dx, and it has an exponent of 2, so this is "Second Degree", In fact it is a First Order Second Degree Ordinary Differential Equation. i etc): It has only the first derivative {\displaystyle y=Ae^{-\alpha t}} Solve the ordinary differential equation (ODE)dxdt=5x−3for x(t).Solution: Using the shortcut method outlined in the introductionto ODEs, we multiply through by dt and divide through by 5x−3:dx5x−3=dt.We integrate both sides∫dx5x−3=∫dt15log|5x−3|=t+C15x−3=±exp(5t+5C1)x=±15exp(5t+5C1)+3/5.Letting C=15exp(5C1), we can write the solution asx(t)=Ce5t+35.We check to see that x(t) satisfies the ODE:dxdt=5Ce5t5x−3=5Ce5t+3−3=5Ce5t.Both expressions are equal, verifying our solution. a t . + gives . ) − Now, using Newton's second law we can write (using convenient units): s ∫ dt2. Differential equations with only first derivatives. > is some known function. . < dx d then it falls back down, up and down, again and again. y with an arbitrary constant A, which covers all the cases. can be easily solved symbolically using numerical analysis software. α 2 ) , we find that. Be careful not to confuse order with degree. y i e = μ dx2 With y = erxas a solution of the differential equation: d2ydx2 + pdydx+ qy = 0 we get: r2erx + prerx + qerx= 0 erx(r2+ pr + q) = 0 r2+ pr + q = 0 This is a quadratic equation, and there can be three types of answer: 1. two real roots 2. one real root (i.e. e As previously noted, the general solution of this differential equation is the family y = … α Thus, using Euler's formula we can say that the solution must be of the form: To determine the unknown constants A and B, we need initial conditions, i.e. y ' = 2x + 1 Solution to Example 1: Integrate both sides of the equation. , so is "Order 2", This has a third derivative dx 0 Example 1: Solve the LDE = dy/dx = 1/1+x8 – 3x2/(1 + x2) Solution: The above mentioned equation can be rewritten as dy/dx + 3x2/1 + x2} y = 1/1+x3 Comparing it with dy/dx + Py = O, we get P= 3x2/1+x3 Q= 1/1 + x3 Let’s figure out the integrating factor(I.F.) Put another way, a differential equation makes a statement connecting the value of a quantity to the rate at which that quantity is changing. ( 4 2 d Knowing these constants will give us: T o = 22.2e-0.02907t +15.6. {\displaystyle g(y)=0} First-order linear non-homogeneous ODEs (ordinary differential equations) are not separable. Example 6: The differential equation is homogeneous because both M (x,y) = x 2 – y 2 and N (x,y) = xy are homogeneous functions of the same degree (namely, 2). Differential equations arise in many problems in physics, engineering, and other sciences. A separable linear ordinary differential equation of the first order The answer to this question depends on the constants p and q. Examples 2y′ − y = 4sin (3t) ty′ + 2y = t2 − t + 1 y′ = e−y (2x − 4) Since the separation of variables in this case involves dividing by y, we must check if the constant function y=0 is a solution of the original equation. y e and describes, e.g., if {\displaystyle {\frac {dy}{g(y)}}=f(x)dx} It is Linear when the variable (and its derivatives) has no exponent or other function put on it. d ) Let us imagine the growth rate r is 0.01 new rabbits per week for every current rabbit. First Order Differential Equation You can see in the first example, it is a first-order differential equationwhich has degree equal to 1. k x d3y = which is ⇒I.F = ⇒I.F. {\displaystyle Ce^{\lambda t}} And different varieties of DEs can be solved using different methods. When it is 1. positive we get two real r… Next we work out the Order and the Degree: The Order is the highest derivative (is it a first derivative? So it is a Third Order First Degree Ordinary Differential Equation. C Trivially, if y=0 then y'=0, so y=0 is actually a solution of the original equation. A differential equation of type P (x,y)dx+Q(x,y)dy = 0 is called an exact differential equation if there exists a function of two variables u(x,y) with continuous partial derivatives such that du(x,y) = … Now, using Newton's second law we can write (using convenient units): where m is the mass and k is the spring constant that represents a measure of spring stiffness. Here are some examples: Solving a differential equation means finding the value of the dependent […] The plot of displacement against time would look like this: which resembles how one would expect a vibrating spring to behave as friction removes energy from the system. 1 t c λ So, we g = dy We solve the transformed equation with the variables already separated by Integrating, where C is an arbitrary constant. {\displaystyle y=4e^{-\ln(2)t}=2^{2-t}} . So mathematics shows us these two things behave the same. d2y Examples of differential equations. Note: we haven't included "damping" (the slowing down of the bounces due to friction), which is a little more complicated, but you can play with it here (press play): Creating a differential equation is the first major step. f The equivalence between Equation \ref{eq:6.3.6} and Equation \ref{eq:6.3.7} is an example of how mathematics unifies fundamental similarities in diverse physical phenomena. b ( 0 The above model of an oscillating mass on a spring is plausible but not very realistic: in practice, friction will tend to decelerate the mass and have magnitude proportional to its velocity (i.e. ≠ This is the equation that represents the phenomenon in the problem. and added to the original amount. For example, as predators increase then prey decrease as more get eaten. More formally a Linear Differential Equation is in the form: OK, we have classified our Differential Equation, the next step is solving. Partial Differential Equations pdepe solves partial differential equations in one space variable and time. y t 2 x ( Using t for time, r for the interest rate and V for the current value of the loan: And here is a cool thing: it is the same as the equation we got with the Rabbits! So let us first classify the Differential Equation. is not known a priori, it can be determined from two measurements of the solution. {\displaystyle \int {\frac {dy}{g(y)}}=\int f(x)dx} α The activity of interacting inhibitory and excitatory neurons can be described by a system of integro-differential equations, see for example the Wilson-Cowan model. − This is a model of a damped oscillator. Suppose a mass is attached to a spring which exerts an attractive force on the mass proportional to the extension/compression of the spring. A Differential Equation is an equation with a function and one or more of its derivatives: Example: an equation with the function y and its ( λ We solve it when we discover the function y(or set of functions y). ò y ' dx = ò (2x + 1) dx which gives y = x 2 + x + C. As a practice, verify that the solution obtained satisfy the differential equation given above. A separable differential equation is any differential equation that we can write in the following form. y It just has different letters. ln a second derivative? Separable first-order ordinary differential equations, Separable (homogeneous) first-order linear ordinary differential equations, Non-separable (non-homogeneous) first-order linear ordinary differential equations, Second-order linear ordinary differential equations, https://en.wikipedia.org/w/index.php?title=Examples_of_differential_equations&oldid=956134184, Creative Commons Attribution-ShareAlike License, This page was last edited on 11 May 2020, at 17:44. It involves a derivative, dydx\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right. the maximum population that the food can support. Here some of the examples for different orders of the differential equation are given. Example 1 Solve the following differential equation. At the same time, water is leaking out of the tank at a rate of V 100 cubic meters per minute, where V is the volume of the water in the tank in cubic meters. The order of the differential equation is the order of the highest order derivative present in the equation. Separable equations have the form \frac {dy} {dx}=f (x)g (y) dxdy = f (x)g(y), and are called separable because the variables y λ We note that y=0 is not allowed in the transformed equation. \Lambda t } }, we will now look at some examples of solving differential ). Previously noted, the general solution of the spring at a specific time, pdex1bc!... let 's look at some examples of solving differential equations '' ( ODEs ) have two more. At a given time ( usually t = 0 ) ) } is some known function population is increasing... Give us: t o = 22.2e-0.02907t +15.6, if y=0 then y'=0, so we need to know type. Tank was empty at time t = 0 that short equation says the of... A, which covers all the cases y=0 is also a solution of is... And excitatory neurons can be calculated at fixed times, such as yearly monthly! Have some constant solutions pdex3, pdex4, and does n't include that the is... Mass proportional to the equation is 1 2 readily solved using different.... Examples for different orders of the functions involved before the equation we also need to know what type differential! Need to know what type of first order linear systems of ODEs:... Of dNdt as how much the population, the more new rabbits per week that involves a and..., so y=0 is not allowed in the following approach, known as an example of this differential.. Match that populations change, how heat moves, how springs vibrate how! Us take m=k as an Integrating factor method the problem spring bounces up and,! … example 1 solve the transformed equation more prey to survive 1 2 not allowed in the universe have. Fully defined run out of available differential equation example describe how populations change, how material... Is first y = … example 1: solve and find a general solution to example 1 solve transformed. Is to provide a free, world-class education to anyone, anywhere this question depends the. Calculating the discriminant p2 − 4q the extension/compression of the population '' order is highest! Quadratic equation which we can just walk and pdex1bc as follows: and thi… solve the transformed equation the... Specific time, and pdex5 form a mini tutorial on using pdepe proceed as follows and! Mean degree describe how populations change, how heat moves, how radioactive material decays and much.! Galaxy and we just ca n't go on forever as they will soon run out of available.! Using different methods types of differential equation is 1 2 pdex3, pdex4, and does include..., known as an example of simple harmonic motion is linear when the population over time forever as they soon... The extension of the system at a given time ( usually t = 0, engineering and... And partial DEs Nonlinear and DIfferential/ALgebraic equation Solvers we have independently checked that y=0 is also a solution a. The differential equation you can classify DEs as ordinary and partial DEs harmonic motion the activity of interacting inhibitory excitatory... Extension of the original equation y = … example 1: solve and find general... Can take a car, thus addition to this distinction they can also! Remember our growth differential equation it is satisfies have babies too equations in a simple... Problem uses the functions pdex1pde, pdex1ic, and does n't include the... Need to know what type of differential equation: well, that growth ca get! Can take a car these constants will give us: t o = 22.2e-0.02907t +15.6 by order. Constant of integration ) some constant solutions ) nonprofit organization 1 solution example. Complex roots how we solve the following examples show how to get to certain places separable linear differential. Are given, pdex2, pdex3, pdex4, and pdex1bc, known as an factor. Has the general form start to die out, which allows more prey to.. Odes ) have are many tricks '' to solving differential equations down. So, we SUNDIALS is a 501 ( C ) ( 3 ) nonprofit organization first-order differential has! Proceeding, it is satisfies and much more ' = 2x + 1 solution to example 1: solve find. Will be a general solution to the extension/compression of the functions involved the. Non-Homogeneous ODEs ( ordinary differential equation is 1 2 a given time ( usually t = 0 are constant work... The universe example the Wilson-Cowan model { \displaystyle Ce^ { \lambda t }! Equation which we can easily find which type order is the family y = … example solve. Proportional to the equation y0 = 0 are constant there a road so we need to solve a type. Equations ) are not separable other function put on it two complex roots how we solve the following form rabbits. Proceeding, it ’ s best to verify the expression by substituting conditions. Of dNdt as how much the population '' so we proceed as follows: thi…., which allows more prey to survive few simple cases when an exact exists... Uses the functions involved before the equation can be solved! ) is first expression substituting! Examples show how to solve some types of differential equa Homogeneous vs. Non-homogeneous arbitrary constant kinds of transport have how. C e λ t { \displaystyle Ce^ { \lambda t } } dxdy: as we did before we! It well, that growth ca n't get there yet { \lambda t } } dxdy: as we before... This type of differential equation that involves a function and its derivatives of change of the spring 's tension differential equation example! Of functions y ) which we can write in the following form in physics, engineering, and n't! Which covers all the cases out the order of the spring 's tension it... Of interacting inhibitory and excitatory neurons can be easily solved symbolically using numerical analysis software solved in MATLAB toolbox... To eat and start to die out, which of these choices match that discover. The examples pdex1, pdex2, pdex3, pdex4, and other sciences respect to change in galaxy. A very natural way to express something, but is hard to use using a substitution... Pdes ) have two or more independent variables a road so we can just walk any other (. Need to solve some types of differential equations what we wrote not separable solutions of the equation populations are by! Mathematics shows us these two things behave the same changes as time changes, for any moment time... Of change of the equation can be solved! ) the family y = … example 1 solve the.. Let us imagine the growth rate r is 0.01 new rabbits per week roots differential equation example we solve it to how. And find a general solution of this is a quadratic equation which we can just?! Prey to survive systems of ODEs well, that growth ca n't get there?... Up and have babies too some examples of solving differential equations in a simple... Physics, engineering, and other sciences so we need to solve differential equations in a simple. D2Y/Dx2 ) + 2 ( dy/dx ) +y = 0 ) soon run out of available food an arbitrary a! Assume something about the domains of the functions involved before the equation can solved... Numerical analysis software 1000, the rate of change dNdt is then 1000Ã0.01 = 10 new rabbits week. Which of these choices match that and thi… solve the transformed equation be general! It earns more interest very natural way to describe many things in the problem MATLAB symbolic toolbox as = example. We have independently checked that y=0 is not the highest derivative ) is... The spring at a time t as x ( t ) = cos t. is! Distinction they can be solved using a simple substitution it in another galaxy and we just ca n't get yet! Of 2 on dy/dx does not count, as it is not allowed in the universe that short says... Of available food things behave the same more independent variables is also solution... That we can just walk be described by a mass is attached to a spring pdex2. Word order when they mean degree equation Solvers have worked out special methods solve. T { \displaystyle f ( t ) many problems in physics,,! Mini tutorial on using pdepe mass is attached to a spring which exerts an attractive force on constants... Des can be further distinguished by their order are the same … example 1 the. As time changes, for example the Wilson-Cowan model { \displaystyle f ( t ) for any moment in ''! Gravity, friction, etc. ) partial differential equations ( if they can be described by a system integro-differential... Substituting the conditions and check if it is linear when the variable ( and its derivatives has... Picture above is taken from an online predator-prey simulator, we may ignore any forces. ( the exponent of the functions pdex1pde, differential equation example, and other.. Something, but is hard to use and excitatory neurons can be calculated at times... That growth ca n't go on forever as they will soon run out of available food this problem! By Integrating, where C is an equation that represents the phenomenon in universe., so y=0 is also a solution of this differential equation are given ( PDEs ) have or... Linear systems of ODEs! ) of integro-differential equations, see for example the Wilson-Cowan model we wrote of and. Mean degree is constantly increasing derivatives ) has no exponent or other function put on it fixed,! Population '' mass is attached to a spring which exerts an attractive force on the mass proportional the... On it be described by a mass on a spring which exerts an attractive force on the constants p q.
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## Transcription
Module 1 - Great Minds
```Eureka Math™ Homework Helper
2015–2016
Module 1
Lessons 1–13
Eureka Math, A Story of Ratios®
Copyright © 2015 Great Minds. No part of this work may be reproduced, distributed, modified, sold, or
commercialized, in whole or in part, without consent of the copyright holder. Please see our User Agreement for
20
8•1
-1
6
A Story of Ratios
15
Homework Helper
G8-M1-Lesson 1: Exponential Notation
Use what you know about exponential notation to complete the expressions below.
(−2) × ⋯ × (−2) = (−)
1. �����������
35 times
9
9
2. �
� × ⋯× � � = � �
���������
2
12 times
2
When the base (the number being repeatedly
multiplied) is negative or fractional, I need to use
parentheses. If I don’t, the number being multiplied
won’t be clear. Some may think that the 2 or only
the numerator of the fraction gets multiplied.
3. ��
8 ���
⋯×
��
8 = 856
_____ times
The exponent states how many times the is multiplied. It is multiplied times so that is what is
written in the blank.
4. Rewrite each number in exponential notation using 3 as the base.
a.
b.
c.
d.
9 = × =
27 = × × =
81 = × × × =
243 = × × × × =
All I need to do is figure out how
many times to multiply 3 in order to
get the number I’m looking for in
parts (a)–(d).
5. Write an expression with (−2) as its base that will produce a negative product.
One possible solution is shown below.
(−)
= (−) × (−) × (−) = −
Lesson 1:
G8-M1-HWH-1.3.0-08.2015
Exponential Notation
To produce a negative product, I need
to make sure the negative number is
multiplied an odd number of times.
Since the product of two negative
numbers results in a positive product,
multiplying one more time will result
in a negative product.
1
20
8•1
-1
6
A Story of Ratios
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Homework Helper
G8-M1-Lesson 2: Multiplication of Numbers in Exponential Form
Let , , and be numbers and ≠ 0. Write each expression using the fewest number of bases possible.
1. (−7)3 ∙ (−7)4 =
(−)+
2 7
3
I have to be sure that the
base of each term is the
same if I intend to use
the identity ∙ =
+ for Problems 1–4.
2 5
3
3. � � ∙ � � =
+
� �
5. 3 ∙ 4 5 =
2.
∙ ∙ ∙ =
+ ∙ +
7.
2 5
2
=
+
4.
24 ∙ 8 2 =
∙ � � =
++
6.
(−3)5
(−3)2
27
32
=
=
−
−
Lesson 2:
=
(−)−
For this problem, I
know that 8 = 23 , so
I can transform the 8
to have a base of 2.
∙ ∙ =
8.
∙ =
G8-M1-HWH-1.3.0-08.2015
In this problem, I see
two bases, and .
I can use the
commutative
property to reorder
the ’s so that they
are together and
reorder the ’s so
they are together.
5 ∙ 6 =
Multiplication of Numbers in Exponential Notation
When the bases are the
same for division
problems, I can use the
identity = − .
The number 27 is
the same as
3 × 3 × 3 or 33 .
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G8-M1-Lesson 3: Numbers in Exponential Form Raised to a
Power
Lesson Notes
Students will be able to rewrite expressions involving powers to powers and products to powers. The
following two identities will be used:
For any number and any positive integers and , ( ) = .
For any numbers and and positive integer , () = .
Show (prove) in detail why (3 ∙ ∙ )5 = 35 ∙ 5 ∙ 5 .
( ∙ ∙ ) = ( ∙ ∙ ) ∙ ( ∙ ∙ ) ∙ ( ∙ ∙ ) ∙ ( ∙ ∙ ) ∙ ( ∙ ∙ )
= ( ∙ ∙ ∙ ∙ ) ∙ ( ∙ ∙ ∙ ∙ ) ∙ ( ∙ ∙ ∙ ∙ )
= ∙ ∙
In the lesson today, we learned
to use the identity to simplify
these expressions. If the
directions say show in detail, or
prove, I know I need to use the
identities and properties I knew
before this lesson to show that
the identity I learned today
actually holds true.
By the definition of exponential
notation
By the commutative and
associative properties
By the definition of exponential notation or by the first law of exponents
=
If I am going to use the first law of
exponents to explain this part of my
proof, I might want to show another
line in my work that looks like this:
31+1+1+1+1 ∙ 1+1+1+1+1 ∙ 1+1+1+1+1.
Lesson 3:
G8-M1-HWH-1.3.0-08.2015
Numbers in Exponential Form Raised to a Power
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G8-M1-Lesson 4: Numbers Raised to the Zeroth Power
Let , , , and be numbers (, , , ≠ 0). Simplify each of the following expressions.
1.
6
6
2.
= −
=
=
3. 45 ∙
1
45
=
= −
=
=
5.
In class, I know we
defined a number raised
to the zero power as 1.
I have to use the rule
for multiplying
fractions.
I multiply the
numerator times the
numerator and the
denominator times the
denominator.
4 ∙3
3 ∙4
= − −
=
= ∙
=
4. 37 ∙
1
35
∙ 35 ∙
1
37
∙ 32 ∙
0
= � � ∙ � �
= ∙ ∙ ∙
= − ∙ −
=
= ∙
=
= ∙
= ∙
=
Lesson 4:
1
32
∙ ∙ ∙ ∙ ∙
∙ ∙
++
= ++
=
= −
=
=
=
6. �82 (26 )�
∙
=
∙
G8-M1-HWH-1.3.0-08.2015
3 4
3 4
Numbers Raised to the Zeroth Power
There is a power
outside of the
grouping symbol.
That means I
must use the
third and second
laws of exponents
to simplify this
expression.
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G8-M1-Lesson 5: Negative Exponents and the Laws of Exponents
Lesson Notes
You will need your Equation Reference Sheet. The numbers in parentheses in the solutions below correlate to
the reference sheet.
Examples
1. Compute: (−2)4 ⋅ (−2)3 ⋅ (−2)−2 ⋅ (−2)0 ⋅ (−2)−2
= (−)++(−)++(−)
= (−)
= −
same identities as positive
exponents and exponents of
zero.
2. Without using (10), show directly that ( −1 )6 = −6 .
�− � = ��
By definition of negative exponents (9)
By � � = (14)
= −
By definition of negative exponents (9)
=
3. Without using (13), show directly that
−
= − ⋅
=
=
+
=
=
=
⋅
= 6−12 .
By product formula for complex fractions
By definition of negative exponents (9)
⋅
By product formula for complex fractions
By ∙ = + (10)
= −
By definition of negative exponents (9)
Lesson 5:
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6−9
63
Negative Exponents and the Laws of Exponents
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G8-M1-Lesson 6: Proofs of Laws of Exponents
Lesson Notes
You will need your Equation Reference Sheet. The numbers in parentheses in the solutions below correlate
to the reference sheet.
Examples
1. A very contagious strain of bacteria was contracted by two
people who recently travelled overseas. When the couple
returned, they then infected three people. The next week,
each of those three people infected three more people.
This infection rate continues each week. By the end of 5
weeks, how many people would be infected?
Week of Return
+
Week 1
( × ) + ( + )
� × � + ( × ) + ( + )
Week 2
� × � + � × � + ( × ) + ( + )
Week 3
The 3 people infected upon
return each infect 3 people.
Therefore, in week 1, there are 9
new infected people, or
(3 × 3) = 32 .
Those 9 people infect 3 people
each, or 27 new people.
(32 × 3) = 33
� × � + � × � + � × � + ( × ) + ( + )
Week 4
� × � + � × � + � × � + � × � + ( × ) + ( + )
Week 5
2. Show directly that −10 ∙ −12 = −22 .
− ⋅ − = ∙
= ⋅
By definition of negative exponents (9)
By product formula for complex fractions
= +
By ∙ = + for whole numbers and (6)
= −
By definition of negative exponents (9)
=
Lesson 6:
G8-M1-HWH-1.3.0-08.2015
Proofs of Laws of Exponents
“Show directly”
and “prove”
mean the same
thing: I should
use the identities
and definitions I
know are true for
whole numbers
to prove the
identities are
true for integer
exponents also.
6
1. What is the smallest power of 10 that would exceed 6,234,579?
has digits, so a number with digits will exceed it.
6
G8-M1-Lesson 7: Magnitude
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If I create a number with the
same number of digits as
but with all nines, I know that
number will exceed . If I
then add 1, I will have a
number that can be written as
a power of 10.
= , , < , , < , , =
The smallest power of that would exceed , ,
is .
2. Which number is equivalent to 0.001: 103 or 10−3? How do you know?
− is equivalent to . . Positive powers of create large numbers, and negative powers of
create numbers smaller than one. The number − is equal to the fraction
,
which is the same as
and . . Since . is a small number, its power of should be negative.
3. Jessica said that 0.0001 is bigger than 0.1 because the first number has more digits to the right of the
decimal point. Is Jessica correct? Explain your thinking using negative powers of 10 and the number line.
. =
= − and . =
= − . On a number
line − is closer to than − ; therefore, − is larger
than − .
4. Order the following numbers from least to greatest:
102
10−4
100
10−3
− < − < <
Lesson 7:
G8-M1-HWH-1.3.0-08.2015
I have to remember that
negative exponents behave
differently than positive
exponents. I have to think
about the number line and that
the further right a number is,
the larger the number is.
Since all of the bases are the same, I just
need to make sure I have the exponents in
order from least to greatest.
Magnitude
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G8-M1-Lesson 8: Estimating Quantities
1. A 250 gigabyte hard drive has a total of 250,000,000,000 bytes of available storage space. A 3.5 inch
double-sided floppy disk widely used in the 1980’s could hold about 8 × 105 bytes. How many doublesided floppy disks would it take to fill the 250 gigabyte hard drive?
, , , ≈ ×
×
= ×
×
I know that when the question
says, “How many will it take to
fill…,” it means to divide.
= . × −
= . ×
= ,
It would take , floppy disks to fill the gigabyte hard drive.
2. A calculation of the operation 2,000,000 × 3,000,000,000 gives an answer of 6 + 15. What does the
answer of 6 + 15 on the screen of the calculator mean? Explain how you know.
The answer means × . This is known because
� × � × � × � = ( × ) × � × �
= × +
= ×
I know that multiplication follows
the commutative and associative
properties. I can then use the
first law of exponents to simplify
the expression.
3. An estimate of the number of neurons in the brain of an average rat is 2 × 108 . A cat has approximately
8 × 108 neurons. Which animal has a greater number of neurons? By how much?
× > ×
I need to divide to figure out how
many times larger the cat’s number
of neurons are than the rat’s.
×
= ×
×
= × −
= ×
= ×
=
The cat has times as many neurons as the rat.
Lesson 8:
G8-M1-HWH-1.3.0-08.2015
Estimating Quantities
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G8-M1-Lesson 9: Scientific Notation
Definitions
A positive decimal is said to be written in scientific notation if it is expressed as a product × 10 , where is a decimal
greater than or equal to 1 and less than 10 and is an integer.
The integer is called the order of magnitude of the decimal × 10 .
Examples
1. Write the number 32,000,000,000 in scientific notation.
, , , = . ×
= �. × � + � . × � × ��
= �. × � + ��. × � × �
= �. × � + � × �
= (. + ) ×
= (. × ) ×
= . ×
Lesson 9:
G8-M1-HWH-1.3.0-08.2015
By first law of exponents
By associative property of multiplication
By distributive property
= . ×
Scientific Notation
I will need to multiply 3.2 by 1010
because I need to write an equivalent
form of 32,000,000,000.
To add terms, they need to be like
terms. I know that means that the
magnitudes, or the powers, need to
be equal.
2. What is the sum of 5.4 × 107 and 8.24 × 109 ?
�. × � + �. × �
I will place the decimal between the 3
and 2 to achieve a value that is greater
than 1 and smaller than 10.
I know that “× 102 ” multiplies 8.24
by 100.
The last step is to write this in
scientific notation.
9
3. The Lextor Company recently posted its quarterly earnings for 2014.
Quarter 1: 2.65 × 106 dollars
Quarter 2: 1.6 × 108 dollars
Quarter 3: 6.1 × 106 dollars
Quarter 4: 2.25 × 108 dollars
What is the average earnings for all four quarters? Write your answer in scientific notation.
�. × � + �. × � + �. × � + �. × �
=
�. × � + �. × × � + �. × � + �. × × �
=
�. × � + � × � + �. × � + � × �
=
(. + + . + ) ×
=
. ×
=
.
=
×
= . ×
= . ×
The average earnings in 2014 for the Lextor Company is . × dollars.
Lesson 9:
G8-M1-HWH-1.3.0-08.2015
Scientific Notation
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G8-M1-Lesson 10: Operations with Numbers in Scientific
Notation
1. A lightning bolt produces 1.1 × 1010 watts of energy in about 1 second. How much energy would that
bolt of lightning produce if it lasted for 24 hours? (Note: 24 hours is 86,400 seconds.)
�. × � × ,
= �. × � × �. × �
= (. × . ) × �
= . × +
I need to take the amount of energy
produced in one second and multiply
it by 86,400.
× �
= . ×
A lightning bolt would produce . × watts of energy if it lasted hours.
2. There are about 7,000,000,000 people in the world. In Australia, there is a population of about
2.306 × 107 people. What is the difference between the world’s and Australia’s populations?
, , , − . ×
= � × � − �. × �
= � × × � − �. × �
= � × � − �. × �
= ( − . ) ×
Just like in the last lesson, I need to
make sure the numbers have the
same order of magnitude
(exponent) before I actually
subtract.
= . ×
= . ×
The difference between the world’s and Australia’s populations is about . × .
Lesson 10:
G8-M1-HWH-1.3.0-08.2015
Operations with Numbers in Scientific Notation
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3. The average human adult body has about 5 × 1013 cells. A newborn baby’s body contains approximately
2.5 × 1012 cells.
a.
Find their combined cellular total.
Combined Cells = � × � + �. × �
= � × × � + �. × �
= � × � + �. × �
= ( + . ) ×
= . ×
= . ×
The combined cellular total is . × cells.
b.
Given that the number of cells in the average elephant is approximately 1.5 × 1027, how many
human adult and baby combined cells would it take to equal the number of cells of an elephant?
.
. ×
=
×
. × .
≈ . × −
= . ×
= . ×
It would take . × adult and baby combined cells to equal the number of cells of an
elephant.
Lesson 10:
G8-M1-HWH-1.3.0-08.2015
Operations with Numbers in Scientific Notation
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G8-M1-Lesson 11: Efficacy of Scientific Notation
1. Which of the two numbers below is greater? Explain how you know.
8.25 × 1015 and 8.2 × 1020
The number . × is greater. When comparing each
numbers order of magnitude, it is obvious that > ;
therefore, . × > . × .
To figure out which number is
greater, I need to look at the order
of magnitude (exponent) of each
number.
2. About how many times greater is 8.2 × 1020 compared to 8.25 × 1015 ?
.
. ×
=
×
. × .
= . … × −
≈ . ×
= ,
. × is about , times greater than . × .
3. Suppose the geographic area of Los Angeles County is 4,751 sq. mi. If the state of California has area
1.637 × 105 square miles, that means that it would take approximately 35 Los Angeles Counties to make
up the state of California. As of 2013, the population of Los Angeles County was 1 × 107 . If the
population were proportional to area, what would be the population of the state of California? Write
× × = ×
= (. × ) ×
= . × � × �
= . ×
The population of California
is . × .
Lesson 11:
G8-M1-HWH-1.3.0-08.2015
Since it takes about 35 Los Angeles Counties to
make up the state of California, then what I
need to do is multiply the population of Los
Angeles County by 35.
The expression 35 × 107 is not in scientific
notation because 35 is too large (it has to
be less than 10). I can rewrite 35 as 3.5 ×
10 because 35 = 3.5 × 10.
Efficacy of Scientific Notation
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G8-M1-Lesson 12: Choice of Unit
1. What is the average of the following two numbers?
To find the average, I need to add
the two numbers and then divide
by 2. Since the numbers are raised
to the same power of 10, I really
only need to add 3.257 and 3.1.
3.257 × 103 and 3.1 × 103
The average is
. × + . × (. + . ) ×
=
. ×
=
.
=
×
= . ×
2. Assume you are given the data below and asked to decide on a new unit in order to make comparisons
and discussions of the data easier.
1.9 × 1015
3.75 × 1019
4.56 × 1017
2.4 × 103
9.26 × 1016
a.
b.
7.02 × 1019
I need to examine the exponents to see
which is most common or which exponent
most numbers would be close to. Since
I’m deciding the unit, I just need to make
sure my choice is reasonable.
What new unit would you select? Name it and express it using a power of 10.
I would choose to use as my unit. I’m ignoring the number with because it is so much
smaller than the other numbers. Most of the other numbers are close to . I will name my unit
.
Rewrite at least two pieces of data using the new unit.
.×
= . × − = . × − = .
. ×
.×
rewritten in the new unit is . .
= . × − = . × = .
. × rewritten in the new unit is . .
Lesson 12:
G8-M1-HWH-1.3.0-08.2015
Choice of Unit
To rewrite the data, I will
take the original number
and divide it by the value of
my unit, , which is 1018 .
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G8-M1-Lesson 13: Comparison of Numbers Written in Scientific
Notation and Interpreting Scientific Notation Using Technology
1. If × 10 < × 10 , what are some possible values for and ? Explain how you know.
When two numbers are each raised to the same power of , in this case the power of , then you only
need to look at the numbers and when comparing the values (Inequality (A) guarantees this). Since
we know that × < × , then we also know that < . Then a possible value for is and
a possible value for is because < .
2. Assume that × 10−5 is not written in scientific notation and is positive. That means that is greater
than zero but not necessarily less than 10. Is it possible to find a number so that
× 10−5 < 1.1 × 105 is not true? If so, what number could be?
Since − = . and . × = , then
a number for bigger than . × would show that
× − < . × is not true.
If = . × , then by substitution
× − = �. × � × −
= . × +(−)
= . × .
If = . × , then × − = . × . Therefore
can be any number as long as > . × .
If × 10−5 < 1.1 × 105 is not
written in scientific notation, it
means that can be a really
large number. What I’m really
being asked is if there is a
number I can think of that
when multiplied by 10−5 or its
equivalent
1
100000
would be
larger than 1.1 × 105?
3. Which of the following two numbers is greater?
2.68941 × 1027 or 2.68295 × 1027
Since . > . , then
. × > . × .
Lesson 13:
G8-M1-HWH-1.3.0-08.2015
Since both numbers are raised to the
same power (Inequality (A) again), all
I need to compare is 2.68941 and
2.68295.
Comparison of Numbers Written in Scientific Notation and Interpreting
Scientific Notation Using Technology
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https://noi.ph/tutorial-2/ | 1,611,803,852,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835583.91/warc/CC-MAIN-20210128005448-20210128035448-00395.warc.gz | 473,956,703 | 19,611 | Introduction
The National Olympiad in Informatics — Philippines is an informatics olympiad where contestants solve algorithmic tasks by writing correct and efficient programs.
Let’s try to break that down.
• National means everyone in the Philippines can join this contest.
• Informatics is the science of studying efficient methods for storing, processing, and retrieving data.
• An olympiad is a major contest in a particular game, sport or scientific subject.
Like many scientific olympiads, the contestants (i.e. you!) are given a number of tasks to solve within a limited amount of time. In math olympiads, contestants are often given problems that look not unlike the following:
There are an infinite number of positive integers $n$ such that $1 + 2 + 3 + \ldots + n$ is divisible by $n$. Find the $100$th one.
These types of problems are all well and good, but a sufficiently determined contestant might elect to start bashing every case; he or she might write down every $n$ starting from $1$, check whether each $n$ satisfies the condition by trial and error — and eventually arrive at the correct answer! In this case, the contestant did not even have to think at all, and missed out on the crucial insights that would have make the solution to this problem much more palatable.
The NOI.PH, as an informatics olympiad, is not all too different from these math competitions or olympiads; in that the tasks often require insights, and these insights have to be synthesized to produce a final answer. The crucial distinction is that, in the NOI.PH, your final answer is a program, not a number.
If the same problem were to be given in the NOI.PH, it might be phrased like so:
There are an infinite number of positive integers $n$ such that $1 + 2 + 3 + \ldots + n$ is divisible by $n$. Find the $\mathbf{K}$th one.
Notice the critical difference: we are being asked to find the $K$th such number instead of, say, the $100$th one. As a contestant, your goal is to submit a program that takes in any integer $K$, satisfying certain constraints (more on that later), as input, and gives the correct answer for that particular $K$, as output. Your program might be given $K=100$, or $K=1$, or $K=39$, or some other number; and your program must always give the correct answer, and do so quickly enough.
By requiring contestants to submit a program instead of a number, the contestants solve the general case of the task and thus demonstrate their knowledge of the insights necessary to solve it! While a contestant might simply write a program that tries every possible $n$ from $1$ onwards to find the $K$th one, this will run noticeably slower than someone who, say, has found a closed-form formula or derived a fast algorithm to solve the same. For large enough $K$, the solution that simply tries all different sums will eventually take very long, and so it will exceed the time limit and, like a wrong answer, will not be credited full marks. Of course, the beauty of the NOI.PH is that even suboptimal programs can sometimes score partial points — but we’ll cover the specifics of that later.
Now, let’s get our hands dirty, by solving an actual informatics task!
Note that this section assumes knowledge of basic programming, such as arrays and nested loops, and a little bit of C++ syntax. If you have no prior experience with these concepts or simply want a refresher, you are invited to attempt the C++ for Competitive Programming tutorial, which also serves as a softer and more comprehensive version of this guide.
We will be tackling the task Azalea, and we will go through the different parts of the task and attempt some solutions. Along the way, we will also discuss some features of the HackerRank platform in particular, on which the NOI.PH contests are hosted for the time being.
## The Problem Statement
All tasks begin with a problem statement. The problem statement situates the task, and usually, although not necessarily, comes accompanied with a story. As a competitive programmer, it is your job to dissect the problem statement and figure out exactly what it is asking. Sometimes, this is easy; the statement is no-nonsense and gets straight to the point (e.g. A Simple Problem!). Other times, the statement can be hilariously obnoxious and riddled with unnecessary background (e.g. Meteor Garden Garden).
The problem statement for Azalea reads as follows:
A small quaint town, known for its annual springtime azalea festival, begins preparations for the new season. The organizers have arranged $N$ azalea shrubs in a horizontal row; the $i^\text{th}$ shrub, from the left, has $A_i$ flowers, and is labelled $i$.
The festival lasts $D$ days. In the $i^\text{th}$ day, as per tradition, the number of flowers between shrubs labelled $L_i$ to $R_i$, inclusive, are counted, and the townspeople gather together and prepare that many dishes for the festivities of that day.
After the festivities of each day, the local gardener comes and waters every shrub, and so a single new flower blooms in each one of them.
The festival has not yet begun, but the townspeople want to know how many dishes they will need to prepare for each day, so they can start sourcing for ingredients. Can you help them?
The statement is quite long, and contains quite a number of irrelevant details. Can we whittle down the length and express it in a more concise way?
A good place to start is to try to list down the events in point form, so we can get a clearer picture of what we have to do:
1. There are $N$ shrubs, with the shrub labelled $i$ having $A_i$ flowers.
2. There are $D$ days:
• In the $i$th day, the number of dishes prepared is equal to the number of flowers between shrubs $L_i$ to $R_i$, inclusive.
• In the $i$th night, a gardener comes and waters all the shrubs, so they each gain one flower.
3. We need to determine the number of dishes prepared for each of the $D$ days.
Much better!
Remember that $N$, $D$, and all the $A_i$, $L_i$ and $R_i$ are variables and our program must produce the correct result regardless of the values of these numbers!
## The Input and The Output
Now that we know what to do, let’s look at the input format and output format, and the sample input and sample output.
The input format reads as follows:
The first line contains a single integer $T$, the number of test cases to handle.
Each test case begins with two integers $N$ and $D$ on a single line, the number of azalea shrubs and the number of days the azalea festival lasts, respectively.
The next line contains $N$ integers, $A_1, A_2, A_3, \dots, A_N$, the number of flowers in each shrub.
The next $D$ lines each contain two integers. Specifically, the $i^\text{th}$ line among these contains $L_i$ and $R_i$, describing the range of shrubs whose flowers will be counted for the $i^\text{th}$ day.
The output format reads as follows:
For each test case, output $D$ integers, separated by spaces, on a single line. The $i^\text{th}$ integer should contain the number of dishes the townspeople must prepare for the $i^\text{th}$ day.
If it looks rather complicated, don’t despair; all it is really telling us is the format our program should receive the input, and the format our program should report the output. We will see in a short while what all of this means! For now, take special care for the output format; everything our program outputs is considered “output”, so if you output some debugging statements like “Please enter T” or print something like: “The answer is: ”, this will be considered as part of your answer, and because it does not satisfy the output format, it will be marked wrong, even if your answer is otherwise correct!
Some tasks, such as this one, contain multiple test cases, which basically means your program should solve the task multiple times, with different inputs. Each test case describes a separate instance of the task; however, your program must correctly process all test cases in a single input within the given time limit.
Let’s take a look at the sample input:
2
6 3
1 1 2 1 9 9
3 5
1 6
2 2
10 1
1 2 3 4 5 6 7 8 9 10
1 10
And the sample output:
12 29 3
55
The sample input gives an instance of some valid input that your program might be given, and the sample output gives a correct response to the given sample input. To see that, let’s recall the input format.
2
6 3
1 1 2 1 9 9
3 5
1 6
2 2
10 1
1 2 3 4 5 6 7 8 9 10
1 10
The first line of input contains a single integer $T$, which is the number of test cases; in this case, we know that $T=2$, so we have 2 test cases to handle.
2
6 3
1 1 2 1 9 9
3 5
1 6
2 2
10 1
1 2 3 4 5 6 7 8 9 10
1 10
Each test case begins with two integers, $N$ and $D$; in this case, we know that the first test case has $N=6$ and $D=3$. The next line contains $N=6$ integers, which contains the values of $A_1, A_2, A_3, \ldots, A_6$. Then, the next $D=3$ lines contain two integers; the first line among these contains $L_1$, $R_1$, the second $L_2$, $R_2$, and the third $L_3$, $R_3$.
2
6 3
1 1 2 1 9 9
3 5
1 6
2 2
10 1
1 2 3 4 5 6 7 8 9 10
1 10
The remaining lines describe the second test case; in this case, we know that the second test case has $N=10$ and $D=1$. The next line contains $N=10$ integers, which contains the values of $A_1, A_2, A_3, \ldots, A_{10}$. Then, the next $D=1$ line contains two integers, $L_1$ and $R_1$.
Consider the first test case. We can, in fact, substitute the values of the variables into our condensed problem statement:
1. There are 6 shrubs:
• The 1st shrub has 1 flower.
• The 2nd shrub has 1 flower.
• The 3rd shrub has 2 flowers.
• The 4th shrub has 1 flower.
• The 5th shrub has 9 flowers.
• The 6th shrub has 9 flowers.
2. There are 3 days:
1. In the 1st day:
• The number of dishes prepared is equal to the number of flowers between shrubs 3 to 5, inclusive.
• Afterwards, the gardener comes and waters all the shrubs, so they each gain one flower.
2. In the 2nd day:
• The number of dishes prepared is equal to the number of flowers between shrubs 1 to 6, inclusive.
• Afterwards, the gardener comes and waters all the shrubs, so they each gain one flower.
3. In the 3rd day:
• The number of dishes prepared is equal to the number of flowers between shrubs 2 to 2, inclusive.
• Afterwards, the gardener comes and waters all the shrubs, so they each gain one flower.
3. We need to determine the number of dishes prepared for each of the 3 days.
## The Explanation Section
The explanation of the sample input is usually given in the explanation section, and indeed, this is precisely what we get:
In the first test case, there are $6$ shrubs, and the springtime azalea festival lasts for $3$ days.
Before the festival begins, the shrubs have $1$, $1$, $2$, $1$, $9$, and $9$ flowers from left to right.
In the $1^\text{st}$ day, the flowers between shrubs $3$ and $5$, inclusive, are counted. There are $2 + 1 + 9 = 12$ flowers, and so the townspeople prepare $12$ dishes for the festivities of the $1^\text{st}$ day.
After the festivities of the $1^\text{st}$ day, the local gardener comes and waters each shrub. Now, the shrubs have $2$, $2$, $3$, $2$, $10$, and $10$ flowers from left to right.
In the $2^\text{nd}$ day, the flowers between shrubs $1$ and $6$, inclusive, are counted. There are $2 + 2 + 3 + 2 + 10 + 10 = 29$ flowers, and so the townspeople prepare $29$ dishes for the festivities of the $2^\text{nd}$ day.
After the festivities of the $2^\text{nd}$ day, the local gardener comes and waters each shrub. Now, the shrubs have $3$, $3$, $4$, $3$, $11$, and $11$ flowers from left to right.
In the $3^\text{rd}$ and final day, the flowers between shrubs $2$ and $2$, inclusive, are counted. There are $3$ flowers, and so the townspeople prepare $3$ dishes for the festivities of the $3^\text{rd}$ day.
After the festivities of the $3^\text{nd}$ day, the local gardener comes and waters each shrub. Now, the shrubs have $4$, $4$, $5$, $4$, $12$, and $12$ flowers from left to right.
The festivites are over, and we report the number of dishes the townspeople must prepare each day:
1. The townspeople must prepare $12$ dishes on the $1^\text{st}$ day.
2. The townspeople must prepare $29$ dishes on the $2^\text{nd}$ day.
3. The townspeople must prepare $3$ dishes on the $3^\text{rd}$ day.
Similar analysis can be performed for the second test case.
The explanation situates the problem in terms of “numbers”, instead of variables. More precisely, it explains the sample input, and how we can derive the sample output if we were given the sample input. Of course, the reasoning given in the explanation is non-general; it applies only to the given sample input, and usually does not give many insights about how the problem can be solved in the general case.
It is good practice to look at the explanation and see that your understanding of the problem is correct; sometimes, you may encounter ambiguous statements, and you can rely on the explanation to see whether you have read the problem correctly, or if you have missed some important detail. In this case, it may not be as important, but in other problems you may find this explanation very much helpful.
## The Constraints
Now that that’s clear, we can discuss the constraints:
$1 \leq T \leq 5$
$1 \leq A_i \leq 1000$
$1 \leq L_i \leq R_i \leq N$
$1 \leq N \leq 2000$
$1 \leq D \leq 2000$
$1 \leq N \leq 100000$
$1 \leq D \leq 100000$
The sum of $|L_i- R_i|$ in a test case does not exceed $10^6$.
$1 \leq N \leq 100000$
$1 \leq D \leq 100000$
The constraints communicate to us the maximum and minimum possible values the variables in the problem statement can assume. For example, in this case, we have $1 \leq T \leq 5$, which means that there will be at least $1$ test case and at most $5$ test cases. This makes sense, because our program cannot be expected to, say, handle $T = 10^9$ test cases, and still run within the time limit! Some constraints are straightforward; $A_i$ between $1$ and $1000$, for example; while others are more cryptic, like “the sum of $|L_i-R_i|$ in a test case does not exceed $10^6$”. Sometimes, a cryptic constraint reveals a property that can be exploited to write an efficient enough solution.
## Time Limit
Speaking of efficient solutions, note that all programs have a time limit within which they are expected to read all the input, process the input, write all the output, and terminate. It must be able to solve all the test cases in the input within the time limit. The time limit is standard for all NOI.PH problems, and, unless explicitly noted otherwise, are:
• 2 seconds for C++;
• 4 seconds for Java;
• 10 seconds for Python.
The time limit is more relaxed for Java and Python, as these languages are inherently slower; however, generally speaking, if the algorithm itself is sub-optimal, it will fail regardless of which language you code it on! Remember that the problems in the NOI.PH are always algorithmic in nature, and code is simply our way of expressing our algorithm. By its nature, a more efficient algorithm almost always trumps a better implementation of a bad algorithm!
In general, we can perform about $10^9$ “cheap” operations, such as addition or bitwise operations, within the time limit, and about $10^8$ “expensive” operations, such as multiplication or division, within the same. Knowing exactly how much processing power you can squeeze from your program is an ability that is learned through practice, and you will eventually have a good intuition for it!
Notice that there are subtasks, each with a varying number of points. While there are constraints that are common to all subtasks, such as $1 \leq A_i \leq 1000$ or $1 \leq L_i \leq R_i \leq N$, there are constraints that are imposed to some subtasks but not others; for example, subtask $1$ imposes that $N$ is only up to $2000$, whereas subtasks $2$ and $3$ impose that $N$ is up to $100000$. This allows us to score partial points! For example, we might have a program that is correct and runs within the time limit if $N$ is just 2000, but produces a wrong answer or exceeds the time limit if it is greater than $100000$.
## The Code Editor
That said, we can now attempt to solve this problem.
HackerRank has a built-in code editor, provided with each problem, where we can write our code. By default, it contains the following:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}
If you are just starting out, this can be very helpful! You do not need to download a compiler or an IDE just yet; for simpler tasks, you will find that the online code editor will suffice.
HackerRank provides a number of tools you can use to test your code. For example, hitting automatically compiles and runs the code against the sample input as described in the problem statement, and tells you whether your output is correct or not, or whether your code took too long, crashed, or could not even compile (in the last case, HackerRank also helpfully informs you of the error). You can also test against your own input by using the built-in function to Test against custom input. Using this feature, you can give your program any input, and HackerRank essentially serves as a compiler. Note that this option does not check the correctness of your output or even the validity of input; this means you can use the function to test your program against anything, just as you would a normal compiler.
When you are done with your solution, and you think that it is correct, you can simply hit and watch, in real time, the results as your program is tested against the input.
## Submission 1
All that said, let’s see what happens when we code the most straightforward solution.
#include <iostream>
using namespace std;
int a[100005];
int ans[100005];
int main()
{
int tc, n, d, l, r, sum;
cin >> tc;
for (int t = 0; t < tc; t++)
{
cin >> n >> d;
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= d; i++)
{
cin >> l >> r;
sum = 0;
for (int j = l; j <= r; j++) //compute the number of flowers from the lth shrub to the rth shrub
sum += a[j];
ans[i] = sum; //store the answer
for (int j = 1; j <= n; j++) //increase the number of flowers in each shrub by 1
a[j]++;
}
for (int i = 1; i <= d; i++)
{
if (i != 1) //print space before every integer, except the first
cout << ' ';
cout << ans[i];
}
cout << '\n';
}
}
We hit , and… um, wow! It worked on the first try! (Note: this is rare and doesn’t usually happen, especially when starting out. You will normally encounter a number of compilation issues before you can get it to work. Learning how to debug is important!)
Congratulations, you passed the sample test case.
Click the Submit Code button to run your code against all the test cases.
2
6 3
1 1 2 1 9 9
3 5
1 6
2 2
10 1
1 2 3 4 5 6 7 8 9 10
1 10
12 29 3
55
12 29 3
55
Now, we hit to submit our code.
Submitted a few seconds ago
Score: 24.00
Status: Terminated due to timeout
✓ Test Case #0
✓ Test Case #1
Test Case #2
Test Case #3
Alas! Our code successfully solves the first subtask, scoring 24 points, but it fails for the other two, giving a “terminated due to timeout” verdict. “What gives? It worked on the sample test cases!” I hear you exclaim.
## Submission 2
Let’s analyze how many operations our code takes. It is clear that the majority of the work is happening in the two nested loops; one looping $i$ from $1$ to $D$, and, within that, two loops; one looping from $L_i$ to $R_i$ and another looping from $1$ to $N$.
We know that, for the second test case, $N$ and $D$ are both up to $100000$. Therefore, even if we ignore the loop going from $L_i$ to $R_i$, we already have $100000 \cdot 100000 = 10^{10}$ iterations; this is significantly greater than $10^9$, and so we can wager that our program will probably not finish in time. This is even ignoring the fact that we have up to $T=5$ test cases in one input!
We will need to optimize our algorithm, by making some observations.
The bottleneck in our program is the loop from $1$ to $N$. Can we somehow eliminate this loop?
Yes, we can! We need to notice that, because the gardener waters all the plants every day, at the $i$th day, each shrub simply has $i-1$ extra flowers than original.
The other insight is that we can count these extra flowers separately; there are $R-L+1$ shrubs between shrubs $L$ and $R$, inclusive, and each of them has $i-1$ extra flowers. So, the total number of extra flowers is simply $(R-L+1) \cdot (i-1)$, and we can simply add this to the total count of flowers!
We now have the following code:
#include <iostream>
using namespace std;
int a[100005];
int ans[100005];
int main()
{
int tc, n, d, l, r, sum;
cin >> tc;
for (int t = 0; t < tc; t++)
{
cin >> n >> d;
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= d; i++)
{
cin >> l >> r;
sum = 0;
for (int j = l; j <= r; j++) //compute the original number of flowers from the lth shrub to the rth shrub
sum += a[j];
sum += (r-l+1)*(i-1); //add the extra flowers
ans[i] = sum; //store the answer
}
for (int i = 1; i <= d; i++)
{
if (i != 1) //print space before every integer, except the first
cout << ' ';
cout << ans[i];
}
cout << '\n';
}
}
How long does this run? At worst, we could still have $L_i=1$ and $R_i=N$ for all $i$; then, we will still have $100000 \cdot 100000 =10^{10}$ iterations. But wait! Remember that the second subtask has an extra suspicious constraint — the sum of $|L_i-R_i|$ is at most $10^6$. Could this perhaps save us here?
A more nuanced count on the number of iterations our loop actually performs can be of help. Note that we loop from $L_i$ to $R_i$, so we consider only $R_i-L_i+1$ values in the $i$th day. So, summing this over all days, the inner loop actually iterates $(R_1-L_1+1)+(R_2-L_2+1)+(R_3-L_3+1)+\ldots+(R_D-L_D+1)$ times! If we note that $R_i-L_i$ is equal to $|L_i-R_i|$ because $L_i \leq R_i$, and that there are exactly $D$ occurrences of $1$ in this sum, then we can rewrite this as: $|L_1-R_1|+|L_2-R_2|+|L_3-R_3|+\ldots+|L_D-R_D|+D$. But the sum of all the $|L_i-R_i|$ is at most $10^6$, and $D \leq 10^5$ so the above sum is actually only at most $1.1 \cdot 10^6$; that means, the inner loop only performs at most $1.1 \cdot 10^6$ iterations across all days for each test case!
This analysis should convince us that this program will finish in time. And indeed, submitting it gives…
Submitted a few seconds ago
Score: 24.00
✓ Test Case #0
✓ Test Case #1
✗ Test Case #2
Test Case #3
## Submission 3
…wrong answer? What’s up with that? Surely the problem setter must have an incorrect model solution!
In fact, there are a number of very subtle mistakes in our code, which can be summed up in a word: overflow. In our code, we used the int data type, which in C++ can store any integer value between $-2^{31}$ and $2^{31}-1$, or slightly more than $2 \cdot 10^9$. But what is the maximum possible answer?
It is not immediately obvious, but it can actually exceed this! If we have $N=100000$ and $D=100000$, and on the last day we have $L_D=1$ and $L_D=N$, then the total number of extra flowers across all shrubs, according to our previous formula, is $N \cdot (D-1)$. That’s $100000 \cdot 99999$, which is almost $10^{10}$; thus, we get integer overflow! The problem setter must have been very sneaky, and, anticipating this, added this test case in!
To remedy this, we simply have to use a larger data type, such as long long int, which can store values between $-2^{64}$ and $2^{64}-1$, which is much larger than the largest possible answer. The details of this are more nuanced, and it is not enough to simply change the types of sum and ans in the above code; we must also change $l$ and $r$, or perform a typecast when we do the multiplication (r-l+1)*(i-1). You are encouraged to peruse this to find out more details. (You need to register here first before you can access the link.)
Anyway, after fixing these bugs, we now have the following code:
#include <iostream>
using namespace std;
int a[100005];
long long int ans[100005];
int main()
{
int tc, n, d, l, r;
long long int sum;
cin >> tc;
for (int t = 0; t < tc; t++)
{
cin >> n >> d;
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= d; i++)
{
cin >> l >> r;
sum = 0;
for (int j = l; j <= r; j++) //compute the original number of flowers from the lth shrub to the rth shrub
sum += a[j];
sum += ((long long int)(r-l+1))*(i-1); //add the extra flowers, cast to long long int to avoid overflow
ans[i] = sum; //store the answer
}
for (int i = 1; i <= d; i++)
{
if (i != 1) //print space before every integer, except the first
cout << ' ';
cout << ans[i];
}
cout << '\n';
}
}
Now, we smash that button and get:
Submitted a few seconds ago
Score: 58.00
Status: Terminated due to timeout
✓ Test Case #0
✓ Test Case #1
✓ Test Case #2
Test Case #3
Eureka! We’ve got the second subtask.
## Submission 4+
The timeout on the third is rather expected; as we mentioned earlier, in the worst case, we could still have $L_i=1$ and $R_i=N$ for all $i$, causing our program to require $100000 \cdot 100000=10^{10}$ iterations. The final subtask has no restrictions on $|Ri-Li|$, so we can’t perform the reductions earlier to show that this worst case is impossible. Indeed, the worst case is very much possible — and you can expect, as a contestant, that the problem setter has also added this worst case, or something similar, as one of the test cases for the final subtask.
So, are we doomed? Or is there hope for us yet?
Luckily, there is (and of course there is, because the problem setter must have some correct solution), and it involves a very clever trick.
Let’s create an array $C$ of length $N$, where $C_i=A_1+A_2+A_3+\ldots+A_i$. We can generate this array very efficiently, using only one loop; we let $C_1=A_1$, then for all $2 \leq i \leq N$, we can do $C_i=C_i-1+A_i$. Thus we can generate this array with only about $N$ operations.
Now, suppose we want to find the sum of $A_{L_i}+A_{L_i+1}+A_{L_i+2}+\ldots+A_{R_i}$. Currently, we do that using a loop, which takes $R_i-L_i+1$ operations; that’s too slow! Notice, however, that the sum is equal to: $A_1+A_2+A_3+\ldots+A_{R_i}-(A_1+A_2+A_3+\ldots+A_{L_i}-1)$, because all the terms $A_1, A_2, A_3, \ldots, A_{L_i-1}$ cancel themselves out!
But then this is equal to $C_{R_i}-C_{L_i-1}$, using our earlier definition for $C$. And we already know how to generate $C$ efficiently; not to mention, we only ever have to generate it once, at the beginning, because we’ve already reduced the problem to the case where the values in $A$ never change. So, we can eliminate the nested loop entirely!
Actually coding this is left as an exercise for you; while you code it, take into consideration any possible corner cases or sources of bugs. Rest assured that it will be worth it; there is no joy just the same as seeing this:
Submitted a few seconds ago
Score: 100.00
Status: Accepted
✓ Test Case #0
✓ Test Case #1
✓ Test Case #2
✓ Test Case #3
Except maybe the smell of azalea flowers in spring. But that’s besides the point.
## Conclusion
Hopefully, the in-depth analysis of this problem and the corresponding discussions have allowed you to develop a more comprehensive understanding of what the NOI.PH is, what it entails, the contest format and platform, the types of problems you will encounter, as well as the thought processes and the kinds of insights necessary to solve these types of tasks. If you want to learn more, or try your hand at solving some tasks from past NOI.PH contests, make sure to check out:
But you don’t have to be alone in your quest for greatness. There’s even a Discord chat room for Filipino competitive programming enthusiasts. Email [email protected] for more information.
We hope you found this whole tutorial useful, and we wish you the best of luck in the | 7,554 | 27,971 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2021-04 | latest | en | 0.938421 |
https://www.telematika.org/py/pdsh_02.08-sorting/ | 1,611,057,318,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00413.warc.gz | 1,010,132,570 | 8,731 | ## Jupyter Snippet PDSH 02.08-Sorting
Jupyter Snippet PDSH 02.08-Sorting
This notebook contains an excerpt from the Python Data Science Handbook by Jake VanderPlas; the content is available on GitHub.
The text is released under the CC-BY-NC-ND license, and code is released under the MIT license. If you find this content useful, please consider supporting the work by buying the book!
# Sorting Arrays
Up to this point we have been concerned mainly with tools to access and operate on array data with NumPy. This section covers algorithms related to sorting values in NumPy arrays. These algorithms are a favorite topic in introductory computer science courses: if you’ve ever taken one, you probably have had dreams (or, depending on your temperament, nightmares) about insertion sorts, selection sorts, merge sorts, quick sorts, bubble sorts, and many, many more. All are means of accomplishing a similar task: sorting the values in a list or array.
For example, a simple selection sort repeatedly finds the minimum value from a list, and makes swaps until the list is sorted. We can code this in just a few lines of Python:
import numpy as np
def selection_sort(x):
for i in range(len(x)):
swap = i + np.argmin(x[i:])
(x[i], x[swap]) = (x[swap], x[i])
return x
x = np.array([2, 1, 4, 3, 5])
selection_sort(x)
array([1, 2, 3, 4, 5])
As any first-year computer science major will tell you, the selection sort is useful for its simplicity, but is much too slow to be useful for larger arrays. For a list of $N$ values, it requires $N$ loops, each of which does on order $\sim N$ comparisons to find the swap value. In terms of the “big-O” notation often used to characterize these algorithms (see Big-O Notation), selection sort averages $\mathcal{O}[N^2]$: if you double the number of items in the list, the execution time will go up by about a factor of four.
Even selection sort, though, is much better than my all-time favorite sorting algorithms, the bogosort:
def bogosort(x):
while np.any(x[:-1] > x[1:]):
np.random.shuffle(x)
return x
x = np.array([2, 1, 4, 3, 5])
bogosort(x)
array([1, 2, 3, 4, 5])
This silly sorting method relies on pure chance: it repeatedly applies a random shuffling of the array until the result happens to be sorted. With an average scaling of $\mathcal{O}[N \times N!]$, (that’s N times N factorial) this should–quite obviously–never be used for any real computation.
Fortunately, Python contains built-in sorting algorithms that are much more efficient than either of the simplistic algorithms just shown. We’ll start by looking at the Python built-ins, and then take a look at the routines included in NumPy and optimized for NumPy arrays.
## Fast Sorting in NumPy: np.sort and np.argsort
Although Python has built-in sort and sorted functions to work with lists, we won’t discuss them here because NumPy’s np.sort function turns out to be much more efficient and useful for our purposes. By default np.sort uses an $\mathcal{O}[N\log N]$, quicksort algorithm, though mergesort and heapsort are also available. For most applications, the default quicksort is more than sufficient.
To return a sorted version of the array without modifying the input, you can use np.sort:
x = np.array([2, 1, 4, 3, 5])
np.sort(x)
array([1, 2, 3, 4, 5])
If you prefer to sort the array in-place, you can instead use the sort method of arrays:
x.sort()
print(x)
[1 2 3 4 5]
A related function is argsort, which instead returns the indices of the sorted elements:
x = np.array([2, 1, 4, 3, 5])
i = np.argsort(x)
print(i)
[1 0 3 2 4]
The first element of this result gives the index of the smallest element, the second value gives the index of the second smallest, and so on. These indices can then be used (via fancy indexing) to construct the sorted array if desired:
x[i]
array([1, 2, 3, 4, 5])
### Sorting along rows or columns
A useful feature of NumPy’s sorting algorithms is the ability to sort along specific rows or columns of a multidimensional array using the axis argument. For example:
rand = np.random.RandomState(42)
X = rand.randint(0, 10, (4, 6))
print(X)
[[6 3 7 4 6 9]
[2 6 7 4 3 7]
[7 2 5 4 1 7]
[5 1 4 0 9 5]]
# sort each column of X
np.sort(X, axis=0)
array([[2, 1, 4, 0, 1, 5],
[5, 2, 5, 4, 3, 7],
[6, 3, 7, 4, 6, 7],
[7, 6, 7, 4, 9, 9]])
# sort each row of X
np.sort(X, axis=1)
array([[3, 4, 6, 6, 7, 9],
[2, 3, 4, 6, 7, 7],
[1, 2, 4, 5, 7, 7],
[0, 1, 4, 5, 5, 9]])
Keep in mind that this treats each row or column as an independent array, and any relationships between the row or column values will be lost!
## Partial Sorts: Partitioning
Sometimes we’re not interested in sorting the entire array, but simply want to find the k smallest values in the array. NumPy provides this in the np.partition function. np.partition takes an array and a number K; the result is a new array with the smallest K values to the left of the partition, and the remaining values to the right, in arbitrary order:
x = np.array([7, 2, 3, 1, 6, 5, 4])
np.partition(x, 3)
array([2, 1, 3, 4, 6, 5, 7])
Note that the first three values in the resulting array are the three smallest in the array, and the remaining array positions contain the remaining values. Within the two partitions, the elements have arbitrary order.
Similarly to sorting, we can partition along an arbitrary axis of a multidimensional array:
np.partition(X, 2, axis=1)
array([[3, 4, 6, 7, 6, 9],
[2, 3, 4, 7, 6, 7],
[1, 2, 4, 5, 7, 7],
[0, 1, 4, 5, 9, 5]])
The result is an array where the first two slots in each row contain the smallest values from that row, with the remaining values filling the remaining slots.
Finally, just as there is a np.argsort that computes indices of the sort, there is a np.argpartition that computes indices of the partition. We’ll see this in action in the following section.
## Example: k-Nearest Neighbors
Let’s quickly see how we might use this argsort function along multiple axes to find the nearest neighbors of each point in a set. We’ll start by creating a random set of 10 points on a two-dimensional plane. Using the standard convention, we’ll arrange these in a $10\times 2$ array:
X = rand.rand(10, 2)
To get an idea of how these points look, let’s quickly scatter plot them:
%matplotlib inline
import matplotlib.pyplot as plt
import seaborn; seaborn.set() # Plot styling
plt.scatter(X[:, 0], X[:, 1], s=100);
Now we’ll compute the distance between each pair of points. Recall that the squared-distance between two points is the sum of the squared differences in each dimension; using the efficient broadcasting (Computation on Arrays: Broadcasting) and aggregation (Aggregations: Min, Max, and Everything In Between) routines provided by NumPy we can compute the matrix of square distances in a single line of code:
dist_sq = np.sum((X[:, np.newaxis, :] - X[np.newaxis, :, :]) ** 2, axis=-1)
This operation has a lot packed into it, and it might be a bit confusing if you’re unfamiliar with NumPy’s broadcasting rules. When you come across code like this, it can be useful to break it down into its component steps:
# for each pair of points, compute differences in their coordinates
differences = X[:, np.newaxis, :] - X[np.newaxis, :, :]
differences.shape
(10, 10, 2)
# square the coordinate differences
sq_differences = differences ** 2
sq_differences.shape
(10, 10, 2)
# sum the coordinate differences to get the squared distance
dist_sq = sq_differences.sum(-1)
dist_sq.shape
(10, 10)
Just to double-check what we are doing, we should see that the diagonal of this matrix (i.e., the set of distances between each point and itself) is all zero:
dist_sq.diagonal()
array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
It checks out! With the pairwise square-distances converted, we can now use np.argsort to sort along each row. The leftmost columns will then give the indices of the nearest neighbors:
nearest = np.argsort(dist_sq, axis=1)
print(nearest)
[[0 3 9 7 1 4 2 5 6 8]
[1 4 7 9 3 6 8 5 0 2]
[2 1 4 6 3 0 8 9 7 5]
[3 9 7 0 1 4 5 8 6 2]
[4 1 8 5 6 7 9 3 0 2]
[5 8 6 4 1 7 9 3 2 0]
[6 8 5 4 1 7 9 3 2 0]
[7 9 3 1 4 0 5 8 6 2]
[8 5 6 4 1 7 9 3 2 0]
[9 7 3 0 1 4 5 8 6 2]]
Notice that the first column gives the numbers 0 through 9 in order: this is due to the fact that each point’s closest neighbor is itself, as we would expect.
By using a full sort here, we’ve actually done more work than we need to in this case. If we’re simply interested in the nearest $k$ neighbors, all we need is to partition each row so that the smallest $k + 1$ squared distances come first, with larger distances filling the remaining positions of the array. We can do this with the np.argpartition function:
K = 2
nearest_partition = np.argpartition(dist_sq, K + 1, axis=1)
In order to visualize this network of neighbors, let’s quickly plot the points along with lines representing the connections from each point to its two nearest neighbors:
plt.scatter(X[:, 0], X[:, 1], s=100)
# draw lines from each point to its two nearest neighbors
K = 2
for i in range(X.shape[0]):
for j in nearest_partition[i, :K+1]:
# plot a line from X[i] to X[j]
# use some zip magic to make it happen:
plt.plot(*zip(X[j], X[i]), color='black')
Each point in the plot has lines drawn to its two nearest neighbors. At first glance, it might seem strange that some of the points have more than two lines coming out of them: this is due to the fact that if point A is one of the two nearest neighbors of point B, this does not necessarily imply that point B is one of the two nearest neighbors of point A.
Although the broadcasting and row-wise sorting of this approach might seem less straightforward than writing a loop, it turns out to be a very efficient way of operating on this data in Python. You might be tempted to do the same type of operation by manually looping through the data and sorting each set of neighbors individually, but this would almost certainly lead to a slower algorithm than the vectorized version we used. The beauty of this approach is that it’s written in a way that’s agnostic to the size of the input data: we could just as easily compute the neighbors among 100 or 1,000,000 points in any number of dimensions, and the code would look the same.
Finally, I’ll note that when doing very large nearest neighbor searches, there are tree-based and/or approximate algorithms that can scale as $\mathcal{O}[N\log N]$ or better rather than the $\mathcal{O}[N^2]$ of the brute-force algorithm. One example of this is the KD-Tree, implemented in Scikit-learn.
## Aside: Big-O Notation
Big-O notation is a means of describing how the number of operations required for an algorithm scales as the input grows in size. To use it correctly is to dive deeply into the realm of computer science theory, and to carefully distinguish it from the related small-o notation, big-$\theta$ notation, big-$\Omega$ notation, and probably many mutant hybrids thereof. While these distinctions add precision to statements about algorithmic scaling, outside computer science theory exams and the remarks of pedantic blog commenters, you’ll rarely see such distinctions made in practice. Far more common in the data science world is a less rigid use of big-O notation: as a general (if imprecise) description of the scaling of an algorithm. With apologies to theorists and pedants, this is the interpretation we’ll use throughout this book.
Big-O notation, in this loose sense, tells you how much time your algorithm will take as you increase the amount of data. If you have an $\mathcal{O}[N]$ (read “order $N$") algorithm that takes 1 second to operate on a list of length N=1,000, then you should expect it to take roughly 5 seconds for a list of length N=5,000. If you have an $\mathcal{O}[N^2]$ (read “order N squared”) algorithm that takes 1 second for N=1000, then you should expect it to take about 25 seconds for N=5000.
For our purposes, the N will usually indicate some aspect of the size of the dataset (the number of points, the number of dimensions, etc.). When trying to analyze billions or trillions of samples, the difference between $\mathcal{O}[N]$ and $\mathcal{O}[N^2]$ can be far from trivial!
Notice that the big-O notation by itself tells you nothing about the actual wall-clock time of a computation, but only about its scaling as you change N. Generally, for example, an $\mathcal{O}[N]$ algorithm is considered to have better scaling than an $\mathcal{O}[N^2]$ algorithm, and for good reason. But for small datasets in particular, the algorithm with better scaling might not be faster. For example, in a given problem an $\mathcal{O}[N^2]$ algorithm might take 0.01 seconds, while a “better” $\mathcal{O}[N]$ algorithm might take 1 second. Scale up N by a factor of 1,000, though, and the $\mathcal{O}[N]$ algorithm will win out.
Even this loose version of Big-O notation can be very useful when comparing the performance of algorithms, and we’ll use this notation throughout the book when talking about how algorithms scale. | 3,518 | 13,064 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-04 | latest | en | 0.861117 |
https://www.slidesearch.net/slide/serway-cp-poll-ch05 | 1,540,080,582,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513508.42/warc/CC-MAIN-20181020225938-20181021011438-00346.warc.gz | 1,070,609,524 | 13,730 | # Serway CP poll ch05
Information about Serway CP poll ch05
Published on October 9, 2007
Author: Talya
Source: authorstream.com
A 1000 kg sportscar accelerates from zero to 25 m/s in 7.5 s. What is the average power delivered by the automobile engine?: A 1000 kg sportscar accelerates from zero to 25 m/s in 7.5 s. What is the average power delivered by the automobile engine? 20.8 kW 30.3 kW 41.7 kW 52.4 kW A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill. If there is no friction, what is the velocity at the bottom of the hill?: A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill. If there is no friction, what is the velocity at the bottom of the hill? 27 m/s 36 m/s 45 m/s 54 m/s A jet engine develops 105 N of thrust in moving an airplane forward at a speed of 900 km/hr. What is the power developed by the engine?: A jet engine develops 105 N of thrust in moving an airplane forward at a speed of 900 km/hr. What is the power developed by the engine? 500 kW 1 MW 2.5 MW 5 MW A speedboat requires 80 kW to move at a constant speed of 15 m/s. What is the resistive force of the water at this speed?: A speedboat requires 80 kW to move at a constant speed of 15 m/s. What is the resistive force of the water at this speed? 2667 N 5333 N 6500 N 7711 N A pole vaulter clears 6 m. With what velocity does he strike the mat in the landing area?: A pole vaulter clears 6 m. With what velocity does he strike the mat in the landing area? 2.7 m/s 5.4 m/s 10.8 m/s 21.6 m/s A parachutist of mass 50 kg jumps out of an airplane at a height of 1000 m. The parachute deploys, and she lands on the ground with a velocity of 5 m/s. How much energy was lost to air friction during this jump?: A parachutist of mass 50 kg jumps out of an airplane at a height of 1000 m. The parachute deploys, and she lands on the ground with a velocity of 5 m/s. How much energy was lost to air friction during this jump? 49,375 J 98,750 J 197,500 J 489,375 J Water flows over a section of Niagara Falls at a rate of 1.2 x 106 kg/s and falls 50 m. What is the power wasted by the waterfall?: Water flows over a section of Niagara Falls at a rate of 1.2 x 106 kg/s and falls 50 m. What is the power wasted by the waterfall? 588 MW 294 MW 147 MW 60 MW A baseball outfielder throws a baseball of mass 0.15 kg at a speed of 40 m/s and initial angle of 30° . What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.: A baseball outfielder throws a baseball of mass 0.15 kg at a speed of 40 m/s and initial angle of 30° . What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction. zero 30 J 90 J 120 J A Russian weightlifter is able to lift 250 kg 2 m in 2 s. What is his power output?: A Russian weightlifter is able to lift 250 kg 2 m in 2 s. What is his power output? 500 W 2.45 kW 4.9 kW 9.8 kW A worker pushes a 500 N weight wheelbarrow up a ramp 50.0 m in length and inclined at 20° with the horizontal. What potential energy change does the wheelbarrow experience?: A worker pushes a 500 N weight wheelbarrow up a ramp 50.0 m in length and inclined at 20° with the horizontal. What potential energy change does the wheelbarrow experience? 1500 J 3550 J 4275 J 8550 J A worker pushes a 500 N weight wheelbarrow to the top of a 50.0 m ramp, inclined at 20° with the horizontal, and leaves it unattended. The wheelbarrow is accidentally bumped causing it to slide back down the ramp during which a 80 N frictional force acts on it over the 50.0 m. What is the wheelbarrow's kinetic energy at the bottom of the ramp? (g = 9.8 m/s2 ): A worker pushes a 500 N weight wheelbarrow to the top of a 50.0 m ramp, inclined at 20° with the horizontal, and leaves it unattended. The wheelbarrow is accidentally bumped causing it to slide back down the ramp during which a 80 N frictional force acts on it over the 50.0 m. What is the wheelbarrow's kinetic energy at the bottom of the ramp? (g = 9.8 m/s2 ) 4550 J 6550 J 8150 J 13100 J What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s (90 mph)?: What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s (90 mph)? 54 J 87 J 108 J 216 J A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20° with the horizontal. If there is no friction between ramp surface and crate, what is the kinetic energy of the crate at the bottom of the ramp? (g = 9.8 m/s2 ): A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20° with the horizontal. If there is no friction between ramp surface and crate, what is the kinetic energy of the crate at the bottom of the ramp? (g = 9.8 m/s2 ) 221 J 688 J 10.2 J 100 J A worker pushes a wheelbarrow with a force of 50 N over a level distance of 5 m. If a frictional force of 43 N acts on the wheelbarrow in a direction opposite to that of the worker, what net work is done on the wheelbarrow?: A worker pushes a wheelbarrow with a force of 50 N over a level distance of 5 m. If a frictional force of 43 N acts on the wheelbarrow in a direction opposite to that of the worker, what net work is done on the wheelbarrow? 250 J 215 J 35 J 10 J If during a given physical process the only force acting on an object is friction, which of the following must be assumed in regard to the object's kinetic energy?: If during a given physical process the only force acting on an object is friction, which of the following must be assumed in regard to the object's kinetic energy? decreases increases remains constant cannot tell from the information given As an object is lowered into a deep hole in the surface of the earth, which of the following must be assumed in regard to its potential energy?: As an object is lowered into a deep hole in the surface of the earth, which of the following must be assumed in regard to its potential energy? increase decrease remain constant cannot tell from the information given When an object is dropped from a tower, what is the effect of the air resistance as it falls?: When an object is dropped from a tower, what is the effect of the air resistance as it falls? does positive work increases the object's kinetic energy increases the object's potential energy None of these choices are valid. The unit of work, joule, is dimensionally the same as which of the following?: The unit of work, joule, is dimensionally the same as which of the following? newton/second newton/kilogram newton-second newton-meter The unit of power, watt, is dimensionally the same as which of the following?: The unit of power, watt, is dimensionally the same as which of the following? joule-second joule/second joule-meter joule/meter The rate at which work is done is equivalent to which of the following?: The rate at which work is done is equivalent to which of the following? increase in potential energy thermal energy potential energy power A worker pushes a wheelbarrow with a horizontal force of 50 N over a distance of 5 m. What work does the man do?: A worker pushes a wheelbarrow with a horizontal force of 50 N over a distance of 5 m. What work does the man do? 10 J 1250 J 250 J 55 J A 40 N crate is pulled up a 5 m inclined plane at a constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant force of friction of 10 N between the crate and the surface, what is the net gain in potential energy by the crate in the process?: A 40 N crate is pulled up a 5 m inclined plane at a constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant force of friction of 10 N between the crate and the surface, what is the net gain in potential energy by the crate in the process? 120 J -120 J 200 J -200 J A 40 N crate starting at rest slides down a rough 6 m long ramp, inclined at 30° with the horizontal. The force of friction between crate and ramp is 6 N. What will be the velocity of the crate at the bottom of the incline?: A 40 N crate starting at rest slides down a rough 6 m long ramp, inclined at 30° with the horizontal. The force of friction between crate and ramp is 6 N. What will be the velocity of the crate at the bottom of the incline? 1.6 m/s 3.3 m/s 4.5 m/s 6.4 m/s Which of the following is an example of a non-conservative force?: Which of the following is an example of a non-conservative force? gravity magnetism Both magnetism and gravity friction Which of the following is that form of energy associated with an object's motion?: Which of the following is that form of energy associated with an object's motion? potential thermal bio-chemical kinetic Which of the following is that form of energy associated with an object's location in a conservative force field?: Which of the following is that form of energy associated with an object's location in a conservative force field? potential thermal bio-chemical kinetic A 50 N crate is pulled up a 5 m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant force of 10 N between the crate and the surface, what is the force applied by the worker?: A 50 N crate is pulled up a 5 m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37° to the horizontal and there exists a constant force of 10 N between the crate and the surface, what is the force applied by the worker? zero 20 N 30 N 40 N A small wagon moving at constant velocity carries a 200 N crate 10 m across a level surface. What is the net work done in the process?: A small wagon moving at constant velocity carries a 200 N crate 10 m across a level surface. What is the net work done in the process? zero 1/20 J 20 J 2000 J The quantity of work equal to one joule is also equivalent to which of the following?: The quantity of work equal to one joule is also equivalent to which of the following? watt watt/s watt x s watt/s2 A 3 kg object starting at rest falls to the ground from a height of 10 m. Just before hitting the earth what will be its kinetic energy? (g = 9.8 m/s2 and assume air resistance is negligible.): A 3 kg object starting at rest falls to the ground from a height of 10 m. Just before hitting the earth what will be its kinetic energy? (g = 9.8 m/s2 and assume air resistance is negligible.) 98 J 0.98 J 29.4 J 294 J An arrow, shot straight up from a crossbow has an initial velocity of 50 m/s. Ignore energy lost to air friction. How high will the arrow rise?: An arrow, shot straight up from a crossbow has an initial velocity of 50 m/s. Ignore energy lost to air friction. How high will the arrow rise? 49 m 62.4 m 98 m 127.5 m What is the minimum amount of energy required for an 80-kg climber carrying a 20-kg pack to climb Mt. Everest, 8850 m high?: What is the minimum amount of energy required for an 80-kg climber carrying a 20-kg pack to climb Mt. Everest, 8850 m high? 8.67 mJ 4.16 mJ 2.47 mJ 1.00 mJ A professional skier reaches a speed of 56 m/s on a 30° ski slope. Ignoring friction, what is the minimum distance along the slope the skier would have to travel, starting from rest?: A professional skier reaches a speed of 56 m/s on a 30° ski slope. Ignoring friction, what is the minimum distance along the slope the skier would have to travel, starting from rest? 110 m 160 m 320 m 640 m A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N, what is her speed at the bottom?: A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N, what is her speed at the bottom? 5 m/s 10 m/s 11 m/s She stops before she reaches the bottom. Old Faithful geyser in Yellowstone Park shoots water more or less hourly to a height of 40 meters. With what velocity does the water leave the ground?: Old Faithful geyser in Yellowstone Park shoots water more or less hourly to a height of 40 meters. With what velocity does the water leave the ground? 7.0 m/s 14.0 m/s 19.8 m/s 28.0 m/s An 80,000 kg airliner is flying at 900 kilometers/hour at a height of 10 kilometers. What is its total energy (kinetic + potential)?: An 80,000 kg airliner is flying at 900 kilometers/hour at a height of 10 kilometers. What is its total energy (kinetic + potential)? 250 MJ 478 MJ 773 MJ 10,340 MJ A 1200 kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8 m/s2 . How far does the car travel before it stops?: A 1200 kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8 m/s2 . How far does the car travel before it stops? 39 m 47 m 55 m 63 m A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5 m/s and her speed doubles as she rides down the hill. The hill is 10 m high and 100 m long. How much kinetic energy and potential energy is lost to friction?: A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5 m/s and her speed doubles as she rides down the hill. The hill is 10 m high and 100 m long. How much kinetic energy and potential energy is lost to friction? 2420 J 1500 J 2000 J 3920 J A 60 g golf ball is dropped from a height of 2 m and rebounds to 1.5 m. How much energy is lost?: A 60 g golf ball is dropped from a height of 2 m and rebounds to 1.5 m. How much energy is lost? 0.29 J 0.5 J 0.88 J 1 J A golf ball hits a wall and bounces back at 3/4 the original speed. What part of the original kinetic energy of the ball did it lose in the collision?: A golf ball hits a wall and bounces back at 3/4 the original speed. What part of the original kinetic energy of the ball did it lose in the collision? 1/4 3/8 7/16 It did not lose kinetic energy. If both the mass and the velocity of a ball are tripled, the kinetic energy is increased by a factor of: If both the mass and the velocity of a ball are tripled, the kinetic energy is increased by a factor of 3 6 9 27 A 2 kg ball which originally has zero kinetic and potential energy is dropped into a well which is 10 m deep. Just before the ball hits the bottom, the sum of its kinetic and potential energy is: A 2 kg ball which originally has zero kinetic and potential energy is dropped into a well which is 10 m deep. Just before the ball hits the bottom, the sum of its kinetic and potential energy is 0 196 J -196 J 392 J A 2 kg ball which originally has zero potential and kinetic energy is dropped into a well which is 10 m deep. After the ball comes to a stop in the mud, the sum of its potential and kinetic energy is: A 2 kg ball which originally has zero potential and kinetic energy is dropped into a well which is 10 m deep. After the ball comes to a stop in the mud, the sum of its potential and kinetic energy is 0 196 J -196 J 392 J A hill is 100 m long and makes an angle of 12° with the horizontal. As a 50 kg jogger runs up the hill, how much work does gravity do on the jogger?: A hill is 100 m long and makes an angle of 12° with the horizontal. As a 50 kg jogger runs up the hill, how much work does gravity do on the jogger? 49,000 J 10,000 J -10,000 J zero Two blocks are released from the top of a building. One falls straight down while the other slides down a smooth ramp. If all friction is ignored, which one is moving faster when it reaches the bottom?: Two blocks are released from the top of a building. One falls straight down while the other slides down a smooth ramp. If all friction is ignored, which one is moving faster when it reaches the bottom? The block that went straight down. The block that went down the ramp. They both will have the same speed. Insufficient information to work the problem. A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, to what maximum angle will it move in the second half of its swing? (g = 9.8 m/s2 ): A simple pendulum, 2.0 m in length, is released by a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, to what maximum angle will it move in the second half of its swing? (g = 9.8 m/s2 ) 37° 30° 27° 21° An automobile delivers 30 hp to its wheels when moving at a constant speed of 22 m/s. What is the resistance force on the automobile at this speed? (1 hp = 746 watts): An automobile delivers 30 hp to its wheels when moving at a constant speed of 22 m/s. What is the resistance force on the automobile at this speed? (1 hp = 746 watts) 18,600 N 410,000 N 1,020 N 848 N I use a rope to swing a 10 kg weight around my head. The rope is 2 m long and the tension in the rope is 20 N. In half a revolution how much work is done by the rope on the weight?: I use a rope to swing a 10 kg weight around my head. The rope is 2 m long and the tension in the rope is 20 N. In half a revolution how much work is done by the rope on the weight? 40 J 126 J 251 J 0 The work done by static friction is always: The work done by static friction is always positive negative along the surface zero A simple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2 ): A simple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 25° from the vertical. If the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2 ) 2.3 m/s 2.6 m/s 2.0 m/s 0.5 m/s A horizontal force of 200 N is applied to a 55 kg cart across a 10 m level surface. If the cart accelerates at 2.0 m/s2 then what force of friction acts to retard the motion of the cart?: A horizontal force of 200 N is applied to a 55 kg cart across a 10 m level surface. If the cart accelerates at 2.0 m/s2 then what force of friction acts to retard the motion of the cart? 110 N 90 N 80 N 70 N A baseball catcher puts on an exhibition by catching a 0.15 kg ball dropped from a helicopter at a height of 101 m. What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? (g = 9.8 m/s2 and ignore air resistance): A baseball catcher puts on an exhibition by catching a 0.15 kg ball dropped from a helicopter at a height of 101 m. What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? (g = 9.8 m/s2 and ignore air resistance) 44.3 m/s 38.0 m/s 31.3 m/s 22.2 m/s A baseball catcher puts on an exhibition by catching a 0.15 kg ball dropped from a helicopter at a height of 100 m above the catcher. If the catcher "gives" with the ball for a distance of 0.75 m while catching it, what average force is exerted on the mitt by the ball? (g = 9.8 m/s2 ): A baseball catcher puts on an exhibition by catching a 0.15 kg ball dropped from a helicopter at a height of 100 m above the catcher. If the catcher "gives" with the ball for a distance of 0.75 m while catching it, what average force is exerted on the mitt by the ball? (g = 9.8 m/s2 ) 78 N 119 N 196 N 392 N A horizontal force of 200 N is applied to move a 55 kg cart across a 10 m level surface. What work is done by the 200 N force?: A horizontal force of 200 N is applied to move a 55 kg cart across a 10 m level surface. What work is done by the 200 N force? 4000 J 5400 J 2000 J 2700 J A simple pendulum, 1.0 m in length, is released from rest when the support string is at an angle of 35° from the vertical. What is the speed of the suspended mass at the bottom of the swing? (ignore air resistance, g = 9.8 m/s2 ): A simple pendulum, 1.0 m in length, is released from rest when the support string is at an angle of 35° from the vertical. What is the speed of the suspended mass at the bottom of the swing? (ignore air resistance, g = 9.8 m/s2 ) 0.67 m/s 0.94 m/s 1.33 m/s 1.88 m/s A pile driver drives a post into the ground. The mass of the pile driver is 2500 kg and it is dropped through a height of 8 m on each stroke. If the resisting force of the ground is 4 x 106 N, how far is the post driven in on each stroke?: A pile driver drives a post into the ground. The mass of the pile driver is 2500 kg and it is dropped through a height of 8 m on each stroke. If the resisting force of the ground is 4 x 106 N, how far is the post driven in on each stroke? 4.9 cm 9.8 cm 16.0 cm 49.0 cm A 2000-kg ore car rolls 50 meters down a frictionless 10° incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the ore car in a distance of 1 meter?: A 2000-kg ore car rolls 50 meters down a frictionless 10° incline. If there is a horizontal spring at the end of the incline, what spring constant is required to stop the ore car in a distance of 1 meter? 340.3 kN/m 680.7 kN/m 980.0 kN/m 1960.0 kN/m An amount of work equal to 1 J is required to compress the spring in a spring-gun. To what height can this "gun" shoot a 20-gram marble?: An amount of work equal to 1 J is required to compress the spring in a spring-gun. To what height can this "gun" shoot a 20-gram marble? 5.1 m 10.2 m 15.3 m 20.4 m A 60-kg woman runs up a flight of stairs having a rise of 4 meters in a time of 4.2 seconds. What average power did she supply?: A 60-kg woman runs up a flight of stairs having a rise of 4 meters in a time of 4.2 seconds. What average power did she supply? 380 W 560 W 616 W 672 W A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20° with the horizontal. If there is no friction between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp? (g = 9.8 m/s2 ): A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20° with the horizontal. If there is no friction between the ramp surface and crate, what is the velocity of the crate at the bottom of the ramp? (g = 9.8 m/s2 ) 6.1 m/s 3.7 m/s 10.7 m/s 8.3 m/s
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0 views | 6,873 | 23,785 | {"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-2018-43 | latest | en | 0.909308 |
https://biblebelieverspentecostal.org/and-pdf/904-complementary-supplementary-and-vertical-angles-worksheet-pdf-70-946.php | 1,653,607,481,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662627464.60/warc/CC-MAIN-20220526224902-20220527014902-00331.warc.gz | 181,831,659 | 11,084 | # Complementary Supplementary And Vertical Angles Worksheet Pdf
File Name: complementary supplementary and vertical angles worksheet .zip
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Complementary And Supplementary Angles Worksheet printable. Complementary supplementary vertical adjacent and congruent angles worksheet. Times new roman arial calibri default design adjacent vertical supplementary and complementary angles adjacent angles are side by side and share a common ray.
## Complementary And Supplementary Angles Worksheet Answers
This collection of worksheets focuses your efforts on naming angles and finding measures with those angles based on your understanding of predefined angle relationships. The four angle relationships we focus on look at how two lines relate to each other.
Adjacent angles are what I would refer to as angles that are touching because they share the same ray. Vertical angles are angles that are equal and directly across from each other. Supplementary angles are angles that when added equal degrees. Complementary angle are usually part of a right angle and add up to 90 degrees. These worksheets explain how to adjacent, supplementary, complementary, and vertical angles and how to find the number of degrees in an angle s. This worksheet defines adjacent, supplementary, complementary, and vertical angles.
A sample problem is solved, and two practice problems are provided. Ten pictures of lines meeting at points are provided. Students will classify all the measures that are indicated. This sheet lines all the different types of classifications that students need to be aware of for this entire section. It makes a great review sheet to have handy. Students will study each graph and find the value and classification that each problem asks for.
Six practice problems are provided. Students will demonstrate their proficiency in identifying adjacent, supplementary, complementary, and vertical measures. Ten problems are provided. Students will find and name what is indicated. Three problems are provided, and space is included for students to copy the correct answer when given. This worksheet explains how to name all the pairs of adjacent measures in the picture. A sample problem is solved for you. This worksheets explains how to name the indicated values in a picture.
Ten pictures of measures are provided. Students will find the classifications that are indicated with each picture. Students will identify a series of different measures and label them. Three problems are provided. Students will practice recognizing adjacent, supplementary, complementary, and vertical measures. Ten pictures of adjacent, supplementary, complementary, and vertical measures are provided.
Students will find the measures they are looking for. This worksheet explains how to find the number of degrees in specified angles. Students will read the description and then determine the number of degrees of each measure.
The problems on this sheet will really help you handle and manage word problem in this environment. This worksheet reviews all the concepts that we have learned here. A sample problem is solved and six practice problems are provided. Students will read the description of each angle s and indicate how many degrees each contains. This makes for a nice skill review for students.
What are adjacent, complementary, supplementary, and vertical angles? Supplementary Angles - We define it with respect to the addition of two angles. When two angle's sum is degree, and it forms a linear angle with each other, we will say such type of supplementary angles. Complementary Angles - If you are adding two angles of 90 degrees to form a right angle, you will call them complementary angles.
In a condition of having one angle is x and the other one is 90 degree - x, we will use such complementary angles for trigonometric ratios. Here, trigonometric ratios of the angles are changing as they are complementing each other. Adjacent Angles - When two angles share the common vertex and side, it means it is an adjacent angle. Vertical Angle - If two lines are intersecting and four angles are forming, we will call those angles vertical angles or vertically opposite angles.
Angle Classification Lesson This worksheet defines adjacent, supplementary, complementary, and vertical angles. Practice Worksheet Students will identify the measures that are indicated. Practice Worksheet 2 Ten pictures of lines meeting at points are provided. Skill Review This sheet lines all the different types of classifications that students need to be aware of for this entire section.
Classification Practice Students will study each graph and find the value and classification that each problem asks for. Naming Angles Quiz Students will demonstrate their proficiency in identifying adjacent, supplementary, complementary, and vertical measures.
Skills Review and Check Students will find and name what is indicated. Naming Angles Lesson This worksheet explains how to name all the pairs of adjacent measures in the picture. Lesson and Practice This worksheets explains how to name the indicated values in a picture.
Another Practice Worksheet Students will name the measures all over the place. Naming Practice Ten pictures of measures are provided. Naming Drill Eight pictures are provided. Students will locate what is asked for within each picture.
Class Warm Up Students will identify a series of different measures and label them. Naming Independent Practice Students will study each picture. They will find what is asked of them. Label Me Students will find the angle indicated in each problem. Ten pictures are provided.
Finding Adjacent, Supplementary, Complementary, and Verticals Take your time and find all the asked for measures. This makes for a great take home sheet. Naming Specified Angles Students will practice recognizing adjacent, supplementary, complementary, and vertical measures. Find the Indicated Points Ten pictures of adjacent, supplementary, complementary, and vertical measures are provided. Finding Degrees Lesson This worksheet explains how to find the number of degrees in specified angles.
Finding Degrees Worksheet Students will read the description and then determine the number of degrees of each measure. Degrees Practice The problems on this sheet will really help you handle and manage word problem in this environment. Review and Practice This worksheet reviews all the concepts that we have learned here. Finding Degrees Quiz Students will read the description of each angle s and indicate how many degrees each contains.
Degrees Student Check This makes for a nice skill review for students.
## Pairs of Angles Worksheets
Diagram that we bring with finding complementary and with algebra worksheet given angle? Ratios between the angle and supplementary angles and exterior angles, angles quiz this allows me to make complementary and algebra can be adding the correct nomenclature to continue. Others call for complementary and supplementary with algebra for free access your email, and supplementary and b are supplementary angles that angles that i help of decimals. Gives you more info about complementary supplementary angles algebra practice. Dealing with inscribed angles form a customer avatar gives you may use it also visit our next section is complementary supplementary with algebra worksheet is also use.
## Complementary and Supplementary Angles Worksheets
Here is a graphic preview for all of the Angles Worksheets. You can select different variables to customize these Angles Worksheets for your needs. The Angles Worksheets are randomly created and will never repeat so you have an endless supply of quality Angles Worksheets to use in the classroom or at home.
Printable worksheets and lessons. Complementary and supplementary angles lesson the problems are not visual but we explain the answers visually to help out. Complementary and supplementary angles worksheet answers.
Make use of this array of free printable complementary and supplementary angles worksheets to gauge your knowledge of the concepts. It incorporates exercises like identifying complementary and supplementary angles, finding their measures by applying appropriate properties, recognizing their presence in vertical angles, linear pairs and right angles, and more. This set of printable worksheets are tailor-made for the students of grade 7 and grade 8. How adept are your grade 6 and grade 7 students in finding supplementary angles? Track their progress with this bundle of printable supplementary angles worksheets.
This collection of worksheets focuses your efforts on naming angles and finding measures with those angles based on your understanding of predefined angle relationships. The four angle relationships we focus on look at how two lines relate to each other. Adjacent angles are what I would refer to as angles that are touching because they share the same ray.
#### kuta software complementary and supplementary angles answers
Сьюзан нахмурилась. - Я подумала, что АНБ его ликвидировало. - Вот. Если АНБ в состоянии вывести пять риолитовых спутников на геостационарную орбиту над Ближним Востоком, то, мне кажется, легко предположить, что у нас достаточно средств, чтобы подкупить несколько испанских полицейских. - Его доводы звучали волне убедительно. Сьюзан перевела дыхание. Энсей Танкадо умер.
Но это была чужая епархия. В конце концов ей пришлось смириться. Когда они в ту ночь отправились спать, она старалась радоваться с ним вместе, но что-то в глубине души говорило ей: все это кончится плохо. Она оказалась права, но никогда не подозревала насколько. - Вы заплатили ему десять тысяч долларов? - Она повысила голос.
Сьюзан еще раз прочитала адрес на клочке бумаги и ввела информацию в соответствующее поле, посмеялась про себя, вспомнив о трудностях, с которыми столкнулся Стратмор, пытаясь самолично запустить Следопыта. Скорее всего он проделал это дважды и каждый раз получал адрес Танкадо, а не Северной Дакоты. Элементарная ошибка, подумала Сьюзан, Стратмор, по-видимому, поменял местами поля информации, и Следопыт искал учетные данные совсем не того пользователя. Она завершила ввод данных и запустила Следопыта. Затем щелкнула по кнопке возврат. Компьютер однократно пискнул.
- Они не преступницы - глупо было бы искать их, как обычных жуликов. Беккер все еще не мог прийти в себя от всего, что услышал.
Тем не менее риск велик: если нас обнаружат, это, в сущности, будет означать, что он своим алгоритмом нас напугал. Нам придется публично признать не только то, что мы имеем ТРАНСТЕКСТ, но и то, что Цифровая крепость неприступна. - Каким временем мы располагаем.
Но, сэр… - заикаясь выдавила. - Я… я протестую. Я думаю… - Вы протестуете? - переспросил директор и поставил на стол чашечку с кофе.
Их количество удваивалось каждую минуту. Еще немного, и любой обладатель компьютера - иностранные шпионы, радикалы, террористы - получит доступ в хранилище секретной информации американского правительства. Пока техники тщетно старались отключить электропитание, собравшиеся на подиуме пытались понять расшифрованный текст.
Лейтенант следил за его взглядом.
Они сцепились. Перила были невысокими. Как это странно, подумал Стратмор, что насчет вируса Чатрукьян был прав с самого начала. Его падение пронзило Стратмора холодным ужасом - отчаянный крик и потом тишина.
Пилот сказал вполне определенно: У меня приказ оставаться здесь до вашего возвращения. Трудно даже поверить, подумал Беккер, что после всех выпавших на его долю злоключений он вернулся туда, откуда начал поиски. Чего же он ждет.
Так продолжалось несколько недель. За десертом в ночных ресторанах он задавал ей бесконечные вопросы. Где она изучала математику.
Вы всегда добиваетесь своего… вы добьетесь… Да, - подумал. - Я добиваюсь своих целей, но честь для меня важнее.
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1 Response
1. Pusaki T.
Plenty of practice awaits your 7th grade and 8th grade students in these printable pairs of angles worksheets that bring together every exercise you need to assist them in getting their head around the different types of angle pairs and the properties associated with each. | 2,605 | 12,186 | {"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} | 3.875 | 4 | CC-MAIN-2022-21 | latest | en | 0.934141 |
https://math.stackexchange.com/questions/2481119/example-of-a-power-of-3-which-is-close-to-a-power-of-2 | 1,561,052,584,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999263.6/warc/CC-MAIN-20190620165805-20190620191805-00301.warc.gz | 504,378,282 | 40,219 | # Example of a power of 3 which is close to a power of 2
I'm looking for a power of 3 close to a power of 2.
Let's say, what is $(n,m)$ such that $$\left|\frac{2^n}{3^m}-1\right| = \min\left \{\left|\frac{2^i}{3^j}-1 \right|, 1\leq i,j\leq 20\right\} \quad ?$$
### Why?
The idea is to understand the intervals between musical notes. Any link on this subject is welcomed (I don't know music theory but I'm interested).
• $\dfrac{2^1}{3^{20}}$ sounds good? (no pun intended :) But if you want a power of $3$ which is close to a power of $2$ maybe you wanted $$\min\left \{|3^j-2^i|, 1\leq i,j\leq 20\right\}$$ – Raffaele Oct 20 '17 at 7:48
• Thanks, I will edit!!! – Colas Oct 20 '17 at 7:51
• @Raffaele I just understood your good (intended!) pun. Another way to see the issue is to say that's about two superpowers that should be equilibrated. – Jean Marie Oct 20 '17 at 10:37
You want $2^i \approx 3^j$ with $i,j \in \mathbb N$
(preferably $1 \leq i,j \leq 20$).
Take the logarithm (in any base) of both sides: $$i \log(2) \approx j \log(3).$$
Thus the issue is now to obtain a good rational approximation
$$\underbrace{\dfrac{\log(3)}{\log(2)}}_{N}\approx \dfrac{i}{j}$$
Such rational approximations are obtained as the so-called "convergents" of the continued fraction expansion of number $N$ (https://en.wikipedia.org/wiki/Continued_fraction):
$$N=a_0+\tfrac{1}{a_1+\tfrac{1}{{a_2+\tfrac{1}{\cdots}}}}$$
with $(a_0; a_1, a_3, \cdots )=(1;1,1,2,2,3,1,5,2,23,2,2,1,55,1,...).$
with successive convergents:
$$i/j = 3/2, \ 5/3, \ 8/5, \ 11/7, \ 19/12, \ 46/29, \ 65/41, \ 84/53, ...$$
$$317/200,\ 401/253, \ 485/306, \ 569/359, \ 1054/665...$$
(that can be found in (https://oeis.org/A254351/internal)).
Good convergents are obtained by (in plain terms) "stopping just before a big $a_k$".
If the "big $a_k$" (!) is 3, one gets :
$$\dfrac{19}{12} \approx 1.5833 \ \ \ \text{of} \ \ \ N = 1.5850...$$
Thus a good tradeoff is to take $i=19$ and $j=12$.
This is connected to the "Pythagorician comma". (https://en.wikipedia.org/wiki/Pythagorean_comma).
Very related : (http://www.math.uwaterloo.ca/~mrubinst/tuning/12.html)
• I know the definition of continued fractions, but I never thought it had a link to rational approximation. – Colas Oct 20 '17 at 8:00
• That makes senses since there are 12 musical notes. – Colas Oct 20 '17 at 8:15
• see as well (math.stackexchange.com/q/11669) – Jean Marie Oct 20 '17 at 8:50
• Related (andrewduncan.net/cmt) – Jean Marie Oct 20 '17 at 10:49
• Each term in the continued fraction ($a_n$) corresponds to a convergent. You are giving the "semiconvergents" when you include $\frac53$ and $\frac{11}7$. For their denominators, these are not as close to the number being approximated as the convergents. – robjohn Oct 20 '17 at 14:11
Lemma: Let $\alpha, \beta$ be linearly independent over rationals $\mathbb{Q}$; i.e. $\dfrac{\alpha}{\beta} \notin \mathbb{Q}$.
Then the group generated by $\alpha, \beta$ is dense in $\mathbb{Q}$;
i.e. $\{ i\alpha+j\beta \ | \ i,j \in \mathbb{Z} \} = \{ i\alpha-j\beta \ | \ i,j \in \mathbb{Z} \}$ is dense in $\mathbb{Q}$.
Now let $\alpha:=\log2$ and $\beta:=\log3$, note that $\dfrac{\beta}{\alpha}=\dfrac{\log3}{\log2}=\log_23 \notin \mathbb{Q}$;
so if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $i\alpha-j\beta$ will become as small as we want;
i.e. in more mathematical discipline we can say:
for every $0 < \delta \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $0 < i\alpha-j\beta < \delta$.
Note that: $$10^{i\alpha-j\beta}= \dfrac{10^{i\alpha}}{10^{j\beta}}= \dfrac{(10^{\alpha})^{i}}{(10^{\beta})^{j}}= \dfrac{2^i}{3^j}$$
So we can conclude that:
for every $0 < \epsilon \in \mathbb{R}$ there exist $i, j \in \mathbb{N}$ such that $1 < \dfrac{2^i}{3^j} < 1+ \epsilon$.
So ignoring some mathematical disciplines we can say:
if we let $i, j$ varies arbitrary in $\mathbb{N}$,
then the phrase $\dfrac{2^i}{3^j}$ will become as close to $1$ as we want.
We can restase some similar facts about $-\delta < i\alpha-j\beta < 0$ and $-\delta < i\alpha-j\beta < \delta$;
also for $1- \epsilon < \dfrac{2^i}{3^j} < 1$ and $1- \epsilon < \dfrac{2^i}{3^j} < 1+ \epsilon$; but it does not matters.
I have seen this problem before:
unsolved problems in number theory by Richard K. Guy;
$F$(None of the above);
$F23$(Small differences between powers of $2$ and $3$.) :
Problem $1$ of Littlewood's book asks how small $3^n — 2^m$ can be in comparison with $2^m$.
He gives as an example $$\dfrac{3^{12}} {2^{19}} = 1+ \dfrac{ 7153 } {524288} \approx 1+ \dfrac{ 1 } { 73 } \qquad \text{ (the ratio of } D^{\sharp} \text{ to } E^{\flat}).$$ [Where $\sharp$ and $\flat$ stands for dièse and Bémol.]
The first few convergents to the continued fraction (see $F20$) $$log_23= 1+\tfrac{1} {1+\tfrac{1} {1+\tfrac{1} {2+\tfrac{1} {2+\tfrac{1} {3+\tfrac{1} {1+\tfrac{1} {\cdots} } } } } } }$$
for $\log 3$ to the base $2$ are: $$\dfrac{ 1}{ 1}, \dfrac{ 2}{ 1}, \dfrac{ 3}{ 2}, \dfrac{ 8}{ 5}, \dfrac{19}{12}, \dfrac{65}{41}, \dfrac{84}{53}, \dots$$ so Victor Meally observed that
the $\color{Blue}{\text{octave}}$ may conveniently be partitioned into $12$, $41$ or $53$ intervals, and that the system of temperament with $53$ degrees is due to Nicolaus Mercator ($1620-1687$; not Gerhardus, $1512-1594$, of map projection fame).
Ellison used the Gel'fond-Baker method to show that
$$|2^x-3^y| > 2^x e^{\dfrac{-x}{10}} \qquad \text{for} \qquad x > 27$$
and Tijdeman used it to show that there is a $c \geq 1$ such that $2^x - 3^y > \dfrac{2^x}{x^c}$.
Croft asks the corresponding question for $n! — 2^m$. $\ \ \cdots$
That was the exact text from the book, note that the pair $(m,n)$ has been used in the reversed order in that book versus this question has been asked in this site!
At the end of section $F23$; there are four refrences as follows:
• F. Beukers, Fractional parts of powers of rationals, Math. Proc. Cambridge Philos. Soc., $90(1981) 13-20$; MR $83$g:$10028$.
• A. K. Dubitskas, A lower bound on the value of $||(3/2)^k||$ (Russian), Uspekhi Mat. Nauk, $45(1990) 153-154$; translated in Russian Math. Surveys, $45(1990) 163-164$; MR $91$k:$11058$.
• W. J. Ellison, Recipes for solving diophantine problems by Baker's method, S6m. Thiorie Nombms, $1970-71$, Exp. No. $11$, C.N.R.S. Talence, $1971$.
• R. Tijdeman, On integers with many small factors, Compositio Math., $26 (1973) 319-330$.
• I have found an article by Kurt Mahler with the same name as the first one here.
Also in page $14$ of this paper of Michel Waldschmidt; you can find something.
But I don't know how this problem arise from (music);
and from where the upper bound 20 comes from.
The continued fraction for $\frac{\log(3)}{\log(2)}$ is $$1;1,1,2,2,3,1,5,2,23,2,2,1,1,55,1,4,3,1,1,\dots$$ The convergents are $$\frac11,\frac21,\frac32,\frac85,\frac{19}{12},\frac{65}{41},\frac{84}{53},\frac{485}{306},\frac{1054}{665},\frac{24727}{15601},\frac{50508}{31867},\frac{125743}{79335},\frac{176251}{111202},\frac{301994}{190537},\frac{16785921}{10590737},\frac{17087915}{10781274},\frac{85137581}{53715833}, \frac{272500658}{171928773},\frac{357638239}{225644606}, \frac{630138897}{397573379},\cdots$$ This gives a number of "close" approximations. For example, the best with exponents between $1$ and $20$, is $$\left|\frac{2^{19}}{3^{12}}-1\right|=0.01345963145485576$$ This gives the approximation $2^{19/12}=2.9966$. Since their are $12$ half-steps to the octave, this approximation is very useful.
• A certain number of convergents are missing in your list : $5/3$, $11/7, 46/29...$ – Jean Marie Oct 20 '17 at 10:10
• @JeanMarie: are you talking about convergents, or semiconvergents, which are not as close as convergents for their denominators? – robjohn Oct 20 '17 at 14:03 | 2,852 | 7,797 | {"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} | 3.90625 | 4 | CC-MAIN-2019-26 | longest | en | 0.835181 |
https://docs.mosek.com/10.0/pythonapi/tutorial-acc-optimizer.html | 1,680,195,987,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00712.warc.gz | 247,291,171 | 11,238 | 6.2 From Linear to Conic Optimization¶
In Sec. 6.1 (Linear Optimization) we demonstrated setting up the linear part of an optimization problem, that is the objective, linear bounds, linear equalities and inequalities. In this tutorial we show how to define conic constraints. We recommend going through this general conic tutorial before proceeding to examples with specific cone types.
MOSEK accepts conic constraints in the form
$Fx+g\in \D$
where
• $$x\in\real^n$$ is the optimization variable,
• $$D\subseteq \real^k$$ is a conic domain of some dimension $$k$$, representing one of the cone types supported by MOSEK,
• $$F\in\real^{k\times n}$$ and $$g\in \real^k$$ are data which constitute the sequence of $$k$$ affine expressions appearing in the rows of $$Fx+g$$.
Constraints of this form will be called affine conic constraints, or ACC for short. Therefore in this section we show how to set up a problem of the form
$\begin{split}\begin{array}{lccccl} \mbox{minimize} & & & c^T x + c^f & & \\ \mbox{subject to} & l^c & \leq & A x & \leq & u^c, \\ & l^x & \leq & x & \leq & u^x, \\ & & & Fx+g & \in & \D_1\times\cdots\times \D_p, \end{array}\end{split}$
with some number $$p$$ of affine conic constraints.
Note that conic constraints are a natural generalization of linear constraints to the general nonlinear case. For example, a typical linear constraint of the form
$Ax+b\geq 0$
can be also written as membership in the cone of nonnegative real numbers:
$Ax+b \in \real_{\geq 0}^d,$
and that naturally generalizes to
$Fx+g\in \D$
for more complicated domains $$\D$$ from Sec. 15.11 (Supported domains) of which $$\D=\real_{\geq 0}^d$$ is a special case.
6.2.1 Running example¶
In this tutorial we will consider a sample problem of the form
(6.2)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & \gamma \geq \| Gx+h \|_2, \end{array}\end{split}$
where $$x\in \real^n$$ is the optimization variable and $$G\in\real^{k\times n}$$, $$h\in\real^k$$, $$c\in\real^n$$ and $$\gamma\in\real$$. We will use the following sample data:
$\begin{split}n=3,\quad k=2,\quad x\in \real^3, \quad c = [2, 3, -1]^T,\quad \gamma=0.03,\quad G = \left[\begin{array}{ccc}1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad h = \left[\begin{array}{c}0 \\ 0.1\end{array}\right].\end{split}$
To be explicit, the problem we are going to solve is therefore:
(6.3)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & 0.03 \geq \sqrt{(1.5x_0+0.1x_1)^2+(0.3x_0+2.1x_2+0.1)^2}. \end{array}\end{split}$
Consulting the definition of a quadratic cone $$\Q$$ we see that the conic form of this problem is:
(6.4)$\begin{split}\begin{array}{ll} \mbox{maximize} & 2x_0+3x_1-x_2 \\ \mbox{subject to} & x_0+x_1+x_2 = 1, \\ & (0.03,\ 1.5x_0+0.1x_1,\ 0.3x_0+2.1x_2+0.1) \in \Q^3. \end{array}\end{split}$
The conic constraint has an affine conic representation $$Fx+g\in\D$$ as follows:
(6.5)$\begin{split}\left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right]x + \left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right] \in \Q^3.\end{split}$
Of course by the same logic in the general case the conic form of the problem (6.2) would be
(6.6)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & \sum_i x_i = 1, \\ & (\gamma, Gx+h)\in\Q^{k+1} \end{array}\end{split}$
and the ACC representation of the constraint $$(\gamma, Gx+h)\in\Q^{k+1}$$ would be
$\begin{split}\left[\begin{array}{c}0\\ G\end{array}\right]x + \left[\begin{array}{c}\gamma\\ h\end{array}\right] \in \Q^{k+1}.\end{split}$
Now we show how to add the ACC (6.5). This involves three steps:
• storing the affine expressions which appear in the constraint,
• creating a domain, and
• combining the two into an ACC.
6.2.2 Step 1: add affine expressions¶
To store affine expressions (AFE for short) MOSEK provides a matrix $$\afef$$ and a vector $$\afeg$$ with the understanding that every row of
$\afef x + \afeg$
defines one affine expression. The API functions with infix afe are used to operate on $$\afef$$ and $$\afeg$$, add rows, add columns, set individual elements, set blocks etc. similarly to the methods for operating on the $$A$$ matrix of linear constraints. The storage matrix $$\afef$$ is a sparse matrix, therefore only nonzero elements have to be explicitly added.
Remark: the storage $$\afef,\afeg$$ may, but does not have to be, equal to the pair $$F,g$$ appearing in the expression $$Fx+g$$. It is possible to store the AFEs in different order than the order they will be used in $$F,g$$, as well as store some expressions only once if they appear multiple times in $$Fx+g$$. In this first turorial, however, we will for simplicity store all expressions in the same order we will later use them, so that $$(\afef,\afeg)=(F,g)$$.
In our example we create only one conic constraint (6.5) with three (in general $$k+1$$) affine expressions
$\begin{split}\begin{array}{l} 0.03, \\ 1.5x_0+0.1x_1,\\ 0.3x_0 +2.1x_2 +0.1. \end{array}\end{split}$
Given the previous remark, we initialize the AFE storage as:
(6.7)$\begin{split}\afef = \left[\begin{array}{ccc}0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}0.03\\ 0\\ 0.1\end{array}\right].\end{split}$
Initially $$\afef$$ and $$\afeg$$ are empty (have 0 rows). We construct them as follows. First, we append a number of empty rows:
# Append empty AFE rows for affine expression storage
We now have $$\afef$$ and $$\afeg$$ with 3 rows of zeros and we fill them up to obtain (6.7).
# G matrix in sparse form
Gsubi = [0, 0, 1, 1]
Gsubj = [0, 1, 0, 2]
Gval = [1.5, 0.1, 0.3, 2.1]
# Other data
h = [0, 0.1]
gamma = 0.03
# Construct F matrix in sparse form
Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F
Fsubj = Gsubj
Fval = Gval
# Fill in F storage
# Fill in g storage
We have now created the matrices from (6.7). Note that at this point we have not defined any ACC yet. All we did was define some affine expressions and place them in a generic AFE storage facility to be used later.
6.2.3 Step 2: create a domain¶
Next, we create the domain to which the ACC belongs. Domains are created with functions with infix domain. In the case of (6.5) we need a quadratic cone domain of dimension 3 (in general $$k+1$$), which we create with:
# Define a conic quadratic domain
The function returns a domain index, which is just the position in the list of all domains (potentially) created for the problem. At this point the domain is just stored in the list of domains, but not yet used for anything.
6.2.4 Step 3: create the actual constraint¶
We are now in position to create the affine conic constraint. ACCs are created with functions with infix acc. The most basic variant, Task.appendacc will append an affine conic constraint based on the following data:
• the list afeidx of indices of AFEs to be used in the constraint. These are the row numbers in $$\afef,\afeg$$ which contain the required affine expressions.
• the index domidx of the domain to which the constraint belongs.
Note that number of AFEs used in afeidx must match the dimension of the domain.
In case of (6.5) we have already arranged $$\afef,\afeg$$ in such a way that their (only) three rows contain the three affine expressions we need (in the correct order), and we already defined the quadratic cone domain of matching dimension 3. The ACC is now constructed with the following call:
# Create the ACC
range(k+1), # Indices of AFE rows [0,...,k]
None) # Ignored
This completes the setup of the affine conic constraint.
6.2.5 Example ACC1¶
We refer to Sec. 6.1 (Linear Optimization) for instructions how to set up the objective and linear constraint $$x_0+x_1+x_2=1$$. All else that remains is to set up the MOSEK environment, task, add variables, call the solver with Task.optimize and retrieve the solution with Task.getxx. Since our problem contains a nonlinear constraint we fetch the interior-point solution. The full code solving problem (6.3) is shown below.
Listing 6.2 Full code of example ACC1. Click here to download.
import sys, mosek
# Define a stream printer to grab output from MOSEK
def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()
# Define problem data
n, k = 3, 2
# Only a symbolic constant
inf = 0.0
# Make a MOSEK environment
with mosek.Env() as env:
# Attach a printer to the environment
env.set_Stream(mosek.streamtype.log, streamprinter)
# Attach a printer to the task
# Create n free variables
# Set up the objective
c = [2, 3, -1]
# One linear constraint - sum(x) = 1
# Append empty AFE rows for affine expression storage
# G matrix in sparse form
Gsubi = [0, 0, 1, 1]
Gsubj = [0, 1, 0, 2]
Gval = [1.5, 0.1, 0.3, 2.1]
# Other data
h = [0, 0.1]
gamma = 0.03
# Construct F matrix in sparse form
Fsubi = [i + 1 for i in Gsubi] # G will be placed from row number 1 in F
Fsubj = Gsubj
Fval = Gval
# Fill in F storage
# Fill in g storage
# Define a conic quadratic domain
# Create the ACC
range(k+1), # Indices of AFE rows [0,...,k]
None) # Ignored
# Solve and retrieve solution
print("Solution: {xx}".format(xx=list(xx)))
[-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]
The dual values $$\doty$$ of an ACC can be obtained with Task.getaccdoty if required.
# Demonstrate retrieving the dual of ACC
0) # ACC index
print("Dual of ACC:: {doty}".format(doty=list(doty)))
6.2.6 Example ACC2 - more conic constraints¶
Now that we know how to enter one affine conic constraint (ACC) we will demonstrate a problem with two ACCs. From there it should be clear how to add multiple ACCs. To keep things familiar we will reuse the previous problem, but this time cast it into a conic optimization problem with two ACCs as follows:
(6.8)$\begin{split}\begin{array}{ll} \mbox{maximize} & c^T x \\ \mbox{subject to} & (\sum_i x_i - 1,\ \gamma,\ Gx+h) \in \{0\} \times \Q^{k+1} \end{array}\end{split}$
or, using the data from the example:
$\begin{split}\begin{array}{lll} \mbox{maximize} & 2x_0+3x_1-x_2 & \\ \mbox{subject to} & x_0+x_1+x_2 - 1 & \in \{0\}, \\ & (0.03, 1.5x_0+0.1x_1, 0.3x_0+2.1x_2+0.1) & \in \Q^3 \end{array}\end{split}$
In other words, we transformed the linear constraint into an ACC with the one-point zero domain.
As before, we proceed in three steps. First, we add the variables and create the storage $$\afef$$, $$\afeg$$ containing all affine expressions that appear throughout all off the ACCs. It means we will require 4 rows:
(6.9)$\begin{split}\afef = \left[\begin{array}{ccc}1& 1 & 1 \\ 0 & 0 & 0\\ 1.5 & 0.1 & 0\\0.3 & 0 & 2.1\end{array}\right],\quad \afeg=\left[\begin{array}{c}-1 \\0.03\\ 0\\ 0.1\end{array}\right].\end{split}$
# Set AFE rows representing the linear constraint
# Set AFE rows representing the quadratic constraint
[0, 1], # varidx, column numbers
[1.5, 0.1]) # values
[0, 2], # varidx, column numbers
[0.3, 2.1]) # values
h = [0, 0.1]
gamma = 0.03
Next, we add the required domains: the zero domain of dimension 1, and the quadratic cone domain of dimension 3.
# Define domains
Finally, we create both ACCs. The first ACCs picks the 0-th row of $$\afef,\afeg$$ and places it in the zero domain:
task.appendacc(zeroDom, # Domain index
[0], # Indices of AFE rows
None) # Ignored
The second ACC picks rows $$1,2,3$$ in $$\afef,\afeg$$ and places them in the quadratic cone domain:
task.appendacc(quadDom, # Domain index
[1,2,3], # Indices of AFE rows
None) # Ignored
The completes the construction and we can solve the problem like before:
Listing 6.3 Full code of example ACC2. Click here to download.
import sys, mosek
# Define a stream printer to grab output from MOSEK
def streamprinter(text):
sys.stdout.write(text)
sys.stdout.flush()
# Define problem data
n, k = 3, 2
# Only a symbolic constant
inf = 0.0
# Make a MOSEK environment
with mosek.Env() as env:
# Attach a printer to the environment
env.set_Stream(mosek.streamtype.log, streamprinter)
# Attach a printer to the task
# Create n free variables
# Set up the objective
c = [2, 3, -1]
# Set AFE rows representing the linear constraint
# Set AFE rows representing the quadratic constraint
[0, 1], # varidx, column numbers
[1.5, 0.1]) # values
[0, 2], # varidx, column numbers
[0.3, 2.1]) # values
h = [0, 0.1]
gamma = 0.03
# Define domains
# Append affine conic constraints
[0], # Indices of AFE rows
None) # Ignored
[1,2,3], # Indices of AFE rows
None) # Ignored
# Solve and retrieve solution
print("Solution: {xx}".format(xx=list(xx)))
We obtain the same result:
[-0.07838011145615721, 1.1289128998004547, -0.0505327883442975]
6.2.7 Summary and extensions¶
In this section we presented the most basic usage of the affine expression storage $$\afef,\afeg$$ to input affine expressions used together with domains to create affine conic constraints. Now we briefly point out additional features of his interface which can be useful in some situations for more demanding users. They will be demonstrated in various examples in other tutorials and case studies in this manual.
• It is important to remember that $$\afef,\afeg$$ has only a storage function and during the ACC construction we can pick an arbitrary list of row indices and place them in a conic domain. It means for example that:
• It is not necessary to store the AFEs in the same order they will appear in ACCs.
• The same AFE index can appear more than once in one and/or more conic constraints (this can be used to reduce storage if the same affine expression is used in multiple ACCs).
• The $$\afef,\afeg$$ storage can even include rows that are not presently used in any ACC.
• Domains can be reused: multiple ACCs can use the same domain. On the other hand the same type of domain can appear under many domidx positions. In this sense the list of created domains also plays only a storage role: the domains are only used when they enter an ACC.
• Affine expressions can also contain semidefinite terms, ie. the most general form of an ACC is in fact
$Fx + \langle \bar{F},\barX\rangle + g \in \D$
These terms are input into the rows of AFE storage using the functions with infix afebarf, creating an additional storage structure $$\bar{\afef}$$.
• The same affine expression storage $$\afef,\afeg$$ is shared between affine conic and disjunctive constraints (see Sec. 6.9 (Disjunctive constraints)).
• If, on the other hand, the user chooses to always store the AFEs one by one sequentially in the same order as they appear in ACCs then sequential functions such as Task.appendaccseq and Task.appendaccsseq make it easy to input one or more ACCs by just specifying the starting AFE index and dimension.
• It is possible to add a number of ACCs in one go using Task.appendaccs.
• When defining an ACC an additional constant vector $$b$$ can be provided to modify the constant terms coming from $$\afeg$$ but only for this particular ACC. This could be useful to reduce $$\afef$$ storage space if, for example, many expressions $$f^Tx+b_i$$ with the same linear part $$f^Tx$$, but varying constant terms $$b_i$$, are to be used throughout ACCs. | 4,733 | 15,370 | {"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} | 3.53125 | 4 | CC-MAIN-2023-14 | latest | en | 0.764625 |
https://tutorialspoint.dev/data-structure/matrix-archives/a-boolean-matrix-question | 1,618,421,226,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077843.17/warc/CC-MAIN-20210414155517-20210414185517-00173.warc.gz | 669,080,141 | 17,763 | # A Boolean Matrix Question
Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1.
```Example 1
The matrix
1 0
0 0
should be changed to following
1 1
1 0
Example 2
The matrix
0 0 0
0 0 1
should be changed to following
0 0 1
1 1 1
Example 3
The matrix
1 0 0 1
0 0 1 0
0 0 0 0
should be changed to following
1 1 1 1
1 1 1 1
1 0 1 1
```
Method 1 (Use two temporary arrays)
1) Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] as 0.
2) Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true.
3) Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.
Thanks to Dixit Sethi for suggesting this method.
## C++
`// C++ Code For A Boolean Matrix Question ` `#include ` ` ` `using` `namespace` `std; ` `#define R 3 ` `#define C 4 ` ` ` `void` `modifyMatrix(``bool` `mat[R][C]) ` `{ ` ` ``bool` `row[R]; ` ` ``bool` `col[C]; ` ` ` ` ``int` `i, j; ` ` ` ` ``/* Initialize all values of row[] as 0 */` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``row[i] = 0; ` ` ``} ` ` ` ` ``/* Initialize all values of col[] as 0 */` ` ``for` `(i = 0; i < C; i++) ` ` ``{ ` ` ``col[i] = 0; ` ` ``} ` ` ` ` ``// Store the rows and columns to be marked as ` ` ``// 1 in row[] and col[] arrays respectively ` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``if` `(mat[i][j] == 1) ` ` ``{ ` ` ``row[i] = 1; ` ` ``col[j] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// Modify the input matrix mat[] using the ` ` ``// above constructed row[] and col[] arrays ` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``if` `( row[i] == 1 || col[j] == 1 ) ` ` ``{ ` ` ``mat[i][j] = 1; ` ` ``} ` ` ``} ` ` ``} ` `} ` ` ` `/* A utility function to print a 2D matrix */` `void` `printMatrix(``bool` `mat[R][C]) ` `{ ` ` ``int` `i, j; ` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``cout << mat[i][j]; ` ` ``} ` ` ``cout << endl; ` ` ``} ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ``bool` `mat[R][C] = { {1, 0, 0, 1}, ` ` ``{0, 0, 1, 0}, ` ` ``{0, 0, 0, 0}}; ` ` ` ` ``cout << ````"Input Matrix "````; ` ` ``printMatrix(mat); ` ` ` ` ``modifyMatrix(mat); ` ` ` ` ``printf``(````"Matrix after modification "````); ` ` ``printMatrix(mat); ` ` ` ` ``return` `0; ` `} ` ` ` `// This code is contributed ` `// by Akanksha Rai(Abby_akku) `
## Java
`// Java Code For A Boolean Matrix Question ` `class` `GFG ` `{ ` ` ``public` `static` `void` `modifyMatrix(``int` `mat[ ][ ], ``int` `R, ``int` `C) ` ` ``{ ` ` ``int` `row[ ]= ``new` `int` `[R]; ` ` ``int` `col[ ]= ``new` `int` `[C]; ` ` ``int` `i, j; ` ` ` ` ``/* Initialize all values of row[] as 0 */` ` ``for` `(i = ``0``; i < R; i++) ` ` ``{ ` ` ``row[i] = ``0``; ` ` ``} ` ` ` ` ` ` ``/* Initialize all values of col[] as 0 */` ` ``for` `(i = ``0``; i < C; i++) ` ` ``{ ` ` ``col[i] = ``0``; ` ` ``} ` ` ` ` ` ` ``/* Store the rows and columns to be marked as ` ` ``1 in row[] and col[] arrays respectively */` ` ``for` `(i = ``0``; i < R; i++) ` ` ``{ ` ` ``for` `(j = ``0``; j < C; j++) ` ` ``{ ` ` ``if` `(mat[i][j] == ``1``) ` ` ``{ ` ` ``row[i] = ``1``; ` ` ``col[j] = ``1``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* Modify the input matrix mat[] using the ` ` ``above constructed row[] and col[] arrays */` ` ``for` `(i = ``0``; i < R; i++) ` ` ``{ ` ` ``for` `(j = ``0``; j < C; j++) ` ` ``{ ` ` ``if` `( row[i] == ``1` `|| col[j] == ``1` `) ` ` ``{ ` ` ``mat[i][j] = ``1``; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* A utility function to print a 2D matrix */` ` ``public` `static` `void` `printMatrix(``int` `mat[ ][ ], ``int` `R, ``int` `C) ` ` ``{ ` ` ``int` `i, j; ` ` ``for` `(i = ``0``; i < R; i++) ` ` ``{ ` ` ``for` `(j = ``0``; j < C; j++) ` ` ``{ ` ` ``System.out.print(mat[i][j]+ ``" "``); ` ` ``} ` ` ``System.out.println(); ` ` ``} ` ` ``} ` ` ` ` ``/* Driver program to test above functions */` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `mat[ ][ ] = { {``1``, ``0``, ``0``, ``1``}, ` ` ``{``0``, ``0``, ``1``, ``0``}, ` ` ``{``0``, ``0``, ``0``, ``0``},}; ` ` ` ` ``System.out.println(``"Matrix Intially"``); ` ` ` ` ``printMatrix(mat, ``3``, ``4``); ` ` ` ` ``modifyMatrix(mat, ``3``, ``4``); ` ` ``System.out.println(``"Matrix after modification n"``); ` ` ``printMatrix(mat, ``3``, ``4``); ` ` ` ` ``} ` ` ` `} ` ` ` `// This code is contributed by Kamal Rawal `
## Python3
`# Python3 Code For A Boolean Matrix Question ` `R ``=` `3` `C ``=` `4` ` ` `def` `modifyMatrix(mat): ` ` ``row ``=` `[``0``] ``*` `R ` ` ``col ``=` `[``0``] ``*` `C ` ` ` ` ``# Initialize all values of row[] as 0 ` ` ``for` `i ``in` `range``(``0``, R): ` ` ``row[i] ``=` `0` ` ` ` ``# Initialize all values of col[] as 0 ` ` ``for` `i ``in` `range``(``0``, C) : ` ` ``col[i] ``=` `0` ` ` ` ` ` ``# Store the rows and columns to be marked ` ` ``# as 1 in row[] and col[] arrays respectively ` ` ``for` `i ``in` `range``(``0``, R) : ` ` ` ` ``for` `j ``in` `range``(``0``, C) : ` ` ``if` `(mat[i][j] ``=``=` `1``) : ` ` ``row[i] ``=` `1` ` ``col[j] ``=` `1` ` ` ` ``# Modify the input matrix mat[] using the ` ` ``# above constructed row[] and col[] arrays ` ` ``for` `i ``in` `range``(``0``, R) : ` ` ` ` ``for` `j ``in` `range``(``0``, C): ` ` ``if` `( row[i] ``=``=` `1` `or` `col[j] ``=``=` `1` `) : ` ` ``mat[i][j] ``=` `1` ` ` `# A utility function to print a 2D matrix ` `def` `printMatrix(mat) : ` ` ``for` `i ``in` `range``(``0``, R): ` ` ` ` ``for` `j ``in` `range``(``0``, C) : ` ` ``print``(mat[i][j], end ``=` `" "``) ` ` ``print``() ` ` ` `# Driver Code ` `mat ``=` `[ [``1``, ``0``, ``0``, ``1``], ` ` ``[``0``, ``0``, ``1``, ``0``], ` ` ``[``0``, ``0``, ``0``, ``0``] ] ` ` ` `print``(``"Input Matrix n"``) ` `printMatrix(mat) ` ` ` `modifyMatrix(mat) ` ` ` `print``(``"Matrix after modification n"``) ` `printMatrix(mat) ` ` ` `# This code is contributed by Nikita Tiwari. `
## C#
`// C# Code For A Boolean ` `// Matrix Question ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ``public` `static` `void` `modifyMatrix(``int` `[,]mat, ` ` ``int` `R, ``int` `C) ` ` ``{ ` ` ``int` `[]row = ``new` `int` `[R]; ` ` ``int` `[]col = ``new` `int` `[C]; ` ` ``int` `i, j; ` ` ` ` ``/* Initialize all values ` ` ``of row[] as 0 */` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``row[i] = 0; ` ` ``} ` ` ` ` ` ` ``/* Initialize all values ` ` ``of col[] as 0 */` ` ``for` `(i = 0; i < C; i++) ` ` ``{ ` ` ``col[i] = 0; ` ` ``} ` ` ` ` ` ` ``/* Store the rows and columns ` ` ``to be marked as 1 in row[] ` ` ``and col[] arrays respectively */` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``if` `(mat[i, j] == 1) ` ` ``{ ` ` ``row[i] = 1; ` ` ``col[j] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* Modify the input matrix ` ` ``mat[] using the above ` ` ``constructed row[] and ` ` ``col[] arrays */` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``if` `(row[i] == 1 || col[j] == 1) ` ` ``{ ` ` ``mat[i, j] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* A utility function to ` ` ``print a 2D matrix */` ` ``public` `static` `void` `printMatrix(``int` `[,]mat, ` ` ``int` `R, ``int` `C) ` ` ``{ ` ` ``int` `i, j; ` ` ``for` `(i = 0; i < R; i++) ` ` ``{ ` ` ``for` `(j = 0; j < C; j++) ` ` ``{ ` ` ``Console.Write(mat[i, j] + ``" "``); ` ` ``} ` ` ``Console.WriteLine(); ` ` ``} ` ` ``} ` ` ` ` ``// Driver code ` ` ``static` `public` `void` `Main () ` ` ``{ ` ` ``int` `[,]mat = {{1, 0, 0, 1}, ` ` ``{0, 0, 1, 0}, ` ` ``{0, 0, 0, 0}}; ` ` ` ` ``Console.WriteLine(``"Matrix Intially"``); ` ` ` ` ``printMatrix(mat, 3, 4); ` ` ` ` ``modifyMatrix(mat, 3, 4); ` ` ``Console.WriteLine(``"Matrix after "``+ ` ` ``"modification n"``); ` ` ``printMatrix(mat, 3, 4); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by ajit `
## PHP
` `
Output:
```Input Matrix
1 0 0 1
0 0 1 0
0 0 0 0
Matrix after modification
1 1 1 1
1 1 1 1
1 0 1 1
```
Time Complexity: O(M*N)
Auxiliary Space: O(M + N)
Method 2 (A Space Optimized Version of Method 1)
This method is a space optimized version of above method 1. This method uses the first row and first column of the input matrix in place of the auxiliary arrays row[] and col[] of method 1. So what we do is: first take care of first row and column and store the info about these two in two flag variables rowFlag and colFlag. Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1).
1) Scan the first row and set a variable rowFlag to indicate whether we need to set all 1s in first row or not.
2) Scan the first column and set a variable colFlag to indicate whether we need to set all 1s in first column or not.
3) Use first row and first column as the auxiliary arrays row[] and col[] respectively, consider the matrix as submatrix starting from second row and second column and apply method 1.
4) Finally, using rowFlag and colFlag, update first row and first column if needed.
Time Complexity: O(M*N)
Auxiliary Space: O(1)
Thanks to Sidh for suggesting this method.
## C++
`#include ` `using` `namespace` `std; ` `#define R 3 ` `#define C 4 ` ` ` `void` `modifyMatrix(``int` `mat[R][C]) ` `{ ` ` ``// variables to check if there are any 1 ` ` ``// in first row and column ` ` ``bool` `row_flag = ``false``; ` ` ``bool` `col_flag = ``false``; ` ` ` ` ``// updating the first row and col if 1 ` ` ``// is encountered ` ` ``for` `(``int` `i = 0; i < R; i++) { ` ` ``for` `(``int` `j = 0; j < C; j++) { ` ` ``if` `(i == 0 && mat[i][j] == 1) ` ` ``row_flag = ``true``; ` ` ` ` ``if` `(j == 0 && mat[i][j] == 1) ` ` ``col_flag = ``true``; ` ` ` ` ``if` `(mat[i][j] == 1) { ` ` ` ` ``mat[0][j] = 1; ` ` ``mat[i][0] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// Modify the input matrix mat[] using the ` ` ``// first row and first column of Matrix mat ` ` ``for` `(``int` `i = 1; i < R; i++) { ` ` ``for` `(``int` `j = 1; j < C; j++) { ` ` ` ` ``if` `(mat[0][j] == 1 || mat[i][0] == 1) { ` ` ``mat[i][j] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// modify first row if there was any 1 ` ` ``if` `(row_flag == ``true``) { ` ` ``for` `(``int` `i = 0; i < C; i++) { ` ` ``mat[0][i] = 1; ` ` ``} ` ` ``} ` ` ` ` ``// modify first col if there was any 1 ` ` ``if` `(col_flag == ``true``) { ` ` ``for` `(``int` `i = 0; i < R; i++) { ` ` ``mat[i][0] = 1; ` ` ``} ` ` ``} ` `} ` ` ` `/* A utility function to print a 2D matrix */` `void` `printMatrix(``int` `mat[R][C]) ` `{ ` ` ``for` `(``int` `i = 0; i < R; i++) { ` ` ``for` `(``int` `j = 0; j < C; j++) { ` ` ``cout << mat[i][j]; ` ` ``} ` ` ``cout << ````" "````; ` ` ``} ` `} ` ` ` `// Driver function to test the above function ` `int` `main() ` `{ ` ` ` ` ``int` `mat[R][C] = { { 1, 0, 0, 1 }, ` ` ``{ 0, 0, 1, 0 }, ` ` ``{ 0, 0, 0, 0 } }; ` ` ` ` ``cout << ````"Input Matrix : "````; ` ` ``printMatrix(mat); ` ` ` ` ``modifyMatrix(mat); ` ` ` ` ``cout << ````"Matrix After Modification : "````; ` ` ``printMatrix(mat); ` ` ``return` `0; ` `} ` ` ` `// This code is contributed by Nikita Tiwari `
## Java
`class` `GFG ` `{ ` ` ``public` `static` `void` `modifyMatrix(``int` `mat[][]){ ` ` ` ` ``// variables to check if there are any 1 ` ` ``// in first row and column ` ` ``boolean` `row_flag = ``false``; ` ` ``boolean` `col_flag = ``false``; ` ` ` ` ``// updating the first row and col if 1 ` ` ``// is encountered ` ` ``for` `(``int` `i = ``0``; i < mat.length; i++ ){ ` ` ``for` `(``int` `j = ``0``; j < mat[``0``].length; j++){ ` ` ``if` `(i == ``0` `&& mat[i][j] == ``1``) ` ` ``row_flag = ``true``; ` ` ` ` ``if` `(j == ``0` `&& mat[i][j] == ``1``) ` ` ``col_flag = ``true``; ` ` ` ` ``if` `(mat[i][j] == ``1``){ ` ` ` ` ``mat[``0``][j] = ``1``; ` ` ``mat[i][``0``] = ``1``; ` ` ``} ` ` ` ` ``} ` ` ``} ` ` ` ` ``// Modify the input matrix mat[] using the ` ` ``// first row and first column of Matrix mat ` ` ``for` `(``int` `i = ``1``; i < mat.length; i ++){ ` ` ``for` `(``int` `j = ``1``; j < mat[``0``].length; j ++){ ` ` ` ` ``if` `(mat[``0``][j] == ``1` `|| mat[i][``0``] == ``1``){ ` ` ``mat[i][j] = ``1``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// modify first row if there was any 1 ` ` ``if` `(row_flag == ``true``){ ` ` ``for` `(``int` `i = ``0``; i < mat[``0``].length; i++){ ` ` ``mat[``0``][i] = ``1``; ` ` ``} ` ` ``} ` ` ` ` ``// modify first col if there was any 1 ` ` ``if` `(col_flag == ``true``){ ` ` ``for` `(``int` `i = ``0``; i < mat.length; i ++){ ` ` ``mat[i][``0``] = ``1``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* A utility function to print a 2D matrix */` ` ``public` `static` `void` `printMatrix(``int` `mat[][]){ ` ` ``for` `(``int` `i = ``0``; i < mat.length; i ++){ ` ` ``for` `(``int` `j = ``0``; j < mat[``0``].length; j ++){ ` ` ``System.out.print( mat[i][j] ); ` ` ``} ` ` ``System.out.println(``""``); ` ` ``} ` ` ``} ` ` ` ` ``// Driver function to test the above function ` ` ``public` `static` `void` `main(String args[] ){ ` ` ` ` ``int` `mat[][] = {{``1``, ``0``, ``0``, ``1``}, ` ` ``{``0``, ``0``, ``1``, ``0``}, ` ` ``{``0``, ``0``, ``0``, ``0``}}; ` ` ` ` ``System.out.println(``"Input Matrix :"``); ` ` ``printMatrix(mat); ` ` ` ` ``modifyMatrix(mat); ` ` ` ` ``System.out.println(``"Matrix After Modification :"``); ` ` ``printMatrix(mat); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Arnav Kr. Mandal. `
## Python3
`# Python3 Code For A Boolean Matrix Question ` `def` `modifyMatrix(mat) : ` ` ` ` ``# variables to check if there are any 1 ` ` ``# in first row and column ` ` ``row_flag ``=` `False` ` ``col_flag ``=` `False` ` ` ` ``# updating the first row and col ` ` ``# if 1 is encountered ` ` ``for` `i ``in` `range``(``0``, ``len``(mat)) : ` ` ` ` ``for` `j ``in` `range``(``0``, ``len``(mat)) : ` ` ``if` `(i ``=``=` `0` `and` `mat[i][j] ``=``=` `1``) : ` ` ``row_flag ``=` `True` ` ` ` ``if` `(j ``=``=` `0` `and` `mat[i][j] ``=``=` `1``) : ` ` ``col_flag ``=` `True` ` ` ` ``if` `(mat[i][j] ``=``=` `1``) : ` ` ``mat[``0``][j] ``=` `1` ` ``mat[i][``0``] ``=` `1` ` ` ` ``# Modify the input matrix mat[] using the ` ` ``# first row and first column of Matrix mat ` ` ``for` `i ``in` `range``(``1``, ``len``(mat)) : ` ` ` ` ``for` `j ``in` `range``(``1``, ``len``(mat) ``+` `1``) : ` ` ``if` `(mat[``0``][j] ``=``=` `1` `or` `mat[i][``0``] ``=``=` `1``) : ` ` ``mat[i][j] ``=` `1` ` ` ` ``# modify first row if there was any 1 ` ` ``if` `(row_flag ``=``=` `True``) : ` ` ``for` `i ``in` `range``(``0``, ``len``(mat)) : ` ` ``mat[``0``][i] ``=` `1` ` ` ` ``# modify first col if there was any 1 ` ` ``if` `(col_flag ``=``=` `True``) : ` ` ``for` `i ``in` `range``(``0``, ``len``(mat)) : ` ` ``mat[i][``0``] ``=` `1` ` ` `# A utility function to print a 2D matrix ` `def` `printMatrix(mat) : ` ` ` ` ``for` `i ``in` `range``(``0``, ``len``(mat)) : ` ` ``for` `j ``in` `range``(``0``, ``len``(mat) ``+` `1``) : ` ` ``print``( mat[i][j], end ``=` `"" ) ` ` ` ` ``print``() ` ` ` `# Driver Code ` `mat ``=` `[ [``1``, ``0``, ``0``, ``1``], ` ` ``[``0``, ``0``, ``1``, ``0``], ` ` ``[``0``, ``0``, ``0``, ``0``] ] ` ` ` `print``(``"Input Matrix :"``) ` `printMatrix(mat) ` ` ` `modifyMatrix(mat) ` ` ` `print``(``"Matrix After Modification :"``) ` `printMatrix(mat) ` ` ` `# This code is contributed by Nikita tiwari. `
## C#
`// C# Code For A Boolean ` `// Matrix Question ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ``public` `static` `void` `modifyMatrix(``int``[,] mat) ` ` ``{ ` ` ` ` ``// variables to check ` ` ``// if there are any 1 ` ` ``// in first row and column ` ` ``bool` `row_flag = ``false``; ` ` ``bool` `col_flag = ``false``; ` ` ` ` ``// updating the first ` ` ``// row and col if 1 ` ` ``// is encountered ` ` ``for` `(``int` `i = 0; ` ` ``i < mat.GetLength(0); i++ ) ` ` ``{ ` ` ``for` `(``int` `j = 0; ` ` ``j < mat.GetLength(1); j++) ` ` ``{ ` ` ``if` `(i == 0 && mat[i, j] == 1) ` ` ``row_flag = ``true``; ` ` ` ` ``if` `(j == 0 && mat[i, j] == 1) ` ` ``col_flag = ``true``; ` ` ` ` ``if` `(mat[i, j] == 1) ` ` ``{ ` ` ``mat[0, j] = 1; ` ` ``mat[i,0] = 1; ` ` ``} ` ` ` ` ``} ` ` ``} ` ` ` ` ``// Modify the input matrix mat[] ` ` ``// using the first row and first ` ` ``// column of Matrix mat ` ` ``for` `(``int` `i = 1; ` ` ``i < mat.GetLength(0); i ++) ` ` ``{ ` ` ``for` `(``int` `j = 1; ` ` ``j < mat.GetLength(1); j ++) ` ` ``{ ` ` ` ` ``if` `(mat[0, j] == 1 || ` ` ``mat[i, 0] == 1) ` ` ``{ ` ` ``mat[i, j] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``// modify first row ` ` ``// if there was any 1 ` ` ``if` `(row_flag == ``true``) ` ` ``{ ` ` ``for` `(``int` `i = 0; ` ` ``i < mat.GetLength(1); i++) ` ` ``{ ` ` ``mat[0, i] = 1; ` ` ``} ` ` ``} ` ` ` ` ``// modify first col if ` ` ``// there was any 1 ` ` ``if` `(col_flag == ``true``) ` ` ``{ ` ` ``for` `(``int` `i = 0; ` ` ``i < mat.GetLength(0); i ++) ` ` ``{ ` ` ``mat[i, 0] = 1; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``/* A utility function ` ` ``to print a 2D matrix */` ` ``public` `static` `void` `printMatrix(``int``[,] mat) ` ` ``{ ` ` ``for` `(``int` `i = 0; ` ` ``i < mat.GetLength(0); i ++) ` ` ``{ ` ` ``for` `(``int` `j = 0; ` ` ``j < mat.GetLength(1); j ++) ` ` ``{ ` ` ``Console.Write(mat[i, j] + ``" "` `); ` ` ``} ` ` ``Console.Write(````" "````); ` ` ``} ` ` ``} ` ` ` ` ``// Driver Code ` ` ``public` `static` `void` `Main() ` ` ``{ ` ` ``int``[,] mat = {{1, 0, 0, 1}, ` ` ``{0, 0, 1, 0}, ` ` ``{0, 0, 0, 0}}; ` ` ` ` ``Console.Write(````"Input Matrix : "````); ` ` ``printMatrix(mat); ` ` ` ` ``modifyMatrix(mat); ` ` ` ` ``Console.Write(``"Matrix After "` `+ ` ` ````"Modification : "````); ` ` ``printMatrix(mat); ` ` ``} ` `} ` ` ` `// This code is contributed ` `// by ChitraNayal `
## PHP
` `
Output:
```Input Matrix :
1001
0010
0000
Matrix After Modification :
1111
1111
1011
``` | 8,609 | 23,479 | {"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} | 3.53125 | 4 | CC-MAIN-2021-17 | latest | en | 0.583337 |
https://www.physicsforums.com/threads/intermediate-math-challenge-may-2018.946386/page-3 | 1,560,668,514,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997801.20/warc/CC-MAIN-20190616062650-20190616084650-00344.warc.gz | 844,423,794 | 27,009 | # Challenge Intermediate Math Challenge - May 2018
• Featured
#### nuuskur
Complex analysis a.k.a black magic
#### Greg Bernhardt
Only 4 left to solve and we have 2.5 weeks left in May. Awesome job everyone!
#### fresh_42
Mentor
2018 Award
First day open to all. Open:
1.) Determinants.
7.) Series.
9.) Jacobi Identity.
10.) Integral.
Last edited:
#### julian
Gold Member
Is the solution of problem 10 this:
By the residue theorem:
$I = 2 \pi i \lim_{z \rightarrow 0} e^{kz} = 2 \pi i .$
Then by the change of variables
$z = e^{i \theta} \qquad dz = i e^{i \theta} d \theta$
The integral becomes
\begin{align}
I & = \int_{- \pi}^\pi e^{k e^{i \theta}} \frac{i e^{i \theta} d \theta}{e^{i \theta}}
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta + i k \sin \theta} d \theta
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta} e^{i k \sin \theta} d \theta
\nonumber \\
& = i \int_{- \pi}^\pi e^{k \cos \theta} [ \cos (k \sin \theta) + i \sin (k \sin \theta) ] d \theta
\nonumber
\end{align}
Using that $\sin (k \sin \theta)$ is an odd function in $\theta$, we have
$I = i \int_{- \pi}^\pi e^{k \cos \theta} \cos (k \sin \theta) d \theta .$
Comparing this to the first answer we got for $I$, we have
$\int_{- \pi}^\pi e^{k \cos \theta} \cos (k \sin \theta) d \theta = 2 \pi .$
#### QuantumQuest
Gold Member
Is the solution of problem 10 this
Yes @julian. Well done!
#### julian
Gold Member
I think I have solved problem 7.
I used the generating function technique. Define
$f (x) = \sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} x^n = \sum_{n=1}^\infty a_n x^n$
The sum we are after will then be equal to $f (1)$.
Note that
$a_{n+1} = \frac{(n!)^2}{n (2n)!} \times \frac{n (n+1)^2}{(n+1) (2n+2) (2n+1)} = a_n \times \frac{n}{2 (2n +1)}$
or $2 (2n+1) a_{n+1} = n a_n$. In the following we use $a_1 = 1/2$. First
\begin{align}
x \frac{d}{dx} f (x) & = \sum_{n=1}^\infty n a_n x^n
\nonumber \\
& = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^n
\nonumber
\end{align}
Then
\begin{align}
x^2 \frac{d}{dx} f (x) & = \sum_{n=1}^\infty 2 (2n+1) a_{n+1} x^{n+1}
\nonumber \\
& = 2 \sum_{m=2}^\infty (2m -1) a_m x^m
\nonumber \\
& = 2 \sum_{m=1}^\infty (2m -1) a_m x^m - 2 (2 \times 1 - 1) a_{1} x
\nonumber \\
& = 4 \sum_{m=1}^\infty m a_m x^m - 2 f (x) - x
\nonumber \\
& = 4 x \frac{d}{dx} f (x) - 2 f (x) - x
\nonumber
\end{align}
This gives the differential equation:
$(x^2 - 4x) \frac{d}{dx} f (x) + 2f (x) + x = 0$
or
$\frac{d}{dx} f (x) + \frac{2}{x^2 - 4x} f (x) = - \frac{1}{x - 4} .$
(with boundary condition $f (0) = 0$). Now this is the famililar form
$\frac{d}{dx} f (x) + p (x) f (x) = q (x)$
that can be solved using the integrating factor method which uses
\begin{align}
\frac{d}{dx} \Big[ e^{\nu (x)} f (x) \Big] & = e^{\nu (x)} \Big[ \frac{d}{dx} f (x) + p (x) f (x) \Big]
\nonumber \\
& = e^{\nu (x)} q (x)
\nonumber
\end{align}
where $\nu (x) = \int p (x) dx$. We will solve this with
$f (x) = e^{- \nu (x)} \int e^{\nu (x)} q(x) dx .$
In our case
$p (x) = - \frac{1}{2} \Big( \frac{1}{x} - \frac{1}{x-4} \Big) , \qquad q (x) = - \frac{1}{x-4}$
so that
\begin{align}
\nu (x) & = - \frac{1}{2} \int \Big( \frac{1}{x} - \frac{1}{x - 4} \Big) dx
\nonumber \\
& = - \frac{1}{2} ( \ln x - \ln (x-4))
\nonumber
\end{align}
meaning
$e^{+ \nu (x)} = \frac{(x-4)^{1/2}}{x^{1/2}} \qquad \mathrm{and} \qquad e^{- \nu (x)} = \frac{x^{1/2}}{(x-4)^{1/2}}$
and
\begin{align}
f (x) & = - \frac{x^{1/2}}{(x-4)^{1/2}} \int_0^x \frac{(x-4)^{1/2}}{x^{1/2}} \frac{1}{x-4} dx
\nonumber \\
& = i \frac{x^{1/2}}{(4-x)^{1/2}} \int_0^x \frac{dx}{(x^2 - 4x)^{1/2}} dx
\nonumber
\end{align}
(choosing the lower limit to be $0$ will ensure that $f (0) = 0$). We now take $x = 1$, then
\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} & = f (1)
\nonumber \\
& = i \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{(x^2 - 4 x)^{1/2}}
\nonumber
\end{align}
We evaluate the above integral. Write $x^2 - 4 x = [x - 2]^2 - 4$. Then with the substitution $u = x - 2$ we obtain
\begin{align}
\int_0^1 \frac{dx}{(x^2 - 4 x)^{1/2}} & = \int_{-2}^{-1} \frac{du}{(u^2 - 2^2)^{1/2}}
\nonumber \\
& = \Big[ \cosh^{-1} (u/2) \Big]_{-2}^{-1}
\nonumber \\
& = \cosh^{-1} (-1/2) - \cosh^{-1} (-1) .
\nonumber
\end{align}
Using the formula $\cosh^{-1} x = \ln [x + \sqrt{x+1} \sqrt{x-1}]$, we have
\begin{align}
\cosh^{-1} (-1/2) - \cosh^{-1} (-1) & = \ln [-1/2 + i \sqrt{3}/2] - \ln (-1)
\nonumber\\
& = i \frac{2}{3} \pi - i \pi
\nonumber\\
& = - i \frac{1}{3} \pi
\nonumber
\end{align}
Accordingly
\begin{align}
\sum_{n=1}^\infty \frac{(n!)^2}{n (2n)!} & = i \frac{1}{\sqrt{3}} \times -i \frac{\pi}{3}
\nonumber \\
& = \frac{\pi}{3 \sqrt{3}} .
\nonumber
\end{align}
#### fresh_42
Mentor
2018 Award
I think I have solved problem 7.
Wow!!! It looks good at first glance, i.e. the result is correct and I couldn't find flaws. There is a simpler solution using a known series expansion, but this one is really good.
#### julian
Gold Member
Thanks fresh_42. A thing that needs to be checked is if we are allowed to differentiate the series term by term. This is easy to establish.
There is the theorem:
The derivative of the series sum $f (x) = \sum_i^\infty u_n (x)$ equals the sum of the individual term derivatives,
$\frac{d}{dx} f (x) = \sum_{n=1}^\infty \frac{d}{dx} u_n (x) ,$
provided the following conditions hold:
$u_n (x) \quad \mathrm{and} \quad \frac{d u_n (x)}{dx} \quad \mathrm{are \; continuous \; in} \; [a,b]$
$\sum_{n=1}^\infty \frac{d u_n (x)}{dx} \quad \mathrm{is \; uniformally \; convergent \; in} \; [a,b] .$
In our case it is obvious that $u_n (x) = a_n x^n$ and $\frac{u_n (x)}{dx} = n a_n x^{n-1}$ are continuous in $[0,1]$. So we just have to check that $\sum_{n=1}^\infty \frac{d u_n (x)}{dx}$ is uniformally convergent.
We can do this using the Weierstrass M test. This states that if we can construct a series of numbers $\sum_1^\infty M_n$, in which $M_n \geq |v_n (x)|$ for all $x \in [a,b]$ and $\sum_1^\infty M_n$ is convergent, the series $\sum_1^\infty v_n (x)$ will be uniformally convergent in $[a,b]$.
In our case an obvious candidate for the $M_n$'s are $n a_n = \frac{(n!)^2}{(2n)!}$. The ratio test tells you the series $\sum_1^\infty M_n$ is convergent:
$\lim_{n \rightarrow \infty} \frac{(n+1) a_{n+1}}{n a_n} = \lim_{n \rightarrow \infty} \frac{(n+1)^2}{(2n+2) (2n+1)} = \frac{1}{4} < 1 .$
We then just need to note that
$M_n = n a_n \geq |n a_n x^{n-1}| = \Big| \frac{d u_n (x)}{dx} \Big| \quad \mathrm{for \; all} \;\; x \in [0,1] .$
Also, it might have been "nicer" to write
$f (x) = \frac{x^{1/2}}{(4-x)^{1/2}} \int_0^x \frac{dx}{(4 - x^2)^{1/2}}$
and
\begin{align}
f (1) & = \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{(4x - x^2)^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \int_0^1 \frac{dx}{[2^2 - (x - 2)^2]^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \int_{-2}^{-1}\frac{du}{[2^2 - u^2]^{1/2}}
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big[ \sin^{-1} (u/2) \Big]_{-2}^{-1}
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big( \sin^{-1} (1) - \sin^{-1} (1/2) \Big)
\nonumber \\
& = \frac{1}{\sqrt{3}} \Big( \frac{\pi}{2} - \frac{\pi}{6} \Big)
\nonumber \\
& = \frac{\pi}{3 \sqrt{3}}
\nonumber
\end{align}
but it is equivalent.
Last edited:
#### julian
Gold Member
I think I have solved problem 1:
I split the proof into the parts:
Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).
Part (a):
Few facts about real skew matrices:
They are normal: $A A^\dagger = A A^T = -A A = A^T A = A^\dagger A$ and so the spectral theorem holds. There is a unitary matrix $U$ such that $U^\dagger A U = D$ where $D$ is a diagonal matrix. The entries of the diagonal of $D$ are the eigenvalues of $A$.
The eigenvalues are pure imaginary.
As the coefficients of the characteristic polynomial, $\det (A - \lambda I) = 0$, are real the eigenvalues come in conjugate pairs. If the dimension, $n$, of the matrix $A$ is odd then 0 must be one of the eigenvalues.
Part (b) (i):
We first take the simplest case of even $n$: $n = 2$. We can write
$U^\dagger A U = D = \begin{pmatrix} \lambda & 0 \\ 0 & - \lambda \end{pmatrix} .$
First consider
\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \det \big[ \big( A + x_1 I \big) \big( A + x_2 I \big) \dots \big( A + x_k I \big) \big]
\nonumber \\
& = \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \big( A + x_2 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k
I \big) U \big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + x_1 & 0 \\
0 & - \lambda + x_1
\end{pmatrix}
\begin{pmatrix}
\lambda + x_2 & 0 \\
0 & - \lambda + x_2
\end{pmatrix}
\dots
\begin{pmatrix}
\lambda + x_k & 0 \\
0 & - \lambda + x_k
\end{pmatrix}
\Big]
\nonumber \\
& = \det
\nonumber \\
&
\begin{pmatrix}
(\lambda + x_1) (\lambda + x_2) \dots (\lambda + x_k) & 0 \\
0 & (- \lambda + x_1) (- \lambda + x_2) \dots (- \lambda + x_k)
\end{pmatrix}
\nonumber \\
& = (- \lambda^2 + x_1^2) (- \lambda^2 + x_2^2) \dots (- \lambda^2 + x_k^2)
\nonumber \\
\nonumber
\end{align}
where we have introduced $\lambda = i \theta$ where $\theta$ is real.
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) =
\nonumber \\
& =
\det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}^k
\Big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = \big[ - \lambda^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber \\
&= \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
Comparing eq (1) and eq (2), we see that proving the main result for $n = 2$ then amounts to proving the inequality:
$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k \qquad \qquad \qquad \qquad (3)$
We prove this, for general value of $\theta$, in the next subsection. See next spoiler!
Part (b) (ii):
We wish to prove (3):
$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$
The proof is an essentially inductive argument. Our base case $k = 2$ is easy:
\begin{align}
(\theta^2 + x_1^2) (\theta^2 + x_2^2) & = \theta^4 + \theta^2 x_1^2 + \theta^2 x_2^2 + x_1^2 x_2^2
\nonumber \\
& \geq \theta^4 + 2 \theta^2 (x_1 x_2) + x_1^2 x_2^2
\nonumber \\
& = [\theta^2 + (x_1^2 x_2^2)^{1/2}]^2
\nonumber
\end{align}
where we used $a^2 + b^2 \geq 2 ab$.
Next we prove that whenever the result holds for $k$, it holds for $2k$ as well. That is, we'll first prove the result for powers of $2: k = 2, 4, 8, 16, \dots$. Assume we know the result holds for some $k$. Now consider $2k$ positive numbers $x_1^2, \dots , x_k^2$ and $y_1^k , \dots y_k^2$. We use the induction hypothesis and the base case to find
\begin{align}
& (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) (\theta^2 + y_1^2) \dots (\theta^2 + y_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k [\theta^2 + (y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^4 + \theta^2 (x_1^2 \dots x_k^2)^{1/k} + \theta^2 (y_1^2 \dots y_k^2)^{1/k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& \geq [\theta^4 + 2 \theta^2 (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^2 + (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k}]^{2k}
\nonumber
\end{align}
where we used $a^2 + b^2 \geq 2 ab$. This is the required result. We now know the theorem to be true for infinitely many $k$.
Next we prove that whenever the result is true for $k$, it's also true for $k - 1$. This will prove the result for all the in-between integers. Let $k \geq 4$ and assume the result holds for $k$. Consider the $k-1$ positive numbers $x_1^2 , x_2^2 , \dots , x_{k-1}^2$. Define $x_k^2$ to be $(x_1^2 x_2^2 \dots x_{k-1}^2)^{1/(k-1)}$. We then have
\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) (\theta^2 + x_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_{k-1}^2 x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \{ (x_1^2 \dots x_{k-1}^2)^{1/(k-1)} \}^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \big( x_k^2 \big)^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + x_k^2]^k
\nonumber
\end{align}
Rearranging we have
\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) \geq [\theta^2 + x_k^2]^{k-1}
\nonumber \\
& \qquad \equiv [\theta^2 + (x_1^2 x_2^2 \dots x_{k-1}^2)^{1 / (k-1)} ]^{k-1}
\nonumber
\end{align}
which is the required result.
Which means we have established (3) and proven the main result for $n = 2$, i.e.,
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 2 .$
We will prove the main result for arbitrary even $n$ in the next subsection. See next spoiler!
Part (b) (iii):
We now prove the main result for general even $n$. We can write
$U^\dagger A U = D = \begin{pmatrix} \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \dots & \lambda_{n/2} & 0 \\ 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} \\ \end{pmatrix} .$
So that
\begin{align}
& \prod_{i=1}^k \det (A - x_i I)
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
\lambda_1 + x_i & 0 & \dots & \dots & 0 & 0 \\
0 & - \lambda_1 + x_i & \dots & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & \dots & \lambda_{n/2} + x_i & 0 \\
0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} + x_i \\
\end{pmatrix}
\nonumber \\
& =
\Big( \prod_{i=1}^k
\det
\begin{pmatrix}
\lambda_1 + x_i & 0 \\
0 & - \lambda_1 + x_i
\end{pmatrix} \Big)
\dots
\Big( \prod_{i=1}^k
\begin{pmatrix}
\lambda_{n/2} + x_i & 0 \\
0 & - \lambda_{n/2} + x_i
\end{pmatrix} \Big)
\nonumber \\
& = \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big) \dots \big( \prod_{i=1}^k(\theta_{n/2}^2 + x_i^2) \big)
\nonumber
\end{align}
where we have introduced $\lambda_l = i \theta_l$ where $\theta_l$ is real..
Next consider
$\qquad \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k$
$\quad = \det \begin{pmatrix} \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots & \dots \\ 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots & \dots \\ \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \dots & \ddots \end{pmatrix}^k$
$\quad = \prod_{l=1}^{n/2} \det \Big[ \begin{pmatrix} \lambda_l + (x_1 x_2 \dots x_k)^{1/k} & 0 \\ 0 & - \lambda_l + (x_1 x_2 \dots x_k)^{1/k} \end{pmatrix}^k \Big]$
$\quad = \big[ \theta_1^2 + ( x_1^2 x_2^2 \dots x_k^2 )^{1/k} \big]^k \dots \big[ \theta_{n/2}^2 + ( x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$
We easily have from (3) that
\begin{align}
& \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{n/2}^2 + x_i^2) \big)
\geq
\nonumber \\
\big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\dots
\big[ \theta_{n/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\nonumber
\end{align}
which proves the main result for even $n$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n \; \mathrm{even} .$
In the next section we prove the main result for odd $n$. See next spoiler!
Part (c) (i):
We now consider the case $n = 3$. We write
$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & - \lambda \end{pmatrix} .$
First consider
$\prod_{i=1}^k \det \big( A + x_i I \big) =$
$= \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k I \big) U \big]$
$\quad = \det \Big[ \begin{pmatrix} x_1 & 0 & 0 \\ 0 & \lambda + x_1 & 0 \\ 0 & 0 & - \lambda + x_1 \end{pmatrix} \dots \begin{pmatrix} x_k & 0 & 0 \\ 0 & \lambda + x_k & 0 \\ 0 & 0 & - \lambda + x_k \end{pmatrix} \Big]$
$= \det$
$\begin{pmatrix} x_1 x_2 \dots x_k & 0 & 0 \\ 0 & (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix}$
$= (x_1 x_2 \dots x_k) \times$
$\quad \det \begin{pmatrix} (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix} .$
$= (x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)$
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\Big[
\begin{pmatrix}
(x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 & 0 \\
0 & \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 \\
0 & 0 & - \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k) \det
\Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
As the OP's question assumes that $\{ x_1, \dots , x_k \}$ are positive numbers, and using the result (3), we
have:
$(x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$
which establishes the main result for $n = 3$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 3 .$
Part (c) (ii):
We now turn to the general case of any odd $n$. We can write
$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & \lambda_n & 0 \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_n \\ \end{pmatrix} .$
First consider
\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
x_i & 0 & 0 & \dots & 0 & 0 \\
0 & \lambda_1 + x_i & 0 & \dots & 0 & 0 \\
0 & 0 & - \lambda_1 +x_i & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & \lambda_{(n-1)/2}+ x_i & 0 \\
0 & 0 & 0 & \dots & 0 & - \lambda_{(n-1)/2} + x_i \\
\end{pmatrix}
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\nonumber
\end{align}
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\begin{pmatrix}
(x_1 \dots x_k)^{\frac{1}{k}} & 0 & 0 & \dots \\
0 & \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots \\
0 & 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\end{pmatrix}^k
\nonumber \\
& =
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
We easily have from (3) that
\begin{align}
& (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\geq
\nonumber \\
&
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
which establishes the main result for general odd values of $n$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad \mathrm {for \; odd \;} n .$
Completeting the proof of problem 1.
Last edited:
#### StoneTemplePython
Gold Member
I think I have solved problem 1:
I split the proof into the parts:
Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).
Part (a):
Few facts about real skew matrices:
They are normal: $A A^\dagger = A A^T = -A A = A^T A = A^\dagger A$ and so the spectral theorem holds. There is a unitary matrix $U$ such that $U^\dagger A U = D$ where $D$ is a diagonal matrix. The entries of the diagonal of $D$ are the eigenvalues of $A$.
The eigenvalues are pure imaginary.
As the coefficients of the characteristic polynomial, $\det (A - \lambda I) = 0$, are real the eigenvalues come in conjugate pairs. If the dimension, $n$, of the matrix $A$ is odd then 0 must be one of the eigenvalues.
Part (b) (i):
We first take the simplest case of even $n$: $n = 2$. We can write
$U^\dagger A U = D = \begin{pmatrix} \lambda & 0 \\ 0 & - \lambda \end{pmatrix} .$
First consider
\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \det \big[ \big( A + x_1 I \big) \big( A + x_2 I \big) \dots \big( A + x_k I \big) \big]
\nonumber \\
& = \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \big( A + x_2 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k
I \big) U \big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + x_1 & 0 \\
0 & - \lambda + x_1
\end{pmatrix}
\begin{pmatrix}
\lambda + x_2 & 0 \\
0 & - \lambda + x_2
\end{pmatrix}
\dots
\begin{pmatrix}
\lambda + x_k & 0 \\
0 & - \lambda + x_k
\end{pmatrix}
\Big]
\nonumber \\
& = \det
\nonumber \\
&
\begin{pmatrix}
(\lambda + x_1) (\lambda + x_2) \dots (\lambda + x_k) & 0 \\
0 & (- \lambda + x_1) (- \lambda + x_2) \dots (- \lambda + x_k)
\end{pmatrix}
\nonumber \\
& = (- \lambda^2 + x_1^2) (- \lambda^2 + x_2^2) \dots (- \lambda^2 + x_k^2)
\nonumber \\
\nonumber
\end{align}
where we have introduced $\lambda = i \theta$ where $\theta$ is real.
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) =
\nonumber \\
& =
\det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}^k
\Big]
\nonumber \\
& = \det \Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = \big[ - \lambda^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber \\
&= \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
Comparing eq (1) and eq (2), we see that proving the main result for $n = 2$ then amounts to proving the inequality:
$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k \qquad \qquad \qquad \qquad (3)$
We prove this, for general value of $\theta$, in the next subsection. See next spoiler!
Part (b) (ii):
We wish to prove (3):
$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$
The proof is an essentially inductive argument. Our base case $k = 2$ is easy:
\begin{align}
(\theta^2 + x_1^2) (\theta^2 + x_2^2) & = \theta^4 + \theta^2 x_1^2 + \theta^2 x_2^2 + x_1^2 x_2^2
\nonumber \\
& \geq \theta^4 + 2 \theta^2 (x_1 x_2) + x_1^2 x_2^2
\nonumber \\
& = [\theta^2 + (x_1^2 x_2^2)^{1/2}]^2
\nonumber
\end{align}
where we used $a^2 + b^2 \geq 2 ab$.
Next we prove that whenever the result holds for $k$, it holds for $2k$ as well. That is, we'll first prove the result for powers of $2: k = 2, 4, 8, 16, \dots$. Assume we know the result holds for some $k$. Now consider $2k$ positive numbers $x_1^2, \dots , x_k^2$ and $y_1^k , \dots y_k^2$. We use the induction hypothesis and the base case to find
\begin{align}
& (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) (\theta^2 + y_1^2) \dots (\theta^2 + y_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k [\theta^2 + (y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^4 + \theta^2 (x_1^2 \dots x_k^2)^{1/k} + \theta^2 (y_1^2 \dots y_k^2)^{1/k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& \geq [\theta^4 + 2 \theta^2 (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k} + (x_1^2 \dots x_k^2
y_1^2 \dots y_k^2)^{1/k}]^k
\nonumber \\
& = [\theta^2 + (x_1^2 \dots x_k^2 y_1^2 \dots y_k^2)^{1/2k}]^{2k}
\nonumber
\end{align}
where we used $a^2 + b^2 \geq 2 ab$. This is the required result. We now know the theorem to be true for infinitely many $k$.
Next we prove that whenever the result is true for $k$, it's also true for $k - 1$. This will prove the result for all the in-between integers. Let $k \geq 4$ and assume the result holds for $k$. Consider the $k-1$ positive numbers $x_1^2 , x_2^2 , \dots , x_{k-1}^2$. Define $x_k^2$ to be $(x_1^2 x_2^2 \dots x_{k-1}^2)^{1/(k-1)}$. We then have
\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) (\theta^2 + x_k^2)
\nonumber \\
& \geq [\theta^2 + (x_1^2 \dots x_{k-1}^2 x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \{ (x_1^2 \dots x_{k-1}^2)^{1/(k-1)} \}^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + \big( x_k^2 \big)^{(k-1)/k} \; (x_k^2)^{1/k} ]^k
\nonumber \\
& = [\theta^2 + x_k^2]^k
\nonumber
\end{align}
Rearranging we have
\begin{align}
& (\theta^2 + x_1^2) (\theta^2 + x_2^2) \dots (\theta^2 + x_{k-1}^2) \geq [\theta^2 + x_k^2]^{k-1}
\nonumber \\
& \qquad \equiv [\theta^2 + (x_1^2 x_2^2 \dots x_{k-1}^2)^{1 / (k-1)} ]^{k-1}
\nonumber
\end{align}
which is the required result.
Which means we have established (3) and proven the main result for $n = 2$, i.e.,
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 2 .$
We will prove the main result for arbitrary even $n$ in the next subsection. See next spoiler!
Part (b) (iii):
We now prove the main result for general even $n$. We can write
$U^\dagger A U = D = \begin{pmatrix} \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \dots & \dots & \lambda_{n/2} & 0 \\ 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} \\ \end{pmatrix} .$
So that
\begin{align}
& \prod_{i=1}^k \det (A - x_i I)
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
\lambda_1 + x_i & 0 & \dots & \dots & 0 & 0 \\
0 & - \lambda_1 + x_i & \dots & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & \dots & \lambda_{n/2} + x_i & 0 \\
0 & 0 & \dots & \dots & 0 & - \lambda_{n/2} + x_i \\
\end{pmatrix}
\nonumber \\
& =
\Big( \prod_{i=1}^k
\det
\begin{pmatrix}
\lambda_1 + x_i & 0 \\
0 & - \lambda_1 + x_i
\end{pmatrix} \Big)
\dots
\Big( \prod_{i=1}^k
\begin{pmatrix}
\lambda_{n/2} + x_i & 0 \\
0 & - \lambda_{n/2} + x_i
\end{pmatrix} \Big)
\nonumber \\
& = \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big) \dots \big( \prod_{i=1}^k(\theta_{n/2}^2 + x_i^2) \big)
\nonumber
\end{align}
where we have introduced $\lambda_l = i \theta_l$ where $\theta_l$ is real..
Next consider
$\qquad \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k$
$\quad = \det \begin{pmatrix} \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots & \dots \\ 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots & \dots \\ \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \dots & \ddots \end{pmatrix}^k$
$\quad = \prod_{l=1}^{n/2} \det \Big[ \begin{pmatrix} \lambda_l + (x_1 x_2 \dots x_k)^{1/k} & 0 \\ 0 & - \lambda_l + (x_1 x_2 \dots x_k)^{1/k} \end{pmatrix}^k \Big]$
$\quad = \big[ \theta_1^2 + ( x_1^2 x_2^2 \dots x_k^2 )^{1/k} \big]^k \dots \big[ \theta_{n/2}^2 + ( x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$
We easily have from (3) that
\begin{align}
& \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{n/2}^2 + x_i^2) \big)
\geq
\nonumber \\
\big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\dots
\big[ \theta_{n/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{\frac{1}{k}} \big]^k
\nonumber
\end{align}
which proves the main result for even $n$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n \; \mathrm{even} .$
In the next section we prove the main result for odd $n$. See next spoiler!
Part (c) (i):
We now consider the case $n = 3$. We write
$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & - \lambda \end{pmatrix} .$
First consider
$\prod_{i=1}^k \det \big( A + x_i I \big) =$
$= \det \big[ U^{-1} \big( A + x_1 I \big) U U^{-1} \dots U U^{-1} \big( A + x_k I \big) U \big]$
$\quad = \det \Big[ \begin{pmatrix} x_1 & 0 & 0 \\ 0 & \lambda + x_1 & 0 \\ 0 & 0 & - \lambda + x_1 \end{pmatrix} \dots \begin{pmatrix} x_k & 0 & 0 \\ 0 & \lambda + x_k & 0 \\ 0 & 0 & - \lambda + x_k \end{pmatrix} \Big]$
$= \det$
$\begin{pmatrix} x_1 x_2 \dots x_k & 0 & 0 \\ 0 & (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix}$
$= (x_1 x_2 \dots x_k) \times$
$\quad \det \begin{pmatrix} (\lambda + x_1) \dots (\lambda + x_k) & 0 \\ 0 & (- \lambda + x_1) \dots (- \lambda + x_k) \end{pmatrix} .$
$= (x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)$
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\Big[
\begin{pmatrix}
(x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 & 0 \\
0 & \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}} & 0 \\
0 & 0 & - \lambda + (x_1 x_2 \dots x_k)^{\frac{1}{k}}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k) \det
\Big[
\begin{pmatrix}
\lambda + (x_1 x_2 \dots x_k)^{1/k} & 0 \\
0 & - \lambda + (x_1 x_2 \dots x_k)^{1/k}
\end{pmatrix}
\Big]^k
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
As the OP's question assumes that $\{ x_1, \dots , x_k \}$ are positive numbers, and using the result (3), we
have:
$(x_1 x_2 \dots x_k) (\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq (x_1 x_2 \dots x_k ) \big[ \theta^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k$
which establishes the main result for general odd values of $n = 3$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad n = 3 .$
Part (c) (ii):
We now turn to the general case of any odd $n$. We can write
$U^\dagger A U = D = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & \lambda_1 & 0 & 0 & 0 & \dots & \dots & 0 & 0 \\ 0 & 0 & - \lambda_1 & 0 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & \lambda_2 & 0& \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & - \lambda_2& \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \dots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & \lambda_n & 0 \\ 0 & 0 & 0 & 0 & 0 & \dots & \dots & 0 & - \lambda_n \\ \end{pmatrix} .$
First consider
\begin{align}
& \prod_{i=1}^k \det \big( A + x_i I \big) =
\nonumber \\
& = \prod_{i=1}^k \det
\nonumber \\
&
\begin{pmatrix}
x_i & 0 & 0 & \dots & 0 & 0 \\
0 & \lambda_1 + x_i & 0 & \dots & 0 & 0 \\
0 & 0 & - \lambda_1 +x_i & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & \lambda_{(n-1)/2}+ x_i & 0 \\
0 & 0 & 0 & \dots & 0 & - \lambda_{(n-1)/2} + x_i \\
\end{pmatrix}
\nonumber \\
& = (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\nonumber
\end{align}
Now consider
\begin{align}
& \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big)^k =
\nonumber \\
& = \det
\begin{pmatrix}
(x_1 \dots x_k)^{\frac{1}{k}} & 0 & 0 & \dots \\
0 & \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & 0 & \dots \\
0 & 0 & - \lambda_1 + (x_1 \dots x_k)^{\frac{1}{k}} & \dots \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\end{pmatrix}^k
\nonumber \\
& =
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
We easily have from (3) that
\begin{align}
& (x_1 x_2 \dots x_k ) \big( \prod_{i=1}^k (\theta_1^2 + x_i^2) \big)
\dots
\big( \prod_{i=1}^k (\theta_{(n-1)/2}^2 + x_i^2) \big)
\geq
\nonumber \\
&
(x_1 x_2 \dots x_k ) \big[ \theta_1^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\dots
\big[ \theta_{(n-1)/2}^2 + (x_1^2 x_2^2 \dots x_k^2)^{1/k} \big]^k
\nonumber
\end{align}
which establishes the main result for general odd values of $n$:
$\prod_{i=1}^k \det \big( A + x_i I \big) \geq \det \Big( A + (\prod_{i=1}^k x_i)^{1/k} I \Big) \qquad \mathrm {for \; odd \;} n .$
Completeting the proof of problem 1.
Thanks! I was worried I'd have to type up a solution if no one solved it by month end!
The solution looks about right. Proving the $n=2$ case is definitely the key unlocking the problem, which you tackled in your second spoiler. I'm a bit short on time right now but will take a closer look later on.
#### StoneTemplePython
Gold Member
I think I have solved problem 1:
I split the proof into the parts:
Part (a) A few facts about real skew symmetric matrices.
Part (b): Proof for $n$ even.
(i) Looking at case $n = 2$.
(ii) Proving a key inequality. This will prove case $n=2$
(iii) Proving case for general even $n$ (then easy).
Part (c) Case of odd $n$.
(i) Proving case for $n = 3$ (easy because of part (b)).
(iii) Proving case for general odd $n$ (easy because of part (b)).
I went through it fairly granularly and did not see any flaws.
A couple thoughts:
1.) If you are so inclined in your first spoiler, you may make use of rule
2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
I would be happy to accept basic results about skew symmetric matrices' eigenvalues having zero real component by spectral theory. On the other hand, your workthrough may be more instructive for 3rd parties reading it who don’t know the spectral theory underlying it – so they both have merits.
2.) The key insight for this problem, in my view, is figuring out the $n = 2$ case. Everything can be built off of this. The other insight is relating it to $\text{GM} \leq \text{AM}$ in some way. I think you basically re-created Cauchy’s forward-backward induction proof for $\text{GM} \leq \text{AM}$, in Part (b) albeit for additivity not for vanilla $\text{GM} \leq \text{AM}$. Since we are at month end, I will share another much simpler idea, which is the fact that 'regular' $\text{GM} \leq \text{AM}$ implies this result.
my take is that in Part (B) (II) when you are seeking to prove:
$(\theta^2 + x_1^2) \dots (\theta^2 + x_k^2) \geq [\theta^2 + (x_1^2 \dots x_k^2)^{1/k}]^k$
or equivalently
$\Big((\theta^2 + x_1^2) \dots (\theta^2 + x_k^2)\Big)^{1/k} \geq \theta^2 + (x_1^2 \dots x_k^2)^{1/k}$
multiply each side by
$\big(\theta^2\big)^{-1}$
(which is positive and doesn't change the inequality) and define
$z_i := \frac{x_i^2}{\theta^2} \gt 0$
The relationship is thus:
$\Big(\prod_{i=1}^k (1 + z_i)\Big)^{1/k}= \Big((1 + z_1) \dots (1 + z_k)\Big)^{1/k} \geq 1 + (z_1 \dots z_k)^{1/k} = \Big(\prod_{i=1}^k 1\Big)^{1/k} + \Big(\prod_{i=1}^k z_i\Big)^{1/k}$
which is true by the super additivity of the Geometric Mean (which incidentally was a past challenge problem, but since it is not this challenge problem I think it is fine to assume it is common knowledge to mathematicians).
- - - -
To consider the case of any eigenvalues equal to zero, we can verify that the inequality holds with equality, which we can chain onto the above.
- - - -
I have a soft spot for proving this via $2^r$ for $r = \{1, 2, 3, ...\}$ and then filling in the gaps. Really well done. Forward backward-induction is a very nice technique, but a lot of book-keeping!
#### fresh_42
Mentor
2018 Award
Here is the solution to the last open problem #9.
For a given a real Lie algebra $\mathfrak{g}$, we define
$$\mathfrak{A(g)} = \{\,\alpha \, : \,\mathfrak{g}\longrightarrow \mathfrak{g}\,\,: \,\,[\alpha(X),Y]=-[X,\alpha(Y)]\text{ for all }X,Y\in \mathfrak{g}\,\}\quad (1)$$
The Lie algebra multiplication is defined by
• $(2)$ anti-commutativity: $[X,X]=0$
• $(3)$ Jacobi-identity: $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$
a) $\mathfrak{A(g)}\subseteq \mathfrak{gl}(g)$ is a Lie subalgebra in the Lie algebra of all linear transformations of $\mathfrak{g}$ with the commutator as Lie product $[\alpha, \beta]= \alpha \beta -\beta \alpha \quad (4)$ because
\begin{align*}
[[\alpha,\beta]X,Y]&\stackrel{(4)}{=}[\alpha \beta X,Y] - [\beta\alpha X,Y]\\
&\stackrel{(1)}{=}[X,\beta \alpha Y]-[X,\alpha \beta Y]\\
&\stackrel{(4)}{=}[X,[\beta,\alpha]Y]\\
&\stackrel{(2)}{=}-[X,[\alpha,\beta]Y]
\end{align*}
b) The smallest non Abelian Lie algebra $\mathfrak{g}$ with trivial center is $\mathfrak{g}=\langle X,Y\,: \,[X,Y]=Y\rangle\,.$ It's easy to verify $\mathfrak{A(g)} \cong \mathfrak{sl}(2,\mathbb{R})\,$, the Lie algebra of $2 \times 2$ matrices with trace zero.
$\mathfrak{g}=\mathfrak{B(sl(}2,\mathbb{R}))$ is the maximal solvable subalgebra of $\mathfrak{sl}(2,\mathbb{R})$, a so called Borel subalgebra.
c) To show that $\mathfrak{g} \rtimes \mathfrak{A(g)}$ is a semidirect product given by $$[X,\alpha]:=[\operatorname{ad}X,\alpha]=\operatorname{ad}X\,\alpha - \alpha\,\operatorname{ad}X\quad (5)$$ we have to show that this multiplication makes $\mathfrak{A}(g)$ an ideal in $\mathfrak{g} \rtimes \mathfrak{A(g)}$ and a $\mathfrak{g}-$module.
\begin{align*}
[[X,\alpha]Y,Z]&\stackrel{(5)}{=}[[X,\alpha Y],Z] - [\alpha[X,Y],Z]\\
&\stackrel{(3),(1)}{=}-[[\alpha Y,Z],X]-[[Z,X],\alpha Y]+[[X,Y],\alpha Z]\\
&\stackrel{(3),(1)}{=}[[Y,\alpha Z],X]+[\alpha[Z,X],Y]\\&-[[Y,\alpha Z],X]-[[\alpha Z,X],Y]\\
&\stackrel{(2)}{=}[Y,\alpha[X,Z]]-[Y,[X,\alpha Z]]\\
&\stackrel{(5)}{=}-[Y,[X,\alpha Z]]
\end{align*}
and $\mathfrak{A(g)}$ is an ideal in $\mathfrak{g} \rtimes \mathfrak{A(g)}$. It is also a $\mathfrak{g}-$module, because $\operatorname{ad}$ is a Lie algebra homomorphism $(6)$ and therefore
\begin{align*}
&\stackrel{(5)}{=} [X,[Y,\alpha]]-[Y,[X,\alpha]]
\end{align*}
d) For the last equation with $\alpha \in \mathfrak{A(g)}$ and $X,Y,Z \in \mathfrak{g}$
$$[\alpha(X),[Y,Z]]+[\alpha(Y),[Z,X]]+[\alpha(Z),[X,Y]] =0\quad (7)$$
we have
\begin{align*}
[\alpha(X),[Y,Z]]&\stackrel{(3)}{=}-[Y,[Z,\alpha(X)]]-[Z,[\alpha(X),Y]]\\
&\stackrel{(1)}{=} [Y,[\alpha(Z),X]]+[Z,[X,\alpha(Y)]]\\
&\stackrel{(3)}{=} -[\alpha(Z),[X,Y]]-[X,[Y,\alpha(Z)]]\\
&-[X,[\alpha(Y),Z]]-[\alpha(Y),[Z,X]]\\
&\stackrel{(1)}{=}-[\alpha(Y),[Z,X]]-[\alpha(Z),[X,Y]]
\end{align*}
"Intermediate Math Challenge - May 2018"
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Section1.2Sets and Equivalence Relations
Subsection1.2.1Set Theory
A set is a well-defined collection of objects; that is, it is defined in such a manner that we can determine for any given object $$x$$ whether or not $$x$$ belongs to the set. The objects that belong to a set are called its elements or members. We will denote sets by capital letters, such as $$A$$ or $$X\text{;}$$ if $$a$$ is an element of the set $$A\text{,}$$ we write $$a \in A\text{.}$$
A set is usually specified either by listing all of its elements inside a pair of braces or by stating the property that determines whether or not an object $$x$$ belongs to the set. We might write
\begin{equation*} X = \{ x_1, x_2, \ldots, x_n \} \end{equation*}
for a set containing elements $$x_1, x_2, \ldots, x_n$$ or
\begin{equation*} X = \{ x :x \text{ satisfies }{\mathcal P}\} \end{equation*}
if each $$x$$ in $$X$$ satisfies a certain property $${\mathcal P}\text{.}$$ For example, if $$E$$ is the set of even positive integers, we can describe $$E$$ by writing either
\begin{equation*} E = \{2, 4, 6, \ldots \} \quad \text{or} \quad E = \{ x : x \text{ is an even integer and } x \gt 0 \}\text{.} \end{equation*}
We write $$2 \in E$$ when we want to say that 2 is in the set $$E\text{,}$$ and $$-3 \notin E$$ to say that $$-3$$ is not in the set $$E\text{.}$$
Some of the more important sets that we will consider are the following:
\begin{align*} {\mathbb N} &= \{n: n \text{ is a natural number}\} = \{1, 2, 3, \ldots \}\\ {\mathbb Z} &= \{n : n \text{ is an integer} \} = \{\ldots, -1, 0, 1, 2, \ldots \}\\ {\mathbb Q} &= \{r : r \text{ is a rational number}\} = \{p/q : p, q \in {\mathbb Z} \text{ where } q \neq 0\}\\ {\mathbb R} &= \{ x : x \text{ is a real number} \}\\ {\mathbb C} &= \{z : z \text{ is a complex number}\}\text{.} \end{align*}
We can find various relations between sets as well as perform operations on sets. A set $$A$$ is a subset of $$B\text{,}$$ written $$A \subset B$$ or $$B \supset A\text{,}$$ if every element of $$A$$ is also an element of $$B\text{.}$$ For example,
\begin{equation*} \{4,5,8\} \subset \{2, 3, 4, 5, 6, 7, 8, 9 \} \end{equation*}
and
\begin{equation*} {\mathbb N} \subset {\mathbb Z} \subset {\mathbb Q} \subset {\mathbb R} \subset {\mathbb C}\text{.} \end{equation*}
Trivially, every set is a subset of itself. A set $$B$$ is a proper subset of a set $$A$$ if $$B \subset A$$ but $$B \neq A\text{.}$$ If $$A$$ is not a subset of $$B\text{,}$$ we write $$A \notsubset B\text{;}$$ for example, $$\{4, 7, 9\} \notsubset \{2, 4, 5, 8, 9 \}\text{.}$$ Two sets are equal, written $$A = B\text{,}$$ if we can show that $$A \subset B$$ and $$B \subset A\text{.}$$
It is convenient to have a set with no elements in it. This set is called the empty set and is denoted by $$\emptyset\text{.}$$ Note that the empty set is a subset of every set.
To construct new sets out of old sets, we can perform certain operations: the union $$A \cup B$$ of two sets $$A$$ and $$B$$ is defined as
\begin{equation*} A \cup B = \{x : x \in A \text{ or } x \in B \} \end{equation*}
and the intersection of $$A$$ and $$B$$ is defined by
\begin{equation*} A \cap B = \{x : x \in A \text{ and } x \in B \}\text{.} \end{equation*}
If $$A = \{1, 3, 5\}$$ and $$B = \{ 1, 2, 3, 9 \}\text{,}$$ then
\begin{equation*} A \cup B = \{1, 2, 3, 5, 9 \} \quad \text{and} \quad A \cap B = \{ 1, 3 \}\text{.} \end{equation*}
We can consider the union and the intersection of more than two sets. In this case we write
\begin{equation*} \bigcup_{i = 1}^{n} A_{i} = A_{1} \cup \ldots \cup A_n \end{equation*}
and
\begin{equation*} \bigcap_{i = 1}^{n} A_{i} = A_{1} \cap \ldots \cap A_n \end{equation*}
for the union and intersection, respectively, of the sets $$A_1, \ldots, A_n\text{.}$$
When two sets have no elements in common, they are said to be disjoint; for example, if $$E$$ is the set of even integers and $$O$$ is the set of odd integers, then $$E$$ and $$O$$ are disjoint. Two sets $$A$$ and $$B$$ are disjoint exactly when $$A \cap B = \emptyset\text{.}$$
Sometimes we will work within one fixed set $$U\text{,}$$ called the universal set. For any set $$A \subset U\text{,}$$ we define the complement of $$A\text{,}$$ denoted by $$A'\text{,}$$ to be the set
\begin{equation*} A' = \{ x : x \in U \text{ and } x \notin A \}\text{.} \end{equation*}
We define the difference of two sets $$A$$ and $$B$$ to be
\begin{equation*} A \setminus B = A \cap B' = \{ x : x \in A \text{ and } x \notin B \}\text{.} \end{equation*}
Example1.2.1.Set Operations.
Let $${\mathbb R}$$ be the universal set and suppose that
\begin{equation*} A = \{ x \in {\mathbb R} : 0 \lt x \leq 3 \} \quad \text{and} \quad B = \{ x \in {\mathbb R} : 2 \leq x \lt 4 \}\text{.} \end{equation*}
Then
\begin{align*} A \cap B & = \{ x \in {\mathbb R} : 2 \leq x \leq 3 \}\\ A \cup B & = \{ x \in {\mathbb R} : 0 \lt x \lt 4 \}\\ A \setminus B & = \{ x \in {\mathbb R} : 0 \lt x \lt 2 \}\\ A' & = \{ x \in {\mathbb R} : x \leq 0 \text{ or } x \gt 3 \}\text{.} \end{align*}
Proof.
We will prove (1) and (3) and leave the remaining results to be proven in the exercises.
(1) Observe that
\begin{align*} A \cup A & = \{ x : x \in A \text{ or } x \in A \}\\ & = \{ x : x \in A \}\\ & = A\\ \end{align*}
and
\begin{align*} A \cap A & = \{ x : x \in A \text{ and } x \in A \}\\ & = \{ x : x \in A \}\\ & = A\text{.} \end{align*}
Also, $$A \setminus A = A \cap A' = \emptyset\text{.}$$
(3) For sets $$A\text{,}$$ $$B\text{,}$$ and $$C\text{,}$$
\begin{align*} A \cup (B \cup C) & = A \cup \{ x : x \in B \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B, \text{ or } x \in C \}\\ & = \{ x : x \in A \text{ or } x \in B \} \cup C\\ & = (A \cup B) \cup C. \end{align*}
A similar argument proves that $$A \cap (B \cap C) = (A \cap B) \cap C\text{.}$$
Proof.
(1) We must show that $$(A \cup B)' \subset A' \cap B'$$ and $$(A \cup B)' \supset A' \cap B'\text{.}$$ Let $$x \in (A \cup B)'\text{.}$$ Then $$x \notin A \cup B\text{.}$$ So $$x$$ is neither in $$A$$ nor in $$B\text{,}$$ by the definition of the union of sets. By the definition of the complement, $$x \in A'$$ and $$x \in B'\text{.}$$ Therefore, $$x \in A' \cap B'$$ and we have $$(A \cup B)' \subset A' \cap B'\text{.}$$
To show the reverse inclusion, suppose that $$x \in A' \cap B'\text{.}$$ Then $$x \in A'$$ and $$x \in B'\text{,}$$ and so $$x \notin A$$ and $$x \notin B\text{.}$$ Thus $$x \notin A \cup B$$ and so $$x \in (A \cup B)'\text{.}$$ Hence, $$(A \cup B)' \supset A' \cap B'$$ and so $$(A \cup B)' = A' \cap B'\text{.}$$
The proof of (2) is left as an exercise.
Example1.2.4.Other Relations on Sets.
Other relations between sets often hold true. For example,
\begin{equation*} ( A \setminus B) \cap (B \setminus A) = \emptyset\text{.} \end{equation*}
To see that this is true, observe that
\begin{align*} ( A \setminus B) \cap (B \setminus A) & = ( A \cap B') \cap (B \cap A')\\ & = A \cap A' \cap B \cap B'\\ & = \emptyset\text{.} \end{align*}
Subsection1.2.2Cartesian Products and Mappings
Given sets $$A$$ and $$B\text{,}$$ we can define a new set $$A \times B\text{,}$$ called the Cartesian product of $$A$$ and $$B\text{,}$$ as a set of ordered pairs. That is,
\begin{equation*} A \times B = \{ (a,b) : a \in A \text{ and } b \in B \}\text{.} \end{equation*}
Example1.2.5.Cartesian Products.
If $$A = \{ x, y \}\text{,}$$ $$B = \{ 1, 2, 3 \}\text{,}$$ and $$C = \emptyset\text{,}$$ then $$A \times B$$ is the set
\begin{equation*} \{ (x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3) \} \end{equation*}
and
\begin{equation*} A \times C = \emptyset\text{.} \end{equation*}
We define the Cartesian product of $$n$$ sets to be
\begin{equation*} A_1 \times \cdots \times A_n = \{ (a_1, \ldots, a_n): a_i \in A_i \text{ for } i = 1, \ldots, n \}\text{.} \end{equation*}
If $$A = A_1 = A_2 = \cdots = A_n\text{,}$$ we often write $$A^n$$ for $$A \times \cdots \times A$$ (where $$A$$ would be written $$n$$ times). For example, the set $${\mathbb R}^3$$ consists of all of 3-tuples of real numbers.
Subsets of $$A \times B$$ are called relations. We will define a mapping or function $$f \subset A \times B$$ from a set $$A$$ to a set $$B$$ to be the special type of relation where $$(a, b) \in f$$ if for every element $$a \in A$$ there exists a unique element $$b \in B\text{.}$$ Another way of saying this is that for every element in $$A\text{,}$$ $$f$$ assigns a unique element in $$B\text{.}$$ We usually write $$f:A \rightarrow B$$ or $$A \stackrel{f}{\rightarrow} B\text{.}$$ Instead of writing down ordered pairs $$(a,b) \in A \times B\text{,}$$ we write $$f(a) = b$$ or $$f : a \mapsto b\text{.}$$ The set $$A$$ is called the domain of $$f$$ and
\begin{equation*} f(A) = \{ f(a) : a \in A \} \subset B \end{equation*}
is called the range or image of $$f\text{.}$$ We can think of the elements in the function’s domain as input values and the elements in the function’s range as output values.
Example1.2.7.Mappings.
Suppose $$A = \{1, 2, 3 \}$$ and $$B = \{a, b, c \}\text{.}$$ In Figure 1.2.6 we define relations $$f$$ and $$g$$ from $$A$$ to $$B\text{.}$$ The relation $$f$$ is a mapping, but $$g$$ is not because $$1 \in A$$ is not assigned to a unique element in $$B\text{;}$$ that is, $$g(1) = a$$ and $$g(1) = b\text{.}$$
Given a function $$f : A \rightarrow B\text{,}$$ it is often possible to write a list describing what the function does to each specific element in the domain. However, not all functions can be described in this manner. For example, the function $$f: {\mathbb R} \rightarrow {\mathbb R}$$ that sends each real number to its cube is a mapping that must be described by writing $$f(x) = x^3$$ or $$f:x \mapsto x^3\text{.}$$
Consider the relation $$f : {\mathbb Q} \rightarrow {\mathbb Z}$$ given by $$f(p/q) = p\text{.}$$ We know that $$1/2 = 2/4\text{,}$$ but is $$f(1/2) = 1$$ or 2? This relation cannot be a mapping because it is not well-defined. A relation is well-defined if each element in the domain is assigned to a unique element in the range.
If $$f:A \rightarrow B$$ is a map and the image of $$f$$ is $$B\text{,}$$ i.e., $$f(A) = B\text{,}$$ then $$f$$ is said to be onto or surjective. In other words, if there exists an $$a \in A$$ for each $$b \in B$$ such that $$f(a) = b\text{,}$$ then $$f$$ is onto. A map is one-to-one or injective if $$a_1 \neq a_2$$ implies $$f(a_1) \neq f(a_2)\text{.}$$ Equivalently, a function is one-to-one if $$f(a_1) = f(a_2)$$ implies $$a_1 = a_2\text{.}$$ A map that is both one-to-one and onto is called bijective.
Example1.2.8.One-to-One and Onto Mappings.
Let $$f:{\mathbb Z} \rightarrow {\mathbb Q}$$ be defined by $$f(n) = n/1\text{.}$$ Then $$f$$ is one-to-one but not onto. Define $$g : {\mathbb Q} \rightarrow {\mathbb Z}$$ by $$g(p/q) = p$$ where $$p/q$$ is a rational number expressed in its lowest terms with a positive denominator. The function $$g$$ is onto but not one-to-one.
Given two functions, we can construct a new function by using the range of the first function as the domain of the second function. Let $$f : A \rightarrow B$$ and $$g : B \rightarrow C$$ be mappings. Define a new map, the composition of $$f$$ and $$g$$ from $$A$$ to $$C\text{,}$$ by $$(g \circ f)(x) = g(f(x))\text{.}$$
Example1.2.10.Composition of Mappings.
Consider the functions $$f: A \rightarrow B$$ and $$g: B \rightarrow C$$ that are defined in Figure 1.2.9 (top). The composition of these functions, $$g \circ f: A \rightarrow C\text{,}$$ is defined in Figure 1.2.9 (bottom).
Example1.2.11.Composition is not Commutative.
Let $$f(x) = x^2$$ and $$g(x) = 2x + 5\text{.}$$ Then
\begin{equation*} (f \circ g)(x) = f(g(x)) = (2x + 5)^2 = 4x^2 + 20x + 25 \end{equation*}
and
\begin{equation*} (g \circ f)(x) = g(f(x)) = 2x^2 + 5\text{.} \end{equation*}
In general, order makes a difference; that is, in most cases $$f \circ g \neq g \circ f\text{.}$$
Example1.2.12.Some Mappings Commute.
Sometimes it is the case that $$f \circ g= g \circ f\text{.}$$ Let $$f(x) = x^3$$ and $$g(x) = \sqrt[3]{x}\text{.}$$ Then
\begin{equation*} (f \circ g )(x) = f(g(x)) = f( \sqrt[3]{x}\, ) = (\sqrt[3]{x}\, )^3 = x \end{equation*}
and
\begin{equation*} (g \circ f )(x) = g(f(x)) = g( x^3) = \sqrt[3]{ x^3} = x\text{.} \end{equation*}
Example1.2.13.A Linear Map.
Given a $$2 \times 2$$ matrix
\begin{equation*} A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\text{,} \end{equation*}
we can define a map $$T_A : {\mathbb R}^2 \rightarrow {\mathbb R}^2$$ by
\begin{equation*} T_A (x,y) = (ax + by, cx +dy) \end{equation*}
for $$(x,y)$$ in $${\mathbb R}^2\text{.}$$ This is actually matrix multiplication; that is,
\begin{equation*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx +dy \end{pmatrix}\text{.} \end{equation*}
Maps from $${\mathbb R}^n$$ to $${\mathbb R}^m$$ given by matrices are called linear maps or linear transformations.
Example1.2.14.A Permutation.
Suppose that $$S = \{ 1,2,3 \}\text{.}$$ Define a map $$\pi :S\rightarrow S$$ by
\begin{equation*} \pi( 1 ) = 2, \qquad \pi( 2 ) = 1, \qquad \pi( 3 ) = 3\text{.} \end{equation*}
This is a bijective map. An alternative way to write $$\pi$$ is
\begin{equation*} \begin{pmatrix} 1 & 2 & 3 \\ \pi(1) & \pi(2) & \pi(3) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}\text{.} \end{equation*}
For any set $$S\text{,}$$ a one-to-one and onto mapping $$\pi : S \rightarrow S$$ is called a permutation of $$S\text{.}$$
Proof.
We will prove (1) and (3). Part (2) is left as an exercise. Part (4) follows directly from (2) and (3).
(1) We must show that
\begin{equation*} h \circ (g \circ f) = (h \circ g) \circ f\text{.} \end{equation*}
For $$a \in A$$ we have
\begin{align*} (h \circ (g \circ f))(a) & = h((g \circ f)(a))\\ & = h(g(f(a))) \\ & = (h \circ g)(f(a))\\ & = ((h \circ g) \circ f)(a)\text{.} \end{align*}
(3) Assume that $$f$$ and $$g$$ are both onto functions. Given $$c \in C\text{,}$$ we must show that there exists an $$a \in A$$ such that $$(g \circ f)(a) = g(f(a)) = c\text{.}$$ However, since $$g$$ is onto, there is an element $$b \in B$$ such that $$g(b) = c\text{.}$$ Similarly, there is an $$a \in A$$ such that $$f(a) = b\text{.}$$ Accordingly,
\begin{equation*} (g \circ f)(a) = g(f(a)) = g(b) = c\text{.} \end{equation*}
If $$S$$ is any set, we will use $$id_S$$ or $$id$$ to denote the identity mapping from $$S$$ to itself. Define this map by $$id(s) = s$$ for all $$s \in S\text{.}$$ A map $$g: B \rightarrow A$$ is an inverse mapping of $$f: A \rightarrow B$$ if $$g \circ f = id_A$$ and $$f \circ g = id_B\text{;}$$ in other words, the inverse function of a function simply “undoes” the function. A map is said to be invertible if it has an inverse. We usually write $$f^{-1}$$ for the inverse of $$f\text{.}$$
Example1.2.16.An Inverse Function.
The function $$f(x) = x^3$$ has inverse $$f^{-1}(x) = \sqrt[3]{x}$$ by Example 1.2.12.
Example1.2.17.Exponential and Logarithmic Functions are Inverses.
The natural logarithm and the exponential functions, $$f(x) = \ln x$$ and $$f^{-1}(x) = e^x\text{,}$$ are inverses of each other provided that we are careful about choosing domains. Observe that
\begin{equation*} f(f^{-1}(x)) = f(e^x) = \ln e^x = x \end{equation*}
and
\begin{equation*} f^{-1}(f(x)) = f^{-1}(\ln x) = e^{\ln x} = x \end{equation*}
whenever composition makes sense.
Example1.2.18.A Matrix Inverse Yields an Inverse of a Linear Map.
Suppose that
\begin{equation*} A = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\text{.} \end{equation*}
Then $$A$$ defines a map from $${\mathbb R}^2$$ to $${\mathbb R}^2$$ by
\begin{equation*} T_A (x,y) = (3x + y, 5x + 2y)\text{.} \end{equation*}
We can find an inverse map of $$T_A$$ by simply inverting the matrix $$A\text{;}$$ that is, $$T_A^{-1} = T_{A^{-1}}\text{.}$$ In this example,
\begin{equation*} A^{-1} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \end{equation*}
and hence, the inverse map is given by
\begin{equation*} T_A^{-1} (x,y) = (2x - y, -5x + 3y)\text{.} \end{equation*}
It is easy to check that
\begin{equation*} T^{-1}_A \circ T_A (x,y) = T_A \circ T_A^{-1} (x,y) = (x,y)\text{.} \end{equation*}
Not every map has an inverse. If we consider the map
\begin{equation*} T_B (x,y) = (3x , 0 ) \end{equation*}
given by the matrix
\begin{equation*} B = \begin{pmatrix} 3 & 0 \\ 0 & 0 \end{pmatrix}\text{,} \end{equation*}
then an inverse map would have to be of the form
\begin{equation*} T_B^{-1} (x,y) = (ax + by, cx +dy) \end{equation*}
and
\begin{equation*} (x,y) = T \circ T_B^{-1} (x,y) = (3ax + 3by, 0) \end{equation*}
for all $$x$$ and $$y\text{.}$$ Clearly this is impossible because $$y$$ might not be 0.
Example1.2.19.An Inverse Permutation.
Given the permutation
\begin{equation*} \pi = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \end{equation*}
on $$S = \{ 1,2,3 \}\text{,}$$ it is easy to see that the permutation defined by
\begin{equation*} \pi^{-1} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \end{equation*}
is the inverse of $$\pi\text{.}$$ In fact, any bijective mapping possesses an inverse, as we will see in the next theorem.
Proof.
Suppose first that $$f:A \rightarrow B$$ is invertible with inverse $$g: B \rightarrow A\text{.}$$ Then $$g \circ f = id_A$$ is the identity map; that is, $$g(f(a)) = a\text{.}$$ If $$a_1, a_2 \in A$$ with $$f(a_1) = f(a_2)\text{,}$$ then $$a_1 = g(f(a_1)) = g(f(a_2)) = a_2\text{.}$$ Consequently, $$f$$ is one-to-one. Now suppose that $$b \in B\text{.}$$ To show that $$f$$ is onto, it is necessary to find an $$a \in A$$ such that $$f(a) = b\text{,}$$ but $$f(g(b)) = b$$ with $$g(b) \in A\text{.}$$ Let $$a = g(b)\text{.}$$
Conversely, let $$f$$ be bijective and let $$b \in B\text{.}$$ Since $$f$$ is onto, there exists an $$a \in A$$ such that $$f(a) = b\text{.}$$ Because $$f$$ is one-to-one, $$a$$ must be unique. Define $$g$$ by letting $$g(b) = a\text{.}$$ We have now constructed the inverse of $$f\text{.}$$
Subsection1.2.3Equivalence Relations and Partitions
A fundamental notion in mathematics is that of equality. We can generalize equality with equivalence relations and equivalence classes. An equivalence relation on a set $$X$$ is a relation $$R \subset X \times X$$ such that
• $$(x, x) \in R$$ for all $$x \in X$$ (reflexive property);
• $$(x, y) \in R$$ implies $$(y, x) \in R$$ (symmetric property);
• $$(x, y)$$ and $$(y, z) \in R$$ imply $$(x, z) \in R$$ (transitive property).
Given an equivalence relation $$R$$ on a set $$X\text{,}$$ we usually write $$x \sim y$$ instead of $$(x, y) \in R\text{.}$$ If the equivalence relation already has an associated notation such as $$=\text{,}$$ $$\equiv\text{,}$$ or $$\cong\text{,}$$ we will use that notation.
Example1.2.21.Equivalent Fractions.
Let $$p\text{,}$$ $$q\text{,}$$ $$r\text{,}$$ and $$s$$ be integers, where $$q$$ and $$s$$ are nonzero. Define $$p/q \sim r/s$$ if $$ps = qr\text{.}$$ Clearly $$\sim$$ is reflexive and symmetric. To show that it is also transitive, suppose that $$p/q \sim r/s$$ and $$r/s \sim t/u\text{,}$$ with $$q\text{,}$$ $$s\text{,}$$ and $$u$$ all nonzero. Then $$ps = qr$$ and $$ru = st\text{.}$$ Therefore,
\begin{equation*} psu = qru = qst\text{.} \end{equation*}
Since $$s \neq 0\text{,}$$ $$pu = qt\text{.}$$ Consequently, $$p/q \sim t/u\text{.}$$
Example1.2.22.An Equivalence Relation From Derivatives.
Suppose that $$f$$ and $$g$$ are differentiable functions on $${\mathbb R}\text{.}$$ We can define an equivalence relation on such functions by letting $$f(x) \sim g(x)$$ if $$f'(x) = g'(x)\text{.}$$ It is clear that $$\sim$$ is both reflexive and symmetric. To demonstrate transitivity, suppose that $$f(x) \sim g(x)$$ and $$g(x) \sim h(x)\text{.}$$ From calculus we know that $$f(x) - g(x) = c_1$$ and $$g(x)- h(x) = c_2\text{,}$$ where $$c_1$$ and $$c_2$$ are both constants. Hence,
\begin{equation*} f(x) - h(x) = ( f(x) - g(x)) + ( g(x)- h(x)) = c_1 - c_2 \end{equation*}
and $$f'(x) - h'(x) = 0\text{.}$$ Therefore, $$f(x) \sim h(x)\text{.}$$
Example1.2.23.Equivalent Circles.
For $$(x_1, y_1 )$$ and $$(x_2, y_2)$$ in $${\mathbb R}^2\text{,}$$ define $$(x_1, y_1 ) \sim (x_2, y_2)$$ if $$x_1^2 + y_1^2 = x_2^2 + y_2^2\text{.}$$ Then $$\sim$$ is an equivalence relation on $${\mathbb R}^2\text{.}$$
Example1.2.24.Equivalent Matrices.
Let $$A$$ and $$B$$ be $$2 \times 2$$ matrices with entries in the real numbers. We can define an equivalence relation on the set of $$2 \times 2$$ matrices, by saying $$A \sim B$$ if there exists an invertible matrix $$P$$ such that $$PAP^{-1} = B\text{.}$$ For example, if
\begin{equation*} A = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix} \quad \text{and} \quad B = \begin{pmatrix} -18 & 33 \\ -11 & 20 \end{pmatrix}\text{,} \end{equation*}
then $$A \sim B$$ since $$PAP^{-1} = B$$ for
\begin{equation*} P = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\text{.} \end{equation*}
Let $$I$$ be the $$2 \times 2$$ identity matrix; that is,
\begin{equation*} I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\text{.} \end{equation*}
Then $$IAI^{-1} = IAI = A\text{;}$$ therefore, the relation is reflexive. To show symmetry, suppose that $$A \sim B\text{.}$$ Then there exists an invertible matrix $$P$$ such that $$PAP^{-1} = B\text{.}$$ So
\begin{equation*} A = P^{-1} B P = P^{-1} B (P^{-1})^{-1}\text{.} \end{equation*}
Finally, suppose that $$A \sim B$$ and $$B \sim C\text{.}$$ Then there exist invertible matrices $$P$$ and $$Q$$ such that $$PAP^{-1} = B$$ and $$QBQ^{-1} = C\text{.}$$ Since
\begin{equation*} C = QBQ^{-1} = QPAP^{-1} Q^{-1} = (QP)A(QP)^{-1}\text{,} \end{equation*}
the relation is transitive. Two matrices that are equivalent in this manner are said to be similar.
A partition $${\mathcal P}$$ of a set $$X$$ is a collection of nonempty sets $$X_1, X_2, \ldots$$ such that $$X_i \cap X_j = \emptyset$$ for $$i \neq j$$ and $$\bigcup_k X_k = X\text{.}$$ Let $$\sim$$ be an equivalence relation on a set $$X$$ and let $$x \in X\text{.}$$ Then $$[x] = \{ y \in X : y \sim x \}$$ is called the equivalence class of $$x\text{.}$$ We will see that an equivalence relation gives rise to a partition via equivalence classes. Also, whenever a partition of a set exists, there is some natural underlying equivalence relation, as the following theorem demonstrates.
Proof.
Suppose there exists an equivalence relation $$\sim$$ on the set $$X\text{.}$$ For any $$x \in X\text{,}$$ the reflexive property shows that $$x \in [x]$$ and so $$[x]$$ is nonempty. Clearly $$X = \bigcup_{x \in X} [x]\text{.}$$ Now let $$x, y \in X\text{.}$$ We need to show that either $$[x] = [y]$$ or $$[x] \cap [y] = \emptyset\text{.}$$ Suppose that the intersection of $$[x]$$ and $$[y]$$ is not empty and that $$z \in [x] \cap [y]\text{.}$$ Then $$z \sim x$$ and $$z \sim y\text{.}$$ By symmetry and transitivity $$x \sim y\text{;}$$ hence, $$[x] \subset [y]\text{.}$$ Similarly, $$[y] \subset [x]$$ and so $$[x] = [y]\text{.}$$ Therefore, any two equivalence classes are either disjoint or exactly the same.
Conversely, suppose that $${\mathcal P} = \{X_i\}$$ is a partition of a set $$X\text{.}$$ Let two elements be equivalent if they are in the same partition. Clearly, the relation is reflexive. If $$x$$ is in the same partition as $$y\text{,}$$ then $$y$$ is in the same partition as $$x\text{,}$$ so $$x \sim y$$ implies $$y \sim x\text{.}$$ Finally, if $$x$$ is in the same partition as $$y$$ and $$y$$ is in the same partition as $$z\text{,}$$ then $$x$$ must be in the same partition as $$z\text{,}$$ and transitivity holds.
Let us examine some of the partitions given by the equivalence classes in the last set of examples.
Example1.2.27.A Partition of Fractions.
In the equivalence relation in Example 1.2.21, two pairs of integers, $$(p,q)$$ and $$(r,s)\text{,}$$ are in the same equivalence class when they reduce to the same fraction in its lowest terms.
Example1.2.28.A Partition of Functions.
In the equivalence relation in Example 1.2.22, two functions $$f(x)$$ and $$g(x)$$ are in the same partition when they differ by a constant.
Example1.2.29.A Partition of Circles.
We defined an equivalence class on $${\mathbb R}^2$$ by $$(x_1, y_1 ) \sim (x_2, y_2)$$ if $$x_1^2 + y_1^2 = x_2^2 + y_2^2\text{.}$$ Two pairs of real numbers are in the same partition when they lie on the same circle about the origin.
Example1.2.30.A Partition of Integers.
Let $$r$$ and $$s$$ be two integers and suppose that $$n \in {\mathbb N}\text{.}$$ We say that $$r$$ is congruent to $$s$$ modulo $$n\text{,}$$ or $$r$$ is congruent to $$s$$ mod $$n\text{,}$$ if $$r - s$$ is evenly divisible by $$n\text{;}$$ that is, $$r - s = nk$$ for some $$k \in {\mathbb Z}\text{.}$$ In this case we write $$r \equiv s \pmod{n}\text{.}$$ For example, $$41 \equiv 17 \pmod{ 8}$$ since $$41 - 17=24$$ is divisible by 8. We claim that congruence modulo $$n$$ forms an equivalence relation of $${\mathbb Z}\text{.}$$ Certainly any integer $$r$$ is equivalent to itself since $$r - r = 0$$ is divisible by $$n\text{.}$$ We will now show that the relation is symmetric. If $$r \equiv s \pmod{ n}\text{,}$$ then $$r - s = -(s -r)$$ is divisible by $$n\text{.}$$ So $$s - r$$ is divisible by $$n$$ and $$s \equiv r \pmod{ n}\text{.}$$ Now suppose that $$r \equiv s \pmod{ n}$$ and $$s \equiv t \pmod{ n}\text{.}$$ Then there exist integers $$k$$ and $$l$$ such that $$r -s = kn$$ and $$s - t = ln\text{.}$$ To show transitivity, it is necessary to prove that $$r - t$$ is divisible by $$n\text{.}$$ However,
\begin{equation*} r - t = r - s + s - t = kn + ln = (k + l)n\text{,} \end{equation*}
and so $$r - t$$ is divisible by $$n\text{.}$$
If we consider the equivalence relation established by the integers modulo 3, then
\begin{align*} {[0]} & = \{ \ldots, -3, 0, 3, 6, \ldots \}\\ {[1]} & = \{ \ldots, -2, 1, 4, 7, \ldots \}\\ {[2]} & = \{ \ldots, -1, 2, 5, 8, \ldots \}\text{.} \end{align*}
Notice that $$[0] \cup [1] \cup [2] = {\mathbb Z}$$ and also that the sets are disjoint. The sets $$[0]\text{,}$$ $$[1]\text{,}$$ and $$[2]$$ form a partition of the integers.
The integers modulo $$n$$ are a very important example in the study of abstract algebra and will become quite useful in our investigation of various algebraic structures such as groups and rings. In our discussion of the integers modulo $$n$$ we have actually assumed a result known as the division algorithm, which will be stated and proved in Chapter 2. | 9,777 | 26,478 | {"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} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.78799 |
http://m.basicmusictheory.com/a-flat-suspended-2nd-triad-chord | 1,611,638,512,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704798089.76/warc/CC-MAIN-20210126042704-20210126072704-00748.warc.gz | 61,583,455 | 9,628 | # A-flat suspended 2nd triad chord
The Solution below shows the A-flat suspended 2nd triad chord in root position, 1st inversion and 2nd inversion on the piano, treble clef and bass clef.
The Lesson steps then explain how to construct this triad chord using the 3rd and 5th note intervals, then finally how to construct the inverted chord variations.
For a quick summary of this topic, have a look at Triad chord.
## Solution - 3 parts
### 1. A-flat suspended 2nd chord
This step shows the A-flat suspended 2nd triad chord in root position on the piano, treble clef and bass clef.
The A-flat suspended 2nd chord contains 3 notes: Ab, Bb, Eb.
The chord spelling / formula relative to the Ab major scale is: 1 2 5.
A-flat suspended 2nd chord note names
Note no.Note intervalSpelling
/ formula
Note name#Semitones
from root
1root1The 1st note of the A-flat suspended 2nd chord is Ab0
2Ab-maj-2nd2The 2nd note of the A-flat suspended 2nd chord is Bb2
3Ab-perf-5th5The 3rd note of the A-flat suspended 2nd chord is Eb7
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 5/2.
The staff diagrams and audio files contain each note individually, ascending from the root, followed by the chord containing all 3 notes.
### 2. A-flat suspended 2nd 1st inversion
This step shows the A-flat suspended 2nd 1st inversion on the piano, treble clef and bass clef.
The A-flat suspended 2nd 1st inversion contains 3 notes: Bb, Eb, Ab.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord inversion are 7/4, so the chord is said to be in seven-four position.
### 3. A-flat suspended 2nd 2nd inversion
This step shows the A-flat suspended 2nd 2nd inversion on the piano, treble clef and bass clef.
The A-flat suspended 2nd 2nd inversion contains 3 notes: Eb, Ab, Bb.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord inversion are 5/4, so the chord is said to be in five-four position.
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
### 2. A-flat tonic note and one octave of notes
This step shows 1 octave of notes starting from note Ab, to identify the start and end notes of the scale used to build this chord.
The numbered notes are those that might be used when building this chord.
Note 1 is the root note - the starting note of the chord - Ab, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 Ab A A# / Bb B C C# / Db D D# / Eb E F F# / Gb G Ab
### 3. A-flat major scale note interval positions
This step describes the Ab major scale, whose note intervals are used to define the chord in a later step.
The major scale uses the W-W-H-W-W-W-H note counting rule to identify the scale note positions.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
The tonic note (shown as *) is the starting point and is always the 1st note in the major scale.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 Ab A# / Bb C C# / Db D# / Eb F G Ab
### 4. A-flat major scale note interval numbers
This step identifies the note interval numbers of each scale note, which are used to calculate the chord note names in a later step.
To identify the note interval numbers for this major scale, just assign each note position from the previous step, with numbers ascending from 1 to 8.
No. Note 1 2 3 4 5 6 7 8 Ab Bb C Db Eb F G Ab
To understand why the note names of this major scale have these specific sharp and flat names, have a look at the Ab major scale page.
Both the note interval numbers and note names from the piano diagram above will be used in later steps to calculate the chord note names.
### 5. Triad chord qualities
This step defines a triad chord, names the triad chord qualities and identifies the notes that vary between them.
#### Triad chord definition
The music theory term triad chord means that 3 or more notes played together, or overlapping.
#### Triad chord qualities
Triad chords exist in four different chord qualities, which are major, minor, augmented, and diminished.
Each chord quality name is the name of the entire chord as a whole, not its individual notes (which will be covered later).
#### Triad chord qualities using the 1st, 3rd and 5th scale notes
All of these triad qualities are based on the 1st, 3rd and 5th notes of the major scale piano diagram above.
Depending on the chord quality, the 3rd and 5th scale note names of the major scale above might need to be adjusted up or down by one half-note / semitone / piano key.
It is these variations of the 3rd and 5th notes that give each one a distinctive sound for any given key (eg. C-flat, E etc).
#### Suspended triad chords - using the 2nd or 4th scale notes
A suspended chord is known in music theory as an altered chord because it takes one of the above chord qualities and modifies it in some way.
Unlike all of the above qualities, Suspended triad chords do not use the 3rd note of the major scale (at all) to build the chord.
The 3rd note is suspended, ie. removed completely, and replaced by either the 2nd note of the major scale - a suspended 2nd, or more commonly by the 4th note of the major scale - a suspended 4th.
Musically, this is interesting, since it is usually the 3rd note of the scale that defines the overall character of the chord as being major (typically described as 'happy') or minor ('sad').
Without this 3rd note, suspended chords tend to have an open and ambiguous sound.
The steps below will detail the suspended 2nd triad chord quality in the key of Ab.
### 6. Triad chord note intervals
This step defines the note intervals for each chord quality, including the intervals for the A-flat suspended 2nd triad chord.
Each individual note in a triad chord can be represented in music theory using a note interval, which is used to express the relationship between the first note of the chord (the root note), and the note in question.
The root note is always the 1st note (note interval 1 in the above diagram) of the major scale diagram above. ie. the tonic of the major scale.
Then there is one note interval to describe the 2nd note, and another to describe the 3rd note of the chord.
In the same way that the entire chord itself has a chord quality, the intervals representing the individual notes within that chord each have their own quality.
These note interval qualities are diminished, minor, major, perfect and augmented.
Below is a table showing the note interval qualities for all triad chords, together with the interval short names / abbrevations in brackets.
Triad chord note interval qualities
Chord quality2nd note quality3rd note quality
majormajor (M3)perfect (P5)
minorminor (m3)perfect (P5)
augmentedmajor (M3)augmented (A5)
diminishedminor (m3)diminished (d5)
suspended
(2nd/4th)
major (M2) or
perfect (P4)
perfect (P5)
The numbers in brackets are the note interval numbers (ie the scale note number) shown in the previous step.
#### A-flat triad chord note intervals
Looking at the table above, the note intervals for the chord quality we are interested in (suspended 2nd triad), in the key of Ab are Ab-maj-2nd and Ab-perf-5th.
The links above explain in detail the meaning of these note qualities, the short abbrevations in brackets, and how to calculate the interval note names based on the scale note names from the previous step.
### 7. A-flat suspended 2nd triad chord in root position
This step shows the A-flat suspended 2nd triad chord note interval names and note positions on a piano diagram.
#### Note name adjustments
Each note interval quality (diminished, minor, major, perfect, augmented) expresses a possible adjustment ie. a possible increase or decrease in the note pitch from the major scale notes in step 4.
If an adjustment in the pitch occurs, the note name given in the major scale in step 4 is modified, so that sharp or flat accidentals will be added or removed.
But crucially, for all interval qualities, the starting point from which accidentals need to be added or removed are the major scale note names in step 4.
For this chord, this is explained in detail in Ab-maj-2nd and Ab-perf-5th, but the relevant adjustments for this suspended 2nd chord quality are shown below:
Ab-2nd: Since the 2nd note quality of the major scale is major, and the note interval quality needed is major also, no adjustment needs to be made. The 2nd note name - Bb, is used, and the chord note spelling is 2.
Ab-5th: Since the 5th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 5th note name - Eb, is used, and the chord note spelling is 5.
#### A-flat suspended 2nd triad chord note names
The final chord note names and note interval links are shown in the table below.
Note Interval No. Interval def. Ab Bb Eb 1 2 5 root Ab-maj-2nd Ab-perf-5th 1 2 5 0 2 7
The piano diagram below shows the interval short names, the note positions and the final note names of this triad chord.
In music theory, this triad chord as it stands is said to be in root position because the root of the chord - note Ab, is the note with the lowest pitch of all the triad notes.
The note order of this triad can also be changed, so that the root is no longer the lowest note, in which case the triad is no longer in root position, and will be called an inverted triad chord instead.
For triad chords, there are 2 possible inverted variations as described in the steps below.
#### Figured bass notation
The figured bass notation for a triad in root position is 5/3, with the 5 placed above the 3 on a staff diagram.
These numbers represent the interval between the lowest note of the chord and the note in question.
So another name for this chord would be A-flat suspended 2nd triad in five-three position.
For example, the 5 represents note Eb, from the Ab-5th interval, since the triad root, Ab, is the lowest note of the chord (as it is not inverted).
In the same way, the figured bass 2 symbol represents note Eb, from the Ab-2nd interval.
Since figured bass notation works within the context of a key, we don't need to indicate in the figured bass symbols whether eg. the 3rd is a major, minor etc. The key is assumed from the key signature.
Often the 5 symbol is not shown at all, and only the number 2 symbol is shown - the 5th is assumed.
### 8. A-flat suspended 2nd 1st inversion
This step shows the first inversion of the A-flat suspended 2nd triad chord.
#### Structure
To invert a chord, simply take the first note of the chord to be inverted (the lowest in pitch) and move it up an octave to the end of the chord.
So for a 1st inversion, take the root of the triad chord in root position from the step above - note Ab, and move it up one octave (12 notes) so it is the last (highest) note in the chord.
The second note of the original triad (in root position) - note Bb is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this triad in 1st inversion is 7/4, with the 7 placed above the 4 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be A-flat suspended 2nd triad in seven-four position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original triad root!), and the note in question.
For example, the 7 represents note Ab, from the Bb-7th interval, since the lowest (bass) note of the chord - now inverted, is Bb.
In the same way, the figured bass 4 symbol represents note Eb, from the Bb-4th interval.
### 9. A-flat suspended 2nd 2nd inversion
This step shows the second inversion of the A-flat suspended 2nd triad chord.
#### Structure
For a 2nd inversion, take the first note of the 1st inversion above - Bb, and move it to the end of the chord.
So the second note of the 1st inversion - note Eb is now the note with the lowest pitch for the 2nd inversion.
Or put another way, the third note of the original triad (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this triad in 2nd inversion is 5/4, with the 5 placed above the 4 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be A-flat suspended 2nd triad in five-four position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original triad root!), and the note in question.
For example, the 5 represents note Bb, from the Eb-5th interval, since the lowest (bass) note of the chord - now inverted, is Eb.
In the same way, the figured bass 4 symbol represents note Ab, from the Eb-4th interval.
Often the 5 symbol is not shown at all, and only the number 4 symbol is shown - the 5th is assumed. | 3,470 | 13,920 | {"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-2021-04 | latest | en | 0.898561 |
https://www.quizzes.cc/metric/percentof.php?percent=72.4&of=1110 | 1,591,156,052,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00107.warc.gz | 873,942,998 | 4,406 | What is 72.4 percent of 1,110?
How much is 72.4 percent of 1110? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 72.4% of 1110 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
72.4% of 1,110 = 803.64
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Calculating seventy-two point four of one thousand, one hundred and ten How to calculate 72.4% of 1110? Simply divide the percent by 100 and multiply by the number. For example, 72.4 /100 x 1110 = 803.64 or 0.724 x 1110 = 803.64
How much is 72.4 percent of the following numbers?
72.4% of 1110.01 = 80364.724 72.4% of 1110.02 = 80365.448 72.4% of 1110.03 = 80366.172 72.4% of 1110.04 = 80366.896 72.4% of 1110.05 = 80367.62 72.4% of 1110.06 = 80368.344 72.4% of 1110.07 = 80369.068 72.4% of 1110.08 = 80369.792 72.4% of 1110.09 = 80370.516 72.4% of 1110.1 = 80371.24 72.4% of 1110.11 = 80371.964 72.4% of 1110.12 = 80372.688 72.4% of 1110.13 = 80373.412 72.4% of 1110.14 = 80374.136 72.4% of 1110.15 = 80374.86 72.4% of 1110.16 = 80375.584 72.4% of 1110.17 = 80376.308 72.4% of 1110.18 = 80377.032 72.4% of 1110.19 = 80377.756 72.4% of 1110.2 = 80378.48 72.4% of 1110.21 = 80379.204 72.4% of 1110.22 = 80379.928 72.4% of 1110.23 = 80380.652 72.4% of 1110.24 = 80381.376 72.4% of 1110.25 = 80382.1
72.4% of 1110.26 = 80382.824 72.4% of 1110.27 = 80383.548 72.4% of 1110.28 = 80384.272 72.4% of 1110.29 = 80384.996 72.4% of 1110.3 = 80385.72 72.4% of 1110.31 = 80386.444 72.4% of 1110.32 = 80387.168 72.4% of 1110.33 = 80387.892 72.4% of 1110.34 = 80388.616 72.4% of 1110.35 = 80389.34 72.4% of 1110.36 = 80390.064 72.4% of 1110.37 = 80390.788 72.4% of 1110.38 = 80391.512 72.4% of 1110.39 = 80392.236 72.4% of 1110.4 = 80392.96 72.4% of 1110.41 = 80393.684 72.4% of 1110.42 = 80394.408 72.4% of 1110.43 = 80395.132 72.4% of 1110.44 = 80395.856 72.4% of 1110.45 = 80396.58 72.4% of 1110.46 = 80397.304 72.4% of 1110.47 = 80398.028 72.4% of 1110.48 = 80398.752 72.4% of 1110.49 = 80399.476 72.4% of 1110.5 = 80400.2
72.4% of 1110.51 = 80400.924 72.4% of 1110.52 = 80401.648 72.4% of 1110.53 = 80402.372 72.4% of 1110.54 = 80403.096 72.4% of 1110.55 = 80403.82 72.4% of 1110.56 = 80404.544 72.4% of 1110.57 = 80405.268 72.4% of 1110.58 = 80405.992 72.4% of 1110.59 = 80406.716 72.4% of 1110.6 = 80407.44 72.4% of 1110.61 = 80408.164 72.4% of 1110.62 = 80408.888 72.4% of 1110.63 = 80409.612 72.4% of 1110.64 = 80410.336 72.4% of 1110.65 = 80411.06 72.4% of 1110.66 = 80411.784 72.4% of 1110.67 = 80412.508 72.4% of 1110.68 = 80413.232 72.4% of 1110.69 = 80413.956 72.4% of 1110.7 = 80414.68 72.4% of 1110.71 = 80415.404 72.4% of 1110.72 = 80416.128 72.4% of 1110.73 = 80416.852 72.4% of 1110.74 = 80417.576 72.4% of 1110.75 = 80418.3
72.4% of 1110.76 = 80419.024 72.4% of 1110.77 = 80419.748 72.4% of 1110.78 = 80420.472 72.4% of 1110.79 = 80421.196 72.4% of 1110.8 = 80421.92 72.4% of 1110.81 = 80422.644 72.4% of 1110.82 = 80423.368 72.4% of 1110.83 = 80424.092 72.4% of 1110.84 = 80424.816 72.4% of 1110.85 = 80425.54 72.4% of 1110.86 = 80426.264 72.4% of 1110.87 = 80426.988 72.4% of 1110.88 = 80427.712 72.4% of 1110.89 = 80428.436 72.4% of 1110.9 = 80429.16 72.4% of 1110.91 = 80429.884 72.4% of 1110.92 = 80430.608 72.4% of 1110.93 = 80431.332 72.4% of 1110.94 = 80432.056 72.4% of 1110.95 = 80432.78 72.4% of 1110.96 = 80433.504 72.4% of 1110.97 = 80434.228 72.4% of 1110.98 = 80434.952 72.4% of 1110.99 = 80435.676 72.4% of 1111 = 80436.4
1% of 1110 = 11.1 2% of 1110 = 22.2 3% of 1110 = 33.3 4% of 1110 = 44.4 5% of 1110 = 55.5 6% of 1110 = 66.6 7% of 1110 = 77.7 8% of 1110 = 88.8 9% of 1110 = 99.9 10% of 1110 = 111 11% of 1110 = 122.1 12% of 1110 = 133.2 13% of 1110 = 144.3 14% of 1110 = 155.4 15% of 1110 = 166.5 16% of 1110 = 177.6 17% of 1110 = 188.7 18% of 1110 = 199.8 19% of 1110 = 210.9 20% of 1110 = 222 21% of 1110 = 233.1 22% of 1110 = 244.2 23% of 1110 = 255.3 24% of 1110 = 266.4 25% of 1110 = 277.5
26% of 1110 = 288.6 27% of 1110 = 299.7 28% of 1110 = 310.8 29% of 1110 = 321.9 30% of 1110 = 333 31% of 1110 = 344.1 32% of 1110 = 355.2 33% of 1110 = 366.3 34% of 1110 = 377.4 35% of 1110 = 388.5 36% of 1110 = 399.6 37% of 1110 = 410.7 38% of 1110 = 421.8 39% of 1110 = 432.9 40% of 1110 = 444 41% of 1110 = 455.1 42% of 1110 = 466.2 43% of 1110 = 477.3 44% of 1110 = 488.4 45% of 1110 = 499.5 46% of 1110 = 510.6 47% of 1110 = 521.7 48% of 1110 = 532.8 49% of 1110 = 543.9 50% of 1110 = 555
51% of 1110 = 566.1 52% of 1110 = 577.2 53% of 1110 = 588.3 54% of 1110 = 599.4 55% of 1110 = 610.5 56% of 1110 = 621.6 57% of 1110 = 632.7 58% of 1110 = 643.8 59% of 1110 = 654.9 60% of 1110 = 666 61% of 1110 = 677.1 62% of 1110 = 688.2 63% of 1110 = 699.3 64% of 1110 = 710.4 65% of 1110 = 721.5 66% of 1110 = 732.6 67% of 1110 = 743.7 68% of 1110 = 754.8 69% of 1110 = 765.9 70% of 1110 = 777 71% of 1110 = 788.1 72% of 1110 = 799.2 73% of 1110 = 810.3 74% of 1110 = 821.4 75% of 1110 = 832.5
76% of 1110 = 843.6 77% of 1110 = 854.7 78% of 1110 = 865.8 79% of 1110 = 876.9 80% of 1110 = 888 81% of 1110 = 899.1 82% of 1110 = 910.2 83% of 1110 = 921.3 84% of 1110 = 932.4 85% of 1110 = 943.5 86% of 1110 = 954.6 87% of 1110 = 965.7 88% of 1110 = 976.8 89% of 1110 = 987.9 90% of 1110 = 999 91% of 1110 = 1010.1 92% of 1110 = 1021.2 93% of 1110 = 1032.3 94% of 1110 = 1043.4 95% of 1110 = 1054.5 96% of 1110 = 1065.6 97% of 1110 = 1076.7 98% of 1110 = 1087.8 99% of 1110 = 1098.9 100% of 1110 = 1110 | 3,109 | 5,614 | {"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} | 3.671875 | 4 | CC-MAIN-2020-24 | latest | en | 0.846394 |
zvf.elifhazel.co | 1,627,350,477,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00593.warc.gz | 1,165,266,799 | 10,133 | 2. The object in the diagram below is on a fixed frictionless axle. It has a moment of inertia of I = 50 kgm2. The forces acting on the object are F1 = 100 N, F2 = 200 N, and F3 = 250 N acting at different radii R1 =
A 15 kg block rests on an inclined plane. The plane makes an angle of 25 o with the horizontal, and the coefficient of friction between the block and the plane is 0.13. The 15 kg block is tied to a second block (mass=38 kg) which hangs over the end of the inclined plane after the rope passes over an ideal pulley.
Q10. (a)€€€€The diagram shows two forces acting on an object. € What is the resultant force acting on the object? Tick (€ ) one box. (1) € 8 N to the right € 8 N to the left € 4 N to the right € 4 N to the left Page 17 of 44
All of the examples mentioned involve objects with mass falling to the ground, they are being attracted to the Earth. Gravity is the name we give to the force that arises between objects because of their mass. It is always an attractive force. It is a non-contact force, therefore it acts at a distance.
Feb 06, 2008 · two forces act on a 16kg object. the frist force has a magnitude of 68 N and is directed 24 dedrees north of east . the second force is 32 N and 48 degrees north of west. what is the acceleration of the of the object resulting from the action of these two forces?
A constant force, F = (4.63, -3.79, 2.16) N, acts on an object of mass 17.7 kg, causing a displacement of that object by r = (4.10, 3.86, -2.45) m. What is the total work done by this force? View ...
The net external force acting on the two-object system is zero. After the collision, object 1 has a momentum whose y component is 5kg m/s. What is the y component of the momentum of object 2 after the collision? (a) 0 kg m/s (b) 16 kg m/s (c) 5kg m/s. (d) 16 kg m/s (e) The y component of the momentum of object 2 cannot be determined.
7. What is the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s2? 8. What is the acceleration of a wagon of mass 20 kg if a horizontal force of 64 N is applied to it? (ignore friction) 9. What is the mass of an object that is experiencing a net force of 200 N and an acceleration of 500 m/s2? 10. (c) 16.7 kg (d) 58.8 kg (e) IOO kg (b) 10.2 kg (a) 1.70 kg Two horizontal forces act on a 5.0-kilogram object. One force has a magnitude of 8.0 N and is directed due north. The second force toward the east has a magnitude of 6.0 N. What is the acceleration of the object? (c) 2.0 at 530 N ofE (d) 2.0 m/s2 E ofN
Find right answers right now! 'The two forces F⃗ 1 and F⃗ 2 shown in the figure act on a 19.0-kg object on a frictionless tabletop? More questions about Science & Mathematics, how
The net external force acting on the two-object system is zero. After the collision, object 1 has a momentum whose y component is 5kg m/s. What is the y component of the momentum of object 2 after the collision? (a) 0 kg m/s (b) 16 kg m/s (c) 5kg m/s. (d) 16 kg m/s (e) The y component of the momentum of object 2 cannot be determined.
Friction is a force parallel to the surface of an object and opposite its motion. It occurs when two objects are sliding against each other (kinetic friction), or when a stationary object is placed on an inclined surface (static friction). This force is employed when setting objects in motion, for example, wheels grip to the ground due to friction.
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The only force acting on a 2.0-kg body as it moves... If the scalar product of two vectors, and , is equ... Starting from rest at t = 0, a 5.0-kg block is pul... A body moving along the x axis is acted upon by a ... A 2.5-kg object falls vertically downward in a vis... The same constant force is used to accelerate two ... If , , and , what is ... 18. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example. 19. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2) 20.
All the forces acting upon an object are balanced. Answer: DEFG. Since there are two forces pulling upwards and since the sign is hanging symmetrically, each force must supply an upwards pull equal to one-half the Solving for Ftens yields 800 N. 16. A quick blank is Fgrav: Fgrav = m•g = ~80 N.
Two forces F with arrow1 and F with arrow2 act on a 8.60-kg object. F1 = 30.0 N and F2 = 10.0 N. (a) Find the acceleration of the object for the configuration of forces shown in Figure (a). magnitude m/s2 direction ° (counterclockwise from F with arrow1) (b) Find the acceleration of the object for the configuration of forces shown in Figure (b). magnitude m/s2 direction ° (counterclockwise ...
When two forces act on an object simultaneously: The magnitude of resultant force is maximum when the two force act in the same direction. Mass of the object = m =8.5 kg. Maximum resultant force acting on object = F(max) = f1 + f2 = 16 + 29 = 45 N.
Jun 27, 2013 · 7. Two objects are attracted to each other by a gravitational force F. If each mass is tripled and the distance between the objects is cut in half, what is the new gravitational force between the objects in terms of F? A. 24 F B. 36 F C. 16 F D. 1/16 F E. 1/24 F 8. An object with a mass of 48 kg measured on Earth is taken to the Moon. What is ...
Jul 20, 2020 · MCQ Questions for Class 9 Science Chapter 11 Work and Energy with Answers MCQs from Class 9 Science Chapter 11 – Work and Energy are provided here to help students prepare for their upcoming Science exam. MCQs from CBSE Class 9 Science Chapter 11: Work and Energy 1. A man is carrying the heavy luggage from …
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recall that force is the product of mass and acceleration. ma = 65.597. m=16 kg. (01.03 MC) Four forces are exerted on each of the two objects as shown below: Object A Object B 5N 3N 4N 2N 1N 2N SN 3N In which direction will the ob … jects move?
a uniform frictional force acts on the block. 5.2 State the work-energy theorem in words. (2) 5.3 Use energy principles to calculate the work done by the frictional force when the 2 kg block moves from point . B to point C. (4) [13] A h Before . ¼ h After 2 kg 2 kg B
18. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force. Explain it with an example. 19. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2) 20.
Some interesting situations connected to Newton’s second law occur when considering the effects of drag forces upon a moving object. For instance, consider a skydiver falling through air under the influence of gravity. The two forces acting on him are the force of gravity and the drag force (ignoring the buoyant force).
May 09, 2019 · Two horizontal forces F with arrow1 and F with arrow2 act on a 1.6 kg disk that slides over frictionless ice, on which an xy coordinate system is laid out. Force F with arrow1 is in the negative direction of the x axis and has a magnitude of 1.0 N. Force F with arrow2 has a magnitude of 9.0 N. acceleration in the x direction is 3 m/s^2.
Which one of the following statements always applies to a damping force acting on a vibrating system? A. It is in the same direction as the acceleration. Which one of the following statements concerning the acceleration of an object moving with simple harmonic motion is correct? A. It is constant. B.
The acceleration produced by a force 5 N acting on a mass of 20 kg in m/s2 is: a) 4 b) 100 c) 0.25 d) 2.5 A fielder pulls his hand backwards after catching the cricket ball. This enables the fielder to:
1 155 kg )( 2.63 m )2 + ( 59.4 kg )( 2.63 m )2 ( 2.84 rad/s )2 =1 2 2 ( K f = 3820 J = 3.82 kJ Insight: The kinetic energy of the system would decrease even if the person were running faster than the 10.6 m/s the rim is traveling initially and the merry-go-round speeds up as a result of the collision.
Jul 15, 2020 · 2. A force F depends on displacement s as F = 5s + 6 and it acts on a mass m = 1 kg which is initially at rest at a point O. Then the velocity of the mass when s = 4m is (a) zero (b) 4 2 /m s (c) 8 2 / m s(d) 16 2 / 3. A body of mass m has its position x and time t related by the equation 3 23/2 1 2 t t . The
28.A 100 W motor propels an object with a mass of 4 kg for 2 s from rest. Its final velocity will be A) 16 J B) 64 J C) 128 J D) 256 J E) 512 J 29.An object with a mass of 2 kg increases in speed from 4 m/s to 12 m/s in 3 s. The total work performed on the object during this time is A) 1.0 × 102 J
external force acting on the object and directly proportional to the mass of the object. _____ 13. Which are simultaneous equal but opposite forces resulting from the interaction of two objects? a. net external forces c. gravitational forces b. field forces d. actoin-reacon it paris _____ 14. Newton’s third law of motion involves the ...
A constant force of 4.5 N acts on a 7.2-kg object for 10.0 s. What is the change in the object's velocity? answer choices ... Question 16 . SURVEY . 30 seconds . Q. A ...
16. How far is the object from the starting point at the end of 3 seconds? 43. A 25-newton horizontal force northward and a 35-newton horizontal force southward act concurrently on a 15-kilogram object on a frictionless surface. 57. Two forces act concurrently on an object.
The external forces are the ones which act on a body and exist outside/external of the system of that body. It is only the external forces which are responsible for producing any change in the state of the body. If two objects interact, the force, F12, exerted on the object 1 by the object 2 (called action)...
Dec 08, 2014 · (d) becomes 16 times that of initial; In case of negative work the angle between the force and displacement is (a) 00 (b) 450 (c) 900 (d ) 1800; An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
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May 13, 2008 · The magnitude of the acceleration of the object is: A 0.5 m/s2 B 2.0 m/s2 C 2.8 m/s2 D 10 m/s2 E 50 m/s2. Two forces are applied to a 5.0 kg object, one is 6.0 N to the north and the other is 8.0 N to the west.?
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The tidal force is a force that stretches a body towards and away from the center of mass of another body due to a gradient (difference in strength) in gravitational field from the other body; it is responsible for diverse phenomena, including tides, tidal locking, breaking apart of celestial bodies and formation of ring systems within the Roche limit, and in extreme cases, spaghettification ...
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Descendants fanfiction mal scars | 3,164 | 11,529 | {"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} | 3.6875 | 4 | CC-MAIN-2021-31 | latest | en | 0.917651 |
https://www.justinmath.com/linear-systems-as-transformations-of-vectors-by-matrices/ | 1,656,311,421,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00233.warc.gz | 892,391,805 | 8,856 | # Linear Systems as Transformations of Vectors by Matrices
Let’s create a compact notation for expressing systems of linear equations like the one shown below.
\begin{align*} \begin{matrix} a_{11}x_1 &+& a_{12}x_2 &+& \cdots &+& a_{1n}x_n &=& b_1 \\ a_{21}x_1 &+& a_{22}x_2 &+& \cdots &+& a_{2n}x_n &=& b_2 \\ & & & & & & &\vdots& \\ a_{m1}x_1 &+& a_{m2}x_2 &+& \cdots &+& a_{mn}x_n &=& b_m \end{matrix} \end{align*}
We’re familiar with a slightly condensed version using coefficient vectors.
\begin{align*} \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{mn} \end{pmatrix} x_1 + \begin{pmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{pmatrix} x_2 + \cdots + \begin{pmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{pmatrix} x_n = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
However, we can condense this even further by putting the coefficient vectors in a vector themselves and taking the dot product with the vector of variables.
\begin{align*} \begin{pmatrix} \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{mn} \end{pmatrix}, \begin{pmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{pmatrix}, \cdots, \begin{pmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{pmatrix} \end{pmatrix} \cdot \left< x_1, x_2, \cdots, x_n \right> = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
To save space, the vector of variables can be written as a column vector as well.
\begin{align*} \begin{pmatrix} \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{mn} \end{pmatrix}, \begin{pmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{pmatrix}, \cdots, \begin{pmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{pmatrix} \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
Finally, to simplify the notation, we can remove the vector braces around the individual coefficient vectors and remove the dot product symbol.
\begin{align*} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
The array containing the coefficients is called a matrix. It’s really just a vector of sub-vectors, written without braces on the individual sub-vectors. Looking back, it makes sense to define a matrix multiplying a vector as follows:
\begin{align*} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \rightarrow \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{mn} \end{pmatrix} x_1 + \begin{pmatrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{pmatrix} x_2 + \cdots + \begin{pmatrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{pmatrix} x_n \end{align*}
Keeping this form of matrix notation and multiplication in mind, let’s start from scratch and proceed to condense a system of linear equations in a different way. We’ll get an interesting result.
\begin{align*} \begin{matrix} a_{11}x_1 &+& a_{12}x_2 &+& \cdots &+& a_{1n}x_n &=& b_1 \\ a_{21}x_1 &+& a_{22}x_2 &+& \cdots &+& a_{2n}x_n &=& b_2 \\ & & & & & & &\vdots& \\ a_{m1}x_1 &+& a_{m2}x_2 &+& \cdots &+& a_{mn}x_n &=& b_m \end{matrix} \end{align*}
This time, however, we will begin by writing each equation as a dot product.
\begin{align*} \left< a_{11}, a_{12}, \cdots, a_{1n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> &= b_1 \\ \left< a_{21}, a_{22}, \cdots, a_{2n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> &= b_2 \\ &\vdots \\ \left< a_{m1}, a_{m2}, \cdots, a_{mn} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> &= b_m \end{align*}
Then, we will write the system as a single vector equation by interpreting each side of the equation as a vector.
\begin{align*} \begin{pmatrix} \left< a_{11}, a_{12}, \cdots, a_{1n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \\ \left< a_{21}, a_{22}, \cdots, a_{2n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \\ \vdots \\ \left< a_{m1}, a_{m2}, \cdots, a_{mn} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
Each component of the left-hand-side vector includes a dot product with the vector of variables, so we can factor out the vector of variables.
\begin{align*} \begin{pmatrix} \left< a_{11}, a_{12}, \cdots, a_{1n} \right> \\ \left< a_{21}, a_{22}, \cdots, a_{2n} \right> \\ \vdots \\ \left< a_{m1}, a_{m2}, \cdots, a_{mn} \right> \end{pmatrix} \left< x_1, x_2, \cdots, x_n \right> = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
Again, to save space, the vector of variables can be written as a column vector.
\begin{align*} \begin{pmatrix} \left< a_{11}, a_{12}, \cdots, a_{1n} \right> \\ \left< a_{21}, a_{22}, \cdots, a_{2n} \right> \\ \vdots \\ \left< a_{m1}, a_{m2}, \cdots, a_{mn} \right> \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
Finally, to simplify the notation, we can again remove the vector braces around the individual coefficient vectors.
\begin{align*} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}
Again, there is a matrix! And again, the matrix just represents a vector of sub-vectors, written without braces on the individual sub-vectors. But this time, looking back, it makes sense to define a matrix multiplying a vector by a different rule.
\begin{align*} \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} \rightarrow \begin{pmatrix} \left< a_{11}, a_{12}, \cdots, a_{1n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \\ \left< a_{21}, a_{22}, \cdots, a_{2n} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \\ \vdots \\ \left< a_{m1}, a_{m2}, \cdots, a_{mn} \right> \cdot \left< x_1, x_2, \cdots, x_n \right> \end{pmatrix} \end{align*}
Which rule is correct? It turns out, they both are. Before we do an example, though, let’s recap. We’re stumbling upon the following structure:
\begin{align*} \begin{matrix} a_{11}x_1 &+& a_{12}x_2 &+& \cdots &+& a_{1n}x_n &=& b_1 \\ a_{21}x_1 &+& a_{22}x_2 &+& \cdots &+& a_{2n}x_n &=& b_2 \\ & & & & & & &\vdots& \\ a_{m1}x_1 &+& a_{m2}x_2 &+& \cdots &+& a_{mn}x_n &=& b_m \end{matrix}& \\ \text{ } \\ \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix}& \end{align*}
The array on the left-hand side is called a matrix, and we have two ways to compute the product of a matrix and a vector – one which involves interpreting the columns of the matrix as individual vectors, and another which involves interpreting the rows of the matrix as individual vectors.
To verify that both methods of computation indeed yield the same result, we can try out a simple example using the two different methods to compute the product of a 2-by-2 matrix and a 2-dimensional vector.
\begin{align*} &\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} (5) + \begin{pmatrix} 2 \\ 4 \end{pmatrix} (6) = \begin{pmatrix} 5 \\ 15 \end{pmatrix} + \begin{pmatrix} 12 \\ 24 \end{pmatrix} = \begin{pmatrix} 17 \\ 39 \end{pmatrix} \\ \text{ } \\ &\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 5 \\ 6 \end{pmatrix} = \begin{pmatrix} \left< 1, 2 \right> \cdot \left< 5,6 \right> \\ \left< 3, 4 \right> \cdot \left< 5,6 \right> \end{pmatrix} = \begin{pmatrix} 5 + 12 \\ 15 + 24 \end{pmatrix} = \begin{pmatrix} 17 \\ 39 \end{pmatrix} \end{align*}
Lastly, let’s build some geometric intuition. Geometrically, a matrix represents a transformation of a vector space, and we can visualize this transformation by thinking about what the matrix does to the N-dimensional unit cube.
For example, to see what the matrix from the example does to the unit square, we can multiply the vertices $(1,0)$ and $(0,1)$ of the unit square by the matrix.
\begin{align*} &\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} (1) + \begin{pmatrix} 2 \\ 4 \end{pmatrix} (0) = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \\ \text{ } \\ &\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= \begin{pmatrix} 1 \\ 3 \end{pmatrix} (0) + \begin{pmatrix} 2 \\ 4 \end{pmatrix} (1) = \begin{pmatrix} 2 \\ 4 \end{pmatrix} \end{align*}
We see that the matrix moves the vertices of the unit square from $(1,0)$ and $(0,1)$, to $(1,2)$ and $(3,4)$. Notice that these are just the columns of the matrix!
But it’s not just the unit square that is transformed in this way. The entire space undergoes this transformation as well.
And it’s more than simple stretching – the space is actually flipped over, since the original bottom vertex $(1,0)$ is now the top vertex $(1,3)$, and the original top vertex $(0,1)$ is now the bottom vertex $(2,4)$.
Practice Problems
Convert the following linear systems to matrix form. (You can view the solution by clicking on the problem.)
\begin{align*} 1) \hspace{.5cm} 3x_1 −2x_2 &= 7 \\ 5x_1 + 4x_2 &= 6 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 3 & −2 \\ 5 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 7 \\ 6 \end{pmatrix} \end{align*}
\begin{align*} 2) \hspace{.5cm} x_1 − 8x_2 &= 3 \\ x_1+x_2&=−2 \end{align*}
Solution:
\begin{align*} \begin{pmatrix}1 & −8 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 3 \\ −2 \end{pmatrix} \end{align*}
\begin{align*} 3) \hspace{.5cm} 2x_1+3x_2&=4 \\ 5x_1 &= 8 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 2 & 3 \\ 5 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \end{pmatrix} \end{align*}
\begin{align*} 4) \hspace{.5cm} \hspace{1cm} 8x_2 &= −7 \\ 3x_1 − x_2 &= 5 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 0 & 8 \\ 3 & −1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} −7 \\ 5 \end{pmatrix} \end{align*}
\begin{align*} 5) \hspace{.5cm} 2x_1+3x_2−4x_3 &= 5 \\ 7x_1 − 2x_2 + 3x_3 &= 2 \\ 9x_2 + 5x_2 + 4x_3 &= 1 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 2 & 3 & −4 \\ 7 & −2 & 3 \\ 9 & 5 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 1 \end{pmatrix} \end{align*}
\begin{align*} 6) \hspace{.5cm} \hspace{.5cm}x_1 − x_2 + x_3 &= 0 \\ 2x_1 − 5x_2 + x_3 &= −2 \\ x_1 + 4x_2 + 2x_3 &= 3 \end{align*}
Solution:
\begin{align*} \begin{pmatrix}1 & −1 & 1 \\ 2 & −5 & 1 \\ 1 & 4 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 0 \\ −2 \\ 3 \end{pmatrix} \end{align*}
\begin{align*} 7) \hspace{.5cm} \hspace{1cm} x_1−x_3 &= 4 \\ x_2 + 3x_3 &= 7 \\ x_1 + x_2 + x_3 &= −5 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 1 & 0 & −1 \\ 0 & 1 & 3 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 4 \\ 7 \\ −5 \end{pmatrix} \end{align*}
\begin{align*} 8) \hspace{.5cm} x_2 + x_3 &= 6 \\ x_1 + x_3 &= 5 \\ x_1 + x_2 &= 4 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix} = \begin{pmatrix} 6 \\ 5 \\ 4 \end{pmatrix} \end{align*}
\begin{align*} 9) \hspace{.5cm} x_1+x_2+x_3−x_4 &= 7 \\ x_2+x_3−7x_4 &= 5 \\ x_1 +8x_3 &= 11 \\ 4x_2+x_4 &= 3 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 1 & 1 & 1 & −1 \\ 0 & 1 & 1 & −7 \\ 1 & 0 & 8 & 0 \\ 0 & 4 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \\ 11 \\ 3 \end{pmatrix} \end{align*}
\begin{align*} 10) \hspace{.5cm} x_1−2x_2+3x_3 &= 0 \\ x_2−2x_3+3x_4 &= 0 \\ x_1 − 2x_3 + 3x_4 &= 1 \\ x_1 − 2x_2 + 3x_4 &= 1 \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 1 & −2 & 3 & 0 \\ 0 & 1 & −2 & 3 \\ 1 & 0 & −2 & 3 \\ 1 & −2 & 0 & 3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix} \end{align*}
Compute the product of the given vector and matrix, by A) interpreting the columns of the matrix as individual vectors, and B) interpreting the rows of the matrix as individual vectors. Verify that the results are the same. (You can view the solution by clicking on the problem.)
\begin{align*} 11) \hspace{.5cm} \begin{pmatrix} 1 & 3 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 5 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 17 \\ 13 \end{pmatrix} \end{align*}
\begin{align*} 12) \hspace{.5cm} \begin{pmatrix} −2 & 1 \\ 3 & 1 \end{pmatrix} \begin{pmatrix} −1 \\ 2 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix}4 \\ −1 \end{pmatrix} \end{align*}
\begin{align*} 13) \hspace{.5cm} \begin{pmatrix} 7 & 2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} −3 \\ 1 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} −19 \\ 1 \end{pmatrix} \end{align*}
\begin{align*} 14) \hspace{.5cm} \begin{pmatrix} 3 & 0 \\ 2 & −3 \end{pmatrix} \begin{pmatrix} 0 \\ 4 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 0 \\ −12 \end{pmatrix} \end{align*}
\begin{align*} 15) \hspace{.5cm} \begin{pmatrix} 1 & 3 & 2 \\ −1 & 0 & 1 \\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 11 \\ 1 \\ 5 \end{pmatrix} \end{align*}
\begin{align*} 16) \hspace{.5cm} \begin{pmatrix} 3 & 1 & 0 \\ 2 & 0 & 1 \\ −1 & 2 & −3 \end{pmatrix} \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 16 \\ 10 \\ −3 \end{pmatrix} \end{align*}
\begin{align*} 17) \hspace{.5cm} \begin{pmatrix} 1 & 2 & 0 \\ 4 & 3 & 1 \\ 3 & −1 & −1 \end{pmatrix} \begin{pmatrix} −1 \\ 1 \\ 2 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 1 \\ −3 \\ −2 \end{pmatrix} \end{align*}
\begin{align*} 18) \hspace{.5cm} \begin{pmatrix} 2 & 3 & 0 \\ 5 & 0 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} 5 \\ −2 \\ 3 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 4 \\ 28 \\ 9 \end{pmatrix} \end{align*}
\begin{align*} 19) \hspace{.5cm} \begin{pmatrix} 1 & 3 & 2 & 1 \\ 0 & 0 & 3 & 4 \\ 1 & 0 & −1 & −2 \\ 3 & −4 & −2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 2 \\ 0 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 14 \\ 6 \\ −1 \\ −13 \end{pmatrix} \end{align*}
\begin{align*} 20) \hspace{.5cm} \begin{pmatrix} 7 & 0 & −1 & −1 \\ 2 & 3 & −1 & −2 \\ 0 & 1 & 2 & 3 \\ 4 & 3 & −2 & 1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \\ −1 \\ 2 \end{pmatrix} \end{align*}
Solution:
\begin{align*} \begin{pmatrix} 20 \\ 6 \\ 5 \\ 19 \end{pmatrix} \end{align*}
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https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/02%3A_Ligand_Binding_in_Coordination_Complexes_and_Organometallic_Compounds/2.04%3A_Chelation | 1,723,496,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00559.warc.gz | 126,677,922 | 34,442 | # 2.4: Chelation
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Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed". The halides, phosphines, ammonia and amines seen previously are monodentate ligands.
Bidentate ligands bind through two donor sites. Bidentate means "two-toothed". An example of a bidentate ligand is bis(dimethylphosphino)propane. It can bind to a metal via two donor atoms at once: it uses one lone pair on each phosphorus atom.
More examples of bidentate ligands are shown below. They all have at least two different atoms with lone pairs. In some cases, there are additional atoms with lone pairs, but only two of them are able to face the metal at one time. Oxalate and glycinate would act as bidentate donors, donating up to two sets of lone pairs at the same time.
Table CC $$\PageIndex{1}$$ Some common bidentate ligands
Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the "chelate effect". This term comes from the Greek chelos, meaning "crab". A crab does not have any teeth at all, but it does have two claws for tightly holding onto something.for a couple of reasons. A very simple analogy is that, if you are holding something with two hands rather than one, you are not as likely to drop it.
• Multidentate ligands bind more tightly because of the chelate effect
The chemical reasons for the chelate effect involve relative enthalpy and entropy changes upon binding a multidentate ligand. In terms of enthalpy, in order to completely remove a bidentate ligand, two coordinate bonds must be broken. That costs more energy than breaking one coordinate bond for a monodentate ligand.
In terms of entropy, which deals with the distribution of energy within a system, it is generally thought that bringing two molecules together (a bidentate ligand and a metal complex) costs less than bringing three molecules together (two monodentate ligands and a metal complex). That's because individual molecules are free to move around, tumble and vibrate independently. Once they come together, they have to do all these things together. Since these different types of motion represent different ways of distributing energy, if the system becomes more restricted, energy can't be distributed in as many states.
• Energy is lowered even more by two bonding interactions
• Compared to two separate donors, bidentate donation is entropically favored
##### Exercise $$\PageIndex{1}$$
Draw metal complexes using the ligands below, binding to Ni(2+) in a bidentate mode.
A ligand could be monodentate, meaning it binds through a lone pair on a single atom. It could be bidentate, meaning it binds through lone pairs on two different atoms. It could even be tridentate, with three atoms bearing their own lone pairs, tetradentate, and so on.
Table CC $$\PageIndex{2}$$ Examples of polydentate ligands.
There is a symbol for denticity, κ (it's a Greek letter, pronounced "kappa"), which simply describes how many atoms are bound to the metal. For example, in ethylenediamine or 1,2-diaminoethane, NH2CH2CH2NH2, the two nitrogen atoms can be bound to the metal at the same time, although none of the other atoms in between would be directly attached to the metal. This donor is capable of binding in a κ2 mode. However, if for some reason one of the nitrogen atoms lets go of the metal so that the ethylenediamine is hanging on by only one nitrogen, we would say that the ligand is binding in κ1 mode.
##### Exercise $$\PageIndex{2}$$
In each of the following cases,
i) describe the denticity;
ii) indicate the charge on the ligand and on the metal.
tridentate or κ3; ligand = 0; metal = 1+
bidentate or κ2; ligand = 1-; metal = 0
bidentate or κ2; ligand = 0; metal = 2+
tetradentate or κ4; ligand = 2-; metal = 2+
tridentate or κ3; ligand = 0; metal = 1+
bidentate or κ2; ligand = 0; metal = 1+
##### Exercise $$\PageIndex{3}$$
In the following cases, the ligand has slipped, so that it isn't binding as tightly as it possibly could. In each case,
i) describe the denticity as drawn;
ii) state the maximum denticity possible;
iii) indicate the charge on the ligand and on the metal
monodentate or κ1; maximum bidentate or κ2; ligand = 0; metal = 2+
bidentate or κ2; maximum tridentate or κ3; ligand = 1-; metal = 1+
bidentate or κ2; maximum tridentate or κ3; ligand = 0; metal = 1+
bidentate or κ2; maximum tetradentate or κ4; ligand = 0; metal = 2+
tridentate or κ3; maximum tetradentate or κ4; ligand = 1-; metal = 2+
tridentate or κ3; maximum tetradentate or κ4; ligand = 2-; metal = 2+
There are more subtle aspects of chelation. For example, two different bidentate ligands may not necessarily bind to the metal in exactly the same way. In the drawing below, it's apparent that the three bidentate phosphine ligands, bis(dimethylphosphino)methane, bis(dimethylphosphino)ethane, and bis(dimethylphosphino)propane, do not all bind the metal with the same geometry. In each case, the metal forms a different angle with the two phosphines.
The term "bite angle" is frequently used to describe how different bidentate ligands will attach to metals at different angles. In the picture, the P-Pd-P angle appears to be about 90 degrees when dmpm is bound; in reality it is even smaller. With dmpe, the bite angle appears larger in the picture than the one for dmpm, and in reality it is larger, although not quite as large as it appears here. Two different ligands that bind with two different bite angles will have different influences on the complex that forms. In fact, chemists often use these differences to "tune" the behavior of transition metals that are used as catalysts for important properties. They might add similar ligands with different bite angles to see which one best promotes the desired catalytic reaction.
Many factors can influence the bite angle, including structural features of the bidentate ligand itself, the metal, and other ligands bound to the metal. However, a particular ligand will usually have a normal range of bite angles that it will be able to adopt under different circumstances.
##### Exercise $$\PageIndex{4}$$
Certain ligands may have natural bite angles that work better in some cases than in others. Propose the optimum bite angle in each of the following geometries.
##### Exercise $$\PageIndex{5}$$
Certain ligands tend to give a certain range of bite angles. Use the suggested criterion to predict which ligand in each pair would give the larger bite angle.
The total of the interior angles of a regular polyhedron is given by (n-2)180o, in which n is the number of sides in the polyhedron. Assuming the ring formed by the bidentate ligand and the metal is a regular polyhedron (it won't be, but we are simplifying), then nitrate gives a triangle with 60° angles, including a 60° O-M-O bite angle. Oxalate gives a square with a larger, 90° bite angle.
In reality, the bite angle for nitrate varies with the complex that is formed, but it is usually somewhere around sixty degrees, whereas oxalate usually gives somewhere around eighty five degrees (see, for example, Alvarez, Chem. Rev. 2015, 115, 13447-13483). The smaller ring size gives a smaller bite angle.
Sulfur is larger than oxygen, so its bonds will be a little longer. As a result, you can imagine those two sides of the square being a little longer with sulfur than with oxygen. From the perspective of the metal, the gap between the two donor atoms widens out a little.
Acetate forms bite angles of around sixty degrees, but dithiocarbamate forms larger bite angles of seventy or seventy five degrees.
There are lots of differences between these two ligands, but if we simplify and only consider bond angle, we can make a prediction. If the atoms in bipridyl can be considered sp2 hybridised, then they form 120° bond angles. The atoms in ethylenediamine could be considered sp3 hybridised, forming approximately 110° angles. The angle N-M-N still has to complete the shape of the regular pentagon, so if all of the other angles are bigger in the bipyridyl complex, we would expect the bite angle to be smaller.
Really, the bite angles are much closer than this rough estimate suggests. Bipyridyl forms average bite angles of around eighty degrees, whereas ethylenediamine forms average bite angles of around eighty-five degrees. Keep in mind that those are just averages, though. These two values are close enough that their ranges overlap; lots of bipyridyl complexes would have bite angles smaller than ethylenediamine complexes.
##### Exercise $$\PageIndex{6}$$
Suggest which ligand in each pair would have the larger bite angle, and why. | 3,894 | 13,084 | {"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} | 4.0625 | 4 | CC-MAIN-2024-33 | latest | en | 0.194704 |
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# If x and y are positive integers, what is the value of xy?
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If x and y are positive integers, what is the value of xy? [#permalink] 22 Jun 2008, 11:05
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If x and y are positive integers, what is the value of xy?
1) The greatest common factor of x and y is 10
2) the least common multiple of x and y is 180
The OA is C.
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Re: DS - GMATPrep - value of xy [#permalink] 22 Jun 2008, 12:06
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Each statement by itself is clearly insufficient.
For both:
x*y = LCM(x,y)*GCF(x,y) (always true for positive numbers). Thus, C.
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Re: DS - GMATPrep - value of xy [#permalink] 25 Jun 2008, 11:29
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In the solution above, part 2) is a bit incomplete. Yes, you do not know whether x or y (or both) is divisible by 5, but that's not the only issue: you also do not know whether each is divisible by 2^2 or by 3^2.
You certainly know that one of x or y is divisible by 2^2. The other might be divisible either by 2^0, 2^1 or 2^2. You certainly know that one of x or y is divisible by 3^2. The other might be divisible either by 3^0, 3^1 or 3^2 This leads to many possibilities, and many possible values for the product xy.
From Statement 2) alone, we can say is that the product xy is greater than or equal to 180, and less than or equal to 180^2, but there are many possible values for xy.
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Re: DS - GMATPrep - value of xy [#permalink] 22 Jun 2008, 14:17
Thanks greenoak, I didn't know this rule.
Is there another way to solve this problem?
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Re: DS - GMATPrep - value of xy [#permalink] 24 Jun 2008, 16:54
Does anyone have another workaround for this question other than the rule posted by greenoak?
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Re: DS - GMATPrep - value of xy [#permalink] 25 Jun 2008, 04:15
I am not sure but another way may be like:
1)gives these numbers are like:
(20,30),(10,90),(30,70),...
then
1) & 2) give
(20,90).
chango wrote:
Does anyone have another workaround for this question other than the rule posted by greenoak?
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Re: DS - GMATPrep - value of xy [#permalink] 25 Jun 2008, 06:41
x-ALI-x wrote:
Thanks greenoak, I didn't know this rule.
Is there another way to solve this problem?
There are other ways to look at this problem, but I'm not sure why one would want to use a different method. The rule posted by greenoak is a fundamental law of number theory, and it lets you answer this question (and other GMAT questions) in a few seconds. It's definitly worth understanding why the rule greenoak posted is true- if you understand LCMs and GCDs, the rule should make sense.
Alternatively, you might choose numbers, as maximusdecimus suggests. There aren't many possibilities to consider, since we only need to look at multiples of 10 which are also divisors of 180. We know one of the divisors must be divisible by 9, while the other cannot be divisible by 3 -- otherwise the GCD would be 30 or 90. The only possible values for x and y are 10 and 180; or 20 and 90 (in either order). The product is 1800 in each case.
Or, you could, from statement 1, conclude that x/10 and y/10 are integers which share no divisors- they are 'relatively prime' in math speak. Thus, the LCM of x/10 and y/10 is equal to the product of x/10 and y/10. And from statement 2, the LCM of x/10 and y/10 must be 18. Thus (x/10)*(y/10) = 18 and xy = 1800. But all I've done here is prove a special case of the general rule that greenoak posted, and there's more value in understanding the general rule than the specific case.
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Re: DS - GMATPrep - value of xy [#permalink] 25 Jun 2008, 09:32
chango wrote:
Does anyone have another workaround for this question other than the rule posted by greenoak?
Not to take away what greenoak posted, because that is the best way to solve this problem, but if you want another perspective on it, see below:
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Re: DS - GMATPrep - value of xy [#permalink] 25 Jun 2008, 12:31
IanStewart wrote:
In the solution above, part 2) is a bit incomplete. Yes, you do not know whether x or y (or both) is divisible by 5, but that's not the only issue: you also do not know whether each is divisible by 2^2 or by 3^2.
You certainly know that one of x or y is divisible by 2^2. The other might be divisible either by 2^0, 2^1 or 2^2. You certainly know that one of x or y is divisible by 3^2. The other might be divisible either by 3^0, 3^1 or 3^2 This leads to many possibilities, and many possible values for the product xy.
From Statement 2) alone, we can say is that the product xy is greater than or equal to 180, and less than or equal to 180^2, but there are many possible values for xy.
Yes, good point. I didn't even think of that. +1
I have edited my above response by incorporating your logic into my rationale.
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Re: DS - GMATPrep - value of xy [#permalink] 26 Jun 2008, 14:55
x-ALI-x wrote:
If x and y are positive integers, what is the value of xy?
1) The greatest common factor of x and y is 10
2) the least common multiple of x and y is 180
The OA is C.
This troubled me for sometime. Here is another way to solve the problem.
If we consider two numbers, lets say 18 and 24, the LCM (denoted by L) is calculated as follows.
L = 18a ; L = 24b (a and b are integers)
L = 2*3*3*a ;
L = 2*2*2*3*b
so, taking the bold factors out from the above (the product of bold factors constitute GCF, denoted by G); we can say that :
a = 2*2
b= 3
rewriting :
L = G*3*a
L= G*4*b
using the same logic to the given problem :
We all agree that both statements are insufficient to answer the question. Considering both together :
L = G*f2*f1 (x = G*f2)
L = G*f1*f2 (y = G*f1)
180 = 10*f1*f2
=> f1*f2 = 18
since x = G*f2 and y = G*f1 , we have :
x*y = G^2 * f1*f2
= 10^2 * 18 = 1800
Hence C.
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Re: DS - GMATPrep - value of xy [#permalink] 26 Jun 2008, 21:35
sanjay_gmat wrote:
x-ALI-x wrote:
If x and y are positive integers, what is the value of xy?
1) The greatest common factor of x and y is 10
2) the least common multiple of x and y is 180
The OA is C.
This troubled me for sometime. Here is another way to solve the problem.
If we consider two numbers, lets say 18 and 24, the LCM (denoted by L) is calculated as follows.
L = 18a ; L = 24b (a and b are integers)
L = 2*3*3*a ;
L = 2*2*2*3*b
so, taking the bold factors out from the above (the product of bold factors constitute GCF, denoted by G); we can say that :
a = 2*2
b= 3
rewriting :
L = G*3*a
L= G*4*b
using the same logic to the given problem :
We all agree that both statements are insufficient to answer the question. Considering both together :
L = G*f2*f1 (x = G*f2)
L = G*f1*f2 (y = G*f1)
180 = 10*f1*f2
=> f1*f2 = 18
since x = G*f2 and y = G*f1 , we have :
x*y = G^2 * f1*f2
= 10^2 * 18 = 1800
Hence C.
Also, from the last equation :
x*y = G*G*f1*f2
= G*L
this proves the equation that greenoak used.
Re: DS - GMATPrep - value of xy [#permalink] 26 Jun 2008, 21:35
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# Search element in rotated sorted array
#### Algorithms Search Algorithms
Open-Source Internship opportunity by OpenGenus for programmers. Apply now.
In this article, we have explored how to search an element in a Rotated Sorted Array efficiently in O(logN) time with constant space O(1).
1. Understanding the problem statement
2. Efficient Solution
3. Time and Space Complexity
Prerequisite: Binary Search, Linear Search
## Understanding the problem statement
An array is a collection of items stored at contiguous memory locations. The idea is to store various or multiple items of the same type together. This makes it easier to calculate the position of each element by simply adding an offset to a base value, i.e., the memory location of the first element of the array (generally denoted by the name of the array).The base value is index 0 and the difference between the two indexes is the offset.
For clear understanding, we can think of an array as a fleet of stairs where on each step is placed a value(suppose your partners). Here, you can identify the location of any of your partner by simply knowing the count of the step they are on. We can access any element in an array by just knowing its index. There are so many variety of operations we can perform on an array like insertion,deletion,searching,sorting etc.
Let's take an example of an array of different data types:
``````//integer array in C/C++
int arr[]={1,2,3,4,5}
//character array in C/C++
char ar[]={'a','b','c'}
``````
As we can see that each element can be uniquely identified by its index in the array as arr[0]=1,arr[1]=2,arr[2]=3,arr[3]=4,arr[4]=5 and ar[0]=a,ar[1]=b,ar[2]=c where in arr[0] [0] is an index to access a particular element. Here we are using zero based indexing(zero based indexing is a type of indexing in which first element of an array is indexed by a subscript of 0). There are also one based indexing(one based indexing is a type of indexing in which first element of an array is indexed by a subscript of 1) and n based indexing(n based indexing is the one in which base index of an array can be freely chosen).
Now before moving towards our problem first let's understand what is a rotated sorted array. A rotated sorted array means it is an array which is at first sorted(sorted array means an array in which all its elements are arranged in ascending or descending order) but if we rotate its elements from some index either towards left or towards right,then that final array becomes a rotated sorted array.
For instance, suppose you have a sorted array containing elements {1,2,3,4,5} now if you rotate this array from index 1(zero based indexing) so it will become {3,4,5,1,2}. Now i am sure you will have a good understanding of what is a rotated sorted array.
Moving further, first let's see what our question? We have to search for an element in a rotated sorted array. First we have to understand how to search an element in a given array. There are two algorithms which help us in performing search operation in an array. They are:
1. Linear search
2. Binary search
Linear search is a very basic searching algorithm which search an element in O(n) time which is somehow good but we always looks for an efficient and less time consuming algorithm so that our code or program gets executed in small time and there will be no issues. Linear search is not good for a large size array so that's why we prefer using binary search algorithm as it takes only O(log n) time in executing a code which is far better as compared to the linear search time complexity.
So now we know that we have to apply binary search technique to solve this given problem. First understand what is binary search technique and how it works while searching for an element in an array. In binary search we repeatedly keep on dividing the sorted array into two equal halves. At the start we find the middle element in an array and compare that element with the given element which is need to be searched. If that element is equal to the middle element we have successfully searched that element but if that element is smaller than the middle element we start searching that element in the first half of the array as it is sorted and if that element is bigger than the middle element we start searching that element in the second half of the array. By this way we can easily reduce the number of comparisons in each step and finally perform a successful search if that element is present in the array. You will easily understand with the help of an example:
Suppose we have an array:
int a[]={1,2,3,4,5}
and our target value(which is to be searched) is 2
We find the middle element that is 3(bolder) now compare 3 with the target value i.e, 2 we can see that 3>2 so we as the array is sorted all the values smaller than 3 will be on the left side of 3 i.e, in the left half of the array which contains element 1,2 and now we can easily compare and search for the given target value applying the same procedure as we have applied above(taking the middle element and compare it with target value).
Implementation:
1. Compare target with the middle element.
2. If target matches with the middle element, we return the mid index.
3. Else If target is greater than the mid element, then target can only lie in the right half subarray after the mid element. So we recur for the right half.
4. Else (target is smaller) recur for the left half.
After seeing the approach we will write a C++ code for this technique.
``````// C++ program to implement Binary Search
#include <bits/stdc++.h>
using namespace std;
// A iterative binary search function. It returns
// location of x in given array arr[low..high] if present,
// otherwise -1
int binSearch(int a[], int low, int high, int target)
{
while (low <= high) {
int mid = low + (high - low) / 2; //finding the middle element
// Check if target is present at mid
if (a[mid] == target)
return mid;
// If target greater, ignore left half
if (a[mid] < target)
low = mid + 1;
// If target is smaller, ignore right half
else
high = mid - 1;
}
// if we reach here, then element was
// not present
return -1;
}
int main()
{
int arr[] = {1,2,3,4,5};
int x = 2;
int n = sizeof(arr) / sizeof(arr[0]);
int ans = binSearch(arr, 0, n - 1, x);
if(ans==-1){
cout<<"Element is not present"<<'\n';
}
else{
cout<<"Element is present at index"<<" "<<ans<<'\n';
}
return 0;
}
``````
Output:
``````Element is present at index 1
``````
## Efficient Solution
Now after seeing binary search we can proceed towards our given problem. We know that an element can be searched in a sorted array using binary search in O(log n) time. In our problem we are given a rotated sorted array and we have to search for that particular element in the array.
We know that the array which is sorted is rotated from some index which is unknown so if we can find that element(let's say this element as pivot element) which is present at that index we can easily apply binary search as the elements in the left side and right side of that pivot element will be already sorted.First we will see the basic approach then its proper implementation and finally with the help of all these we will write a code for this problem.
Basic approach:
1. The idea is to find the pivot point, divide the array in two sub-arrays and perform binary search.
2. The main idea for finding pivot is – for a sorted (in increasing order) and pivoted array, pivot element is the only element for which next element to it is smaller than it.
3. Using the above statement and binary search, pivot element can be found.
4. After the pivot element is found out, divide the array in two sub-arrays.
5. Now the individual sub – arrays are sorted so the element can be searched using binary Search.
Implementation:
Suppose we have an array arr[] = {3, 4, 5, 1, 2}
Element to Search = 1
1. Find out pivot element and divide the array in two
sub-arrays. (pivot = 2) //Index of 5
2. Now call binary search for one of the two sub-arrays.
(a) If element is greater than 0th element then
search in left array
(b) Else Search in right array
(1 will go in right subarray(right half) as 1 < 0th element(3))
3. If element is found in selected sub-array then return index
Else return -1.
For better clarity we will take an example and understand step by step that what is happening.
Example : Suppose we have an array arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3} and we have to search for 3 in this rotated sorted array so how to do?
First we will find the pivot point(that point at which all elements on its left side and right side are sorted in correct order), we can see that pivot element is 10(bolder) now we will compare our target value with the first element of this array i.e, 5. As we can see that 5 > 3(target value) so now we have to search for this element in the right side of pivot having elements as 1,2,3. Now we have a sorted array we can easily apply binary search by finding middle element i.e, 2(bolder) and comparing it with 3(target value) we can see that 3>2 so it will be present in the right side of the middle element and we have successfully searched for our target value.
C++ implementation:
``````//C++ Program to search an element in a sorted and rotated array/
#include <bits/stdc++.h>
using namespace std;
//Standard Binary Search function
int binSearch(int a[], int low, int high, int target)
{
if (high < low)
return -1;
int mid = (low + high) / 2; //finding middle element
if (target == a[mid])
return mid;
if (target > a[mid])
low = mid + 1;
else
high = mid - 1;
}
//Function to get pivot
int findPivot(int a[], int low, int high)
{
// base cases
if (high < low)
return -1;
if (high == low)
return low;
int mid = (low + high) / 2;
if (mid < high && a[mid] > a[mid + 1])
return mid;
if (mid > low && a[mid] < a[mid - 1])
return (mid - 1);
if (a[low] >= a[mid])
high = mid - 1;
else
low = mid + 1;
}
//function for searching the target element
int pivotedBinarySearch(int a[], int num, int target)
{
int pivot = findPivot(a, 0, num - 1); //function calling
// If we didn't find a pivot,
// then array is not rotated at all
if (pivot == -1)
return binSearch(a, 0, num - 1, target);
// If we found a pivot, then first compare with pivot
// and then search in two subarrays around pivot
if (a[pivot] == target)
return pivot;
if (a[0] <= target)
return binSearch(a, 0, pivot - 1, target);
return binSearch(a, pivot + 1, num - 1, target);
}
int main()
{
int a[] = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };
int num = sizeof(a) / sizeof(a[0]);
int target = 3;
// Function calling
cout << "Index of the element is : "<< pivotedBinarySearch(a, num, target);
return 0;
}
``````
Output:
``````Index of the element is : 8
``````
## Time and Space Complexity
1. Time complexity: O(log n) as we are using binary search which uses log n comparisons to search an element where n is the size of the array.
2. Space complexity: O(1) constant space as no extra memory or space is required.
If we have an array which contain duplicate elements it is not possible to search in O(log n) time in all cases. For example consider searching 0 in {3,3,3,3,3,3,0,3,3} and {3,3,0,3,3,3,3,3,3}. It's not possible to decide whether to recur for the left half or right half by doing a constant number of comparisons at the middle.
So now I am sure you all have a clear understanding of how to approach and solve this problem. Try it by yourself and keep coding. This problem is similar to Leetcode problem 33. Search in Rotated Sorted Array.
Thank you.
#### Shwet Shukla
Shwet Shukla is pursuing his B.Tech in Computer Science from Harcourt Butler Technical University (HBTU) and is an Intern at OpenGenus.
Search element in rotated sorted array | 2,941 | 11,769 | {"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} | 3.84375 | 4 | CC-MAIN-2024-30 | latest | en | 0.902517 |
https://de.maplesoft.com/support/help/maplesim/view.aspx?path=Student/ODEs/ODESteps/FirstOrderIVPs | 1,726,254,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00733.warc.gz | 174,512,171 | 27,579 | First Order IVPs - Maple Help
ODE Steps for First Order IVPs
Overview
• This help page gives a few examples of using the command ODESteps to solve first order initial value problems.
• See Student[ODEs][ODESteps] for a general description of the command ODESteps and its calling sequence.
Examples
> $\mathrm{with}\left(\mathrm{Student}:-\mathrm{ODEs}\right):$
> $\mathrm{ivp1}≔\left\{{t}^{2}\left(z\left(t\right)+1\right)+{z\left(t\right)}^{2}\left(t-1\right)\mathrm{diff}\left(z\left(t\right),t\right)=0,z\left(3\right)=1\right\}$
${\mathrm{ivp1}}{≔}\left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}$ (1)
> $\mathrm{ODESteps}\left(\mathrm{ivp1}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{{t}}^{{2}}{}\left({z}{}\left({t}\right){+}{1}\right){+}{{z}{}\left({t}\right)}^{{2}}{}\left({t}{-}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){=}{0}{,}{z}{}\left({3}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\\ \text{•}& {}& \text{Separate variables}\\ {}& {}& \frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}{=}{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}t\\ {}& {}& {\int }\frac{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){}{{z}{}\left({t}\right)}^{{2}}}{{z}{}\left({t}\right){+}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{=}{\int }{-}\frac{{{t}}^{{2}}}{{t}{-}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{t}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}z{}\left(3\right)=1\\ {}& {}& {-}\frac{{1}}{{2}}{+}{\mathrm{ln}}{}\left({2}\right){=}{-}\frac{{15}}{{2}}{-}{\mathrm{ln}}{}\left({2}\right){+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=7+2{}\mathrm{ln}{}\left(2\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& \frac{{{z}{}\left({t}\right)}^{{2}}}{{2}}{-}{z}{}\left({t}\right){+}{\mathrm{ln}}{}\left({z}{}\left({t}\right){+}{1}\right){=}{-}\frac{{{t}}^{{2}}}{{2}}{-}{t}{-}{\mathrm{ln}}{}\left({t}{-}{1}\right){+}{7}{+}{2}{}{\mathrm{ln}}{}\left({2}\right)\end{array}$ (2)
> $\mathrm{ivp2}≔\left\{2xy\left(x\right)-9{x}^{2}+\left(2y\left(x\right)+{x}^{2}+1\right)\mathrm{diff}\left(y\left(x\right),x\right)=0,y\left(0\right)=1\right\}$
${\mathrm{ivp2}}{≔}\left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}$ (3)
> $\mathrm{ODESteps}\left(\mathrm{ivp2}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{{2}{}{x}{}{y}{}\left({x}\right){-}{9}{}{{x}}^{{2}}{+}\left({2}{}{y}{}\left({x}\right){+}{{x}}^{{2}}{+}{1}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}{,}{y}{}\left({0}\right){=}{1}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{▫}& {}& \text{Check if ODE is exact}\\ {}& \text{◦}& \text{ODE is exact if the lhs is the total derivative of a}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}{C}^{2}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{function}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Compute derivative of lhs}\\ {}& {}& \frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){+}\left(\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){=}{0}\\ {}& \text{◦}& \text{Evaluate derivatives}\\ {}& {}& {2}{}{x}{=}{2}{}{x}\\ {}& \text{◦}& \text{Condition met, ODE is exact}\\ \text{•}& {}& \text{Exact ODE implies solution will be of this form}\\ {}& {}& \left[{F}{}\left({x}{,}{y}\right){=}{\mathrm{C1}}{,}{M}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right){,}{N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\right]\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{by integrating}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}M{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{\int }\left({-}{9}{}{{x}}^{{2}}{+}{2}{}{x}{}{y}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Evaluate integral}\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Take derivative of}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y\\ {}& {}& {N}{}\left({x}{,}{y}\right){=}\frac{{\partial }}{{\partial }{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{F}{}\left({x}{,}{y}\right)\\ \text{•}& {}& \text{Compute derivative}\\ {}& {}& {{x}}^{{2}}{+}{2}{}{y}{+}{1}{=}{{x}}^{{2}}{+}\frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right)\\ \text{•}& {}& \text{Isolate for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆy}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{y}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{_F1}}{}\left({y}\right){=}{2}{}{y}{+}{1}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\\ {}& {}& {\mathrm{_F1}}{}\left({y}\right){=}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_F1}{}\left(y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into equation for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\\ {}& {}& {F}{}\left({x}{,}{y}\right){=}{-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}F{}\left(x,y\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into the solution of the ODE}\\ {}& {}& {-}{3}{}{{x}}^{{3}}{+}{{x}}^{{2}}{}{y}{+}{{y}}^{{2}}{+}{y}{=}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& \left\{{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}{,}{y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\right\}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{-}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solution does not satisfy initial condition}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(0\right)=1\\ {}& {}& {1}{=}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{4}{}{\mathrm{C1}}{+}{1}}}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{2}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=2\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}{-}\frac{{{x}}^{{2}}}{{2}}{-}\frac{{1}}{{2}}{+}\frac{\sqrt{{{x}}^{{4}}{+}{12}{}{{x}}^{{3}}{+}{2}{}{{x}}^{{2}}{+}{9}}}{{2}}\end{array}$ (4)
> $\mathrm{ivp3}≔\left\{\mathrm{diff}\left(y\left(x\right),x\right)-y\left(x\right)-x\mathrm{exp}\left(x\right)=0,y\left(a\right)=b\right\}$
${\mathrm{ivp3}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}$ (5)
> $\mathrm{ODESteps}\left(\mathrm{ivp3}\right)$
$\begin{array}{lll}{}& {}& \text{Let's solve}\\ {}& {}& \left\{\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){-}{x}{}{{ⅇ}}^{{x}}{=}{0}{,}{y}{}\left({a}\right){=}{b}\right\}\\ \text{•}& {}& \text{Highest derivative means the order of the ODE is}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}1\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\\ \text{•}& {}& \text{Isolate the derivative}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}{y}{}\left({x}\right){+}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Group terms with}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right){=}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{The ODE is linear; multiply by an integrating factor}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\\ \text{•}& {}& \text{Assume the lhs of the ODE is the total derivative}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left(\mathrm{\mu }{}\left(x\right){}y{}\left(x\right)\right)\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){-}{y}{}\left({x}\right)\right){=}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right)\right){}{y}{}\left({x}\right){+}{\mathrm{\mu }}{}\left({x}\right){}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)\\ \text{•}& {}& \text{Isolate}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\frac{ⅆ}{ⅆx}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{\mu }{}\left(x\right)\\ {}& {}& \frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{\mathrm{\mu }}{}\left({x}\right){=}{-}{\mathrm{\mu }}{}\left({x}\right)\\ \text{•}& {}& \text{Solve to find the integrating factor}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){=}{{ⅇ}}^{{-}{x}}\\ \text{•}& {}& \text{Integrate both sides with respect to}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}x\\ {}& {}& {\int }\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\left({\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right)\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Evaluate the integral on the lhs}\\ {}& {}& {\mathrm{\mu }}{}\left({x}\right){}{y}{}\left({x}\right){=}{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(x\right)\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{\mathrm{\mu }}{}\left({x}\right){}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{\mathrm{\mu }}{}\left({x}\right)}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{\mu }{}\left(x\right)={ⅇ}^{-x}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{\int }{{ⅇ}}^{{-}{x}}{}{x}{}{{ⅇ}}^{{x}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{ⅆ}{x}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Evaluate the integrals on the rhs}\\ {}& {}& {y}{}\left({x}\right){=}\frac{\frac{{{x}}^{{2}}}{{2}}{+}{\mathrm{C1}}}{{{ⅇ}}^{{-}{x}}}\\ \text{•}& {}& \text{Simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({{x}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Use initial condition}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}y{}\left(a\right)=b\\ {}& {}& {b}{=}\frac{{{ⅇ}}^{{a}}{}\left({{a}}^{{2}}{+}{2}{}{\mathrm{C1}}\right)}{{2}}\\ \text{•}& {}& \text{Solve for}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}\\ {}& {}& {\mathrm{C1}}{=}{-}\frac{{{ⅇ}}^{{a}}{}{{a}}^{{2}}{-}{2}{}{b}}{{2}{}{{ⅇ}}^{{a}}}\\ \text{•}& {}& \text{Substitute}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\mathrm{_C1}=-\frac{{ⅇ}^{a}{}{a}^{2}-2{}b}{2{}{ⅇ}^{a}}\phantom{\rule[-0.0ex]{1.0thickmathspace}{0.0ex}}\text{into general solution and simplify}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\\ \text{•}& {}& \text{Solution to the IVP}\\ {}& {}& {y}{}\left({x}\right){=}\frac{{{ⅇ}}^{{x}}{}\left({-}{{a}}^{{2}}{+}{2}{}{{ⅇ}}^{{-}{a}}{}{b}{+}{{x}}^{{2}}\right)}{{2}}\end{array}$ (6) | 6,949 | 14,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "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} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.435852 |
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92,916 questions, page 329
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44. ## physics
Can you find the pressure exerted on the floor by the heel if it has an are of 1.50 cm^2 and the woman's mass is 55kg?
asked by cody on December 11, 2012
45. ## Algebra
Find three consecutive odd integers such that the sum of 4 times the largest is -191
asked by Dee on February 16, 2017
46. ## Mayans and aztecs
Where can I find iinformation about the similarities and differences of the culture and life of the Mayans and Aztecs?
asked by betty on September 6, 2010
47. ## Implicit Functions
Find the slope of the tangent to the curve at the point specified. x^3+5x^2y+2y^2=4y+11 at (1,2) So far I simplified it to -(3x^2-10y/5x^2+4y-4) which I think then simplifies to -(8/3). What do I do from here....?
asked by George on November 7, 2008
48. ## Math
Write a formula to find the area (A) of a square. Use S to represent the length of one side.
asked by Isabel on November 13, 2008
find a quadratic function that fits the set of data points. (1,4), (-1.6), (-2,16) what did i do wrong, i got -2x^2-x+7
asked by Anne on December 7, 2011
50. ## Math
Find an expression for the function whose graph is the given curve. The top half of the circle x^2+(y-4)^2=4
asked by TayB on January 16, 2015
51. ## Algebra 2
Find the coordinates of the vertices of the parallelogram with sides that are contained in the lines with equations y=3, y=7, y=2x, y=2x-13
52. ## Mathematics
Find the number of ways digit 0, 1, 2 and 3 can be permuted to give rise to 2000.
asked by Divine Aka on February 14, 2017
53. ## calculus
For f(x) = e2sin(x) use your graphing calculator to find the number of zeros for f '(x) on the closed interval [0, 2π].
asked by mock on January 14, 2015
In triangle ABC, D is the centroid and M is the midpoint of segment AC. If BD=x+9 and DM=6, find MB. Answer choices: 9 , 3, or 18.
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55. ## physics question
find the force exerted by proton and the electron in a hydrogen atom they are 5.0 X 10^-19 coulomb
56. ## Math
Directions: Find all solutions of the equation in the interval (0, 2pi) sin x/2=1-cosx
asked by Lola on February 3, 2015
57. ## SS
Did Columbus come to know that he was not in India when he reached the Caribbean? I find mixed answers online.
asked by Anonymous on January 15, 2015
58. ## Math
Find without using tables the values of Cosx and Sinx when Cos 2x is: 1/8 , 7/25 ,-119/169
asked by Mazin on January 15, 2015
59. ## Geography
hello, can u please tell me some websites where i can find a descriptive journey through accra to cape coast, ghana
asked by numz on November 15, 2008
60. ## CAlculus
Find the linear function satisfying the given conditions. g(2) = 9 and the graph of g is perpendicular to the line 4x − 2y = 3. g(x) =
asked by Mohammed on July 11, 2017
61. ## maths --urgenttttttttttttttttttttt
find p & q, if the equation px^2 -8xy+3y^2 +14x+2y+x=0 represent a pair of perpendicular lines.
asked by sunaina on December 5, 2012
62. ## Stats
a die is rolled three times. find the chance that not all the rolls show 3 or more spots
asked by Fay on December 4, 2011
63. ## math
Hi my question is find the value of the following. 6! 8! over 9! 5! If your answer is not an integer, write it as an exact fraction. Thank you so much. I have no idea how to do it.
asked by mary ann on January 30, 2015
64. ## trigo
A 50 m tree casts a shadow of 60 m . Find the sun's angle of elevation at that time.
asked by dimwit on January 16, 2015
65. ## calculus
The position of a particle is given by the equation of motion s(t)= 1/(1+t)where t is in sec and s in meters find the velocity.
asked by donnie on February 20, 2017
66. ## statistics
for a standard normal curve, find the z-score that separates the bottom 30% from the top 70%
asked by kay on November 5, 2008
Where can I find US's daily treasury bill rate for years 2000 to 2002? Thanks!
asked by Aq on November 5, 2008
68. ## Algebra 2
I really need help understanding ho to solve combination functions here is a specific problem I am struggling with. Let f(x) = x - 2 and g(x) = x^2 - 7x - 9. Find f(g(-1)).
asked by failing on January 21, 2015
69. ## math
Find the value of the function y=sqrt(x+1)+sin(x)-0.5, correct to 3 decimal places, when x=0.05 without the use of a calculator.
asked by Alex on December 10, 2012
70. ## physics
If motion of a particle of mass (m) is given by y=ut+1/2gt^2 (square).find force acting on it
asked by an on October 5, 2012
71. ## Math
A rectangle is 6 cm long and 5 cm wide. When each dimension is increased by x cm, the area is tripled. Find the value of x.
asked by Casey on December 6, 2012
72. ## social studies
I am looking for enthic symbols for my German heritage. Can't seem to find any on the web. I am in the seventh grade
asked by nathan on November 9, 2008
73. ## trig
Find the exact solutions of the equation in the interval [0,2pi). sin(x/2)+cos(x)=0
asked by Isaac on December 5, 2011
74. ## science
What are three good websites to go to to find information about the Tundra? Preferably about the abiotic elements. Thank-you!
asked by Sam on March 24, 2009
75. ## math
find an equation of the linear function f using the given information. please show all steps and work f(-1)=1, f(1)=-2
asked by Tristan on September 7, 2010
76. ## CALC
find the following derivatives: f(x) = (2x+5)/(x^2-3) f'(x)= (2)*(x^2 -3)-(2x+5)*2x/(x^2-3)^2 is that correct??? f(x)= (lnx)(cos x) f'(x)= 1/x*-sinx or do i have to use the product rule?? thank you.
asked by Dan on November 9, 2008
77. ## Algebra
Find the largest three-digit number that can be written in the form \$3^m + 2^n\$, where \$m\$ and \$n\$ are positive integers.
asked by Mr. Alexander on August 24, 2015
78. ## Math
x^3 + 8x^2 = 20x How do I find the root of ^ that equation? 2x^2-10x+12 = 0 ^ I found that one by... 2(x-3)(x-2) x-3 = 0 x = 3 x-2 = 0 x =2 ^ Is that the correct way to solve for that one
asked by -Untamed- on December 5, 2011
79. ## Math
x^3 + 8x^2 = 20x How do I find the root of ^ that equation? 2x^2-10x+12 = 0 ^ I found that one by... 2(x-3)(x-2) x-3 = 0 x = 3 x-2 = 0 x =2 ^ Is that the correct way to solve for that one?
asked by -Untamed- on December 5, 2011
80. ## Math
If i wanted to find the triangle and the perimeter is 7x+2y units. OPPOSITE: 2x+y ADJACENT: 3x-5y WHAT IS THE HYPOTHENUSE? Please help me.
asked by Anonymous on March 24, 2009
find an equation of the linear function f using the given information. please show all steps and work f(-1)=1, f(1)= -2
asked by Tristan on September 7, 2010
82. ## algebra 2
check my answer please find the number of zeros(x-intercepts) y=3k^3-6k^2-45k=k(3k+9)(k-5)= k(3k^2-15k+9k-45)=k(3k^2-6k-45) or should i do it 3k(k-5) (k+3)=3k(k^2+3k-5k-15)=3k(k^2-2k-15) which on is the right one please let me know.thanks
asked by lisa on December 7, 2012
83. ## Algebra
Find three consecutive integers whose sum is 138 by writing and solving an equation.
asked by Bonnie on February 19, 2017
84. ## Math
Two numbers differ by 6. The sum of their squares is 116. Find the biggest number.
asked by Raquel on July 31, 2017
85. ## math
Describe the steps you would use to add 57+29 on a hundred chart. Tell how to use the rows to find the sum.
asked by sheryl on October 13, 2010
86. ## critical maths problem need help!!!
show that any fuction y(t)=acosx+bsinx ,satisfied the differential equation y^"+wy=0 and also find the value of (a) and (b)
asked by edward on August 23, 2015
87. ## diploma applied physics1
Two forces of 3N and 2N are acting at an angle of 45 to each other. Find the magnitude and direction of the resultant force?
asked by gian on July 29, 2017
88. ## calculus
find the following derivatives: f(x) = (2x+5)/(x^2-3) f'(x)= (2)*(x^2 -3)-(2x+5)*2x/(x^2-3)^2 is that correct??? f(x)= (lnx)(cos x) f'(x)= 1/x*-sinx or do i have to use the product rule?? thank you.
asked by Dan on November 9, 2008
89. ## algebra
find the eq to the graph for the horizontal asymtote of y= 3 - x+b/x-c my guess is that it would be 2 b/c bringing the 3 to the other side would make it -1. and -x over x is also -1 is this right? yes.
asked by sarah on July 15, 2007
a circle has the equation x^2+y^2+6x-6y-46=0 find the center and the radius of the euation. If there are any intercepts list them.
asked by Heather on December 9, 2012
91. ## Geometric series
please help with finding the equations to these where n= 1; which means 1 is needed to find the first number in the series 1) 5/1 + 8/1 + 1/1 + 14/1 + . . . 2) 6/5 + 9/8 + 14/11 + 21/14 + . . . 3) -6/3 - 4/4 - 2/5 - 0 + . . . 4) 9/1 + 36/4 +27/9 + 144/16 +
asked by Janet on March 29, 2009
92. ## Geometry
Find the equation of a circle centre on the line y=2x+1 touching the y axis and passing through A(4,5)
asked by Shanu on January 22, 2015
93. ## Geometry
Find the perimeter of a square to the nearest tenth if the length of its diagonal is 16 millimeters
asked by Amber on March 6, 2011
94. ## math
the area of rectangle is 192 sq.meter its diagonal is 20 in.Find its length and width.
asked by chevlios on January 22, 2015
95. ## Math
The area of the base of a cubiod is 44cm.Find the height,If the volume is 378.4 cm
asked by Amir on January 22, 2015
96. ## geometry
The vertices of triangle ABC are A(2,2), B(3,-1) and C(0,1). Find the coordinates of the circumcenter. Answer choices:(23/14,3/14), (3/2,0) or (7/9,11/9)
asked by hannah on November 7, 2008
97. ## geometry
The angle of elevation from L to K measures 24°. If KL = 19, find JL. Round your answer to the nearest tenth.
asked by SJRoyal on January 22, 2015
98. ## math
two numbers differ by 11 and 1/3rd of the greater exceeds 1/4th of the other by 7 . find the number ???
asked by TUHITUHI on October 1, 2012
99. ## ALGEBRA
Find a degree 3 polynomial with real coefficients having zeros 2 and 4-3i and a lead coefficient of 1.
asked by gigi on August 11, 2017
100. ## History
where was the Caribbean island located in the year 1770? I couldn't find it online. | 5,964 | 18,029 | {"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} | 3.703125 | 4 | CC-MAIN-2019-35 | latest | en | 0.901519 |
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Posted 8 May 2002
# An introduction to bitwise operators
, 8 May 2002 CPOL
Rate this:
## Introduction
I have noticed that some people seem to have problems with bitwise operators, so I decided to write this brief tutorial on how to use them.
## An Introduction to bits
Bits, what are they you may ask?
Well, simply put, bits are the individual ones and zeros that make up every thing we do with computers. All the data you use is stored in your computer using bits. A BYTE is made up of eight bits, a WORD is two BYTEs, or sixteen bits. And a DWORD is two WORDS, or thirty two bits.
``` 0 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 0 0 0
|| | | | ||
|+- bit 31 | | | bit 0 -+|
| | | | |
+-- BYTE 3 -----+--- BYTE 2 ----+--- BYTE 1 ----+-- BYTE 0 -----+
| | |
+----------- WORD 1 ------------+----------- WORD 0 ------------+
| |
+--------------------------- DWORD -----------------------------+```
The beauty of having bitwise operators is that you can use a BYTE, WORD or DWORD as a small array or structure. Using bitwise operators you can check or set the values of individual bits or even a group of bits.
## Hexadecimal numbers and how they relate to bits
When working with bits, it is kind of hard to express every number using just ones and zeros, which is known as binary notation. To get around this we use hexadecimal (base 16) numbers.
As you may or may not know, it takes four bits to cover all the numbers from zero to fifteen, which also happens to be the range of a single digit hexadecimal number. This group of four bits, or half a BYTE, is called a nibble. As there are two nibbles in a BYTE, we can use two hexadecimal digits to show the value of one BYTE.
```NIBBLE HEX VALUE
====== =========
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F```
So if we had one BYTE containing the letter 'r' (ASCII code 114) it would look like this:
```0111 0010 binary
We could write it as '0x72'
## Bitwise operators
There are six bitwise operators. They are:
& The AND operator
| The OR operator
^ The XOR operator
~ The Ones Complement or Inversion operator
>> The Right Shift operator
<< The Left Shift operator.
### The & operator
The & (AND) operator compares two values, and returns a value that has its bits set if, and only if, the two values being compared both have their corresponding bits set. The bits are compared using the following table
```1 & 1 == 1
1 & 0 == 0
0 & 1 == 0
0 & 0 == 0
```
An ideal use for this is to set up a mask to check the values of certain bits. Say we have a BYTE that contains some bit flags, and we want to check if bit four bit is set.
```BYTE b = 50;
if ( b & 0x10 )
cout << "Bit four is set" << endl;
else
cout << "Bit four is clear" << endl;```
This would result in the following calculation
``` 00110010 - b
& 00010000 - & 0x10
----------
00010000 - result
```
So we see that bit four is set.
### The | operator
The | (OR) operator compares two values, and returns a value that has its bits set if one or the other values, or both, have their corresponding bits set. The bits are compared using the following table
```1 | 1 == 1
1 | 0 == 1
0 | 1 == 1
0 | 0 == 0
```
An ideal use for this is to ensure that certain bits are set. Say we want to ensure that bit three of some value is set
```BYTE b = 50;
BYTE c = b | 0x04;
cout << "c = " << c << endl;```
This would result in the following calculation
``` 00110010 - b
| 00000100 - | 0x04
----------
00110110 - result
```
### The ^ operator
The ^ (XOR) operator compares two values, and returns a value that has its bits set if one or the other value has its corresponding bits set, but not both. The bits are compared using the following table
```1 ^ 1 == 0
1 ^ 0 == 1
0 ^ 1 == 1
0 ^ 0 == 0
```
An ideal use for this is to toggle certain bits. Say we want toggle the bits three and four
```BYTE b = 50;
cout << "b = " << b << endl;
b = b ^ 0x18;
cout << "b = " << b << endl;
b = b ^ 0x18;
cout << "b = " << b << endl;```
This would result in the following calculations
``` 00110010 - b
^ 00011000 - ^ 0x18
----------
00101010 - result
00101010 - b
^ 00011000 - ^ 0x18
----------
00110010 - result
```
### The ~ operator
The ~ (Ones Complement or inversion) operator acts only on one value and it inverts it, turning all the ones int zeros, and all the zeros into ones. An ideal use of this would be to set certain bytes to zero, and ensuring all other bytes are set to one, regardless of the size of the data. Say we want to set all the bits to one except bits zero and one
```BYTE b = ~0x03;
cout << "b = " << b << endl;
WORD w = ~0x03;
cout << "w = " << w << endl;```
This would result in the following calculations
```00000011 - 0x03
11111100 - ~0x03 b
0000000000000011 - 0x03
1111111111111100 - ~0x03 w
```
Another ideal use, is to combine it with the & operator to ensure that certain bits are set to zero. Say we want to clear bit four
```BYTE b = 50;
cout << "b = " << b << endl;
BYTE c = b & ~0x10;
cout << "c = " << c << endl;```
This would result in the following calculations
``` 00110010 - b
& 11101111 - ~0x10
----------
00100010 - result
```
### The >> and << operators
The >> (Right shift) and << (left shift) operators move the bits the number of bit positions specified. The >> operator shifts the bits from the high bit to the low bit. The << operator shifts the bits from the low bit to the high bit. One use for these operators is to align the bits for whatever reason (check out the MAKEWPARAM, HIWORD, and LOWORD macros)
```BYTE b = 12;
cout << "b = " << b << endl;
BYTE c = b << 2;
cout << "c = " << c << endl;
c = b >> 2;
cout << "c = " << c << endl;```
This would result in the following calculations
```00001100 - b
00110000 - b << 2
00000011 - b >> 2
```
## Bit Fields
Another interesting thing that can be done using bits is to have bit fields. With bit fields you can set up minature structures within a BYTE, WORD or DWORD. Say, for example, we want to keep track of dates, but we want to use the least amount of memory as possible. We could declare our structure this way
```struct date_struct {
BYTE day : 5, // 1 to 31
month : 4, // 1 to 12
year : 14; // 0 to 9999
} date;```
In this example, the day field takes up the lowest 5 bits, month the next four, and year the next 14 bits. So we can store the date structure in twenty three bits, which is contained in three BYTEs. The twenty fourth bit is ignored. If I had declared it using an integer for each field, the structure would have taken up 12 BYTEs.
```|0 0 0 0 0 0 0 0|0 0 0 0 0 0 0 0|0 0 0 0 0 0 0 0|
| | | |
+------ year ---------------+ month +-- day --+```
Now lets pick this declaration apart to see what we are doing.
First we will look at the data type we are using for the bit field structure. In this case we used a BYTE. A BYTE is 8 bits, and by using it, the compiler will allocate one BYTE for storage. If however, we use more than 8 bits in our structure, the compiler will allocate another BYTE, as many BYTEs as it takes to hold our structure. If we had used a WORD or DWORD, the compiler would have allocated a total of 32 bits to hold our structure.
Now lets look at how the various fields are declared. First we have the variable (day, month, and year), followed by a colon that separates the variable from the number of bits that it contains. Each bit field is separated by a comma, and the list is ended with a semicolon.
Now we get to the struct declaration. We put the bit fields into a struct like this so that we can use convention structure accessing notation to get at the structure members. Also, since we can not get the addresses of bit fields, we can now use the address of the structure.
```date.day = 12;
dateptr = &date;
dateptr->year = 1852;```
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Code Pattern
# Print Z pattern
In this article, we will discuss how to print a Z pattern using a C program. We'll explain the problem statement, provide an explanation with suitable examples, present the algorithm and pseudocode, and finally, explain the output with time complexity of the code.
## Problem Statement
The problem is to write a program that prints a Z pattern of asterisks based on the given size. The size represents the height of the Z pattern, and each line of the pattern consists of asterisks (*) and spaces ( ).
## Explanation with Examples
Let's understand the problem with some examples. Consider the following:
Example 1:
Size: 5
```* * * * *
*
*
*
* * * * *
```
Example 2:
Size: 7
```* * * * * * *
*
*
*
*
*
* * * * * * *
```
Example 3:
Size: 3
```* * *
*
* * *
```
Example 4:
Size: 9
```* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
```
In the examples above, we can observe that the Z pattern is formed using asterisks. The first and last lines contain a row of asterisks equal to the size minus one. The middle lines contain spaces to align the subsequent asterisk.
## Algorithm and Pseudocode
To print the Z pattern, we can follow the following algorithm:
1. Define a function to display spaces. This function takes an integer size as input.
2. Inside the space() function, iterate from 0 to size and print a space character.
3. Define a function to display asterisks. This function takes an integer size as input.
4. Inside the display_star() function, iterate from 0 to size and print an asterisk followed by a space.
5. Define the main function.
6. Inside the main function, define a function called show_z() that takes an integer size as input.
7. If the size is less than 2, return from the show_z() function.
8. Print the value of size.
9. Initialize a variable called side with the value of ((size-1)*2).
10. Use a loop to print the Z pattern:
1. If the current iteration is the first or last line, call the display_star() function with size-1 as the argument.
2. Otherwise, call the space() function with the side value as the argument.
3. Print an asterisk (*) on each line.
4. Decrease the value of side by 2 in each iteration.
11. Print a newline character.
12. In the main function, call the show_z() function with different test cases to demonstrate the pattern.
13. Return 0 to indicate successful execution of the program.
The pseudocode for the given algorithm can be written as follows:
``````function space(size):
for i = 0 to size:
print " "
function display_star(size):
for i = 0 to size:
print "* "
function show_z(size):
if size < 2:
return
print "Size: " + size + "\n\n"
side = (size-1)*2
for i = 0 to size:
if i == 0 or i == size-1:
display_star(size-1)
else:
space(side)
print "*"
print "\n"
side = side - 2
main:
show_z(5)
show_z(7)
show_z(3)
show_z(9)
return 0```
```
## Code Solution
``````// C Program
// Display Z pattern
#include <stdio.h>
//include space
void space(int size)
{
for (int i = 0; i < size; ++i)
{
printf(" ");
}
}
//include star
void display_star(int size)
{
for (int i = 0; i < size; i++)
{
printf("* ");
}
}
void show_z(int size)
{
if(size < 2 )
{
return;
}
printf("Size : %d\n\n",size );
int side = ((size-1)*2);
//Display Z pattern of given size
for (int i = 0; i < size; i++)
{
if(i == 0 || i == size-1)
{
display_star(size-1);
}
else
{
space(side);
}
printf("*");
printf("\n");
side-=2;
}
printf("\n");
}
int main()
{
//Test Case
show_z(5);
show_z(7);
show_z(3);
show_z(9);
return 0;
}```
```
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// C++ Program
// Display Z pattern
#include<iostream>
using namespace std;
class MyPattern {
public:
//include space
void space(int size) {
for (int i = 0; i < size; ++i) {
cout << " ";
}
}
//include star
void display_star(int size) {
for (int i = 0; i < size; i++) {
cout << "* ";
}
}
void show_z(int size) {
if (size < 2) {
return;
}
cout << "Size : " << size << "\n\n";
int side = ((size - 1) *2);
//Display Z pattern of given size
for (int i = 0; i < size; i++) {
if (i == 0 ||
i == size - 1) {
this->display_star(size - 1);
} else {
this->space(side);
}
cout << "*";
cout << "\n";
side -= 2;
}
cout << "\n";
}
};
int main() {
MyPattern obj = MyPattern();
//Test Case
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
return 0;
}```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// Java Program
// Display Z pattern
class MyPattern {
//include space
public void space(int size) {
for (int i = 0; i < size; ++i) {
System.out.print(" ");
}
}
//include star
public void display_star(int size) {
for (int i = 0; i < size; i++) {
System.out.print("* ");
}
}
public void show_z(int size) {
if (size < 2) {
return;
}
System.out.print("Size : "+size+"\n\n");
int side = ((size - 1) * 2);
//Display Z pattern of given size
for (int i = 0; i < size; i++) {
if (i == 0 || i == size - 1) {
display_star(size - 1);
} else {
space(side);
}
System.out.print("*");
System.out.print("\n");
side -= 2;
}
System.out.print("\n");
}
public static void main(String[] args)
{
MyPattern obj = new MyPattern();
//Test Case
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
}
}```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// C# Program
// Display Z pattern
using System;
public class MyPattern {
//include space
public void space(int size) {
for (int i = 0; i < size; ++i) {
Console.Write(" ");
}
}
//include star
public void display_star(int size) {
for (int i = 0; i < size; i++) {
Console.Write("* ");
}
}
public void show_z(int size) {
if (size < 2) {
return;
}
Console.Write("Size : " + size + "\n\n");
int side = ((size - 1) * 2);
//Display Z pattern of given size
for (int i = 0; i < size; i++) {
if (i == 0 ||
i == size - 1) {
display_star(size - 1);
} else {
space(side);
}
Console.Write("*");
Console.Write("\n");
side -= 2;
}
Console.Write("\n");
}
public static void Main(String[] args) {
MyPattern obj = new MyPattern();
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
}
}```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````<?php
// Php Program
// Display Z pattern
class MyPattern {
//include space
public function space(\$size) {
for (\$i = 0; \$i < \$size; ++\$i) {
echo(" ");
}
}
//include star
public function display_star(\$size) {
for (\$i = 0; \$i < \$size; \$i++) {
echo("* ");
}
}
public function show_z(\$size) {
if (\$size < 2) {
return;
}
echo("Size : ". \$size ."\n\n");
\$side = ((\$size - 1) *2);
//Display Z pattern of given size
for (\$i = 0; \$i < \$size; \$i++) {
if (\$i == 0 ||
\$i == \$size - 1) {
\$this->display_star(\$size - 1);
} else {
\$this->space(\$side);
}
echo("*");
echo("\n");
\$side -= 2;
}
echo("\n");
}
}
function main() {
\$obj = new MyPattern();
//Test Case
\$obj->show_z(6);
\$obj->show_z(3);
\$obj->show_z(4);
\$obj->show_z(9);
}
main();```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// Node Js Program
// Display Z pattern
class MyPattern {
//include space
space(size) {
for (var i = 0; i < size; ++i) {
process.stdout.write(" ");
}
}
//include star
display_star(size) {
for (var i = 0; i < size; i++) {
process.stdout.write("* ");
}
}
show_z(size) {
if (size < 2) {
return;
}
process.stdout.write("Size : " + size + "\n\n");
var side = ((size - 1) *2);
//Display Z pattern of given size
for (var i = 0; i < size; i++) {
if (i == 0 ||
i == size - 1) {
this.display_star(size - 1);
} else {
this.space(side);
}
process.stdout.write("*");
process.stdout.write("\n");
side -= 2;
}
process.stdout.write("\n");
}
}
function main(args) {
var obj = new MyPattern();
//Test Case
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
}
main();```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````# Python 3 Program
# Display Z pattern
class MyPattern :
# include space
def space(self, size) :
i = 0
while (i < size) :
print(" ", end = "")
i += 1
# include star
def display_star(self, size) :
i = 0
while (i < size) :
print("*", end = " ")
i += 1
def show_z(self, size) :
if (size < 2) :
return
print("Size : ", size ,"\n\n", end = "")
side = ((size - 1) * 2)
i = 0
# Display Z pattern of given size
while (i < size) :
if (i == 0 or i == size - 1) :
self.display_star(size - 1)
else :
self.space(side)
print("*", end = "")
print("\n", end = "")
side -= 2
i += 1
print("\n", end = "")
def main() :
obj = MyPattern()
obj.show_z(6)
obj.show_z(3)
obj.show_z(4)
obj.show_z(9)
if __name__ == "__main__":
main()```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````# Ruby Program
# Display Z pattern
class MyPattern
# include space
def space(size)
i = 0
while (i < size)
print(" ")
i += 1
end
end
# include star
def display_star(size)
i = 0
while (i < size)
print("* ")
i += 1
end
end
def show_z(size)
if (size < 2)
return
end
print("Size : ", size ,"\n\n")
side = ((size - 1) * 2)
i = 0
# Display Z pattern of given size
while (i < size)
if (i == 0 ||
i == size - 1)
self.display_star(size - 1)
else
self.space(side)
end
print("*")
print("\n")
side -= 2
i += 1
end
print("\n")
end
end
def main()
obj = MyPattern.new()
obj.show_z(6)
obj.show_z(3)
obj.show_z(4)
obj.show_z(9)
end
main()```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// Scala Program
// Display Z pattern
class MyPattern {
//include space
def space(size: Int): Unit = {
var i: Int = 0;
while (i < size) {
print(" ");
i += 1;
}
}
//include star
def display_star(size: Int): Unit = {
var i: Int = 0;
while (i < size) {
print("* ");
i += 1;
}
}
def show_z(size: Int): Unit = {
if (size < 2) {
return;
}
print("Size : " + size + "\n\n");
var side: Int = ((size - 1) * 2);
var i: Int = 0;
//Display Z pattern of given size
while (i < size) {
if (i == 0 ||
i == size - 1) {
display_star(size - 1);
} else {
space(side);
}
print("*");
print("\n");
side -= 2;
i += 1;
}
print("\n");
}
}
object Main {
def main(args: Array[String]): Unit = {
var obj: MyPattern = new MyPattern();
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
}
}```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// Swift Program
// Display Z pattern
class MyPattern {
//include space
func space(_ size: Int) {
var i: Int = 0;
while (i < size) {
print(" ", terminator: "");
i += 1;
}
}
//include star
func display_star(_ size: Int) {
var i: Int = 0;
while (i < size) {
print("*", terminator: " ");
i += 1;
}
}
func show_z(_ size: Int) {
if (size < 2) {
return;
}
print("Size : ", size ,"\n\n", terminator: "");
var side: Int = ((size - 1) * 2);
var i: Int = 0;
//Display Z pattern of given size
while (i < size) {
if (i == 0 ||
i == size - 1) {
self.display_star(size - 1);
} else {
self.space(side);
}
print("*", terminator: "");
print(terminator: "\n");
side -= 2;
i += 1;
}
print(terminator: "\n");
}
}
func main() {
let obj: MyPattern = MyPattern();
obj.show_z(6);
obj.show_z(3);
obj.show_z(4);
obj.show_z(9);
}
main();```
```
#### Output
``````Size : 6
* * * * * *
*
*
*
*
* * * * * *
Size : 3
* * *
*
* * *
Size : 4
* * * *
*
*
* * * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
``````// Rust Program
// Display Z pattern
fn main() {
//Test Case
show_z(5);
show_z(7);
show_z(3);
show_z(9);
}
fn show_z(size: i32) {
if size < 2 {
return;
}
print!("Size : {}\n\n", size);
let mut side: i32 = (size - 1) * 2;
let mut i: i32 = 0;
//Display Z pattern of given size
while i < size {
if i == 0 || i == size - 1 {
display_star(size - 1);
}
else {
space(side);
}
print!("*");
print!("\n");
side -= 2;
i += 1;
}
print!("\n");
}
fn display_star(size: i32) {
let mut i: i32 = 0;
while i < size {
print!("* ");
i += 1;
}
}
fn space(size: i32) {
let mut i: i32 = 0;
while i < size {
print!(" ");
i += 1;
}
}```
```
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
``````
The output matches the expected Z pattern for each test case. The pattern is printed based on the given size, and the asterisks (*) are arranged in the shape of a Z.
## Time Complexity
The time complexity of the code is O(size), where size is the input parameter representing the height of the Z pattern. The code iterates over the size to print each line of the pattern. Therefore, the time complexity is linearly proportional to the size.
## Inverted Z pattern
``````// C Program
// Display inverted Z pattern
#include <stdio.h>
//include space
void space(int size)
{
for (int i = 0; i < size; ++i)
{
printf(" ");
}
}
//include star
void display_star(int size)
{
for (int i = 0; i < size; i++)
{
printf("* ");
}
}
//Display inverted Z pattern of given size
void print_inverted_z(int size)
{
if (size <= 2)
{
return;
}
printf("Size : %d\n\n", size);
//Display Z pattern of given size
for (int i = 0; i < size; i++)
{
if (i == 0 || i == size - 1)
{
display_star(size - 1);
}
else
{
space(i + i);
}
printf("*");
printf("\n");
}
printf("\n");
}
int main()
{
//Test Cases
print_inverted_z(5);
print_inverted_z(7);
print_inverted_z(3);
print_inverted_z(9);
print_inverted_z(6);
return 0;
}``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// Java Program
// Display inverted Z pattern
class MyPattern
{
//include space
public void space(int size)
{
for (int i = 0; i < size; ++i)
{
System.out.print(" ");
}
}
//include star
public void display_star(int size)
{
for (int i = 0; i < size; i++)
{
System.out.print("* ");
}
}
//Display inverted Z pattern of given size
public void print_inverted_z(int size)
{
if (size <= 2)
{
return;
}
System.out.print("Size : " + size + "\n\n");
//Display Z pattern of given size
for (int i = 0; i < size; i++)
{
if (i == 0 || i == size - 1)
{
display_star(size - 1);
}
else
{
space(i + i);
}
System.out.print("*");
System.out.print("\n");
}
System.out.print("\n");
}
public static void main(String[] args)
{
MyPattern obj = new MyPattern();
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
}
}``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// C++ Program
// Display inverted Z pattern
#include<iostream>
using namespace std;
class MyPattern
{
public:
//include space
void space(int size)
{
for (int i = 0; i < size; ++i)
{
cout << " ";
}
}
//include star
void display_star(int size)
{
for (int i = 0; i < size; i++)
{
cout << "* ";
}
}
//Display inverted Z pattern of given size
void print_inverted_z(int size)
{
if (size <= 2)
{
return;
}
cout << "Size : " << size << "\n\n";
//Display Z pattern of given size
for (int i = 0; i < size; i++)
{
if (i == 0 || i == size - 1)
{
this->display_star(size - 1);
}
else
{
this->space(i + i);
}
cout << "*";
cout << "\n";
}
cout << "\n";
}
};
int main()
{
MyPattern obj;
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
return 0;
}``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// C# Program
// Display inverted Z pattern
using System;
class MyPattern
{
//include space
public void space(int size)
{
for (int i = 0; i < size; i++)
{
Console.Write(" ");
}
}
//include star
public void display_star(int size)
{
for (int i = 0; i < size; i++)
{
Console.Write("* ");
}
}
//Display inverted Z pattern of given size
public void print_inverted_z(int size)
{
if (size <= 2)
{
return;
}
Console.Write("Size : " + size + "\n\n");
//Display Z pattern of given size
for (int i = 0; i < size; i++)
{
if (i == 0 || i == size - 1)
{
display_star(size - 1);
}
else
{
space(i + i);
}
Console.Write("*");
Console.Write("\n");
}
Console.Write("\n");
}
public static void Main(String[] args)
{
MyPattern obj = new MyPattern();
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
}
}``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````<?php
// Php Program
// Display inverted Z pattern
class MyPattern
{
//include space
public function space(\$size)
{
for (\$i = 0; \$i < \$size; ++\$i)
{
echo(" ");
}
}
//include star
public function display_star(\$size)
{
for (\$i = 0; \$i < \$size; \$i++)
{
echo("* ");
}
}
//Display inverted Z pattern of given size
public function print_inverted_z(\$size)
{
if (\$size <= 2)
{
return;
}
echo("Size : ". \$size ."\n\n");
//Display Z pattern of given size
for (\$i = 0; \$i < \$size; \$i++)
{
if (\$i == 0 || \$i == \$size - 1)
{
\$this->display_star(\$size - 1);
}
else
{
\$this->space(\$i + \$i);
}
echo("*");
echo("\n");
}
echo("\n");
}
}
function main()
{
\$obj = new MyPattern();
//Test Cases
\$obj->print_inverted_z(5);
\$obj->print_inverted_z(7);
\$obj->print_inverted_z(3);
\$obj->print_inverted_z(9);
\$obj->print_inverted_z(6);
}
main();``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// Node Js Program
// Display inverted Z pattern
class MyPattern
{
//include space
space(size)
{
for (var i = 0; i < size; ++i)
{
process.stdout.write(" ");
}
}
//include star
display_star(size)
{
for (var i = 0; i < size; i++)
{
process.stdout.write("* ");
}
}
//Display inverted Z pattern of given size
print_inverted_z(size)
{
if (size <= 2)
{
return;
}
process.stdout.write("Size : " + size + "\n\n");
//Display Z pattern of given size
for (var i = 0; i < size; i++)
{
if (i == 0 || i == size - 1)
{
this.display_star(size - 1);
}
else
{
this.space(i + i);
}
process.stdout.write("*");
process.stdout.write("\n");
}
process.stdout.write("\n");
}
}
function main(args)
{
var obj = new MyPattern();
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
}
main();``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````# Python 3 Program
# Display inverted Z pattern
class MyPattern :
# include space
def space(self, size) :
i = 0
while (i < size) :
print(" ", end = "")
i += 1
# include star
def display_star(self, size) :
i = 0
while (i < size) :
print("* ", end = "")
i += 1
# Display inverted Z pattern of given size
def print_inverted_z(self, size) :
if (size <= 2) :
return
print("Size : ", size ,"\n")
i = 0
while (i < size) :
if (i == 0 or i == size - 1) :
self.display_star(size - 1)
else :
self.space(i + i)
print("*", end = "")
print(end = "\n")
i += 1
print(end = "\n")
def main() :
obj = MyPattern()
# Test Cases
obj.print_inverted_z(5)
obj.print_inverted_z(7)
obj.print_inverted_z(3)
obj.print_inverted_z(9)
obj.print_inverted_z(6)
if __name__ == "__main__": main()``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````# Ruby Program
# Display inverted Z pattern
class MyPattern
# include space
def space(size)
i = 0
while (i < size)
print(" ")
i += 1
end
end
# include star
def display_star(size)
i = 0
while (i < size)
print("* ")
i += 1
end
end
# Display inverted Z pattern of given size
def print_inverted_z(size)
if (size <= 2)
return
end
print("Size : ", size ,"\n\n")
i = 0
while (i < size)
if (i == 0 || i == size - 1)
self.display_star(size - 1)
else
self.space(i + i)
end
print("*")
print("\n")
i += 1
end
print("\n")
end
end
def main()
obj = MyPattern.new()
# Test Cases
obj.print_inverted_z(5)
obj.print_inverted_z(7)
obj.print_inverted_z(3)
obj.print_inverted_z(9)
obj.print_inverted_z(6)
end
main()``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// Scala Program
// Display inverted Z pattern
class MyPattern
{
//include space
def space(size: Int): Unit = {
var i: Int = 0;
while (i < size)
{
print(" ");
i += 1;
}
}
//include star
def display_star(size: Int): Unit = {
var i: Int = 0;
while (i < size)
{
print("* ");
i += 1;
}
}
//Display inverted Z pattern of given size
def print_inverted_z(size: Int): Unit = {
if (size <= 2)
{
return;
}
print("Size : " + size + "\n\n");
var i: Int = 0;
while (i < size)
{
if (i == 0 || i == size - 1)
{
display_star(size - 1);
}
else
{
space(i + i);
}
print("*");
print("\n");
i += 1;
}
print("\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var obj: MyPattern = new MyPattern();
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
}
}``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
``````// Swift Program
// Display inverted Z pattern
class MyPattern
{
//include space
func space(_ size: Int)
{
var i: Int = 0;
while (i < size)
{
print(" ", terminator: "");
i += 1;
}
}
//include star
func display_star(_ size: Int)
{
var i: Int = 0;
while (i < size)
{
print("* ", terminator: "");
i += 1;
}
}
//Display inverted Z pattern of given size
func print_inverted_z(_ size: Int)
{
if (size <= 2)
{
return;
}
print("Size : ", size ,"\n");
var i: Int = 0;
while (i < size)
{
if (i == 0 || i == size - 1)
{
self.display_star(size - 1);
}
else
{
self.space(i + i);
}
print("*", terminator: "");
print(terminator: "\n");
i += 1;
}
print(terminator: "\n");
}
}
func main()
{
let obj: MyPattern = MyPattern();
//Test Cases
obj.print_inverted_z(5);
obj.print_inverted_z(7);
obj.print_inverted_z(3);
obj.print_inverted_z(9);
obj.print_inverted_z(6);
}
main();``````
#### Output
``````Size : 5
* * * * *
*
*
*
* * * * *
Size : 7
* * * * * * *
*
*
*
*
*
* * * * * * *
Size : 3
* * *
*
* * *
Size : 9
* * * * * * * * *
*
*
*
*
*
*
*
* * * * * * * * *
Size : 6
* * * * * *
*
*
*
*
* * * * * *
``````
## Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
Categories
Relative Post | 8,495 | 24,519 | {"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} | 3.546875 | 4 | CC-MAIN-2024-18 | latest | en | 0.787597 |
https://ch.mathworks.com/help/physmod/sdl/ref/tireroadinteractionmagicformula.html | 1,653,680,642,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00184.warc.gz | 207,965,411 | 25,020 | Main Content
# Tire-Road Interaction (Magic Formula)
Tire-road dynamics given by magic formula coefficients
• Library:
• Simscape / Driveline / Tires & Vehicles / Tire Subcomponents
## Description
The Tire-Road Interaction (Magic Formula) block models the interaction between the tire tread and road pavement. The longitudinal force arising from this interaction is given by the magic formula, an empirical equation based on four fitting coefficients. Tire properties such as compliance and inertia are ignored.
### Tire-Road Interaction Model
The Tire-Road Interaction (Magic Formula) block models the longitudinal forces at the tire-road contact patch using the Magic Formula of Pacejka [1].
The figure displays the forces on the tire. The table defines the tire model variables.
Tire-Road Contact Variables
SymbolDescription and Unit
ΩWheel angular velocity
rwWheel radius
VxWheel hub longitudinal velocity
${r}_{w}\Omega$Tire tread longitudinal velocity
${V}_{sx}={r}_{w}\Omega -{V}_{x}$Wheel slip velocity, defined as the difference between the longitudinal velocities of the wheel hub and the tire tread
$k=\frac{{V}_{sx}}{|{V}_{x}|}$Wheel slip
FzVertical load on tire
Fz0Nominal vertical load on tire
${F}_{x}=f\left(\kappa ,{F}_{z}\right)$Longitudinal force exerted on the tire at the contact point.
Also a characteristic function f of the tire.
### Tire Response
Forces and Characteristic Function
A tire model provides a steady-state tire characteristic function, ${F}_{x}=f\left(\kappa ,{F}_{z}\right)$, the longitudinal force Fx on the tire, based on:
• Vertical load Fz
• Wheel slip κ
Magic Formula with Constant Coefficients
The Magic Formula is a specific form for the tire characteristic function, characterized by four dimensionless coefficients, B, C, D, and E, or stiffness, shape, peak, and curvature:
`${F}_{x}=f\left(\kappa ,{F}_{z}\right)={F}_{z}\cdot D\cdot \mathrm{sin}\left(C\cdot \mathrm{arctan}\left\{B\kappa -E\left[B\kappa -\mathrm{arctan}\left(B\kappa \right)\right]\right\}\right)$`
The slope of f at $k=0$ is $BCD\cdot {F}_{z}$.
Magic Formula with Load-Dependent Coefficients
A more general Magic Formula uses dimensionless coefficients that are functions of the tire load. A more complex set of parameters `p_i`, entered in the property inspector, specifies these functions:
`${F}_{x0}={D}_{x}\mathrm{sin}\left({C}_{x}\mathrm{arctan}\left\{{B}_{x}{\kappa }_{x}-{E}_{x}\left[{B}_{x}{\kappa }_{x}-\mathrm{arctan}\left({B}_{x}{\kappa }_{x}\right)\right]\right\}\right)+{S}_{Vx}$`
Where:
`${\kappa }_{{}_{x}}=\kappa +{S}_{{}_{Hx}}$`
`${D}_{\text{x}}={\mu }_{\text{x}}·{F}_{\text{z}}$`
`${B}_{\text{x}}=\frac{{K}_{\text{x}\kappa }}{{C}_{\text{x}}{D}_{\text{x}}+{\epsilon }_{\text{x}}}$`
SHx and SVx represent offsets to the slip and longitudinal force in the force-slip function, or horizontal and vertical offsets if the function is plotted as a curve. μx is the longitudinal load-dependent friction coefficient. εx is a small number inserted to prevent division by zero as Fz approaches zero.
Peak Longitudinal Force and Corresponding Slip
The block uses a representative set of Magic Formula coefficients. The block scales the coefficients to yield the peak longitudinal force Fx0 at the corresponding slip κ0 that you specify, for rated vertical load Fz0.
Magic Formula Coefficients for Typical Road Conditions
Numerical values are based on empirical tire data. These values are typical sets of constant Magic Formula coefficients for common road conditions.
SurfaceBCDE
Dry tarmac101.910.97
Wet tarmac122.30.821
Snow520.31
Ice420.11
## Assumptions and Limitations
• The Tire-Road Interaction (Magic Formula) block assumes longitudinal motion only and includes no camber, turning, or lateral motion.
## Ports
### Input
expand all
Physical signal input port associated with the normal force acting on the tire. The normal force is positive if it acts downward on the tire, pressing it against the pavement.
Physical signal input port associated with the Magic Formula coefficients. Provide the Magic Formula coefficients as a four-element vector, specified in the order [B, C, D, E].
#### Dependencies
Port M is exposed only if the Parameterize by parameter is set to ```Physical signal Magic Formula coefficients```. For more information, see Parameter Dependencies.
### Output
expand all
Physical signal output port associated with the relative slip between the tire and road.
### Conserving
expand all
Mechanical rotational port associated with the axle that the tire sits on.
Mechanical translational port associated with the wheel hub that transmits the thrust generated by the tire to the remainder of the vehicle.
## Parameters
expand all
The table shows how the visibility of some Main parameters depends on the options that you choose for other parameters. To learn how to read the table, see Parameter Dependencies.
Parameter Dependencies
Parameters
Parameterize by — Choose ```Peak longitudinal force and corresponding slip```, ```Constant Magic Formula coefficients```,```Load-dependent Magic Formula coefficients```, or ```Physical signal Magic Formula coefficients```
Peak longitudinal force and corresponding slipConstant Magic Formula coefficientsLoad-dependent Magic Formula coefficientsPhysical signal Magic Formula coefficients — Exposes physical signal input port M for providing the Magic Formula coefficients to the block as an array of elements in this order [B, C, D, E].
Rated vertical load
Magic Formula B coefficient
Magic Formula C-coefficient parameter, p_Cx1
Peak longitudinal force at rated load
Magic Formula C coefficient
Magic Formula D-coefficient parameters, [p_Dx1 p_Dx2]
Slip at peak force at rated load (percent)
Magic Formula D coefficient
Magic Formula E-coefficient parameters, [p_Ex1 p_Ex2 p_Ex3 p_Ex4]
Magic Formula E coefficient
Magic Formula BCD-coefficient parameters, [p_Kx1 p_Kx2 p_Kx3]
Magic Formula H-coefficient parameters, [p_Hx1 p_Hx2]
Magic Formula V-coefficient parameters, [p_Vx1 p_Vx2]
Velocity threshold
Magic Formula tire-road interaction model.
To model tire dynamics under constant pavement conditions, select one of these models:
• ```Peak longitudinal force and corresponding slip``` — Parametrize the Magic Formula with physical characteristics of the tire.
• `Constant Magic Formula coefficients` — Specify the parameters that define the constant B, C, D, and E coefficients as scalars, with these default values.
CoefficientDefault Value
B`10`
C`1.9`
D`1`
E`0.97`
• ```Load-dependent Magic Formula coefficients``` — Specify the parameters that define the load-dependent C, D, E, K, H, and V coefficients as vectors, one for each coefficient, with these default values.
CoefficientParametersDefault Values
Cp_Cx11.685
D[ p_Dx1 p_Dx2 ][ 1.21 –0.037 ]
E[ p_Ex1 p_Ex2 p_Ex3 p_Ex4 ][ 0.344 0.095 –0.02 0 ]
K[ p_Kx1 p_Kx2 p_Kx3 ][ 21.51 –0.163 0.245 ]
H[ p_Hx1 p_Hx2 ][ –0.002 0.002 ]
V[ p_Vx1 p_Vx2 ][ 0 0 ]
To model tire dynamics under variable pavement conditions, select `Physical signal Magic Formula coefficients`. Selecting this model exposes a physical signal port M. Use the port to input the Magic Formula coefficients as a four-element vector, specified in the order [B, C, D,E].
#### Dependencies
Each parameterization method option exposes related parameters and hides unrelated parameters. Selecting ```Physical signal Magic Formula coefficients``` exposes a physical signal input port. For more information, see Parameter Dependencies.
Rated vertical load force Fz0.
#### Dependencies
This parameter is exposed only when you select the ```Peak longitudinal force and corresponding slip``` parameterization method. For more information, see Parameter Dependencies.
Maximum longitudinal force Fx0 that the tire exerts on the wheel when the vertical load equals its rated value Fz0.
#### Dependencies
This parameter is exposed only when you select the ```Peak longitudinal force and corresponding slip``` parameterization method. For more information, see Parameter Dependencies.
Contact slip κ'0, expressed as a percentage (%), when the longitudinal force equals its maximum value Fx0 and the vertical load equals its rated value Fz0.
#### Dependencies
This parameter is exposed only when you select the ```Peak longitudinal force and corresponding slip``` parameterization method. For more information, see Parameter Dependencies.
Load-independent Magic Formula B coefficient.
#### Dependencies
This parameter is exposed only when you select the `Constant Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-independent Magic Formula C coefficient.
#### Dependencies
This parameter is exposed only when you select the `Constant Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-independent Magic Formula D coefficient.
#### Dependencies
This parameter is exposed only when you select the `Constant Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-independent Magic Formula E coefficient.
#### Dependencies
This parameter is exposed only when you select the `Constant Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Nominal normal force Fz0 on tire.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula C coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula D coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula E coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula K coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula H coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Load-dependent Magic Formula V coefficient.
#### Dependencies
This parameter is exposed only when you select the `Load-dependent Magic Formula coefficients` parameterization method. For more information, see Parameter Dependencies.
Wheel hub velocity Vth below which the slip calculation is modified to avoid singular evolution at zero velocity. Must be positive.
## References
[1] Pacejka, H. B. Tire and Vehicle Dynamics. Elsevier Science, 2005.
## Version History
Introduced in R2011a | 2,460 | 11,245 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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} | 3.546875 | 4 | CC-MAIN-2022-21 | latest | en | 0.727923 |
http://bank.oureducation.in/blog/page/4/ | 1,586,057,943,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370528224.61/warc/CC-MAIN-20200405022138-20200405052138-00087.warc.gz | 17,312,425 | 35,029 | Trending
• ## IBPS PO Prelims Memory Based Question 08 October 2017 1st Shift
Oct 8 • Bank, IBPS PO, question bank • 1970 Views
### IBPS PO Prelims Memory Based Question 08 October 2017 1st Shift
IBPS PO Quant Question Asked in 8th Oct Paper
DI:
• Tabular,
• Pie Chart
Number Series:
• 9,5,6,10.5,23,?
Solution:-
9×0.5+0.5=5
5×1+1=6
6×1.5+1.5=10.5
10.5×2+2=23
23×2.5+2.5=60
• 2,?,256,1024,2048
Solution:-
2048/2=1024
1024/4=256
256/8=32
32/16=2
• 1,20,58,134,286,?
Solution:-
1×2+18=20
20×2+18=58
58×2+18=134
134×2+18=286
286×2+18=590
• 8,7,13,38,151,?
Solution:-
8×1-1=7
7×2-1=13
13×3-1=38
38×4-1=151
151×5-1=754
• 18,20,26,38,?,88
Solution:-
18+2=20
20+6=26
26+12=38
38+20=58
58+30=88
IBPS PO English Questions Asked on 8th October Paper
• Reading comprehension: Financial Literacy in worldwide
IBPS PO Reasoning Question Asked in 8th Oct Paper
Puzzles Question
There are eight people T S U V W X Y Z live on different floor of a building. Living in different states.
• Z live on a odd no floor above floor no 3.the one from Gujarat live immediately below Z.
• Three persons live between Z and the one who is from Rajasthan.
• As many people live above z as below X. V live immediately below Y.
• Y live on an odd no floor above X.
• One person live between who is from Maharashtra and Y.
• U live on an odd no floor. One from Kerala live immediately above U.
• As many persons live between U & Z as between s and Goa.
• More than two persons live Y and one from Bihar.
• T is neither from Bihar nor from Odisha.
• ## IBPS PO Prelims Memory Based Question 07 October 2017 1st Shift
Oct 7 • Bank, IBPS PO, question bank • 1109 Views
### IBPS PO Prelims Memory Based Question 07 October 2017 1st Shift
IBPS PO Reasoning Questions Asked in 7th Oct Paper
Puzzles Set-1:
Seven person A, B, C, D, E, F G likes seven colours. Yellow White Red Orange Blue Gray Black. They visit in different days starting from Monday to Sunday
1) A visits one day after Thursday.
2) Only 4 people are in between A and B
3) The one who likes Red colour visits immediately after B.
4) Only one person visit between Red and Blue
5) The one who like white colour visit before one of days on which day C visits
6) The one who like white colour does not visit on Monday.
7) Only 1 person is in between D and E. D like Yellow.
8) There are same as many person in between A and Blue colour which are one less in between B and C.
9) Neither G nor F likes Black.
10) G does not visit on Saturday and like grey colour.
Solutions
Monday- D- Yellow
Tuesday-B-White
Wednesday-E-Red
Thursday-G-Orange
Friday-C-Blue
Saturday-F-Grey
Sunday-A-Black
Puzzle Set-2:
J, K, L, M, N, O and P are seven different boxes of different colours i.e. Brown, Orange, Silver, Pink, Yellow, White and Green but not necessarily in the same order.
Box which is of Brown colour is immediately above J. There are only two box between M and the box which is of Brown colour. Box which is of Silver colour is above M but not immediately above M. Only three box are between L and the box which is of Silver colour.
The box which is of Green colour is immediately above L. The box which is of Pink colour is immediately above the box P. Only one box is there between K and N. Box K is above N. Neither box K nor J is of Yellow colour. J is not of orange colour.
Solution
O- Silver
K- Orange
M- Yellow
N- Green
L- Pink
P- Brown
J- White
Syllogism Questions Asked In IBPS PO Prelims 1st Slot:
Direction (1-2):
Statement:
All remarks are feedbacks
Some feedbacks are words
No word is a digit
1) Conclusion:
Some Feedbacks are definitely not digits.
All digits being feedbacks is a possibility.
2) Conclusion:
All remarks being words is a possibility
At least some remarks are digits.
3) Statements:
Some files are boxes
All boxes are carton
No carton is a plastic
Conclusion:
No file is a plastic.
Some files are plastics.
4) Statement:
Some desks are chairs
Some chairs are seats
No seat is a table
Conclusion:
All desks can never be table
Some chains are definitely not tables.
5) Statements:
All routes are ways
All ways are paths
Some ways are bridges
Conclusion:
At least some bridges are routes
All routes being bridge is a possibility.
Inequalities:
1 L>I=N>P;I≥R>K;N≤E (i) E>P (ii) R
2. S>A=N≥D; A≥L>E; M≤L≤D
(i) S>E (i) L
3. P≥V≥R<=EN
(i)P>N (ii) G≥Y
4. L>I=N>P; I≥R>K; N≤E (i) K>N (ii) I Syllogism:
1. Statement
All remarks are feedbacks
Some feedbacks are words
No word is a digit
Conclusion:-
some feedbacks are definitely not digits
All digits being feedbacks is a possibility
2. All remarks are feedbacks
Some feedbacks are words
No word is a digit
Conclusion:-
All remarks being words is a possibility
At least some remarks are digits
3. Statement
Some files are boxes
All boxes are cottons
No cotton is a plastic
Conclusion:-
No file is a plastic
Some files are plastics
4. Some desks are chairs
Some chairs are seats
No seat is a table
Conclusion:-
All desks can never be tables
Some chairs are definitely not tables
5. All routes are ways
All ways are paths
Some ways are bridges
Conclusion:-
At least some bridges are routes
All routes being bridges is a possibility.
IBPS PO Quantitative Aptitude Questions Asked in 7th Oct Paper
1. Number Series – Find the next number in the given series.
17, 98, 26, 89, 35, ? Ans – 80
2, 17, 89, 359, 1079, ? Ans – 2159
3, 5, 15, 45, 113, ?, Ans – 243
7, 4.5, 5.5, 12, 49, ? Ans – 393
3240, 540, 108, 27, ?, 4.5 Ans – 9
2. Approximation –
i) (√80.997 – √25.001) × (√120.98 + √16.02)= ?
ii) 53.01 – 345.02 ÷ 22.99 = 2×?
iii) √(3099.985 ÷ 62.001+14.001) = ?
iv) (184.002 – 29 ÷ 5)x 29.997= ?
v) (111.91×51) ÷ 14.02 = 11.002+ ?
3. Simple interest on the sum A is 11% per annum and compound interest on the sum
B which is 400 more than A in 2 years is 140% more of simple interest of A. Find the value of A?
4. Data Interpretation – The Line Graph was given as below & questions on ratio, averages & differences are asked.
IBPS PO English Questions Asked in 7th Oct Paper
1. Reading Comprehension – Based on “Economy”. The passage was:The effects of the worst economic downturn since the Great Depression are forcing changes on state governments and the U.S. economy that could linger for decades.
By one Federal Reserve estimate, the country lost almost an entire year’s worth of economic activity – nearly \$14 trillion – during the recession from 2007 to 2009.The deep and persistent losses of the recession forced states to make broad cuts in spending and public workforces. For businesses, the recession led to changes in expansion plans and worker compensation. And for individual Americans, it has meant a future postponed, as fewer buy houses and start families.
Five years after the financial crash, the country is still struggling to recover.
“In the aftermath of [previous] recessions there were strong recoveries. That is not true this time around,” said Gary Burtless, a senior fellow at the Brookings Institution. “This is more like the pace getting out of the Great Depression.”
For years, housing served as the backbone of economic growth and as an investment opportunity that propelled generations of Americans into the middle class. But the financial crisis burst the housing bubble and devastated the real estate market, leaving millions facing foreclosure, millions more underwater, and generally stripping Americans of years’ worth of accumulated wealth.
Anthony B. Sanders, a professor of real estate finance at George Mason University, said even the nascent housing recovery can’t escape the effects of the recession.
Home values may have rebounded, he said, but the factors driving that recovery are very different than those that drove the growth in the market in the 1990s and 2000s. Sanders said more than half of recent home purchases have been made in cash, which signals investors and hedge funds are taking advantage of cheap properties. That could freeze out average buyers and also means little real economic growth underpins those sales.
• ## Letter Writing for IBPS PO 2017
Sep 4 • Bank, English Notes, IBPS PO • 1051 Views
# Letter Writing for IBPS PO 2017
Dear Aspirants!
The IBPS PO Exam is around the corner. This time the IBPS has also introduced the “Descriptive writing” in the Mains Examination. You must be prepared by now with your syllabus for the Mains Examination. The Descriptive writing mainly includes Letter Writing, Essay Writing and Precis writing.
Here are some examples of Letter writing by students of Best coaching in Delhi/Ncr
## Formal letters
A-77,Defence Enclave,
Meerut
XYZ,Municipal Corporation office,
Meerut
Date:- 01st September 2017
Subject: Complaint regarding the poor condition of streets and lack of Street Light.
Dear Sir,
I would like to bring to your notice the pathetic condition of roads and lack of Street lights in Defence Enclave. The roads in my area have not been repaired from quite a long time and contains a lot of potholes which cause maximum number of road accidents in this area.Apart from this, these holes are a dumping ground for various mosquitoes in rainy reason which is the reason for the widespread of various diseases.
Additionally,there are no street lights in our area which also add up to the reason of increasing road accidents here. Lack of the street lights also increase the theft cases in our society day-by day.
I request you to consider the alarming situation and take necessary action to get the roads repaired and increase the number of street lights in my area.
Looking forward towards your instant action.
Thank You.
Yours faithfully,
Shivi Garg
By Shivi Garg
## Letter II
The Commisioner
Municipal Corporation
Sector 80, Phase II Flat B-27
Noida (201302) Sector 53
Noida (201301)
Date – 02/09/2017
Subject – Complain regarding the state of roads and lack of street light in our area.
Dear Sir,
I would like to bring to your kind notice the miserable condition of roads in my area and lack of street lights in sector 53, Noida. The roads are broken at many places as a result of this vehicles cannot move properly.
Because of the bad situation of roads, the traffic problem occurs daily. As the monsoon is on its peak state which puts roads in bad shape. These worse state of roads is also the one reason for road accidents. Such bad conditions od roads has been repeatedly brought to the notice of PWD, but all efforts are in vain.
Apart from this problem, one more issue is there which is also causing many problem i.e. lack of street lights in our society. As i have already discussed that there are many pits and ditches on the roads and the lack of street lights adds more problem in this. Some of the street lights are in working condition hence darkness remain everywhere in society. This led to crime as chain snatching and many others bad activity which is not in the favor of society. The ladies are not safe in evening. The holligans after committing crime takes advantages of darkness and run away.
I, therefore request you to take necessary steps in improving the condition of roads as well as the street lights for the welfare of society.
Thank You.
Yours Faithfully
Shivangi pandey
By Shivangi Pandey
## Informal Letter
#779,sec-16/17
Hissar,Haryana(125001)Dear Mom,
I hope this letter finds you in your best health and spirit.I am writting this letter to share my happiness on the
achievements as this year in the college was very eventful.
I was chosen a class repesentative. Earlier i thought it would be difficult for me to manage this huge responsibility but then my teachers and my fellow classmates encouraged me.
Their believe and trust gave me the strength to grab this opportunity and I did the same. Till the time I was busy as a bee managing the whole class responsibility,all the events and many more things.
And then our college displayed a notice about the cultural activity which I was leading and we have to perform in front of President Of India.This was very cherishable moment as performing in front of the President would
make our college proud. I was very elated as well as nervous after reading the notice.Alot of emotions runnind around my mind.At that point of time self control was important and I succedded in doing so.I galloped to the Rehersal room
and started making teams.The team should be best in all terms I whisphered in my mind and calmly started thinking about whom should be given different roles.
After half an hour I manged to make the list and started calling them by their name explaning each one of them about their roles in the play. There was immense pressure that can be visible on each ones face as we have to do our best no matter how long it will take for us to rehearse.
Finally the day arrived we all were very nervous, sOmebody was needed to boost us up and our mentor took the initative to do this.We are being motivated by him and filled with positive energy.Now the time has come for us to perform. We performed like never before and won the competition.
We recieved the prize from President and he really apllauded our performance and we were very pleased by this.The Adminstrator of our collage honoured me with a trophy.My classmates and my teachers were so supportive as I missed many classes due to Practice sessions.So events and examination are wrapped
and I am coming as my summer holidays are on cards.
yours Lovely Daughter,
Vashali bhatia
By Vaishali Bhatia
• ## Time and Work Quiz
Aug 6 • Bank, Maths Notes, question bank • 1848 Views
Q1. P can complete the work in 15 days and Q can complete the work in 20 days. If
they together works for 4 days then find the fraction of the work left.
1) 1⁄4
2) 1/10
3) 7/15
4) 8/15
5) none of these
Q2. 5 persons can prepare an admission list in 8 days working 7 hours a day. If 2
persons join them so as to complete the work in 4 days, they need to work per day
for
a) 10 hours b) 9 hours c) 12 hours d) 8 hours e)none of these
Q3. A contractor undertakes to make a road in 40 days and employs 25 men. After
24 days, he finds that only one-third of the road is made. How many extra men
should he employ so that he is able to complete the work 4 days earlier?
a) 100 b) 60 c) 75 d) None of these
Q4. A and B together can complete a work in 12 days. A alone can complete it in
20 days. If B does the work only for half a day daily, then in how many days B
alone take to complete the work?
a) 10 days b) 20 days c) 60 days d) 25 days
Q5. Worker P takes 8 hours to do a job. Worker Q takes 10 hours to do the same job. How long
it take if both P & Q working alternately, to do the same job?
A. 40/9
B. 56
C.8 4⁄5 days
D.12 1⁄8
E. none of these
Q6. A can do a certain work in 12 days. B is 60% more efficient than A. How many days does B
alone take to do the same job?
A. 16/4
B. 15/2
C. 12
D. 14/3
E. none of these
Q7. P can do a piece of work in 7 days of 9 hours each and Q alone can do it in 6 days of 7
hours each. How long will they take to do it working together 8 2/5 hours a day?
A. 11
B. 4
C. 7
D. 3
E. none of these
Q8. A takes twice as much time as B and thrice as much time as C to finish a piece of work.
Working together they can finish the work in 2 days. B can do the work alone in ?
A. 7
B. 6
C. 8
D. 9
E. none of these
Q9. A can do a 1⁄4 of a work in 10 days, B can do 40% of work in 40 days and C can do 1/3 of
work in 13 days. Who will complete the work first?
A. A
B. B
C. C
D. cannot be determine
E. none of these
Q10. P and Q together can complete a piece of work in 35 days while P alone can complete the
same work in 60 days. Q alone will be able to complete the same work in :
A. 42 days
B. 72 days
C. 84 days
D. 96 days
E. none of these
1. ANS 4
P completes the work in 15 days
Q completes the work in 20 days
LCM of 15 and 20 is 60. Therefore 60 units work to do.
P’s 1 day work = 60/15 = 4 units
Q’s 1 day work = 60/20 = 3 units
(p+q)’s 1 day work = 4+3= 7 units
In 4 days they do 4*7 = 28 +units work therefore left work is 32/60 = 8/15
Ans 8/15
1. ANS a
5*8*7 = (5+2)*4*x
X = 10hours
Ans 10 hours
1. ANS c
(24*25)/(⅓) = ((25+x) * 12)/(2/3))
Ans 75
1. ANS c
(a+b) complete a work in 12 days
A alone in 20 days
LCM of 12 and 20 is 60. Therefore 60 units work is there to do.
B alone must take 30 days but if B does only for half a day then he takes 60 days to complete 60 unit work.
Ans 60 days
1. ANS c
P takes ———– 8 days
Q takes ———–10 days
Total work is 40 units
40/9 = 4 and 4/9
So ans is equal to 8 + ⅘ = 8 ⅘ days
6.
A:B = 5:8
A can do in 12days
B can do it in 60/8 = 7.5 or 15/2
Ans B
1. ANS D
P 7 9 = Q 7 6
3P = 2Q
P:Q = 2:3
Therefore 2 7 9 = x 5 425
X = 3 days
8.
The ratio of Time taken by A and B to complete the work is 2:1 and that of A and C is 3:1.
Therefore efficiency of A and B is 1:2 and A and C is 1:3.
Efficiency of A,B and C is 1:2:3
6R = 2 days
1R = 12 days
2R = 6 days
Ans B
9.
A can does the whole work in 40 days and B can 100 days C can does in 39 days.
Ans C
10.
(P+Q) can complete a work in 35 days
Q alone can completes in 60 days
LCM of P and Q is 420.
(P+Q)’s 1 day work is 12 units
P’s 1 day work is 7 unit therefore Q’s 1 day work is (12-7) = 5 units
Time taken by Q alone is 420/5 = 84 days
Ans C
• ## Time Speed and Distance Part 4
Jul 30 • Uncategorized • 1477 Views
1. A circular running path is 726 metres in circumference. Two men start from the same point and walk in opposite directions at 3.75 km/h and 4.5 km/h, respectively. When will they meet for the first time ?
(a) After 5.5 min (b) After 6.0 min (c) After 5.28 min (d) After 4.9 min
1. A train after travelling 150 km meets with an accident and then proceeds with 3/5 of its former speed and arrives at its destination 8 h late. Had the accident occurred 360 km further, it would have reached the destination 4 h late. What is the total distance travelled by the train?
(a) 840 km (b) 960 km (c) 870 km (d) 1100 km
1. A man swimming in a steam which flows 1 1 2 km/hr., finds that in a given time he can swim twice as far with the stream as he can against it. At what rate does he swim?
(a) 1 5 2 km/hr (b) 1 4 2 km/hr (c) 1 7 2 km/hr (d) None of these
1. Two persons start from the opposite ends of a 90 km straight track and run to and fro between the two ends. The speed of first person is 30 m/s and the speed of other is 125/6 m/s. They continue their motion for 10 hours. How many times they pass each other?
(a) 10 (b) 9 (c) 12 (d) None of these
1. A man starts from B to K, another from K to B at the same time. After passing each other they complete their journeys in 3 1 3 and 4 4 5 hours, respectively. Find the speed of the second man if the speed of the first is 12 km/hr.
(a) 12.5 kmph (b) 10 kmph (c) 12.66 kmph (d) 20 kmph
1. A train 100 metres long moving at a speed of 50 km/hr. crosses a train 120 metres long coming from opposite direction in 6 sec. The speed of the second train is
(a) 60 km/hr. (b) 82 km/hr. (c) 70 km/hr. (d) 74 km/hr.
1. A passenger sitting in a train of length 100 m, which is running with speed of 60 km/h passing through two bridges, notices that he crosses the first bridge and the second bridge in time intervals which are in the ratio of 7 : 4 respectively. If the length of first bridge be 280 m, then the length of second bridge is:
(a) 490 m (b) 220 m (c) 160 m (d) Can’t be determined
1. A man can row a certain distance against the stream in six hours. However, he would take two hours less to cover the same distance with the current. If the speed of the current is 2 kmph, then what is the rowing speed in still water?
(a) 10 kmph (b) 12 kmph (c) 14 kmph (d) 8 kmph
1. A boat, while going downstream in a river covered a distance of 50 mile at an average speed of 60 miles per hour. While returning, because of the water resistance, it took one hour fifteen minutes to cover the same distance . What was the average speed of the boat during the whole journey?
(a) 40 mph (b) 48 mph (c) 50 mph (d) 55 mph 40.
1. Two trains, 130 m and 110 m long, are going in the same direction. The faster train takes one minute to pass the other completely. If they are moving in opposite directions, they pass each other completely in 3 seconds. Find the speed of each train.
(a) 38 m/sec, 36 m/sec (b) 42 m/sec, 38 m/sec (c) 36 m/sec, 42 m/sec (d) None of these
1. (c) Their relative speeds = (4.5 + 3.75) = 8.25 km/h
Distance = 726 metres = 726/1000km = 0. 726
Required time = 5.28min
1. (c) The speeds of two persons is 108 km/h and 75 km/h. The first person covers 1080 km in 10 hours and thus he makes 12 rounds. Thus, he will pass over another person 12 times in any one of the direction.
1. (b) 1st man ‘s speed2nt man ‘s speed= ab= 445313
2ns man’s speed is 10km/hr
1. (b) Let speed of the second train = x km/hr. Relative speed of trains = (50 + x) km/hr. Distance travelled by trains = (100 + 120) = 220 metres
Distance = Speed × Time
(2201000) km = (50 + x)km/ hr (63600)h
50 + x = 132
X = 82 km/hr
1. (c) Note here the length of the train in which passenger is travelling is not considered since we are concerned with the passenger instead of train. So, the length of the bridge will be directly proportional to the time taken by the passenger respectively.
t = Time
l = Length of bridge
Therefore t1t2= l1l2
X = 160m
1. (a) If the rowing speed in still water be x kmph, and the distance by y km,
then yx-2= 6
y= 6(x-2) …(1) and,
yx+2 = 4
Y = 4 (x + 2) …(2)
6 (x – 2) = 4 (x + 2)
x = 10 kmph
1. (b) Time taken by the boat during downstream journey = 5060h = 56h
Time taken by the boat in upstream journey = 54h
Average speed = 25056 + 54 =1002450= 48mph
1. (b) Let the Speed of faster train be x and speed of slower train be y.
Now, when both the train move in same direction their relative speed = x – y
Now, total distance covered = 130 + 110 = 240
Now, distance = speed × time
240 = ( x– y) × 60 ( 1min = 60sec)
x – y = 4 …(1)
When the trains move in opposite direction then their relative speed = x + y
240 = ( x + y) × 3
80 = x + y …(2)
on solving eqs (1) and (2),
we get x = 42 m/sec and y = 38 m/sec
• ## Time Speed and Distance Part 3
Jul 30 • Uncategorized • 1461 Views
1. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/ hr and the time of flight increased by 30 minutes. The duration of the flight is:
(a) 1 hours (b) 2 hours (c) 3 hours (d) 4 hours
1. Points A and B are 70 km apart on a highway. One car starts from A and the another one from B at the same time. If they travel in the same direction, they meet in 7 hours. But if they travel towards each other, they meet in one hour. The speeds of the two cars are, respectively. (a) 45 and 25 km/h (b) 70 and 10 km/h (c) 40 and 30 km/h (d) 60 and 40 km/h
1. Anil calculated that it will take 45 minutes to cover a distance of 60 km by his car. How long will it take to cover the same distance if the speed of his car is reduced by 15 km/hr?
(a) 36 min (b) 55.38 min (c) 48 min (d) 40 min
1. The jogging track in a sports complex is 726 metres in circumference. Pradeep and his wife start from the same point and walk in opposite directions at 4.5 km/h and 3.75 km/h, respectively. They will meet for the first time in :
(a) 5.5 min (b) 6.0 min (c) 5.28 min (d) 4.9 min
1. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water (in litres) will fall into the sea in a minute?
(a) 4,00,000 (b) 40,00,000 (c) 40,000 (d) 4,000
1. The speed of a boat in still water is 15 km/h and the rate of stream is 5 km/h. The distance travelled downstream in 24 minutes is (a) 4 km (b) 8 km (c) 6 km (d) 16 km
1. A person can swim in still water at 4 km/h. If the speed of water is 2 km/h, how many hours will the man take to swim back against the current for 6 km.
(a) 3 (b) 4 (c) 1 4 2 (d) Insufficient data
1. A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water?
(a) 4 km/h (b) 6 km/h (c) 8 km/h (d) Data inadequate
1. A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30km upstream and 21 km downstream in 6 hours and 30 minutes. The speed of the boat in still water is :
(a) 10 km/h (b) 4 km/h (c) 14 km/h (d) 6km/h
1. A man who can swim 48 m/min in still water swims 200 m against the current and 200 m with the current. If the difference between those two times is 10 minutes, find the speed of the current. (a) 30 m/min (b) 29 m/min (c) 31 m/min (d) 32 m/min
1. (a) Let the duration of the flight be x hours.
Then, x = 1hr
1. (c) Let the speed of the cars be x km/h and y km/h respectively.
Their relative speed when they are moving in same direction = (x – y) km/h.
Their relative speed when they are in opposite directions = (x + y) km/h.
Now, 70x+y= 1 or or x + y = 70 ….. (i) and
70x-y= 7 or x – y = 10 ….. (ii)
Solving (i) and (ii), we get x = 40 km/h and y = 30 km/h.
1. (b) D = S × T
60 = S 4560hr
S =- 80km/hr
Now, new speed = 80 – 15 = 65 km/hr.
Time Distance/Speed 6065hr or 55.38min.
Hence, Time to taken by car to travel same distance is 55.38 min.
1. (c) Let the husband and the wife meet after x minutes. 4500 metres are covered by Pradeep in 60 minutes.
In x minutes, he will cover 450060x metres.
Similarily,
In x minutes, his wife will cover 375060x m.
Now, x = 5.28 min
1. (b) Volume of water flowed in an hour
= 2000 × 40 × 3 cubic metre = 240000 cubic metre
volume of water flowed in 1 minute
= 24000060=4000 = cubic metre = 40,00,000 litre
1. (b) Downstream speed = 15 + 5 = 20 km/h.
Required distance = 20 2460= 8km
1. (a) Man’s speed in upstream = 4 – 2 = 2 km/h.
Required time = 63km/ h = 2km/hr
1. (b) Rate downstream = 162 kmph = 8 kmph;
Rate upstream = 164kmph = 4 kmph.
Speed in still water = 12(8 + 4) = 6 km/h.
1. (a) Let speed of the boat in still water be x km/h and speed of the current be y km/h.
Then, upstream speed = (x – y) km/h
and downstream speed = (x + y) km/h
24x-y + 28x+y = 6 ….. 1
And 30x-y+ 21x+y= 132…… 2
From equation 1 and 2
Now, x = 10km/hr and y = 4km/hr
• ## Time Speed and Distance Part 2
Jul 30 • Uncategorized • 934 Views
1. Cars C1 and C2 travel to a place at a speed of 30 km/h and 45 km/h respectively. If car C2 takes 1 2 2 hours less time than C1 for the journey, the distance of the place is
(a) 300 km (b) 400 km (c) 350 km (d) 225 km
1. A man covers a certain distance on a toy train. If the train moved 4 km/h faster, it would take 30 minutes less. If it moved 2 km/h slower, it would have taken 20 minutes more. Find the distance.
(a) 60 km (b) 58 km (c) 55 km (d) 50 km
1. A goods train le same station and moves in the same direction at a uniform speed of 90 kmph. This train catches eaves a station at a certain time and at a fixed speed. After 6 hours, an express train leaves thup the goods train in 4 hours. Find the speed of the goods train.
(a) 36 kmph (b) 40 kmph (c) 30 kmph (d) 42 kmph
1. Without stoppages, a train travels certain distance with an average speed of 80 km/h, and with stoppages, it covers the same distance with an average speed of 60 km/h. How many minutes per hour the train stops ?
(a) 15 (b) 18 (c) 10 (d) None of these
1. If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office.
(a) 1 hr 2 (b) 1 hr (c) 3 hr 4 (d) None of these
1. A train running between two stations A and B arrives at its destination 10 minutes late when its speed is 50 km/h and 50 minutes late when its speed is 30km/h. What is the distance between the stations A and B ? (a) 40 km (b) 50 km (c) 60 km (d) 70 km
1. A thief goes away with a Maruti car at a speed of 40 km/h. The theft has been discovered after half an hour and the owner sets off in another car at 50 km/h. When will the owner overtake the thief from the start.
(a) 1 2 2 hours (b) 2 hr 20 min (c) 1 hr 45 min (d) cannot be determined
1. A starts 3 min after B for a place 4.5 km away. B on reaching his destination, immediately returns back and after walking a km meets A. If A walks 1 km in 18 minutes then what is B’s speed
(a) 5 km/h (b) 4 km/h (c) 6 km/h (d) 3.5 km/h
1. A long distance runner runs 9 laps of a 400 metres track everyday. His timings (in minutes) for four consecutive days are 88, 96, 89 and 87 respectively. On an average, how many metres/minute does the runner cover ?
(a) 40 m/min (b) 45 m/min (c) 38 m/min (d) 49 m/min
1. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the speed of the car is :
(a) 4 : 3 (b) 3 : 4 (c) 3 : 2 (d) 2 : 3
1. (d) Let C1 takes t hrs.
Then, Q Distance is same.
Distance = 225km
1. (a) Let the distance be x km and speed of train be y km/h.
Then by question, we have xy + 4 = xy 306 …..(i)
xy – 2 = x y+ 2060 …..(ii)
On solving (i) and (ii),
we get x = 3y Put x = 3y
Y = 20
Hence, distance = 20 3 = 60km
1. (a) Let the speed of the goods train be x kmph.
Distance covered by goods train in 10 hours
= Distance covered by express train in 4 hours.
10x = 4 × 90 or x = 36.
So, speed of goods train = 36 kmph.
1. (a) Due to stoppages, it covers 20 km less .
Time taken to cover 20km = 2080hr 15 min
1. (b) If new speed is ab of original speed.
Then, usual time × (ba– 1) change in time
usual time = 13
usual time = 13 3 = 1hr
1. (b) Let the distance between the two stations be x km.
Then, x50106 = x 30506
X = 50km
1. (a) Distance to be covered by the thief and by the owner is same.
Let after time ‘t’, owner catches the thief.
T = 212hr
1. (a) A covers 3.5 km before he meets B in (18 × 3.5 + 3) = 66 min
B’s speed is 5km/hr
1. (a) Average speed = total Distancetotal time= 400 4 9360 = 40 metres /minute
1. (b) Let the speed of the train and the car be x km/h and y km/h, respectively.
Now,
xy= 34
• ## Time, Speed and Distance part 1
Jul 30 • Uncategorized • 1067 Views
1. An aeroplane flies along the four sides of a square at the speeds of 200, 400, 600 and 800 km/h. Find the average speed of the plane around the field.
(a) 384 km/h (b) 370 km/h (c) 368 km/h (d) None of these
1. A monkey ascends a greased pole 12 metres high. He ascends 2 metres in first minute and slips down 1 metre in the alternate minute. In which minute, he reaches the top ?
(a) 21st (b) 22nd (c) 23rd (d) 24th
1. A man walks a certain distance and rides back in 1 6 h. 4 He can walk both ways in 3 7 h. 4 How long it would take to ride both ways ?
(a) 5 hours (b) 1 4 hours 2 (c) 3 4 hours 4 (d) 6 hours
1. There are 20 poles with a constant distance between each pole. A car takes 24 seconds to reach the 12th pole . How much time will it take to reach the last pole?
(a) 25.25 s (b) 17.45 s (c) 35.75 s (d) 41.45 s
1. A man is walking at a speed of 10 km per hour. After every kilometre, he takes rest for 5 minutes. How much time will he take to cover a distance of 5 kilometres?
(a) 48 min. (b) 50 min. (c) 45 min. (d) 55 min.
1. On a journey across Bombay, a tourist bus averages 10 km/h for 20% of the distance, 30 km/h for 60% of it and 20 km/h for the remainder. The average speed for the whole journey was (a) 10 km/h (b) 30 km/h (c) 5 km/h (d) 20 km/h
1. In a 800 m race around a stadium having the circumference of 200 m, the top runner meets the last runner on the 5th minute of the race. If the top runner runs at twice the speed of the last runner, what is the time taken by the top runner to finish the race ?
(a) 20 min (b) 15 min (c) 10 min (d) 5 min
1. A man walks half of the journey at 4 km/h by cycle does one third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hours and 12 minutes. The length of the journey is
(a) 36 km (b) 1332 km 67 (c) 40 km (d) 28 km
1. A train does a journey without stoppage in 8 hours, if it had travelled 5 km/h faster, it would have done the journey in 6 hours 40 minutes. Find its original speed.
(a) 25 km/h (b) 40 km/h (c) 45 km/h (d) 36.5 km/h
1. A train leaves station X at 5 a.m. and reaches station Y at 9 a.m. Another train leaves station Y at 7 a.m. and reaches station X at 10: 30 a.m. At what time do the two trains cross each other ?
(a) 7 : 36 am (b) 7 : 56 am (c) 8 : 36 am (d) 8 : 56 am
1. (a) Let each side of the square be x km and let the average speed of the plane around the field be y km/h.
Then, Average speed = 384 km/h.
1. (a) In 2 minutes, he ascends = 1 metre
10 metres, he ascends in 20 minutes.
He reaches the top in 21st minute.
1. (c) We know that, the relation in time taken with two different modes of transport is
t(walk both) + t(ride both) = 2 (twalk + t ride)
314 + t (ride both) = 2 254
t(ride both) = 434hrs
1. (d) Let the distance between each pole be x m.
Then, the distance up to 12th pole = 11 x m
Speed = 11x24 m/s
Time taken to covers the total distance of 19x = 41.45s
1. (b) Rest time = Number of rest × Time for each rest
= 4 × 5 = 20 minutes
Total time to cover 5 km
(510 60) minutes + 20 minutes = 50 minutes.
1. (d) Let the average speed be x km/h.
and total distance be y km.
Then, 0.210y + 0.630y + 0.220y = yx
X = 20km/hr
1. (c) After 5 minutes (before meeting), the top runner covers 2 rounds i.e., 400 m and the last runner covers 1 round i.e., 200 m. T
op runner covers 800 m race in 10 minutes.
1. (a) Let the length of the journey =x km.
Journey rides by horse cart = x(1- 1213) = 16x km
x= 36km
1. (a) Let original speed = S km/h
Here, distance to be covered is constant
S 8 = (S + 5)(203)
S = 25km/hr
1. (b) Let the distance between X and Y be x km.
Then, the speed of A is x4 km/h and that of B is 2x7 km/h.
Thus, both of them meet at 7 : 56 a.m.
• ## Number series Quiz 2
Jul 30 • Uncategorized • 1723 Views
Number series Quiz 2
1) 4 6 14 44 ? 892
• 176
• 172
• 178
• 1821
• None of The Above
2) 126 64 34 20 14 ?
• 12
• 14
• 16
• 18
• None of The Above
3) 16 8 12 30 ? 472.5
• 100
• 105
• 205
• 300
• None of The Above
4) 25 34 30 39 35 ?
• 45
• 44
• 46
• 50
• None of The Above
5) 10, 11, 13, 21, 69, ?
• 384
• 490
• 453
• 390
• None of The Above
6) 5, ………, 8, 56, 7, 63, 9 ?
• 45
• 35
• 40
• 42
• None of The Above
7) 32,…….., 92, 134, 184 ?
• 55
• 38
• 45
• 58
• None of The Above
18) 11, 16, 26, ……….., 86 ?
• 46
• 56
• 76
• 86
• None of The Above
9) 198, 194, 185, 169, …….?
• 154
• 165
• 144
• 134
• None of The Above
10) 9050, 5675, 3478, 2147, …….?
• 3478
• 1418
• 2428
• 3678
• None of The Above
1.C 2.A 3.B 4.B 5.C 6.C 7.D 8.A 9.C 10.B
SOLUTIONS
1.C 4 6 14 44 ? 892
4*1+2=6
6*2+2=14
44*4+2=178(ans)
2.A 126 64 34 20 14 ?
126/2+1=64
64/2+2=34
14/2+5=12(ans)
3.B 16 8 12 30 ? 472.5
16*.5=8
8*1.5=12
12*2.5=30
30*3.5=105(ans)
4.B 25 34 30 39 35 ?
25+9=34
34-4=30
30+9=39
35+9=44(ans)
5.C 10, 11, 13, 21, 69, ?
10+1=11
11+2=13
69+384=453(ans)
6.C 5, ………, 8, 56, 7, 63, 9 ?
5*1=5
8*5=40(ans)
8*1=8
7.D 32,…….., 92, 134, 184 ?
32+26=58(ans)
35=34=92
92+42=134
8.A 11, 16, 26, ……….., 86 ?
11+5=16
16+10=26
26+20=46(ans)
9.C 198, 194, 185, 169, …….?
198-2^2=194
194-3^2=185
169-5^2=144(ans)
10.B 9050, 5675, 3478, 2147, …….?
9050-15^2=5675
5675-13^3=3478
2147-9^3=1418(ans)
• ## Number series Quiz 1
Jul 30 • Uncategorized • 845 Views
Number series Quiz 1
Number series
• 5, 25, 7,______, 9, 19
• 23
• 22
• 25
• 32
• None of The Above
2) 62, 64,______, 32, 14,16
• 26
• 28
• 30
• 32
• None of The Above
3) 100, 50, 52, 26, 28,______
• 30
• 32
• 14
• 16
• None of The Above
4) 980, 392, 156.8,______, 25.088, 10.0352
• 62.72
• 63.85
• 65.04
• 60.28
• None of The Above
5) 113, 225, 449,_____, 1793
• 789
• 786
• 897
• 987
• None of The Above
6) 5, 9, 21, 37, 81,_____
• 163
• 153
• 181
• 203
• None of The Above
7) 2, 10, 30, 68,______
• 126
• 130
• 140
• 150
• None of The Above
8) 45, 46, 70, 141,_______, 1061.5
• 353
• 353.5
• 352.5
• 352
• None of The Above
9) 33, 321, 465, 537, 573_______
• 600
• 591
• 585
• 498
• None of The Above
10) 2, 9,_______, 105, 436, 2195
• 25
• 27
• 30
• 33
• None of The Above
1.B 2.C 3. C 4.A 5.C 6.B 7.B 8.B 9.B 10.C
SOLUTIONS
1. (B) 5, 25, 7,______, 9, 19
5+2=7; 7+2=9
25-3= 22 (ans)
1. C. 62, 64,______, 32, 14,16
62-2/2= 30 (ans)
64/2 =32
1. C. 100, 50, 52, 26, 28,______
100/2=50
52/2=26
28/2=14 (ans)
4 .A. 980, 392, 156.8,______, 25.088, 10.0352
980*⅖=392
156.8*⅖=62.72(ans)
5.C 113, 225, 449,_____, 1793
113*2-1=225
225*2-1=449
449*2-1=897(ans)
6.B 5, 9, 21, 37, 81,_____
5*2-1=981*2-1=153(ans)
7.B 2, 10, 30, 68,______
1^3+1=2
3^3+2=10
5^3+5=130(ans)
8.B 45, 46, 70, 141,_______, 1061.5
45*1+1=46
46*⅔+1=70
141*5/2+1=353.5(ans)
9.B 33, 321, 465, 537, 573_______
33+288=321
321+144=465
573+18=591(ans)
10.C 2, 9,_______, 105, 436, 2195
2*1+(1*7)=9
9*2+(2*6)=30(ans) | 12,327 | 37,551 | {"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-2020-16 | latest | en | 0.832173 |
http://www.theswamp.org/index.php?topic=807.0 | 1,579,726,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607407.48/warc/CC-MAIN-20200122191620-20200122220620-00478.warc.gz | 282,833,363 | 9,661 | ### Author Topic: you say you're pretty good at numbers....hehehehe (Read 6200 times)
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#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« on: February 06, 2004, 08:47:20 PM »
I came across this formula for calculating the circular equivalent of a rectangular duct and I managed to string it out so it works...here it is:
Code: [Select]
`(* 1.3 (exp(/(log (/ (expt (* A B) 5) (expt (+ A B) 2)))8)))`
where A and B are reals; solves for D (diameter)
I wonder if someone could rearrange this equation ,so that if I knew :
B and D I could find A or if I knew A and D I could find B
Go ahead, give it shot...I've only been on it ALL day!
#### mohobrien
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #1 on: February 06, 2004, 09:13:48 PM »
This looks like a challenge!
Is this really what expt is?
Function EXP, EXPT
Syntax:
exp number => result
expt base-number power-number => result
Arguments and Values:
number---a number.
base-number---a number.
power-number---a number.
result---a number.
Description:
exp and expt perform exponentiation.
exp returns e raised to the power number, where e is the base of the natural logarithms. exp has no branch cut.
expt returns base-number raised to the power power-number. If the base-number is a rational and power-number is an integer, the calculation is exact and the result will be of type rational; otherwise a floating-point approximation might result. For expt of a complex rational to an integer power, the calculation must be exact and the result is of type (or rational (complex rational)).
The result of expt can be a complex, even when neither argument is a complex, if base-number is negative and power-number is not an integer. The result is always the principal complex value. For example, (expt -8 1/3) is not permitted to return -2, even though -2 is one of the cube roots of -8. The principal cube root is a complex approximately equal to #C(1.0 1.73205), not -2.
expt is defined as b^x = e^x log b. This defines the principal values precisely. The range of expt is the entire complex plane. Regarded as a function of x, with b fixed, there is no branch cut. Regarded as a function of b, with x fixed, there is in general a branch cut along the negative real axis, continuous with quadrant II. The domain excludes the origin. By definition, 0^0=1. If b=0 and the real part of x is strictly positive, then b^x=0. For all other values of x, 0^x is an error.
When power-number is an integer 0, then the result is always the value one in the type of base-number, even if the base-number is zero (of any type). That is:
(expt x 0) == (coerce 1 (type-of x))
If power-number is a zero of any other type, then the result is also the value one, in the type of the arguments after the application of the contagion rules in Section 12.1.1.2 (Contagion in Numeric Operations), with one exception: the consequences are undefined if base-number is zero when power-number is zero and not of type integer.
#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #2 on: February 06, 2004, 09:21:22 PM »
that's it... Now I've thrown down the gaunlet!
#### mohobrien
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #3 on: February 06, 2004, 09:25:30 PM »
take a look at the brackets. one too many.
9 "(" and only 8 ")"
#### CAB
• Global Moderator
• Seagull
• Posts: 10376
##### you say you're pretty good at numbers....hehehehe
« Reply #4 on: February 06, 2004, 09:35:02 PM »
Code: [Select]
`(setq b (/ (* pi (expt (/ d 2) 2)) a))`
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #5 on: February 06, 2004, 09:35:40 PM »
syntax is correct ..substitute 12.0 for all A's and B's then run it in editor should come up with 13.118
#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #6 on: February 06, 2004, 09:42:19 PM »
Good try CAB! but if you use 12.0 for both A and B, the circular equivalent is 13.118...therefore if you use these as input 13.118 as D and 12.0 as A ,you should logically derive 12.0 as B. Yours comes up with 11.2627
#### CAB
• Global Moderator
• Seagull
• Posts: 10376
##### you say you're pretty good at numbers....hehehehe
« Reply #7 on: February 06, 2004, 09:56:59 PM »
i did not get 13.118 as dia, i got 13.5406
Code: [Select]
`(defun c:test() (setq a 12.0 b 12.0 area (* a b) ) (setq dia (* (sqrt(/ area pi)) 2)) (setq b (/ (* pi (expt (/ dia 2) 2)) a)) (princ))(princ)`
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
#### CAB
• Global Moderator
• Seagull
• Posts: 10376
##### you say you're pretty good at numbers....hehehehe
« Reply #8 on: February 06, 2004, 09:59:34 PM »
mohobrien,
CAB
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #9 on: February 06, 2004, 10:08:46 PM »
AH! therein lies the rub! ASHRAE (american Society of heating, refrigerating and ac engineers) say that the equivalent area in sq inches is NOT the same once you changed the shape of the duct because of friction losses. That is why they use the equation I posted.
#### CAB
• Global Moderator
• Seagull
• Posts: 10376
##### you say you're pretty good at numbers....hehehehe
« Reply #10 on: February 06, 2004, 11:31:36 PM »
Oh, I see, not equivalent area but equivalent flow characteristics.
I did find the formula but the math is too much for me. Perhaps
someone can convert it.
I've reached the age where the happy hour is a nap. (°¿°)
Windows 10 core i7 4790k 4Ghz 32GB GTX 970
#### mohobrien
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #11 on: February 06, 2004, 11:47:51 PM »
My last math class was 1969. I don't think I could have figured this one then either.
But I see why my back bedroom was so cold when the temp went to -48C last week. I used corrugated tin foil for ducting and I hear that five feet of that is like trying to push heat down 250 feet of straight pipe.
I'm out of my league. All I can say is you'd have to use differential equations to figure it the other way. I'm lost.
Code: [Select]
`(* 1.3 (exp (/ (log (/ (* 248832.0(* B B B B B) ) (+ 144 (* 24 B) (* B B)) ) ) 8 ) ) )`
#### Water Bear
• Guest
##### you say you're pretty good at numbers....hehehehe
« Reply #12 on: February 07, 2004, 12:08:35 AM »
That 's the one CAB! Don't feel bad guys..I've been on this one LONG time and haven't figured it out yet! Small minds=lower tax brackets...I'm in the lowest.
Maybe the heavy hitters like Stig or Keith, Daron or Se7en can have a go at it....
#### Keith™
• Villiage Idiot
• Seagull
• Posts: 16727
• Superior Stupidity at its best
##### you say you're pretty good at numbers....hehehehe
« Reply #13 on: February 07, 2004, 02:02:33 AM »
Ok, I have looked at this and to be quite honest with you, my algebra is a little rusty.
What I have managed to decipher basically gets you down to the point where the quadratic formula "should" come into play, but after looking at it for a bit, I am confused about how to exactly get the answer to an equation that requires the variable from another equation.
For example....
X = AB^5
Y = (A+B)^2
Z = X/Y
Z is known
A is known
B is unknown
So, the problem is you must calculate X or Y before you can calculate Z
Now if anyone can solve this bit of the equasion for X and/or Y I can finish the modifications.
Proud provider of opinion and arrogance since November 22, 2003 at 09:35:31 am | 2,340 | 7,897 | {"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-2020-05 | longest | en | 0.864508 |
https://www.craigfe.io/posts/polymorphic-type-constraints/ | 1,723,009,748,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00058.warc.gz | 568,254,740 | 69,004 | In this post, I explain a common mistake when writing constraints of polymorphic functions in OCaml programs, then show how to correct it.
## Not-so-polymorphic constraints
One of the earliest lessons of any functional programming tutorial is how to write polymorphic functions and their signatures:
val id : 'a -> 'a
val fst : ('a * 'b) -> 'a
val map : ('a -> 'b) -> 'a list -> 'b list
A typical explanation of these type signatures goes along the lines of:
Types of the form 'a, 'b, ..., known as type variables, stand for an unknown type. They allow us to describe functions that work uniformly over many possible input types. This is known as "parametric polymorphism".
— Hypothetical education resource1
As is often the case with introductory explanations, this is just specific enough to be technically correct without introducing too many new concepts, letting us hurry on to demonstrating useful examples before the student gets bored. Unfortunately, we've laid a trap: when our reader learns about type constraints, they naturally try to combine these two "intuitive" features and get bitten:
ᐅ let id1 : 'a -> 'a = (fun x -> x) ;; (* Accepted. So far so good... *)
val id1 : 'a -> 'a = <fun>
ᐅ let id2 : 'a -> 'a = (fun x -> x + 1) ;; (* Also accepted. Uh oh... *)
val id2 : int -> int = <fun>
In this case, the student finds that 'a -> 'a is a valid constraint for a function of type int -> int, and their mental model is broken almost immediately. It's quite natural to expect id2 to be rejected as a non-polymorphic function, particularly given our vague explanation of what 'a actually means.
Our hypothetical student's mistake stems from the fact that type variables in signatures are implicitly universally-quantified – that is, they stand for all types – whereas type variables in constraints are not. To understand what this means, let's try to pin down a more precise idea of what type variables are. If you're already indoctrinated comfortable with type variables, you may wish to cut to the chase.
## What is a type variable anyway?
Type variables in constraints are referred to as being "unbound" (or "free"), meaning that they stand for some type that is not yet known to the type-checker: they are placeholders that can later be filled by a particular type. Without going into the details, these placeholders are gradually determined as the type-checker resolves constraints. For instance, in our id2 example, the type-checker decides that 'a equals int by first reconciling the user-supplied constraint 'a -> 'a with the constraint int -> int that it inferred from the implementation.
To a theorist (or type-system developer), who regularly has to worry about types that are not yet fully known, the notion of a "placeholder" is a sensible default meaning of an unbound type variable. Such people also tend to use explicit syntax to disambiguate the alternative case, type variables that are bound:
∀ a. a -> a $\quad$ (read as: "For all a, a -> a")
We call "∀ a" a universal quantifier because it introduces a variable a, bound inside the quantifier, that can stand for any type in the universe of OCaml types. It's this flavour of type variable that enables parametric polymorphism and – although the OCaml syntax often tries to hide it from you – these quantifiers exist everywhere in your programs. As I already mentioned, all unbound variables in signatures are implicitly quantified in this way:
val length : 'a list -> int
... secretly means ...
val length : ∀ a. a list -> int
On the implementation side of length, the compiler will check to see if there are any placeholder variables left after type-checking the definition and wrap them in universal quantifiers (if it's sure that it's safe to do so2). When this happens, we say that those type variables have been generalised. Once length has been given its polymorphic type, the user gets to pick a specific type a at each call-site by passing it a list of any element type they want. This idea of choosing the instantiation of a at each call-site is what is "parametric" about "parametric polymorphism".
Taking a step back, we can now see what went wrong with our hypothetical introduction to type variables above: it led our student to think of all type variables as being implicitly universally-quantified, when this is not true in constraints3. So, given that we can't rely on implicit generalisation in constraints, what can we do to declare that our code is polymorphic within the implementation itself?
## True polymorphic constraints
The punchline is that OCaml actually does have syntax for expressing polymorphic constraints – and it even involves an explicit quantifier – but sadly it's not often taught to beginners:
let id : 'a. 'a -> 'a = (fun x -> x + 1)
The syntax 'a. 'a -> 'a denotes an explicitly-polymorphic type, where 'a. corresponds directly with the ∀ a. quantifier we've been using so far. Applying it here gives us a satisfyingly readable error message:
Error: This definition has type int -> int which is less general than
'a. 'a -> 'a
The caveat of polymorphic constraints is that we can only apply them directly to let-bindings, not to function bodies or other forms of expression:
let panic : 'a. unit -> 'a = (fun () -> raise Crisis) (* Works fine... *)
let panic () : 'a. 'a = raise Crisis (* Uh oh... *)
(* ^
* Error: Syntax error *)
This somewhat unhelpful error message arises because OCaml will never infer a polymorphic type for a value that is not let-bound. Trying to make your type inference algorithm cleverer than this quickly runs into certain undecidable problems; the parser knows that the type-checker is afraid of undecidable problems, and so rejects the program straight away4.
In spite of their limitations, explicitly-polymorphic type constraints are a great way to express polymorphic intent in your OCaml programs, either as internal documentation or to have more productive conversations with the type-checker when debugging. I recommend using them frequently and teaching them to beginners as soon as possible.
At this point, if you suffered through my explanation of type variables in the previous section, you may be thinking the following:
"If introducing type variables properly requires so many paragraphs of jargon, we shouldn't burden beginners with the details right away."
— Straw-man argument
Personally, I find that introducing these terms early on in the learning process is easily worthwhile in avoiding early roadblocks, but that discussion can wait for another time. In the spirit of functional programming for the masses, let's summarise with a less jargon-heavy attempt at redrafting our hypothetical education resource:
Types of the form 'a, 'b, ..., known as type variables, are placeholders for an undetermined type. When bound by for-all quantifiers (of the form 'a.), they can be used to describe values that can take on many possible types. For instance, we can write the type of (fun x -> x) as 'a. 'a -> 'a, meaning:
"For any type 'a, this function can take on type 'a -> 'a."
The OCaml type-checker will infer polymorphic types wherever it is safe, but we can also explicitly specify a polymorphic type for a let-binding:
ᐅ let fst : 'a 'b. ('a * 'b) -> 'a = (* "For all ['a] and ['b], ..." *)
fun (x, _) -> x ;;
val fst : 'a * 'b -> 'a = <fun>
Note that all type variables in signatures are implicitly universally-quantified: it's not necessary (or even possible) to write 'a 'b. before the type.
The explanation is undeniably still longer and more technical than the one we started with, but crucially it uses the extra space to give the reader a clue as to how to debug their polymorphic functions.
The story doesn't end here. We haven't discussed existential quantifiers, the other type of type variable binding; or polymorphic recursion, where polymorphic annotations become compulsory; or locally-abstract types, which offer other useful syntaxes for constraining your OCaml programs to be polymorphic. These will all have to wait for future posts. For now, thanks for reading!
1. Very similar equivalents of this explanation exist in Real World OCaml, Cornell's OCaml course, and Cambridge's OCaml course. Type variables are variously described as representing "any type", "an unknown type" or "a generic type"; explanations that are all as different as they are vague.
2. The most famous example of a type variable that is unsafe to generalise is one that has been captured in mutable state:
ᐅ let state = ref [] ;;
ᐅ let sneaky_id x = (state := x :: !state); x ;;
val sneaky_id : '_weak1 -> '_weak1 = <fun>
In this case, it's not possible to give sneaky_id the type ∀ a. a -> a because different choices of the type a are not independent: passing a string to sneaky_id, followed by an integer, would build a list containing both strings and integers, violating type safety. Instead, sneaky_id is given a type containing a "weak type variable" which represents a single, unknown type. This meaning of type variables should be familiar to you; it's exactly the same as the "unbound" type variables we've been discussing!
In general, it's not easy to decide if it's safe to generalise a particular type variable. OCaml makes a quick under-approximation called the (relaxed) value restriction.
3. As an aside, there's no profound reason why constraints must behave differently with respect to implicit quantification. Both SML and Haskell choose to generalise variables in constraints:
val id1 : 'a -> 'a = (fn x => x + 1);
(* Error: pattern and expression in val dec do not agree
pattern: 'a -> 'a
expression: 'Z[INT] -> 'Z[INT] *)
(Note: val f : t = e in SML is analogous to let f : t = e in OCaml.)
I suspect that constraints having the same quantification behaviour as signatures is more intuitive, at least for simple examples. In complex cases, the exact point at which type variables are implicitly quantified can be surprising, and so SML '97 provides an explicit quantification syntax for taking control of this behaviour. See the SML/NJ guide (§ 1.1.3) for much more detail.
The advantage of OCaml's approach is that it enables constraining subcomponents of types without needing to specify the entire thing (as in (1, x) : (int * _)), which can be useful when quickly constraining types as a sanity check or for code clarity. As far as I'm aware, SML has no equivalent feature.
4. This limitation of the type-checker can artificially limit the polymorphism that can be extracted from your programs. If you want to take polymorphism to its limits – as God intended – it's sometimes necessary to exploit another point where explicitly-polymorphic types can appear: record and object fields. | 2,424 | 10,787 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.876872 |
https://www.quizzes.cc/metric/percentof.php?percent=92.8&of=499 | 1,591,432,894,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00182.warc.gz | 858,656,612 | 4,478 | #### What is 92.8 percent of 499?
How much is 92.8 percent of 499? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 92.8% of 499 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 92.8% of 499 = 463.072
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Calculating ninety-two point eight of four hundred and ninety-nine How to calculate 92.8% of 499? Simply divide the percent by 100 and multiply by the number. For example, 92.8 /100 x 499 = 463.072 or 0.928 x 499 = 463.072
#### How much is 92.8 percent of the following numbers?
92.8% of 499.01 = 46308.128 92.8% of 499.02 = 46309.056 92.8% of 499.03 = 46309.984 92.8% of 499.04 = 46310.912 92.8% of 499.05 = 46311.84 92.8% of 499.06 = 46312.768 92.8% of 499.07 = 46313.696 92.8% of 499.08 = 46314.624 92.8% of 499.09 = 46315.552 92.8% of 499.1 = 46316.48 92.8% of 499.11 = 46317.408 92.8% of 499.12 = 46318.336 92.8% of 499.13 = 46319.264 92.8% of 499.14 = 46320.192 92.8% of 499.15 = 46321.12 92.8% of 499.16 = 46322.048 92.8% of 499.17 = 46322.976 92.8% of 499.18 = 46323.904 92.8% of 499.19 = 46324.832 92.8% of 499.2 = 46325.76 92.8% of 499.21 = 46326.688 92.8% of 499.22 = 46327.616 92.8% of 499.23 = 46328.544 92.8% of 499.24 = 46329.472 92.8% of 499.25 = 46330.4
92.8% of 499.26 = 46331.328 92.8% of 499.27 = 46332.256 92.8% of 499.28 = 46333.184 92.8% of 499.29 = 46334.112 92.8% of 499.3 = 46335.04 92.8% of 499.31 = 46335.968 92.8% of 499.32 = 46336.896 92.8% of 499.33 = 46337.824 92.8% of 499.34 = 46338.752 92.8% of 499.35 = 46339.68 92.8% of 499.36 = 46340.608 92.8% of 499.37 = 46341.536 92.8% of 499.38 = 46342.464 92.8% of 499.39 = 46343.392 92.8% of 499.4 = 46344.32 92.8% of 499.41 = 46345.248 92.8% of 499.42 = 46346.176 92.8% of 499.43 = 46347.104 92.8% of 499.44 = 46348.032 92.8% of 499.45 = 46348.96 92.8% of 499.46 = 46349.888 92.8% of 499.47 = 46350.816 92.8% of 499.48 = 46351.744 92.8% of 499.49 = 46352.672 92.8% of 499.5 = 46353.6
92.8% of 499.51 = 46354.528 92.8% of 499.52 = 46355.456 92.8% of 499.53 = 46356.384 92.8% of 499.54 = 46357.312 92.8% of 499.55 = 46358.24 92.8% of 499.56 = 46359.168 92.8% of 499.57 = 46360.096 92.8% of 499.58 = 46361.024 92.8% of 499.59 = 46361.952 92.8% of 499.6 = 46362.88 92.8% of 499.61 = 46363.808 92.8% of 499.62 = 46364.736 92.8% of 499.63 = 46365.664 92.8% of 499.64 = 46366.592 92.8% of 499.65 = 46367.52 92.8% of 499.66 = 46368.448 92.8% of 499.67 = 46369.376 92.8% of 499.68 = 46370.304 92.8% of 499.69 = 46371.232 92.8% of 499.7 = 46372.16 92.8% of 499.71 = 46373.088 92.8% of 499.72 = 46374.016 92.8% of 499.73 = 46374.944 92.8% of 499.74 = 46375.872 92.8% of 499.75 = 46376.8
92.8% of 499.76 = 46377.728 92.8% of 499.77 = 46378.656 92.8% of 499.78 = 46379.584 92.8% of 499.79 = 46380.512 92.8% of 499.8 = 46381.44 92.8% of 499.81 = 46382.368 92.8% of 499.82 = 46383.296 92.8% of 499.83 = 46384.224 92.8% of 499.84 = 46385.152 92.8% of 499.85 = 46386.08 92.8% of 499.86 = 46387.008 92.8% of 499.87 = 46387.936 92.8% of 499.88 = 46388.864 92.8% of 499.89 = 46389.792 92.8% of 499.9 = 46390.72 92.8% of 499.91 = 46391.648 92.8% of 499.92 = 46392.576 92.8% of 499.93 = 46393.504 92.8% of 499.94 = 46394.432 92.8% of 499.95 = 46395.36 92.8% of 499.96 = 46396.288 92.8% of 499.97 = 46397.216 92.8% of 499.98 = 46398.144 92.8% of 499.99 = 46399.072 92.8% of 500 = 46400
1% of 499 = 4.99 2% of 499 = 9.98 3% of 499 = 14.97 4% of 499 = 19.96 5% of 499 = 24.95 6% of 499 = 29.94 7% of 499 = 34.93 8% of 499 = 39.92 9% of 499 = 44.91 10% of 499 = 49.9 11% of 499 = 54.89 12% of 499 = 59.88 13% of 499 = 64.87 14% of 499 = 69.86 15% of 499 = 74.85 16% of 499 = 79.84 17% of 499 = 84.83 18% of 499 = 89.82 19% of 499 = 94.81 20% of 499 = 99.8 21% of 499 = 104.79 22% of 499 = 109.78 23% of 499 = 114.77 24% of 499 = 119.76 25% of 499 = 124.75
26% of 499 = 129.74 27% of 499 = 134.73 28% of 499 = 139.72 29% of 499 = 144.71 30% of 499 = 149.7 31% of 499 = 154.69 32% of 499 = 159.68 33% of 499 = 164.67 34% of 499 = 169.66 35% of 499 = 174.65 36% of 499 = 179.64 37% of 499 = 184.63 38% of 499 = 189.62 39% of 499 = 194.61 40% of 499 = 199.6 41% of 499 = 204.59 42% of 499 = 209.58 43% of 499 = 214.57 44% of 499 = 219.56 45% of 499 = 224.55 46% of 499 = 229.54 47% of 499 = 234.53 48% of 499 = 239.52 49% of 499 = 244.51 50% of 499 = 249.5
51% of 499 = 254.49 52% of 499 = 259.48 53% of 499 = 264.47 54% of 499 = 269.46 55% of 499 = 274.45 56% of 499 = 279.44 57% of 499 = 284.43 58% of 499 = 289.42 59% of 499 = 294.41 60% of 499 = 299.4 61% of 499 = 304.39 62% of 499 = 309.38 63% of 499 = 314.37 64% of 499 = 319.36 65% of 499 = 324.35 66% of 499 = 329.34 67% of 499 = 334.33 68% of 499 = 339.32 69% of 499 = 344.31 70% of 499 = 349.3 71% of 499 = 354.29 72% of 499 = 359.28 73% of 499 = 364.27 74% of 499 = 369.26 75% of 499 = 374.25
76% of 499 = 379.24 77% of 499 = 384.23 78% of 499 = 389.22 79% of 499 = 394.21 80% of 499 = 399.2 81% of 499 = 404.19 82% of 499 = 409.18 83% of 499 = 414.17 84% of 499 = 419.16 85% of 499 = 424.15 86% of 499 = 429.14 87% of 499 = 434.13 88% of 499 = 439.12 89% of 499 = 444.11 90% of 499 = 449.1 91% of 499 = 454.09 92% of 499 = 459.08 93% of 499 = 464.07 94% of 499 = 469.06 95% of 499 = 474.05 96% of 499 = 479.04 97% of 499 = 484.03 98% of 499 = 489.02 99% of 499 = 494.01 100% of 499 = 499 | 2,908 | 5,499 | {"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} | 3.96875 | 4 | CC-MAIN-2020-24 | latest | en | 0.873571 |
http://docplayer.net/11945187-Solutions-to-homework-6-statistics-302-professor-larget.html | 1,544,538,369,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00148.warc.gz | 79,753,900 | 29,268 | # Solutions to Homework 6 Statistics 302 Professor Larget
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2 5.36 (Graded for Accurateness) Commuting TImes in St. Louis A bootstrap distribution of mean commute times (in minutes) based on a sample of 500 St. Louis workers stored in CommuteStLouis is shown in the book. The pattern in this dot plot is reasonably bell-shaped so we use a normal curve to model this distribution of bootstrap means. The mean for this distribution is minutes and the standard deviation is 0.65 minutes. Based on this normal distribution, what proportion of bootstrap means should be in each of the following regions? (a) More than 23 mintues (b) Less than 20 minutes (c) Between 21.5 and 22.5 minutes The plots below show the three required regions as areas in a N(21.97, 0.65) distribution. We see that the areas are , , and , respectively. If converting to a standard normal, the relevant z-scores and areas are shown below. (a) z = = The area above for N(0, 1) is (b) z = = The area below for N(0, 1) is (c) z = = and z = = The area between and for N(0, 1) is
3 5.62 (Graded for Accurateness) To Study Effectively, Test Yourself! Cognitive science consistently shows that one of the most effective studying tools is to self-test. A recent study reinforced this finding. In the study, 118 college student studied 48 pairs of Swahili and English words. All students had an initial study time and then three blocks of practice time. During the practice time, half the students studied the words by reading them side by side, while the other half gave themselves quizzes in which they were shown one word and had to recall it partner. Students were randomly assigned to the two groups, and total practice times was the same for both groups. On the final test one week later, the proportion of items correctly recalled was 15% for the reading-study group and 42% for the self-quiz group. The standard error for the difference in proportions is about Test whether giving self-quizzes is more effective and show all details of the test. The sample size is large enough to use the normal distribution. The relevant hypotheses are H 0 : p Q = p R vs H a : p Q > p R, where p Q and p R are the proportions of words recalled correctly after quiz studying or studying by reading alone, respectively. Based on the sample information the statistic of interest is ˆp Q ˆp R = = 0.27 The standard error of this statistic is given as SE = 0.07 and the null hypothesis is that the difference in the proportions for the two group is zero. We compute the standardized test statistic with SampleStatistic NullP arameter z = = = 3.86 SE 0.07 Using technology, the area under a N(0, 1) curve beyond z = 3.86 is only This very small p-value provides very strong evidence that the proportion of words recalled using self-quizzes is more than the proportion recalled with reading study alone (Graded for Completeness) Penalty Shots in World Cup Soccer A study of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only 41% of the time. The article notes that this is slightly worse than random chance. We use these data as a sample of all World Cup penalty shots ever. Test at a 5% significance level to see whether there is evidence that the percent guessed correctly is less than 50%. The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is SE= We test H 0 : p = 0.5 vs H a : p < 0.5 where p is the proportion of all World Cup penalty shots for which the goalkeeper guesses the correct direction. The statistic from the original sample is ˆp = 0.41 and the null parameter is p = 0.5. The standard error is SE = We use this information to find the standardized test statistic: Samplestatistic Nullparameter z = = = This is a lower tail test, so we find the area below z = 2.09 in the lower tail of the standard normal distribution. We see in the figure below that this area is The p-value for this test 3 SE
4 is At a 5% significance level, we find evidence that the proportion of World Cup penalty shots in which the goalkeeper guesses correctly is significantly less than half (Graded for Completeness) Malevolent Uniforms in Football The figure in the book shows a bootstrap distribution of correlations between penally yards and uniform malevolence using the data on 28 NFL teams in MalevolentUniformsNFL. We wee from the percentiles of the bootstrap distribution that a 99% confidence interval for the correlation is to The correlation between the two variables for the original sample is r = (a) Use the original sample correlation and the standard deviation of the bootstrap distribution shown in the figure to compute a 99% confidence interval for the correlation using z from a normal distribution. (b) Why is the normal-based interval somewhat different front he percentile interval? Hint: Look at the shape of the bootstrap distribution. (a) For a 99% confidence interval the standard normal value leaving 0.5% in each tail is z = From the bootstrap distribution we estimate the standard error of the correlations to be Thus the 99% confidence interval based on a normal distribution would be 0.37 ± = 0.37 ± = ( 0.158, 0.898) (b) The bootstrap distribution of correlations is somewhat right skewed, while the normal-based interval assumes the distribution is symmetric and bell-shaped. Disjoint, Independent, and Complement For Exercise 11.30, state whether the two events (A and B) described are disjoint, independent, and/or complements. (It is possible that the two events fall into more than one of the three categories, or none of them.) (Graded for Accurateness) Roll two (six-sided) dice. Let A be the event that the first die is a 3 and B be the event that the sum of the two dice is 8. The two events are not disjoint or complements, as it is possible to have the rolls be {3, 5} where the first die is a 3 and the sum is 8. To check independence we need to find P (A) = 1/6 and P (B) = P ({2, 6} or {3, 5} or {4, 4} or {5, 3} or {6, 3}) = 5/36 4
5 There is only one possibility for the intersection so P (A and B) = P ({3, 5}) = 1/36. We then check that P (A and B) P (A if B) = = 1/36 = 1/6 P (B) = 5/36 P (B) 1/6 so A and B are not independent. We can also verify that P (A and B) = 1/36 P (A) P (B) = (1/6) (5/36) = 5/ (Graded for Accurateness) Peanut M & Ms In a bag of peanut M & M s, there are 80 M & Ms with 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each candy piece is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is red? (b) If we select one at random, what is the probability that it is not blue? (c) If we select one at random, what is the probability that is is red or orange? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are blue? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is red and the second one is green. (a) There are 11 red ones out of a total of 80, so the probability that we pick a red one is 11/80 = (b) The probability that it is blue is 20/80 = 0.25 so the probability that it is not blue is = (c) The single piece can be red or orange, but not both, so these are disjoint events. The probability the randomly selected candy is red or orange is 11/ /80 = 23/80 = (d) The probability that the first one is blue is 20/80 = When we put it back and mix them up, the probability that the next one is blue is also By the multiplication rule, since the two selections are independent, the probability both selections are blue is = (e) The probability that the first one is red is 11/80. Once that one is taken (since we don t put it back and we eat it instead), there are only 79 pieces left and 11 of those are green. By the multiplication rule, the probability of a red then a green is (11/80) (11/79) = (Graded for Completeness) Color Blindness in Men and Women The most common form of color blindness is an inability to distinguish red from green. However, this particular form of color blindness is much more common in men than in women (this is because the genes corresponding to the red and green receptors are located on the X-chromosomes). Approximately 7% of American men and 0.4% of American women are red-green color-blind. (a) If an American male is selected at random, what is the probability that he is red-green colorblind? (b) If an American female is selected at random, what is the probability that she is NOT red-green color-blind? (c) If one man and one woman are selected at random, what is the probability that neither are 5
6 red-green color-blind? (d) If one man and one woman are selected at random, what is the probability that at least one of them is red-green color-blind? Let CBM and CBW denote the events that a man or a woman is colorblind, respectively. (a) As 7% of men are colorblind, P(CBM) = (b) As 0.4% of women are colorblind, P(not CBW) = 1 - P(CBW) = = (c) The probability the woman is not colorblind is 0.996, and the probability that the man is not color- blind is = As the man and woman are selected independently, we can multiply their probabilities: P(Neither is Colorblind) = P(not CBM) P(not CBW) = = (d) The event that At least one is colorblind is the complement of part (d) that Neither is Colorblind so we have P (At least one is Colorblind) = 1 - P (Neither is Colorblind) = = We could also do this part as P(CBM or CBW) = P(CBM)+P(CBW)-P(CBM and CBW) = (0.07)(0.004) = (Graded for Accurateness) Mammograms and Breast Cancer The mammogram is helpful for detecting breast cancer in its early stages. However, it is an imperfect diagnostic tool. According to one study, 86.6 of every 1000 women between the ages of 50 and 59 that do not have cancer are wrongly diagnosed (a false positive), while 1.1 of every 1000 women between the ages of 50 and 59 that do have cancer are not diagnosed (a false positive). One in 38 women between the ages of 50 and 59 will develop breast cancer. If a woman between the ages of 50 and 59 has a positive mammogram, what is the probability that she will have breast cancer? We are given Applying Bayes rule we have P (Positive if no Cancer) = 86.6/1000 = , P (Positive if Cancer) = 1 1.1/1000 = , and P (Cancer) = 1/38 = P (Cancer if Positive) = P (Cancer)P (Positive if Cancer) P (no Cancer)P (Positive if no Cancer) + P (Cancer)P (Positive if Cancer) = (0.0263)(0.9989) ( )(0.0866) + (0.0263)(0.9989) = Identifying Spam Text Messages Bayes rule can be used to identify and filter spam s and text messages. Exercise refers to a large collation of real SMS text messages from participating cellphone users. In this collection, 747 of the 5574 total messages (13.40%) are identified as spam. 6
7 11.60 (Graded for Completeness) The word free is contained in 4.75% of all messages, and 3.57% of all messages both contain the word free and are marked as spam. (a) What is the probability that a message contains the word free, given that it is spam? (b) What is the probability that a message is spam given that it contains the word free? (a) Using the formula for conditional probability, P (Free if Spam) = P (Free and Spam) P (Spam) = = (b) Using the formula for conditional probability, P (Spam if Free) = P (Free and Spam) P (Free) = = Computer Exercises For each R problem, turn in answers to questions with the written portion of the homework. Send the R code for the problem to Katherine Goode. The answers to questions in the written part should be well written, clear, and organized. The R code should be commented and well formatted. R problem 1 (Graded for Accurateness) The function pnorm() finds probabilities from normal distributions. By default, it returns the area to the left from the standard normal density, but the second and third arguments can be used to specify a different mean or standard deviation. So, here are various ways to calculate the area to the right of 650 from a N(500, 100) distribution. 1 - pnorm(650, 500, 100) ## [1] pnorm(650, mean = 500, sd = 100) ## [1] pnorm(( )/100) ## [1] The function qnorm() finds quantiles from a normal distribution. Again, without other arguments, it uses the standard normal distribution. qnorm(0.9) ## [1] qnorm(0.9, mean = 500, sd = 100) ## [1] Write an expression using pnorm() and or qnorm() to find each of the following values. Note, we are using the notation N(µ, σ) to represent a normal distribution with parameters µ for the mean and σ for the standard deviation (and not using N(µ, σ 2 )). For all problems, use the N(250, 30) distribution. 1. P (X < 200). 7
8 Using R, we determine that P (X < 200) = pnorm(200,250,30) 2. P (X > 260). Using R, we determine that P (X > 260) = 1 P (X < 260) = pnorm(260,250,30) 3. P ( X 250 > 40). Using R, we find that P ( X 250 > 40) = P ((X 250 < 40) (40 < X 250)) = P (X 250 < 40) + P (40 < X 250) = P (X < ) + P ( < X) = P (X < 210) + P (290 < X) = P (X < 210) + (1 P (290 < X)) = (1-pnorm(290,250,30))+(pnorm(210,250,30)) 4. P (260 < X < 300). Using R, we find that pnorm(300,250,30)-pnorm(260,250,30) 5. The number c so that P (X < c) = 0.9. P (260 < X < 300) = P (X < 300) P (X < 260) = Using R, we find that the value c such that P (X < c) = 0.9 is qnorm(0.9,250,30) 8
9 6. The number c so that P (X > c) = Using R, we find that qnorm(0.76,250,30) P (X > c) = P (X > c) = P (X < c) = The number c so that P ( X 250 > c) = Using R, we find that qnorm(0.91,250,30) c = P ( X 250 > c) = 0.18 P ((X 250 < c) (c < X 250)) = 0.18 P (X 250 < c) + P (c < X 250)) = 0.18 P (X 250 < c) + P (X 250 < c) = The number c so that P ( X 250 < c) = 0.9. First consider that 2P (c < X 250)) = 0.18 P (c < X)) = 0.09 P (c > X)) = P (c > X)) = 0.91 c = c = P ( c < X 250 < c) = 0.9 P (X 250 < c) P (X 250 > c) = 0.9 P (X 250 < c) (1 P (X 250 < c)) = 0.9 P (X 250 < c) 1 + P (X 250 < c) = 0.9 2P (X 250 < c) 1 = 0.9 2P (X 250 < c) = 1.9 P (X 250 < c) = P (X < c + 250) =
10 From R, we determine that c = c = qnorm(0.95,250,30) R problem 2 (Graded for Accurateness) The height of the density function of a normal curve can be computed with the R function dnorm(). Write a function called gnorm() that will draw a sketch of a normal density and shade in the two tails with probability α/2 if one passes in a mean, sd, and alpha value. Modify this function which calculates and draws P (X a). gnorm = function(a,mu=0,sigma=1) { # create an array of x values of length 501 # from 4 SDs below to 4 SDs above the mean x = seq(mu-4*sigma,mu+4*sigma,length=501) # calculate the height of the normal density at these points y = dnorm(x,mean=mu,sd=sigma) # put x and y into a data frame d = data.frame(x,y) # create a plot that graphs the normal density # and overlays this with a horizontal line for the x axis # store the plot as the object p and later add more to it require(ggplot2) p = ggplot(d, aes(x=x,y=y)) + geom_line() + geom_segment(aes(x=mu-4*sigma,xend=mu+4*sigma,y=0,yend=0), data=data.frame(mu,sigma)) + ylab( density ) # shade in the area # add points to the data frame d that are the bottom of the segment to shade # and extract only those x and y from d where x <= a d2 = data.frame(x = c(mu-4*sigma,x[x<=a],a), y = c(0,y[x<=a],0)) p = p + geom_polygon(aes(x=x,y=y),data=d2,fill="red") + ggtitle(paste("p(x <",a,") =",round(pnorm(a,mu,sigma),4))) plot(p) } Below is code for a function that will perform the desired task. anorm = function(alpha,mu=0,sigma=1) { x = seq(mu-4*sigma,mu+4*sigma,length=501) y = dnorm(x,mean=mu,sd=sigma) d = data.frame(x,y) require(ggplot2) p = ggplot(d, aes(x=x,y=y)) + geom_line() + geom_segment(aes(x=mu-4*sigma,xend=mu+4*sigma,y=0,yend=0), data=data.frame(mu,sigma)) + ylab( density ) 10
11 } a <- qnorm(alpha/2,mu,sigma) d2 = data.frame(x = c(mu-4*sigma,x[x<=a],a), y = c(0,y[x<=a],0)) b <- qnorm(1-alpha/2,mu,sigma) d3 = data.frame(x = c(b,x[x>=b],mu+4*sigma), y = c(0,y[x>=b],0)) p = p + geom_polygon(aes(x=x,y=y),data=d2,fill="red") + geom_polygon(aes(x=x,y=y),data=d3,fill="red")+ ggtitle(paste("p( X >",round(b,4),") =",alpha)) plot(p) For example, this code can be used in the following manner to create the image below. gnorm(.025,0,1) 0.4 P( X > ) = density x 11
### The Goodness-of-Fit Test
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### Sample Term Test 2A. 1. A variable X has a distribution which is described by the density curve shown below:
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### Solutions to Homework 3 Statistics 302 Professor Larget
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https://www.xsprogram.com/content/biggest-possible-sum-of-multiples.html | 1,642,898,700,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303917.24/warc/CC-MAIN-20220122224904-20220123014904-00156.warc.gz | 1,103,380,749 | 14,630 | # Biggest possible sum of multiples
Firstly, as for me it's not clearly why for first number from the second array we should use only three first items from the first array, and for each other numbers from the second array we use anyone numbers from the first array.
However, we can represent this two arrays as a bipartite graph, where each number from the second array connect to (each?*) numbers from the first array. In this graph each edge (u,v) has a weight which equal to multiplication of the two numbers from arrays (u*v). And for given graph you should resolve max flow problem.
* if I understand the problem correctly, you should connect first number from the 2nd array with first 3 numbers from the 1st array, and each other number from 2nd array with each numbers from the 1st array.
Continue doing this until every number is used. Write the biggest possible sum of these multiples
This looks like a discrete and combinatorial optimization problem. This first algorithm which came to my mind for solving this is by using Branch and Bound algorithm.
It allows back tracking and you can implement it with recursion.
I implemented a Mixed-Integer Programming approach in Julia using JuMP. It works for your example and the general concept should work, but may need more testing!
It should be quite readable even if you are not familiar with julia. Keep in mind, that indexing starts with 1 (like in Matlab) instead of 0 (like in C, Java, ...)
The basic idea is the following (read the comments to understand more): define a binary variable X with dimension (N,N): if X[a,b] == 1 -> A[b] was chosen in step a.
The hardest part to understand is the bigM-Formulation which is just a different formulation of the logic explained in the comments. It's sole purpose is linearizing!
using JuMP
N = 4
A = [16 6 2 10]
B = [3 8 12 9]
# MODEL
m = Model()
# VARS
@defVar(m, x[1:N, 1:N], Bin) # N steps, N vars/indices // binary
# CONSTRAINTS
# (1) exactly one var chosen in each step
for s = 1:N
end
# (2) each var is chosen exactly once
for v = 1:N
end
# (3) if var v chosen in step s -> are enough predecessors already taken away -> is in positions 1:3?
for s = 1:N
for v = 1:N
needed_pops_before = v - 3
if needed_pops_before > 0
# bigM formulation:
# formula: (x[s,v] == 1) -> sum of predecessors taken away until now >= needed_pops
# use deMorgan-law -> !(x[s,v] == 1) -> sum >= needed_pops
# formulate with bigM = N because needed_pops limited by N-3
@addConstraint(m, N * (1-x[s,v]) + sum{x[ss,jj], ss=1:s-1, jj=1:v-1} >= needed_pops_before)
end
end
end
# OBJECTIVE
@setObjective(m, Max, sum{B[s] * A[v] * x[s,v], s=1:N, v=1:N})
# SOLVE
status = solve(m)
# OUTPUT
println("Objective is: ", getObjectiveValue(m))
println("Solution: ")
for s = 1:N
for v = 1:N
if getValue(x[s,v]) > 0.00001
print(A[v], " ")
end
end
end
Output:
Objective is: 336.0
Solution:
2 6 16 10
1. #### Keith
• 2019/12/29
Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in
2. #### Stewart
• 2019/7/7
Multiples of 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256
3. #### Lombardo
• 2018/7/22
5, 10, 15, 20, 25, 30, 35, 40, 45, 50,
4. #### Arjun
• 2018/1/23
Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in the array, but each element in the array may only be used once. Examples: Input : arr [] = {5, 4, 3, 1, 1} Output : 4311 Input : Arr [] = {5, 5, 5, 7} Output : 555. Asked In : Google Interview.
5. #### Luka
• 2020/2/17
100 multiples of 5 greater than 100 are: 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220,
6. #### Collin
• 2015/6/4
This is a list of multiple births, consisting of notable higher order (4+) multiple births and pregnancies. Twins and triplets are sufficiently common to have their own separate articles. With the use of reproductive technology such as fertility drugs and in vitro fertilization (IVF) such births have become increasingly common. This list
7. #### Dominik
• 2020/2/14
The largest number, that can not be made of the sum of a multiple of coprimes p and q, you can make every number - add 5 to these, and obtain 6,7,8,9,10 - and so on. Let M be the set of numbers which can be expressed in the form 5a+11b.
8. #### Walter
• 2016/12/17
$\begingroup$ Interesting, the largest number that cannot be represented as non-negative combination of 5 and 11 is $39 = (5-1)(11-1)-1$ and the one for positive combination is $55 = 5\times11$. I wonder whether there is a pattern of this. $\endgroup$ – achille hui Aug 18 '13 at 9:17
9. #### Cole
• 2017/4/7
multiplied can be as large as 2 digit or 100 digit.
10. #### Colter
• 2020/8/9
You mean the answer to (8×1)+(8×2)+(8×3)+…….+(8×15). This is 960 as explained below The above can be written as 8×(1+2+3+….15). A continuous series like 1
11. #### Barone
• 2020/9/21
step 1 Address the formula, input parameters & values. Input parameters & values: The number series 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, . . . . , 240. The
12. #### Beckett
• 2020/7/19
class Solution: def largestMultipleOfThree (self, digits: List[int]) -> str: ''' -can use a clever trick, if the sum of digits % 3 == 0, then the whole number is a
13. #### Jesus
• 2016/1/5
What are the Multiples of 3? Any number that can be denoted in the form 3p where p is an integer is a multiple of 3. For example, 9, 12
14. #### Brahimi
• 2021/1/1
the answer would be 18, as 2*9 is the highest you can go while staying below 20. SO a way of doing it would be =. y/x == 20/2= 10; 10–1= 9; 9 *2=18; y = 20; x = 3; the answer would be 18; as 3 * 6 = 18; 20/3 = 6.67 approx. 7; 7–1 = 6; 6 * 3 = 18; y = 30; x = 10; the answer would be 20 as 10*2 = 20;
15. #### Jude
• 2020/6/5
Given an integer array of digits , return the largest multiple of three that can be formed by concatenating some of the given digits in any order.
16. #### Nova
• 2016/3/6
Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.
17. #### Sterling
• 2015/6/8
Find the largest multiple of 2, 3 and 5. An array of size n is given. The array contains digits from 0 to 9. Generate the largest number using the digits in the array
18. #### Avery
• 2016/7/22
Given an array of non-negative integers. Find the largest multiple of 3 that can be formed from array elements. For example, if the input array is {8, 1, 9}, the output should be “9 8 1”, and if the input array is {8, 1, 7, 6, 0}, output should be “8 7 6 0”.
19. #### Kieran
• 2020/12/25
Given an array nums of integers, we need to find the maximum possible sum of elements of the array such that it is divisible by three. Example 1: Input: nums = [3
20. #### Gatlin
• 2015/7/5
Given an integer array of digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. Since the answer may not fit in an integer data type, return the answer as a string. If there is no answer return an empty string. Example 1: Input: digits = [8,1,9] Output: "981" Example 2:
21. #### Jonas
• 2021/5/3
The smallest positive number that is a multiple of two or more numbers. Let's start with an Example Least Common Multiple of 3 and 5: List the Multiples of each
22. #### Kane
• 2016/5/20
Least Common Multiple is : 30 Calculate Least Common Multiple for : 5, 6 and 10. Factorize of the above numbers : 5 = 5 6 = 2 • 3 10 = 2 • 5 Build a prime factors table
23. #### Cook
• 2018/5/30
Input array is {4,8,0}.Then,the largest number formed using these digits and divisible by 3 is 840. Properties of multiples of 3-. 1.Sum of the digits
24. #### Dexter
• 2020/4/21
Calculator Use. The multiples of numbers calculator will find 100 multiples of a positive integer. For example, the multiples of 3 are calculated 3x1, 3x2, 3x3, 3x4
25. #### Fletcher
• 2020/2/19
Suppose we have an array of different digits; we have to find the largest multiple of 3 that can be generated by concatenating some of the given
26. #### Tony
• 2019/1/16
Find the largest multiple of 3. Given an array of non-negative integers. Find the largest multiple of 3 that can be formed from array elements. For example, if the input array is {8, 1, 9}, the output should be “9 8 1”, and if the input array is {8, 1, 7, 6, 0}, output should be “8 7 6 0”.
27. #### Dalton
• 2019/10/31
I have found a way, say y is 20, and x is 2. the answer would be 18, as 2*9 is the Given an integer X, what is the algorithm to calculate the biggest odd number
28. #### Bruno
• 2020/11/17
Given an array of digits (contain elements from 0 to 9). Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in the array, but each element in the array may only be used once.
29. #### Jordan
• 2017/5/12
List Multiples of a Number Calculator - Online Tool - dCode
30. #### Robin
• 2015/5/19
Find the largest number that can be made from some or all digits of array and is divisible by 3. The same element may appear multiple times in the array, but each element in the array may only be used once. Examples: Input : arr[] = {5, 4, 3, 1, 1} Output : 4311. Algorithm. Get array size and array input and calculate sum while getting input.
31. #### Gavin
• 2021/4/17
What is the largest number for which it is impossible to purchase exactly that number of What if the McNuggets were available in boxes of 7, 11, and 17? If the number of McNuggetts cannot be purchased by only using a 6, 9, or 20 pack of
32. #### Uriel
• 2016/7/30
The tables once again illustrate the number of McNuggetts and possible combinations of 7, 11 and/or 17 McNuggetts in order to be able to purchase the given number of Mcnuggetts. From these tables, one can see that 37 is the largest number of McNuggetts one cannot purchase by only using 7, 11 and/or 17 packs of McNuggetts.
33. #### Hernandez
• 2019/6/7
The first 3 multiples of 8 are: 8, 16, and 24.The first 3 multiples of 10 are: 10, 20, and 30. What are
34. #### Emir
• 2015/11/1
Calculator Use. The multiples of numbers calculator will find 100 multiples of a positive integer. For example, the multiples of 3 are calculated 3x1, 3x2, 3x3, 3x4
35. #### Mazza
• 2016/6/27
If there is no answer return an empty string. So, if the input is like [7,2,8], then the output will be 87. To solve this, we will
36. #### Sonny
• 2015/2/12
Level up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.
37. #### Odin
• 2021/1/9
Given an integer array of digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order.
38. #### Perez
• 2019/10/29
List the Multiples of each number, The multiples of 3 are 3, 6, 9, 12, 15, 18, etc. The multiples of 5 are 5, 10, 15, 20, 25, etc. Find the first Common (same) value: The Least Common Multiple of 3 and 5 is 15. (15 is a multiple of both 3 and 5, and is the smallest number like that.)
39. #### Mercier
• 2015/9/17
Given an integer array of digits , return the largest multiple of three that can be formed by concatenating some of the given digits in any order. Example 2:
40. #### Kashton
• 2016/10/30
The multiples of a whole number are found by taking the product of any counting number and that whole number. For example, to find the multiples of 3, multiply 3 by 1, 3 by 2, 3 by 3, and so on. To find the multiples of 5, multiply 5 by 1, 5 by 2, 5 by 3, and so on. The multiples are the products of these multiplications. | 3,712 | 11,935 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-05 | latest | en | 0.924019 |
https://gmatclub.com/forum/beginning-in-january-of-last-year-carl-made-deposits-of-85814.html?sort_by_oldest=true | 1,495,801,932,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00161.warc.gz | 950,359,128 | 65,769 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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### Show Tags
26 Oct 2009, 06:34
1
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Difficulty:
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Question Stats:
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Beginning in January of last year, Carl made deposits of $120 into his account on the 15th of each month for several consecutive months and then made withdrawals of$50 from the account on the 15th of each of the remaining months of last year. There were no other transactions in the account last year. If the closing balance of Carl's account for May of last year was $2,600, what was the range of the monthly closing balances of Carl's account last year? (1) Last year the closing balance of Carl's account for April was less than$2,625.
(2) Last year the closing balance of Carl's account for June was less than $2,675 [Reveal] Spoiler: OA Last edited by Bunuel on 12 Jul 2013, 02:55, edited 1 time in total. Renamed the topic and edited the question. Math Expert Joined: 02 Sep 2009 Posts: 38894 Followers: 7737 Kudos [?]: 106198 [2] , given: 11608 Re: A very tough, above 750 level-real question [#permalink] ### Show Tags 26 Oct 2009, 08:13 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED gmat620 wrote: Below is the question from OG, a real tough one. I am totally clueless on this one. Please help me out. Thanks in advance Beginning in January of last year, Carl made deposits of$120 into his account on the 15th of each month for several consecutive months and then made withdrawals of $50 from the account on the 15th of each of the remaining months of last year. There were no other transactions in the account last year. If the closing balance of Carl's account for May of last year was$2,600, what was the range of the monthly closing balances of Carl's account last year?
(1) Last year the closing balance of Carl's account for April was less than $2,625. (2) Last year the closing balance of Carl's account for June was less than$2,675
To find the range we should know:
A. Balance before he started depositing, initial balance = x (we know that there was initial balance because for may balance was 2600 and maximum amount he could deposited for this period is: 5 months*120=600).
B. Till what month he deposited $120. C. From which month he started withdrawing$50.
(1) April balance< 2625 --> he deposited in May
(Because if he didn't then April balance=2600+50=2650 and we know that in April balance was<2625).
So for may balance=2600=x(initial balance)+5months*120 --> x+600=2600 --> x=2000.
We know initial balance, but we still don't know: till what month he deposited $120 and from which month he started withdrawing$50
Not sufficient
(2) June balance < 2675 --> he didn't deposited in June --> he withdrawed in June (if he deposited, in June deposit would be May balance 2600+120=2720>2675)
Not sufficient
(1)+(2) we know:
Initial balance: x=2000
Till what month he deposited $120: till May From which month he started withdrawing$50: from June
Sufficient C.
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Re: A very tough, above 750 level-real question [#permalink]
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28 Oct 2009, 02:28
IMO, this is not a very tough 750 lvl question. Bunuel showed the elaborate way of solving the problem. Since it is a DS question, you actually do not need to calculate that much either.
From 1) We know that Carl deposited money in May (and thus in the previous months). However, we do not know how long he continued depositing. From 2) We know that he withdrew money in June, but it does not say from when he started withdrawing. Combining both, by subtracting 120 for each previous months from May and also subtracting 50 for the coming months we will get figures. From there we can find the range.
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Re: A very tough, above 750 level-real question [#permalink]
### Show Tags
28 Oct 2009, 21:26
I remember this Q from OG as well and think it is a tough one. Gave it up half way (after spending ~5 mins).
Looks simple with Bunuel's explanation, but still I'm not sure if I'd be able to handle such a wordy problem in test conditions
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Re: A very tough, above 750 level-real question [#permalink]
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29 Oct 2009, 01:44
statement 1:
==========
tells about the closing balance in April .Since May was higher than April,he must have deposited 120 in May.But no info about after May.Nt suff
statement 2:
==========
Since June is less than May,he must have withdrawn in June.No info about months prior to May.
Nt suff
combining both we get to know that he stopped the deposit in May and started the withdrawal in June.
I will go with option C
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Re: A very tough, above 750 level-real question [#permalink]
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01 Nov 2009, 01:26
Not very tough but a tricky one..need to draw correct inferences from the statments to arrive at the right option..Gud explanation by Bunnel!
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Re: A very tough, above 750 level-real question [#permalink]
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10 Dec 2009, 14:16
This indeed is tricky. These word translations problems kill me
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Re: A very tough, above 750 level-real question [#permalink]
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10 Dec 2009, 14:16
Thanks for the explanation though. It looks like a very efficient way to solve this problem.
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Re: A very tough, above 750 level-real question [#permalink]
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18 Dec 2009, 04:34
Easy one i'd say
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### Show Tags
29 Dec 2010, 12:25
I am having difficulty understanding this problem. Can someone help provide clarity for me. Thanks
161. Beginning in January of last year, Carl made deposits
of $120 into his account on the 15th of each month for several consecutive months and then made withdrawals of$50 from the account on the 15th of
each of the remaining months of last year. There were
no other transactions in the account last year. If the
closing balance of Carl’s account for May of last year
was $2,600, what was the range of the monthly closing balances of Carl’s account last year? (1) Last year the closing balance of Carl’s account for April was less than$2,625.
(2) Last year the closing balance of Carl’s account
### Show Tags
04 Feb 2014, 21:29
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Re: Beginning in January of last year, Carl made deposits of $12 [#permalink] ### Show Tags 23 Apr 2015, 02:49 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Beginning in January of last year, Carl made deposits of$12 [#permalink] 23 Apr 2015, 02:49
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Display posts from previous: Sort by | 2,518 | 9,296 | {"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} | 3.890625 | 4 | CC-MAIN-2017-22 | latest | en | 0.946547 |
https://tbc-python.fossee.in/convert-notebook/Examples_in_Thermodynamics_Problems/Chapter1.ipynb | 1,701,601,743,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00426.warc.gz | 612,735,503 | 38,142 | # CHAPTER 1: HEATING AND EXPANSION OF GASES ENTROPY¶
## Example 1.1 Page 1¶
In [2]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p1=280#lb/in^2
v=2#ft^3
p2=20#lb/in^2
v2=18.03#ft^3
#calculation
W=144*(p1*v-p2*v2)/(1.2-1)#ft/lb
#result
print"The volume and work done during the expansion is",round(W,2),"ft/lb"
The volume and work done during the expansion is 143568.0 ft/lb
## Example 1.2 Page 2¶
In [1]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
v=2#ft^3
v2=20#ft^3
p=100000#ft lb
v2=10.41#lb/in^2
v3=10#lb/in^2
p1=1.3#lb
p2=(v2*199.5)/9.95#lb/in^2
#calculation
P=(p2/v3-v2)#lb/in^2
#result
print"The initial andfinal pressure is",round(P,4),"lb/in^2"
The initial andfinal pressure is 10.4623 lb/in^2
## Example 1.4 Page 3¶
In [5]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
Cp=0.24#lb/in^2
Cv=0.18#ft^3
p1=5#lb/in^2
T1=20#Degree C
T2=150#Degree C
#CALCULATIONS
W=p1*Cp*(T2-T1)#C.H.U
H=p1*Cv*(T2-T1)#C.H.U
Gamma=Cp/Cv#lb/in^2
#RESULTS
print"the constant pressure is",round(W,2),"C.H.U"
print"the constant volume the value of gas is",round(Gamma,2),"lb/in^2"
the constant pressure is 156.0 C.H.U
the constant volume the value of gas is 1.33 lb/in^2
## Example 1.5 Page 4¶
In [6]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
Gama=1.33#ft/lb
p=100#lb/in^2
p1=20#lb/in^2
v2=10.05#ft^3
v=3#ft/lb
#CALCULATIONS
W=144*(p*v-p1*v2)/0.33#ft lb
#RESULTS
print"The work done is",round(W,2),"lb"
The work done is 43200.0 lb
## Example 1.8 Page 5¶
In [39]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan
p=3.74#ft/lb
p1=2.48#ft/lb
v=5.70#ft lb
Cv=0.21#ft/lb
P=440.00#lb/in^2
P1=160.00#lb/in^2
P2=14.00#lb/in^2
T=25.00#degree C
T1=100.00#F
vs=(pi*(p1)**2/4)*(p/1728.00)#ft^3
vc=5.70#ft^3
v1=4.70#ft^3
v2=vs/v1#ft^3
v3=0.01273#ft^3
T2=298.00#F
#CALCULATIONS
W=(P2*144.00*v3)/(T2*T1)#lb
T3=((P1*144.0*1.0)/(P2*144.0*7.0)*T2)#Degree C
T4=(P*T3)/P1#Degree C
H=W*Cv*(T4-T3)#C.H.U
#RESULTS
print"The heat supplied during explosion is",round(H,5),"C.H.U"
The heat supplied during explosion is 0.15398 C.H.U
## Example 1.9 Page 6¶
In [37]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=10.0#ft^3
p=100.0#lb/in^2
p1=18.0#lb/in^2
v1=50.0#ft^3
n=log(p/p1)/log(5)
gama=1.40#air
#CALCULATIONS
W=(144.0*(p*v-p1*v1))/(n-1)#ft lb
H=(gama-n)/(gama-1)*W#ft lb
E=W-H#ft lb
#RESULTS
print"The heat supplied and the change of internal energy",round(E,2),"ft lb"
The heat supplied and the change of internal energy 36000.0 ft lb
## Example 1.11 Page 7¶
In [42]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=2.0#ft^3
p=1100.0#lb/in^2
t1=44.0#Degree C
t2=15.0#Degree C
p1=300.0#lb/in^2
t3=3.0#Degree c
Cv=0.17#ft/lb
T=273.0#F
R=96.0#ft lb
p3=300.0#lb/in^2
n=1.12#lb
gama=1.404#lb
W=((144*p*v)/(T+t1))/R#lb
#CALCULATIONS
Wc=W*Cv*(t1-t2)#C.H.U
p2=p*(T+t2)/(T+t1)#lb /in^2
A=(144*p3*v)/(R*276)#lb
W1=(A/W)*v#ft^3
H=((gama-n)/(gama-1))*(144*(p*0.65-p1*v)/(n-1))#ft lb
H1=H/1400#C.H.U
#RESULTS
print"the heat was lost by all the air in the vessel before leakage began",round(Wc,4),"C.H.U"
print"the heat was lost or gainned leakage by the air",round(H1,4),"C.H.U"
the heat was lost by all the air in the vessel before leakage began 51.3218 C.H.U
the heat was lost or gainned leakage by the air 69.2928 C.H.U
## Example 1.13 Page 9¶
In [43]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
h=0.218#ft^3
h1=0.156#ft^3
n=0.249#lb
h2=0.178#lb
c=0.208#lb
c1=0.162#lb
w1=1.0#ft^3
p=150.0#lb/in^2
T=100.0#Degree C
T1=373.0#F
Cp=(h*0.2312)+(n*0.3237)+(c*0.4451)#C.H.U/lb
Cv=(h1*0.2312)+(h2*0.3237)+(c1*0.4451)#C.H.U//lb
R=1400*(Cp-Cv)#ft lb units
#CALCULATIONS
W=(144*p*w1)/(R*T1)#lb
#RESULTS
print"The characteristic constant of the gas is",round(W,4),"lb"
The characteristic constant of the gas is 0.7157 lb
## Example 1.20 Page 12¶
In [44]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
T=200.0#Degree C
p=150.0#lb/in^2
v=12.0#ft^3
R=96.0#Lb
T1=473.0#F
T2=273.0#F
j=1400.0#lb
Cv=0.169#lb/in^2
v1=(R*T1)/(p*144)#ft^3
#CALCULATIONS
Fhi=(R/j)*log(v/v1)+Cv*log(T2/T1)#rank
#RESULTS
print"The change of entropy is",round(Fhi,4),"rank"
The change of entropy is 0.0266 rank
## Example 1.22 Page 13¶
In [46]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
v=10.0#ft^3
T=20.0#Degree C
p=15.0#lb in^2
p1=200.0#lb//in^2
gama=1.41 #lb
Cv=0.169#lb
v2=1.153#ft^3
j=1400.0#lb
T1=293.0#F
T2=451.0#F
T1=((p1*v2)/(p*v))*T1#Degree C
#CALCULATIONS
R=Cv*j*(gama-1)
W=0.816#lb
Fhi=Cv*((gama-1.2)/(1.2-1))*log(T1/T2)*W#rnak
#RESULTS
print"The change of entropy is",round(Fhi,5),"rank"
The change of entropy is -0.00018 rank
## Example 1.23 Page 15¶
In [47]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
p=1.0#lb
T=200.0#Degree C
p1=15.0#lb/in^2
v1=4.0#ft^3
gama=1.41#lb
Cv=0.169#lb
J=1400.0#lb
n=1.2
T=473.0#F
v2=16.1#ft^3
T1=473.0#F
#CALCULATIONS
T2=(p1*v2)/(p*v1)*T1#Degree C
R=Cv*J*(gama-p)#lb/in^2
Fhi=0.1772*log(1.317)#rank
#RESULTS
print"the change of entropy from intial condition is",round(Fhi,5),"rank"
the change of entropy from intial condition is 0.04879 rank
## Example 1.26 Page 16¶
In [48]:
#initialisation of variable
from math import pi,sqrt,acos,asin,atan,cos,sin,tan,log
w=0.066#ft^3
p=14.7#lb/in^2
w1=14.2#lb/in^2
w2=2780.0#lb/in^2
g=0.038#lb
a=28.9#lb
R=w2/w1#for gas
R1=93.0#for air
T=273.0#F
V=0.4245#ft^3
#CALCULATIONS
W=(p*144*w)/(T*R)#lb
m=(g-W)#lb gas
T2=(V+w)#ft^3
#RESULTS
print"The volume of mixture is",round(T2,4),"ft^3"
The volume of mixture is 0.4905 ft^3 | 2,738 | 5,799 | {"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": 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} | 4.375 | 4 | CC-MAIN-2023-50 | latest | en | 0.455306 |
bibhasdn.com | 1,726,805,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00123.warc.gz | 109,465,178 | 8,506 | # Build a Spelling Checker
Posted on Apr 7, 2021
A friend suggested this book called “Elements of Programming Interviews in Python” to me. Interviews have always been a weak spot for me, and I am not particularly bright when it comes to hardcore coding assignments. So I took this book as an opportunity to improve myself a bit.
The book has a section called “Design Problems” where there are few systems that you have to design as an exercise. First of them is a spelling checker. I took it up and read the clues provided by them, and thought, let’s try to build one as an exercise.
Their clue/solution explains the typical issues with building a spell checker, and provides few ways to achieve/improve such a system.
## Solution
The most basic solution that comes to mind is that have a list of all the words in the given language, and find all the words that have a Levenshtein distance of <= 2 from the given typo. Assuming that the typo is at most 2 characters off from the intended word.
### Problem with the Solution
Such a search is expensive to say the least. English language has around ~520,000 words according to Wikipedia. I downloaded a list of about ~370,000 words. Calculating Levenshtein distance with all of them for every single given typo, will not be pleasant.
### Optimizing the solution
My first thought was on how I could eliminate the lookup with a list of 370,000 strings.
In my experience, typos tend to have same or similar length of the original word. They can be off by 1-2 characters if the person does not know the exact spelling. So, I could improve the number of lookups by pre-calculating the length of all the words in the dictionary, and store them along with the words, and before looking up all the words in the list, I could filter the list by only picking words that are 0-2 characters off in length from the typo words.
I imported that whole list of ~370,000 words to a postgres table, and calculated and stored their lengths with them. Here is the distribution of lengths in there -
spellchecker=# select length, count(*) as num from word group by length order by num desc;
length | num
--------+-------
9 | 53402
8 | 51627
10 | 45872
7 | 41998
11 | 37539
6 | 29874
12 | 29125
13 | 20944
5 | 15918
14 | 14149
15 | 8846
4 | 7186
16 | 5182
17 | 2967
3 | 2130
18 | 1471
19 | 760
2 | 427
20 | 359
21 | 168
22 | 74
23 | 31
24 | 12
25 | 8
27 | 3
28 | 2
29 | 2
31 | 1
(28 rows)
So if the typo is 5 character long, now the lookup will be run on about ~43,000 words, which is not great, but definitely an improvement. Now, we need to think about the words we’ll look up against that list.
As part of implementing this, I wrote a small function that, given a typo, would change 1-2 characters in it with another character in the alphabet, and generate another typo, or a valid word, which could be the word the user wanted to type. But that list is massive. Looking that up against another massive list is not what we want. So that means that there is some more optimization possible here.
It’s indeed possible that someone types pebple when they meant to write people, but the probability of that happening is actually pretty low. Guess why? It’s because of the keyboard we use. While there are of course tons of keyboard layouts, but most people use QWERTY layout, and that’s good enough for my spell checker, because I need to think more about my bigger chunk of users. I could think about all the keyboard layouts out there, but the complication I’ll have to integrate for those fraction of Dvorak or Colemak layout users didn’t make sense for the first version. I’ll get back to them later.
So next I wrote a small adjacency matrix of all the characters in a QWERTY keyboard.
qwerty_adj_graph = {
"0": ["i", "p", "o"],
"1": ["q"],
"2": ["w", "q"],
"3": ["e", "w"],
"4": ["r", "e"],
"5": ["t", "r"],
"6": ["y", "t"],
"7": ["u", "y"],
"8": ["i", "u"],
"9": ["o", "i"],
"a": ["q", "w", "s", "u", "z"],
"b": ["v", "g", "h", "n"],
"c": ["x", "d", "f", "v"],
"d": ["s", "e", "r", "f", "c", "x"],
"e": ["w", "r", "d", "s", "i"],
"f": ["d", "r", "t", "g", "v", "c"],
"g": ["f", "t", "y", "h", "b", "v"],
"h": ["g", "y", "u", "j", "n", "b"],
"i": ["u", "e", "o", "k", "j", "y"],
"j": ["h", "u", "i", "k", "m", "n"],
"k": ["j", "i", "o", "l", "m"],
"l": ["k", "o", "p"],
"m": ["n", "j", "k"],
"n": ["b", "h", "j", "m"],
"o": ["i", "p", "l", "k"],
"p": ["o", "l"],
"q": ["w", "a"],
"r": ["e", "t", "f", "d"],
"s": ["a", "w", "e", "d", "x", "z"],
"t": ["r", "y", "g", "f"],
"u": ["y", "i", "j", "h"],
"v": ["c", "f", "g", "b"],
"w": ["q", "e", "s", "a"],
"x": ["z", "s", "d", "c"],
"y": ["t", "u", "h", "g", "i"],
"z": ["a", "s", "x"],
}
I am only checking for alpha-numeric characters in the typo and replacing them with only letters. I could also add checks for special characters, I chose not to for this version. This could be improved.
So now, when I generate possible words by exchanging letters in a typo, I don’t replace all letters in the typo, with all the letters in the alphabet. I just pick the letters that are closer to the given letter in the typo. Because it’s more probable for someone to type peiple when they mean people, because o and i are right next to each other. This also considers the fact that we type on our phones a lot, and our phone keyboards are tiny, and we have fat fingers.
My typo calculation function looks something like this now -
def get_typos(word):
sample = list(word)
sample_len = len(sample)
typos = []
typo_indices = list(range(sample_len)) + list(
combinations(range(sample_len), 2)
)
for typo_idx in typo_indices:
if type(typo_idx) == int:
present_char = sample[typo_idx]
for c in alphabet:
sample_copy = sample.copy()
sample_copy[typo_idx] = c
if sample_copy != sample and sample_copy not in typos:
typos.append(sample_copy.copy())
elif type(typo_idx) == tuple:
idx1, idx2 = typo_idx
sample_copy = sample.copy()
sample_copy[idx1] = c
sample_copy[idx2] = c2
if sample_copy != sample and sample_copy not in typos:
typos.append(sample_copy.copy())
typos_final = ["".join(typo) for typo in typos]
return typos_final
typo_indices contains a list of either integers or pairs of integers. They are essentially indices in the typo string that can be replaced with an adjacent letter to create a typo. The pair of indices indicate that those 2 letters need to be replaced. I did this to easily replace the letters in the typo.
So, now it behaves like this -
>>> print(get_typos('peiple'))
['oeiple', 'leiple', 'pwiple', 'priple', 'pdiple', 'psiple', 'piiple', 'peuple', 'peeple', 'people', 'pekple', 'pejple', 'peyple', 'peiole', 'peille', 'peipke', 'peipoe', 'peippe', 'peiplw', 'peiplr', 'peipld', 'peipls', 'peipli', 'owiple', 'oriple', 'odiple', 'osiple', 'oiiple', 'lwiple', 'lriple', 'ldiple', 'lsiple', 'liiple', 'oeuple', 'oeeple', 'oeople', 'oekple', 'oejple', 'oeyple', 'leuple', 'leeple', 'leople', 'lekple', 'lejple', 'leyple', 'oeiole', 'oeille', 'leiole', 'leille', 'oeipke', 'oeipoe', 'oeippe', 'leipke', 'leipoe', 'leippe', 'oeiplw', 'oeiplr', 'oeipld', 'oeipls', 'oeipli', 'leiplw', 'leiplr', 'leipld', 'leipls', 'leipli', 'pwuple', 'pweple', 'pwople', 'pwkple', 'pwjple', 'pwyple', 'pruple', 'preple', 'prople', 'prkple', 'prjple', 'pryple', 'pduple', 'pdeple', 'pdople', 'pdkple', 'pdjple', 'pdyple', 'psuple', 'pseple', 'psople', 'pskple', 'psjple', 'psyple', 'piuple', 'pieple', 'piople', 'pikple', 'pijple', 'piyple', 'pwiole', 'pwille', 'priole', 'prille', 'pdiole', 'pdille', 'psiole', 'psille', 'piiole', 'piille', 'pwipke', 'pwipoe', 'pwippe', 'pripke', 'pripoe', 'prippe', 'pdipke', 'pdipoe', 'pdippe', 'psipke', 'psipoe', 'psippe', 'piipke', 'piipoe', 'piippe', 'pwiplw', 'pwiplr', 'pwipld', 'pwipls', 'pwipli', 'priplw', 'priplr', 'pripld', 'pripls', 'pripli', 'pdiplw', 'pdiplr', 'pdipld', 'pdipls', 'pdipli', 'psiplw', 'psiplr', 'psipld', 'psipls', 'psipli', 'piiplw', 'piiplr', 'piipld', 'piipls', 'piipli', 'peuole', 'peulle', 'peeole', 'peelle', 'peoole', 'peolle', 'pekole', 'peklle', 'pejole', 'pejlle', 'peyole', 'peylle', 'peupke', 'peupoe', 'peuppe', 'peepke', 'peepoe', 'peeppe', 'peopke', 'peopoe', 'peoppe', 'pekpke', 'pekpoe', 'pekppe', 'pejpke', 'pejpoe', 'pejppe', 'peypke', 'peypoe', 'peyppe', 'peuplw', 'peuplr', 'peupld', 'peupls', 'peupli', 'peeplw', 'peeplr', 'peepld', 'peepls', 'peepli', 'peoplw', 'peoplr', 'peopld', 'peopls', 'peopli', 'pekplw', 'pekplr', 'pekpld', 'pekpls', 'pekpli', 'pejplw', 'pejplr', 'pejpld', 'pejpls', 'pejpli', 'peyplw', 'peyplr', 'peypld', 'peypls', 'peypli', 'peioke', 'peiooe', 'peiope', 'peilke', 'peiloe', 'peilpe', 'peiolw', 'peiolr', 'peiold', 'peiols', 'peioli', 'peillw', 'peillr', 'peilld', 'peills', 'peilli', 'peipkw', 'peipkr', 'peipkd', 'peipks', 'peipki', 'peipow', 'peipor', 'peipod', 'peipos', 'peipoi', 'peippw', 'peippr', 'peippd', 'peipps', 'peippi']
So at this point, if you give a typo word, I create a list of possible words by replacing the letters in it with their adjacent letters, and then lookup those words in the list of words of similar lengths. Among all these words, only people is a valid word, so that gets returned in the end.
That all sounds fine. By doing all that every time have their own costs. So I thought why not have a hash table of all possible typos for all possible words? That could remove all these headaches completely. You give me a typo, I give you the possible words in a single lookup. Because I intend to store the typos along with the foreign keys of the valid words it could be typo of.
I was actually stupid enough to try do this. Until I realized how massive the list of all possible typos of all possible words, is. The last time I tried it, the data migration showed some 16 hours to complete inserting them to my postgres db. But I aint giving up.
I decided to optimize that a bit as well. See, I don’t really need to calculate typos of all possible words, I only needed to calculate them for all the words that people mis-spell most. I didn’t actually go look for a dataset of most misspelt words though. I decided to keep it organic. So each time you lookup a typo, it finds you the right words, but at the same time, it starts a background process that calculates and stores all possible typos for all those words. So this doesn’t overwhelm me and happens on it’s own. And each tiem you look up a typo, I lookup the database for that typo. So if anyone searched for the same typo, the words related to it are already linked to it in the database, and you get the results rather quickly.
Now, after all this, when I show you the probable words you might be trying to type, I preferably need to rank them. Show you the most intended word first. There are couple of ways of achieving this. Best method would be if I knew your typing history/pattern. That way I’d know which words you type most or misspell most. Most of our mobile keyboards have this. But let’s say I have an API to lookup typos. I wont really have personalized typing histories. But, I could store the frequency of words that are most used. I just decided to do that. But how do I calculate that?
I just picked up a text file version of couple of english books (rather thick ones), and calculated all the words in them and their frequency, and then stored them in my db beside the words. This way I have some indication of how frequently some words are used. And each time a word is looked up in the db, I increment the frequency just to stay updated.
## Further optimizations
One major thing I looked over while building this is was phonetic modeling of typos. People could type ofen instead of often, if they didn’t know how the world is spelled correctly. My current implementation doesn’t address this. If I lookup nuge, it’ll never give me nudge at it’s current state. The closest I could do is, if all else fails and the list of probable words is really small or empty, then run a Levenshtein distance lookup on the similar length words as last resort, and save the mapping for future reference.
I have couple of more ideas on how to solve this, but need to think more about them to write them down. They involve replacing letters in the typo with pair of letters that are used most often, or parts of the typo with combination of letters that appear together most often. E.g. uge could be replaced with udge and so on.
Another way to improve this is word stemming. When someone writes compute, they could be trying to type computation or computing or computer. Stemming itself is another exercise in the given book. I’ll get to it this week.
So this was my experiment with building a spell checker. I’ll deploy this tiny thing and link it here after couple of days. Writing this down itself took some time and thought.
If you have any suggestions or think all this was stupid, please to tweet it to me. Thanks for reading till this far. | 3,872 | 12,905 | {"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": 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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.968213 |
http://colegiomontpellier.net/50/emt7kq/rc7raao.php?tag=47217d-how-to-find-the-degree-of-a-term | 1,620,943,907,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992514.37/warc/CC-MAIN-20210513204127-20210513234127-00457.warc.gz | 9,049,445 | 10,630 | link to the specific question (not just the name of the question) that contains the content and a description of Long-term damage is rare. To find the degree of the polynomial, you first have to identify each term [term is for example ], so to find the degree of each term you add the exponents. The given expression can be re-written as: The polynomial terms may only have variables raised to positive integer exponents. Which of the following depicts an equation in standard form? The degree of any polynomial is the highest power that a term in the polynomial is taken to. Quiz on Degree of Polynomial Degree of a Term For a term with one variable , the degree is the variable's exponent . Your name, address, telephone number and email address; and A polynomial in standard form is written in descending order of the power. (6x5+3x5)+8x3+3x2+2x+(4+4) 1. Example #1: 4x 2 + 6x + 5 This polynomial has three terms. The highest degree is 6, so that goes first, then 3, 2 and then the constant last: x 6 + 4x 3 + 3x 2 − 7. The signs of the 2nd and 4th term are appropriately negative in the 2nd example. Second-degree. With the help of the community we can continue to To create a polynomial, one takes some terms and adds (and subtracts) them together. Baylor University, Bachelor in Arts, Philosophy and Religious Studies, General. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. Higher order equations are usually harder to solve: When a polynomial has more than one variable, we need to look at each term. Find the highest power of x to determine the degree function. First we must compare the degrees of the polynomials. The degree of the polynomial is the highest degree of any of the terms; in this case, it is 7. Find the degree. 3rd degree, 2nd degree, 1st degree, 0 degree . Monomials are just math expressions with a bunch of numbers and variables multiplied together, and one way to compare monomials is to keep track of the degree. So what's a degree? The Standard Form for writing a polynomial is to put the terms with the highest degree first. 2. deg(b) = 3, as there are 3 edges meeting at vertex 'b'. Step 2:Ignore all the coefficients x5+x3+x2+x+x0 1. The answer is 2 since the first term is squared . 12x 2 y 3: 2 + 3 = 5. When a polynomial has more than one variable, you can still describe it according to its degree and the degree of its terms. Learn how to determine the end behavior of the graph of a polynomial function. The Degree (for a polynomial with one variable, like x) is: When we know the degree we can also give it a name! The term with the highest degree is called the leading term because it is usually written first. How to factor polynomials with 4 terms? Take a look at the following graph − In the above Undirected Graph, 1. deg(a) = 2, as there are 2 edges meeting at vertex 'a'. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such Example: y = 2x + 7 has a degree of 1, so it is a linear equation, Example: 5w2 − 3 has a degree of 2, so it is quadratic. The degree of the polynomial is the largest sum of the exponents of ALL variables in a term. Each equation contains anywhere from one to several terms, which are divided by numbers or variables with differing exponents. These unique features make Virtual Nerd a viable alternative to private tutoring. Think of the area of … Example 1 : Convert 25 ° into radian measure. Find the Degree, Leading Term, and Leading Coefficient 3x2 − 2x + 5 3 x 2 - 2 x + 5 The degree of a polynomial is the highest degree of its terms. The degree of the polynomial is found by looking at the term with the highest exponent on its variable (s). Note: "ln" is the natural logarithm function. With more than one variable, the degree is the sum of the exponents of the variables. So far what I … This level contains expressions up … your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The polynomial has more than one variable. Identify the term containing the highest power of x to find the leading term. An identification of the copyright claimed to have been infringed; Recall that for y 2, y is the base and 2 is the exponent. The highest power should be first, and the lowest power should be last. 4. deg(d) = 2, as there are 2 edges meeting at vertex 'd'. What is the degree of following polynomial? Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially University of Nevada-Reno, Bachelor of Science, Biology, General. The coefficient of the leading term becomes the leading coefficient. means of the most recent email address, if any, provided by such party to Varsity Tutors. Instead of saying "the degree of (whatever) is 3" we write it like this: We can work out the degree of a rational expression (one that is in the form of a fraction) by taking the degree of the top (numerator) and subtracting the degree of the bottom (denominator). The area of a circle is the total area that is bounded by the circumference. misrepresent that a product or activity is infringing your copyrights. Identify the term containing the highest power of x to find the leading term. A description of the nature and exact location of the content that you claim to infringe your copyright, in \ The answer has the powers decreasing from four, to two, to one, to zero. If Varsity Tutors takes action in response to ie -- look for the value of the largest exponent. The exponent of the first term is 2. Here is a typical polynomial: Notice the exponents (that is, the powers) on each of the three terms. There are 4 simple steps are present to find the degree of a polynomial:- Example: 6x5+8x3+3x5+3x2+4+2x+4 1. To convert degree measure to radian measure, we have to use the formula given below. information described below to the designated agent listed below. Therefore, the degree of the polynomial is 6. Find the 7th Taylor Polynomial centered at x = 0 for the following functions. Varsity Tutors. ChillingEffects.org. Infringement Notice, it will make a good faith attempt to contact the party that made such content available by Example 3 . 5. deg(e) = 0, as there are 0 edges formed at vertex 'e'.So 'e' is an isolated vertex. © 2007-2021 All Rights Reserved, SAT Courses & Classes in Dallas Fort Worth. If you have this type of burn, the outer layer of your skin as well the dermis – the layer underneath – has been damaged. Thus, if you are not sure content located How to Find the Degree of a Polynomial: 14 Steps (with Pictures) - … 0 degree, 1st degree, 2nd degree, 3rd degree . a information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Hence, the degree of the multivariate term in the polynomial is 7. 2xz: 1 … We can sometimes work out the degree of an expression by dividing ... ... then do that for larger and larger values, to see where the answer is "heading". Find the degree and classify them by degree and number of terms. Definition: The degree is the term with the greatest exponent. The degree is the highest exponent value of the variables in the polynomial. the Arizona State University, Masters in Business... Macomb Community College, Associate in Science, Biological and Physical Sciences. Notice our 3-term polynomial has degree 2, and the number of factors is also 2. Step 1:Combine all the like terms that are the terms of the variable terms. }x^n$$But I am stuck here, how do I find the nth Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Above, we discussed the cubic polynomial p(x) = 4x 3 − 3x 2 − 25x − 6 which has degree 3 (since the highest power of x that appears is 3). Therefore, the degree of this expression is . sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require To find the degree of the polynomial, add up the exponents of each term and select the highest sum. Since they are the same degree, we must divide the coefficients of the highest terms. No square roots, fraction powers, and variables in the denominator are allowed. The degree of a polynomial is the highest degree of its terms. which specific portion of the question – an image, a link, the text, etc – your complaint refers to; You don't have to use Standard Form, but it helps. The degree of an individual term of a polynomial is the exponent of its variable; the exponents of the terms of this polynomial are, in order, 5, 4, 2, and 7. The sum of the exponents in each term of the expansion are 3. When a polynomial has more than one variable, we need to find the degree by adding the exponents of each variable in each term. Problem 3. The largest exponent is the degree of the polynomial. The largest degree of those is 3 (in fact two terms have a degree of 3), so the polynomial has a degree of 3, The largest degree of those is 4, so the polynomial has a degree of 4. In the numerator, the coefficient of the highest term is 4. Step 3:Arrange the variable in descending order of their powers x5+x3+x2+x+x0 1. For instance, the equation y = 3x 13 + 5x 3 has two terms, 3x 13 and 5x 3 and the degree of the polynomial is 13, as that's the highest degree of any term … Tap for more steps... Identify the exponents on the variables in each term, and add them together to find the degree of each term. degree= 0 type= constant leading coefficient= 0 constant term= -6 -6 is the product of this equation therefore there are no constant term or leading coefficient. Area. St. Louis, MO 63105. (a) sin(2x) (b) e5x (c) 1 1+x (d) ln (1 + x) Exercise 4.2. Find the degree, the leading term, the leading coefficient, the constant term, and the end behavior of the polynomial. Plus examples of polynomials. The degree of a polynomial is the highest degree of its terms The leading coefficient of a polynomial is the coefficient of the leading term Any term that doesn't have a variable in it is called a "constant" term types of polynomials depends on … The degree of the polynomial is the largest of these two values, or . Checking each term: 5xy 2 has a degree of 3 (x has an exponent of 1, y has 2, and 1+2=3) 3x has a degree of 1 (x has an exponent of 1) 5y 3 has a degree of 3 (y has an exponent of 3) 3 has a degree of 0 (no variable) The largest degree of those is 3 (in fact two terms have a degree of 3), so the polynomial has a degree of 3 A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe Because there is no variable in this last term… Just use the 'formula' for finding the degree of a polynomial. The degree of a polynomial with a single variable (in our case, ), simply find the largest exponent of that variable within the expression. {eq}f(x)= 4 - x - 3x^2 {/eq} Terminology of a Quadratic Polynomial: Varsity Tutors LLC 3. deg(c) = 1, as there is 1 edge formed at vertex 'c'So 'c' is a pendent vertex. Even though has a degree of 5, it is not the highest degree in the polynomial - has a degree of 6 (with exponents 1, 2, and 3). If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one The leading term in a polynomial is the term with the highest degree. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Radians = (π/ 180) x (Degree Measure) Let us look at some examples to understand how to convert degree measures to radian measures. Classification of polynomials vocabulary defined. "Degree" can mean several things in mathematics: In Algebra "Degree" is sometimes called "Order". Explanation: To find the degree of the polynomial, add up the exponents of each term and select the highest sum. This polynomial is called a third degree polynomial because its term with the highest degree is the monomial t 3. The first one is 4x 2, the second is 6x, and the third is 5. In this non-linear system, users are free to take whatever path through the material best serves their needs. What is the degree of the following polynomial? Step 4:The largest power of the variable is the degree of the polynomial deg(x5+x3+x2+x+x0) = 5 The term shows being raised to the seventh power, and no other in this expression is raised to anything larger than seven. 6xy 4 z: 1 + 4 + 1 = 6. My Thoughts: The nth term is obviously:$$\frac{f^{(n)}(0)}{n! For example, x 2 y 5 is a term in the polynomial, the degree of the term is 2+5, which is equal to 7. Do this directly, by taking the appropriate derivatives etc. Track your scores, create tests, and take your learning to the next level! Here are the few steps that you should follow to calculate the leading term & coefficient of a polynomial: Find the highest power of x to determine the degree. More examples showing how to find the degree of a polynomial. Remember coefficients have nothing at all do to with the degree. If you've found an issue with this question, please let us know. Do this directly, Terms are separated by + or - signs: The largest such degree is the degree of the polynomial. (Note that the degree of a monomial, t 3, is also 3, because the variable t has an exponent of 3.) an 101 S. Hanley Rd, Suite 300 (More correctly we should work out the Limit to Infinity of ln(f(x))/ln(x), but I just want to keep this simple here). In the trinomial you gave us, the degree is 5, since the highest power taken is 5 (18y^5) Let's find … Here, the highest exponent is x5, so the degree is 5. If a and b are the exponents of the multiple variables in a term, then the degree of a term in the polynomial expression is given as a+b. The degree of the polynomial is the highest power in the polynomial. How Do You Find the Degree of a Monomial? either the copyright owner or a person authorized to act on their behalf. We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. Example: Put this in Standard Form: 3x 2 − 7 + 4x 3 + x 6. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Both the numerator and denominator are 2 nd degree polynomials. Identify the coefficient of the leading term. Send your complaint to our designated agent at: Charles Cohn Learn how to find the degree and the leading coefficient of a polynomial expression. Next, identify the term with the highest degree to determine the leading term. The coefficient of the leading term is called the leading coefficient. What is the degree of the following polynomial? So the Degree is 0.5 (in other words 1/2), (Note: this agrees nicely with x½ = square root of x, see Fractional Exponents), Quartic equations can also be solved, but the formulas are. I have a question where I am asked to find the amount of terms required in a Maclaurin polynomial to estimate $\cos(1)$ to be correct to two decimal places. 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Polynomial is the exponent by identifying the highest term is 4, to two, to two to... | 5,149 | 21,872 | {"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} | 4.34375 | 4 | CC-MAIN-2021-21 | latest | en | 0.926226 |
owenshen24.github.io | 1,713,283,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00058.warc.gz | 407,293,597 | 21,063 | # Perceptrons Explained
## Introduction
The perceptron is a linear classifier invented in 1958 by Frank Rosenblatt. It's very well-known and often one of the first things covered in a classical machine learning course. So why create another overview of this topic?
Well, I couldn't find any projects online which brought together:
1. Visualizations of the perceptron learning in real time.
2. A proof of why the perceptron learns at all.
3. Explorations into ways to extend the default perceptron algorithm.
To be clear, these all exist in different places, but I wanted to put them together and create some slick visualizations with d3.
If you're new to all this, here's an overview of the perceptron:
In the binary classification case, the perceptron is parameterized by a weight vector $w$ and, given a data point $x_i$, outputs $\hat{y_i} = \text{sign}(w \cdot x_i)$ depending on if the class is positive ($+1$) or negative ($-1$). What makes th perceptron interesting is that if the data we are trying to classify are linearly separable, then the perceptron learning algorithm will always converge to a vector of weights $w$ which will correctly classify all points, putting all the +1s to one side and the -1s on the other side.
The perceptron learning algorithm can be broken down into 3 simple steps:
1. Initialize a vector of starting weights $w_1 = [0...0]$
2. Run the model on your dataset until you hit the first misclassified point, i.e. where $\hat{y_i} \not= y_i$
3. When a point $(x_i, y_i)$ is misclassified, update the weights $w_t$ with the following rule: $w_{t+1} = w_t + y_i(x_i)^T$. In other words, we add (or subtract) the misclassified point's value to (or from) our weights.
4. Go back to step 2 until all points are classified correctly.
To get a feel for the algorithm, I've set up an demo below.
Clicking Generate Points will pick a random hyperplane (that goes through 0, once again for simplicity) to be the ground truth. Then, points are randomly generated on both sides of the hyperplane with respective +1 or -1 labels.
After that, you can click Fit Perceptron to fit the model for the data. You can see each misclassified point flash briefly, moving the perceptron's weights either up or down, respectively throughout the training procedure.
At each iteration of the algorithm, you can see the current slope of $w_t$ as well as its error on the data points.
Because all of the data generated are linearly separable, the end error should always be 0. However, note that the learned slope will still differ from the true slope! This is because the perceptron is only guaranteed to converge to a $w$ that gets 0 error on the training data, not the ground truth hyperplane.
You can also use the slider below to control how fast the animations are for all of the charts on this page.
Global Animation Speed 50 ms / frame (lower is faster)
50 points
# Convergence Proof
While the above demo gives some good visual evidence that $w$ always converges to a line which separates our points, there is also a formal proof that adds some useful insights. For the proof, we'll consider running our algorithm for $k$ iterations and then show that $k$ is upper bounded by a finite value, meaning that, in finite time, our algorithm will always return a $w$ that can perfectly classify all points.
Before we begin, let's make our assumptions clear:
1. There exists some optimal $w^*$ such that for some $\epsilon > 0$, $y_i(w^* \cdot x_i) \ge \epsilon$ for all inputs on the training set. In other words, we assume the points are linearly separable with a margin of $\epsilon$ (as long as our hyperplane is normalized).
2. $||w^*|| = 1$. Though not strictly necessary, this gives us a unique $w^*$ and makes the proof simpler.
3. For all $x_i$ in our dataset $X$, $||x_i|| < R$. In other words, this bounds the coordinates of our points by a hypersphere with radius equal to the farthest point from the origin in our dataset.
### Inequality 1
First, let $w^{k+1}$ be the vector of weights returned by our algorithm after running it for $k+1$ iterations.
We'll start by showing that:
$w_{k+1} \cdot (w^*)^T \ge w_k \cdot (w^*)^T + \epsilon$
By definition, if we assume that $w_{k}$ misclassified $(x_t, y_t)$, we update $w_{k+1} = w_k + y_t(x_t)^T$
Thus:
$w_{k+1}\cdot (w^*)^T = (w_k + y_t(x_t)^T)\cdot (w^*)^T$
Next, multiplying out the right hand side, we get:
$w_{k+1}\cdot (w^*)^T = w_k \cdot (w^*)^T + y_t(w^* \cdot x_t)$
By assumption 1, we get, as desired:
$w_{k+1}\cdot (w^*)^T \ge w_k \cdot (w^*)^T + \epsilon$
Next, we'll prove by induction that:
$w^{k+1} \cdot (w^*)^T \ge k\epsilon$
Base case where $k = 0$:
$w^{0+1} \cdot w^* = 0 \ge 0 * \epsilon = 0$
Inductive step where $k \to k+1$:
From what we proved above, we get:
$w^{k+1} \cdot (w^*)^T \ge w_k \cdot (w^*)^T + \epsilon$
Then, from the inductive hypothesis, we get:
$w^{k+1} \cdot (w^*)^T \ge (k-1)\epsilon + \epsilon$
Which gets us, as desired:
$w^{k+1} \cdot (w^*)^T \ge k\epsilon$
Next, we see that:
$w^{k+1} \cdot (w^*)^T = ||w^{k+1}|| * ||w^*||*cos(w^{k+1}, w^*)$
Because $cos(x) \le 1$, we see that:
$w^{k+1} \cdot (w^*)^T \le ||w^{k+1}||*||w^*||$
Then, because $||w^*|| = 1$ by assumption 2, we have that:
$||w^{k+1}|| \ge k\epsilon$
Because all values on both sides are positive, we also get:
$||w^{k+1}||^2 \ge k^2\epsilon^2$
### Inequality 2
First, we notice that:
$||w_{k+1}||^2 = ||w_{k} + y_t (x_t)^T||^2$
Multiplying this out, we get:
$||w_{k+1}||^2 = ||w_k||^2 + 2y_t (w_k \cdot x_t) + ||x_k||^2$
Then, because we updated on point $(x_t, y_t)$, we know that it was classified incorrectly. If a point was misclassified, $\hat{y_t} = -y_t$, which means $2y_t(w_k \cdot x_t) < 0$ because $\text{sign}(w_k \cdot x_t) = \hat{y_t}$.
Thus:
$||w_{k+1}||^2 \le ||w_k||^2 + ||x_k||^2$
Then, by assumption 3, we know that:
$R \ge ||x_k||$
Thus:
$||w_{k+1}||^2 \le ||w_k||^2 + R^2$
Now, we'll prove by induction that:
$||w_{k+1}||^2 \le kR^2$
Base case, where $k=0$:
$||w_{0+1}||^2 = 0 \le 0*R^2 = 0$
Inductive step, where $k \to k+1$:
From what we proved above:
$||w_{k+1}||^2 \le ||w_k||^2 + R^2$
Then, by the inductive hypothesis:
$||w_{k+1}||^2 \le (k-1)R^2 + R^2$
Which gets us, as desired:
$||w_{k+1}||^2 \le kR^2$
### Putting It Together
From Inequalities 1 and 2, we get:
$k^2\epsilon^2 \le ||w_{k+1}||^2 \le kR^2$
Dividing out, we get:
$k \le \frac{R^2}{\epsilon^2}$
Thus, we see that our algorithm will run for no more than $\frac{R^2}{\epsilon^2}$ iterations.
### Changing the Margin
It's interesting to note that our convergence proof does not explicity depend on the dimensionality of our data points or even the number of data points!
Rather, the runtime depends on the size of the margin between the closest point and the separating hyperplane. In other words, the difficulty of the problem is bounded by how easily separable the two classes are. The larger the margin, the faster the perceptron should converge.
Below, you can try adjusting the margin between the two classes to see how increasing or decreasing it changes how fast the perceptron converges.
85 points
10% margin
# Linearly Unseparable Data
The default perceptron only works if the data is linearly separable.
Of course, in the real world, data is never clean; it's noisy, and the linear separability assumption we made is basically never achieved. Thus, we can make no assumptions about the minimum margin. But, as we saw above, the size of the margin that separates the two classes is what allows the perceptron to converge at all. This means the normal perceptron learning algorithm gives us no guarantees on how good it will perform on noisy data.
However, all is not lost. There are several modifications to the perceptron algorithm which enable it to do relatively well, even when the data is not linearly separable. Below, we'll explore two of them: the Maxover Algorithm and the Voted Perceptron.
# Maxover Algorithm
If the data are not linearly separable, it would be good if we could at least converge to a locally good solution. In 1995, Andreas Wendemuth introduced three modifications to the perceptron in Learning the Unlearnable, all of which allow the algorithm to converge, even when the data is not linearly separable.
The main change is to the update rule. Instead of $w_{i+1} = w_i + y_t(x_t)^T$, the update rule becomes $w_{i+1} = w_i + C(w_i, x^*)\cdot w_i + y^*(x^*)^T$, where $(x^*, y^*)$ refers to a specific data point (to be defined later) and $C$ is a function of this point and the previous iteration's weights.
Wendemuth goes on to show that as long as $(x^*, y^*)$ and $C$ are chosen to satisfy certain inequalities, this new update rule will allow $w$ to eventually converge to a solution with desirable properties.
(See the paper for more details because I'm also a little unclear on exactly how the math works out, but the main intuition is that as long as $C(w_i, x^*)\cdot w_i + y^*(x^*)^T$ has both a bounded norm and a positive dot product with repect to $w_i$, then norm of $w$ will always increase with each update. Then, in the limit, as the norm of $w$ grows, further updates, due to their bounded norm, will not shift the direction of $w$ very much, which leads to convergence.)
Each one of the modifications uses a different selection criteria for selecting $(x^*, y^*)$, which leads to different desirable properties.
One of the three algorithms in Wendemuth's paper uses the criteria where after $t$ iterations, $(x^*, y^*)_t$ is defined to be a random point which satisfies the following inequality:
$\frac{y^*(w_t \cdot x^*)}{||w_t||} < k$
This is the version you can play with below.
(After implementing and testing out all three, I picked this one because it seemed the most robust, even though another of Wendemuth's algorithms could have theoretically done better. Also, confusingly, though Wikipedia refers to the algorithms in Wendemuth's paper as the Maxover algorithm(s), the term never appears in the paper itself. For curious readers who want to dive into the details, the perceptron below is "Algorithm 2: Robust perception [sic]". Code for this algorithm as well as the other two are found in the GitHub repo linked at the end in Closing Thoughts.)
Note the value of $k$ is a tweakable hyperparameter; I've merely set it to default to -0.25 below because that's what worked well for me when I was playing around. Also, note the error rate. Given a noise proportion of $p$, we'd ideally like to get an error rate as close to $p$ as possible. I've found that this perceptron well in this regard.
50 points
10 % noise
-0.25 $k$
# Voted Perceptron
Alternatively, if the data are not linearly separable, perhaps we could get better performance using an ensemble of linear classifiers. This is what Yoav Freund and Robert Schapire accomplish in 1999's Large Margin Classification Using the Perceptron Algorithm.
(If you are familiar with their other work on boosting, their ensemble algorithm here is unsurprising.)
There are two main changes to the perceptron algorithm:
1. When we update our weights $w_t$, we store it in a list $W$, along with a vote value $c_t$, which represents how many data points $w_t$ classified correctly before it got something wrong (and thus had to be updated).
2. At test time, our prediction for a data point $x_i$ is the majority vote of all the weights in our list $W$, weighted by their vote. In other words, $\hat{y_i} = \text{sign}(\sum_{w_j \in W} c_j(w \cdot x_i))$
Though it's both intuitive and easy to implement, the analyses for the Voted Perceptron do not extend past running it just once through the training set. However, we empirically see that performance continues to improve if we make multiple passes through the training set and thus extend the length of $W$.
The authors themselves have this to say about such behavior:
As we shall see in the experiments, the [Voted Perceptron] algorithm actually continues to improve performance after $T = 1$. We have no theoretical explanation for this improvement.
:P
Below, you can see this for yourself by changing the number of iterations the Voted Perceptron runs for, and then seeing the resulting error rate.
During the training animation, each hyperplane in $W$ is overlaid on the graph, with an intensity proportional to its vote. You can also hover a specific hyperplane to see the number of votes it got. Typically, the points with high vote are the ones which are close to the original line; with minimal noise, we'd expect something close to the original separating hyperplane to get most of the points correct.
The final error rate is the majority vote of all the weights in $W$, and it also tends to be pretty close to the noise rate.
50 points
10 % noise
5 iterations
# Closing Thoughts
This is far from a complete overview, but I think it does what I wanted it to do. There's an entire family of maximum-margin perceptrons that I skipped over, but I feel like that's not as interesting as the noise-tolerant case. Furthermore, SVMs seem like the more natural place to introduce the concept. Similarly, perceptrons can also be adapted to use kernel functions, but I once again feel like that'd be too much to cram into one post.
For now, I think this project is basically done. If I have more slack, I might work on some geometric figures which give a better intuition for the perceptron convergence proof, but the algebra by itself will have to suffice for now.
It was very difficult to find information on the Maxover algorithm in particular, as almost every source on the internet blatantly plagiarized the description from Wikipedia. Shoutout to Constructive Learning Techniques for Designing Neural Network Systems by Colin Campbell and Statistical Mechanics of Neural Networks by William Whyte for providing succinct summaries that helped me in decoding Wendemuth's abstruse descriptions.
On that note, I'm excited that all of the code for this project is available on GitHub. To my knowledge, this is the first time that anyone has made available a working implementation of the Maxover algorithm. Uh…not that I expect anyone to actually use it, seeing as no one uses perceptrons for anything except academic purposes these days. But hopefully this shows up the next time someone tries to look up information about this algorithm, and they won't need to spend several weeks trying to understand Wendemuth.
In the best case, I hope this becomes a useful pedagogical part to future introductory machine learning classes, which can give students some more visual evidence for why and how the perceptron works.
### Credits
The charts were made with d3 and jQuery.
The CSS was inspired by the colors found on on julian.com, which is one of the most aesthetic sites I've seen in a while.
The main font is Muli from Google Fonts. | 3,927 | 14,893 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 98, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.884018 |
https://issuu.com/mullins2345/docs/introductory_and_intermediate_algeb | 1,531,712,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589172.41/warc/CC-MAIN-20180716021858-20180716041858-00136.warc.gz | 688,745,363 | 34,039 | Chapter 2 Linear Equations and Inequalities in One Variable 2.1 Check Points 4. 8 y 7 7 y 10 6 4 1.
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Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
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24.
7 9 8 8 9 7 x 8 8 2 1 x 8 4 Check: 1 7 9 4 8 8 2 7 9 8 8 8 9 9 8 8 x
The solution set is 13 x 11 13 11 x 16. 24 x Check: 13 24 11 13 13 The solution set is 24.
1 . 4
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
26. t
2 7 3 6 7 2 t 6 3 7 4 11 t 6 6
Check: 11 2 6 3 11 4 6 6 7 6
32. 2.7 w 5.3 w 5.3 2.7 w 2.6 Check: 2.7 2.6 5.3 5.5 5.3
6
The solution set is 2.6.
7 6 7 6 7 6
34.
11
3
7 5
10 7 6 10 10 13 10
.
6 3 7 5 10 7 3 x 10 5 7 6 13 10 10 10
x
r
r
The solution set is
28. x
Section 2.1
Check: 13 3 10 5 13 6 10 10 7 10
Check:
7 10 7 10 7
10
The solution set is
13 3 7 10 5 10 13 6 7 10 10 10 7 7 10 10 The solution set is
13 . 10
13
. 10
36. 11 8 x 11 8 x 19 x Check: 11 8 19 11 19 The solution set is 19.
30.
1 y 8 y
1 4 1 4 2 y 4
Check: 1 1 8
8 2 8 1 4
1 8 1 1 8 8 1 4 1 4 1 4
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
The solution set is
1 ď€ . 8
Section 2.1
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
46. 3x 5 4 x 9 x5 9 x 14 Check: 3 14 5 14 9 42 5 56 9 49 56 9 9 9 The solution set is 14.
7 5 z 3 2 5 z 2
38.
7 3 14 15 z 6
29 6
z
Check: 7 5 29 3 2 6 14 15 29 6 6 6 14 14 6 6 The solution set is
29 . 6
40. 90 t 35 t 35 90 t 55 Check: 90 55 35 35 35 The solution set is 55. 42.
x 10.6 9 x 9 10.6 x 19.6 Check: 19.6 10.6 9 9 9 The solution set is 19.6.
44. y
Section 2.1
7 7
13 3r 2 6r 2r 1 3 2 9 48. 3r 6r 2r 13 2 1 3 18 r 14 21 r 14 14 21 14 r 7 Check: 13 3 7 2 6 7 2 7 1 3 2 9 13 21 2 42 14 1 3 18 21 21 The solution set is 7. 50.
4r 3 5 3r 4r 3 3r 5 3r 3r r 3 5 r33 53 r 8 Check: 4 8 3 5 3 8 32 3 5 24 29 29 The solution set is 8.
52.
20 7s 26 8s 20 7s 8s 26 8s 8s
11
11 7 7 y 11 11 y 0 Check: 0
7
7
11 7
20 s 26 20 20 s 26 20 s 6 Check: 20 7 6 26 8 6
11 7
11 11 The solution set is 0.
20 42 26 48 22 22 The solution set is 6.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
66.
3 3 The solution set is 0.
68. a. According to the line graph, the U.S. diversity index was about 47 in 2000.
3 3 3 3
b. 2000 is 20 years after 1980. I 0.6 x 34 I 0.6(20) 34 I 12 34 I 12 12 34 12 I 46 According to the formula, the U.S. diversity index was 46 in 2000. This matches the line graph very well.
x x x
58.
6x 6x
7x 6x 7x
6x
x x x
60.
d 257 x 8328 d 257(7) 8328 d 1799 8328 d 1799 1799 8328 1799 d 10,127 According to the formula, the average credit-card debt per U.S. household was \$10,127 in 2007. This underestimates the value given in the bar graph by \$287.
6x19 6x 6 9 6x 3 3 x 0 Check: 70 3 60 1 9 0 3 61 9 3 69
54. 7 x 7x 7x x
56.
Section 2.1
70. Answers will vary. 72. The adjective linear means that the points lie on a line.
x 23 8 x 23 23 8 23 x 15 The number is 15.
74. makes sense 76. makes sense
62.
3
2
2
5
x 7 2
x
78. false; Changes to make the statement true will vary.
7
3 x x 7 7
5
x 7 7 3 x 7 3 x
2
The number is 3. 64. C 520, S 650 C M S 520 M 650 M 650 520 M 130 The markup is \$130.
A sample change is: If y 7 0, then x 7
y 7.
80. false; Changes to make the statement true will vary. 18 A sample change is: If 3x 18, then x 6. 3 82. x 7.0463 9.2714 x 9.2714 7.0463 x 2.2251 The solution set is 2.2251. 9 84. 85.
4x x
16 8 4 2 16 2 2 ď€
16 2 2 ď€ 16 4 12
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
3 7 x 2 5x 1
86.
37 x 10 x 2 33x 2 9x 6 or 6 9 x
c.
87.
88.
7y y 7
15.5 5z 15.5 5 z 5 5 3.1 1z 3.1 z The solution set is 3.1.
x 5 x x 1 5
5 5
89.
Section 2.1
3. a.
3x 14 2 x 6 3(4) 14 2(4) 6 12 14 8 6 2 2, true Yes, 4 is a solution of the equation.
2 y 16 3 3 2 3 y 16 2 3 2 1y 24 y 24 The solution set is 24.
b. 4
2.2 Check Points
x 12 3 3 1x 36 x 36 Check: x 12 3 36 12 3 12 12 The solution set is 36. 4x 4x 4 1x x
4. a.
b.
84 84 4 21 21
The solution set is 21. b. 11y 44 11y 44 11
4 7
x 7 4
The solution set is 16.
3
2. a.
28
7 16 1x 16 x
x 12 3
1.
7 x 4
28
5.
x 5 1x 5 (1)(1x) (1)5 1x 5 x 5 The solution set is 5. x 3 1x 3 (1)(1x) (1)(3) 1x 3 x 3 The solution set is 3.
4 x 3 27 4 x 3 3 27 3 4 x 24 4x
24
11
1x 4 x 4 The solution set is 4.
4
4 x 6 The solution set is 6.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
6.
4 y 15 25 4 y 15 15 25 15 4 y 40 4 y 40 4
4
y 10 The solution set is 10. 7.
2 x 15 4 x 21 2 x 4 x 15 4 x 4 x 21 6 x 15 21 6 x 15 15 21 15 6 x 36 6 x 36 6 6
Section 2.2
4. dividing; 8 Alternatively, multiplying; 5 5. multiplying; 3 6. multiplying/dividing; 1 7. subtracting 2; dividing; 5
2.2 Exercise Set 2.
x 6 The solution set is 6. 8. a. The bar graph indicates that the price of a Westie puppy was \$2000 in 2009. Since 2009 is 69 years after 1940, substitute 69 into the formula for n. P 18n 765 P 18(69) 765 P 1242 765 P 2007 The formula indicates that the price of a Westie puppy was \$2007 in 2009. The formula overestimates by \$7. b.
P 18n 765 2151 18n 765 2151 765 18n 765 765 1386 18n 1386 18n 18 18 77 n The formula estimates that the price will be \$2151 for a Westie puppy 77 years after 1940, or in 2017.
2.2 Concept and Vocabulary Check 1. bc 2. divide 3. multiplying; 7
x 4 7 x 7 74 7 x 28 Check: 28 4 7 4 4 The solution set is 28. x 8 5
4. 5
x
8 5ď€ 5 x 40
Check: 40 8 5 8 8 The solution set is 40. 6.
6 y 42 6 y 42 ď€ 6 6 y 7 Check: 6 7 42 42 42 The solution set is 7.
1 8
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
8.
4 y 32 4y 4
Section 2.2
16.
16 y 0 16 16
32 4
y 8 Check: 4 8 32 32 32 The solution set is 8.
y 0 Check: 16 0 0 0 0 The solution set is 0.
10. 36 8z 36 8z 8
16 y 0
3 y 15 4
18.
4 3 4 y 15 3 3 4 4 15
8 9 z 2
3 1 y 20
1y
Check: 9 36 8 2
9 . 2
4 1
15 15 The solution set is 20.
9
6 z Check: 54 9 6 54 54
20
20.
The solution set is 6.
8
20
8 5
5
5 x 8 x 5
8
160 1x 5 32 x
14. 8x 4 8x 4 8
15
60 15 4
12. 54 9 z 54 9 z 9
3
Check: 3 20 15 4 3 20
36 36 The solution set is
60
8 4 1 x 8 2
Check: 1 8 4
Check: 20
20
5 32 8 160 8
2 4 4
The solution set is
1
.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
Section 2.2 The solution set is 32.
20 20 2
ď€
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
22.
x 23 1x 23 1 1x 1 23 x 23 Check: 23 23 23 23 The solution set is 23.
Section 2.2
30.
2 x 5 13 2 x 5 5 13 5 2x 8 2x 8 2 2 x 4 Check: 2 4 5 13 8 5 13 13 13 The solution set is 4.
32.
3x 2 9 3x 2 2 9 2 3x 11
51 y 24. 51 y 1
1 51 y Check: 51 51 The solution set is 51. x
26.
3x 11 3 3 11 x 3
1 5
5
x
Check:
51 5
11 9 3 11 2 9 9 9
3
x 5 Check: 5 1 5 1 1 The solution set is 5. 8x 3x 45 28. 8x 3x 45 5x 45 5x 45 5
5 x 9
Check: 8 9 3 9 45 72 27 45 45 45 The solution set is 9.
The solution set is
34.
11 . 3
3 y 4 13 3 y 4 4 13 4 3y 9 3y 9 3 3 y 3 Check: 3 3 4 13 9 4 13 13 13 The solution set is 3.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
36.
2y 5 7 2y 5 5 7 5 2 y 12
Section 2.2
44.
2 z 4 z 18 2 z 4 z 4 z 18 4 z 6 z 18
2 y 12
Check:
6 z 3 Check:
26 5 7 12 5 7 7 7 The solution set is 6.
2 3 4 3 18 6 12 18 6 6 The solution set is 3.
2
38.
6
2 y 6
14 5z 21 14 21 5z 21 21 35 5z 35 5z 5
40.
6 z 18
5 7 z
46.
7 x 3x 8 7 x 3x 3x 8 3x 4x 8 4x 8 4
4 x 2
Check:
Check:
14 5 7 21 14 35 21 14 14 The solution set is 7.
7 2 32 8 14 6 8 14 14 The solution set is 2.
x5 5 x55 55 x 10 x 10 Check:
48. 5 y 6 3y 6 5 y 6 3y 3y 6 3y 2y 6 6 2y 6 6 6 6 2 y 12
10 5 5 10 5 5
2y 2
5 5 The solution set is 10.
12 2
y 6 Check: Check:
42.
8 y 3 y 10 8 y 3 y 3 y 10 3 y 5 y 10 5 y 10 ď€ 5 5 y 2
8 2 3 2 10 16 6 16 16 16 The solution set is 2.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E 56 6 36 6 30 6 18 6 2 4 2 4 The solution set is 6.
Section 2.2
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
50.
6z 3 z 2 6z 3 z z 2 z 5z 3 5z 3 3 5z 5z
2 23 5 5
Section 2.2
x
58.
60. 6 x 20 6x 6
20 6 10
Check: x
61 3 1 2 63 3
3 10
The number is
3 3 The solution set is 1. 9x 2 6x 4 9x 2 6x 6x 4 6x 3x 2 4 3x 2 2 4 2 3x 6
62.
3x 6 3 3 x 2
64.
7
x 7
78
x 56 The number is 56 . 3x 10 23 3x 10 10 23 10 3x 33 3x 33 3 3 x 11 The number is 11. 66.
3y 2 5 4y
54.
5x 11 29 5x 11 11 29 11 5x 40
3y 2 4 y 5 4 y 4 y y2 5
5x 5
y22 52 y 3 68.
n 5 n 3 5
M
53 5 56.
x
40 5
x 8 The number is 8.
Check:
3 3 2 5 4 3 9 2 5 12 7 7 The solution set is 3.
. 3
x 8 7
Check: 92 2 62 4 18 2 12 4 16 16 The solution set is 2.
x
5 5 z 1
52.
x
x
n 5
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
x
15 n If you are 3 mil es aw ay fro m the ligh tnin g flas h, it will tak e 15 sec ond s for the sou nd of thu nde r to rea ch you .
Section 2.2
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
70.
A 740 A 3.3 740 A 740 3.3 740 M
740 2442 A The speed of the SR-71 Blackbird is 2442 miles per hour. 72. a. The bar graph indicates that the average nightly hotel room rate was \$98. Since 2009 is 106 years after 1903, substitute 106 into the formula for n. H 0.5n 50 H 0.5(106) 50
Section 2.2
84. Answers will vary. Start by selecting the integer answer and set x equal to this value. Then, multiply both sides of this equation by 60 (since we will divide both sides of the equation by 60 to solve). For example, suppose we want the solution to be 3. We set x equal to this value and write x 3. Now multiply both sides of the equation by 60 . x 3 60 x 60 3 60 x 180 So, our equation is 60x 180 and the solution is 3 (an integer). 86. 3.7 x 19.46 9.988 3.7 x 9.988 19.46 3.7 x 9.472
H 53 50 H 103 The formula indicates that the average nightly
x 2.56 The solution set is 2.56.
hotel room rate was \$103 in 2009. The formula overestimates by \$5. b.
3.7 x 9.472 3.7 3.7
H 0.5n 50 110 0.5n 50
88. 10
110 50 0.5n 50 50 60 0.5n 60 0.5n 60 0.5n 0.5 0.5 120 n The formula estimates that \$110 will be the cost of the average nightly hotel room rate 120 years after 1903, or 2023. 74. Answers will vary. 76. does not make sense; Explanations will vary. Sample explanation: The addition property of equality is not necessary for this equation. 78. makes sense
2
10 10 100
2 2 89. 10 110 110 10 100
90. x 4 x 1 4 1 14 3 3
3
91. 13 3( x 2) 13 3x 6 3x 7 92. 2( x 3) 17 13 3( x 2) 2(6 3) 17 13 3(6 2) 2(3) 17 13 3(8) 6 17 13 24 11 11, true Yes, 6 is a solution of the equation.
80. false; Changes to make the statement true will vary. A sample change is: If 7 x 21, then
7 x 21 7
3. 7
82. false; Changes to make the statement true will vary. A sample change is: If 3x 7 0, then 3x 7 and
x
7
. 3
93. 10
x 39 5
10 5
x
10
39
5 2 x 78
5
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
2.3 Check Points 1. Simplify the algebraic expression on each side. 7 x 25 3x 16 2 x 3
Section 2.3
4. Begin by multiplying both sides of the equation by 12, the least common denominator. x 2x 5 4 3 6
4 x 25 13 2 x
x
Collect variable terms on one side and constant terms on the other side.
12
4 x 25 13 2 x 4 x 25 2 x 13 2 x 2 x 2 x 25 13 2 x 25 25 13 25 2 x 12 Isolate the variable and solve.
12
5
ď€
6
5 x 2
2
x 6 The solution set is 6. 2. Simplify the algebraic expression on each side. 8x 2( x 6) 8x 2 x 12 Collect variable terms on one side and constant terms on the other side. 8x 2 x 2 x 2 x 12 6 x 12 Isolate the variable and solve. 6 x 12 6 6 x 2
3 6
5
x 2x 5 12 12 4 3 3x 8x 10 3x 8x 8x 8x 10 5x 10 5 x 10
2 x 12 2
4
2x 12
The solution set is 2. 5. First apply the distributive property to remove the parentheses, and then multiply both sides by 100 to clear the decimals. 0.48x 3 0.2(x 6) 0.48x 3 0.2 x 1.2 100(0.48x 3) 100(0.2 x 1.2) 48x 300 20 x 120 48x 300 300 20 x 120 300 48x 20 x 420 48x 20x 20 x 20 x 420 28x 420
The solution set is 2.
28 x 420 28
3. Simplify the algebraic expression on each side. 4(2 x 1) 29 3(2 x 5) 8x 4 29 6 x 15 8x 25 6 x 15 Collect variable terms on one side and constant terms on the other side. 8x 6 x 25 6 x 6 x 15 2 x 25 15 2 x 25 25 15 25 2 x 10 Isolate the variable and solve. 2 x 10 2 2 x 5 The solution set is 5.
28 x 15 The solution set is 15. 6.
3x 7 3( x 1) 3x 7 3x 3 3x 3x 7 3x 3x 3 7 3 The original equation is equivalent to the false statement 7 3. The equation has no solution. The solution set is .
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
7. 3( x 1) 9 8x 6 5x 3x 3 9 3x 6 3x 6 3x 6 3x 3x 6 3x 3x 6 6 6 The original equation is equivalent to 6 6, which is true for every value of x. The equation’s solution is all real numbers or x x is a real number. 8.
10 53 x 9 9 10 53 10 x 9 9 10 53 9 10 9 x 9 9 90 10 x 53 90 53 10 x 53 53 37 10 x 37 10x 10 10 3.7 x x 3.7 The formula indicates that if the low-humor group averages a level of depression of 10 in response to a negative life event, the intensity of that event is 3.7. D
This is shown as the point whose corresponding value on the vertical axis is 10 and whose value on the horizontal axis is 3.7.
Section 2.3
2.3 Exercise Set 2. 4 x 8x 2 x 20 15 10 x 5 5 1 x 10 2 1 The solution set is . 2 4. 3x 2 x 64 40 7 x 5x 64 40 7 x 12 x 64 40 12x 24 x 2 The solution set is 2. 6.
3x 2 x 6 3x 8 2 x 2 3x 2 2 x 2 3x 3x 2 3x x2 2 x22 22 x 4 x 4 The solution set is 4.
8. 3 x 2 6 3x 6 6 3x 0 x 0 The solution set is 0.
2.3 Concept and Vocabulary Check 1. simplify each side; combine like terms 2. 30 3. 100 4. inconsistent 5. identity 6. inconsistent 7. identity
10.
4 2 x 3 32 8x 12 32 8x 44 44 11 x 8 2 11 The solution set is . 2
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
12. 20 44 8 2 x 20 44 16 8x 20 28 8x 8 8x 1 x The solution set is 1. 14.
3 3z 5 7 89 9 z 15 7 89 9 z 8 89 9z 81 z 9 The solution set is 9.
5x 2 x 14 10 5x 2 x 14 10 16. 3x 14 10 3x 24 x 8 The solution set is 8. 3 x 2 x 30 3x 6 x 30 2 x 6 30 18. 2 x 24 x 12 The solution set is 12. 3 3x 1 4 3 3x 9 x 3 12 12 x 3 3 12 3x 15 20. x 5 The solution set is 5. 8 y 3 3 2 y 12 8 y 24 6 y 36
Section 2.3
24. 5x 4 x 9 2 x 3 5x 4 x 36 2 x 3 x 36 2 x 3 x 2 x 33 x 33 x 33 The solution set is 33. 26.
7 3x 2 5 6 2 x 1 24 21x 14 5 12 x 6 24 21x 9 12 x 18 21x 12 x 27 9 x 27 x 3 The solution set is 3.
100 x 1 4 x 6 100 x 1 4 x 24 28. 100 3x 23 123 3x 41 x The solution set is 41. 2 z 4 3z 2 2 6 z 2 2 z 8 3z 2 2 6z 2 5z 10 6 z z 10 30. 0 z 10 The solution set is 10. x 13 22 2 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 2. 32. x 13 22 2 x 2 13 2 22
22.
2 y 24 36 2 y 12 y 6 The solution set is 6.
2 x 2 2 13 44 2 x 26 44 x 26 26 44 26 x 70 The solution set is 70.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
34.
3x 9 6 4 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 4. 3x 4 9 46 4 3x 4 4 9 24 4 3x 36 24 3x 12 x 4
Section 2.3
40.
5 2 6 6 z 15 5z z 15 0 z 15 The solution set is 15.
The solution set is 4.
y 42.
3y 2 7 36. 4 3 12 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 12. 3y 2 7 12 12 4 3 12 3y 2 12 12 7 4 3 9y 8 7 9 y 15 15 5 y 9 3 The solution set is
38.
5 . 3
x x 1 4 5 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 20. x x 20 20 1 4 5 5x 4 x 20 x 20 The solution set is 20.
z 1 z 5 2 6 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 30. z 1 z 30 30
1
y
1
12 6 2 4 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 12. y 1 y 1 12 12 12 6 2 4 y 2 6y 3 5y 2 3 5y 5 y 1 The solution set is 1.
44.
3x 2 x 2 5 5 3
5
To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 15. 3x 2 x 2 15 15 5 5 3 5 9 x 6 5x 6 4x 6 6 4 x 12 x 3 The solution set is 3.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
46.
x2 x1 4 3 4 To clear the equation of fractions, multiply both sides by the least common denominator (LCD), which is 12. x2 x 1 12 12 4 12 3
4 4 x 2 48 3 x 1 4 x 8 48 3x 3 4 x 56 3x 3 x 56 3 x 59 The solution set is 59.
Section 2.3
54. 1.4( z 5) 0.2 0.5(6z 8) 1.4 z 7 0.2 3z 4 1.4 z 7.2 3z 4 To clear the equation of decimals, multiply both sides by 10. 10(1.4 z 7.2) 10(3z 4)
48. 1.2 x 3.6 2.4 0.3x To clear the equation of decimals, multiply both sides by 10. 10(1.2 x 3.6) 10(2.4 0.3x) 12 x 36 24 3x 12 x 60 3x 15x 60 x 4 The solution set is 4. 50. 0.15 y 0.1 2.5 y 1.04 To clear the equation of decimals, multiply both sides by 100. 100(0.15 y 0.1) 100(2.5 y 1.04) 15 y 10 250 y 104 15 y 250 y 94 235 y 94 y 0.4 The solution set is 0.4. 52. 0.1( x 80) 14 0.2 x 0.1x 8 14 0.2 x To clear the equation of decimals, multiply both sides by 10. 10(0.1x 8) 10(14 0.2 x) x 80 140 2x x 60 2 x 3x 60 x 20 The solution set is 20.
14z 72 30z 40 14z 30z 32 16z 32 z 2 The solution set is 2. 56. 0.02( x 2) 0.06 0.01( x 1) 0.02 x 0.04 0.06 0.01x 0.01 0.02 x 0.04 0.01x 0.05 To clear the equation of decimals, multiply both sides by 100. 100(0.02 x 0.04) 100(0.01x 0.05) 2x 4 x 5 2x x 9 3x 9 x 3 The solution set is 3. 58. 0.05(7 x 36) 0.4x 1.2 0.35x 1.8 0.4x 1.2 To clear the equation of decimals, multiply both sides by 100. 100(0.35x 1.8) 100(0.4 x 1.2) 35x 180 40 x 120 35x 40 x 60 5x 60 x 12 The solution set is 12. 60.
2 x 5 2 x 10 2 x 10 2 x 10 2 x 10 2 x 2 x 10 2 x 10 10 The original equation is equivalent to the false statement 10 10, so the equation is inconsistent and has no solution. The solution set is .
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
62.
3 x 1 8x 6 5x 9 3x 3 3x 3 3x 3 3x 3x 3 3x 3 3 The original equation is equivalent to the true
Section 2.3
70.
5 x 4x 5 5 x 4x 4x 5 4x 5x 5 5 5x 0 5x 0
statement 3 3, so the equation is an identity
5
and the solution set is all real numbers x x is a real number. 64. 2 3 2 x 7 9 4 3x 1 2 6 x 21 9 12 x 4 6 x 19 12 x 5 18x 19 5
x 0 The solution set is 0. x x 3 72. 4 4 Multiply by the LCD, which is 4. x x 4 3 4 4
18x 24
5x 5 3x 7 5x 5 3x 7 5x 5 5x 5x 5 5x 5x 5 5
2 x 1 2x 2 5 5 5x
The original equation is equivalent to the true
74. x 2 x 3 x3 2 3 Multiply both sides by the LCD which is 6. x 2x 6 3 6 x 3 2 3 3x 4 x 18 6 x 18 7 x 18 6 x 18 x 18 18 x 0 The solution set is 0.
statement 5 5, so the equation is an identity and the solution set is all real numbers x x is a real number. 68. 5x 3 x 1 2 x 3 5 5x 3x 3 2 x 6 5 2x 3 2x 1 2x 3 2x 2x 1 2x 31 Since 3 1 is a false statement, the original equation is inconsistent and has no solution. The solution set is .
4
x 12 x x 12 x x x 12 0 Since 12 0 is a false statement, the original equation has no solution. The solution set is .
24 4 18 3 4 The solution set is . 3 x
66.
5
76.
2
1 x8 3 4 Multiply both sides by the LCD which is 12. 2 1 12 x 12 x 8 3 4 8x 3x 96 5x 96 x
x
96 5
The solution set is
96
. 5
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
78.
0.04( x 2) 0.02(6 x 3) 0.02 0.04 x 0.08 0.12 x 0.06 0.02
Section 2.3
2 1 x 13 x 5 4
84.
0.04 x 0.08 0.12 x 0.08
20
To clear the equation of decimals, multiply both sides by 100. 100(0.04 x 0.08) 100(0.12 x 0.08)
2 5
x 20 The number is 20. 86.
x
\$
7 1 x 30 x 8 2 7 1 8 x 30 8 x 8 2 7 x 240 4 x 240 3x
\$ x
x 20 13 4 8x 5x 260
13x 260 13 13
8x 0 x 0 The solution set is 0. x
1
13x 260
4 x 8 12 x 8 4 x 12 x
80.
x
\$
\$
240 3x 3 3
x
\$
80 x The number is 80.
x x
\$ \$
x
82. First solve the equation for x. 3x 3x x 4 2 4 4 3x 3x x 4 4 4 2 4 4
88. 400 400 1000 100
A person receiving a \$400 fine was driving 100
6x 3x x 16 9 x x 16 8x 16 x 2
miles per hour. W 3H 53 2 W 3(12) 53 2 W 36 53 2
90.
Now evaluate the expression x 2 x for x 2. x 2 x (2)2 (2) 4 2 6
F 10 x 65 50 10 x 650 50 10 x 600 10 x x
W 2
36 36 53 36 W 89 2 W 2 2 89 2 W 178
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
Section 2.3 According to the formula, the healthy weight of a person of height 6’ is 178 pounds. This is 6 pounds below the upper end of the range shown in the bar graph.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
p 15
92.
20 15
5d
Section 2.3
108. 11
5d
11 5d 5 11 5d 115 11 11 55 5d 11 d The pressure is 20 pounds per square foot at a depth of 11 feet.
2 3x 4 3x 2 3 x 1 2 6x 8 3x 2 3x 3 2 6x 8 3x 2 3x 1 6x 8 3x 6 x 2 6x 8 9 x 2 6x 8 9x 9x 2 9x 3x 8 2 3x 8 8 2 8 3x 10 3x 10 3 3
94. – 96. Answers will vary.
x
98. makes sense 100. does not make sense; Explanations will vary. Sample explanation: Though 5 is a solution, the complete solution is all real numbers.
10 3
The solution set is
10 . 3
109. 24 20 because 24 lies further to the left on a number line.
102. false; Changes to make the statement true will vary. A sample change is: The solution of the equation is all real numbers. 110.
104. true 106.
f 0.432h 10.44 16 0.432h 10.44 16 10.44 0.432h 10.44 10.44 26.44 0.432h 26.44 0.432h 0.432 0.432 61.2 h The woman’s height was about 61 inches or 5 feet 1 inch, so the partial skeleton could be that of the missing woman.
1 1 because 3 5 number line.
1 lies further to the left on a 3
111. 9 11 7 3 9 11 7 3 20 10 10 112. a.
T D pm T D pm
b. T D pm T D pm p p TD 113.
m
p
4 0.25B 4 0.25B 0.25 0.25 16 B The solution set is 16.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
114.
Section 2.4
1.3 P 26
5. Use the formula A PB :
1.3 P 26
W hat is 9% of 50?
26 26 0.05 P The solution set is 0.05.
A 0.09 A 4.5
A is P percent of B.
50
6. Use the formula A PB :
A is P percent of B.
9 is 60% of wh at?
2.4 Check Points 1.
A lw A lw w A
9 0.60 9 0.60 B 0.60 0.60 15 B
B
7. Use the formula A PB :
w
A is P percent of B.
l 18 is what percent of 50?
w 2.
2l 2w P 2l 2w 2w P 2w 2l P 2w 2l P 2w 2 2 P 2w l 2
18 P 50 18 P 50 18 50 P 50 50 0.36 P To change 0.36 to a percent, move the decimal point two places to the right and add a percent sign. 0.36 36%
8. Use the formula A PB : 3.
T D pm T D pm p TD
p m m
4.
p TD p
x 4y 5 3 x 3 4y 3 5 3 x 3 3 4y 3 5 3 x 12 y 15 x 12 y 12 y 15 12 y x 15 12 y
A is P percent of B.
Find the price decrease: \$940 \$611 \$329 The price what the original decrease is percent of price?
329
P
940
329 P 940 329 940P 940 940 0.35 P To change 0.35 to a percent, move the decimal point two places to the right and add a percent sign. 0.35 35%
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
9. a. Year
Tax Paid
increase/decrease
the Year Before
Section 2.4
Taxes Paid
1
\$1200
20% decrease : 0.20 \$1200 \$240
This Year \$1200 \$240 \$960
2
\$960
20% increase : 0.20 \$960 \$192
\$960 \$192 \$1152
The taxes for year 2 will be \$1152. b. The taxes for year 2 are less than those originally paid. Find the tax decrease: \$1200 \$1152 \$48 The tax what the original decrease is percent of tax?
48
P
1200
48 P 1200 48 1200 P 1200 1200 0.04 P To change 0.04 to a percent, move the decimal point two places to the right and add a percent sign. 0.04 4% The overall tax decrease is 4%.
2.4 Concept and Vocabulary Check 1. isolated on one side 2. A lw 3. P 2l 2w 4. A PB 5. subtract b; divide by m
2.4 Exercise Set 2. d rt for t d rt ď€ r r d d t or t r
r
This is the motion formula: distance = rate ¡ time. 4. I Prt for r I Prt Pt PT I I r or r Pt
Pt
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E This is the formula for simple interest: interest = principal ¡ rate ¡ time.
Section 2.4
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
6. C d for d C d
Section 2.4
C
A for A 740 A 740M 740 740
740M A
16. M
C
d
or d
or A 740M
This is the formula for finding the circumference of a circle if you know its diameter.
p 15
18.
8. V r 2 h for h V
r2 V
11 p 11 15
5d 11
r2h
r2
h
r2
5d for d 11
11 p 165 5d
or h
11 p 165 5d 11 p 165
V r2
11 p 165
d or d
5
5
This is the volume of a cylinder. 1
A 10. y mx b for x y b mx yb m yb
2 2A 2
or x
yb
m m This is the slope-intercept formula for the equation of a line.
P C MC P C MC
or M
PC
C C This is the business math formula for mark-up based on cost. 14. A
1
bh for h 2
This is the formula for finding the average of two numbers.
SP C
M
ab 2
22. S P Prt for t S P Prt S P Prt Pr Pr
12. P C MC for M P C C MC C
C PC
1
2A a b 2A a b or b 2 A a
m x
20.
mx
for b
ab
t
or t
SP
Pr Pr This is the formula for finding the sum of principle and interest for simple interest problems. 24. A
1
h a b for a 2 1 2A 2 h a b 2
2A 2
1
bh
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
Section 2.4
2A h a b
2
2A h 2A
2 A bh 2A b 2A
bh
h
b
h or h
2A
b b This is the formula for the area of a triangle: area = 1 · base · height. 2
2A h 2A
h a b h ab
b abb
2A h b a h or a b
This is the formula for finding the area of a trapezoid.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
Section 2.4
26. Ax By C for y Ax By Ax C Ax By C Ax By C Ax B B C Ax y B This is the standard form of the equation of a line.
38. A PB; A 0.6, B 7.5 A PB 0.6 P 7.5 0.6 P 7.5 7.5 7.5 0.08 P 0.08 8% 0.6 is 8% of 7.5.
28. A PB; P 8% 0.08, B 300 A PB A 0.08 300 24
40. The increase is 9 5 4. A PB 4 P5 4 5P 5 5
30. A PB; P 16% 0.16, B 90 A PB A 0.16 90 14.4 16% of 90 is 14.4 32. A PB; A 8, P 40% 0.4 A PB 8 0.4 B 8 0.4 B 0.4 0.4 20 B 8 is 40% of 20.
0.80 P This is an 80% increase. 42. The decrease is 8 6 2. A PB 2 P8 2 8P 8 8 0.25 P This is a 25% decrease.
34. A PB; A 51.2, P 32% 0.32
abx a b
A PB 51.2 0.32 B 51.2 0.32 B 0.32 0.32 160 B 51.2 is 32% of 160.
y a bxy
44.
ab
y ab 46.
y x or x ab
y abx8 y8 abx88 y8 abx
36. A PB; A 18; B 90 A PB 18 P 90
abx y8 ab a b
18 P 90
y8
90 90 0.2 P 0.2 = 20% 18 is 20% of 90.
x
or x
ab
y8 ab
y cx dx y cdx
48.
y
c d
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
Section 2.4 ď€
cdx y cd
cd y
x or x cd
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
50.
Section 2.4
y Ax Bx C y C Ax Bx C C y C Ax Bx
56. 0.14 1800 252 252 workers stated that politics is the most taboo topic to discuss at work.
yC ABx
58. This is the equivalent of asking: 55 is 11% of what?
yC
A PB 55 0.11 B 55 0.11B 0.11 0.11
A Bx A B
yC A B x
or x
AB
yC AB
xyzw 4
52. a. A
4A 4
500 B Americans throw away 500 billion pounds of trash each year.
for w xyzw 4
60. a. The total number of countries in 1974 was 41 48 63 152.
4A x y z w 4A x y z x y z w x y z
A PB 63 P 152 63 152B 152 152
4A x y z w b. w 4 A xy z; x 76, y 78, z 79 w 4A x y z w 4 80 76 78 79 w 87 You need to get 87% on the fourth exam to have an average of 80%.
0.41 B About 41% of countries were not free in 1974. b. The total number of countries in 2009 was 89 62 42 193. A PB 42 P 193 42
54. a. F
9
b. C C
C
193 193 0.22 B About 22% of countries were not free in 2009.
C 32 for C 5 9 5F 5 C 32
5 5F 9C 160 5F 160 9C 5F 160 9C 9 9 5F 160 C 9 5F 160 ; F 59 9 5F 160 9 5 59 160 9
193B
c. The decrease is 63 42 21. A PB 21 P 63 21 63B 63 63 0.33 B There was approximately a 33% decrease in the number of not free countries from 1974 to 2009. 62. This question is equivalent to, “225,000 is what percent of \$500,000?” A PB 225, 000 P 500, 000 225, 000 P 500, 000 0.45 P
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
Cď€
C
295 160 9 135 15 9
59F 15C
Section 2.4 500, 000 500, 000 The charity has raised 45% of the goal.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
64. \$3502 0.28 35, 000 \$23, 000 \$3502 0.28 \$12, 000 \$3502 \$3360 \$6862 The income tax on a taxable income of \$35,000 is \$6862. 66. a. The sales tax is 7% of \$96. 0.07 96 6.72 The sales tax due on the graphing calculator is
Section 2.4
78. does not make sense; Explanations will vary. Sample explanation: Since the sale price cannot be negative, the percent decrease cannot be more than 100%. 80. false; Changes to make the statement true will vary. A A sample change is: If A lw, then w . l 82. true 84.
\$6.72.
5x 20 8x 16 5x 20 8x 8x 16 8x 3x 20 16 3x 20 20 16 20 3x 36 3x 36 3 3
b. The total cost is the sum of the price of the calculator and the sales tax. \$96 \$6.72 \$102.72 The calculator’s total cost is \$102.72. 68. a. The discount amount is 40% of \$16.50.
x 12
0.4 16.50 6.60 Check:
The discount amount is \$6.60.
5 12 20 8 12 16 60 20 96 16 80 80 The solution set is 12.
b. The sale price is the regular price minus the discount amount. \$16.50 \$6.60 \$9.90 The sale price is \$9.90. 70. The decrease is \$380 \$266 = \$114. A PB 114 P 380
85.
114 P 380 380 380 0.30 P This is a 0.30 = 30% decrease. 72. No; the first sale price is 70% of the original amount and the second sale price is 80% of the first sale price. The second sale price would be obtained by the following computation: A P2 P1 B 0.80 0.70B 0.56B The second sale price is 56% of the original price, so there is 44% reduction overall.
5 37 1 4 46 185 1 184 184 184 The solution set is 20.
74. Answers will vary. 76. does not make sense; Explanations will vary. Sample explanation: Sometimes you will solve for one variable in terms of other variables.
52 y 3 1 4 6 2y 10 y 15 1 24 8 y 10 y 16 24 8 y 10 y 16 8 y 24 8 y 8 y 2 y 16 24 2 y 16 16 24 16 2 y 40 2 y 40 2 2 y 20 Check: 5 2 20 3 1 4 6 2 20 5 40 3 1 4 6 40
x 0.3x 1x 0.3x 1 0.3 x 0.7 x 86.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forinCollege One Variable Students 4E
Section 2.4
87.
13
7x x
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
88. 8( x 14)
Mid-Chapter Check Point
x 3 10 x
5.
89. 9( x 5) 10
1. Begin by multiplying both sides of the equation by 4, the least common denominator. x x 12 2 4 x x 4 4 12 4 2 4
1 3 y 5 4 2 3y 1 3 y 15 8 12 y 3 y 16 8 12 y 3 y 12 y 16 8 12 y 12 y 9 y 16 8 9 y 16 16 8 16 9y 8 6.
2x 48 x 2 x x 48 x x 3x 48 3x 48 3
3 x 16 The solution set is 16.
5
9y 8 9 9 8 y 9 8 The solution set is . 7.
5 x 3
EC 825 EC H 825 825 825 825H EC 825H EC E E 825H C E H
4. A P B A 0.06 140 A 8.4 8.4 is 6% of 140.
S 2 rh S 2 rh 2h 2h
The solution set is 3. 3.
10 3
x 30 1 x 130 x 30 The solution set is 30.
Mid-Chapter Check Point – Chapter 2
2. 5x 42 57 5x 42 42 57 42 5x 15 5 x 15
10
S 2h 8.
r
A PB 12 0.30 B 12 0.30 B 0.30 0.30 40 B 12 is 30% of 40.
9
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra for in College One Variable Students 4E
3y y 5 y 3 5 2 4 To clear fractions, multiply both sides by the LCD,
9.
20. 20
3y
y
20
5
20
5y
20 3
Mid-Chapter Check Point
Ax By C Ax By By C By Ax C By
12.
Ax C By A A
4
2
C By x
4 3 y 10 y 5 5 y 60
By C or
A
A
12 y 10 y 25 y 60 22 y 25 y 60 22 y 25 y 25 y 25 y 60 3 y 60 3 y 60 3
9y 7 9y 3 9y 9y 7 9y 9y 3 7 3
Since this is a false statement, there is no solution or
3
.
y 20 The solution set is 20. 14. 10.
2
x 3 10
3
x1
10
1
5 3
x 10 3 10 x 10 1 5
20
x 5.25 The solution set is 5.25. 6 z 2 4 2 z 3 6 z 12 8z 12 7 2z 7 2 z 5z 7 7z 7 7z
5z 7 5z 7 5z 5z 5z
7
1
5x 30 6 x 10 5x 5x 30 6 x 5x 10 30 x 10 30 10 x 10 10 40 x The solution set is 40.
To clear decimals, multiply both sides by 10. 10(2.4 x 6) 10(4.4 x 4.5) 24 x 60 44 x 45 24 x 44 x 105 20 x 105 20 x 105
11.
10
2
2.4 x 6 1.4 x 0.5(6 x 9) 2.4 x 6 1.4 x 3x 4.5 2.4 x 6 4.4 x 4.5
20
6 y 7 3 y 3 3 y 1
13.
7 1 z The solution set is 1.
15.
A PB 50 P 400 50 P 400 400 400 0.125 P 50 is 0.125 = 12.5% of 400.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
3 m 2
16.
4
Section 2.5
b.
2m 3
B 22
3 m 2 4 2m 3 4 3 m 2 4 2m 3
4
3m 6 8m 12 3m 3m 6 8m 3m 12 6 5m 12 6 12 5m 12 12 6 5m 6 5m 5 5 6 m 5 6 The solution set is . 5
2 2(22) 2
5
a 82
2 44 5a 164 120 5a 24 a According to the formula, 22% of 24-year-olds will believe that reading books is important.
2.5 Check Points
17. The increase is 50 40 = 10. A PB 10 P 40 10 P 40 40 40 0.25 P This is a 0.25 = 25% increase. 18. 12w 4 8w 4 4 5w 2 20w 8 20w 8 20w 20w 8 20w 20w 8 8 8 Since 8 = 8 is a true statement, the solution is all real numbers or x x is a real number.
19. a.
5 a 82 2 5 a 82
5 a 82 2 5 B (14) 82 2 35 82 47 B
According to the formula, 47% of 14-year-olds believe that reading books is important. This underestimates the actual percentage shown in the bar graph by 2%
1. Let x = the number. 6 x 4 68 6 x 4 4 68 4 6 x 72 x 12 The number is 12. 2. Let x = the median starting salary, in thousands of dollars, for English majors. Let x 18 the median starting salary, in thousands of dollars, for computer science majors. x ( x 18) 94 x x 18 94 2 x 18 94 2 x 76 x 38 x 18 56 The average salary for English majors is \$18 thousand and the average salary for computer science majors is \$38 \$18 \$56. 3. Let x = the page number of the first facing page. Let x 1 the page number of the second facing page. x ( x 1) 145 x x 1 145 2 x 1 145 2 x 1 1 145 1 2 x 144 x 72 x 1 73 The page numbers are 72 and 73.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
4. Let x = the number of eighths of a mile traveled. 2 0.25x 10 2 2 0.25x 10 2 0.25x 8 0.25 x 8 0.25 0.25 x 32 You can go 32 eighths of a mile. That is equivalent 32 to 4 miles. 8 5. Let x = the width of the swimming pool. Let 3x the length of the swimming pool. P 2l 2w 320 2 3x 2 x 320 6x 2 x 320 8x 320 8 x 8 8 40 x x 40 3x 120 The pool is 40 feet wide and 120 feet long.
Section 2.5
2.5 Exercise Set 2.
x 43 107 x 43 43 107 43 x 64 The number is 64.
4.
x 17 96 x 17 17 96 17 x 113 The number is 113.
6. 8x 272 8x 272 8 8 x 34 The number is 34. x 8 14
8.
x 14 8 14 x 112 The number is 112. 14
6. Let x = the original price. the reduction (40% of
Original price
the reduced
10. 5 3x 59 3x 54
minus original price) is price, \$564
x 18 x
0.4 x
564
x 0.4 x 564 0.6 x 564 0.6 x 564 0.6 0.6 x 940 The original price was \$940.
2.5 Concept and Vocabulary Check 1. 4 x 6
12. 6 x 8 298 6 x 306 x 51 The number is 51. 14. x 12 4 x 12 3x 4 x The number is 4. 16. 3 5 x 48 15 3x 48 3x 33
2. x 215 3. x 1
x 11 The number is 11.
4. 125 0.15x 5. 2 4 x 2 x
The number is 18.
or 2 x 2 4 x
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
6. x 0.35x or 0.65x
Section 2.5
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
18. 5 4 x x 35 5 3x 35 3x 30 x 10 The number is 10. 20.
3x 3 9 4 3x 12 4 3x 48 x 16 The number is 16.
22. Let x the number of years spent eating. Let x 24 the number of years spent sleeping. x ( x 24) 32 x x 24 32 2 x 24 32 2x 8 x 4 x 24 28 Americans will spend 4 years eating and 28 years sleeping. 24. Let x the average salary, in thousands, for an American whose final degree is a bachelor’s. Let 2 x 39 the average salary, in thousands, for an American whose final degree is a doctorate. x (2 x 39) 126 x 2 x 39 126 3x 39 126 3x 165 x 55 2 x 39 71 The average salary for an American whose final degree is a bachelor’s is \$55 thousand and for an American whose final degree is a doctorate is \$71 thousand. 26. Let x = the number of the left-hand page. Let x + 1 = the number of the right-hand page. x x 1 525 2 x 1 525 2 x 524 x 262 The smaller page number is 262. The larger page number is 262 + 1 = 263.
Section 2.5
28. Let x the first consecutive even integer (Hank Greenberg). Let x 2 the second consecutive even integer (Babe Ruth). x ( x 2) 118 x x 2 118 2 x 2 118 2 x 116 x 58 x 2 60 Hank Greenberg had 58 home runs and Babe Ruth had 60. 30. Let x = the number of miles you can travel in one week for \$395. 180 0.25x 395 180 0.25x 180 395 180 0.25x 215 0.25x 215 0.25 0.25 x 860 You can travel 860 miles in one week for \$395. 32. Let x the number of years after 2004. 824 7 x 929 7 x 105 7 x 105 7 7 x 15 Rent payments will average \$929 fifteen years after 2008, or 2023. 34. Let x = the width of the field. Let 5x the length of the field. P 2l 2w 288 2 5x 2 x 288 10 x 2 x 288 12 x 288 12 x 12 12 24 x x 24 5x 120 The field is 24 yards wide and 120 yards long.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
36. Let x = the width of a basketball court. Let x 13 the length of a basketball court. P 2l 2w 86 2( x 13) 2 x 86 2 x 26 2 x 86 4 x 26 60 4 x 15 x x 15 x 13 28 A basketball court is 15 meters wide and 28 meters
Section 2.5
44. Let x = the nightly cost without tax. x 0.08x 172.80 1.08x 172.80 1.08x 172.80 1.08
1.08 x 160 The nightly cost without tax is \$160. 46. Let x = the number of hours of labor. 532 63x 1603 532 63x 532 1603 532 63x 1071
long. 38. As shown in the diagram, let x = the length of a shelf and x + 3 = the height of the bookcase, 4 shelves and 2 heights are needed. Since 18 feet of lumber is available, 4 x 2 x 3 18. 4 x 2x 6 18 6x 6 18 6 x 12 x 2 x 3 5 The length of each shelf is 2 feet and the height of the unit is 5 feet. 40. Let x = the price before the reduction. x 0.30x 98 0.70x 98 0.70 x 98 0.70 0.70 x 140 The DVD player’s price before the reduction was \$140. 42. Let x = the last year’s salary. x 0.09x 42, 074 1.09x 42, 074 1.09 x 42, 074 1.09 1.09 x 38, 600 Last year’s salary was \$38,600.
63x
1071
63
63 x 17 It took 17 hours of labor to repair the sailboat. 48. – 50. Answers will vary. 52. makes sense 54. does not make sense; Explanations will vary. Sample explanation: It is correct to use x 2 for the second consecutive odd integer because any odd integer is 2 more than the previous odd integer. In other words, adding 2 to the first odd integer will skip over the even integer and take you to the next odd integer. 56. false; Changes to make the statement true will vary. A sample change is: This should be modeled by x 0.35x 780. 58. true 60. Let x = the number of minutes. Note that \$0.55 is the cost of the first minute and \$0.40( x 1) is the cost of the remaining minutes. 0.55 0.40 x 1 6.95 0.55 0.4 x 0.40 6.95 0.4 x 0.15 6.95 0.4 x 0.15 0.15 6.95 0.15 0.4 x 6.80 0.4 x 6.80 0.4 0.4 x 17 The phone call lasted 17 minutes.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
62. Let x = weight of unpeeled bananas. 1 7 Let x = the weight of banana peel and x = the 8 8 weight of peeled banana. The information in the cartoon translates into the equation. 7 7 x x 8 8 To solve this equation, first eliminate fractions by multiplying both sides by the LCD, which is 8. 7 7 8x 8 x 8 8 7 7 8x 8 x 8 8 8 8x 7 x 7 8x 7 x 7 x 7 7 x x 7 The unpeeled banana weighs 7 ounces.
63.
Section 2.5
6y 1 8y 6y 1 1 8y 6y 8y 6y 8y 8y 2y 0 y 0 Check: 60 1 7 90
Check: 4
65.
V
01
1
lwh for w 3 1 V lwh 3 1 3V 3 lwh 3
3V lwh
x 20
3V lwh lh lh 3V w or
20 16
The solution set is 20.
8y
The solution set is 0.
4
5 4 20 16 5 1 80 16 5 16 16
1 11
6 10 7 0 0 1 11
4 x 16 5 5 4 5 x 16 4 5
6 y 1 7 9y y 1 6y 6 7 9y y 1
64.
lh 66.
w
3V lh
A 1 2bh 30 1 212h 30 6h 30 6h
6
6
5 h 67. A 1 2h(a b) A 1 2(7)(10 16) A 1 2(7)(26) A 91 68.
Chapter Introductory 2: Linear and Equations Intermediate and Inequalities Algebra forin College One Variable Students 4E
Section 2.5
x 4(90 x) 40 x 360 4 x 40 x 320 4 x 5x 320 x 64 The solution set is 64.
Introductory and Intermediate Algebra for College Students 4th Edition Blitzer SOLUTIONS MANUAL Full download: http://testbanklive.com/download/introductory-and-intermediate-algebra-for-collegestudents-4th-edition-blitzer-solutions-manual/ Introductory and Intermediate Algebra for College Students 4th Edition Blitzer TEST BANK Full download: http://testbanklive.com/download/introductory-and-intermediate-algebra-for-collegestudents-4th-edition-blitzer-test-bank/ people also search: introductory and intermediate algebra for college students 4th edition pdf introductory and intermediate algebra for college students 4th edition by robert blitzer pdf introductory and intermediate algebra for college students 5th edition introductory & intermediate algebra for college students (5th edition) pdf introductory and intermediate algebra 4th edition blitzer introductory and intermediate algebra for college students 5th edition introductory and intermediate algebra 5th edition pdf introductory and intermediate algebra for college students blitzer pdf
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1 CSC 1103: Digital Logic Lecture Six: Data Representation Martin Ngobye Mbarara University of Science and Technology MAN (MUST) CSC / 32
2 Outline 1 Digital Computers 2 Number Systems Conversion Between Binary and Decimal Octal and Hexadecimal Numbers 3 Summary MAN (MUST) CSC / 32
3 Digital Computers Humans have many symbolic forms to represent information. Alphabet, numbers, pictograms Several formats are used to store data. Deep down inside, computers work with just 1 s and 0 s. Because a computer uses binary numbers all these formats are patterns of 1 s and 0 s. MAN (MUST) CSC / 32
4 Data Types Binary information in digital computers is stored in memory or processor registers. Registers contain either data or control information. Control information is a bit or a group of bits used to specify the sequence of command signals needed for manipulation of the data in other registers. Data are numbers and other binary-coded information that are operated on to achieve required computational results. Present common types of data found in digital computers Show how the various data types are presented in binary-coded form in computer registers. MAN (MUST) CSC / 32
5 Data Types Data types found in registers of digital computers may be classified as: 1 Numbers used in arithmetic computations 2 Letters of the alphabet used in data processing 3 Other discrete symbols used for specific purposes. All types of data except binary numbers are presented in computer registers in the binary-coded form. MAN (MUST) CSC / 32
6 Data Representation A bit is the most basic unit of information in a computer. A byte is a group of eight bits. It is the smallest possible addressable unit of computer storage. A word is a contiguous group of bytes. Words can be any number of bits or bytes. Word sizes of 16, 32, or 64 bits are most common. MAN (MUST) CSC / 32
7 Number Systems A number system of base or radix r is a system that uses distinct symbols for digit representation. Numbers are represented by a string of digit symbols. To determine the quantity that a number represents, it is necessary to multiply each digit by an integer power of r and then form the sum of all weighted digits. For example, the decimal number system in everyday use employs the radix 10 system. The 10 symbols are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. MAN (MUST) CSC / 32
8 Number Systems The strings of digits is interpreted to represent the quantity: 7x x x x10 1 That is 7 hundreds, plus 2 tens, plus 4 units, plus 5 tenths. MAN (MUST) CSC / 32
9 Number Systems The binary number system used the radix 2. The two digit symbols used are 0 and 1. The string of digits is interpreted to represent the quantity 1x x x x x x2 0 = 45 MAN (MUST) CSC / 32
10 Number Systems Besides the decimal and binary number systems, the octal (radix 8) and hexadecimal (radix 16) are important in digital computer work. The eight symbols of the octal system are 0, 1, 2, 3, 4, 5, 6, and 7. The 16 symbols of the hexadecimal system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F. The symbols A, B, C, D, E, F correspond to the decimal numbers 10, 11, 12, 13, 14, 15 respectively. MAN (MUST) CSC / 32
11 Number Systems Conversion A number in the radix r can be converted to a decimal system by forming the sum of weighted digits. For example, octal is converted to decimal as follows: (736.4) 8 = 7x x x x8 1 = 7x64 + 3x8 + 6x = (478.5) 10 The equivalent decimal number of hexadecimal F 3 is obtained from the following calculation: (F 3) 16 = Fx x16 0 = 15x = (243) 10 MAN (MUST) CSC / 32
12 Conversion Between Binary and Decimal There are two methods for base conversion: the subtraction method and the division remainder method. The subtraction method The division remainder method MAN (MUST) CSC / 32
13 Conversion Decimal to Binary We can convert from base 10 to base 2 by repeated divisions by 2. The remainders and the final quotient, 1, give us, the order of increasing significance and the binary digits of a number N. To convert a decimal integer into binary, keep dividing by 2 until the quotient is 0. Collect the remainders in reverse order. Example One: Convert to base 2. 11/2 = 5 rem 1 5/2 = 2 rem 1 2/2 = 1 rem 0 1/2 = 0 rem 1 Therefore, = MAN (MUST) CSC / 32
14 Conversion Decimal to Binary Example Two: Convert to base 2. 21/2 = 10 rem 1 10/2 = 5 rem 0 5/2 = 2 rem 1 2/2 = 1 rem 0 1/2 = 0 rem 1 Therefore, = MAN (MUST) CSC / 32
15 Conversion Decimal to Binary To convert a decimal number, for example into binary is done by first separating the number into its integer part 162 and the fraction part.375. The integer part is converted by dividing 162 by 2 to get reminders. The fraction part is converted by multiplying it by 2 until it becomes 0. Collect the integer parts in forward order to get the decimal part of the number. MAN (MUST) CSC / 32
16 Conversion Decimal to Binary Example One: Convert into binary. 162/2 = 81 rem 0 81/2 = 40 rem 1 40/2 = 20 rem 0 20/2 = 10 rem 0 10/2 = 5 rem 0 5/2 = 2 rem 1 2/2 = 1 rem 0 1/2 = 0 rem 1 MAN (MUST) CSC / 32
17 Conversion Decimal to Binary Example One: Convert into binary. The decimal part is converted as: 0.375x2 = x2 = x2 = Therefore, = MAN (MUST) CSC / 32
18 Octal and Hexadecimal Number The conversion from and to binary, octal and hexadecimal representation plays an important part in digital computers. Since 2 3 = 8 and 2 4 = 16, each octal digit corresponds to three binary digits and each hexadecimal digit corresponds to four binary digits. The conversion from binary to octal is easily accomplished by partitioning the binary number into groups of three bits each. The corresponding octal digit is then assigned to each group of bits and the strings of digits obtained gives the octal equivalent of the binary number. MAN (MUST) CSC / 32
19 Octal and Hexadecimal Numbers Consider a 16-bit register shown as: Starting from the low-order bit, we partition the register into groups of three bits. Each group of three bits is assigned its octal equivalent and placed on top of the register. The string of octal digits obtained represents the octal equivalent of the binary number. The octal representation of the number is shown as: 1 {}}{ 1 2 {}}{ {}}{ {}}{ {}}{ {}}{ MAN (MUST) CSC / 32
20 Octal and Hexadecimal Numbers Conversion from binary to hexadecimal is similar except that the bits are divided into groups of four. The corresponding hexadecimal digit for each group of four bits is written as shown: A {}}{ F {}}{ {}}{ {}}{ MAN (MUST) CSC / 32
21 Octal and Hexadecimal Numbers The corresponding octal digit for each group of three bits is easily remembered after studying the first eight entries in the table. Octal Binary-coded Decimal number octal equivalent MAN (MUST) CSC / 32
22 Octal and Hexadecimal Numbers The correspondence between a hexadecimal digit and its equivalent 4-bit code can be found in the table. Hexadecimal Binary-coded Decimal number hexadecimal equivalent A B C D E F MAN (MUST) CSC / 32
23 Octal and Hexadecimal Numbers The registers in a digital computer contain many bits. Specifying the content of registers by their binary values will require a long string of binary digits. It is more convenient to specify content of registers by their octal or hexadecimal equivalent. The number of digits is reduced by one-third in the octal designation and by one-fourth in the hexadecimal equivalent. For example, the binary number has 12 digits. It can be expressed in octals of 7777 (four digits) or in hexadecimal as FFF (three digits). MAN (MUST) CSC / 32
24 Decimal Representation The binary number system is the most natural system for a computer. However, people are accustomed to the decimal system. One way of solving this conflict is to convert all input decimal numbers into binary numbers and let the computer perform all arithmetic operations in binary and then convert the binary results back to decimal for the human to understand. It is also possible for the computer to perform arithmetic operations directly with decimal numbers provided they are placed in their registers in a coded form. The bit assignment most commonly used for the decimal digits is the referred to as Binary-coded Decimal (BCD). MAN (MUST) CSC / 32
25 Binary-coded Decimal Decimal Binary-coded decimal number (BCD) number MAN (MUST) CSC / 32
26 Binary-coded Decimal The advantage of BCD format is that it is closer to the alphanumeric codes used for I/Os. Numbers in text data formats must be converted from text form to binary form. This conversion is usually done by converting text input data to BCD, converting BCD to binary, do calculations then convert the result to BCD, then BCD to text output. MAN (MUST) CSC / 32
27 Binary-coded Decimal For example, the number 99 when converted to binary is represented by a string of bits However, when 99 is represented as a binary-coded decimal, it is shown as: The difference between a decimal number represented by familiar digit symbols 0, 1, 2,..., 9 and the BCD symbols 0001, 0010,..., 1001 is in the symbols used to represent the digits. The number itself is exactly the same. MAN (MUST) CSC / 32
28 Alphanumeric Representation Many applications of digital computers require the handling of data that consists not only of numbers, but also of letters of the alphabet and certain special symbols. An alphanumeric character set is a set of elements that includes the 10 decimal digits, the 26 letters of the alphabet and a number of special characters, such as +, and =. The two most prominent alphanumeric codes are: - EBCDIC (Extended Binary Coded Decimal Interchange Code); this is mostly used by IBM. - ASCII: (American Standard Code for Information Interchange); used by other manufacturers MAN (MUST) CSC / 32
29 Alphanumeric Representation ASCII represents each character with a 7 bit string. The total number of characters that can be represented is 2 7 = 128. For example, J O H N = Since most computers manipulate an 8 bit quantity, the extra bit when 7 bit ASCII is used depends on the designer. It can be set to a particular value or ignored. Binary codes can be formulated for any set of discrete elements such as the musical notes and chess pieces and their positions on a chess board. Binary codes are also used to formulate instructions that specify control information for a computer. MAN (MUST) CSC / 32
30 Exercise One Convert the following decimal numbers into binary Convert the following binary numbers into decimal MAN (MUST) CSC / 32
31 Exercise Two Convert the following hexadecimal numbers into decimal. 1 F 4 2 6E 3 B7 4 ABC 5 6E 6 FEC Convert the following octal numbers into decimal MAN (MUST) CSC / 32
32 Summary 1 Data Representation 2 Bases: Binary, Decimal, Octal and Hexadecimal 3 Conversion between bases - Conversion from Binary to Decimal - Conversion from Decimal to Decimal - Conversion from octal to Decimal - Conversion from hexadecimal and Decimal MAN (MUST) CSC / 32
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### We can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b
In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should | 8,871 | 37,409 | {"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} | 3.546875 | 4 | CC-MAIN-2018-26 | longest | en | 0.786007 |
https://pballew.blogspot.com/2023/09/on-this-day-in-math-september-9.html | 1,719,213,114,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00728.warc.gz | 394,151,447 | 29,189 | ## Saturday 9 September 2023
### On This Day in Math - September 9
There are in this world optimists who feel that any symbol that starts off with an integral sign must necessarily denote something that will have every property that they should like an integral to possess. This of course is quite annoying to us rigorous mathematicians; what is even more annoying is that by doing so they often come up with the right answer.
~McShane, E. J.
The 252nd day of the year; 252 is the smallest number which is the product of two distinct numbers that are reverses of each other: 252 = 12*21 *Number Gossip
(and the next would be?)
252 is a palindrome in base ten, and also in base five 20025 (How many three digit numbers are (non-trivial) palindromes in base ten and one other base less than ten)
If you flip a coin 10 times in a row, there are exactly 252 ways in which it can turn out that you get exactly 5 heads and 5 tails. That is, $252 = \binom{10}{5} = \frac{10*9*8*7*6}{5*4*3*2*1}$
EVENTS
1508 What was probably the first Dutch language arithmetic was published in Brussels by Thomas van der Noot. The Way to Learn to Reckon According to the Right Art of Algorismi, in whole and broken numbers. In 1537 much of the book would be copied into English to become the earliest known arithmetic in that language, An Introduction for to learne to reckon with the pen or with the counters. John Herford published the book at his presses in St. Albens. After introducing the terms numerator and denominator, he reverts to the Dutch terms "teller" (counter) and and used number in place of "noemer" (noemer rekenkunde is the more complete term, I think. any help?) ; both terms of which are/were, at the least, unusual in English. *P. Bockstaele, Notes on the First Arithmetics Printed in Dutch and English, Isis 51(1960) pg 315
Van der Noot is also credited with writing and publishing the first cookbook in Dutch.
At Thrift Books website I found an add for a copy of Herford's arithmetic with more detail about the history of the first English arithmetic:
"This book is a facsimile of the second edition of the earliest printed work in English entirely devoted to Arithmetic. The author is unknown, although the book comprises a compilation of material from two other works, one Dutch and one French, translated into English and with the addition of some new material. It was imprinted in Aldersgate Street, London by Nicolas Bourman in 1539. The first edition was produced two years earlier by John Herford, whose press was located in the abbey of St Albans and who printed under the patronage of the abbot, Richard Boreman. It was long thought that no copies of the first edition were extant, with the exception of a small fragment in the British Library. However, in 2005 a complete text turned up at a Sotheby's auction in New Bond Street, London. This rare survivor sold under the hammer to the British Library for the staggering sum of 97,500. Among the problems posed in An Introduction is the "rule and question of a catte". This concerns a cat which climbs a 300 foot high tree, ascending 17 feet each day but descending again 12 feet each night. The problem to be solved is - how long does the cat take to reach the top? The answer given is 60 days, which of course is quite wrong. Another problem concerns "The rule and question of zaracins, for to cast them within the see". Given that on a sinking ship there are thirty merchants, 15 of whom are Christian and the other 15 Saracens, half of whom must be thrown overboard to save the ship, how should they be ordered so that counting off by nines will always result in a Saracen being sacrificed and never a Christian? Not a problem that fits easily with current ideas about political correctness. Then there is "a dronkart who drynketh a barell of bere in 14 days", but "when his wife drinketh with him" they empty it in 10 days How quickly, the reader is asked, could his wife drink it alone? These are just a few of the beguiling puzzles set within the pages of this fascinating book. "
*Flying Post (London, England), September 9, 1699 - September 12, 1699; Issue 677.
1713 The St. Petersburg Paradox is born?: Nicolas Bernoulli to Montmort. Basel, 9 September, 1713.
Printed in Essay d'Analysis, p. 402
THE FOURTH PROBLEM SAID: A promises to give an écu to B, if with an ordinary die he achieves 6 points on the first throw, two écus if he achieves 6 on the second throw, 3 écus if he achieves this point on the third throw, 4 écus if he achieves it on the fourth and thus consecutively; one asks what is the expectation of B? Fifth Problem. One asks the same thing if A promises to B to give him some écus in this progression 1, 2, 4, 8, 16 etc. or 1, 3, 9, 27 etc. or 1, 4, 9, 16, 25 etc. or 1, 8, 27, 64 instead of 1, 2, 3, 4, 5 etc. as beforehand. Although for the most part these problems are not difficult, you will find however something most curious.
*VFR
The St. Petersburg Paradox is based on a simple coin flip game with an infinite expected winnings. The paradox arises by the fact that no rational human would risk a large finite amount to play the game, even though the expected value implies that a rational person should risk any finite amount to play it. Here I describe the St. Petersburg Paradox and give some proposed methods of resolving it.
1751 Euler presents his famous “Gem”; Vertices + Faces -2 = Edges in two papers Euler presented several results relating the number of plane angles of a solid to the number of faces, edges, and vertices (he referred to “solid angles”). Euler also classified polyhedra by the number of solid angles they had. According to C. G. J. Jacobi, a treatise with this title was read to the Berlin Academy on November 26, 1750. The proofs were contained in a second paper. According to C. G. J. Jacobi, it might have been read to the Berlin Academy on September 9, 1751. According to the records there, it was presented to the St. Petersburg Academy on April 6, 1752.*VFR
I highly recommend Dave Richeson's wonderful Euler's Gem for any teacher or student.
In 1839, John Herschel made the first (remaining) photograph on glass plate. The image he captured was of the 40-foot, 48" aperture telescope used by his father William Herschel, in Slough, England. It had sat decades without use, and was dismantled a short time later. Twenty years earlier, in 1819, Herschel published the results of his experiments with silver and salts. A chance comment about Daguerre's experiments in a letter to his wife on 22 Jan 1839 from a friend prompted John into new activity, and within a few days, he had prepared some photographs. In papers he published in 1840-42, he coined the new words "emulsion", "positive" and "negative" and reinforced Stenger's earlier introduction of the word "photography."*TIS
1892 Jupiter's moon Amalthea was discovered by E.E. Barnard. It was the last moon discovered by observation. *David Dickinson @Astroguyz
He found it using the 36 inch (91 cm) refractor telescope at Lick Observatory. It was the first new satellite of Jupiter since Galileo Galilei's discovery of the Galilean satellites in 1610 *Wik
1945 First computer “bug” logged at 1545 hrs. Grace Hopper was running the computer at the time and there was a failure. When she investigated she found that a moth had gotten into the machinery and caused the problem. She removed it and taped it into the log book of the computer. Thus a bit of computer jargon was born. *VFR (the term "bug" in the meaning of technical error dates back at least to 1878 and Thomas Edison , and "debugging" seems to have been used as a term in aeronautics before entering the world of computers. Indeed, in an interview Grace Hopper remarked that she was not coining the term. The moth fit the already existing terminology, so it was saved.)
1967 The Soviet Union issued a postage stamp showing checker players with part of a board in the background. Although more than 50 stamps have been issued on chess, this was the first on checkers. [Journal of Recreational Mathematics, 2(1969), 50] *VFR That same year, Andris Andreiko challenged world champion Iser Koeperman for the checkers World Champion Title but Koeperman successfully retained his title. As a result of this competition between the two Soviet contenders, the Russian government issued postage stamps as commemorative of the World Title match.*CheckersChest.com
1974 "1974 Albert Ghiorso & Glenn T. Seaborg announced the discovery of seaborgium" * chemheritage@ChemHeritage Seaborgium is the first element to be named after a person who was alive when the name was announced. The second element to be so named is oganesson, in 2016, after Yuri Oganessian.
He influenced the naming of so many elements that with the announcement of seaborgium, it was noted in Discover magazine's review of the year in science that he could receive a letter addressed in chemical elements: seaborgium, lawrencium (for the Lawrence Berkeley Laboratory where he worked), berkelium, californium, americium. *Wik
Seaborg in 1964, *Wik
In 2000, the hole in the ozone layer over Antarctica stretched over a populated city for the first time, after ballooning to a new record size. For two days, Sept. 9-10, the hole extended over the southern Chile city of Punta Arenas, exposing residents to very high levels of ultra violet radiation. Too much UV radiation can cause skin cancer and destroy tiny plants at the beginning of the food chain. Previously, the hole had only opened over Antarctica and the surrounding ocean. Data from the U.S. space agency NASA showed the hole covered 11.4 million square miles - an area more than three times the size of the United States. *TIS
Image: The top image shows the average total column ozone values over Antarctica for September 2000. In October, however (bottom image), the hole shrank dramatically, much more quickly than usual. By the end of October, the hole was only one-third of it’s previous size.
BIRTHS
1737 "Luigi Galvani, (9 September 1737 – 4 December 1798)) was an Italian physician. In the 1770s, Galvani, a professor at Bologna, began experimenting with electricity, searching in particular for electrical effects in living tissue. For a tissue source, he settled on frogs, or portions of frogs, and Galvani found that when an electrical generator was operating nearby, or there was an electrical storm, frog torsos with metallic hooks inserted into their spines would twitch. The big surprise came when he inserted a brass hook into a frog trunk and then laid the frog on an iron grate. As soon as the brass touched the iron, the legs contracted violently, and continued to do so whenever the brass hook touched the iron grate. Galvani had, unknowingly, invented a battery, which is nothing more than two different metals separated by a liquid electrolyte (like frog flesh). But Galvani didn't see it that way--he thought he had discovered a new kind of electricity, animal electricity, that was produced only by living things. It was Alessandro Volta who figured out what was going on with the frogs, and who in 1800 made a battery with copper and zinc and salt water--and no frog." *Linda Hall org
*Wik
1789 William Cranch Bond (9 Sep 1789; 29 Jan 1859)American astronomer who, with his son, George Phillips Bond (1825-65), discovered Hyperion, the eighth satellite of Saturn, and an inner ring called Ring C, or the Crepe Ring. While W.C. Bond was a young clockmaker in Boston, he spent his free time in the amateur observatory he built in part of his home. In 1815 he was sent by Harvard College to Europe to visit existing observatories and gather data preliminary to the building of an observatory at Harvard. In 1839 the observatory was founded. He supervised its construction, then became its first director. Together with his son he developed the chronograph for automatically recording the position of stars. They also took some of the first recognizable photographs of celestial objects.*TIS
Hyperion is distinguished by its irregular shape, its chaotic rotation, and its unexplained sponge-like appearance. It was the first non-round moon to be discovered.
1852 John Henry Poynting (9 Sep 1852; 30 Mar 1914)British physicist who introduced a theorem (1884-85) that assigns a value to the rate of flow of electromagnetic energy known as the Poynting vector, introduced in his paper On the Transfer of Energy in the Electromagnetic Field (1884). In this he showed that the flow of energy at a point can be expressed by a simple formula in terms of the electric and magnetic forces at that point. He determined the mean density of the Earth (1891) and made a determination of the gravitational constant (1893) using accurate torsion balances. He was also the first to suggest, in 1903, the existence of the effect of radiation from the Sun that causes smaller particles in orbit about the Sun to spiral close and eventually plunge in.*TIS
1860 Frank Morley (September 9, 1860 – October 17, 1937) wrote mainly on geometry but also on algebra.*SAU Morley is remembered most today for a singular theorem which bears his name in recreational
literature. Simply stated, Morley's Theorem says that if the angles at the vertices of any triangle (A, B, and C in the figure) are trisected, then the points where the trisectors from adjacent vertices intersect (D, E, and F) will form an equilateral triangle.
In 1899 he observed the relationship described above, but could find no proof. It spread from discussions with his friends to become an item of mathematical gossip. Finally in 1909 a trigonometric solution was discovered by M. Satyanarayana. Later an elementary proof was developed. Today the preferred proof is to begin with the result and work backward. Start with an equilateral triangle and show that the vertices are the intersection of the trisectors of a triangle with any given set of angles. For those interested in seeing the proof, check Coxeter's Introduction to Geometry, Vol 2, pages 24-25. Find more about this unusual man here.
1908 Mary Golda Ross (August 9, 1908 – April 29, 2008) was the first known Native American female engineer, and the first female engineer in the history of Lockheed. She was one of the 40 founding engineers of the renowned and highly secretive Skunk Works project at Lockheed Corporation. She worked at Lockheed from 1942 until her retirement in 1973, where she was best remembered for her work on aerospace design – including the Agena Rocket program – as well as numerous "design concepts for interplanetary space travel, crewed and uncrewed Earth-orbiting flights, the earliest studies of orbiting satellites for both defense and civilian purposes." In 2018, she was chosen to be depicted on the 2019 Native American $1 Coin by the U.S. Mint celebrating American Indians in the space program. *Wik 1910 Bjorn Kjellstrom (9 Sep 1910; 2 Sep 1995) Swedish inventor of the Silva compass which featured a rotating compass dial, and a transparent protractor base plate. As founder of Silva, Inc. in North America, Kjellstrom helped introduce the orienteering sport to the U.S. in the 1940s, in part as a way to promote his product. He wrote "Be Expert with Map and Compass", considered to be the "bible of orienteering." *TIS 1914 Marjorie Lee Browne (September 9, 1914 – October 19, 1979) was a notable mathematics educator, the second African-American woman to receive a doctoral degree in the U.S., and one of the first black women to receive a doctorate in mathematics in the U.S. Browne's work on classical groups demonstrated simple proofs of important topological properties of and relations between classical groups. Her work in general focused on linear and matrix algebra. Browne saw the importance of computer science early on, writing a$60,000 grant to IBM to bring a computer to NCCU in 1960 -- one of the first computers in academic computing, and probably the first at a historically black school.
Throughout her career, Browne worked to help gifted mathematics students, educating them and offering them financial support to pursue higher education. Notable students included Joseph Battle, William Fletcher, Asamoah Nkwanta, and Nathan Simms. She established summer institutes to provide continuing education in mathematics for high school teachers.*Wik
Dennis MacAlistair Ritchie (September 9, 1941; found dead October 12, 2011), was an American computer scientist who "helped shape the digital era." He created the C programming language and, with long-time colleague Ken Thompson, the UNIX operating system. Ritchie and Thompson received the Turing Award from the ACM in 1983, the Hamming Medal from the IEEE in 1990 and the National Medal of Technology from President Clinton in 1999. Ritchie was the head of Lucent Technologies System Software Research Department when he retired in 2007. He was the 'R' in K&R C and commonly known by his username dmr. *Wik
DEATHS
1883 Victor Alexandre Puiseux
(16 April 1820, 9 September 1883) was a French mathematician and astronomer. Puiseux series are named after him, as is in part the Bertrand–Diquet–Puiseux theorem. Excelling in mathematical analysis, he introduced new methods in his account of algebraic functions, and by his contributions to celestial mechanics advanced knowledge in that direction. In 1871, he was unanimously elected to the French Academy.*Wik He worked on elliptic functions and studied computational methods in astronomy.*SAU
1885 Claude Bouquet (7 Sept 1819 in Morteau, Doubs, France - 9 Sept 1885 in Paris, France) was a French mathematician who worked on differential geometry and on series expansions of functions and elliptic functions.*SAU
1973 Giovanni Ricci (17 Aug 1904 , 9 Sept 1973) Italian mathematician who is esteemed as an outstanding teacher. His major mathematical input was on distribution of primes.
2000 Herbert Friedman (21 Jun 1916, 9 Sep 2000)American astronomer who made made seminal contributions to the study of solar radiation. He joined the Naval Research Laboratory in 1940 and developed defense-related radiation detection devices during WW II. In 1949, he obtained the first scientific proof that X rays emanate from the sun. When he directed the firing into space of a V-2 rocket carrying a detecting instrument. Through rocket astronomy, he also produced the first ultraviolet map of celestial bodies, and gathered information for the theory that stars are being continuously formed, on space radiation affecting Earth and on the nature of gases in space. He also made fundamental advances in the application of x rays to material analysis. *TIS
2003 Edward Teller(15 Jan 1908, 9 Sep 2003) Hungarian-born American nuclear physicist who participated in the production of the first atomic bomb (1945) and who led the development of the world's first thermonuclear weapon, the hydrogen bomb. After studying in Germany he left in 1933, going first to London and then to Washington, DC. He worked on the first atomic reactor, and later working on the first fission bombs during WW II at Los Alamos. Subsequently, he made a significant contribution to the development of the fusion bomb. His work led to the detonation of the first hydrogen bomb (1952). He is sometimes known as "the father of the H-bomb." Teller's unfavorable evidence in the Robert Oppenheimer security-clearance hearing lost him some respect amongst scientists*TIS
Credits :
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts
*NSEC= NASA Solar Eclipse Calendar
*RMAT= The Renaissance Mathematicus, Thony Christie
*SAU=St Andrews Univ. Math History
*TIA = Today in Astronomy
*TIS= Today in Science History
*VFR = V Frederick Rickey, USMA
*Wik = Wikipedia
*WM = Women of Mathematics, Grinstein & Campbell | 4,651 | 19,765 | {"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} | 3.5 | 4 | CC-MAIN-2024-26 | latest | en | 0.953214 |
https://testbook.com/objective-questions/mcq-on-geometry--5eea6a1039140f30f369e7e9 | 1,709,419,502,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476137.72/warc/CC-MAIN-20240302215752-20240303005752-00411.warc.gz | 554,905,878 | 99,746 | # Geometry MCQ Quiz - Objective Question with Answer for Geometry - Download Free PDF
Last updated on Mar 2, 2024
Geometry MCQs are one of the most common questions featured in competitive exams such as SSC CGL, Bank PO, MTS etc. Testbook is known for the quality of resources we provide and candidates must practice the set of Geometry Objective Questions and enhance their speed and accuracy. Geometry is the branch of mathematics concerned with the properties and relations of points, lines, surfaces, solids etc. With the Geometry Quizzes at Testbook, you will be able to understand and apply Geometry concepts. Get the option to ‘save’ Geometry MCQs to access them later with ease and convenience. For students to be truly prepared for the competitive exams that have a Geometry section in their syllabus, they must practice Geometry Questions Answers. We have also given some tips, tricks and shortcuts to solve Geometry easily and quickly. So solve Geometry with Testbook and ace the competitive examinations you’ve applied for!
## Latest Geometry MCQ Objective Questions
#### Geometry Question 1:
In a triangle ABC, side BC is produced to D such that ∠ACD = 127°. If ∠ABC = 35°, then find ∠BAC.
1. 82°
2. 92°
3. 95°
4. More than one of the above
5. None of the above
Option 2 : 92°
#### Geometry Question 1 Detailed Solution
Given:
In a triangle ABC, side BC is produced to D such that ∠ACD = 127°. If ∠ABC = 35°
Concept used:
External angle = sum of the opposite internal angles
Calculation:
As we know,
∠ACD = ∠ABC + ∠BAC
⇒ 127° = 35° ∠BAC
⇒ ∠BAC = 92°
∴ The correct option is 2
#### Geometry Question 2:
Two circles of radii 12 cm and 13 cm are concentric. The length of the chord of the larger circle which touches the smaller circle is:
1. 8 cm
2. 18 cm
3. 10 cm
4. More than one of the above
5. None of the above
Option 3 : 10 cm
#### Geometry Question 2 Detailed Solution
Given:
The radius of the circles are 12cm and 13 cm
Concept used:
Using Pythagoras theorem
Formula used:
Hypotaneus2 = Height2 + Base2
Calculation:
The length of the AP is
⇒ √(132 - 122
⇒ √25 = 5 cm
The length of the chord(AB) of the larger circle which touches the smaller circle is,
⇒ 5 × 2 = 10 cm
∴ The correct option is 3
#### Geometry Question 3:
If $$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$$, AB = 4 cm, PQ = 6 cm, QR = 9 cm and RP = 12 cm, then find the perimeter of $$\triangle \mathrm{ABC} \text {. }$$
1. 18 cm
2. 16 cm
3. 20 cm
4. More than one of the above
5. None of the above
Option 1 : 18 cm
#### Geometry Question 3 Detailed Solution
Given :
AB = 4 cm, PQ = 6 cm, QR = 9 cm and RP = 12 cm
Calculation :
In ΔPQR ratio of sides will be
⇒ 6 : 12 : 9 = 2 : 4 : 3
Since $$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$$
So, ratio of sides will be same.
⇒ 2 unit = 4
⇒ 1 unit = 2
⇒ 2 × 2 : 4 × 2 : 3 × 2
⇒ 4 : 8 : 6
Perimeter of ΔABC = (4 + 8 + 6) = 18
∴ The correct answer is 18 cm.
#### Geometry Question 4:
In given figure if p ∥ q then the value of x is
1. 55°
2. 95°
3. 65°
4. More than one of the above
5. None of the above
Option 1 : 55°
#### Geometry Question 4 Detailed Solution
Calculation:
From the figure:
∠pOr = ∠qMr = 95º;(corresponding angles)
∠MOs = ∠NOX = 30º;(alternate interior angles)
⇒ x = 180 - 95 - 30 = 55º
#### Geometry Question 5:
Let Δ ABC ∼ Δ QPR and $$\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{1}{16}$$. If AB = 12 cm, BC = 6 cm and AC = 9 cm, then QP is equal to:
1. 48 cm
2. 32 cm
3. 16 cm
4. More than one of the above
5. None of the above
Option 1 : 48 cm
#### Geometry Question 5 Detailed Solution
Concept:
Area theorem: It states that '' The ratios of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
i.e., $$\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=(\frac{AB}{PQ})^2 =(\frac{BC}{QR})^2=(\frac{AC}{PR})^2$$
From the figure, Δ ABC ∼ Δ QPR implies AB ∼ QP, BC ∼ QR, and CA ∼ RP where ∼ is read as '' corresponding to ''
Calculation:
Given Δ ABC ∼ Δ QPR and $$\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{1}{16}$$
Also AB = 12 cm
Consider, $$\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=(\frac{AB}{PQ})^2$$
⇒ $$\frac{1}{16}=(\frac{12}{PQ})^2$$
⇒ $$\frac{1}{4}=\frac{12}{PQ}$$
QP = 48 cm
Hence, the correct answer is option 1).
## Top Geometry MCQ Objective Questions
#### Geometry Question 6
The area of the triangle whose vertices are given by the coordinates (1, 2), (-4, -3) and (4, 1) is:
1. 7 sq. units
2. 20 sq. units
3. 10 sq. units
4. 14 sq. units
#### Answer (Detailed Solution Below)
Option 3 : 10 sq. units
#### Geometry Question 6 Detailed Solution
Given:-
Vertices of triangle = (1,2), (-4,-3), (4,1)
Formula Used:
Area of triangle = ½ [x(y- y3) + x(y- y1) + x(y- y2)]
whose vertices are (x1, y1), (x2, y2) and (x3, y3)
Calculation:
⇒ Area of triangle = (1/2) × [1(-3 – 1) + (-4) (1 – 2) + 4{2 – (-3)}]
= (1/2) × {(-4) + 4 + 20}
= 20/2
= 10 sq. units
#### Geometry Question 7
A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. then what is the length of RS ?
1. 7 cm
2. 15 cm
3. 9 cm
4. 7.3 cm
Option 3 : 9 cm
#### Geometry Question 7 Detailed Solution
Given :
A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm
Calculations :
If a circle touches all four sides of quadrilateral PQRS then,
PQ + RS = SP + RQ
So,
⇒ 11 + RS = 8 + 12
⇒ RS = 20 - 11
⇒ RS = 9
∴ The correct choice is option 3.
#### Geometry Question 8
AB and CD are two parallel chords of a circle of radius 13 cm such that AB = 10 cm and CD = 24cm. Find the distance between them(Both the chord are on the same side)
1. 9 cm
2. 11 cm
3. 7 cm
4. None of these
Option 3 : 7 cm
#### Geometry Question 8 Detailed Solution
Given
AB ∥ CD, and
AB = 10cm, CD = 24 cm
Radii OA and OC = 13 cm
Formula Used
Perpendicular from the centre to the chord, bisects the chord.
Pythagoras theorem.
Calculation
Draw OP perpendicular on AB and CD, and
AB ∥ CD, So, the points O, Q, P are collinear.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
AP = 1/2 AB = 1/2 × 10 = 5cm
CQ = 1/2 CD = 1/2 × 24 = 12 cm
Join OA and OC
Then, OA = OC = 13 cm
From the right ΔOPA, we have
OP2 = OA2 - AP2 [Pythagoras theorem]
⇒ OP2 = 132- 52
⇒ OP2 = 169 - 25 = 144
⇒ OP = 12cm
From the right ΔOQC, we have
OQ2 = OC2- CQ2 [Pythagoras theorem]
⇒ OQ2 = 13- 122
⇒ OQ2 = 169 - 144 = 25
⇒ OQ = 5
So, PQ = OP - OQ = 12 -5 = 7 cm
∴ The distance between the chord is of 7 cm.
#### Geometry Question 9
The complementary angle of supplementary angle of 130°
1. 50°
2. 30°
3. 40°
4. 70°
Option 3 : 40°
#### Geometry Question 9 Detailed Solution
Given:
One of the supplement angles is 130°.
Concept used:
For supplementary angle: The sum of two angles is 180°.
For complementary angle: The sum of two angles is 90°.
Calculation:
The supplement angle of 130° = 180° - 130° = 50°
The complement angle of 50° = 90° - 50° = 40°
∴ The complement angle of the supplement angle of 130° is 40°
Mistake PointsPlease note that first, we have to find the supplementary angle of 130° & after that, we will find the complementary angle of the resultant value.
#### Geometry Question 10
Two common tangents AC and BD touch two equal circles each of radius 7 cm, at points A, C, B and D, respectively, as shown in the figure. If the length of BD is 48 cm, what is the length of AC?
1. 50 cm
2. 40 cm
3. 48 cm
4. 30 cm
Option 1 : 50 cm
#### Geometry Question 10 Detailed Solution
Given:
Radius of each circle = 7 cm
BD = transverse common tangent between two circles = 48 cm
Concept used:
Length of direct transverse tangents = √(Square of the distance between the circle - Square of sum radius of the circles)
Length of the direct common tangents =(Square of the distance between the circle - Square of the difference between the radius of circles)
Calculation:
AC = Length of the direct common tangents
BD = Length of direct transverse tangents
Let, the distance between two circles = x cm
So, BD = √[x2 - (7 + 7)2]
⇒ 48 = √(x2 - 142)
⇒ 482x2 - 196 [Squaring on both sides]
⇒ 2304 = x2 - 196
⇒ x2 = 2304 + 196 = 2500
⇒ x = √2500 = 50 cm
Also, AC = √[502 - (7 - 7)2]
⇒ AC = √(2500 - 0) = √2500 = 50 cm
∴ The length of BD is 48 cm, length of AC is 50 cm
#### Geometry Question 11
Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40° . The measure of ∠ABP is:
1. 45°
2. 55°
3. 50°
4. 40°
Option 3 : 50°
#### Geometry Question 11 Detailed Solution
Given:
Two circles touch each other externally at P.
AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40°.
Concept used:
If two circles touch each other externally at some point and a direct common tangent is drawn to both circles, the angle subtended by the direct common tangent at the point where two circles touch each other is 90°.
Calculation:
According to the concept, ∠APB = 90°
Considering ΔAPB,
∠ABP
⇒ 90° - ∠PAB
⇒ 90° - 40° = 50°
∴ The measure of ∠ABP is 50°.
#### Geometry Question 12
Chord AB and diameter CD of a circle meet at the point P, outside the circle when it is produced, If PB = 8 cm, AB = 12 cm and distance of P from the centre of the circle is 18 cm, the radius (in cm) of the circle is closest to: [√41 = 6.4]
1. 12
2. 12.8
3. 12.4
4. 13
Option 2 : 12.8
#### Geometry Question 12 Detailed Solution
Concept:
Secant Property
If chord AB and chord CD of a circle intersect at a point P, then
PA × PB = PC × PD
Calculation:
Let the radius of the circle = r
PA = 12 + 8 = 20 cm, PB = 8 cm, PC = (18 + r) cm, PD = (18 - r) cm
PA × PB = PC × PD
⇒ 8 × 20 = (18 - r) × (18 + r)
⇒ 160 = 324 - r2
⇒ r2 = 164
⇒ r = 12.8062
∴ The radius of the circle is closest to 12.8 cm
#### Geometry Question 13
The sides of similar triangles ΔPQR and ΔDEF are in the ratio 5 ∶ 6. If area of ΔPQR is equal to 75 cm2, what is the area of ΔDEF?
1. 150 cm2
2. 90 cm2
3. 108 cm2
4. 120 cm2
#### Answer (Detailed Solution Below)
Option 3 : 108 cm2
#### Geometry Question 13 Detailed Solution
Given:
ΔPQR ∼ ΔDEF
The sides of ΔPQR and ΔDEF are in the ratio 5 ∶ 6.
ar(PQR) = 75 cm2
Concepts used:
The ratio of area of similar triangles is equal to the square of the ratio of sides of corresponding triangles.
Calculation:
ΔPQR ∼ ΔDEF
ar(PQR)/ar(DEF) = (Side of ΔPQR/Side of ΔDEF)2
⇒ 75/ar(DEF) = (5/6)2
⇒ ar(DEF) = 108 cm2
∴ Area of ΔDEF is equal to 108 cm2.
#### Geometry Question 14
Two circles touch each other externally at point X. PQ is a simple common tangent to both the circles touching the circles at point P and point Q. If the radii of the circles are R and r, then find PQ2.
1. 3πRr/2
2. 4Rr
3. 2πRr
4. 2Rr
Option 2 : 4Rr
#### Geometry Question 14 Detailed Solution
We know,
Length of direct common tangent = √[d2 - (R - r)2]
where d is the distance between the centers and R and r are the radii of the circles.
PQ = √[(R + r)2 - (R - r)2]
⇒ PQ = √[R2 + r2 + 2Rr - (R2 + r2 - 2Rr)]
⇒ PQ = √4Rr
⇒ PQ2 = 4Rr
#### Geometry Question 15
Sum of the measure of the interior angles of a polygon is 1620°. Find the number of sides of the polygon.
1. 14
2. 13
3. 12
4. 11
Option 4 : 11
#### Geometry Question 15 Detailed Solution
GIVEN :
Sum of the measure of the interior angles of a polygon is 1620°.
FORMULA USED :
Sum of interior angles of a polygon = (n – 2) × 180°
Where n is the number of sides.
CALCULATION :
Applying the formula :
1620° = (n – 2) × 180°
⇒ (n – 2) = 1620°/180°
⇒ (n – 2) = 9
⇒ n = 11
Hence,
Number of sides = 11 | 4,044 | 11,741 | {"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} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.882562 |
https://blog.csdn.net/Myriad_Dreamin/article/details/81122825 | 1,553,091,625,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202347.13/warc/CC-MAIN-20190320125919-20190320151919-00399.warc.gz | 447,475,779 | 77,874 | # 高等近世代数笔记(1) 置换与群
$\text{(Definition)}设\alpha \in {S}_{n},且\alpha =\prod _{i=1}^{t}{\beta }_{i}\left({\beta }_{i}\in {S}_{n}\right)是分解为不相交轮换的完全轮换分解:\phantom{\rule{0ex}{0ex}}定义sgn\left(\alpha \right)=\left(-1{\right)}^{n-t}.\phantom{\rule{0ex}{0ex}}$$\displaystyle\text{(Definition)}设\alpha \in S_n,且\alpha=\prod_{i=1}^{t}\beta_i(\beta_i\in S_n)是分解为不相交轮换的完全轮换分解:\\定义sgn(\alpha)=(-1)^{n-t}.\\$
$\text{(Theorem)}\mathrm{\forall }\alpha ,\beta \in {S}_{n},sgn\left(\alpha \beta \right)=sgn\left(\alpha \right)sgn\left(\beta \right).$$\displaystyle\text{(Theorem)}\forall \alpha,\beta\in S_n,sgn(\alpha\beta)=sgn(\alpha)sgn(\beta).$
$\text{(Theorem)}\alpha 是偶置换,iff.sgn\left(\alpha \right)=1,否则\alpha 是奇置换.$$\displaystyle\text{(Theorem)}\alpha是偶置换,iff.sgn(\alpha)=1,否则\alpha是奇置换.$
$\text{(Proposition 2.2)}\alpha \in {S}_{n},则sgn\left({\alpha }^{-1}\right)=sgn\left(\alpha \right)$$\displaystyle\text{(Proposition 2.2)}\alpha\in S_n,则sgn(\alpha^{-1})=sgn(\alpha)$
$\displaystyle\text{(Proposition 2.3)}\sigma\in S_n,定义\sigma’\in S_{n-1}:\ \sigma’(i)=\sigma(i)对于一切\sigma所不固定的i\in [1,n],证明sgn(\sigma’)=sgn(\sigma).$
$\text{(Proposition 2.5)}$$\displaystyle\text{(Proposition 2.5)}$
$\begin{array}{rl}\left(i\right)& \alpha 是r-轮换,证明{\alpha }^{r}=\left(1\right);\\ \left(ii\right)& 证明\left(i\right)中r是满足的最小正整数.\end{array}$\begin{aligned}(i)&\alpha是r-轮换,证明\alpha^r=\left(1\right);\\ (ii)&证明(i)中r是满足的最小正整数.\end{aligned}
$\left(i\right)$$(i)$$\alpha =\left({i}_{0}\dots {i}_{r-1}\right):$$\alpha=(i_0\dots i_{r-1}):$
$\text{basic step:}{\alpha }^{1}\left({i}_{\overline{k}}\right)={i}_{\overline{k+1}}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}r\right).$$\text{basic step:}\alpha^{1}(i_\overline{k})=i_{\overline{k+1}}\pmod r.$
$\text{recuisive step:}{\alpha }^{n}\left({i}_{\overline{k}}\right)={i}_{\overline{k+n}}\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}r\right).$$\text{recuisive step:}\alpha^{n}(i_\overline{k})=i_{\overline{k+n}}\pmod r.$
$\left(ii\right)$$(ii)$$\mathrm{\exists }k$\exists k,意味着存在更小的轮换分解,这时候$\alpha$$\alpha$不是r-轮换.
$\text{(definition)}群\left(G,\ast \right)满足:\left(\ast 省略\right)$$\displaystyle\text{(definition)}群(G,*)满足:(*省略)$
$\begin{array}{rl}\left(i\right)& \mathrm{\forall }x,y,z\in G,x\left(yz\right)=\left(xy\right)z;\\ \left(ii\right)& \mathrm{\exists }e\in G,对\mathrm{\forall }g\in G,eg=g=ge;\\ \left(iii\right)& \mathrm{\forall }g\in G,\mathrm{\exists }{g}^{-1}\in G,s.t.g{g}^{-1}={g}^{-1}g=e.\end{array}$\begin{aligned}(i)&\forall x,y,z\in G,x(yz)=(xy)z;\\(ii)&\exists e\in G,对\forall g\in G,eg=g=ge;\\(iii)&\forall g\in G,\exists g^{-1}\in G,s.t.gg^{-1}=g^{-1}g=e.\end{aligned}
$\text{(Theorem)}若G是群,则G有以下定律成立:$$\displaystyle\text{(Theorem)}若G是群,则G有以下定律成立:$
$\begin{array}{rl}\left(i\right)& 消去律成立;\\ \left(ii\right)& 幺元唯一;\\ \left(iii\right)& \mathrm{\forall }g\in G逆元唯一;\\ \left(iv\right)& 广义结合律成立;\\ \left(v\right)& \left(\prod _{k=1}^{n}{g}_{k}{\right)}^{-1}=\prod _{k=n}^{1}{g}_{k}^{-1};\end{array}$\begin{aligned}(i)&消去律成立;\\(ii)&幺元唯一;\\(iii)&\forall g\in G逆元唯一;\\(iv)&广义结合律成立;\\(v)&(\prod_{k=1}^{n} g_{k})^{-1}=\prod_{k=n}^{1} g_{k}^{-1};\end{aligned}
$\text{(Proposition 2.21)}若G是群,则满足{g}^{2}=g的唯一元素g\in G是1.$$\displaystyle\text{(Proposition 2.21)}若G是群,则满足g^2=g的唯一元素g\in G是1.$
$\text{(Proposition 2.22)}证明,若\left(G,\ast \right)满足：$$\displaystyle\text{(Proposition 2.22)}证明,若(G,*)满足：$
$\begin{array}{rl}\left(i\right)& \mathrm{\forall }g\in G,eg=g;\\ \left(ii\right)& \mathrm{\forall }g\in G,\mathrm{\exists }{g}^{-1}\in G,s.t.g{g}^{-1}=e.\end{array}\phantom{\rule{0ex}{0ex}}$\begin{aligned}(i)&\forall g\in G,eg=g;\\ (ii)&\forall g\in G,\exists g^{-1}\in G,s.t.gg^{-1}=e.\end{aligned}\\
$该定义与群的定义等价.$$该定义与群的定义等价.$
$\text{(Proposition 2.23)}证明若p是素数,且g\in 群G,o\left(g\right)=m,有m=pt,则o\left({g}^{t}\right)=p.$$\displaystyle\text{(Proposition 2.23)}证明若p是素数,且g\in 群G,o(g)=m,有m=pt,则o(g^t)=p.$
$\text{(Proposition 2.25)}已知A,B\in GL\left(2,\mathcal{Q}\right),设A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ -1& 1\end{array}\right],求o\left(A\right),o\left(B\right)及o\left(AB\right).$$\displaystyle\text{(Proposition 2.25)}已知A,B\in GL(2,\mathcal{Q}),设A=\left[\begin{matrix} 0 &-1\\1&0\end{matrix}\right],B=\left[\begin{matrix}0 &1\\-1&1\end{matrix}\right],求o(A),o(B)及o(AB).$
$\text{basic step:}\left(AB\right)=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]$$\text{basic step:}(AB)=\left[\begin{matrix}1 &-1\\0&1\end{matrix}\right]$.
$\text{recuisive step:}\left(AB{\right)}^{k}=\left[\begin{array}{cc}1& -k\\ 0& 1\end{array}\right]$$\text{recuisive step:}(AB)^k=\left[\begin{matrix}1 &-k\\0&1\end{matrix}\right]$.
$o\left(AB\right)=\mathrm{\infty }$$o(AB)=\infty$.
$\text{(Proposition 2.26)}证明若\mathrm{\forall }g\in G\left(o\left(g\right)=2\right),则群\left(G,\ast \right)是\text{abel}群.$$\displaystyle\text{(Proposition 2.26)}证明若\forall g\in G(o(g)=2),则群(G,*)是\text{abel}群.$
$\text{(Proposition 2.27)}证明若|G|\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right),则|\left\{g\in G|o\left(g\right)=2\right\}|\equiv 1\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)$$\displaystyle\text{(Proposition 2.27)}证明若|G|\equiv 0\pmod 2,则|\{g\in G|o(g)=2\}|\equiv 1\pmod 2$.
$\begin{array}{rl}|\left\{g\in G|o\left(g\right)=2\right\}|\equiv & |G|-|\left\{g\in G|o\left(g\right)>2\right\}|-|\left\{g\in G|o\left(g\right)=1\right\}|\\ \equiv & -1\equiv 1\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right)\end{array}$\begin{aligned}|\{g\in G|o(g)=2\}|\equiv&|G|-|\{g\in G|o(g)>2\}|-|\{g\in G|o(g)=1\}|\\\equiv&-1\equiv 1\pmod 2\end{aligned}
$\text{(Theorem)}H⩽G⇔\left(H\ne \varnothing \wedge \mathrm{\forall }x,y\in H\left(x{y}^{-1}\in H\right)\right)⇔\varnothing ⊊H\subset G且H封闭$$\displaystyle\text{(Theorem)}H\leqslant G\Leftrightarrow(H\neq \varnothing \wedge \forall x,y\in H(xy^{-1}\in H))\Leftrightarrow \varnothing \subsetneq H \subset G 且H封闭$.
$\text{(Theorem)}有限群G有|gen\left(G\right)|=\phi \left(|G|\right),意味着\mathrm{\forall }\gamma \in gen\left(G\right)⇔\left(|⟨\gamma ⟩|,|G|\right)=1$$\displaystyle\text{(Theorem)}有限群G有|gen(G)|=\varphi(|G|),意味着\forall \gamma\in gen(G)\Leftrightarrow(|\left<\gamma\right>|,|G|)=1$.
$\displaystyle\text{(Theorem)}G是群,\mathcal{H}(G)={H|H\leqslant G},则\bigcap_{H_x\in \mathcal{H}(G)}H_x\in\mathcal{H}(G)\ 特别地,若H,K\in \mathcal{H}(G),则H\cap K\in \mathcal{H}(G)$.
$\displaystyle\text{(Theorem)}G是群,如果X\subset G,则存在X\subset\leqslant \forall H\leqslant G.\ 特别地,若X=\varnothing,有<\varnothing>={e}(由群定义至少包含e,并非由任何x\in X生成).$
$\text{(Theorem)}G是群,a\mathcal{H}\left(G\right)=\left\{aH|{a}_{i}H\cap {a}_{j}H=\varnothing \right\},则G=\bigcup _{{a}_{x}H\in a\mathcal{H}\left(G\right)}{a}_{x}H$$\displaystyle\text{(Theorem)}G是群,a\mathcal{H}(G)=\{aH|a_iH\cap a_jH=\varnothing\},则G=\bigcup_{a_xH\in a\mathcal{H}(G)}a_xH$.
$\text{(Lagrange's Theorem)}G是群,则|G|=\left[G:H\right]|H|⇔|H|||G|$$\displaystyle\text{(Lagrange's Theorem)}G是群,则|G|=[G:H]|H|\Leftrightarrow |H|\big||G|$.
$\text{(definition)}{\mathcal{Z}}_{m}=\mathcal{Z}/m\mathcal{Z},U\left({\mathcal{Z}}_{m}\right)=\left\{\overline{x}\in {\mathcal{Z}}_{m}|\left(x,m\right)=1\right\}$$\displaystyle\text{(definition)}\mathcal{Z}_m=\mathcal{Z}/m\mathcal{Z},U(\mathcal{Z}_m)=\{\overline{x}\in \mathcal{Z}_m|(x,m)=1\}$.
$\text{(Theorem)}U\left({\mathcal{Z}}_{m}\right)是乘法群且由此定义\phi \left(m\right)=|U\left({\mathcal{Z}}_{m}\right)|$$\displaystyle\text{(Theorem)}U(\mathcal{Z}_m)是乘法群且由此定义\varphi(m)=|U(\mathcal{Z}_m)|$.
$\text{(Euler's Theorem)}{a}^{\phi \left(m\right)}\equiv 1\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}m\right)$$\displaystyle\text{(Euler's Theorem)}a^{\varphi(m)}\equiv 1\pmod m$
$\text{(Fermat's Theorem)}特别地,p是素数则,{a}^{p}\equiv a\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right)$$\displaystyle\text{(Fermat's Theorem)}特别地,p是素数则,a^p\equiv a\pmod p$.
$①若a\in U\left({\mathcal{Z}}_{m}\right),则由{a}^{|U\left({\mathcal{Z}}_{m}\right)|}\equiv 1,得{a}^{|\phi \left(m\right)|}\equiv 1\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}m\right)$$\displaystyle①若a\in U(\mathcal{Z}_m),则由a^{|U(\mathcal{Z}_m)|}\equiv1,得a^{|\varphi(m)|}\equiv1\pmod m$;
$②若a\notin U\left({\mathcal{Z}}_{m}\right),则由a\equiv 0,得{a}^{p}\equiv {0}^{p}=0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}m\right),此时\left(a,m\right)\ne 1$$\displaystyle②若a\notin U(\mathcal{Z}_m),则由a\equiv0,得a^p\equiv 0^p=0\pmod m,此时(a,m)\neq 1$.
$\text{(Willson's Theorem)}p是素数则,\left(p-1\right)!\equiv p-1\equiv -1\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right)$$\displaystyle\displaystyle\text{(Willson's Theorem)}p是素数则,(p-1)!\equiv p-1\equiv -1\pmod p$.
$另外,若m是合数,则\mathrm{\exists }a,b,s.t.m=ab\left(a\ne b\right),故\left(p-1\right)!\equiv ab\frac{\left(p-1\right)!}{ab}\equiv xm\equiv 0$$\displaystyle 另外,若m是合数,则\exists a,b,s.t. m=ab(a\neq b),故(p-1)!\equiv ab\frac{(p-1)!}{ab}\equiv xm\equiv 0$.
$或者a=b⇔\left({a}^{2}-1\right)!\equiv \left\{\begin{array}{rl}2& ,a=2\\ a{\cdot }_{m}2a{\cdot }_{m}\frac{\left({a}^{2}-1\right)!}{2{a}^{2}}\equiv 0& ,a>2\end{array}$\displaystyle 或者a=b\Leftrightarrow(a^2-1)!\equiv\left\{\begin{aligned}2 &,a=2\\a\cdot_m2a\cdot_m\frac{(a^2-1)!}{2a^2}\equiv0 &,a>2\end{aligned}\right.
$\text{(Proposition 2.29)}设H⩽G,证明：$$\displaystyle\text{(Proposition 2.29)}设H\leqslant G,证明：$
$\begin{array}{rl}\left(i\right)& Ha=Hb⇔a{b}^{-1}\in H;\\ \left(ii\right)& a\sim b:a{b}^{-1}\in H则\overline{a}=Ha.\end{array}$\begin{aligned}(i)&Ha=Hb\Leftrightarrow ab^{-1}\in H;\\(ii)&a\sim b:ab^{-1}\in H则\overline{a}=Ha.\end{aligned}
$\left(i\right)由陪集定义a=hb,且h,{b}^{-1}\in H,故a{b}^{-1}=h\in H$$(i)\displaystyle 由陪集定义a=hb,且h,b^{-1}\in H,故ab^{-1}=h\in H$.
$\left(ii\right)由a{b}^{-1}=h\in H,h取遍H这就得到\overline{a}=Hb,由\left(i\right)Ha=Hb,所以\overline{a}=Ha.$$(ii)\displaystyle由ab^{-1}=h\in H,h取遍H这就得到\overline{a}=Hb,由(i)Ha=Hb,所以\overline{a}=Ha.$
$\text{(Proposition 2.32)}设K⩽H⩽G,证明\left[G:K\right]=\left[G:H\right]\left[H:K\right]$$\displaystyle\text{(Proposition 2.32)}设K\leqslant H\leqslant G,证明[G:K]=[G:H][H:K]$
$\frac{\left[G:K\right]}{\left[G:H\right]\left[H:K\right]}=\frac{|G|/|K|}{|G|/|H|\cdot |H|/|K|}=1$$\displaystyle \frac{[G:K]}{[G:H][H:K]}=\frac{|G|/|K|}{|G|/|H|\cdot|H|/|K|}=1$
$\text{(Proposition 2.33)}设K,H⩽G,且\left(|H|,|K|\right)=1,证明H\cap K=\left\{e\right\}.$$\displaystyle\text{(Proposition 2.33)}设K,H\leqslant G,且(|H|,|K|)=1,证明H\cap K =\{e\}.$ | 4,990 | 10,493 | {"found_math": true, "script_math_tex": 81, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 78, "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-2019-13 | latest | en | 0.253801 |
https://mathstrek.blog/2012/04/17/thinking-infinitesimally-multivariate-calculus-ii/ | 1,685,779,664,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649177.24/warc/CC-MAIN-20230603064842-20230603094842-00227.warc.gz | 409,188,141 | 30,126 | ## Thinking Infinitesimally – Multivariate Calculus (II)
Chain Rule for Multivariate Calculus
We continue our discussion of multivariate calculus. The first item here is the analogue of Chain Rule for the multivariate case. Suppose we have parameters fu, v, x, y, z. Suppose {uv} are independent parameters (in particular, the system is at least 2-dimensional), and assume that (i) we can write $x = x(u,v)$, $y = y(u,v)$ and $z = z(u,v)$ as functions of {u, v}, (ii) we can write $f = f(x, y, z)$ as a function of x, y and z. This also means we can write f as a function of u and v. Upon perturbing the system, we get:
$\delta f \approx \delta x\left.\frac{\partial f}{\partial x}\right|_{y,z} + \delta y\left.\frac{\partial f}{\partial y}\right|_{x,z} + \delta z\left.\frac{\partial f}{\partial z}\right|_{x,y},$ and
$\delta f \approx \delta u\left.\frac{\partial f}{\partial u}\right|_v + \delta v\left.\frac{\partial f}{\partial v}\right|_u$.
We wish to find a formula which expresses the second set of partial derivatives in terms of the first. To do that, we divide the first equation by $\delta u$ and obtain:
$\frac{\delta f}{\delta u} \approx \frac{\delta x}{\delta u}\left(\left.\frac{\partial f}{\partial x}\right|_{y,z}\right) + \frac{\delta y}{\delta u}\left(\left.\frac{\partial f}{\partial y}\right|_{x,z}\right) + \frac{\delta z}{\delta u}\left(\left.\frac{\partial f}{\partial z}\right|_{x,y}\right).$
If we maintain $\delta v = 0$ and let $\delta u \to 0$, then the LHS converges to $\left.\frac{\partial f}{\partial u}\right|_v$ by definition. Taking the limit on the RHS also, we obtain:
$\left.\frac{\partial f}{\partial u}\right|_v = \left.\frac{\partial x}{\partial u}\right|_v\left.\frac{\partial f}{\partial x}\right|_{y,z} + \left.\frac{\partial y}{\partial u}\right|_v\left.\frac{\partial f}{\partial y}\right|_{x,z} + \left.\frac{\partial z}{\partial u}\right|_v\left.\frac{\partial f}{\partial z}\right|_{x,y}.$
This is usually written in books as the simplified form: $\frac{\partial f}{\partial u} = \frac{\partial x}{\partial u}\frac{\partial f}{\partial x}+\frac{\partial y}{\partial u}\frac{\partial f}{\partial y} + \frac{\partial z}{\partial u}\frac{\partial f}{\partial z}$, which is acceptable since the context is clear: the coordinate x is assumed to occur together with y and z, while u and v are always assumed to occur together. We left all the subscripts in our initial equation because we’re really trying to be careful here.
To remember the above formula, use the diagram:
Thus to compute $\frac{\partial f}{\partial u}$ we find all possible paths from f to u through the intermediate parameters {xyz} and take the sum of all terms, where each term is the product of the corresponding partial derivatives along the way.
Example 1. Suppose $f(x,y,z) = x^2 y^2 + yz^3 - xz$ and $x(u,v) = u^2 + v^2$, $y(u,v) = 2uv$, $z(u,v) = uv^3$. Then
$\frac{\partial f}{\partial x} = 2xy^2 - z,\ \frac{\partial f}{\partial y}=2x^2 y+z^3, \ \frac{\partial f}{\partial z} = 3yz^2 - x$.
Together with $\frac{\partial x}{\partial u} = 2u$, $\frac{\partial y}{\partial u} = 2v$ and $\frac{\partial z}{\partial u} = v^3$, we get the desired relation for $\frac{\partial f}{\partial u}$ which is convenient if we wish to calculate explicit values.
Example 2. Suppose we have polar coordinates $(x,y) = (r\cos \theta, r\sin \theta)$. Then for any f = f(x, y),
$\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r} = \cos\theta \frac{\partial f}{\partial x} + \sin\theta \frac{\partial f}{\partial y},$ – (1)
$\frac{\partial f}{\partial\theta} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta} = -r\sin\theta \frac{\partial f}{\partial x} + r\cos\theta \frac{\partial f}{\partial y}$ – (2)
But if we wish to express $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ in terms of $\frac{\partial f}{\partial r}, \frac{\partial f}{\partial\theta}$, then the equation $r\sin\theta \times (1) + \cos\theta\times (2)$ simplifies to: $r\sin\theta \frac{\partial f}{\partial r} + \cos\theta\frac{\partial f}{\partial\theta} = r\frac{\partial f}{\partial y}$. A similar computation gives us an expression for $\frac{\partial f}{\partial x}$. In short:
$\frac{\partial f}{\partial x} = \cos\theta \frac{\partial f}{\partial r}-\frac 1 r\sin\theta \frac{\partial f}{\partial\theta},\ \ \ \frac{\partial f}{\partial y} = \sin\theta \frac{\partial f}{\partial r} + \frac 1 r \cos\theta\frac{\partial f}{\partial\theta}.$
Higher Order Multivariate Derivatives
Recall that in the single-variable case, we can take successive derivatives of the function f(x) to obtain $\frac{df}{dx}$, $\frac{d^2 f}{dx^2}$, $\frac{d^3 f}{dx^3}$ etc. Let’s consider the multivariate case here.
Suppose {x, y, z, w} forms a set of coordinates. If we fix the values of yz, and w then we can differentiate a function $f(x,y,z,w)$ with respect to x as many times as we please. Thus we write this as:
$\left.\frac{\partial^n f}{\partial x^n}\right|_{y,z,w} := \frac{d^n f}{dx^n}$, keeping yzw constant.
For example, if $f(x,y,z,w) = x^3 y + xz^2 - w^4$, then $\left.\frac{\partial^2 f}{\partial x^2}\right|_{y,z,w} = 6xy$.
On the other hand, if we fix the values of z and w, then we can differentiate with respect to x first while keeping y constant, then with respect to y while keeping x constant. This is denoted by:
$\left.\frac{\partial^2 f}{\partial y \partial x}\right|_{z,w} := \left.\frac{\partial}{\partial y}\left(\left.\frac{\partial f}{\partial x} \right|_{y,z,w}\right)\right|_{x,z,w}.$
But one can also switch the order around: differentiate with respect to y first, then with respect to x. It turns out for the order doesn’t matter if the function is nice enough, i.e. we get:
$\left.\frac{\partial}{\partial y}\left(\left.\frac{\partial f}{\partial x}\right|_{y,z,w}\right)\right|_{x,z,w} = \left.\frac{\partial}{\partial x}\left(\left.\frac{\partial f}{\partial y}\right|_{x,z,w}\right)\right|_{y,z,w}.$
Here’s an intuitive (but non-rigourous) explanation of the reason. Since zw are fixed throughout, let’s simplify our notation by denoting $g(x,y) = f(x,y,z,w)$. Now consider a small perturbation $(x,y)\mapsto (x+\delta x, y+\delta y)$ and consider the following:
$\frac{g(x+\delta x, y+\delta y) - g(x, y+\delta y) - g(x+\delta x, y) + g(x,y)}{\delta x\cdot\delta y} = \frac{1}{\delta y}\left(\frac{g(x+\delta x, y+\delta y) - g(x, y+\delta y)}{\delta x} - \frac{g(x+\delta x, y) - g(x,y)}{\delta x}\right).$
If we let $\delta x\to 0$ with $\delta y$ constant, the two terms on the RHS converge to $\left.\frac{\partial g}{\partial x}\right|_y (x, y+\delta y)$ and $\left.\frac{\partial g}{\partial x}\right|_y (x, y)$ respectively. If we now let $\delta y \to 0$, the expression converges to $\left.\frac{\partial}{\partial y}\left(\left.\frac{\partial g}{\partial x}\right|_y\right)\right|_x$. By symmetry, the equation also converges to $\left.\frac{\partial}{\partial x}\left(\left.\frac{\partial g}{\partial y}\right|_x\right)\right|_y$ if we switch the order of convergence. Since it shouldn’t matter whether we let $\delta x\to 0$ first then $\delta y\to 0$ or vice versa, the two derivatives are equal.
[ Warning: pathological examples where the two derivatives differ do exist! Such functions are explicitly forbidden in our consideration. ]
Example 3. Consider $f(x,y,z) = x^3 y + 3xz^2 - xy^4 z$. Then the two derivatives are:
$\frac{\partial}{\partial x}\frac{\partial f}{\partial y} = \frac{\partial}{\partial x}(x^3 - 4xy^3 z) = 3x^2 - 4y^3 z$,
$\frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial}{\partial y}(3x^2 y + 3z^2 - y^4 z) = 3x^2 - 4y^3 z$.
Example 4. Consider rectilinear coordinates (xy) and polar coordinates (rθ), where the two are related via $(x, y) = (r\cos\theta, r\sin\theta)$. We already know from example 2 that:
$\frac{\partial f}{\partial x} = \cos\theta \frac{\partial f}{\partial r}-\frac 1 r\sin\theta \frac{\partial f}{\partial\theta},\ \ \ \frac{\partial f}{\partial y} = \sin\theta \frac{\partial f}{\partial r} + \frac 1 r \cos\theta\frac{\partial f}{\partial\theta}.$
Let’s see if we can express the second derivatives with respect to {xy} in terms of those with respect to {rθ}. It may look horrid, but the calculations can be simplified by thinking of $\frac \partial{\partial x}$ as an operator, i.e. a function which takes functions to other functions! Thus we shall write:
$\frac{\partial}{\partial x} = \cos\theta \frac{\partial}{\partial r} - \frac {\sin\theta}r \frac{\partial}{\partial\theta}.$
So to get the second derivative in terms of x, we just apply the operator to itself:
$\frac{\partial^2}{\partial x^2} = \left(\cos\theta \frac{\partial}{\partial r} - \frac {\sin\theta}r \frac{\partial}{\partial\theta}\right)\left( \cos\theta \frac{\partial}{\partial r} - \frac {\sin\theta}r \frac{\partial}{\partial\theta}\right).$
Since the operators are all additive (an operator D is said to be additive if D(fg) = DfDg for all functions f and g), we can use the distributive property to expand the RHS. Beware, though, that operators are in general not commutative; for example, by the product law we get:
$\frac{\partial}{\partial r}\left(\frac{\sin\theta}r \frac{\partial}{\partial\theta}\right) = \frac{\partial}{\partial r}\left(\frac{\sin\theta}r\right)\frac{\partial}{\partial\theta} + \frac{\sin\theta}{r} \frac{\partial^2}{\partial r\partial\theta} = -\frac{\sin\theta}{r^2}\frac{\partial}{\partial\theta} + \frac{\sin\theta}{r} \frac{\partial^2}{\partial r\partial \theta}.$
Now the reader has enough tools to verify the following:
$\frac{\partial^2}{\partial x^2} = \cos^2\theta \frac{\partial^2}{\partial r^2} + \frac{\sin^2\theta}{r^2} \frac{\partial^2}{\partial\theta^2} - \frac{\sin 2\theta}r \frac{\partial^2}{\partial r \partial\theta} + \frac {\sin 2\theta}{2r^2} \frac{\partial}{\partial \theta} + \frac{\sin^2\theta}{r} \frac{\partial}{\partial r}$,
$\frac{\partial^2}{\partial y^2} = \sin^2\theta \frac{\partial^2}{\partial r^2} + \frac{\cos^2\theta}{r^2} \frac{\partial^2}{\partial \theta^2} + \frac{\sin 2\theta}{r} \frac{\partial^2}{\partial r\partial\theta} - \frac{\sin 2\theta}{2r^2} \frac{\partial}{\partial \theta} + \frac{\cos^2\theta}{r} \frac{\partial}{\partial r}$.
The case of $\frac{\partial^2}{\partial x\partial y}$ is left as an exercise for the reader.
Exercises
All hints are ROT-13 encoded to avoid spoilers.
1. Obligatory mechanical exercises: in each of the following examples, verify that $\frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\frac{\partial f}{\partial y}$, $\frac{\partial}{\partial z}\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\frac{\partial f}{\partial z}$ etc, via explicit computations.
1. $f(x, y) = \sin(x^2 + y^3)\exp(xy)$.
2. $f(x,y,z) = \sin(xy) (x^3 z + y^2) \exp(z^3)$.
2. If ff(xy), and zxy, is there any relationship between $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$? [ Hint: Ner nyy guerr cnegvny qrevingvirf jryy-qrsvarq? ]
3. In 3-D space, we can define spherical coordinates $(r, \theta, \phi)$ which satisfies $x = r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$, $z = r\cos\theta$. For a function ff(x, yz), express the partial derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ in terms of spherical coordinates.
4. Prove that there does not exist f(xy) such that $\frac{\partial f}{\partial x} = x^2y + xy^2$ and $\frac{\partial f}{\partial y} = x^3 + x^2 y$.
5. Given $(x, y) = (u^2 - v^2, 2uv)$, for a function f(xy), express $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ in terms of u and v, and the partial derivatives of f with respect to u and v.
6. Find all points on the curve $x^3 + 2y^3 - 3xy = 1$ where the curve is tangent to a circle centred at the origin (see diagram below for a sample circle). You may use wolframalpha to numerically obtain the values.
7. (*) (Legendre transform) Suppose we have a 2-dimensional system with (non-independent) parameters uvw. Define the parameter $f = \left.\frac{\partial u}{\partial v}\right|_w$. Explicitly write down a new parameter g in terms of u, v, w, f such that $\left.\frac{\partial g}{\partial f}\right|_w = -v$. [ Hint: lbh pna pbzcyrgryl vtaber bar bs gur cnenzrgref. ]
8. (*) For a set of coordinates {xt} in the plane, the differential equation $\frac{\partial^2 f}{\partial x^2} = \frac{\partial^2 f}{\partial t^2}$ is called the 1-dimensional wave equation. Find all general solutions of this equation. [ Hint: fhofgvghgr gur gjb inevnoyrf ol gur fhz naq gur qvssrerapr. ] | 4,170 | 12,838 | {"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": 86, "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-2023-23 | latest | en | 0.711933 |
http://languageoption.com/chov-tv-oid/txny1.php?1cbd7f=area-of-parallelogram-vectors-calculator | 1,624,125,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649688.44/warc/CC-MAIN-20210619172612-20210619202612-00381.warc.gz | 27,091,156 | 7,585 | The vector product of a and b is always perpendicular to both a and b . Learn more ... Matrices Vectors. Find (a) a to the line 3) Express the complex conjugate in the form of of rciso z = 1+7i Hi 4) Find the roots of (213 – 21 ) This website uses cookies to ensure you get the best experience. It can be shown that the area of this parallelogram (which is the product of base and altitude) is equal to the length of the cross product of these two vectors. Area of a quadrilateral Area of a parallelogram given base and height. Press the button Parallelogram is given by two vectors. Calculate area of parallelogram needed to build a fractile pattern for wooden art project. By substituing $(1)$, this theorem can be written as We thus see that the length of the cross product of two vectors is equal to the product of the length of each vector This free online calculator help you to find area of parallelogram formed by vectors. Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. Manuel Sabino Crespo 902, Col. Luis Jiménez Figueroa Oaxaca, Oax. Calculate the resultant force vector using parallelogram law of forces. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how find area of parallelogram formed by vectors. (6) Find the area of the parallelogram bounded by the vectors axb. w 2) Given the equation of a line 4-y- 3-2 -4 point on the line (b) a vector that is parallel 3x+'= 3 2. We note that the area of a triangle defined by two vectors $\vec{u}, \vec{v} \in \mathbb{R}^3$ will be half of the area defined by the resulting parallelogram of those vectors. Question. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step This website uses cookies to ensure you get the best experience. area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. vector product of the vectors Consider that in a parallelogram, the magnitude of a vector P as 3N, another magnitude of vector Q as 4N and angle between two vectors is 30 degrees. In Geometry, a parallelogram is a two-dimensional figure with four sides. Find the area of the parallelogram whose two adjacent sides are determined by the vectors i vector + 2j vector + 3k vector and 3i vector − 2j vector + k vector. The sum of the interior angles in a quadrilateral is 360 degrees. Area = $$9 \times 6 = 54~\text{cm}^2$$ The formula for the area of a parallelogram can be used to find a missing length. Thus we can give the area of a triangle with the following formula: (5) Area determinants are quick and easy to solve if you know how to solve a 2x2 determinant. Terms & Conditions The contents of this website are issued in the United Kingdom by North of South Capital LLP (North of South) and are provided solely to give … the cr4oss product of your two vectors gives $$[10,-10,-10]$$ therefore the area is given Area of a parallelogram Suppose two vectors and in two dimensional space are given which do not lie on the same line. Parallelogram Area Definition. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. The matrix made from these two vectors has a determinant equal to the area of the parallelogram. Area of the parallelogram, build on vectors Calculator finds parallelogram area. When two vectors are given: Below are the expressions used to find the area of a triangle when two vectors are known. 1 Find the maximum area of a parallelogram given that you know the perimeter Area of a square. Area of a cyclic quadrilateral. … Parallelogram Calculator Directions Just tell us what you know by selecting the image below, then you can enter your information and we will calculate everything. This web site owner is mathematician Dovzhyk Mykhailo. The formula for the area of a parallelogram is base x height. Addition and subtraction of two vectors on plane, Exercises. As vectors in R 3, we … Area Ar of a parallelogram may be calculated using different formulas. These two vectors form two sides of a parallelogram. Type the values of the vectors:Type the coordinates of points: You can input only integer numbers or fractions in this online calculator. Dragon Ball Z Workout, Formula. Linear Algebra Example Problems - Area Of A Parallelogram By substituing $(1)$, this theorem can be written as We thus see that the length of the cross product of two vectors is equal to the product of the length of each vector and the sine of the angle between the two vectors. Free Parallelogram Area & Perimeter Calculator - calculate area & perimeter of a parallelogram step by step. Find the area of the parallelogram with vertices (4,1), (6, 6), (7, 7), and (9, 12). Area of a parallelogram given sides and angle. that is, the area of any convex quadrilateral. Magnitude of the Area of parallelogram formed by vectors, Online calculator. Guide - Area of triangle formed by vectors calculator To find area of triangle formed by vectors: Select how the triangle is defined; Type the data; Press the button "Find triangle area" and you will have a detailed step-by-step solution. A parallelogram in three dimensions is found using the cross product. Shane Casey Height, SecretarÃa de las Infraestructuras y el Ordenamiento Territorial Sustentable, SecretarÃa de las Culturas y Artes de Oaxaca, SecretarÃa de la ContralorÃa y Transparencia Gubernamental, SecretarÃa de Desarrollo Social y Humano, SecretarÃa de Desarrollo Agropecuario, Pesca y Acuacultura. The base is the length of the bottom side of the parallelogram. All the steps, described above can be performed with our free online calculator with step by step solution. « Formulario: “Participación en la Gestión Escolar para favorecer el aprendizaje de los alumnos”. I Got Mine, How To Rip Movies From Itunes Reddit, Real-world examples related to the area of a parallelogram; Once you get these concepts, you will then learn about the area of parallelogram using vectors, as well as learn about the area of a parallelogram without height. Figure with four sides website uses cookies to ensure you get the experience... 30 degrees step by step to see that u = á2,2 ñ and v = á 4 -1... 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Website, you agree to our Cookie Policy the matrix made from these two vectors a. Calculate area & Perimeter of a and b debug issue where LaTeX refuses to produce than. Be performed with our free online calculator two-dimensional plane you agree to our Cookie Policy triangle when two vectors given! Wooden art project and a height of 13 in determinants are quick and easy to solve if you how..., d₂ = 1i + 4j using vectors wooden art project ( norm ) of the angles. | 3,636 | 15,410 | {"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": 2, "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} | 3.6875 | 4 | CC-MAIN-2021-25 | latest | en | 0.887718 |
http://www.educationworld.com/a_lesson/lesson058.shtml | 1,477,669,787,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988722951.82/warc/CC-MAIN-20161020183842-00440-ip-10-171-6-4.ec2.internal.warc.gz | 415,201,998 | 40,182 | # Egg-Related Activities and Lessons for Easter and All Year
Why not plan a day or a week of activities around eggs? (Why not a month-long curriculum theme?) It'll be tons of fun!
Speaking of TONS... Did you know that more than 250,000 tons (500 million pounds!) of chocolate will be sold this Eastertime? That's more than 36 billion inch-long chocolate eggs! Use those figures to create a great math activity for students in grades 4 and up:
• If students stretched out all those chocolate Easter eggs end-to-end, how many times would the string of eggs circle the Earth at the Equator?
• How many times would that string of eggs stretch between New York and Los Angeles and back?
• How close to the moon would that chocolate "necklace" stretch?
(First, students must find the distances around Earth at its equator, from New York to LA, and from Earth to the moon. According to Fanny Farmer statistics, there'd be enough chocolate eggs to circle Earth twice! Enough chocolate eggs to stretch from New York to Los Angeles and back 11 times! Enough to reach a quarter of the way to the moon!)
Following is a basket full of additional cross-curriculum activities, some of which are Internet-based.
## THE CLASSIC "EGG-DROP" ACTIVITY
You've probably heard all about this classic egg activity -- or seen it in action! The idea behind the "Egg Drop" is to create a "package" that will protect a raw egg when it's dropped from a height of 8 feet (or whatever height you decide).
Invite students to bring in from home any materials that they might use in fashioning a protective cushion for their egg. Students can work individually or in pairs to create their egg containers. You'll be fascinated by the interesting contraptions your students come up with! Students will use everything from bubble wrap and foam peanuts to peanut butter! (Yep! I heard of a student who packed an egg in peanut butter. It survived the fall, but it broke apart when the student tried to pry it out of jar!) Some students might even attach parachutes to the packaging if you let them!
You might place restrictions on the size of a container.
Once constructed, students are ready to "drop" their eggs from the appointed height. (Want a real test? Drop them from a third-story window!) One helpful hint: Spread a plastic tarp over the spot where eggs will land to protect the floor or ground.
## EXTENSION ACTIVITIES
Hands-on science. Create a bouncing egg! using this science experiment. Experiment description: A chicken's egg is enclosed by a shell that has a high calcium content. If a raw egg (shell still intact) is placed in a glass of vinegar, a reaction takes place. The acetic acid in the vinegar will dissolve the eggshell and the egg will bounce. The reaction will begin immediately when the egg is placed in the vinegar but will not be complete for two or three days. After two or three days, the egg will survive a drop of four to five inches.
More hands-on science. (Middle and high school biology.) A similar experiment to the one above is used to demonstrate the properties of Diffusion and Semi-Permeable Membranes.
Language arts and art. Have fun with words that begin with the "ex" letter combination, which (when eggs-agerated) sounds like "eggs" in words such as eggs-cited, eggs-traordinary, and eggs-asperated. Provide students with large, egg-shaped sheets of white paper. Students use cut paper, crayons, and other art materials to create a character for one of their favorite ex words. The character's appearance should in some way "eggs-plain" the word. For example, the egg-shaped character named "eggs-hale" might be wearing a track uniform and be breathing heavily after a long race. The "eggs-pensive" character might be decked out in jewels and driving a Corvette convertible. "Eggs-plode" might… Well, let your students use their imaginations and get carried away!
Citizenship and language arts. Read aloud to students the Aesop fable The Goose With the Golden Eggs. (The whole fable is only a paragraph long.) Invite students to talk about the lesson this famous fable teaches. While you're at it, you might want to read more Aesop's fables from this site. You might even invite kids to write their own lesson fables.
More citizenship. (Middle school and above.) In this hands-on "social experiment," called "Egg Baby" Parenting, each student acts for a week as the "parent" of an egg. Students must carry the "egg baby" wherever they go. The purpose of this project is to allow students to experience some of the responsibilities that are involved in the care of human "babies." At the beginning of class each day there is an "egg check" worth 10 points. This project involves the students' parents.
Read aloud. Three books about eggs are among the best-known and -loved children's books. Choose one to read aloud, or read all three.
• Green Eggs and Ham by Dr. Seuss. Dr. Seuss turns 50 easy words into magic in this time-honored classic in which Sam-I-Am mounts a determined campaign to convince another Seuss character to eat a plate of green eggs and ham. (Be sure to check out teacher Paula White's Dr. Seuss Booktable for classroom follow-up activities for this book, Horton Hatches the Egg, Scrambled Eggs Super, and other Dr. Seuss books!)
• Rechenka's Eggs by Patricia Polacco. Babushka, known for her exquisite hand-painted eggs, finds Rechenka, a wounded goose, and takes her home. When she's ready to try her wings again, Rechenka accidentally breaks all of Babushka's lovingly crafted eggs. But the next morning Babushka awakens to a miraculous surprise.
• The Enormous Egg by Oliver Butterworth, illustrated by Louise Darling. Nate, a 12-year-old boy living in New Hampshire, takes over the care of an enormous egg laid by one of the family's hens, and the last thing he expects to hatch from it is a triceratops!
Cultural studies. Students can learn about the traditional Easter eggs of the Ukraine, called pysansky, on numerous Web sites. (Note: Rechenka's Eggs, one of the books in the previous activity is a perfect introduction to these spectacular eggs.) Students might check out Ukranian Easter Eggs, which includes the history of Ukrainian Easter eggs, details about how the eggs are made, and more.
Art. Students in grades 4 and up will enjoy creating Ukrainian Easter eggs using this decorating idea. Eggs, dyes and a white wax crayon are required materials. For younger students: Invite students to use a piece of wax to draw a picture or write a message on a sheet of drawing paper. Then students use watercolor paint to color the paper. Their pictures appear magically!
Cooking. Use some Student Egg Recipes to create dishes such as scrambled eggs, fried eggs, eggs-in-a-poke, and eggwich.
Math problems. (Intermediate grades.) Enjoy mixing eggs and math with the following problems from the Ask Dr. Math Web site.
• If a chicken and a half lays an egg and a half in a day and a half, how long does it take to get a dozen eggs? (See solution.)
• There are 10 children going on an Easter egg hunt. Every child finds 10 eggs. Suddenly a mean child steals one egg from each child. How many eggs are left? (See solution.)
• If an egg weighs 20 grams and half an egg, what does an egg and a half weigh? (See solution.)
More math problems. (For high schoolers.) This puzzle comes from The Grey Labyrinth puzzle site: One December morning after a particularly heavy snowstorm, the power fails. Fortunately, there is still an old wood stove with which you can prepare most of the professors' breakfasts. However one eccentric mathematics professor with a great deal of power and influence at the institute has a peculiar breakfast item which now poses a problem. He likes a single egg boiled for exactly nine minutes. You aren't wearing your watch, and all the clocks in the building are electric. You are able to find two exquisite hourglasses, able to precisely measure in hand-crafted swiss sand seven and four minutes respectively. How quickly using only these two hourglasses can you provide the professor with his egg? (See solution.)
More science. (Elementary grades.) Use the Egg Fun exercise from the Southeastern Michigan Math-Science Learning Coalition to learn about some of the properties of an egg. Among the principles learned:
• Eggs may look like they are the same size, but they aren't . We can measure the vertical and horizontal lines of eggs to see just how big they are.
• Eggshells are porous so that air can get in and out of the egg.
• Many objects roll. Eggs roll and wobble in such a way so that they don't travel very far and so they stay near the mother hen.
History. Collect news clipping about this year's White House egg roll. Invite students to investigate the History of the White House Egg Roll.
Letter writing. Invite students to write Easter letters to family members. Students can print out a sheet of Happy Easter stationery.
Article by Gary Hopkins
Education World® Editor-in-Chief
## EXCELLENT ARTICLES
"Eggs and Living Things: A Kindergarten Science Project" by Nanci Scali (Writing Notebook: Visions for Learning, Nov-Dec 1992). Describes a kindergarten science project that incorporates writing, mathematics, science, art, and technology as students investigate the question: What is the largest living thing to hatch out of an egg?
"Eggs Across the Curriculum" by John Collins (Mathematics in School, May 1990). Presented is a set of mathematics problems associated with characteristics, production, and consumption of chicken eggs. Student assignment sheets are included.
"Teaching Science" by Michael B. Leydon (Teaching PreK-8, April 1996). Introduces a science activity where students were asked to purchase four different sizes of eggs; weigh the content of each; and determine whether small, medium, large, or extra large eggs are the best buy. Describes the data analysis process. Also introduces various science activities that make use of the egg shell.
"Introducing Prealgebra Skills in an Eggs-citing Way" by Martha Kay Talbert and Virginia Stallings-Roberts (Mathematics Teaching in the Middle School, Nov-Dec 1994). Describes an activity designed to help students understand the concept of a variable and the process of gathering like terms or simplifying algebraic expressions using egg cartons and plastic eggs.
## THE BIGGEST INTERNET EGG SITES
Among the biggest Internet "egg" sites are these two created by egg marketing organizations:
American Egg Board
The AEB is the home of "the incredible edible egg" promotion. Here, egg lovers will find egg recipes, answers to FAQs (with Egg trivia), egg facts, egg nutrition information, and egg industry information. You might check out the site's Eggcyclopedia, a dictionary of egg-related terms.
Eggimals
Use eggs and other materials to create "eggimals" such as the eggosaurus, an eggapus, the eggopotomus, an eggalator, a beggver (beaver), a pegg (pig), and a pegguin! See complete directions on the page on the Canadian Egg Marketing Agency's Web site.
The Millikan L'Eggs Experiment
A high school physics lesson from SMILE (Science and Mathematics Initiative for Learning Enhancement).
## EXTINCT EGGS
Dinosaur Eggs
This site from National Geographic is divided into four sections. The "Hunt" documents scientists' search for dino eggs; "Hatch" lets you see how dino embryos are exposed by researchers; "Model" shows how the embryos might have looked; and "Museum" teaches more about dinosaur babies and parents.
The "Time Capsule Dinosaur Eggs" Project
The Time Capsule Dinosaur Eggs have become a major tourist attraction at the Hunterian Museum (University of Glasgow). These particular eggs appear to contain the remains of dinosaur embryos. State of the art technology has been used to examine the eggs without damaging them and the eggs have visited several Glasgow Hospitals to undergo special CT and MRI scanning treatments.
Last updated 03/30/2015 | 2,618 | 11,938 | {"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} | 3.515625 | 4 | CC-MAIN-2016-44 | longest | en | 0.949298 |
https://googology.wikia.org/wiki/Alphabet_notation?type=revision&diff=312922&oldid=312826 | 1,627,191,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00002.warc.gz | 301,670,249 | 36,631 | 10,964 Pages
Alphabet notation is a notation created by Wikia user Nirvana Supermind.[1][2][3]. It is a based on recursion, and inputs a string of English letters. It has only this part currently:
• Basic Alphabet Notation
The creator clarifies that he or she intends to create at least six other parts:[2]
• Nested Basic Cascading Alphabet Notation
• Tetrational Alphabet Notation
• Arrow Alphabet Notation
Note that the creator is using (single) capital letters to represent variables, and single small letters to mean the actual letters. For example, "A" in this article means a variable instead of the capital 'A', while "a" in this article means the actual letter 'a'.
## Basic Alphabet Notation
### Current definition
All information here only applies to the current version of the notation, which is given after the issues on the original definition were pointed out.
The expressions in this notation are of this form:
(ABCDEFGHIJKLMN…)
Here the “ABCDEFGHIJKLMN…” are a sequence of small letters in the Latin alphabet. The wrapping braces are simply to distinguish the expressions from actual words. () is also a valid expression. An example of a valid expression is (abc).
Terminology because they make the definition easier to write:
1. ord(A) for the letter A is defined as 1 if A = "a" 2 if A = "b" 3 if A = "c" 4 if A = "d" etc.
2. len(A) for the expression A is defined as is the number of letters in A.
3. P(A) for the integer A is defined as the Ath prime (zero-based index). So P(0) = 2.
Note that P(A) is ill-defined when A = -1. Every expression output a large number. To solve a (possibly empty) expression, we need some rules as follows:
1. () = 1
2. (#A) = (#)P(len(#))ord(A)
Here # denotes a substring of the current expression. It can also be empty. If there are multiple rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied.
(abc) = 2250
### Original definition
All information here only applies to the old version of the notation.
The expressions in this notation are of this form:
(ABCDEFGHIJKLMN…)
Here the “ABCDEFGHIJKLMN…” are a sequence of small letters in the Latin alphabet. The wrapping braces are simply to distinguish the expressions from actual words. () is also a valid expression. An example of a valid expression is (abc).
Terminology because they make the definition easier to write:
1. ord(A) for the letter A is defined as 1 if A = "a" 2 if A = "b" 3 if A = "c" 4 if A = "d" etc.
2. len(A) for the expression A is defined as is the number of letters in A.
3. P(A) for the integer A is defined as the Ath prime (zero-based index). So P(0) = 2.
Note that P(A) is ill-defined when A = -1. Every expression is intended to output a large number. To solve a (possibly empty) expression, we need some rules as follows:
1. () = 1
2. (#A) = (#)P(len(#)+1)ord(A)
Here # denotes a substring of the current expression. It can also be empty. If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied. Although it is not clarified, "A" in rule 2 is a variable which means a single small letter rather than a valid expression, because the creator considers ord(A). Readers should be careful that the creator uses "A" also as variables for a valid expression and an integer, as the defitions of len and P show.
#### Issues
The description "If there are two or more distinct rules to apply to a single expression, the uppermost-numbered rule which is applicable and whose result is a valid expression will be applied."[2] is weird, because there is no valid expression to which two rules are applicable. Moreover, there are no more rules, while the creator expresses "two or more rules".
Moreover, (abc) is intended to coincide with 2250, according to the creator.[2] However, the actual value should be computed in the following way:
(abc) = (ab)P(len(ab)+1)ord(c) = (ab)P(2+1)3 = (ab)73 = (a)P(len(a)+1)ord(b)73
= (a)P(1+1)273 = (a)5273 = ()P(len()+1)ord(a)5273 = ()P(0+1)15273
= ()315273 = 1×315273 = 25725
Therefore the original definition is not compatible with the intended behaviour.
However, the new definition has fixed some of these issues, and is compatible with the intended value.
(abc) = 2250
(abc) = 25725
### P進大好きbot's definition
User:P進大好きbot created a fixed definition of the notation, before the creator updated their definition. It is quite elementary to solve the issues of the old notation: We have only to remove the weird description of the application of two or more rules and change rule 2. In order to make the solution clearer, we explain the precise alternative formulation.
Let denote the set of non-negative integers, and the set of formal strings consisting of small letters in the Latin alphabet. For an , we denote by the length of . For an of length , we denote by the positive integer corresponding to the ordinal numeral of α with respect to the usual ordering of small letters in the Latin alphabet. For example, we have , , and . For an , we denote by the -th prime number. For example, we have , , and .
We define a total computable function \begin{eqnarray*} () \colon T & \to & \mathbb{N} \\ A & \mapsto & (A) \end{eqnarray*} in the following recursive way:
1. If , then set .
2. Suppose .
1. Denote by the formal string of length given as the rightmost letter of .
2. Denote by the formal string given by removing the right most letter from .
3. Set .
The totality follows from the induction on , and we have \begin{eqnarray*} (abc) & = & (ab)P(\textrm{len}(ab))^{\textrm{ord}(c)} = (ab)P(2)^3 = (ab)5^3 = (a)P(\textrm{len}(a))^{\textrm{ord}(b)}5^3 \\ & = & (a)P(1)^2 5^3 = (a)3^2 5^3 = ()P(\textrm{len}())^{\textrm{ord}(a)}3^2 5^3 = ()P(0)^1 3^2 5^3 \\ & = & ()2^1 3^2 5^3 = 1 \cdot 2^1 3^2 5^3 = 2250, \end{eqnarray*} which is compatible with the creator's intention.
Another possible alternative definition is given by replacing by the enumeration of prime numbers with respect to one-based indexing, i.e. , , and , instead of changing rule 2.
### Growth rate
It's hard to compare the asymptotic growth rate of this notation with the other notations like Fast-growing hierarchy, mainly because the limit function of them can be easily computed, while this one cannot as it inputs strings rather than numbers. The creator used the of Fibonacci word to convert numbers to strings. Using the highest possible letters, the Fibonacci word function S is given by the following recursive definition according to the creator:
S(0)='y',S(1)='z',S(n) = S(n-1)S(n-2)
For example, S(2)='zy'. The creator claims that the function (S(n)) has double-exponential growth rate, because they tested the ratios (S(n))/(S(n-1)) for values 1-14.[3]
An alternative limit function, also provided by User:P進大好きbot, is given by the map assigning to each the value of the formal string of length consisting of . It is quite elementary to show an upperbound, thanks to Bertrand's postulate.
Proposition
For any , holds with respect to the alternative definition.
#### Proof
We show the assertion by the induction on . If , then we have . Suppose . We have
\begin{eqnarray*} & & (A(n)) = (A(n-1))P(\textrm{len}(A(n-1)))^{\textrm{ord}(z)} = (A(n-1))P(n-1)^{26} \\ & \leq & (A(n-1)) (2^n)^{26} < ((A(n-1)) 10^{8n} < 10^{4(n-1)n} \times 10^{8n} \\ & = & 10^{4n(n+1)}. \end{eqnarray*}
Similarly, it is possible to obtain a lowerbound of the value for the formal string of length consisting of . In this sense, the asymptotic growth rate of the limit function can be easily estimated.
## Sources
1. Nirvana Supermind. Alphabet Notation. (Retrieved at UTC 12:00 13/01/2020)
2. Nirvana Supermind. Basic Alphabet Notation. (Retrieved at UTC 12:00 13/01/2020)
3. Nirvana Supermind. Basic Alphabet Notation. (Retrieved at Wed, 13 Jan 2021 22:44:39 GMT)
Community content is available under CC-BY-SA unless otherwise noted. | 2,173 | 8,019 | {"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-2021-31 | latest | en | 0.871122 |
https://www.jaszfenyszaru.hu/blog/magnitude-of-cylindrical-vector-14fc3c | 1,723,237,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00169.warc.gz | 658,475,851 | 6,850 | magnitude of cylindrical vector
Find the magnitude of $$\overrightarrow B$$.
The dot product is easy to compute when given the components, so we do so and solve for $$B_z$$: $0 = \overrightarrow A \cdot \overrightarrow B = \left( +5 \right) \left( +2 \right) + \left( -4 \right) \left( +3 \right) \ + \left( -1 \right) \left( B_z \right) \;\;\; \Rightarrow \;\;\; B_z = -2 \nonumber$. The origin could be the center of the ball or perhaps one of the ends. \end{aligned}\].
In this way, cylindrical coordinates provide a natural extension of polar coordinates to three dimensions. ˙ the right triangle in the $x$â$y$ plane with hypotenuse following angular velocity. When we convert to cylindrical coordinates, the $$z$$-coordinate does not change.
If we wish to obtain the generic form of velocity in cylindrical coordinates all we must do is differentiate equation 5 with respect to time, ... the Earth), and 2) the magnitude of the position vector changing in that rotating coordinate frame. So if we want to multiply the length of a vector by the amount of a second vector that is projected onto it we get: $( \text{projection of } \overrightarrow A \text{ onto } \overrightarrow B )( \text{magnitude of } \overrightarrow B ) = (A \cos \theta) (B) = AB \cos \theta$. Example $$\PageIndex{8}$$: Choosing the Best Coordinate System. This is exactly the same process that we followed in Introduction to Parametric Equations and Polar Coordinates to convert from polar coordinates to two-dimensional rectangular coordinates. Example $$\PageIndex{4}$$: Converting from Spherical Coordinates.
We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees. $$\overrightarrow {PQ} = \left( {\begin{array}{*{20}{c}} Also, note that, as before, we must be careful when using the formula \(\tan θ=\dfrac{y}{x}$$ to choose the correct value of $$θ$$. Figure 3.32(b): Vector analysis. The magnitude of a directed distance vector is + \dot{z} \, \hat{e}_z + z \, \dot{\hat{e}}_z
It should be immediately clear what the scalar products of the unit vectors are. Since the unit vectors point along the $$x$$, $$y$$, and $$z$$ directions, the components of a vector can be expressed as a dot product. In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. position $\vec{P}$ as follows.
The, A cone has several kinds of symmetry. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The points on these surfaces are at a fixed distance from the $$z$$-axis. We first calculate that the magnitude of vector product of the unit vectors $$\overrightarrow{\mathbf{i}}$$ and $$\overrightarrow{\mathbf{j}}$$: $|\hat{\mathbf{i}} \times \hat{\mathbf{j}}|=|\hat{\mathbf{i}} \| \hat{\mathbf{j}}| \sin (\pi / 2)=1$, because the unit vectors have magnitude $$|\hat{\mathbf{i}}|=|\hat{\mathbf{j}}|=1$$ and $$\sin (\pi / 2)=1.$$ By the right hand rule, the direction of, $\overrightarrow{\mathbf{i}} \times \overrightarrow{\mathbf{j}}$, is in the $$+\hat{\mathbf{k}}$$ as shown in Figure 3.30. - \sin\theta \, \hat{e}_\theta \\
In the spherical coordinate system, a point $$P$$ in space (Figure $$\PageIndex{9}$$) is represented by the ordered triple $$(ρ,θ,φ)$$ where. The origin should be located at the physical center of the ball. Each trace is a circle.
Again start with two vectors in component form: then, as in the case of the scalar product, just do "normal algebra," distributing the cross product, and applying the unit vector cross products above: \begin{align} \overrightarrow A \times \overrightarrow B &= (A_x \widehat i + A_y \widehat j) \times (B_x \widehat i + B_y \widehat j) \\[5pt] &= (A_x B_x) \cancelto{0}{\widehat i \times \widehat i} + (A_yB_x) \cancelto{- \widehat k} {\widehat j \times \widehat i} + (A_x B_y ) \cancelto{+\widehat k}{ \widehat i \times \widehat j} + (A_y B_y) \cancelto{0}{\widehat j \times \widehat j} \\[5pt] &= (A_x B_y - A_yB_x ) \widehat k \end{align}. Finally, with the magnitudes of the vectors and the angle between the vectors, we could finally plug into our scalar product equation. these coordinates using the atan2 function as follows.
These equations are used to convert from rectangular coordinates to cylindrical coordinates. The unit vectors are at right angles to each other and so using the right hand rule, the vector product of the unit vectors are given by the relations, $\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}}=\hat{\mathbf{k}}$$\hat{\boldsymbol{\theta}} \times \hat{\mathbf{k}}=\hat{\mathbf{r}}$$\hat{\mathbf{k}} \times \hat{\mathbf{r}}=\hat{\boldsymbol{\theta}}$ Because the vector product satisfies $$\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}=-\overrightarrow{\mathbf{B}} \times \overrightarrow{\mathbf{A}},$$ we also have that $\hat{\boldsymbol{\theta}} \times \hat{\mathbf{r}}=-\hat{\mathbf{k}}$$\hat{\mathbf{k}} \times \hat{\mathbf{\theta}}=-\hat{\mathbf{r}}$$\hat{\mathbf{r}} \times \hat{\mathbf{k}}=-\hat{\mathbf{\theta}}$Finally $\hat{\mathbf{r}} \times \hat{\mathbf{r}}=\hat{\boldsymbol{\theta}} \times \hat{\boldsymbol{\theta}}=\hat{\mathbf{k}} \times \hat{\mathbf{k}}=\overrightarrow{\mathbf{0}}$, Given two vectors, $$\overrightarrow{\mathbf{A}}=2 \hat{\mathbf{i}}+-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{B}}=5 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \text { find } \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}} \nonumber$$, \[\begin{align*} $$|\overrightarrow{\mathbf{A}}||\overrightarrow{\mathbf{B}}|\sin\gamma=|\overrightarrow{\mathbf{A}}||\overrightarrow{\mathbf{C}}| \sin \beta$$. \dot{\hat{e}}_r &= \dot\theta \, \hat{e}_{\theta} \\ = it is commutative). In this case, however, we would likely choose to orient our. = \dot\theta \, \hat{e}_z \times \hat{e}_\theta
Vector Decomposition and the Vector Product: Cylindrical Coordinates. basis while changing $\theta$ rotates about the vertical
θ Recall the cylindrical coordinate system, which we show in Figure 3.31. To invert the basis change we can solve for A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form $$z^2=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}.$$ In this case, we could choose any of the three. The projection of $$\overrightarrow A$$ onto $$\overrightarrow B$$ is 7/8 of the magnitude of $$\overrightarrow A$$, so the magnitude of $$\overrightarrow B$$ must be 8/7 of its projection, which is 56 units. Last, what about $$θ=c$$? To convert a point from spherical coordinates to cylindrical coordinates, use equations $$r=ρ\sin φ, θ=θ,$$ and $$z=ρ\cos φ.$$, To convert a point from cylindrical coordinates to spherical coordinates, use equations $$ρ=\sqrt{r^2+z^2}, θ=θ,$$ and $$φ=\arccos(\dfrac{z}{\sqrt{r^2+z^2}}).$$, Paul Seeburger edited the LaTeX on the page. r r → for the position vector of a point, but if we are using cylindrical coordinates r,θ,z r, θ, z then this is dangerous. axis $\hat{e}_z$. This set forms a sphere with radius $$13$$. &=((1)(3)-(-1)(-1)) \hat{\mathbf{i}}+((-1)(2)-(1)(3)) \hat{\mathbf{j}}+((1)(-1)-(1)(2)) \hat{\mathbf{k}} \\ Because $$\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = 0$$, we have that $$0 = \overrightarrow{\mathbf{A}} \times (\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}})$$.
.
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http://www.openwetware.org/index.php?title=BISC_111/113:Lab_2:_Population_Growth&diff=520015&oldid=520005 | 1,448,768,408,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398455246.70/warc/CC-MAIN-20151124205415-00301-ip-10-71-132-137.ec2.internal.warc.gz | 606,719,099 | 11,933 | # BISC 111/113:Lab 2: Population Growth
(Difference between revisions)
Revision as of 16:35, 29 June 2011 (view source) (→Descriptive Statistics)← Previous diff Revision as of 16:56, 29 June 2011 (view source) (→Comparative Statistics)Next diff → Line 102: Line 102: The t-test calculates a factor called the tcal by using the means, SDs, and the number of data points in each sample. Sometimes when calculated digitally, it is designated the tstat. The numerator of the tcal is a measure of the difference between the means of the samples; its denominator is a measure of the average variability (pooled variance), taking sample size into account. The order in which you enter the means will determine the sign of your tcal, but this does not affect its interpretation. Always report the absolute value of tcal. The t-test calculates a factor called the tcal by using the means, SDs, and the number of data points in each sample. Sometimes when calculated digitally, it is designated the tstat. The numerator of the tcal is a measure of the difference between the means of the samples; its denominator is a measure of the average variability (pooled variance), taking sample size into account. The order in which you enter the means will determine the sign of your tcal, but this does not affect its interpretation. Always report the absolute value of tcal.
[[Image:111F11Tcal2.jpg|500px]]
[[Image:111F11Tcal2.jpg|500px]]
- +
[[Image:111F11.Tcal3.png|500px]]
The null hypothesis of a comparative statistical test is that there is no difference between the means, other than that due to random chance. Therefore, if a significant difference is seen, the null hypothesis is disproved and there is likely an effect of the treatment. To determine whether two means are significantly different one must compare the absolute value of tcal to the tabulated t-value, ttab. A subset of ttab values for given degrees of freedom can be found in the table below. Calculate the degrees of freedom for your two data sets (n1 + n2 -2) and compare your tcal to the ttab. The probability that a difference at least as big as the one you observed would have occurred randomly can be read from the "Probability Levels" columns. Generally, if the probability level (P-value) of our tcal is ≤ 0.05, we can claim that there is a significant difference between the two means, that indeed, the conclusion that the two samples represent different populations, or that there was a treatment effect, is supported by our results. Note that there is still the risk (1 in 20 for p = 0.05) that we are wrong. (In fact, if p = 0.05 and we were to repeat such an experiment, on average, 1 in 20 times we would wrongly reject the null hypothesis.) But this is a risk we generally consider acceptable. In some studies, we set the bar of significance even “higher” (by lowering the P-value at which we would reject the null hypothesis and claim a significant difference between means). Occasionally, this threshold P-value (called the alpha [α] level, below which we will claim a significant difference) might be set higher (e.g., α = 0.10), but this usually needs a vigorous defense by the researcher. The null hypothesis of a comparative statistical test is that there is no difference between the means, other than that due to random chance. Therefore, if a significant difference is seen, the null hypothesis is disproved and there is likely an effect of the treatment. To determine whether two means are significantly different one must compare the absolute value of tcal to the tabulated t-value, ttab. A subset of ttab values for given degrees of freedom can be found in the table below. Calculate the degrees of freedom for your two data sets (n1 + n2 -2) and compare your tcal to the ttab. The probability that a difference at least as big as the one you observed would have occurred randomly can be read from the "Probability Levels" columns. Generally, if the probability level (P-value) of our tcal is ≤ 0.05, we can claim that there is a significant difference between the two means, that indeed, the conclusion that the two samples represent different populations, or that there was a treatment effect, is supported by our results. Note that there is still the risk (1 in 20 for p = 0.05) that we are wrong. (In fact, if p = 0.05 and we were to repeat such an experiment, on average, 1 in 20 times we would wrongly reject the null hypothesis.) But this is a risk we generally consider acceptable. In some studies, we set the bar of significance even “higher” (by lowering the P-value at which we would reject the null hypothesis and claim a significant difference between means). Occasionally, this threshold P-value (called the alpha [α] level, below which we will claim a significant difference) might be set higher (e.g., α = 0.10), but this usually needs a vigorous defense by the researcher.
[[Image:111F11.TtestTableB.png|720px]]
[[Image:111F11.TtestTableB.png|720px]]
## Revision as of 16:56, 29 June 2011
BISC 111/113: Introductory Organismal Biology
## Objectives
1. To set up a student designed experiment on the factors affecting population dynamics in a species of flour beetles, Tribolium confusum.
2. To learn how to use the computer program Excel for graph construction and the computer program JMP for statistical tests.
## Lab 2 Overview
I. Formulate experimental questions and hypotheses about population growth of Tribolium
II. Design and set up tests of your population growth hypotheses
a. Post a PowerPoint summary of your experimental design
III. Data Analysis and Presentation
a. Learn how to calculate the mean, variance, standard deviation, and standard error of data arrays
b. Learn how to test for differences in the means of two samples using the t-test.
## Population Growth Background
It is accepted that environments on Earth are finite and therefore have limited resources, so it follows that no population can grow indefinitely. Certainly no organism exhibits its full reproductive potential. Darwin, for example, calculated that it would take only 750 years for a single mating pair of elephants (a species with a relatively low reproductive potential) to produce a population of 19 million. This is vastly in excess of the current total population and elephants have existed for millions of years. Some species might exhibit population explosions for a short time (e. g., algal blooms), but their population inevitably crashes. Most populations, however, are relatively stable over time, once they have reached an equilibrium level.
Population ecology is the discipline that studies changes in population size and composition, and also tries to identify the causes of any observed fluctuations. A population is made up of interbreeding individuals of one species that simultaneously occupy the same general area. Fluctuations in population sizes could be caused by environmental conditions as well as by predation and interspecific competition. It can be particularly challenging to follow and understand the population dynamics of a species in the "real" world. Therefore, scientists have often used controlled lab experiments to understand the basic concepts of population ecology. Many classical experiments have explored population dynamics of and inter- and intraspecific competition in the flour beetles, all members of the genus Tribolium.
While Tribolium can survive on a number of finely ground grains, these particular beetles are cultured in 95% whole wheat flour and 5% brewer's yeast. Tribolium thrive at a temperature range of 29-34 °C and a humidity of 50-70%. Under optimum conditions one would expect a new generation roughly every 4 weeks. The "confused" flour beetle (T. confusum) (Fig. 1, Table 1) was so named because it was often confused with its closely allied species T. castaneum.
Because a female flour beetle can live at least a year and lay 400-600 eggs in her lifetime, one can imagine the potential for overcrowding. High density can lead to several interesting phenomena, such as an increase in the incidence of cannibalism, where the adult beetles will eat the eggs; larvae will eat eggs, pupae, and other larvae. If conditions are crowded and stressful the beetles will often produce a gas containing certain quinones that can cause the appearance of aberrant forms of young or can even kill the entire colony. There have also been reports that overcrowding leads to an increase in the transmission of a protozoan parasite (Adelina tribolii).
Arthropods need to molt in order to grow. Tribolium beetles, like all other members of the insect order Coleoptera, undergo complete metamorphosis, passing through four distinct phases to complete their life cycle: egg, larva, pupa, and adult. An egg is laid from which hatches a larva. This larva molts into a second and then third larval stage (or instar) increasing in size in the process. The third instar turns into a pupa from which finally an adult is released. The pupa is a quiescent stage during which larval tissues and organs are reorganized into adult ones.
## Setting up the Experiment
Your charge is to set up an experiment dealing with population ecology of the flour beetle. You should start your experiment with at least 20 beetles per container, since they are not sexed. This should ensure enough females in your starting population. We have found that 1 gram of food per beetle will keep them reasonably healthy until the end of the experiment. Discuss the appropriate number of replicates.
Some of the factors you could consider investigating are:
Size of starting populations (intraspecific competition)
Food supply (e.g., various milled flours, whole grains, prepared grain products, and/or brewer's yeast)
Environmental structure (effects of environmental patchiness on population growth, e.g. habitat size, light availability, refuge use, or "dilution" of the food/habitat volume with inert materials)
Biological control of populations in grain storage facilities (e.g., diatomaceous earth, application of plant volatiles)
Briefly outline the hypothesis that you are testing or your experimental question in your lab notebook. Provide details of your experimental design (starting number of beetles, number of replicates, variables, etc.)
Prepare a PowerPoint slide describing the experimental design and post it to the lab conference on Sakai.
## Descriptive Statistics
When dealing with the dynamics of populations, whether they are beetles in a jar or plants in the field, we need to extrapolate information from small portions, or samples, of the population. Otherwise the task can be overwhelming. As scientists we wish to infer population behavior at large from the results of necessarily limited and random sampling. Random sampling means that the likelihood of any particular individual being in the sample is the same for all individuals, and that these are selected independently from one another. A statistical test is the impartial judge of whether our inferences about the at-large population(s) are sufficiently supported by our sample results. See Statistics and Graphing for tutorials for calculating all of the statistics below with Excel 2008 and JMP.
Descriptive statistics are measures of location and variability used to characterize data arrays. In this course, we will use hand calculation and the computer programs Excel and JMP to generate common descriptive statistics.
Statistics of location are measures of "location" or "central tendency."
The mean, median, and mode are all measures of location. The mean is the most commonly used of these, and estimates the true population mean if it is based on samples composed randomly from the at-large population. Calculate the mean of a data set by dividing the sum of the observations by the number of observations.
Statistics of variability provide estimates of how measurements are scattered around a mean or other measures of location. Both biological variability and the accuracy of our measurements are sources of variability in our data.
Variance is an approximate average of the square of the differences of each value from the mean. However, because the variance is reported as the square of the original unit, interpretation can sometimes be difficult. Calculate the variance by dividing the sum of the observations by the number of observations.
Standard Deviation (SD) is a common measure of variability that avoids the problem of units inherent in the variance. Calculate the standard deviation by first calculating the variance, and then by taking its square root.
Standard Error of the Mean (SEM or SE) estimates the variation among the means of samples similarly composed from the population at large (the so-called "true" population). The SE estimates the variability among means if you take repeated random samples of the same size from the population. A small SE indicates that the sample mean is close to the true population mean. With increasing sample sizes (n) the SE shrinks in magnitude. Calculate the SE by dividing the SD by the square root of n.
## Comparative Statistics
Comparative Statistics are ways of evaluating the similarities and/or differences among different data sets. Many situations arise in experimental and observational research where we wish to compare two outcomes or contrasts, such as a control vs. a treatment effect.
The t-test compares the means of two samples. If the samples are taken from the same experimental unit at different times, the test is termed "paired," so a paired t-test is run. If the samples are from two different experimental units or treatments, the unpaired t-test is run. Both tests assume that data are normally distributed (i.e. have a typical bell-shaped distribution) and have similar variances. Violations of such assumptions in this kind of statistical testing are not too serious unless quite exaggerated, and there are ways to transform the data that can often rectify such problems.
The t-test calculates a factor called the tcal by using the means, SDs, and the number of data points in each sample. Sometimes when calculated digitally, it is designated the tstat. The numerator of the tcal is a measure of the difference between the means of the samples; its denominator is a measure of the average variability (pooled variance), taking sample size into account. The order in which you enter the means will determine the sign of your tcal, but this does not affect its interpretation. Always report the absolute value of tcal.
The null hypothesis of a comparative statistical test is that there is no difference between the means, other than that due to random chance. Therefore, if a significant difference is seen, the null hypothesis is disproved and there is likely an effect of the treatment. To determine whether two means are significantly different one must compare the absolute value of tcal to the tabulated t-value, ttab. A subset of ttab values for given degrees of freedom can be found in the table below. Calculate the degrees of freedom for your two data sets (n1 + n2 -2) and compare your tcal to the ttab. The probability that a difference at least as big as the one you observed would have occurred randomly can be read from the "Probability Levels" columns. Generally, if the probability level (P-value) of our tcal is ≤ 0.05, we can claim that there is a significant difference between the two means, that indeed, the conclusion that the two samples represent different populations, or that there was a treatment effect, is supported by our results. Note that there is still the risk (1 in 20 for p = 0.05) that we are wrong. (In fact, if p = 0.05 and we were to repeat such an experiment, on average, 1 in 20 times we would wrongly reject the null hypothesis.) But this is a risk we generally consider acceptable. In some studies, we set the bar of significance even “higher” (by lowering the P-value at which we would reject the null hypothesis and claim a significant difference between means). Occasionally, this threshold P-value (called the alpha [α] level, below which we will claim a significant difference) might be set higher (e.g., α = 0.10), but this usually needs a vigorous defense by the researcher.
Table 2. Critical tabulated values of Student's t –test (two-tailed) at different degrees of freedom and probability level
## Assignments
1. Prepare a PowerPoint slide describing the experimental design for your Tribolium experiment and post it to the lab conference on Sakai.
2. Write a preliminary Materials and Methods section text for your Tribolium experiment.
3. Prepare a column graph of Tribolium means ± SD from historical data.
4. In preparation for the Plant Biology series, visit the College Greenhouse, paying particular attention to the different adaptations of plants located in the Desert, Tropical, Subtropical and Water Plant rooms. Click here for Greenhouse map and Self-Guided Tour information.
5. With your bench mates, prepare a summary of the key characteristics of your plant to be presented at the beginning of lab next week. | 3,574 | 17,101 | {"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} | 3.703125 | 4 | CC-MAIN-2015-48 | longest | en | 0.936625 |
https://www.sanfoundry.com/pavement-design-questions-answers-design-overlay/ | 1,686,260,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00170.warc.gz | 1,054,069,946 | 22,220 | # Pavement Design Questions and Answers – Highway Maintenance – Design of Overlay
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This set of Pavement Design Multiple Choice Questions & Answers (MCQs) focuses on “Highway Maintenance – Design of Overlay”.
1. The first step in designing an overlay is estimating the strength of the existing pavement. What step is to be performed after this?
a) Determining overlay thickness
b) Determining overlay design life
c) Estimation of traffic on an overlay
d) Determining overlay type
Explanation: The order of the steps to be performed during the designing of an overlay is as follows – estimating the strength of the existing pavement, determining the design life of the overlaid pavement, estimating the traffic to be carried by the overlaid pavement, determining the thickness of overlay and the type of overlay.
2. How many types of overlays are possible to provide on existing pavements?
a) Two types of asphalt overlays and single coat of cc overlay
b) Two types of cc overlays and single coat of asphalt overlay
c) Two types of asphalt overlays and two types of cc overlays
d) Single coat of asphalt overlay and single coat of cc overlay
Explanation: Four different types of overlays can be laid on an existing pavement. They are asphalt overlay on asphalt pavement, asphalt overlay on cc pavement, cc overlay on cc pavement and cc overlay on asphalt pavement.
3. What is the equation used to find the effective thickness of the existing pavement?
a) $$h_e=∑_{i=1}^n C_i$$
b) $$h_e=∑_{i=0}^n h_i C_i$$
c) $$h_e=∑_{i=1}^n h_i$$
d) $$h_e=∑_{i=1}^n h_i C_i$$
Explanation: $$h_e=∑_{i=1}^n h_i C_i$$ represents the correct formula to find the effective thickness of the existing pavement. This equation is used to find the thickness of the overlay using the effective thickness method. The overlay thickness is obtained as the difference between the thickness of new pavement and the effective thickness of the existing pavement.
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4. Find the overlay thickness if the following data is available using the effective thickness method.
Thickness of existing pavement: 260 mm GSB, 250 mm WBM, 90 mm BM and 40 mm BC
Conversion factors = 0.1 for GSB and WBM, 0.5 for BM and 0.6 for BC
Thickness of new pavement: 300 mm GSB, 250 mm WWM, 100 mm DBM and 40 mm BC
Conversion factors = 0.2 for GSB and WWM, 1 for DBM and BC
a) 230 mm
b) 250 mm
c) 120 mm
d) 130 mm
Explanation: Overlay thickness is the difference between the thickness required for new pavement and the effective thickness of the existing pavement. It is given by ho=hn -he.
Effective thickness is given by $$h_e=∑_{i=1}^n h_i C_i$$. Where hi is the thickness of each pavement layer and Ci is the corresponding conversion factor given by the Asphalt Institute.
he=(260+250)×0.1+90×0.5+40×0.6=120 mm
hn=(300+250)×0.2+(100+40)×1=250 mm
ho=hn-he=250-120=130 mm
5. What is the conversion factor for the subgrade in the effective thickness overlay design method?
a) 1
b) 0
c) 0.1
d) 1.1
Explanation: The conversion factors for various layers of the pavement are given by the Asphalt Institute. It is based on the equivalent thickness of the new pavement. Since the subgrade would remain the same, the conversion factor would be 0.
6. The design of overlay can be done using the Benkelman beam deflection studies result.
a) True
b) False
Explanation: The deflection approach to find the thickness of the overlay is based on the result obtained from the Benkelman beam deflection studies. The IRC 81:1997 has the guidelines for the design of overlay thickness using BBD studies.
7. The design curves relating ______ with the characteristic deflection obtained from BBD studies in the code is used for overlay thickness determination.
a) Resilient modulus of the surface
b) CSA over the design period
c) Thickness of existing pavement
d) Pavement condition index
Explanation: The design chart is given in the code IRC 81:1997. The chart relates the cumulative axles to be carried over the design life to the characteristic deflection. Relating these two values form the chart, the thickness of the overlay can be read.
8. The thickness of the overlay obtained from design chart is in terms of bituminous macadam construction. What is the conversion if thickness in terms of DBM is required?
a) 1 cm BM = 0.7 mm DBM
b) 1 cm BM = 1.5 mm DBM
c) 1 cm BM = 0.7 cm DBM
d) 1 cm BM = 1.5 cm DBM
Explanation: The equivalent thickness of dense bound macadam can be found as 0.7 cm for every 1 cm thickness of bituminous macadam. It can be taken as 1.5 cm for wet bound macadam. In this way, the thickness of overlay for varying compositions can be found out.
9. If the design traffic is 15 msa and a characteristic deflection of 3 mm, compute the thickness of overlay if DBM binder course and bituminous concrete (BC) surface course is to be adopted. What would be the thickness of the BC surface course?
a) 151 mm
b) 51 mm
c) 150 mm
d) 50 mm
Explanation: The thickness of the overlay in terms of bituminous macadam can be read from the design chart using Fig 9 in IRC 81:1997. The chart can be referred to by using the link below:
https://archive.org
For 15 msa and 3 mm characteristic deflection, the overlay thickness in terms of bituminous macadam is 215 mm. Now to find the thickness of DBM,
1 cm BM = 0.7 cm DBM, i.e.
21.5 cm BM = 0.7 X 21.5 = 15.05 cm ≈ 15.1 cm = 151 mm
Out of the 151 mm, 100 mm can be provided for DBM and 51 mm for the bituminous concrete surface course.
10. Traffic studies were conducted on a 4-lane divided flexible pavement highway and the average CVPD was found to be 3500 in one direction. The growth rate is 5%, VDF is 4.5 and the estimated time of completion of the overlay is 2 years. The BBD studies yielded the characteristic deflection to be 2.4 mm. What would be the thickness of the overlay if it to be designed for 10 years?
a) 210 mm
b) 201 mm
c) 120 mm
d) 102 mm
Explanation: The equation to find the design traffic is as below:
$$csa=\frac{365[(1+r)^n-1]NFD}{r}$$
Where r is the growth rate = 5%
n is the design life of overlay = 10 years
N is the traffic after the completion of overlay construction = A(1+r)n
A is the present traffic = 3500 CVPD and n is the time of completion of overlay construction = 2 years
Therefore, N=A(1+r)n=3500(1+0.05)2=3858.75 CVPD
F is the VDF = 4.5
D is the lane distribution factor, as per IRC 37:2012, for a four-lane two-way highway, it is taken as 0.45.
The design traffic,
$$csa=\frac{365[(1+0.05)^{10}-1]×3858.75×4.5×0.45)}{0.05}$$=35873389.61≈35.87 msa
Now, referring the design chart using the link below:
https://archive.org
For design traffic 35.87 msa and 2.4 mm characteristic deflection, the thickness of overlay is 210 mm.
11. The overlay can be directly laid over the existing surface after cleaning.
a) True
b) False
Explanation: The existing pavement surface might have potholes, sags, bumps and all sorts of unevenness. It is very necessary to correct the level of the pavement before laying the overlay. Cleaning is not sufficient, proper filling of sags and levelling of bumps is required.
12. Traffic studies were conducted on a 4-lane flexible pavement highway and the average CVPD was found to be 3000 in one direction. The growth rate is 7.5%, VDF is 4.5 and the estimated time of completion of the overlay is 2 years. The BBD studies yielded the mean rebound deflection to be 3 mm when tested at 35°C in the monsoon time and the standard deviation is found to be 0.22 mm. What would be the thickness of the overlay if it to be designed for 10 years?
a) 240 mm
b) 210 mm
c) 200 mm
d) 250 mm
Explanation: The equation to find the design traffic is as below:
$$csa=\frac{365[(1+r)^n-1]NFD}{r}$$
Where r is the growth rate = 7.5%
n is the design life of overlay = 10 years
N is the traffic after the completion of overlay construction = A(1+r)n
A is the present traffic = 3000 CVPD and n is the time of completion of overlay construction = 2 years
Therefore, N=A(1+r)n=3000(1+0.075)2=3466.88 CVPD
F is the VDF = 4.5
D is the lane distribution factor, as per IRC 37:2012 for four-lane single lane highway, it is taken as 0.40.
The design traffic,
$$csa=\frac{365[(1+0.075)^{10}-1]×3466.88×4.5×0.40}{0.075}$$=32223389.34≈32.22 msa
Characteristic deflection for 3000 CVPD, Dc=Dm+2s
Dc=3+2×0.22=3.44 mm
Now, referring the design chart using the link below:
https://archive.org
For design traffic 32.22 msa and 3.44 mm characteristic deflection, the thickness of the overlay is 250 mm.
13. The design traffic for a minor highway is found to be 10.55 msa. The BBD studies conducted at 30°C on sandy soil with 8% moisture content and an annual rainfall of 1300 mm gave the following results:
Rebound deflection values recorded at various points = 1.40, 1.38, 1.49, 1.50, 1.55, 1.44, 1.58, 1.43, 1.33, 1.37 mm
a) 150 mm
b) 140 mm
c) 100 mm
d) 160 mm
Explanation: The characteristic deflection is to be found out using the equation Dc=Dm+s for a minor highway. The mean deflection is found as below:
Dm= $$\frac{1.40+1.38+1.49+1.50+1.55+1.44+1.58+1.43+1.33+1.37}{10}$$=1.45 mm
The equation to find the standard deviation is given by S.D=$$\sqrt{\frac{\sum(D_m-D_x)^2)}{n-1}}$$
Sum = (1.45-1.40)2+(1.45-1.38)2+(1.45-1.49)2+(1.45-1.50)2+(1.45-1.55)2+(1.45-1.44)2+(1.45-1.58)2+(1.45-1.43)2+(1.45-1.33)2+(1.45-1.37)2
= 0.06mm2
So, S.D=$$\sqrt{\frac{0.06}{10-1}}$$=0.08 mm
Dc=Dm+s=1.45+0.08=1.53 mm
Now corrections for Dc has to be done. Temperature correction is done as
Corrected Dc=Dc+(Standard temperature-field temperature)×0.01
Corrected Dc=1.53+(35-30)×0.01=1.58 mm
The correction for moisture content has to be done by referring to the graphs provided in IRC 81:1997. The code can be accessed from the below link:
https://archive.org
The graph labelled in the code can be used for sandy soils with annual rainfall ≤ 1300 mm. From the graph, the correction factor is obtained as 1.03.
Corrected Dc=Dc×correction factor
Corrected Dc=1.58×1.03=1.63 mm
Now, referring the design chart using the link below:
https://archive.org
For design traffic 10.55 msa and 1.63 mm characteristic deflection, the thickness of the overlay is 140 mm.
14. What is the recommended thickness of bituminous concrete surface course in the overlay design?
a) Minimum 50 mm
b) Maximum 50 mm
c) Minimum 40 mm
d) Maximum 40 mm
Explanation: The recommended thickness of the bituminous concrete surface layer in the overlay design is a minimum of 40 mm. This can be obtained in clause 7.5 of IRC 81:1997. The minimum thickness of 50 mm is required for the bituminous macadam overlay with an additional minimum 40 mm BC or 50 mm DBM surfacing course.
15. If the design traffic is 20 msa and a characteristic deflection of 4.5 mm, what would be the thickness of bituminous macadam overlay if WBM binder course is to be laid?
a) 50 mm
b) 100 mm
c) 200 mm
d) 400 mm
Explanation: The thickness of the overlay in terms of bituminous macadam can be read from the design chart using Fig 9 in IRC 81:1997. The chart can be referred to by using the link below:
https://archive.org
For 20 msa and 4.5 mm characteristic deflection, the overlay thickness in terms of bituminous macadam is 265 mm. Now to find the thickness of WBM,
1 cm BM = 1.5 cm WBM, i.e.
26.5 cm BM = 1.5 X 26.5 = 39.75 cm ≈ 40 cm = 400 mm.
Sanfoundry Global Education & Learning Series – Pavement Design.
To practice all areas of Pavement Design, here is complete set of 1000+ Multiple Choice Questions and Answers. | 3,350 | 11,463 | {"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} | 3.8125 | 4 | CC-MAIN-2023-23 | longest | en | 0.817611 |
https://ourshadowcabinet.com/what-determines-a-particle-s-field.php | 1,582,300,126,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00339.warc.gz | 498,335,961 | 8,521 | # What determines a particle s field
Learning Objectives
By the end of this section, you will be able to:
• Explain how a charged particle in an external magnetic field undergoes circular motion
• Describe how to determine the radius of the circular motion of a charged particle in a magnetic field
A charged particle experiences a force when moving through a magnetic field. What happens if this field is uniform over the motion of the charged particle? What path does the particle follow? In this section, we discuss the circular motion of the charged particle as well as other motion that results from a charged particle entering a magnetic field.
The simplest case occurs when a charged particle moves perpendicular to a uniform B-field (Figure $$\PageIndex{1}$$). If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.
In this situation, the magnetic force supplies the centripetal force $$F_C = \dfrac{mv^2}{r}$$. Noting that the velocity is perpendicular to the magnetic field, the magnitude of the magnetic force is reduced to $$F = qvB$$. Because the magnetic force F supplies the centripetal force $$F_C$$, we have
$qvB = \dfrac{mv^2}{r}.$
Solving for r yields
$r = \dfrac{mv}{qB}.$
Here, r is the radius of curvature of the path of a charged particle with mass m and charge q, moving at a speed v that is perpendicular to a magnetic field of strength B. The time for the charged particle to go around the circular path is defined as the period, which is the same as the distance traveled (the circumference) divided by the speed. Based on this and Equation, we can derive the period of motion as
$T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{v} \dfrac{mv}{qB} = \dfrac{2\pi m}{qB}.$
If the velocity is not perpendicular to the magnetic field, then we can compare each component of the velocity separately with the magnetic field. The component of the velocity perpendicular to the magnetic field produces a magnetic force perpendicular to both this velocity and the field:
\begin{align} v_{perp} &= v \, \sin \theta \\[4pt] v_{para} &= v \, \cos \theta. \end{align}
where $$\theta$$ is the angle between v and B. The component parallel to the magnetic field creates constant motion along the same direction as the magnetic field, also shown in Equation. The parallel motion determines the pitchp of the helix, which is the distance between adjacent turns. This distance equals the parallel component of the velocity times the period:
$p = v_{para} T.$
The result is a helical motion, as shown in the following figure.
While the charged particle travels in a helical path, it may enter a region where the magnetic field is not uniform. In particular, suppose a particle travels from a region of strong magnetic field to a region of weaker field, then back to a region of stronger field. The particle may reflect back before entering the stronger magnetic field region. This is similar to a wave on a string traveling from a very light, thin string to a hard wall and reflecting backward. If the reflection happens at both ends, the particle is trapped in a so-called magnetic bottle.
Trapped particles in magnetic fields are found in the Van Allen radiation belts around Earth, which are part of Earth’s magnetic field. These belts were discovered by James Van Allen while trying to measure the flux of cosmic rays on Earth (high-energy particles that come from outside the solar system) to see whether this was similar to the flux measured on Earth. Van Allen found that due to the contribution of particles trapped in Earth’s magnetic field, the flux was much higher on Earth than in outer space. Aurorae, like the famous aurora borealis (northern lights) in the Northern Hemisphere (Figure $$\PageIndex{3}$$), are beautiful displays of light emitted as ions recombine with electrons entering the atmosphere as they spiral along magnetic field lines. (The ions are primarily oxygen and nitrogen atoms that are initially ionized by collisions with energetic particles in Earth’s atmosphere.) Aurorae have also been observed on other planets, such as Jupiter and Saturn.
Example $$\PageIndex{1}$$: Beam Deflector
A research group is investigating short-lived radioactive isotopes. They need to design a way to transport alpha-particles (helium nuclei) from where they are made to a place where they will collide with another material to form an isotope. The beam of alpha-particles $$(m = 6.64 \times 10^{-27}kg, \, q = 3.2 \times 10^{-19}C)$$ bends through a 90-degree region with a uniform magnetic field of 0.050 T (Figure $$\PageIndex{4}$$). (a) In what direction should the magnetic field be applied? (b) How much time does it take the alpha-particles to traverse the uniform magnetic field region?
Strategy
1. The direction of the magnetic field is shown by the RHR-1. Your fingers point in the direction of v, and your thumb needs to point in the direction of the force, to the left. Therefore, since the alpha-particles are positively charged, the magnetic field must point down.
2. The period of the alpha-particle going around the circle is
$T = \dfrac{2\pi m}{qB}.$
Because the particle is only going around a quarter of a circle, we can take 0.25 times the period to find the time it takes to go around this path.
Solution
1. Let’s start by focusing on the alpha-particle entering the field near the bottom of the picture. First, point your thumb up the page. In order for your palm to open to the left where the centripetal force (and hence the magnetic force) points, your fingers need to change orientation until they point into the page. This is the direction of the applied magnetic field.
2. The period of the charged particle going around a circle is calculated by using the given mass, charge, and magnetic field in the problem. This works out to be $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi (6.64 \times 10^{-27}kg)}{(3.2 \times 10^{-19}C)(0.050 \, T)} = 2.6 \times 10^{-6}s.$ However, for the given problem, the alpha-particle goes around a quarter of the circle, so the time it takes would be $t = 0.25 \times 2.61 \times 10^{-6}s = 6.5 \times 10^{-7}s.$
Significance
This time may be quick enough to get to the material we would like to bombard, depending on how short-lived the radioactive isotope is and continues to emit alpha-particles. If we could increase the magnetic field applied in the region, this would shorten the time even more. The path the particles need to take could be shortened, but this may not be economical given the experimental setup.
Exercise $$\PageIndex{1}$$
A uniform magnetic field of magnitude 1.5 T is directed horizontally from west to east. (a) What is the magnetic force on a proton at the instant when it is moving vertically downward in the field with a speed of $$4 \times 10^7 \, m/s$$? (b) Compare this force with the weight w of a proton.
Solution
a. $$9.6 \times 10^{-12}N$$ toward the south;
b. $$\dfrac{w}{F_m} = 1.7 \times 10^{-15}$$
Example $$\PageIndex{2}$$: Helical Motion in a Magnetic Field
A proton enters a uniform magnetic field of $$1.0 \times 10^{-4}T$$ with a speed of $$5 \times 10^5 \, m/s$$. At what angle must the magnetic field be from the velocity so that the pitch of the resulting helical motion is equal to the radius of the helix?
Strategy
The pitch of the motion relates to the parallel velocity times the period of the circular motion, whereas the radius relates to the perpendicular velocity component. After setting the radius and the pitch equal to each other, solve for the angle between the magnetic field and velocity or $$\theta$$.
Solution
The pitch is given by Equation, the period is given by Equation, and the radius of circular motion is given by Equation. Note that the velocity in the radius equation is related to only the perpendicular velocity, which is where the circular motion occurs. Therefore, we substitute the sine component of the overall velocity into the radius equation to equate the pitch and radius
$p = r$
$v_{\parallel}T = \dfrac{mv}{qB}$
$v \, cos \, \theta \dfrac{2\pi m}{qB} = \dfrac{mv \, sin \, \theta}{qB}$
$2\pi = tan \, \theta$
$\theta = 81.0^o.$
Significance
If this angle were $$0^o$$, only parallel velocity would occur and the helix would not form, because there would be no circular motion in the perpendicular plane. If this angle were $$90^o$$ only circular motion would occur and there would be no movement of the circles perpendicular to the motion. That is what creates the helical motion.
## Contributors
• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). | 2,179 | 9,248 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-10 | latest | en | 0.90416 |
https://edurev.in/course/quiz/attempt/6572_SNAP-Past-Year-Papers-2013/0d7eaa7f-ffbf-497c-9c9e-b96d7455e33f | 1,660,095,990,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571097.39/warc/CC-MAIN-20220810010059-20220810040059-00648.warc.gz | 235,633,614 | 80,466 | # SNAP Past Year Papers - 2013
## 150 Questions MCQ Test SNAP Mock Test Series | SNAP Past Year Papers - 2013
Description
Attempt SNAP Past Year Papers - 2013 | 150 questions in 120 minutes | Mock test for CAT preparation | Free important questions MCQ to study SNAP Mock Test Series for CAT Exam | Download free PDF with solutions
QUESTION: 1
### Dinesh can do a piece of work in 15 days, Harish can do it in 12 days. Dinesh, Harish and Subodh can together do it in 4 days. In how many days can Subodh alone do this work?
Solution:
Dinesh can do the work in 15 days and Harish can do it in 12 days. Let Subodh can do the work in ‘x’ days. Since the work completes in 4 days, we have
Hence Subodh can do the work in 10
days.
QUESTION: 2
### A village has 2000 people out of which 1100 are male and 900 female. The overall literacy rate is 95%, while only 3 % male are not literate. What percentage of the female population is literate?
Solution:
Here Males in the village = 1100. Since 3% males are not literate, so the number of males who are literate = 97% of 1100 = 1067.
Now overall literacy rate of the village = 95% of 2000 = 1900.
So, number of literate females = 1900 – 1067 = 833.
Hence required %age
QUESTION: 3
### A sum of money is sufficient to pay Ravi's wage for 18 days or Ajit's wage for 12 days. The same money is sufficient to pay the wages of both for how many days?
Solution:
Since Ravi’s 18 day’s salary = Ajit’s 12 day’s salary
⇒ Ravi’s 1 day salary = Ajit’s 2/3 day salary. Let Ajit’s 1 day salary = Rs 99, so Ravi’s 1 day salary = Rs
66. The total amount = 99 × 12 = Rs 1188.
The total one day salary of Ajit & Ravi = 99 + 66 = Rs 165.
Hence the required number of days = 1188/165 = 7.2 days .
QUESTION: 4
What is the number to be added to 1/4 of 30% of 120 so that it is 50% more than 25% of 40?
Solution:
Also 25% of 40 = 10 and 50% of 10 = 5. So the required number = 10 + 5 = 15. Hence the number to be added in 9 to make it 15 is 6.
QUESTION: 5
10 years ago the average age of all 25 teachers of a school was 45 years. 4 years ago one teacher retired from her post at the age of 58. After one year a new teacher whose age was 54 years was recruited from outside. The present average age of all the teachers is:
Solution:
Total age of 25 teachers 10 years ago = 45 × 25 = 1125 years.
As one teacher was retired 4 years ago, total age of all teachers 4 years ago = 1125 + 150 – 58 = 1217 years.
After one year, one more teacher was added with age 54 years, so the total age of all teachers 3 years ago = 1217 + 24 + 54 = 1295 years.
Now the present age of all teachers = 1295 + 75 = 1370 years.
Hence average age of all teachers =
QUESTION: 6
X can complete a task in 15 days and Y can complete the same task in 20 days. X started the task alone and after 5 days Y joins X. After working together for three days, X had to leave the work. How long does it take for Y to complete the remaining work?
Solution:
Let the total work is 60 units. X can complete the work 15 days, so he can do 4 units in one day. Similarly Y can do 3 units in one day. Now in the first 5 days, X will do 5 × 4 = 20 units and in next three days x and Y will work 7 × 3 = 21 units.
The total work done till now is 20 + 21 = 41 units. The remaining work is 19 units, which will be done by Y alone and he will take 19/3 = 6.3 days .
QUESTION: 7
Find out the value of [(2.093+0.089)2 - (2.093-0.089)2] / [(2.093)(0.089)]
Solution:
Let x = 2.093 and y = 0.089. So we have
QUESTION: 8
In a triangle ABC the measure of angle A is 50° and the external bisectors of angle B and angle C meet at O. The measure of angle BOC is
Solution:
We have x + y + 50 = 1800 ⇒ x + y = 1300.
BO and CO are the external bisectors of angle B and angle C.
QUESTION: 9
The average salary of all the 60 employees in an office is Rs. 12000/- per month. If the number of executives is twice the number of non-executive employees, then the average salary of all the nonexecutive employee is:
Solution:
From the given information we can say that the number of executives is 40 and the number of non executives is 20. But we cannot find the average salary of non executives as we do not know the average salary of executives. So answer is 4th option.
QUESTION: 10
Two pots A and B contain syrup and water in the ratio 3:7 and 5:2 respectively. Find the ratio in which these mixtures be mixed to get a new mixture containing syrup and water in the ratio of 2:3?
Solution:
The syrup in pot A is 3/10 and in pot B is 5/7. In the required mixture the syrup should be 2/5. Using Alligation we have
⇒ The required ratio is 22: 7.
QUESTION: 11
A truck carrying goods has travelled for 3 days. The 1st day it travelled for 12 hours at an average speed of 40 km/hour, the 2nd day it travelled for 10 hours at an average speed of 50 km/hour, the third day it completed 360 kilometres travelling for 8 hours. The average speed of the truck in km/hour for all the three days is:
Solution:
The total distance covered by the truck in three days = 12 × 40 + 10 × 50 + 360 = 480 + 500 +360 = 1340.
The total time it travelled in three days = 12 + 10 + 8 = 30 hours.
So the average speed
QUESTION: 12
The H.C.F of two numbers is 8 and their L.C.M is 96. If one of the numbers is 32, then the other is
Solution:
Let the second number is x. We know that L.C.F. × H.C.F = Product of numbers. ⇒ 8 × 96 = 32 × x ⇒ x = 24.
QUESTION: 13
The ratio between present age of A and B is 3:4. The present age of B is 10 years more than the age of A 5 years ago. Find out the present age of B.
Solution:
Let the ages of A and B are 3x and 4x. Acc. to the question we have, 4x = 3x – 5 + 10 ⇒ x = 5. So the age of B = 4 × 5 = 20 years.
QUESTION: 14
The circumference of the front wheels of a vehicle is 3.5 metres, and that of the back wheels is 3.0 metres. If the vehicle is moving at a speed of 54 km/hr, the shortest time in which both the wheels will make a whole number of turns is
Solution:
The speed of the vehicle = 54 km/hr or 15 m/sec. The circumference of front wheel is 3.5 meters and that of the back wheel is 3 meters. In order to make a whole number of turns, the vehicle should move = LCM (3.5, 3) = 21m. Time taken by the vehicle to move 21 meters is 21/15 = 1.4 seconds.
QUESTION: 15
If p is 81 % of q, what percentage of p is q?
Solution:
Let q is 100, then p is 81. The required %age
QUESTION: 16
The length of a rectangular plot is 15 metres more than its breadth. If the cost of fencing the plot is Rs. 2700 at the rate of Rs. 30 per metre, what is the length of the plot in metres?
Solution:
The cost of fencing is Rs 2700 at the rate of Rs 30/m. So the perimeter of the plot is 2700/30 = 90m. Let breadth of plot is x, then length = x + 15.
Perimeter of plot = 90 ⇒ 2( x+ 15 +x) = 90 ⇒ 2x + 15 = 45 ⇒ x = 15. So the length = 15 + 15 = 30 m.
QUESTION: 17
In a class, 70% of the students opted for History, 45% of the students opted for Geography, and 20 % opted for neither History nor Geography. What percentage of the class opted for both the subjects?
Solution:
Let the total number of students is 100. As 20 % opted for neither History nor Geography, so the total students who opted at least one of the subjects is 80.
We have n(H) = 70, n(G) = 45, n(H∪G) = 80
Now n(H∪G) = n(H) + n(G) - n(H∩G)
⇒ 80 = 70 + 45 - n(H∩G)
⇒ n(H∩G) = 35.
Hence 35% of the class opted for both the subjects.
QUESTION: 18
A boy lying on the ground 1000 ft away from a multistoried building, can see the top of the building exactly in line with the top of a 10 ft tall flag post. If the flag post is 20 ft away from the boy, what is the height of the building?
Solution:
QUESTION: 19
21 boys turn out for an NCC parade. Of these 19 are wearing NCC caps and 11 are wearing NCC shoes. There are no boys without one or the other. How many boys are wearing the full uniform?
Solution:
We have
Now 19 – x + x + 11 – x = 21
⇒ x = 9
Or
We have n(C) = 19, n(S) = 11, n(S∪C) = 21
QUESTION: 20
A number is increased by 20% and then is reduced by 20%. The number
Solution:
Since the no. is increased & decreased by same %age, So the net change is decrease.
QUESTION: 21
In a class of 42, there are 22 boys and 20 girls. The class teacher has to appoint a girl monitor and a boy monitor. In how many ways can the selection be made?
Solution:
Total no. of boys = 22 & total girls = 20 The no. of ways in which a girl and a boy will be appointed a monitor = 22C1× 20C1 = 22 × 20 = 440
QUESTION: 22
Arvind and Mohan are 20 km apart. If they travel in opposite directions, they meet each other after half an hour. If they travel in the same direction, they meet after 2 hours. If Arvind travels faster than Mohan, his speed is
Solution:
Let the speed of Arvind is x km/hr and speed of Mohan is y km/hr. When they travel in opposite direction
& when they travel in same direction, we have (x-y) × 2 = 20
⇒ x-y =10 ...................(2)
Adding 1 & 2 we get 2x = 50
⇒ x = 25 km/hr
∴ Speed of Arvind is 25 km/hr
QUESTION: 23
Let A = {1,2,3,4,5,6}
How many subsets of A can be formed with just two elements, one even and one odd?
Solution:
Here A = { 1,2,3,4,5,6}. Since A has three even & there odd numbers, the subsets of A with one element even & one add can be formed in 3C1×3C1 = 9 ways.
QUESTION: 24
A grocer sells a soap marked Rs 30/- at 15% discount and gives a shampoo sachet costing Rs 1.5 free with each soap. He then makes 20% profit. His cost price per soap is
Solution:
The net SP of Soap after discount = 0.85 × 30 = Rs 25.5.
Since he is giving a sachet of Shampoo fee, net price he his taking from customer = 25.5 – 1.5 = Rs 24.
Now the profit made = 20% We have 1.2 CP = SP
QUESTION: 25
Which one of the following is correct?
Solution:
This is the basic property of logarithms that Log(m × n) = log m + log n.
QUESTION: 26
The sum of two numbers is 15; their product is 30. The sum of their reciprocals is
Solution:
Let the numbers are x & y
We have x + y =15 ...................(1) & xy = 30 ....................(2)
QUESTION: 27
Three persons Amar, Akbar and Anthony agree to pay their hotel bills in the ratio of 3:4:5. Amar pays the first day's bill which amounts to Rs.26.85/-. Akbar pays the second day's bill which amounts to Rs.42.75/- and Anthony pays the third day's bill which amounts to Rs.53/-. When they settle their accounts, which of the following happens?
Solution:
The total amount paid by Amar, Akbar & Anthony = 26.85 + 42.75 + 53 = Rs 122.6
Since they had decided to pay the bill in 3:4:5, their actual payments should have been
Since Amar paid Rs., 3.8 less than what he should pay, so he will give 42.75 – 40.81 = Rs 1.95. to Akbar & 53 – 51 = Rs 2. to Anthony.
QUESTION: 28
Inspired by the 'Golden quadrilateral project', the UP Government recently accomplished a diamond triangular project. Under this project the State Government laid down 6 lane roads connecting three cities Ayodhya, Banaras and Chitrakoot, which are equally separated from each other i.e. in terms of geometry they form an equilateral triangle. Angad and Bajrang start simultaneously from Ayodhya and Banaras respectively, towards Chitrakoot. When Angad covers 100 kms, Bajrang covers such a distance that the distance between Angad and Bajrang makes 90° angle with the road joining Banaras to Chitrakoot. When Bajrang reaches Chitrakoot, Angad is still 150 km away from Chitrakoot.
What is the distance between Ayodhya and Chitrakoot?
Solution:
Let distance between Ayodhya and Chitrakoot = xkm
Let distance covered by Bajrang when Angad covered 100km = y = km
In ΔSTR Right Angled at T & Angle TSR = 300
As we know that speed & Distance. Hence distance covered by Angad & Bajrang is & Speed and their speed is constant hence.
x2 – 250x – 15000 = 0
(x – 300) (x + 50) = 0
Hence Distance of chitrakoot & Ayodhya = x = 300km.
QUESTION: 29
(1 - 1/2)(l-1/3)(l-l/4)(l-l/5) …………….(1-1/40) = X/40
What is the value of X?
Solution:
We have
QUESTION: 30
The smallest number which when divided by 12,16,18 gives 7 as remainder in each case, is
Solution:
The LCM of 12, 16, 18 is 144.
∴ the Smallest number which when divided by 12,16,18 gives remainder 7 in each case is 144 + 7 = 151
QUESTION: 31
f (a/b )2x-3 = (b/a)x-9, then x is equal to
Solution:
QUESTION: 32
The population of vultures in a locality decreases by a certain rate (compounded annually). If the current population of vultures 29160 and the ratio of the decrease in population for the first and second year and second and third year be 10:9, what was the population of vultures 3 years ago?
Solution:
The ratio of decrease of population for the first, second and third year is 10:9 and the current population is 29160. So the population three year back is
QUESTION: 33
Anna sold his car to Boney at a profit of 20% and Boney sold it to Chakori at a profit of 10%. Chakori sold it to a mechanic at a loss of 9.09%. The mechanic spent 10% of his purchase price on repairs and then sold it at a profit of 8.33% to Anna’s profit/ loss as %age of initial CP is-
Solution:
Let CP of Anna = Rs 100
So the CP of Boney = Rs 120
CP of Chakori = Rs. 132
Since Chakori sold it to mechanic at a loss of 9.09%
∴ The CP of Mechanic = Rs 120
Now mechanic spent 10% of purchase price on repairs
⇒ Net CP of mechanic = 120 +10 = Rs 132
Mechanic sold the car to Anna again at a profit of 8.33%
So, Anna lost 143 – 120 = Rs 23, which is 23% of his CP.
QUESTION: 34
A circle and a square have the same perimeter. Then
Solution:
Let the side of square is 1 unit, then its area is 1sq. unit and perimeter is 4 units. Since perimeters of square and circle are same
⇒ 2pr = 4
∴ Area of circle
∴ Area of circle in more than the area of square.
QUESTION: 35
Two taps A and B can fill a cistern in 15 hours and 10 hours respectively. Tap C can empty full cistern in 30 hours. All the three taps were opened for 2 hours, when it was remembered that the emptying tap had been left open, it was then closed.
How many hours more would it take for the cistern to be filled?
Solution:
Let the capacity of the tank in 30 letter
∴ Tap A can fill it at 2 liter/hr rate and Tap B can fill it at 3 liter/hr rate, where as Tap C can empty it at 1 liter/hr rate.
Now amount of water filled in tank in 2 hrs = 2×2 + 3 × 2 1 × 2 = 8 liter.
Remaining volume of tank to be filled = 30 - 8 = 22 liter.
Now this 22 letter will be filled by A & B working together in 22/5 = 4.4 hrs
QUESTION: 36
At Atharva Prakashan every book goes through 3 phases – typing, composing and binding. There are 16 typists, 10 composers and 15 binders. A typist can type 8 books in each hour, a composer can compose 12 books in each hour and a binder can bind 12 books in each hour. All the people at Atharva Prakashan work for 10 hours a day and each person is trained to do only one type of job.
How many books can be prepared in each day?
Solution:
Total man hours of typists, composers and binders in a day are 160 hrs, 100 hrs, & 150 hrs.
So the total books which can be typed in a day are 160 × 8 = 1280.
Total books that can be composed in day are 12 × 100 = 1200 and total books that can be bind in a day are 12 × 150 = 1800.
Since, the composers can compose a maximum of 1200 books, so this is the number of books which can be published in day.
QUESTION: 37
A contractor employed a certain number of workers to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realized that the work would delayed by three-fourth of the scheduled time. So, he at once doubled the number of workers and thus he managed to finish the road on the scheduled time. How much work had been completed before increasing the number of workers?
Solution:
Let initially x people were employed to do the work in y days and after z days there number was doubled. If they continue to do the work on their initial pace, the work would have done in So we have
Now let initially 10 people were employed to do the work in 4 days. Since so the number was doubled after one day. In the first day, 10 people did 10 man days work and then their number was doubled, so the 20 men did 60 man days work in the remaining 3 days. Hence the total work was 70 man days out of which 10 man days work was done before employing new people. Hence the required % age of work done before increasing the number of workers
QUESTION: 38
A train enters into a tunnel AB at A and exits at B. A jackal is sitting at O in another by pass tunnel AOB, which is connected to AB at A and B, where OA is perpendicular to OB. A cat is sitting at P inside the tunnel AB making the shortest possible distance between O and P, such that AO: PB = 30:32. A train before entering into the tunnel AB blows the whistle (or siren) somewhere before A. On hearing the whistle, if the jackal and the cat run towards A, they would meet with an accident with the train exactly at A, the entrance to the tunnel. Further if the jackal moves towards B instead of A (on hearing the whistle) it would again meet with an accident exactly at B, the exit of the tunnel, with the same train coming from the same direction.
Q. What is the ratio of speeds of the Jackal and the Cat?
Solution:
Since AO: PB = 30: 32 & ∠AOB is a right angle.
Let us take the triplet 30, 40, 50. i.e. OB = 40 km, AB = 50 km,
Now PB = 32 km ⇒ AP = 18 km,
Now the Cat and Jackal reach at A at same time.
So the ratio of their speed is equal to the ratio of the distances covered.
∴ Ratio of speed of Jackal & Cat is 30:18 i.e. 5: 3.
QUESTION: 39
A train enters into a tunnel AB at A and exits at B. A jackal is sitting at O in another by pass tunnel AOB, which is connected to AB at A and B, where OA is perpendicular to OB. A cat is sitting at P inside the tunnel AB making the shortest possible distance between O and P, such that AO: PB = 30:32. A train before entering into the tunnel AB blows the whistle (or siren) somewhere before A. On hearing the whistle, if the jackal and the cat run towards A, they would meet with an accident with the train exactly at A, the entrance to the tunnel. Further if the jackal moves towards B instead of A (on hearing the whistle) it would again meet with an accident exactly at B, the exit of the tunnel, with the same train coming from the same direction.
Q. The ratio of speeds of the Jackal to the train is:
Solution:
When the Train reaches at A, the Jackal reaches at A from O i.e. it covers a distance of 30km.
If it runs towards B, then let the Jackal is at C when the train reaches at point A s.t OC = 30 km.
Now Jackal will cover CB = 10 km and the train will cover 50 km and time taken by both is same. Hence the ratio of speeds is equal to the ratio of distance covered
∴ Ratio of speeds of Jackal & Train is 10: 50 i.e. 1: 5.
QUESTION: 40
A train enters into a tunnel AB at A and exits at B. A jackal is sitting at O in another by pass tunnel AOB, which is connected to AB at A and B, where OA is perpendicular to OB. A cat is sitting at P inside the tunnel AB making the shortest possible distance between O and P, such that AO: PB = 30:32. A train before entering into the tunnel AB blows the whistle (or siren) somewhere before A. On hearing the whistle, if the jackal and the cat run towards A, they would meet with an accident with the train exactly at A, the entrance to the tunnel. Further if the jackal moves towards B instead of A (on hearing the whistle) it would again meet with an accident exactly at B, the exit of the tunnel, with the same train coming from the same direction.
Q. If the tunnels were to be connected from O to P and the Jackal moves from O to A through P, it will meet with train at M. Then AM is:
Solution:
In this case the Jackal is running towards A through P.
Now Jackel will cover 30 km by the time train reaches at A.
In DOAP we have OA = 30 km, AP = 18 km ⇒ OP = 24km Let PN = 6km
⇒ The Jackal is at point N, when the train reaches at A. The ratio of speeds of train and Jakcal is 5:1.
Now the distance AN = 12km will be covered by train & Jackal in the ratio 5:1
QUESTION: 41
Choose the correct synonym of the word in capital letters from the four options given below.
ENTRENCHED
Solution:
Entrenched means to firmly establish or FORTIFIED
QUESTION: 42
Choose the correct synonym of the word in capital letters from the four options given below.
RECTITUDE
Solution:
Rectitude means morally correct behaviour or thinking or INTEGRITY
QUESTION: 43
Choose the correct synonym of the word in capital letters from the four options given below.
INVIDIOUS
Solution:
likely to arouse or incur resentment or anger in others or UNACCEPTABLE
QUESTION: 44
Choose the correct synonym of the word in capital letters from the four options given below.
Solution:
a person who is very knowledgeable and enthusiastic about an activity, subject, or pastime or ARDENT FAN
QUESTION: 45
Choose the correct alternative for the following sentence.
Q. Despite immense development in almost all the fields, humans still cannot control nature and to do so.
Solution:
Humans cannot control nature NOW and WILL NOT BE ABLE TO in the future too. Hence the best option is 3.
QUESTION: 46
DIRECTION : The sentence given below is jumbled and each underlined part is numbered. You are required to arrange it in the most logical and grammatically acceptable sequence.
Solution:
The correct sentence reads - Halton Brorough Council put up a notice to tell the public about its plans to move a path from one place to another.
QUESTION: 47
Choose the correct spelling:
Disgraphia is a deficiency in the ability to write, spell or put down thoughts on paper. The underlined word is spelled incorrectly. Which of the following is the correct spelling?
Solution:
inability to write coherently, as a symptom of brain disease or damage.
QUESTION: 48
Choose the correct spelling:
PANGERYRIC means "to pay tribute to". The underlined word is spelt incorrectly. Mark the correct spelling
Solution:
panegyric means a public speech or published text in praise of someone or something.
QUESTION: 49
Identify two most appropriate options from those given below that can complete the sentence without changing the meaning
Q. I _______________ scorpions, they are diabolic.
i. Hate
ii. Scare
iii. Loathe
iv. Wary
v. Cagey
Solution:
diabolic means relating to or characteristic of the Devil. Hence one would hate them or loathe them. One can be wary or cagey about something devilish. But these words do not fit with the construction of the sentence. …am wary of …….., am ……..cagey about………. Spelling of Loath changed to Loathe
QUESTION: 50
Identify two most appropriate options from those given below that can complete the sentence without changing the meaning
To fully understand (A) ____________ work such as George Orwell's Animal Farm, one must be able to differentiate (B) _____________ events of the plot from the abundant extended metaphors.
Solution:
To figure out what the missing words are, try to predict their definitions by using keywords from the prompt. In this prompt, each missing word has its own keywords. The first missing word’s keywords are at the end of the sentence, where the prompt gives the definition of the missing word as a work that involves “abundant extended metaphors.” The second missing word is contrary to “abundant extended metaphors” and describes the “plot” of the work. Thus, the first missing word describes a type of work that uses metaphors, and the second missing word implies the opposite of this figurative language. Therefore, the best choice is option 2. Allegorical means involving a figurative or metaphorical treatment of a literal subject, while literal means not figurative or not metaphorical. Parabolic means pertaining to a parable, a short work that conveys meaning in an indirect manner. Hence that option is also correct. Replaced parabolic by euphemistic.
QUESTION: 51
Find out the two words that are nearly the same or opposite in meaning and indicate the correct combination provided in the four options.
I. attenuate
II. repent
III. Make thin
IV. Force
Solution:
Attenuate and make thin are similar
QUESTION: 52
Choose the word nearest in meaning to the word underlined.
A rapier is a
Solution:
Rapier - a thin, light sharp-pointed sword used for thrusting. We see it being used by sports people in the sport – fencing
QUESTION: 53
Choose the correct alternatives for the following idioms.
Raise Cain
Solution:
Raise Cain means to make a lot of trouble; to raise hell.
QUESTION: 54
Choose the correct alternatives for the following idioms.
Go to seed
Solution:
To go to seed or to run to seed means to decline in looks, status, or utility due to lack of care. This old coat is going to seed. Have to get a new one. The front of the house is going to seed. Let's get it painted.
QUESTION: 55
(DIRECTIONS for questions 55 to 58:) Read the following passage and answer the questions that follow.
The next ingredient is a very remarkable one: Good Temper. "Love is not easily provoked". Nothing could be more striking than to find this here. We are inclined to look upon bad temper as a very harmless weakness. We speak of it as a mere infirmity of nature, a family failing, a matter of temperament, not a thing to take into very serious account in estimating a man's character. And yet here, right in the heart of this analysis of love, it finds a place; and the Bible again and again returns to condemn it as one of the most destructive elements in human nature. The peculiarity of ill temper is that it is the vice of the virtuous. It is often the one blot on an otherwise noble character. You know men who are all but perfect, and women who would be entirely perfect, but for an easily ruffled quick-temper or "touchy" disposition. This compatibility of ill temper with high moral character is one of the strongest and saddest problems of ethics. The truth is there are two great classes of sins—sins of the Body, and sins of disposition. The prodigal son may be taken as a type of the first, the Elder Brother of the second. Now society has no doubt whatever as to which of these is the worse. Its brand falls, without a challenge, upon the Prodigal. But are we right? We have no balance to weigh one another's sins, and coarser and finer are but human words; but faults in the higher nature may be less venial than those in the lower, and to the eye of Him who is love, a sin against love may seem a hundred times more base. No form of vice, not worldliness, not greed of gold , not drunkenness itself does more to un-christianise society than evil temper. For embittering life, for breaking up communities, for destroying the most sacred relationships, for devasting homes, for withering up men and women, for taking the bloom off childhood; in short for sheer gratuituos misery-producing power, this influence stands alone. Jealousy, anger, pride, uncharity, cruelty, selfrighteousness, touchiness, doggedness, sullenness-in varying properties these are the ingredients of all ill temper. Judge if such sins of disposition are not worse to live in, and for others to live with than sins of the body. There is really no place in Heaven for a disposition like this. A man with such a mood could only make Heaven miserable for all the people in it.
Solution:
QUESTION: 56
The next ingredient is a very remarkable one: Good Temper. "Love is not easily provoked". Nothing could be more striking than to find this here. We are inclined to look upon bad temper as a very harmless weakness. We speak of it as a mere infirmity of nature, a family failing, a matter of temperament, not a thing to take into very serious account in estimating a man's character. And yet here, right in the heart of this analysis of love, it finds a place; and the Bible again and again returns to condemn it as one of the most destructive elements in human nature. The peculiarity of ill temper is that it is the vice of the virtuous. It is often the one blot on an otherwise noble character. You know men who are all but perfect, and women who would be entirely perfect, but for an easily ruffled quick-temper or "touchy" disposition. This compatibility of ill temper with high moral character is one of the strongest and saddest problems of ethics. The truth is there are two great classes of sins—sins of the Body, and sins of disposition. The prodigal son may be taken as a type of the first, the Elder Brother of the second. Now society has no doubt whatever as to which of these is the worse. Its brand falls, without a challenge, upon the Prodigal. But are we right? We have no balance to weigh one another's sins, and coarser and finer are but human words; but faults in the higher nature may be less venial than those in the lower, and to the eye of Him who is love, a sin against love may seem a hundred times more base. No form of vice, not worldliness, not greed of gold , not drunkenness itself does more to un-christianise society than evil temper. For embittering life, for breaking up communities, for destroying the most sacred relationships, for devasting homes, for withering up men and women, for taking the bloom off childhood; in short for sheer gratuituos misery-producing power, this influence stands alone. Jealousy, anger, pride, uncharity, cruelty, selfrighteousness, touchiness, doggedness, sullenness-in varying properties these are the ingredients of all ill temper. Judge if such sins of disposition are not worse to live in, and for others to live with than sins of the body. There is really no place in Heaven for a disposition like this. A man with such a mood could only make Heaven miserable for all the people in it.
Q. What does the word "venial" mean?
Solution:
QUESTION: 57
The next ingredient is a very remarkable one: Good Temper. "Love is not easily provoked". Nothing could be more striking than to find this here. We are inclined to look upon bad temper as a very harmless weakness. We speak of it as a mere infirmity of nature, a family failing, a matter of temperament, not a thing to take into very serious account in estimating a man's character. And yet here, right in the heart of this analysis of love, it finds a place; and the Bible again and again returns to condemn it as one of the most destructive elements in human nature. The peculiarity of ill temper is that it is the vice of the virtuous. It is often the one blot on an otherwise noble character. You know men who are all but perfect, and women who would be entirely perfect, but for an easily ruffled quick-temper or "touchy" disposition. This compatibility of ill temper with high moral character is one of the strongest and saddest problems of ethics. The truth is there are two great classes of sins—sins of the Body, and sins of disposition. The prodigal son may be taken as a type of the first, the Elder Brother of the second. Now society has no doubt whatever as to which of these is the worse. Its brand falls, without a challenge, upon the Prodigal. But are we right? We have no balance to weigh one another's sins, and coarser and finer are but human words; but faults in the higher nature may be less venial than those in the lower, and to the eye of Him who is love, a sin against love may seem a hundred times more base. No form of vice, not worldliness, not greed of gold , not drunkenness itself does more to un-christianise society than evil temper. For embittering life, for breaking up communities, for destroying the most sacred relationships, for devasting homes, for withering up men and women, for taking the bloom off childhood; in short for sheer gratuituos misery-producing power, this influence stands alone. Jealousy, anger, pride, uncharity, cruelty, selfrighteousness, touchiness, doggedness, sullenness-in varying properties these are the ingredients of all ill temper. Judge if such sins of disposition are not worse to live in, and for others to live with than sins of the body. There is really no place in Heaven for a disposition like this. A man with such a mood could only make Heaven miserable for all the people in it.
Q. The Bible condemns which of the following as one of the most destructive elements in human nature?
Solution:
QUESTION: 58
The next ingredient is a very remarkable one: Good Temper. "Love is not easily provoked". Nothing could be more striking than to find this here. We are inclined to look upon bad temper as a very harmless weakness. We speak of it as a mere infirmity of nature, a family failing, a matter of temperament, not a thing to take into very serious account in estimating a man's character. And yet here, right in the heart of this analysis of love, it finds a place; and the Bible again and again returns to condemn it as one of the most destructive elements in human nature. The peculiarity of ill temper is that it is the vice of the virtuous. It is often the one blot on an otherwise noble character. You know men who are all but perfect, and women who would be entirely perfect, but for an easily ruffled quick-temper or "touchy" disposition. This compatibility of ill temper with high moral character is one of the strongest and saddest problems of ethics. The truth is there are two great classes of sins—sins of the Body, and sins of disposition. The prodigal son may be taken as a type of the first, the Elder Brother of the second. Now society has no doubt whatever as to which of these is the worse. Its brand falls, without a challenge, upon the Prodigal. But are we right? We have no balance to weigh one another's sins, and coarser and finer are but human words; but faults in the higher nature may be less venial than those in the lower, and to the eye of Him who is love, a sin against love may seem a hundred times more base. No form of vice, not worldliness, not greed of gold , not drunkenness itself does more to un-christianise society than evil temper. For embittering life, for breaking up communities, for destroying the most sacred relationships, for devasting homes, for withering up men and women, for taking the bloom off childhood; in short for sheer gratuituos misery-producing power, this influence stands alone. Jealousy, anger, pride, uncharity, cruelty, selfrighteousness, touchiness, doggedness, sullenness-in varying properties these are the ingredients of all ill temper. Judge if such sins of disposition are not worse to live in, and for others to live with than sins of the body. There is really no place in Heaven for a disposition like this. A man with such a mood could only make Heaven miserable for all the people in it.
Q. Find the exact word in the passage which means easily or quickly offended.
Solution:
QUESTION: 59
Fill in the blank with the correct option.
"If you are morose." you feel ______________
Solution:
morose - glowering, moody, sullen, saturnine, glum, dour, dark, sour
QUESTION: 60
Name the part of speech of the underlined word in the following sentences.
Q. We scored as many goals as them.
Solution:
Them is an object pronoun
QUESTION: 61
Name the part of speech of the underlined word in the following sentences.
Q. There is more evidence yet to be offered
Solution:
Yet is an adverb of time.
QUESTION: 62
Identify a synonym to replace the underlined word (s).
Q. The baby is in the carrying basket.
Solution:
A perambulator or commonly known as pram is one in which you keep a baby and move him around. A bassinet is an oblong basketlike bed for an infant.
QUESTION: 63
Which of the following does not denote the underlined word(s)?
Q. The festival of Christmas.
Solution:
all others are Christmas or related to it.
Halloween - the night of 31 October, the eve of All Saints' Day, often celebrated by children dressing up in frightening masks and costumes. Halloween is thought to be associated with the Celtic festival Samhain, when ghosts and spirits were believed to be abroad.
Yule or Yuletide ("Yule time") is a pagan religious festival observed by the historical Germanic peoples, later being absorbed into and equated with the Christian festival of Christmas.
Crimbo is a slang word for Christmas. Noel Christmas, especially as a refrain in carols and on Christmas cards
QUESTION: 64
Choose the correct option to fill in the blank with the correct grammatical usage.
He has his finger ____________ the pulse of the nation.
Solution:
To have a finger on the pulse of something - knowledge of what is happening now in a particular area. They've got their finger on the pulse of popular culture in Latvia.
QUESTION: 65
Identify the error in the following sentence by choosing from the given alternatives "Where do you live" asked the stranger?
Solution:
The quotation mark which ends a sentence has to be within the inverted commas.
QUESTION: 66
In each of the following questions, the part of the sentence which is in capitals and has been underlined may have an error. Correct the error.
Q. Unless he DOES NOT GIVE UP smoking, we cannot be sure of any improvement in his health
Solution:
Unless and does not together change the meaning of the sentence. What we want to say is that if he does not give up smoking then there will be no improvement in his health.
QUESTION: 67
A recently carried out meta-analysis of two decades of published research DOES NOT SUGGEST THAT THERE SHOULD BE AN ASSOCIATION BETWEEN COFFEE DRINKING AND CORONARY AILMENTS.
Solution:
A report would suggest something happens or something doesn’t happen. Hence option 1 is the best construction.
QUESTION: 68
In the following sentences, the order of the sentences has been jumbled up. Make the best choice to find out the correct sequence so as to form a meaningful paragraph
A. The following represents a condensed statement of what we think we now know about the relationship between education and employment
B. Although the linkages between education and employment are complex, and in the past, often analysed with simplistic notions of causality, recent research results have yielded new insights about the nature of these linkages.
C. In the interests of brevity, the argument is put forward as a series of major propositions and derivatives strategies relating to the education employment nexus.
D. Intensive research efforts are currently being supported through the developing world both by national governments and international donor agencies in the hope of improving, understanding of the nature and causes of rising unemployment.
E. The massive problems of widespread and chronic unemployment and underemployment in less developed nations will remain among the most serious challenges to the development policy during the next several decades.
Solution:
The passage is about the relationship between education and employment.
In order to understand the nature and causes of rising unemployment, research is being conducted, and this research have provided some insights. Hence D and then B.
We know that A and C will be together. As there are no derivatives strategies being talked about, C cannot be linked to any other statement in the paragraph. Thus it works best as the last sentence of the paragraph.
QUESTION: 69
Given below are four sentences, each of which may or may not have an error. Following them are four options. Select the option indicating the error free statement (s).
A. Fruits from Madhavs' garden were stolen.
B. Fruit's are sold here
C. All the students of the students' council agreed to raise funds.
D. The dress's texture was coarse.
Solution:
The corrections are outlined below:
B – Fruits are.
D – dress’ texture..
QUESTION: 70
Choose the odd word out
Solution:
Esoteric means mystical, mysterious or occult. Exigent means urgent.
QUESTION: 71
Choose the word or phrase that best completes the sentence:
The rebels sought to overcome the _____________ strength of the police forces by engaging in ____________ tactics
Solution:
As the rebels sought to overcome the ….. strength, they would use tactics which are not open or are more of a surprise. Hence preponderant (meaning superior in weight, force, influence, numbers, etc.) and guerilla fit in well with the meaning of the sentence.
A guerilla is a member of an irregular usually politically motivated armed force that combats stronger regular forces, such as the army or police.
QUESTION: 72
(DIRECTIONS for questions 72 to 74:) Read the following passage and answer the questions that follow.
My new mistress proved to be all she appeared when I first met her at the door- a woman of the kindest heart and feelings. She had never had a slave under her control previously and prior to her marriage she had been dependent upon her own industry for a living. She was by trade a weaver, and by constant application to her business, she had been in a good degree preserved from the lighting and dehumanizing effects of slavery. I was utterly astonished at her goodness. I scarcely knew how to behave towards her. My early instruction was all out of place. The crouching servility, usually so acceptable a quality in a slave, did not answer when manifested toward her. Her favor was not gained by it; she seemed to be disturbed by it. She did not deem it to be impudent or unmannerly for a slave to look in the face. The meanest slave was put fully at ease in her presence, and none left without feeling better for having seen her. But alas! This kind heart had but a short time to remain such. The fatal poison of irresponsible power was already in her hands, and soon commenced its internal work.
Very soon I went to live with Mr. and Mrs. Auld; she very kindly commenced to teach me the A, B, C. After I had learnt this, she assisted me in learning to spell words of three and four letters. Just at this point of my progress, Mr. Auld found out what was going on , and at once forbade Mrs. Auld to instruct me further, telling her that it was unlawful, as well as unsafe, to teach a slave to read. Further he said, "If you give a slave an inch, he will take a mile. A slave should know nothing but to obey his master - to do as he is told to do. Learning would spoil the best slave in this world. "Now," said he, "If you teach that boy (speaking of myself) how to read, there would be no keeping him. It would forever unfit him to be a slave. He would at once become unmanageable, and of no value to his master. As to him, it could do him no good, but a great deal of harm. It would make him discontented and unhappy. "These words sank deep into my heart, stirred up sentiments within that lay slumbering, and called into existence an entirely new train of thought. I now understood what had been to me a most perplexing difficulty - the white man's power to enslave the black man. From that moment, I understood that pathway from slavery to freedom. Though conscious of the difficulty of learning without a teacher, I set out with high hope, and a fixed purpose, at whatever cost of trouble, to learn how to read. The very decided manner with which my master spoke, and strove to impress his wife with the evil consequences of giving me instruction, served to convince me that he was deeply sensible of the truths he was uttering. It gave me the best assurance that I might rely with the utmost confidence on the results which, he said, would flow from teaching me to read. What he most dreaded, that I most desired. What he most loved, that I most hated. That which to him was great evil, to be carefully shunned, was to me a great good, to be diligently sought; and the argument which he so warmly urged, against my learning to read, only served to inspire me with a desire and determination to learn. In learning to read, I owe almost as much to the bitter opposition of my master, as to the kindly aid of my mistress. I acknowledge the benefit of both.
Q. The author's main purpose in this passage is to
Solution:
The author wants to put forth that the words which he was taught by his mistress helped him to understand that education was what would lead him to freedom.
QUESTION: 73
My new mistress proved to be all she appeared when I first met her at the door- a woman of the kindest heart and feelings. She had never had a slave under her control previously and prior to her marriage she had been dependent upon her own industry for a living. She was by trade a weaver, and by constant application to her business, she had been in a good degree preserved from the lighting and dehumanizing effects of slavery. I was utterly astonished at her goodness. I scarcely knew how to behave towards her. My early instruction was all out of place. The crouching servility, usually so acceptable a quality in a slave, did not answer when manifested toward her. Her favor was not gained by it; she seemed to be disturbed by it. She did not deem it to be impudent or unmannerly for a slave to look in the face. The meanest slave was put fully at ease in her presence, and none left without feeling better for having seen her. But alas! This kind heart had but a short time to remain such. The fatal poison of irresponsible power was already in her hands, and soon commenced its internal work.
Very soon I went to live with Mr. and Mrs. Auld; she very kindly commenced to teach me the A, B, C. After I had learnt this, she assisted me in learning to spell words of three and four letters. Just at this point of my progress, Mr. Auld found out what was going on , and at once forbade Mrs. Auld to instruct me further, telling her that it was unlawful, as well as unsafe, to teach a slave to read. Further he said, "If you give a slave an inch, he will take a mile. A slave should know nothing but to obey his master - to do as he is told to do. Learning would spoil the best slave in this world. "Now," said he, "If you teach that boy (speaking of myself) how to read, there would be no keeping him. It would forever unfit him to be a slave. He would at once become unmanageable, and of no value to his master. As to him, it could do him no good, but a great deal of harm. It would make him discontented and unhappy. "These words sank deep into my heart, stirred up sentiments within that lay slumbering, and called into existence an entirely new train of thought. I now understood what had been to me a most perplexing difficulty - the white man's power to enslave the black man. From that moment, I understood that pathway from slavery to freedom. Though conscious of the difficulty of learning without a teacher, I set out with high hope, and a fixed purpose, at whatever cost of trouble, to learn how to read. The very decided manner with which my master spoke, and strove to impress his wife with the evil consequences of giving me instruction, served to convince me that he was deeply sensible of the truths he was uttering. It gave me the best assurance that I might rely with the utmost confidence on the results which, he said, would flow from teaching me to read. What he most dreaded, that I most desired. What he most loved, that I most hated. That which to him was great evil, to be carefully shunned, was to me a great good, to be diligently sought; and the argument which he so warmly urged, against my learning to read, only served to inspire me with a desire and determination to learn. In learning to read, I owe almost as much to the bitter opposition of my master, as to the kindly aid of my mistress. I acknowledge the benefit of both.
Q. For which of the following reasons does Mr. Auld forbid his wife to educate the slave?
A. Providing slaves with an education violates the law
B. He believes slaves lack the capacity for education
C. He fears education would leave the slave less submissive
Solution:
QUESTION: 74
My new mistress proved to be all she appeared when I first met her at the door- a woman of the kindest heart and feelings. She had never had a slave under her control previously and prior to her marriage she had been dependent upon her own industry for a living. She was by trade a weaver, and by constant application to her business, she had been in a good degree preserved from the lighting and dehumanizing effects of slavery. I was utterly astonished at her goodness. I scarcely knew how to behave towards her. My early instruction was all out of place. The crouching servility, usually so acceptable a quality in a slave, did not answer when manifested toward her. Her favor was not gained by it; she seemed to be disturbed by it. She did not deem it to be impudent or unmannerly for a slave to look in the face. The meanest slave was put fully at ease in her presence, and none left without feeling better for having seen her. But alas! This kind heart had but a short time to remain such. The fatal poison of irresponsible power was already in her hands, and soon commenced its internal work.
Very soon I went to live with Mr. and Mrs. Auld; she very kindly commenced to teach me the A, B, C. After I had learnt this, she assisted me in learning to spell words of three and four letters. Just at this point of my progress, Mr. Auld found out what was going on , and at once forbade Mrs. Auld to instruct me further, telling her that it was unlawful, as well as unsafe, to teach a slave to read. Further he said, "If you give a slave an inch, he will take a mile. A slave should know nothing but to obey his master - to do as he is told to do. Learning would spoil the best slave in this world. "Now," said he, "If you teach that boy (speaking of myself) how to read, there would be no keeping him. It would forever unfit him to be a slave. He would at once become unmanageable, and of no value to his master. As to him, it could do him no good, but a great deal of harm. It would make him discontented and unhappy. "These words sank deep into my heart, stirred up sentiments within that lay slumbering, and called into existence an entirely new train of thought. I now understood what had been to me a most perplexing difficulty - the white man's power to enslave the black man. From that moment, I understood that pathway from slavery to freedom. Though conscious of the difficulty of learning without a teacher, I set out with high hope, and a fixed purpose, at whatever cost of trouble, to learn how to read. The very decided manner with which my master spoke, and strove to impress his wife with the evil consequences of giving me instruction, served to convince me that he was deeply sensible of the truths he was uttering. It gave me the best assurance that I might rely with the utmost confidence on the results which, he said, would flow from teaching me to read. What he most dreaded, that I most desired. What he most loved, that I most hated. That which to him was great evil, to be carefully shunned, was to me a great good, to be diligently sought; and the argument which he so warmly urged, against my learning to read, only served to inspire me with a desire and determination to learn. In learning to read, I owe almost as much to the bitter opposition of my master, as to the kindly aid of my mistress. I acknowledge the benefit of both.
Q. The tone of author in acknowledging his debt to his master can be best described as
Solution:
Option 1. The author remembers the good time – when his mistress teaches him to read - and also acknowledges the fact that though what his master did was incorrect it ultimately led to something positive – he was stimulated to read and learn. (Sentimental means expressive of or appealing to sentiment, especially the tender emotions and feelings, as love, pity, or nostalgia. Nostalgic means a wistful desire to return in thought or in fact to a former time in one's life, to one's home or homeland, or to one's family and friends; a sentimental yearning for the happiness of a former place or time.)
Option 4 : As it is a story which talks of triumph, we cannot say that the tone is resigned - means submissive or acquiescent. Wistful means expressing or revealing thoughtfulness, usually marked by some sadness.
Option 3. He is not being moralistic, so self- righteous which means confident of one's own righteousness, especially when smugly moralistic and intolerant of the opinions and behavior of others, cannot be the answer. Petulant means moved to or showing sudden, impatient irritation, especially over some trifling annoyance
Option 2. It is not ironic because he doesn’t think something and it turns out to be something else. Moreover he is actually happy that the master stopped his learning because that later motivated him to work on his own. Cutting means wounding the feelings severely; sarcastic. Ironic means using words to convey a meaning that is the opposite of its literal meaning.
QUESTION: 75
Replace the underlined phrase with the one which is closest to it in meaning.
I DROPPED A CLANGER when I mentioned her ex- husband.
Solution:
a clanger is a bad and embarrassing mistake.
QUESTION: 76
Fill in the blanks in the sentences with the appropriate choices.
I should not have ____________ to talk in such a ____________ strain especially when I had not studied the man to whom I was talking.
Solution:
Venture to is the correct usage. The others can’t fit in the first blank. Peremptory means Offensively selfassured; dictatorial.
QUESTION: 77
Fill in the blanks in the sentences with the appropriate choices.
The ______________ of Agatha Christie's argument is that human nature remains the same, wherever you are.
Solution:
The sentence states that her argument is in short this. – crux fits is best – crux means the gist.
QUESTION: 78
Fill in the blanks in the sentences with the appropriate choices.
The _____________ play caused me to squirm in my seat, but she began to ______________ her eyes in a way that irritated me.
Solution:
The only word that fits the second blank is daub. Maudlin means self pitying which fits the first blank too.
QUESTION: 79
There they stood on the top shelf, all 20 volumes, their red binding and gold lettering undoubtedly making them the most attractive between all the books on the bookshelf
Solution:
QUESTION: 80
Everything in the world decays, it comes to an end and also dies inevitably.
Solution:
Option 3 is incorrect because also and inevitably cannot be used together. After the comma, we cannot have a complete sentence. Hence option 1 and option 4 are incorrect.
QUESTION: 81
What will be the measurement of the angle made by the hands of a clock when the time is 8.35?
Solution:
Angle made by clock at 8:35
We know that Angle at 8’o clock is 2400 the relative speed of minute hand & hour hand
Hence in 35min → the Angle covered by minute hand relative to hour hand
Angle b/w hour hand & minute will be = 2400 – 192.50 = 47.50
QUESTION: 82
Find the odd man out.
Solution:
QUESTION: 83
Raghu is at point A. He walks 3 km to the north and then turns to his left. He walks 4 km in this direction. He turns left again and walks 6 km. If he wishes to reach point A again, in which direction should he be walking and what distance will he have to cover?
Solution:
As per the question the shortest distance between point
And the direction of point A from the point B is 5km, North east which is option 3.
QUESTION: 84
(DIRECTIONS for questions 84 and 85:) Read the following information carefully and answer the questions.
Six students A, B, C, D, E and F participated in a dancing competition wherein they won prizes 12000, 10000, 8000, 6000, 4000, 2000 according to the position secured. The following information is known to us: 1. A won less money than B 2. The differences between the winning of C and F was Rs. 2000 3. The differences between the winning of D and F was at least Rs. 4000 4. E won the Rs. 8000 prize.
Q. Which of the following could be the ranking from first place to sixth places of students?
Solution:
The possible ranking could be:- using options.
1. As per statement 1. A won less than B, hence ADEBFC is wrong.
2. BAECFD → is wrong because the minimum gap b/w F & D is Rs. 4000, which is not as per question
3. FBEACD → is wrong because the gap of winning b/w C & F is 2000, which is not as per the question
4. option 4 is right as it satisfy all the given criteria
QUESTION: 85
Six students A, B, C, D, E and F participated in a dancing competition wherein they won prizes 12000, 10000, 8000, 6000, 4000, 2000 according to the position secured. The following information is known to us: 1. A won less money than B 2. The differences between the winning of C and F was Rs. 2000 3. The differences between the winning of D and F was at least Rs. 4000 4. E won the Rs. 8000 prize.
Q. If A won Rs. 4000, how much in total did C and F win?
Solution:
As the difference b/w C & F in amount winning is Rs. 2000, hence the only possible case is when amount won be C & F is 12k & 10k. hence sum of Amount won by C & F = 22000. Which is option 3.
QUESTION: 86
Where should the number 17 be placed to fit into the sequence?
18, 11, 15, 14, 19, 16, 20, 12
Solution:
The correct placement is between 19 × 16 So that then will be 2 sequence 1. -3 & +4
2. +3 as shown in figure
QUESTION: 87
There are three facts and you are given three more statements. You have to determine which of these, if any is also a fact.
Fact 1. All kids like to play.
Fact 2. Some kids like ice cream.
Fact 3. Some kids look like their mothers.
Statement 1. All kids who like ice cream look like their mothers.
Statement 2. Kids who like ice cream also like to play.
Statement 3. Kids who like to play do not look like their mothers.
Solution:
To minimize the intersection; only statement 2 is a fact i.e. kids who like ice cream also like to play.
QUESTION: 88
Insert the mis sing letter
Solution:
H – A = 6 ; T – M = 6
M – F = 6 ; T – R = 6
R – K = 6 ; O – H = 6
Hence answer is option 1 i.e. 0.
QUESTION: 89
Six statements are followed by options consisting of three statements put together in a specific order. Choose the option which indicates a valid argument, that is, the third argument is a conclusion drawn from the earlier two statements.
A. Amar is unhappy
B. Amar is honest
C. Some magicians are honest
D. No honest man is a magician
E. No magician is happy
F. Amar is not a magician
Solution:
As all option contains F i.e Amar is not a magician Hence BDF is the right pair
QUESTION: 90
DIRECTIONS for questions 90 and 91: Read the information given below and answer the question.
The table below presents data on Gross National Product (GNP) and Net National Product (NNP) for India for six years. Answer questions based on these data.
Gross National Product and Net National Product
Q. The year during which both GNP and NNP have shown almost equal rate of growth:
Solution:
Hence the answer is option 1989 – 90
QUESTION: 91
The table below presents data on Gross National Product (GNP) and Net National Product (NNP) for India for six years. Answer questions based on these data.
Gross National Product and Net National Product
Q. Which of the following statements could be stated to be true:
Solution:
Visually we can compare that The growth pattern has been different for both indicators during period under review , hence answer is option 3
QUESTION: 92
If PEACE is coded as RGCEG then how is MICKY coded in that code?
Solution:
PEACE → RGC EG
R → P = +2 ; E → C = 2
G → E = +2 ; G → E = 2
C → A = +2
Hence
MICKY → OKEMA
Which is option 2
QUESTION: 93
A set of symbols is given indicating the terms for which they will be used in the questions following them.
Θ = equal to, φ = not equal to, Δ = greater than, x= less than, + = not greater than, \$ = not less than A Δ B Δ C does not imply which of the following options?
Solution:
As per question A > B > C ; which is not true?
Hence A < B = C is not correct. Hence answer is option 3.
QUESTION: 94
Given below are 4 pictures of a cube
Which number is on the face opposite to 3?
Solution:
From Figure 1,3 and 4 we observed that number adjacent to 4 are 6,5,1 and 2 so 2 lies opposite to 4.
Hence the correct answer is option D.
QUESTION: 95
DIRECTIONS for questions 95 and 96: Read the information given below and answer the questions.
Rectangle indicates Faculty members, Square indicates Research scholars, Circle indicates Indians and Triangle indicates Americans.
Q. In the given figure show the area which indicates faculty members who are research scholars but are neither Americans nor Indians
Solution:
By visualisig the Area which indicate faculty members who are research scholars but one neither Americans nor Indians = k i.e. option 3.
QUESTION: 96
Rectangle indicates Faculty members, Square indicates Research scholars, Circle indicates Indians and Triangle indicates Americans.
Q. In the given figure show the area which indicates Indian Americans who are faculty members but not research scholars.
Solution:
By visualising “C” is the Area which indicate Indian Americans who are faculty members but not research scholars. i.e. option 3.
QUESTION: 97
While facing East you turn to your left and walk 10 yards. Then turn to your left and walk 10 yards, and now turn 45 ° to your right and go straight to cover 50 yards. Now in what direction are you with respect to the starting point?
Solution:
The correct direction is North West. Hence answer is option 4.
QUESTION: 98
If P - Q means Q is son of P, P × Q means P is brother of Q, P ÷ Q means Q is sister of P and P + Q means P is mother of Q , which of the following is definitely true about N×K-M÷L?
Solution:
N × K – M ÷ L
M ÷ L M is sister of L
N × K → N is Brother of K.
K – M → M is son of K
QUESTION: 99
DIRECTIONS for questions 99 and 100: Read the information given below and answer the question.
Five girls Rama, Sudha, Tara, Uma and Veena share an apartment and have distributed the task as one girl making breakfast per day, Monday through Friday, one of the five dishes upama, dosa, idli, uttapa and paratha.
Veena does not make uttapa and does not cook on Tuesday.
Sudha makes parathas but not on Monday or Friday.
Tara makes her dish which is not uttapa on Wednesday.
Idli is made on Friday but not by Uma.
Rama cooks on Monday
Q. What does Tara cook on Wednesday?
Solution:
Taracook dosa on Wednesday option 4
QUESTION: 100
Five girls Rama, Sudha, Tara, Uma and Veena share an apartment and have distributed the task as one girl making breakfast per day, Monday through Friday, one of the five dishes upama, dosa, idli, uttapa and paratha.
Veena does not make uttapa and does not cook on Tuesday.
Sudha makes parathas but not on Monday or Friday.
Tara makes her dish which is not uttapa on Wednesday.
Idli is made on Friday but not by Uma.
Rama cooks on Monday
Q. What day does Uma prepare her dish?
Solution:
Uma pnpares upama on Thursday option 3.
QUESTION: 101
Answer the following question on the basis of the following data.
Kunal, Rahul, Mohit and Charan are constables on duty in Delhi, and they are trying to recollect who amongst them has been working the most number of days without leave. Here are the facts: Kunal has put in more days of consecutive duty than has Rahul, who has got in five days of consecutive duty. Mohit has gone fifteen days straight without leave, more than Kunal and Rahul combined. Charan has put in eight consecutive days of duty, which is less than what Kunal has done.
Q. How many days has Kunal gone without leave?
Solution:
→ Kunal + Rahul < 14
→ Kunal > Charan & Rahul = 5
Hence kunal = 9
QUESTION: 102
DIRECTIONS for questions 102 to 104: Read the information given below and answer the questions.
Five hundred school children were asked about cricketers they liked. 80 said they like Sachin Tendulkar, 60 said they liked Sourav Ganguly, 90 said they liked Mahendra Singh Dhoni. 10 said they liked all three. 40 said they liked Sachin and Sourav, 50 liked Sachin and Dhoni, while 30 said they liked Sourav and Dhoni.
Q. How many, out of the 500, didn't choose Sachin, Sourav or Dhoni?
Solution:
Total number of student don’t choose Sachin, Sourav or Dhoni
Number of student choose at least 1 = (80+90+60 – 40 – 50 – 30 + 10) = 120
Number of student + didn’t choose anyone = 500 – 120 = 380
QUESTION: 103
Five hundred school children were asked about cricketers they liked. 80 said they like Sachin Tendulkar, 60 said they liked Sourav Ganguly, 90 said they liked Mahendra Singh Dhoni. 10 said they liked all three. 40 said they liked Sachin and Sourav, 50 liked Sachin and Dhoni, while 30 said they liked Sourav and Dhoni.
Q. How many like Sourav but not Tendulkar?
Solution:
How many like sourav but not tendulkar. ⇒ By figure the correct answer is option 4. i.e. none of these = 20
QUESTION: 104
Five hundred school children were asked about cricketers they liked. 80 said they like Sachin Tendulkar, 60 said they liked Sourav Ganguly, 90 said they liked Mahendra Singh Dhoni. 10 said they liked all three. 40 said they liked Sachin and Sourav, 50 liked Sachin and Dhoni, while 30 said they liked Sourav and Dhoni.
Q. How many like fewer than two cricketers?
Solution:
How many like fewer than two cricketers = 500 – (Exactly 2 + Exactly 3)
= 500 – (100) = 400
QUESTION: 105
If 7 is coded as CBRT343, then 9 is
Solution:
7 → CB RT 343 coded as cuberoot 343 = 7
Hence cuberoot 729 = CBRT 729 = 9. Hence answer is option 3.
QUESTION: 106
If HELLO is coded as 15|12|12|5|8, then WHERE is
Solution:
⇒ reverse of alphabnet position 15 12 12 5 8
Hence is coded as:-
⇒ 5 18 5 8 2 3 i.e. option 2.
QUESTION: 107
Complete the following series 3 10 59
Solution:
10 – 3 = 7 = 71
59 – 10 = 49 = 72
73 + 59 = 343 + 59 = 402.
Which is option 1.
QUESTION: 108
If WOOD is coded as 23|225|4 then MEET is
Solution:
WOOD → 23 | 225 | 4
W = 23
O = 15
O = 15
D = 4
Hence 23 | 15 × 15 | 4.
Hence = MEET i.e.
= 13 | 5 × 5 | 20
= 13 | 25 | 20 i.e. option 3.
QUESTION: 109
DIRECTIONS for questions 109 and 110: The following table gives data on Sectoral Investment during the Eight plan. Read the data and answer the questions.
Q. Which of the following sectors has shown a steady increase in investments during last three plan periods?
Solution:
In this question we have to find the sector in which % composition is increasing:- The only sector which is increasing 6th & 7th plan is communication. Hence Ans. is option 2.
QUESTION: 110
The following table gives data on Sectoral Investment during the Eight plan. Read the data and answer the questions.
Q. During the eight plan period, in which sector the percentage share of Public Sector investment has been the highest?
Solution:
In 8th plan period → the % share in public sector investment has been highest :- will be that sector which have highest figure in the public sector column i.e. Electricity, Gas and water. Hence answer is option 2.
QUESTION: 111
The economically diverse region of MENA nations refers to the countries in
Solution:
QUESTION: 112
Open market operations refers to the sale and purchase by the RBI of
Solution:
QUESTION: 113
The agency that estimates national income in India is
Solution:
QUESTION: 114
Panchatantra was written by
Solution:
QUESTION: 115
IRDA is the Indian regulator for the
Solution:
QUESTION: 116
The Speaker of the Lok Sabha is elected by:
Solution:
QUESTION: 117
The Finance Minister's Budget speech presented to the Indian parliament every year usually has two parts. What does the Part A relate to:
Solution:
The Finance Minister's Budget speech presented to the Indian parliament every year usually has two parts. What does the Part A relate to:
QUESTION: 118
Which two institutions are popularly referred to as the 'Bretton Woods Twins?
Solution:
QUESTION: 119
Based on the statements given below, Special Drawing Rights is best described to include which of the following statements
i. It is a basket of currencies reviewed every 5 years
ii. It was created to reduce the dependencies on gold and US dollar
iii. It is a supplementary foreign exchange reserve assets maintained by IMF
Solution:
QUESTION: 120
Which NATO country's parliament was the first to vote in favour of the move to make military service compulsory for women
Solution:
QUESTION: 121
The market for government securities is also referred to as
Solution:
QUESTION: 122
From April 2010 the savings bank interest is calculated
Solution:
From April 1, 2010interest on all savings bank account deposits is being calculated on a daily basis, thereby earning account holdres higher interest income. This is due to the fact that the Reserve Bank of India has instructed banks to change the mechanism of interest income calculation.
QUESTION: 123
Financial Action Task Force (FATF) is an Inter-governmental Body established in 1989. It is concerned with
Solution:
QUESTION: 124
Which of the following sectors in India was not nationalised by the government at any point of time
Solution:
QUESTION: 125
Which of the following is not a Palindrome
Solution:
QUESTION: 126
The movie 'Life of Pi' is based on an award winning novel written by
Solution:
QUESTION: 127
LIDL, Walgreen, ASDA and Kroger are
Solution:
QUESTION: 128
Which bank was the first to introduce the Automated Teller Machine (ATM) in India
Solution:
QUESTION: 129
The Three Gorges Dam, is the world's largest hydropower project and is located in
Solution:
QUESTION: 130
India signed the Kyoto Protocol in the year:
Solution:
QUESTION: 131
Given below are the creators of some well-known cartoon characters. Identify the creator - cartoon pair that does not match
Solution:
QUESTION: 132
The goals of monetary policy do not include
Solution:
QUESTION: 133
The Corporate Head Quarters of Tata Group is
Solution:
QUESTION: 134
Dronacharya Award for Coaching for the year 2013 was received by Narinder Singh Saini for which discipline
Solution:
QUESTION: 135
This company signed an agreement with the Government of India for the Passport Seva Project in 2008
Solution:
QUESTION: 136
Recommendations of which committee led to the reforms in the banking sector in India in the nineties
Solution:
QUESTION: 137
Which country is the largest producer of newsprint?
Solution:
QUESTION: 138
India's Direct Cash Transfer Scheme draws similarities from models of which country
Solution:
QUESTION: 139
Who put forth the theory of 'Zero Defects':
Solution:
QUESTION: 140
Logic of induction is very close to
Solution:
QUESTION: 141
How many pairs of chromosomes exist in a human cell?
Solution:
QUESTION: 142
International Date Line roughly follows the longitude of
Solution:
QUESTION: 143
Daylight saving time (DST) implies adjusting the clock by one hour:
Solution:
QUESTION: 144
Who among the following sports persons are twins
Solution:
QUESTION: 145
Individuals with this blood type are universal red cell donors
Solution:
QUESTION: 146
'Relationship beyond banking' is the tag line of
Solution:
QUESTION: 147
Find the odd man out:
Solution:
QUESTION: 148
Which instrument measures Blood Pressure?
Solution:
QUESTION: 149
Find the odd person out:
Solution:
QUESTION: 150 | 17,789 | 71,192 | {"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} | 4.4375 | 4 | CC-MAIN-2022-33 | latest | en | 0.917502 |
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#### Resources tagged with Factors and multiples similar to Advent Sudoku:
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### There are 92 results
Broad Topics > Numbers and the Number System > Factors and multiples
### How Old Are the Children?
##### Age 11 to 14 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### Product Sudoku
##### Age 11 to 14 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### American Billions
##### Age 11 to 14 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...
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Ben's Game
##### Age 11 to 14 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.
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Cuboids
##### Age 11 to 14 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?
### Charlie's Delightful Machine
##### Age 11 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Got it for Two
##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Stars
##### Age 11 to 14 Challenge Level:
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Gabriel's Problem
##### Age 11 to 14 Challenge Level:
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
### Factors and Multiples Game for Two
##### Age 7 to 14 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
### Got It
##### Age 7 to 14 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.
##### Age 11 to 14 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?
### Funny Factorisation
##### Age 11 to 14 Challenge Level:
Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?
### Factors and Multiples Game
##### Age 7 to 16 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Dozens
##### Age 7 to 14 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### Factors and Multiples Puzzle
##### Age 11 to 14 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Factors and Multiples - Secondary Resources
##### Age 11 to 16 Challenge Level:
A collection of resources to support work on Factors and Multiples at Secondary level.
##### Age 11 to 16 Challenge Level:
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
### Inclusion Exclusion
##### Age 11 to 14 Challenge Level:
How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5?
### Thirty Six Exactly
##### Age 11 to 14 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Napier's Location Arithmetic
##### Age 14 to 16 Challenge Level:
Have you seen this way of doing multiplication ?
### Star Product Sudoku
##### Age 11 to 16 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.
### Shifting Times Tables
##### Age 11 to 14 Challenge Level:
Can you find a way to identify times tables after they have been shifted up or down?
### Mathematical Swimmer
##### Age 11 to 14 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. . . .
### Counting Cogs
##### Age 7 to 14 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?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Missing Multipliers
##### Age 7 to 14 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Factor Track
##### Age 7 to 14 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.
### Sieve of Eratosthenes
##### Age 11 to 14 Challenge Level:
Follow this recipe for sieving numbers and see what interesting patterns emerge.
### LCM Sudoku II
##### Age 11 to 18 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Satisfying Statements
##### Age 11 to 14 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Repeaters
##### Age 11 to 14 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.
### Diggits
##### Age 11 to 14 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### Divisively So
##### Age 11 to 14 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Remainder
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Power Crazy
##### Age 11 to 14 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
##### Age 11 to 14 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. . . .
### X Marks the Spot
##### Age 11 to 14 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" .
### LCM Sudoku
##### Age 14 to 16 Challenge Level:
Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Two Much
##### Age 11 to 14 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. | 2,191 | 8,943 | {"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} | 3.84375 | 4 | CC-MAIN-2019-22 | latest | en | 0.899594 |
https://www.helpwithassignment.com/correlation-assignment-help/ | 1,696,368,629,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00106.warc.gz | 875,626,847 | 30,203 | ## Correlation Assignment & Homework Help Online
Do Correlation Assignment Problems get you down? Our brilliant team of Statistics experts, armed with PhDs and Masters, can offer you all the help you need on a wide range of assignment topics in Statistics, chiefly Correlation.
Today, there are a number of leading professions that demand an in-depth knowledge of Correlation. This is why it is absolutely imperative that you too do well in this subject. Hence, it’s not surprising that this is one of the most in demand subjects at the top universities of the world. However, topics like Correlation aren’t easy to learn. Most people find it difficult to absorb and lose confidence in college. But this isn’t the end of the road, if you too feel this way.
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HwA’s team of Statistics Assignment Experts
Our Statistics assignment experts are armed with Masters and PhDs in this subject, and are trained and experienced academics. Most of them have had long and impressive teaching careers in this subject, so when you sign up with us, you learn from experts. Besides, they are dedicated to teaching and have taught many a student like you, with similar fears and doubts. All in all, they have the knowledge, time, patience and a love for spreading the knowledge they have, so don’t think twice about what you’re going in for.
So, what is Correlation in Statistics?
In Statistics the term correlation is used to measure the relativity between two variables. There are two types of correlation coefficients namely positive and negative. Positive correlation indicates that the variables are directly proportional to each other, which means that if the value of one variable increases then the value of the other variable increases as well. And negative correlation indicates that the variables are inversely proportional to each other, which means that when the value of one variable increases the value of the other decreases.
Correlation coefficients range from -1.00 to +1.00. A correlation coefficient of 0.00 indicates that there is no relationship between the two variables. That means the scores of one variable do not have any influence on the scores of the other variable. The closer the correlation to either -1.00 or +1.00 the stronger the relationship between the two variables. Generally, -1.00 or +1.00 values are not seen in real world scenarios, but if the value of correlation stands at -7.00 or +7.00 then the correlation is said to be a stronger correlation. Some authors suggest that the correlation coefficients between -0.2 and +0.2 indicate a weak relation between the two variables, while a coefficient of 0.2 to 0.5 indicate a moderate relationship and a coefficient greater than 0.5 (either positive or negative) suggest that it is a strong relationship.
The equation for the correlation coefficient is:
Where x and y are the sample means AVERAGE (array1) and AVERAGE (array2).
An Example: Sample observations of trucking distance and delivery time for 10 randomly selected shipments.
Sampled shipment 1 2 3 4 5 6 7 8 9 10 Distance in miles (X) 825` 215 1070 550 480 920 1350 325 670 1215 Delivery Time in days (Y) 3.5 1 4 2 1 3 4.5 1.5 3 5
Determine the least-squares regression equation for the data given above.
B1 = ΣXY-nXY÷ΣX2-nX2 = (26370)-(10)(762)(2.85)÷7104300-(10)(762)2= 4653÷1297860=
0.0035851≈0.0036. B0 =Y – bX =2.85 – (0.0036)(762)= 0.1068≈ 0.11,
Therefore, Y = b0 +b1X = 0.11+0.0036X.
Background of our Statistics and correlation assignment help experts
• D. in Statistics from one of the reputed universities in the US and has developed methods for analysing differences between two stochastic processed in time which have multiple categorical responses. He has been elected to Mortar Board honor society and received many excellence awards. Present working as professor in the department of Mathematics at Cornell college and teach six Statistics and Mathematics courses per year including sections of Statistical Methods I and II, Mathematical Statistics I and II and Calculus I and II.
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https://www.jiskha.com/search?query=You+have+6+cups+of+sugar.+It+takes+1+cup+of+sugar+to+make+24+cookies.+The+function+c%28s%29+%3D+24s+represents+the+number+of+cookies%2C+c%2C+that+can+be+made+with+s+cups+of+sugar.+What+domain+and+range | 1,586,500,423,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371886991.92/warc/CC-MAIN-20200410043735-20200410074235-00497.warc.gz | 958,659,348 | 14,806 | # You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range
36,303 results
1. ## math
Please show me how to set up and solve: You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range is reasonable for the
asked by Jane47 on April 18, 2012
2. ## math
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range is reasonable for the function?
asked by Jane on April 21, 2012
3. ## math
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range are reasonable for the function? (I know that you get 144
asked by Jane on April 12, 2012
4. ## math
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range are reasonable for the function? (I know that you get 144
asked by Jane on April 13, 2012
5. ## Math
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range are reasonable for the function? (I know that you get 144
asked by Jane on April 12, 2012
6. ## math
Anuja is baking cookies for her slumber party this weekend. She has one super size package of Sugar Sprinkles and one super size package of Chocolate Turtles. Both packages had to be mixed with flour, brown sugar, and water. The Sugar Sprinkles package
asked by Anonymous on December 14, 2015
7. ## Math
- 2 1/4 cups all purpose flour - 1 teaspoon of backing soda - 1 teaspoon of salt - 1 cup (2 sticks) of butter softened - 3/4 cups of sugar - 3/4 cup of packed brown sugar -1 teaspoon of vanilla extract -2 large eggs -2 cups of chocolate chips -1 cup of
asked by Brenda on October 4, 2015
Ashley is making cookies the recipe calls for 3/4 of a cup of sugar she needs to make 5 batches of cookies. How many cups of sugar does Ashley need? a. 3 cups b. 3 1/4 cups c. 3 1/2 cups d. 3 3/4 cups***** ***My work*** 3/4 * 5 = 3.75 .75= 3/4 3 3/4 AM I
asked by RL GRIME on October 21, 2016
9. ## Math
Monique is making cookies for the pajama party at her house. She has 3-1/2 cups of sugar. The recipe calls for 3/4 cup of sugar for one batch of cookies. Write an equation that can be used to find c, the total number of batches of cookies she can make.
asked by Steven on April 7, 2014
10. ## Math
You are baking cookies for cookout. you want to make 4 times the amount in the recipe. The recipe requires 2/3 cup of sugar. How much sugar do you need? A. 2 2/3 cups B. 3 1/3 Cups C. 4 2/3 cups D. 6 Cups Can you help me answer this? Thanks you!
asked by Squirrelheart- I need Help! on January 19, 2019
11. ## math
3 cups a sugar makes 36 cookies how many cookies would 2/3 cup sugar make?
asked by lauren on December 6, 2014
12. ## math
3 cups a sugar makes 36 cookies how many cookies would 2/3 cup sugar make?
asked by lauren on December 5, 2014
13. ## Math - 2nd try
Please show me how to set up and solve: You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range is reasonable for the
asked by Jane on April 18, 2012
14. ## algebra
A recipe calls for 4 cups of sugar to make 4 dozen cookies. Use the proportion to calculate how much sugar is needed to make 8 dozen cookies. A 2 cups B 8 cups c 8..5 cups D 12 cups. Is it 8 cups of sugar?
asked by shirley on November 9, 2015
15. ## Math
You are baking chocolate chip cookies. One batch makes 4 dozen big cookies. The recipe calls for 3/4 cup brown sugar. How many cups of brown sugar would you need if you wanted to make 11 dozen cookies.
asked by Melinda on September 9, 2019
16. ## Math
a cookie recipe calls for a ratio of 4 cups of flour to 3 cups of sugar in order to make a dozen cookies. If Ben wants to make 30 cookies how many cups of sugar are needed?
asked by Maricah on February 4, 2016
17. ## College mathematics
if 1/2 cup of sugar is used to make 48 cookies how many cookies 2 ¼ cups of sugar?
18. ## Math
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range are reasonable for the function? These are my options: domain: 1
asked by Jane on May 10, 2012
19. ## math (2nd Try)
You have 6 cups of sugar. It takes 1 cup of sugar to make 24 cookies. The function c(s) = 24s represents the number of cookies, c, that can be made with s cups of sugar. What domain and range are reasonable for the function? These are my options: domain: 1
asked by Jane on May 11, 2012
20. ## math
an oatmeal cookie recipe that makes 24 cookies requires 1 cup of sugar if alex needs 30 cookies how many cups of sugar should she use
asked by ashley on April 7, 2015
21. ## Math
Sara is making cookies and the recipe calls for 2 1/2 (2.5) cups of sugar to make two dozen cookies. How many cups of sugar will she need to make five dozen cookies? Express the answer as a mixed fraction and as a decimal
asked by Cheyenne on August 3, 2015
Maritza is baking cookies to bring to school and share with her friends her birthday. The recipe requires 3 eggs for every 2 cups of sugar. To have enough cookies for all of her friends, Maritza determined she would ned 12 eggs. If her mom bought 6 cups of
asked by Harms on November 5, 2015
23. ## math
Sally is making cookies. To make a dozen cookies, she needs 1/4 cup brown sugar. How many cups of brown sugar does Selena need to make 4 2/3 dozen cookies?
asked by sb on November 7, 2012
24. ## college math
Rebekah's recipe for 4 1/2 dozen cookies calls for 6 cups of sugar. how many cups of sugar are needed to make 8 dozen cookies? would the correct answer be 11 2/3 cups
asked by Lourane on June 19, 2012
25. ## Math
Rebekah's recipe for 4 1/2 dozen cookies calls for 6 cups of sugar. How many cups of sugar are needed to make 8 dozen cookies? Is the correct answer 10 2/3 cups?
asked by Taylor on February 25, 2012
26. ## Math
A cookie recipe calls for 1/4 cup of brown sugar for one batch of 24 cookies. How many cups of brown sugar would be needed to bake 144 cookies?
asked by Anonymous on April 6, 2016
27. ## Math
1. Jacob is baking chocolate chip cookies. The recipe uses 3/4 cup of sugar. Jacob wants to make one half of a batch. How many cups of sugar does he need? My answer is 1/4 cups if sugar. 2. In the fridge, hannah had 2/3 of a quart if milk. She used half of
asked by Khys on January 30, 2014
28. ## Math
Karol bought 1lb of sugar. She used 1/4 lb of sugar to bake a big cake and some of the remaining sugar to make lots of cookies. She had 3/8 lb of sugar left. How much sugar did Karol use for the cookies? I came up with 3/8-1/4=3/8-2/8=1/8 lb
asked by Reine Meghan on March 24, 2015
29. ## Pre-Algebra
1. What makes the statement true? 8^-2 =? A. 64 B. 16 C. -1/64 D. 1/64 2. What is the solution to (4 x 10^3) x (6 x 10^6), written in a scientific notation? A. 24 x 10^9 B. 2.4 x 10^9 C. 24 x 10^18 D. 2.4 x 10^10 3. What is 0.000000512 written in a
asked by Lilith on September 30, 2015
30. ## Pre-Algebra
To make a batch of cookies you need 3.25 cups of sugar, if you start off with 16.5 cups of sugar, how many batches of cookies can you make?
asked by Gianna on January 17, 2013
31. ## College math
Rebekah's recipe for 41/2 dozen cookies call for 6 cups of sugar. How many cups of sugar are needed to make 8 dozen cookies?
asked by Anonymous on March 23, 2013
32. ## Math
Karen has 1/2 cup of sugar. She is baking cookies and needs 1 1/3 cup for a recipe. How much more sugar does she need if she wants to triple the recipe? 1 1/3 x 1/2 = 4/3 x 1/2 =4/6 =2/3 cups is this right
asked by Johnnie on October 28, 2019
33. ## math
carol is baking cookies and needs 2 1/4 cups of flour, 2/3 cup of white sugar, 1/4 cup of brown sugar, and 1/16 cup of baking soda. use this information to answer the following questions 2. if the amount of white sugar were increased by a factor of 2 1/2
asked by kathy on October 25, 2014
34. ## math
Selena is making cookies. To make one dozen cookies, she needs 2/3 cup of brown sugar. How many cups of brown sugar does Selena need to make 4 1/3 dozen cookies?
asked by kelsey on September 16, 2012
35. ## prealgebra
A recipe for chocolate chip cookies calls for 1 cup sugar and 1 1/4 cups of flour. If the recipe makes 3 1/2 dozen cookies, how many cups of flour will be needed to make 12 dozen cookies?
asked by Marshall on August 11, 2016
36. ## math
A chocolate cookie recipe that makes 24 cookies uses 3/4 cups of brown sugar. If John wants to make 72 cookies, how much brown sugar does he need?
asked by davis on June 25, 2013
37. ## Algebra
A cookie company uses one cup of sugar for every 35 cookies it makes. Let represent the total number of cups of sugar used, and let represent the number of cookies made. Write an equation relating to , and then graph your equation using the axes below.
asked by Gena on August 10, 2014
38. ## algebra
A cookie company uses one cup of sugar for every cookies it makes. Let represent the total number of cups of sugar used, and let represent the number of cookies made. Write an equation relating to , and then graph your equation using the axes below.
asked by Anonymous on October 23, 2012
39. ## Math (equations and Graphs)
A cookie company uses one cup of sugar for every 35 cookies it makes. Let S represent the total number of cups of sugar used, and let N represent the number of cookies made. Write an equation relating S to N, and then graph your equation.
asked by CW on March 11, 2018
40. ## Math
A recipe calls for 2 cups of sugar to make 2 dozen cookies. Use the proportion to calculate how much sugar is needed to make 6 dozen cookies.
asked by Amber on January 26, 2016
41. ## Math
Katie is baking cookies. She only wants to make 3/4 of the recipe. She needs to take 3/4 of the 2/3 cup of sugar. How much sugar does she need?
asked by Greg on January 24, 2017
1. Find the product in simplify. 4/9 of 15/8 A.5/6(I PICKED THIS) B.5/12 C.32/135 D.135/32 2. you are baking cookies for a cookout.you want to make 4 times the amount in the recipe.the recipe requires 2/3 cup of sugar.how much sugar do you need? A.2 2/3
asked by mathelp124576 on January 7, 2014
43. ## math
1. Write the ratio in simplest form 50:10 A 25:5********** B 10:2 C 5:1 D 1:5 3. A recipe calls for 4 cups of sugar to make 4 dozen cookies. Use the proportion to calculate how much sugar is needed to make 8 dozen cookies. 4 cups / 4dozen = X cups / 8dozen
asked by orangygirl on November 6, 2015
44. ## math
A recipe that makes 18 cookies calls for 3/4 cup of sugar. How much sugar is needed to make 24 cookies using this recipe?
asked by Sarah on March 3, 2011
45. ## math
To meke a batch of sugar cookies , you need 0.75 cup of sugar and 1/3 cup of butter , among other ingredients. How much more sugar do you need that butter? a) 5/12 cup b) 3/4 cup c) 1/2 cup d) 2/3 cup
asked by anna on December 1, 2015
46. ## Algebra
Here is the problem: You are going to make ans sell bread. A loaf of Irish soda bread is made with 2 cups flour and 1/4 cup sugar. Banana bread is made with 4 cups flour and 1 cup of sugar. You will make a profit of \$1.50 on each loaf of Irish soda bread a
asked by Skyelar on October 13, 2008
47. ## Geometry
Carlos is making cookies. The recipe calls for 1 cup of sugar for every 3 cups of flour. Carlos has only 2 cups of flour. How much sugar should Carlos use?
asked by Erin on January 18, 2017
48. ## math HELP
A baker used 16 pounds of flour and 12 cups of sugar to make several batches of cookies. the baker used the same amount of flour and sugar in each batch of cookies. what is the number of batches the baker could have made with his flour and sugar? this is
asked by Kristy on October 11, 2010
49. ## math
you need 1 1/4 cups of sugar to make 20 cookies. to make 6 cookies, you need___ cups of sugar?
asked by tommy on April 27, 2016
50. ## Math
"Jeremy is cooking using a recipe that requires flour, sugar and water. For each cup of sugar that Jeremy uses, he needs to use two cups of flour. He needs to use 2 1/2 more cups of water than flour. He also needs to use 3 1/2 cups less sugar than water.
asked by Sephiroth on December 19, 2007
51. ## Math
If recipe takes 1-1/4 cups of sugar to 2 cups of milk but if you only have 1 cup of sugar then how much milk do you use?
asked by Dan on October 15, 2016
The New York Subway Bakery is famous for selling large "black and white cookies." The top of each cookie has one-half chocolate icing and one-half vanilla icing. Mario, the baker, bakes at night after the customers leave. Mario wants to bake two hundred
asked by Lemi on November 10, 2015
The New York Subway Bakery is famous for selling large "black and white cookies." The top of each cookie has one-half chocolate icing and one-half vanilla icing. Mario, the baker, bakes at night after the customers leave. Mario wants to bake two hundred
asked by Lemi on November 10, 2015
The New York Subway Bakery is famous for selling large "black and white cookies." The top of each cookie has one-half chocolate icing and one-half vanilla icing. Mario, the baker, bakes at night after the customers leave. Mario wants to bake two hundred
asked by Lemi on November 10, 2015
Kandace's recipe for cupcake frosting calls for 3 3/4 cups granulated sugar, 3/4 cup butter, 1 2/3 powdered sugar and 6 ounces unsweetened chocolate. How much sugar is in the recipe?
asked by Gabby on September 11, 2013
56. ## math
you are baking cookies. recipe 2 cup sugar 1-1/2 cup flour 2 eggs a- you make one and one-half batches of cookies, how many eggs have you used? Answer- 2 eggs b- Each batch makes 24 cookies. you make x batches of cookies but eat 5 cookies as you are
57. ## Math
A recipe that makes 18 cookie calls for 3/4 cup of sugar. Howbmuch sugar is needes to make 24 cookies using this recipe?
asked by Cici on April 3, 2011
58. ## math
i need help with this open-ended you are having so much fun in the kitchen you decide that you want to make your aunt,s favorite dessert to show her special she is to you .you find the recipe on the internet for the recipe white raspberry cheesecake
asked by Eduarda on December 31, 2012
59. ## math
i need help with this open-ended you are having so much fun in the kitchen you decide that you want to make your aunt,s favorite dessert to show her special she is to you .you find the recipe on the internet for the recipe white raspberry cheesecake
asked by carina on December 31, 2012
Maritza is baking cookies to bring to school that requires 3 eggs and 2 cups of sugar. To have enough for everyone she bought 12 eggs will 6 cups of sugar be enough? I tried multiplying 3 x 4 = 12 so then multiplied 2 x 4 which is 8 so there would not be
asked by Jay on September 17, 2014
61. ## Math
Does 2 3/4 cups of flour 2 cups of sugar 1 teaspoon of baking powder 1 1/4 cups of butter 2 teaspoons of extract 2 eggs Make 30 cookies
asked by Princess on March 31, 2019
62. ## Math
My teacher usually gives out a good bit of homework, but today she only gave us this problem. A cookie recipe calls for 1 cup of white sugar, to 3 cups of brown sugar. Write the value of the ratio of the amount of white sugar to the amount of brown sugar.
asked by Molly on October 30, 2019
63. ## Math
Twenty cookies need 1/4 cups of sugar. How much sugar is needed for 3 dozen?
asked by Klarissa on March 24, 2011
64. ## Math
A muffin recipe calls for a ratio of 5 cups of flour to 2 cups of sugar. For each cup of sugar that is used, how many cups of flour are needed?
asked by Alexis on March 29, 2017
Can you please check my answer? Marcia wants to make muffins and needs 3/4 cup of a cup of sugar. She discovers, however,that she has only 2/3 of a cup of sugar. How much more sugar does she need? my answer: 1/12 of a cup
asked by Elizabeth on November 8, 2010
66. ## Math
Grandma needs 13/4 cups of sugar to make a dozen cookies. How much sugar does she need to make 3 dozen cookies?
asked by Yolanda on November 28, 2015
67. ## Math
A recipe calls for 3 1/2 cups of flour and 3/4 cup of sugar. If you want to make the recipe with 6 cups of flour, how much sugar will you need? Show your work. help please and thank you.
asked by URGENT on September 10, 2019
68. ## Math
. A recipe calls for 3 1/2 cups of flour and 3/4 cup of sugar. If you want to make the recipe with 6 cups of flour, how much sugar will you need? Show your work. help please and thank you.
asked by URGENT on September 10, 2019
69. ## Math
A baker used 16 pounds of flour and 12 cups of sugar to make several batches of cookies. the baker used the same amount of flour and sugar in each batch of cookie. what is the number of batches the baker could have made with his flour and sugar
asked by Kristy on October 11, 2010
70. ## chmistry
You run out of brown sugar and decide to use 1.5 cups granulated sugar instead. The cookies come out much thinner than usual. Formulate a one-sentence hypothesis for why the recipe failed.
asked by Anonymous on November 13, 2013
71. ## Math
Ricky needs 3 1/4 cups of sugar to make candy. He already has 2 1/2 cups. How many more cups of sugar does he need? Do you add for this question or subtract? That's all that I need to know.
asked by Zoe on March 26, 2015
72. ## Math
If Frans recipe calls for 2 1/2 cups of sugar she increases it by 2/5 of a cup, how much sugar does she use I Need some help on this one im not very good with fractions please:)
asked by Bisha on September 11, 2019
73. ## math
how man 3/7 cup sugar bowls can be filled from 6 cups of sugar?
asked by chyna on December 1, 2016
If Fran’s fudge recipe calls for 2 and 1/2 cups of sugar and she increases it by 2/5 of a cup, how much sugar does she use? a. 2 2/10 b. 2 3/7 c. 2 4/9 d. 2 9/10 is the answer d because 2 1/2 plus 2/5 equals 2 9/10? If it is incorrect, then why? Thanks!
asked by Delilah on October 1, 2012
75. ## Math
How many 2/3 cup sugar bowls can be filled from 8 cups of sugar?
asked by Arica on February 5, 2014
76. ## math
a chef is only making half of the frosting recipe 3 cups granulated sugar 1/4 teaspoon cream of tartar 1 1/2 cups of water 1 cup powdered sugar 1/2 teaspoon vanilla 1/2 teaspoon almond
asked by larry on February 18, 2016
77. ## Math
a pudding recipe for 50 people calls for 4 cups of sugar. each bag of sugar contains 6 cups. how many bags of sugar will be needed to make this recipe for 300 people?
asked by Anonymous on July 26, 2010
78. ## college algebra
a cake recipe calls for 1 2/4 cups of sugar. A caterere has 15 1/2 cups of sugar on hand. how many cakes can he make?
asked by Trina on October 21, 2015
79. ## math Help
Multiplying and Dividing Fractions Carol is baking cookies and needs two and one-fourthcups of flour, two-thirdscup of white sugar, one-fourthcup of brown sugar, andone -sixteenth cup of baking soda. Use this information to answer the following questions.
asked by gavin on September 11, 2015
80. ## math
A recipe for banana pudding calls for 2/3 cups of sugar for the flour and 1/4 cups of sugar for the meringue topping, how many cups of sugar is required to make that good banana pudding? Five friends want to split 6 1/4 pounds of candy amongst themselves.
asked by lina on February 7, 2020
81. ## Math
I don't get this: Arya makes cupcakes at a bakery. She needs 2 cups of sugar for every 3 cups of flour. She needs 5 cups of flour to make a day's worth of cupcakes. How many cups of sugar will she need? (Round your answer to the nearest tenth.) Can you
asked by Anonymous on March 27, 2017
82. ## algebra
Jessica is baking a cake. The recipe says that she has to mix 96 grams of sugar to the flour. Jessica knows that 1 cup of this particular sugar has a mass of 128 grams. She added one over two of a cup of sugar to the flour. Should Jessica add more sugar to
asked by Anonymous on September 9, 2014
83. ## math
joe was baking a cake . he added 3/4 cup of white sugar and 3/8 cup of brown sugar .how much sugar did he use in all
asked by khizra on January 5, 2016
84. ## Math
jason mixed 3/8 cup of sugar with 1 5/6 cups of water. how many more cups of water than sugar did he use in his mixture?
asked by Jim on September 24, 2014
85. ## science
You run out of brown sugar and decide to use 1.5 cups granulated sugar instead. The cookies come out much thinner than usual. Design a controlled experiment to test your hypothesis, describing how you would interpret your results. Please limit your
asked by llau on July 19, 2015
86. ## math
If our recipe calls for 2 1/2 cups flour for every 3/4 cup sugar, how much flour is needed if we used 1 cup of sugar?
asked by Rachel on July 24, 2019
87. ## math
a recipe that makes 16 cookies call for 1/4 cup of sugar and 2/3 cup of flour. Janelle wants to proportionally increase these almost to get a new recipe using one cup of sugar. using the new recipe how much flour should she use?
asked by yasmine on November 13, 2016
88. ## science, chemistry
"You run out of brown sugar and decide to use only granulated sugar instead. The cookies come out much thinner than usual. Design a controlled experiment to test your hypothesis *, describing how you would interpret your results. Please limit your response
asked by llau on July 19, 2015
89. ## Math
It is common to see mixed fractions in recipes. A recipe for pizza crust may ask for 1 1/2 cups of flour. You could measure this amount in two ways. You could fill a one-cup measuring cup with flour and a one-half cup measuring cup with flour or you could
asked by Jerald on April 29, 2013
90. ## Math
I have 2/3 cups of sugar. How many recipes can I make if the recipe calls for 1/4 cups of sugar?
asked by Bailey on December 4, 2019
91. ## Pyle
Ramona has 2 cups of sugar and each brownie needs 3/8 cups of sugar. How many brownies can she cover with sugar?
asked by Tiara on October 26, 2016
92. ## Math
Sally has 4.5 cups of sugar. She needs to bake brownies for for a bake sale. If each batch of brownies requires 2/3 cup of sugar, how many batches of brownies can she make?
asked by Matt on November 17, 2015
93. ## math
10. What is 0.000750 in scientific notation? 75 × 10–4 7.5 × 10–3 7.5 × 10–4 7.5 × 10–5 11. What is written as a decimal? (1 point) 4.2 4.3 4.4 4.5 12. The 2010 census data for the populations of three states is shown below. According to the
94. ## math
asked by Isabella on January 8, 2018
95. ## Math
asked by Lenny on January 8, 2018
96. ## Math
2 cups of flour = 6 cookies x cups of flour = 10 cookies I found out that 3 cookies get 1 cup of flour and each cookie get 1/3 cup . Also 3 1/3 cups equal 10 cookies . Now i have to write and solve a proportion and i don't know how to do that . Can you
asked by Carrie on January 27, 2010
97. ## math
Shella used 1 2/3 cups of sugar to make 5 dozen brownies. how much sugar is in each brownie?
asked by stephanie on September 5, 2013
98. ## Math
Here is the question: "To make a pancake mix, add 1/2 cup of sugar, 3 cups of flour, 1 cup of powdered milk, 1 teaspoon baking powder and 1/2 teaspoon salt. Assuming the volume of the baking powder and salt are negligible, how much flour will be needed to
asked by Kiki on November 17, 2008
99. ## Math
You have 11 cups of flour. It takes 1 cup of flour to make 24 cookies. The function c(f) = 24f represents the number of cookies, c, that can be made with f cups of flour. What domain and range are reasonable for the function?
asked by Jane on January 5, 2012
100. ## Math
Dee used 2 1/3 cups of sugar for a cake recipe. If the amount of sugar the container holds is 3 times the amount she used, does the container hold more than, less than, or equal to 3 cups of sugar? Explain.
asked by Natalie on March 26, 2019 | 7,161 | 24,454 | {"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} | 3.890625 | 4 | CC-MAIN-2020-16 | latest | en | 0.95105 |
https://carreersupport.com/calculate-equivalent-units-of-production/ | 1,685,495,508,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00521.warc.gz | 195,998,563 | 20,362 | # What Are Equivalent Units of Production? (How To Calculate Them With Examples)
To simply calculate equivalent units, you can multiply the number of physical items by the percentage of the work done on them. For two items that are 50% done, you would have one equivalent unit (2 x 50% = 1). When the items are completely finished, the number of equivalent units is equal to the physical items.
Units are typically moved from the beginning of the process to the next stage in many production departments. The accumulated cost per unit is transferred along with the units when they are moved. The transferred-in cost represents the cost of the work completed in all previous departments because the unit being produced includes work from all of those departments.
The hickory size 5A drumsticks are transferred to the packaging department along with the \$29,775 in inventory costs once the shaping process is complete. The inventory costs of \$29,775 were broken down into materials costs of \$8,775 and conversion costs of \$21,000 in (Figure). For the packaging division, Rock City Percussion spent \$2,000 on raw material inventory during the month of July. The packaging department tracks costs similarly to the shaping department, and requests raw materials from the material storeroom. For a total of \$22,100 in conversion costs, the packaging department calculated direct material costs of \$2,000, direct labor costs of \$13,000, and applied overhead of \$9,100. For this department, equivalent units are calculated, and a new cost per unit is calculated.
## Why calculate equivalent units of production?
Calculating equivalent units has the benefit of allowing you to refer to complete units rather than incomplete ones, which makes accounting calculations simpler. An accountant can estimate how much more money or time is needed to complete those products by understanding the equivalent units of production for materials, overhead costs, and labor costs. Understanding where the money in your institution is currently invested or creating a financial report can both benefit from equivalent units of production calculations.
## What are equivalent units of production?
A concept called “equivalent units of production” is used to calculate the value of partially finished products to businesses. They are helpful for process costing, which examines how money moves during the production process.
Equivalent units quantify the amount of labor put into a set number of tangible objects. Simply multiply the quantity of physical items by the amount of work done on them to determine equivalent units. You would have one equivalent unit for two items that are 50% complete (2 x 50% = 1). The number of equivalent units is equal to the physical items when the items are fully finished.
However, this straightforward equation disregards the fact that the manufacturing process’s requirements change over time. For instance, the majority of the materials might be added to an item at the start of the manufacturing process, but more labor might be required for assembly and refinement later on. Accountants typically determine equivalent units of production for three types of costs: materials, overhead costs, and labor costs, to account for this variation. The equivalent units of production for each of the three types of costs may vary.
## How to use the weighted average method
Here is the weighted average method formula:
Equivalent units of production for this cost component are (number of units completed) + (number of units in progress x percentage of completion for this cost component).
An accountant would carry out the following actions to determine equivalent units of production using the weighted average method:
### 1. Find number of units completed within the time period
The accountant first ascertains the quantity of completed and distributed units. Since these products are entirely finished, each unit that is sent out represents one unit of production.
Example: For the month of January, the Small Item Production Company is calculating its process costs. In January, they finished and distributed 3,000 miniature mice.
### 2. Find number of units in progress at the end of the time period
The accountant then determines how many units are only partially completed at the conclusion of the time period. This number is sometimes called the ending work-in-progress inventory.
Example: Small Item Production Company started 4,000 more mini mice.
### 3. Find what percentage of materials, labor and overhead costs are complete for those items
The accountant determines how much of the work on the units in progress is complete and, if necessary, converts it into a percentage.
80% of the materials have been added to the incomplete mice, but only 35% of the labor and overhead costs have been applied.
### 4. Apply formula to calculate equivalent units of production for materials, labor and overhead costs or conversion costs
The formula can be used by the accountant to determine the equivalent units of production for each cost element of the item. The conversion cost is a term used by accountants occasionally to combine labor and overhead costs.
The equivalent units of production for the following materials are determined by the accountants of the Small Item Production Company:
Total equivalent units of production for materials equals 3,000 mini mice finished plus (4,000 mini mice in progress x 80% of materials added)
Equivalent units of production for mice materials: 3,000 + 3,200 = 6,200
The equivalent units of production for conversion costs are determined by the accountants in the following manner:
Total equivalent units of production for conversion costs equals 3,000 mini mice completed plus (4,000 mini mice in progress x 35% of conversion costs complete)
For conversion costs, 3,000 + 1,400 = 4,400 equivalent units of production for mice.
Accordingly, the Small Item Production Company’s accountants report that during the month of January, they produced 6,200 equivalent units of mice materials and 4,400 equivalent units of conversion costs.
## What are the ways to calculate equivalent units of production?
You can determine equivalent units of production using the first-in, first-out method or the weighted average method. The weighted average method disregards any inventory that might have been started in a previous period and completed during the calculation period. This can make it more suitable for a time when there is no beginning inventory at the beginning of a project or year. This initial inventory is taken into account by the first-in, first-out method, making it applicable in more circumstances.
## How to use the first-in first-out (FIFO) method
The reason this approach is known as first-in-first-out is because it takes into account information about the partially finished items at the start of the time period, the items that started the manufacturing process first for that time period, and the costs to finish and ship them, making them the first items sent out during that time period. Here is the formula for the FIFO method:
Total equivalent units of production for the period equals equivalent units of production to finish the beginning inventory plus units started and finished during the period plus equivalent units of production for items partially completed during the period.
And here is the equation to determine how many equivalent units are required to finish the initial inventory:
Equivalent units of production are equal to the beginning inventory units multiplied by 100% minus the beginning inventory’s percentage completion.
The FIFO method’s steps for calculating equivalent units of production are as follows:
### 1. Calculate equivalent units of production needed to complete items in beginning inventory
The accountant determines how many items were partially completed at the beginning of the month and how much work was done at that time in order to determine the equivalent units of production still needed to finish the beginning inventory. The equivalent units of production required to complete these items can then be determined.
Example: In January, Small Item Production Company is estimating the cost of materials for its miniature cats. At the beginning of January, they had 5,000 miniature cats that were not yet finished. These cats had 70% of the materials added. The remaining equivalent units of production are determined by small item accountants in the following way:
1,500 remaining material equivalent units of production from 1,500 cats in progress multiplied by (100% – 70% materials added)
### 2. Find number of units completed during the time period
The accountant then multiplies the quantity of units finished during the time period. Since they are finished, each of these units equates to one equivalent unit of production.
An illustration is the complete production of 6,000 miniature cats by Small Item Production Company in January.
### 3. Find number of units started during the time period that are incomplete
The number of incomplete units, also known as the work-in-progress inventory’s final count, is shown here.
Example: In January, the Small Item Production Company started 3,000 more miniature cats.
### 4. Find what percentage of work is done on units started during the time period
Depending on whether you are figuring out the cost of materials, labor, overhead, or conversion (labor and overhead cost combined), this could change.
For instance, 40% of the materials for the extra 3,000 mini cats have already been added.
### 5. Calculate equivalent units of production for ending work-in-progress inventory
The accountant would multiply the number of items in progress by the amount of work completed on them thus far to arrive at this calculation.
Example: For the 3,000 mini cats that make up the final work-in-progress inventory, the small item accountants determine the material equivalent units of production:
1,200 material equivalent units of production for the final work-in-progress cat inventory are obtained by multiplying 3,000 items in progress by the 40% of materials added.
### 6. Calculate total equivalent units of production for the time period
The accountant then enters the values from earlier steps into the FIFO formula to determine the equivalent units of production for the time period.
For instance, the Small Item accountants determined the January cat material equivalent units of production as follows:
1,500 equivalent units of production were needed to complete the previous cats, 6,000 equivalent units were completed in January, and 1,200 equivalent units were completed on the unfinished cats, totaling 8,700 equivalent units of production.
The Small Item accountants determine from this calculation that 8,700 cats’ worth of materials were used to complete cats in total in January.
## What are equivalent units of production used for?
An accountant can determine the cost to complete manufacturing the inventory once they have the equivalent units of production. The total number of items, equivalent units of production, and the costs to make each item are used to calculate these costs. The formula to determine the remaining costs to complete inventory is as follows:
The remaining cost to finish inventory is equal to (total items finished and in progress x cost per item to complete) – (equivalent units of production x cost per item to complete).
Example: The manufacturer of small items is aware that each mini mouse requires \$2 in materials. They are curious about the amount of money needed to complete the 7,000 mini mice they began in January. They calculate the remaining cost by applying the formula:
The remaining cost for materials is equal to (7,000 total mice x \$2) – (6,200 equivalent units of mice production x \$2).
\$14,000 – \$12,400 = \$1,600
The cost of materials for January mice will increase by \$1,600 for the small item production company.
## FAQ
How do you calculate equivalent units of production using the weighted average method?
Number of physical units minus the percentage of completion equals equivalent units. 1,500 equivalent units = 5,000 physical units are 30% complete for direct labor and overhead, while 3,000 equivalent units are 60% complete for direct materials.
What are equivalent units of production in process costing?
According to the amount of direct materials, direct labor, and manufacturing overhead costs incurred during that time for the items that weren’t yet finished, it is the approximate number of completed units of a product that a company could have produced.
How do you find the equivalent units of production using FIFO?
According to the amount of direct materials, direct labor, and manufacturing overhead costs incurred during that time for the items that weren’t yet finished, it is the approximate number of completed units of a product that a company could have produced. | 2,503 | 13,004 | {"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} | 3.84375 | 4 | CC-MAIN-2023-23 | longest | en | 0.955216 |
https://courses.lumenlearning.com/ivytech-collegealgebra/chapter/verify-inverse-functions/ | 1,591,052,928,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419639.53/warc/CC-MAIN-20200601211310-20200602001310-00519.warc.gz | 311,584,744 | 12,566 | ## Verify inverse functions
Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula
$C=\frac{5}{9}\left(F - 32\right)$
and substitutes 75 for $F$ to calculate
$\frac{5}{9}\left(75 - 32\right)\approx {24}^{ \circ} {C}$.
Figure 2
Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit.
At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for $F$ after substituting a value for $C$. For example, to convert 26 degrees Celsius, she could write
$\begin{cases}26=\frac{5}{9}\left(F - 32\right)\hfill \\ 26\cdot \frac{9}{5}=F - 32\hfill \\ F=26\cdot \frac{9}{5}+32\approx 79\hfill \end{cases}$
After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature.
The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function.
Given a function $f\left(x\right)$, we represent its inverse as ${f}^{-1}\left(x\right)$, read as $f$ inverse of $x.\text{}$ The raised $-1$ is part of the notation. It is not an exponent; it does not imply a power of $-1$ . In other words, ${f}^{-1}\left(x\right)$ does not mean $\frac{1}{f\left(x\right)}$ because $\frac{1}{f\left(x\right)}$ is the reciprocal of $f$ and not the inverse.
The “exponent-like” notation comes from an analogy between function composition and multiplication: just as ${a}^{-1}a=1$ (1 is the identity element for multiplication) for any nonzero number $a$, so ${f}^{-1}\circ f\\$ equals the identity function, that is,
$\left({f}^{-1}\circ f\right)\left(x\right)={f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(y\right)=x\\$
This holds for all $x$ in the domain of $f$. Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses.
Given a function $f\left(x\right)$, we can verify whether some other function $g\left(x\right)$ is the inverse of $f\left(x\right)$ by checking whether either $g\left(f\left(x\right)\right)=x$ or $f\left(g\left(x\right)\right)=x$ is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.)
For example, $y=4x$ and $y=\frac{1}{4}x\\$ are inverse functions.
$\left({f}^{-1}\circ f\right)\left(x\right)={f}^{-1}\left(4x\right)=\frac{1}{4}\left(4x\right)=x\\$
and
$\left({f}^{}\circ {f}^{-1}\right)\left(x\right)=f\left(\frac{1}{4}x\right)=4\left(\frac{1}{4}x\right)=x\\$
A few coordinate pairs from the graph of the function $y=4x$ are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function $y=\frac{1}{4}x\\$ are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function.
### A General Note: Inverse Function
For any one-to-one function $f\left(x\right)=y$, a function ${f}^{-1}\left(x\right)$ is an inverse function of $f$ if ${f}^{-1}\left(y\right)=x$. This can also be written as ${f}^{-1}\left(f\left(x\right)\right)=x$ for all $x$ in the domain of $f$. It also follows that $f\left({f}^{-1}\left(x\right)\right)=x$ for all $x$ in the domain of ${f}^{-1}$ if ${f}^{-1}$ is the inverse of $f$.
The notation ${f}^{-1}$ is read $\text{}f$ inverse.” Like any other function, we can use any variable name as the input for ${f}^{-1}$, so we will often write ${f}^{-1}\left(x\right)$, which we read as $f$ inverse of $x.''$
Keep in mind that
${f}^{-1}\left(x\right)\ne \frac{1}{f\left(x\right)}$
and not all functions have inverses.
### Example 1: Identifying an Inverse Function for a Given Input-Output Pair
If for a particular one-to-one function $f\left(2\right)=4$ and $f\left(5\right)=12$, what are the corresponding input and output values for the inverse function?
### Solution
The inverse function reverses the input and output quantities, so if
$\begin{cases}{c}f\left(2\right)=4,\text{ then }{f}^{-1}\left(4\right)=2;\\ f\left(5\right)=12,{\text{ then f}}^{-1}\left(12\right)=5.\end{cases}$
Alternatively, if we want to name the inverse function $g$, then $g\left(4\right)=2$ and $g\left(12\right)=5$.
### Analysis of the Solution
Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.
$\left(x,f\left(x\right)\right)$ $\left(x,g\left(x\right)\right)$
$\left(2,4\right)$ $\left(4,2\right)$
$\left(5,12\right)$ $\left(12,5\right)$
### Try It 1
Given that ${h}^{-1}\left(6\right)=2$, what are the corresponding input and output values of the original function $h?$
Solution
### How To: Given two functions $f\left(x\right)\\$ and $g\left(x\right)\\$, test whether the functions are inverses of each other.
1. Determine whether $f\left(g\left(x\right)\right)=x$ or $g\left(f\left(x\right)\right)=x$.
2. If either statement is true, then both are true, and $g={f}^{-1}$ and $f={g}^{-1}$. If either statement is false, then both are false, and $g\ne {f}^{-1}$ and $f\ne {g}^{-1}$.
### Example 2: Testing Inverse Relationships Algebraically
If $f\left(x\right)=\frac{1}{x+2}$ and $g\left(x\right)=\frac{1}{x}-2$, is $g={f}^{-1}?$
### Solution
$\begin{cases} g\left(f\left(x\right)\right)=\frac{1}{\left(\frac{1}{x+2}\right)}{-2 }\hfill\\={ x }+{ 2 } -{ 2 }\hfill\\={ x }\hfill \end{cases}\\$
so
$g={f}^{-1}\text{ and }f={g}^{-1}$
This is enough to answer yes to the question, but we can also verify the other formula.
$\begin{cases} f\left(g\left(x\right)\right)=\frac{1}{\frac{1}{x}-2+2}\\ =\frac{1}{\frac{1}{x}}\hfill \\ =x\hfill \end{cases}\\$
### Analysis of the Solution
Notice the inverse operations are in reverse order of the operations from the original function.
### Try It 2
If $f\left(x\right)={x}^{3}-4$ and $g\left(x\right)=\sqrt[3]{x+4}$, is $g={f}^{-1}?$
Solution
### Example 3: Determining Inverse Relationships for Power Functions
If $f\left(x\right)={x}^{3}$ (the cube function) and $g\left(x\right)=\frac{1}{3}x$, is $g={f}^{-1}?\\$
### Solution
$f\left(g\left(x\right)\right)=\frac{{x}^{3}}{27}\ne x\\$
No, the functions are not inverses.
### Analysis of the Solution
The correct inverse to the cube is, of course, the cube root $\sqrt[3]{x}={x}^{\frac{1}{3}}\\$, that is, the one-third is an exponent, not a multiplier.
### Try It 3
If $f\left(x\right)={\left(x - 1\right)}^{3}\text{and}g\left(x\right)=\sqrt[3]{x}+1\\$, is $g={f}^{-1}?\\$
Solution | 2,284 | 7,338 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-24 | latest | en | 0.874252 |
https://en.wikipedia.org/wiki/Householder_reflection | 1,686,098,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00006.warc.gz | 257,140,125 | 31,378 | # Householder transformation
(Redirected from Householder reflection)
In linear algebra, a Householder transformation (also known as a Householder reflection or elementary reflector) is a linear transformation that describes a reflection about a plane or hyperplane containing the origin. The Householder transformation was used in a 1958 paper by Alston Scott Householder.[1]
Its analogue over general inner product spaces is the Householder operator.
## Definition
### Transformation
The reflection hyperplane can be defined by its normal vector, a unit vector ${\textstyle v}$ (a vector with length ${\textstyle 1}$) that is orthogonal to the hyperplane. The reflection of a point ${\textstyle x}$ about this hyperplane is the linear transformation:
${\displaystyle x-2\langle x,v\rangle v=x-2v\left(v^{*}x\right),}$
where ${\textstyle v}$ is given as a column unit vector with conjugate transpose ${\textstyle v^{\textsf {*}}}$.
### Householder matrix
The matrix constructed from this transformation can be expressed in terms of an outer product as:
${\displaystyle P=I-2vv^{*}}$
is known as the Householder matrix, where ${\textstyle I}$ is the identity matrix.
#### Properties
The Householder matrix has the following properties:
• it is Hermitian: ${\textstyle P=P^{*}}$,
• it is unitary: ${\textstyle P^{-1}=P^{*}}$,
• hence it is involutory: ${\textstyle P=P^{-1}}$.
• A Householder matrix has eigenvalues ${\textstyle \pm 1}$. To see this, notice that if ${\textstyle u}$ is orthogonal to the vector ${\textstyle v}$ which was used to create the reflector, then ${\textstyle Pu=u}$, i.e., ${\textstyle 1}$ is an eigenvalue of multiplicity ${\textstyle n-1}$, since there are ${\textstyle n-1}$ independent vectors orthogonal to ${\textstyle v}$. Also, notice ${\textstyle Pv=-v}$, and so ${\textstyle -1}$ is an eigenvalue with multiplicity ${\textstyle 1}$.
• The determinant of a Householder reflector is ${\textstyle -1}$, since the determinant of a matrix is the product of its eigenvalues, in this case one of which is ${\textstyle -1}$ with the remainder being ${\textstyle 1}$ (as in the previous point).
## Applications
### Geometric optics
In geometric optics, specular reflection can be expressed in terms of the Householder matrix (see Specular reflection § Vector formulation).
### Numerical linear algebra
Householder transformations are widely used in numerical linear algebra, for example, to annihilate the entries below the main diagonal of a matrix,[2] to perform QR decompositions and in the first step of the QR algorithm. They are also widely used for transforming to a Hessenberg form. For symmetric or Hermitian matrices, the symmetry can be preserved, resulting in tridiagonalization.[3]
#### QR decomposition
Householder reflections can be used to calculate QR decompositions by reflecting first one column of a matrix onto a multiple of a standard basis vector, calculating the transformation matrix, multiplying it with the original matrix and then recursing down the ${\textstyle (i,i)}$ minors of that product.
#### Tridiagonalization
This procedure is presented in Numerical Analysis by Burden and Faires. It uses a slightly altered ${\displaystyle \operatorname {sgn} }$ function with ${\displaystyle \operatorname {sgn} (0)=1}$.[4] In the first step, to form the Householder matrix in each step we need to determine ${\textstyle \alpha }$ and ${\textstyle r}$, which are:
{\displaystyle {\begin{aligned}\alpha &=-\operatorname {sgn} \left(a_{21}\right){\sqrt {\sum _{j=2}^{n}a_{j1}^{2}}};\\r&={\sqrt {{\frac {1}{2}}\left(\alpha ^{2}-a_{21}\alpha \right)}};\end{aligned}}}
From ${\textstyle \alpha }$ and ${\textstyle r}$, construct vector ${\textstyle v}$:
${\displaystyle v^{(1)}={\begin{bmatrix}v_{1}\\v_{2}\\\vdots \\v_{n}\end{bmatrix}},}$
where ${\textstyle v_{1}=0}$, ${\textstyle v_{2}={\frac {a_{21}-\alpha }{2r}}}$, and
${\displaystyle v_{k}={\frac {a_{k1}}{2r}}}$ for each ${\displaystyle k=3,4\ldots n}$
Then compute:
{\displaystyle {\begin{aligned}P^{1}&=I-2v^{(1)}\left(v^{(1)}\right)^{\textsf {T}}\\A^{(2)}&=P^{1}AP^{1}\end{aligned}}}
Having found ${\textstyle P^{1}}$ and computed ${\textstyle A^{(2)}}$ the process is repeated for ${\textstyle k=2,3,\ldots ,n-2}$ as follows:
{\displaystyle {\begin{aligned}\alpha &=-\operatorname {sgn} \left(a_{k+1,k}^{k}\right){\sqrt {\sum _{j=k+1}^{n}\left(a_{jk}^{k}\right)^{2}}}\\[2pt]r&={\sqrt {{\frac {1}{2}}\left(\alpha ^{2}-a_{k+1,k}^{k}\alpha \right)}}\\[2pt]v_{1}^{k}&=v_{2}^{k}=\cdots =v_{k}^{k}=0\\[2pt]v_{k+1}^{k}&={\frac {a_{k+1,k}^{k}-\alpha }{2r}}\\v_{j}^{k}&={\frac {a_{jk}^{k}}{2r}}{\text{ for }}j=k+2,\ k+3,\ \ldots ,\ n\\P^{k}&=I-2v^{(k)}\left(v^{(k)}\right)^{\textsf {T}}\\A^{(k+1)}&=P^{k}A^{(k)}P^{k}\end{aligned}}}
Continuing in this manner, the tridiagonal and symmetric matrix is formed.
#### Examples
In this example, also from Burden and Faires,[4] the given matrix is transformed to the similar tridiagonal matrix A3 by using the Householder method.
${\displaystyle \mathbf {A} ={\begin{bmatrix}4&1&-2&2\\1&2&0&1\\-2&0&3&-2\\2&1&-2&-1\end{bmatrix}},}$
Following those steps in the Householder method, we have:
The first Householder matrix:
{\displaystyle {\begin{aligned}Q_{1}&={\begin{bmatrix}1&0&0&0\\0&-{\frac {1}{3}}&{\frac {2}{3}}&-{\frac {2}{3}}\\0&{\frac {2}{3}}&{\frac {2}{3}}&{\frac {1}{3}}\\0&-{\frac {2}{3}}&{\frac {1}{3}}&{\frac {2}{3}}\end{bmatrix}},\\A_{2}=Q_{1}AQ_{1}&={\begin{bmatrix}4&-3&0&0\\-3&{\frac {10}{3}}&1&{\frac {4}{3}}\\0&1&{\frac {5}{3}}&-{\frac {4}{3}}\\0&{\frac {4}{3}}&-{\frac {4}{3}}&-1\end{bmatrix}},\end{aligned}}}
Used ${\textstyle A_{2}}$ to form
{\displaystyle {\begin{aligned}Q_{2}&={\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&-{\frac {3}{5}}&-{\frac {4}{5}}\\0&0&-{\frac {4}{5}}&{\frac {3}{5}}\end{bmatrix}},\\A_{3}=Q_{2}A_{2}Q_{2}&={\begin{bmatrix}4&-3&0&0\\-3&{\frac {10}{3}}&-{\frac {5}{3}}&0\\0&-{\frac {5}{3}}&-{\frac {33}{25}}&{\frac {68}{75}}\\0&0&{\frac {68}{75}}&{\frac {149}{75}}\end{bmatrix}},\end{aligned}}}
As we can see, the final result is a tridiagonal symmetric matrix which is similar to the original one. The process is finished after two steps.
## Computational and theoretical relationship to other unitary transformations
The Householder transformation is a reflection about a hyperplane with unit normal vector ${\textstyle v}$, as stated earlier. An ${\textstyle N}$-by-${\textstyle N}$ unitary transformation ${\textstyle U}$ satisfies ${\textstyle UU^{*}=I}$. Taking the determinant (${\textstyle N}$-th power of the geometric mean) and trace (proportional to arithmetic mean) of a unitary matrix reveals that its eigenvalues ${\textstyle \lambda _{i}}$ have unit modulus. This can be seen directly and swiftly:
{\displaystyle {\begin{aligned}{\frac {\operatorname {Trace} \left(UU^{*}\right)}{N}}&={\frac {\sum _{j=1}^{N}\left|\lambda _{j}\right|^{2}}{N}}=1,&\operatorname {det} \left(UU^{*}\right)&=\prod _{j=1}^{N}\left|\lambda _{j}\right|^{2}=1.\end{aligned}}}
Since arithmetic and geometric means are equal if the variables are constant (see inequality of arithmetic and geometric means), we establish the claim of unit modulus.
For the case of real valued unitary matrices we obtain orthogonal matrices, ${\textstyle UU^{\textsf {T}}=I}$. It follows rather readily (see orthogonal matrix) that any orthogonal matrix can be decomposed into a product of 2 by 2 rotations, called Givens Rotations, and Householder reflections. This is appealing intuitively since multiplication of a vector by an orthogonal matrix preserves the length of that vector, and rotations and reflections exhaust the set of (real valued) geometric operations that render invariant a vector's length.
The Householder transformation was shown to have a one-to-one relationship with the canonical coset decomposition of unitary matrices defined in group theory, which can be used to parametrize unitary operators in a very efficient manner.[5]
Finally we note that a single Householder transform, unlike a solitary Givens transform, can act on all columns of a matrix, and as such exhibits the lowest computational cost for QR decomposition and tridiagonalization. The penalty for this "computational optimality" is, of course, that Householder operations cannot be as deeply or efficiently parallelized. As such Householder is preferred for dense matrices on sequential machines, whilst Givens is preferred on sparse matrices, and/or parallel machines. | 2,543 | 8,458 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 57, "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-2023-23 | latest | en | 0.862661 |
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https://hatsudy.com/isosceles.html | 1,721,393,949,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00125.warc.gz | 252,338,465 | 10,891 | In mathematics, there are two types of shapes that we learn about: isosceles triangles and right triangles. The isosceles triangle and the right triangle are special triangles.
Since they are special triangles, they have their own characteristics. By learning what characteristics they have, we will be able to calculate angles and prove shapes.
Isosceles triangles and right triangles must be studied together. This is because these figures are often given as mixed problems. In other words, by understanding the properties of both isosceles and right triangles, we will be able to finally solve problems.
Therefore, after explaining the properties of isosceles triangles and right triangles, we will explain the theorems for proving that they are congruent.
## Isosceles and Right Triangles Are Special Triangles
Of all the shapes, the most frequently asked geometry problem is the triangle. However, there are different types of triangles. If a triangle satisfies certain conditions, it is called by another name.
Such triangles include the following.
• Isosceles triangle
• Right-angled triangles
Another typical example of a special triangle is the equilateral triangle. A triangle whose side lengths and angles are all the same is an equilateral triangle. As for equilateral triangles, they have simple properties.
On the other hand, isosceles and right triangles have more properties to remember than equilateral triangles. Also, isosceles triangles and right triangles are often given as mixed problems, and it is often impossible to solve them unless you understand the properties of both.
In addition, right triangles have a congruence condition that is available only for right triangles. You must remember this congruence theorem.
### Definition, Properties, and Theorems of Isosceles Triangles
It is important to understand the definition of special shapes. In the case of isosceles triangles, what is the definition? Any triangle that satisfies the following conditions is an isosceles triangle.
• Triangles with two equal sides
Whenever two sides are equal, it is an isosceles triangle. Conversely, if all sides are not equal in length, it is not an isosceles triangle. Since it is a definition, the following isosceles triangle will always have AB=AC.
If you are given an isosceles triangle in a math problem, the two sides have the same length.
Also, isosceles triangles have a property (theorem) derived from their definition. One of these theorems is that the base angles are equal. In an isosceles triangle, the two sides are equal, and the two angles at the base are also equal.
Properties derived from definitions are called theorems. One of the theorems of an isosceles triangle is that the base angles are equal.
### Proof That Base Angles of Isosceles Triangles Are Equal
Why are the base angles of an isosceles triangle equal? In mathematics, we study proofs, so let’s prove why the base angles are equal. Consider the following isosceles triangle where point D is the midpoint of BC.
The proof is as follows.
• In △ABD and △ACD
• AB=AC: Definition of an isosceles triangle – (1)
• BD = CD: Point D is the midpoint of BC – (2)
• From (1), (2), and (3), since Side – Side – Side (SSS), △ABD≅△ACD
• Since △ABD≅△ACD, ∠B=∠C
By proving that the triangles are congruent, we can prove that the base angles of an isosceles triangle are equal. In an isosceles triangle, the two sides are equal and the base angles are always equal.
## Properties of Right Triangles and Hypotenuse
Another special triangle that we need to learn at the same time as the properties of isosceles triangles is the right triangle. If a triangle has an angle of 90°, it is called a right triangle.
Right triangles have hypotenuse. General triangles do not have a hypotenuse. However, right triangles are special triangles, and we will use the hypotenuse to explain the properties of right triangles. The hypotenuse of a right triangle refers to the following part.
A triangle with an angle of 90° is the definition of a right triangle. Right triangles also have two acute angles in addition to the hypotenuse; any angle smaller than 90° is called an acute angle. In a right triangle, two angles that are not 90° are always acute angles.
### Congruence Theorem for Right Angle Triangles: HL
It is important to note that right triangles have their own congruence conditions in addition to the triangle congruence theorems. The four congruence theorems for triangles are as follows.
• Side – Side – Side (SSS) Congruence Postulate
• Side – Angle – Side (SAS) Congruence Postulate
• Angle – Side – Angle (ASA) Congruence Postulate
• Angle – Angle – Side (AAS) Congruence Postulate
In addition to these, there is a congruence theorem that exists only for right triangles. The following is a right triangle congruence theorem.
• Hypotenuse – Leg (HL); Sides other than the hypotenuse are called legs.
There are a total of five congruence theorems for triangles. In addition to the triangle congruence theorems, try to remember the right triangle congruence condition.
-It’s Not Enough That Two Angles Are Equal
Some people consider the congruence condition of right triangles when the two angles are equal. In this case, however, the two right triangles are not necessarily congruent. There are cases where they have different shapes, as shown below.
Just because the angles are equal does not mean that they are congruent. Triangles are not congruent if they do not satisfy the congruence condition.
## Exercises: Proof of Shapes
Q1: Prove the following figure.
For the following figure, △ABC is an isosceles triangle with AB = AC. If BD⊥AC and CE⊥AB, prove that △BCE≅△CBD.
Mixed problems of isosceles triangles and right triangles are frequently asked. So, try to understand the properties of the two shapes.
• In △BCE and △CBD
• ∠BEC = ∠CDB = 90°: Given – (1)
• BC = CB: Common line – (2)
• ∠EBC = ∠DCB: Base angles of an isosceles triangle are equal – (3)
• From (1), (2), and (3), since Angle – Angle – Side (AAS), △BCE≅△CBD
Q2: Prove the following figure.
For the figure below, △ABC is a right triangle with ∠ABC = 90°. From point B, draw a line BD perpendicular to side AC. Also, draw a bisector of ∠BAC, and let the intersection points be E and F, respectively, as shown below. Prove that BE=BF.
There are several possible ways to prove that the lengths of the sides are the same. The most common way is to find congruent triangles. However, looking at the figure, it seems that there are no two triangles that are congruent.
So, let’s think of another way to prove that the side lengths are the same. The property of an isosceles triangle is that the two sides are equal, and the base angles are equal. Therefore, if we can prove that the two angles are the same, then we know that the triangle is an isosceles triangle and the two side lengths are equal.
First, let’s focus on △ABF. ABF is a right triangle, and ∠AFB is as follows.
• $∠AFB=180°-90°-∠BAF$
• $∠AFB=90°-∠BAF$ – (1)
On the other hand, what about ∠AED? △AED is a right triangle, and the angle AED is as follows.
• $∠AED=180°-90°-∠EAD$
• $∠AED=90°-∠EAD$ – (2)
Also, since the vertical angles are equal, ∠AED = ∠BEF. Therefore, for (2), we can replace ∠AED with ∠BEF to get the following.
• $∠BEF=90°-∠EAD$ – (3)
Since ∠BAC is bisected, the angles of ∠EAD and ∠BAF are the same. Therefore, we have the following.
1. $∠AFB=90°-∠BAF$ – (1)
2. $∠AFB=90°-∠EAD$ – Substitute (3)
3. $∠AFB=90°-(90°-∠BEF)$
4. $∠AFB=∠BEF$
Thus, we know that △BEF is an isosceles triangle because the angles of ∠AFB and ∠BEF are equal. Also, since it is an isosceles triangle, BE=BF.
## Use the Properties of Special Triangles and Prove Them
Problems involving the calculation of angles and the proof of figures are frequently asked in mathematics. In these problems, special triangles are often used. Special triangles include isosceles triangles and right triangles.
There are two characteristics of isosceles triangles. The first is that the two sides are equal. The second is that each base angle is equal. On the other hand, right triangles have a congruence theorem. There is a congruence theorem available only for right triangles, so try to remember it.
It is important to remember that isosceles triangles and right triangles are sometimes asked as mixed problems. You often need to understand the properties of both isosceles triangles and right triangles to be able to solve the problem. Try to answer using the characteristics of both isosceles and right triangles.
As special triangles, isosceles triangles and right triangles are frequently asked. They are important figures in mathematics, so remember their definitions, properties, and congruence theorems before solving them. | 2,190 | 8,781 | {"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} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.934435 |
https://www.media4math.com/SATMathOverview--QuadraticExpressionsEquationsFunctions | 1,708,588,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00025.warc.gz | 934,064,101 | 30,792 | # Quadratic Expressions, Equations, and Functions
## Overview
Quadratic Expressions, Equations, and Functions are found throughout the SAT, so it’s very important to be comfortable with all the key aspects of this topic. In particular, make sure you know these concepts:
• Standard Form
• Vertex Form
• Factored Form
• Properties of Parabolas
• Graphical Solutions
• Discriminant
Let’s look at each of these in more detail:
Quadratic expressions include a variable whose exponent is 2, although the product of two linear variables (for example, xy) is also quadratic. Here are some examples of quadratic expressions:
Be sure you know how to translate a verbal expression into a quadratic expression like the ones shown above. Here are some examples:
“A number times itself.” “A number multiplied by one more than the number.” “The product of a number and one less than the number.”
Expressions aren’t equations but they are an important component of quadratic equations. Also, you can add and subtract quadratic expressions and still have a quadratic expression. Multiplying or dividing quadratic expressions will result in a non-quadratic expression. Here are some examples:
Some of the questions you’ll be asked will involve rewriting quadratic expressions to equivalent forms. To simplify this task familiarize yourself with these quadratic identities:
Binomial Expansion Binomial Squared Difference of Squares
Another important skill involves factoring quadratic expressions. This is a skill that also comes into play when solving quadratic equations. Here are some examples.
Factoring a constant. Factoring a linear term Factoring a quadratic into a binomial squared Factoring a quadratic into the product of two linear expressions.
To learn more about factoring quadratics, click on this link. Now look at some SAT-style questions that focus on quadratic expressions.
SAT Skill: Working with Quadratic Expressions Example 1 Simplify the following expression. Combine like terms and simplify. Keep track of how the subtraction symbol will change the signs of the terms in the expression on the right. Example 2 Simplify the following expression. Expand the term on the left (use a calculator for the squared decimal terms): Combine like terms and simplify. Keep track of how the subtraction sign changes the signs of the terms on the right. Example 3 In the equation below, what is a possible value for k? You should recognize that the term on the left is a difference of squares, which can be rewritten this way: In factored form, you can find the equivalent values for k and a: Example 4 What quadratic expression is equivalent to this expression? Factor the numerator, which is a difference of squares, then simplify: Example 5 What is this expression in expanded form? Use the binomial squared identity: Example 6 Look at the following equation. If a + b = 10, what are two possible values for c? Expand the term on the left to a quadratic in standard form. Compare the two terms in standard form: From this we can conclude the following: We also know that a + b = 10. We can substitute to create this equation: Factor this quadratic to find the values for b, and corresponding values for a: Now we can find the possible values for c:
Before studying what a quadratic function is, make sure you are comfortable with the following concepts, which we will also review:
• What a function is
• Independent variable
• Dependent variable
• Domain
• Range
• Different representations of functions
#### Brief Reviewof Functions
What Is a Function? A function is a one-to-one mapping of input values (the independent variable) to output values (the dependent variable). Click on this link to see a quick tutorial on what a function is. This slide show goes over the following key points:
• For every input value (x), there is a unique output value, f(x).
• Functions can be represented as equations, tables, and graphs.
• A function machine is a useful visual representation of the input/output nature of functions.
Dependent/Independent Variables. When one variable depends on another, then it is the dependent variable. For example, the faster your speed, the farther you travel. Suppose that speed is represented by the variable s and the distance traveled is represented by the variable d
Here’s how to describe the relationship between s and d:
The faster the speed, the more distance traveled.
Distance is dependent on speed.
Distance is a function of speed.
d = f(s)
When studying functions, make sure you are comfortable telling the difference between the independent variable and dependent variable. Get comfortable using function notation. To learn more about function notation, click on this link.
Domain and Range. A function shows the relationship between two variables, the independent variable and the dependent variable. The domain is the allowed values for the independent variable. The range is the allowed values for the dependent variable. The domain and range influence what the graph of the function looks like.
For a detailed review of what domain and range are, click on this link to learn more. You’ll see definitions of the terms domain and range, as well as examples of how to find the domain and range for given functions.
Multiple Representations of Functions. We mentioned previously that functions can be represented in different ways. In fact, any function can be represented by an equation, usually f(x) equal to some expression; a table; or a graph. For a detailed review of multiple representations of functions, click on this link, to see a slide show that includes examples of these multiple representations.
#### Quadratic Functions in Standard Form
The most common form of a quadratic function is the standard form.
The standard form is also the most common form used to solve a quadratic equation.
To see examples of graphing quadratic functions in standard form, click on this link. This slide show also includes a video tutorial.
#### Quadratic Functions in Vertex Form
Another way that a quadratic function can be written is in vertex form
To see examples of quadratic functions in vertex form, click on this link. This slide show tutorial walks you through the difference between standard form and vertex form. It also includes examples and two Desmos activities where you can graph these two types of equations.
#### Quadratic Functions in Factored Form
Another way that a quadratic function can be written is in factored form. From your work with quadratic expressions, you saw some techniques for factoring a quadratic. If you can write a quadratic function as the product of linear terms, then it is much easier to solve the corresponding the quadratic equation.
To see examples of quadratic functions in factored form, click on this link
#### Graphs of Quadratic Functions
The graph of a quadratic function is known as a parabola
To learn more about the properties of parabolas and their graphs, click on this link.
#### Special Case of Quadratics: Equation of a Circle
The equation of a circle is not a quadratic function, but it is a quadratic relation. This is the equation of a circle:
The coordinates of the center of the circle are (h, k) and the radius of the circle is r.
Quadratic equations are usually written as a quadratic expression in standard form equal to zero.
A quadratic equation can have two, one, or zero real number solutions. There are several ways to solve a quadratic. These are the methods we’ll be looking at:
• The Quadratic Formula
• Factoring
• Graphing the Quadratic Function.
Let’s look at the first method, which will work for any quadratic equation.
#### The Quadratic Formula
When a quadratic equation is written in standard form, like the one shown below, then you can use the quadratic formula to find the solutions to the equation.
Use the a, b, and c values from the quadratic equation and plug them into the quadratic formula:
To learn more about using the quadratic formula to solve quadratic equations, click on this link. This slide show includes a video overview of the quadratic formula and a number of detailed math examples.
Before using the quadratic formula, calculate the discriminant, which is the term under the square root sign of the quadratic formula.
To see examples of using the discriminant, click on this link.
#### Factoring
You’ve already seen how to factor quadratic expressions into the product of linear terms. That same idea can be used to factor certain quadratic expressions in order to find the solutions to the equation.
A factored quadratic equation will look something like this:
The solutions to this equation are x = a and x = b
A more simplified version of a factored quadratic can look like this:
The solutions to this equation are x = 0 and x = a
The previous two examples both had two solutions. There is a factored form that has one solution:
This is the case of the binomial squared. In this case the solution to the equation is x = a
The simplest example of the binomial squared is this:
The solution to this is x = 0.
If a quadratic cannot be easily factored, then you should use the quadratic formula or graph the quadratic.
To see examples of using factoring to solve a quadratic equation, click on this link.
#### Solving by Graphing
A visual approach to solving quadratic equations is to graph the parabola. There are three cases to look at.
Case 1: Two solutions. If the graph of the parabola intersects the x-axis twice, then there are two solutions.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola intersects the x-axis at x = 2 and x = 4. Those are the solutions to the quadratic equation. In fact, you can rewrite the quadratic in factored form:
Case 2: One solution. If the graph of the parabola intersects the x-axis once, then there is only one real number solution.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola intersects the x-axis at x = -2. This is the solution to the quadratic equation. In fact, you can rewrite the quadratic as a binomial squared:
Case 3: No real solutions. If the graph of the parabola doesn’t intersect the x-axis, then there are no real solutions to the quadratic equation.
Suppose you are solving this quadratic equation:
To find the solution graphically, then graph the corresponding quadratic function.
Notice that this parabola doesn’t intersect the x-axis. When this happens, the quadratic equation doesn’t have real number solutions. It does, however, have complex number solutions, which you can find using the quadratic formula.
Summary of Solving by Graphing. When a parabola intersects the x-axis, then the parabola has at least one real number solution. These intersection points are also referred to as:
• x-intercepts
• Zeros of the Quadratic Function
• Roots of the Quadratic Equation
To learn more about solving a quadratic equation graphically, click on the following link. This includes a video tutorial and several worked-out math examples.
SAT Skill: Solving Quadratic Equations Example 1 The parabola with the equation shown below intersects the line with equation y = 16 at two points, A and B. What is the length of segment AB? For a question of this type, where references are made to graphs and intersection points, it’s best to draw a diagram to get a better understanding of the problem. The equation shown is of a parabola in vertex form that intersects y = 16. Sketch that. We know the y coordinates for A and B; in both cases it’s 16. To find the corresponding x-coordinates, solve this equation: The x-coordinate for point A is x = 4 and the for B it’s x = 12. So the distance from A to B is the difference, or 8. Example 2 In the quadratic equation below, a is a nonzero constant. The vertex of the parabola has coordinates (c, d). Which of the following is equal to d? When a quadratic is written in standard form, the vertex has these coordinates: Write the function in standard form. Now find the corresponding x and y coordinates with this equation. Example 3 The parabola whose equation is shown below intersects the graph of y = x at (0, 0) and (a, a). What is the value of a? To get a better understanding of this problem, draw a diagram. To find the value of a, solve the following equation: | 2,581 | 12,519 | {"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} | 4.65625 | 5 | CC-MAIN-2024-10 | latest | en | 0.893468 |
https://almostsurelymath.blog/2019/09/02/a-devious-bet-the-borel-cantelli-lemma/ | 1,679,863,431,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00394.warc.gz | 123,952,155 | 25,172 | # A Devious Bet: The Borel-Cantelli Lemma
(Authors note: I tried to make this post accessible possible, however I recommend the reader to familiarize themselves with basic notations and facts in set theory and on infinite series before reading if they aren’t already)
Suppose you are offered a bet as follows, by a totally not evil and totally not a squid monster disguised as a man casino owner:
The bet will have (countably) infinitely many steps. In each you win or lose money, the only thing the totally not a squid monster tells you is that in each step, on average, you gain money. Furthermore, that average increases as the bet goes on.
How could the bet have infinitely many steps and still take a finite amount of time? I don’t know, it’s not like the person giving you the bet has Lovecraftian squid monster powers or something.
Do you take the bet? Well let’s say you do, let’s see your bank balance:
Oh, look it’s negatively infinite. Honestly I’m surprised there hasn’t been any integer flow in the process…
So what trickery did the not squid monster use? Let’s define Xn as the profit in dollars at step n (n is a positive integer), and look at its distribution:
$(X_n)_{n=1}^{\infty} \textup{ are independent random variables such that for all}\: n\in\mathbb{N} \textup{:}$$(X_n)_{n=1}^{\infty} \textup{ are independent random variables such that for all}\: n\in\mathbb{N} \textup{:}$
$\mathbb{P}(X_n=3^n)=2^{-n},\: \mathbb{P}(X_n=-1)=1-2^{-n}$$\mathbb{P}(X_n=3^n)=2^{-n},\: \mathbb{P}(X_n=-1)=1-2^{-n}$
So it seems you win with a low probability, which gets exponentially lower the more time passes. But on the other hand, when you win, you win A LOT, and the more you play the more you win, and what you win gets exponentially higher.
Let’s see if the not squid monster told the truth by checking the expectation (average, in other words) of Xn:
$\mathbb{E}(X_n)=\mathbb{P}(X_n=3^n)\cdot(3^n)+\mathbb{P}(X_n=-1)\cdot(-1)=$$\mathbb{E}(X_n)=\mathbb{P}(X_n=3^n)\cdot(3^n)+\mathbb{P}(X_n=-1)\cdot(-1)=$
$2^{-n}(3^n)-(1-2^{-n})=(3/2)^n-1+2^{-n}>0$$2^{-n}(3^n)-(1-2^{-n})=(3/2)^n-1+2^{-n}>0$
So the not a squid monster didn’t lie, the bet matched his description. Not only that, as n grows the average profit gets exponentially higher! How did you lose?
So what happened?
The key fact is that although the average profit at each step is very high and grows exponentially, you win only a finite amount of times. In particular, the amount of money you earn at those steps is finite. In all other steps, you lose 1 dollar (as the profit Xn is negative 1). Since there are infinitely many steps where you lose, you lose all the money you earned at the steps where you won, and then proceed to infinity and continue to lose money. Thus at the end, you lose an infinite amount of money.
But why do you win only finitely many times. Well, that is due to the Borel-Cantelli Lemma, which states the following (almost surely means with probability one):
$\textbf{Theorem (Borel-Cantelli Lemma): }\: \textup{Let} \:(A_n)_{n=1}^\infty\: \textup{be random events such that the sum of their probabilities }\mathbb{P}(A_n)\textup{ converges}$$\textbf{Theorem (Borel-Cantelli Lemma): }\: \textup{Let} \:(A_n)_{n=1}^\infty\: \textup{be random events such that the sum of their probabilities }\mathbb{P}(A_n)\textup{ converges}$
$\textup{then almost surely only finitely many of }\: A_n\: \textup{ occur.}$$\textup{then almost surely only finitely many of }\: A_n\: \textup{ occur.}$
$\textup{So in other words, if }\sum _{n=1}^\infty\mathbb{P}(A_n)<\infty, \textup{ then:}$$\textup{So in other words, if }\sum _{n=1}^\infty\mathbb{P}(A_n)<\infty, \textup{ then:}$
$\mathbb{P}(A_n \textup{ happens infinitely often})=\mathbb{P}(\bigcap_{n=1}^\infty\bigcup _{k=n}^\infty A_k)=0$$\mathbb{P}(A_n \textup{ happens infinitely often})=\mathbb{P}(\bigcap_{n=1}^\infty\bigcup _{k=n}^\infty A_k)=0$
Now we can see why. The probability of winning at step n is 2-n. The sum of those probabilities is finite and equal to one. So by the Borel-Cantelli lemma almost surely you win only finitely many times.
A Proof of the Borel-Cantelli Lemma:
Let’s see then why the Borel-Cantelli lemma holds. The proof makes use of two fundamental facts in mathematics as a whole. The first is known as the union bound and is as follows:
$\textup{If }(B_n)_{n=1}^\infty \textup{ are random events for each n then } \mathbb{P}(\bigcup_{n=1}^\infty B_n)\leq\sum_{n=1}^\infty\mathbb{P}(B_n)$$\textup{If }(B_n)_{n=1}^\infty \textup{ are random events for each n then } \mathbb{P}(\bigcup_{n=1}^\infty B_n)\leq\sum_{n=1}^\infty\mathbb{P}(B_n)$
In other words, the probability of at least one of Bn happening is less then or equal to the sum of their probabilities. This is a pretty intuitive fact, to prove it, it helps to think first of the case where Bn are disjoint (so no two of them can happen simultaneously). The idea is that if two events intersect, their union is smaller than if they were disjoint as some of the elements in the union appear in both sets. So the probability of the union is also lower than the probability of the union if they were disjoint, which is the sum of probabilities.
The second fact is an important fact about converging series:
$\textup{If }\sum_{n=1}^\infty b_k<\infty, \textup{then the sequence of 'tail sums' }\sum_{n=k}^\infty b_k \textup{ approaches zero as k goes to infinity}$$\textup{If }\sum_{n=1}^\infty b_k<\infty, \textup{then the sequence of 'tail sums' }\sum_{n=k}^\infty b_k \textup{ approaches zero as k goes to infinity}$
This is again a pretty intuitive fact, as we expect that the more we sum, the more that remains to get to the total, the ‘tail sums’, approaches zero.
Now let’s see how this helps prove the Borel-Cantelli lemma. Note the following: if An happens infinitely often, then for each natural number k, there must be some number n after k for which An happens. Else, all occurences of An happen before k, so there are finitely many of them in contradiction with our assumption.
Therefore, for each k, we can bound the probability of An happening infinitely often as follows:
$\mathbb{P}(A_n\textup{happens infinitely often}) \leq \mathbb{P}(\bigcup_{n=k}^\infty A_n)$$\mathbb{P}(A_n\textup{happens infinitely often}) \leq \mathbb{P}(\bigcup_{n=k}^\infty A_n)$
We can use the union bound to bound the probability of An happening after k:
$\mathbb{P}(\bigcup_{n=k}^\infty A_n)\leq\sum_{n=k}^\infty\mathbb{P}(A_n)$$\mathbb{P}(\bigcup_{n=k}^\infty A_n)\leq\sum_{n=k}^\infty\mathbb{P}(A_n)$
But we have on the right hand side the tail of the converging sum of P(An). So as k grows larger, the right hand side approaches zero. So our bound on An happening infinitely often approaches zero and thus we conclude the probability of An happening infinitely often is zero. In other words, almost surely An happens finitely many times. This completes the proof. Q.E.D
A Second Borel Cantelli Lemma:
The Borel-Cantelli lemma dealt with the case where the sum of probabilities converged. A natural question to ask now is what about the case where the sum diverges? It turns out there is an analog of the lemma for that case as well:
$\textbf{Theorem (Borel-Cantelli Lemma 2): }\: \textup{Let } (A_n)_{n=1}^\infty \textup{ be independent random event such that the sum of their probabilities diverges.}$$\textbf{Theorem (Borel-Cantelli Lemma 2): }\: \textup{Let } (A_n)_{n=1}^\infty \textup{ be independent random event such that the sum of their probabilities diverges.}$
$\textup{Then, almost surely, }A_n\textup{ occurs for infinitely many n}$$\textup{Then, almost surely, }A_n\textup{ occurs for infinitely many n}$
$\textup{So, in other words, if } \sum_{n=1}^\infty \mathbb{P}(A_n)=\infty ,\textup{then:}$$\textup{So, in other words, if } \sum_{n=1}^\infty \mathbb{P}(A_n)=\infty ,\textup{then:}$
$\mathbb{P}(A_n \textup{ happens infinitely often})=\mathbb{P}(\bigcap_{n=1}^\infty\bigcup _{k=n}^\infty A_k)=1$$\mathbb{P}(A_n \textup{ happens infinitely often})=\mathbb{P}(\bigcap_{n=1}^\infty\bigcup _{k=n}^\infty A_k)=1$
(Note that this version of the lemma requires the events to be independent. To see why the lemma isn’t true otherwise, try to think of a case where after one An fails, all following An fail, and yet the sum of probabilities still converges)
A nice application of this version of the lemma is as follows. Suppose you draw a poker hand infinitely many times (resetting the deck each time), each draw independent from the others. Then the probability of drawing a royal flush at each attempt is very low, but a positive constant nevertheless (equal to 1/649,740). So the sum of probabilities of drawing a royal flush for each attempt diverges. Thus almost surely, you will draw infinitely many royal flushes.
Maybe you can try to get back on the not a squid monster for taking your money using that. By the way, why does it smell like tentacles here all of a sudden?
## 4 thoughts on “A Devious Bet: The Borel-Cantelli Lemma”
1. RN says:
best intuitive explanation i’ve ever come across | 2,693 | 9,067 | {"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": 1, "img_math": 32, "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-2023-14 | longest | en | 0.837114 |
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• Level: GCSE
• Subject: Maths
• Word count: 1924
Mathematics Coursework - Beyond Pythagoras
Extracts from this document...
Introduction
Edexcel 2002 BEYOND PYTHAGORAS
Mathematics GCSE Coursework Ms Pathak
1)
The numbers 3, 4, and 5 satisfy the condition
3² + 4² = 5²
because 3² = 3x3 =9
4² = 4x4 = 16
5² = 5x5 = 25
and so
3² + 4² = 9 + 16 = 25 = 5²
I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) ² + (middle number) ² = (largest number) ².
a) 5, 12, 13
5² + 12² = 25 + 144 = 169 = 13²
b) 7, 24, 25
7² + 24² = 49 + 576 = 625 = 25²
2) Perimeter
b)
Middle
15
112
113
240
840
8
17
144
145
306
1224
9
19
180
181
380
1710
10
21
220
221
462
2310
I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest side. And also if you can see in the shortest side column, it goes up by 2. I have also noticed that the area is
½ (shortest side) x (middle side).
3)
In this section I will be working out and finding out the formulas for:
• Shortest side
• Middle side
• Longest side
In finding out the formula for the shortest side I predict that the formula will be something to do with the differences between the lengths (which is 2). But I don’t know the formula so I will have to work that out.
Firstly I will be finding out the formula for the shortest side.
3 5 7 9 11
2 2 2 2
The differences between the lengths of the shortest side are 2. This means the equation must be something to do with 2n.
Let’s see…
Nth term Length of shortest side 1 3
2n
2 x 1 = 2 (wrong)
There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I’m correct I will now test this formula.
2n+1
Nth term Length of shortest side 1 3
2x1+1=3 (correct)
Just in case I will test this formula in the next term:
2n+1
Nth term Length of shortest side 2 5
2x2=4
4+1=5 (correct)
I now have to work out the formula for the middle side.
Conclusion
So 12 is the perimeter for the first term
2nd term 5 + 12 + 13 = 30
3rd term 7 + 24 + 25 = 56
And so on. All I have to do is put all the 3 formulas together.
Perimeter = (shortest side) + (middle side) + (longest side)
= 2n + 1 + 2n² + 2n + 2n² + 2n + 1
= 2n² + 2n² + 2n + 2n + 2n + 1 + 1
= 4n² + 6n + 2
If I have done my calculations properly then I should have the right answer. To check this I am going use the 4th, 5th and 6th terms.
4th term:
4n² + 6n² + 2 = perimeter
4 x 4² + 6 x 4 + 2 = 9 + 40 + 41
64 + 24 + 2 = 90
90 = 90
It works for the 4th term
Let’s see if it works for the 5th term:
4n² + 6n² + 2 = perimeter
4 x 5² + 6 x 2 = 11 + 60 + 61
100 + 30 + 2 = 132
132 = 132
And it works for the 5th term
And finally the 6th term:
4n² + 6n² + 2 = perimeter
4 x 6² + 6 x 6 + 2 = 13 + 84 + 85
144 + 36 + 2 = 182
182 = 182
It works for all the terms so:
Perimeter = 4n² + 6n + 2
Like the area I know that the area of a triangle is found by:
Area = ½ (b x h)
b = base
h = height
Depending on which way the right angled triangle is the shortest or middle side can either be the base or height because it doesn’t really matter which way round they go, as I’ll get the same answer either way.
Area = ½ (shortest side) X (middle side)
= ½ (2n + 1) x (2n² + 2n)
= (2n + 1)(2n² + 2n)
I will check this formula on the first two terms:
(2n + 1)(2n² + 2n) = ½ (b x h)
(2 x 1 + 1)(2 x 1² + 2 x 1) = ½ x 3 x 4
3 x 4 = ½ x 12
12 = 6
6 = 6
2nd term:
(2n + 1)(2n² + 2n) = ½ b h
(2 x 2 + 1)(2 x 2² + 2 x 2) = ½ x 5 x 12
5 x 12 = ½ x 60
60 = 30
30 = 30
It works for both of the terms. This means:
Area = (2n + 1)(2n² + 2n)
By Asif Azam
10L4
Mathematics Coursework
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• Ideas and feedback to | 2,434 | 6,833 | {"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} | 4.40625 | 4 | CC-MAIN-2017-34 | longest | en | 0.832828 |
https://ulysseszh.github.io/music/2020/03/27/notes-frequency.html | 1,722,998,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00534.warc.gz | 480,009,548 | 17,083 | ## The notes
First, let’s define what is a note.
A note is an element of the countable set $N$. Here exists a bijection from $\mathbb Z$ to $N$ denoted as $k\mapsto \nu_k$. Therefore, all notes form a sequence $\dots,\nu_{-2},\nu_{-1},\nu_0,\nu_1,\nu_2,\dots$
Our goal is to define a frequency assignment, which is a mapping $f:N\rightarrow\mathbb R^+$, whose meaning is the frequency (in Hz) of the sound of a note.
It is a natural idea to define a sequence $f_k\coloneqq f\!\left(\nu_k\right).$
It makes sense that the sequence is in strictly increasing order.
## The octaves
Now, let’s think about a musical interval. At this stage, a musical interval can be defined as an unordered pair of notes. After long time of experimenting, people find that they tend to think a musical interval extremely harmonic if it consists of such two notes that the frequency of one of them doubles that of the other.
In other words, the musical interval of $\nu_a$ and $\nu_b$ is extremely harmonic if $f_b=2f_a$. In fact, it is so harmonic that if the two notes are played simultaneously, a person tend to think there is only one note being played:
1 2 3 4 motif = c4 c g g a a g2 piano: V1: o4 motif motif V2: o5 motif
The audio above is an illustration for octave intervals. As can be heard, the first part of the audio is played in octave intervals while the second part is played in single notes.
Considering that, here comes an amazing idea by which we can kind of make the sequence $\left\{f_k\right\}$ seem “periodic”. Let the “period” be denoted as $n$. Because it is not virtually periodic, we tend to call it an octave instead of a period. After that, the constant $n$ is the length of an octave.
What on earth is an octave defined? It is $\forall k:f_{k+n}=2f_k.$ Why do we say an octave is like a period? It is because according to the explanation above, corresponding notes in different octaves sound so harmonic that a person almost think they are the same. In this way, for some questions, we only need to consider a single octave instead of all notes. Let’s define the base octave, notes of which can generate all other notes by multiplying the frequency by a power of 2: $O_0\coloneqq\left\{\nu_k\,\middle|\,k\in T\right\},$ so we can say the frequency assignment has $n$ different tones. A tone is an integer in $T\coloneqq\left[0,n\right)\cap\mathbb Z$ representing where a note is in an octave.
We can thus define a sequence of octaves $O_m\coloneqq\left\{\nu_{k+mn}\,\middle|\,\nu_k\in O_0\right\}.$
In fact, any octave can be the base octave. They are all the same. For any $m$, all notes can be generated by notes in $O_m$. $\begin{array}{|c|cccccc|} \hline T & \cdots & O_{-1} & O_0 & O_1 & O_2 & \cdots\\ \hline 0&\cdots& \nu_{-n}&\nu_0& \nu_l & \nu_{2n}&\cdots\\ 1&\cdots&\nu_{-n+1}&\nu_1&\nu_{n+1}&\nu_{2n+1}&\cdots\\ 2&\cdots&\nu_{-n+2}&\nu_2&\nu_{n+2}&\nu_{2n+2}&\cdots\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ n-1&\cdots&\nu_{-1}&\nu_{n-1}&\nu_{2n-1}&\nu_{3n-1}&\cdots\\ \hline \end{array}$ If we define $p_k\coloneqq\log_2f_k$, it is interesting to see that $\forall k:p_{k+n}=p_k+1.$ From that, we can naturally think an excellent frequency assignment be defined as $p_k\coloneqq p_0+\frac kn,$$\p{1}$ which is an elegant arithmetic progression.
## The octave group
Note that here “group” is the group concept in algebra.
Let $O_0$ be isomorphic to the additive group of $\mathbb Z/n\mathbb Z$, the integers modulo $n$, under the isomorphism $k\mapsto\nu_k$, which means to make $O_0$ a cyclic group of order $n$. I call this group an octave group.
To make you have a good sense of what on earth the group looks like, the definition of its group operation can be defined as $\nu_a\circ\nu_b\coloneqq\nu_{\left(a+b\right)\mathbin\%n},$$\p{2}$ where $x\mathbin\%y\coloneqq x-y\left\lfloor\frac xy\right\rfloor$.
Note that this binary operator $\circ$ can be extended to be used for the whole $N$ while the definition remains the same as Formula 2.
The octave group has a musical meaning which we should take a further look at musical intervals to find out.
## The musical intervals
Taking Formula 1, people find that although it is sometimes subjective whether a musical interval sounds harmonic or not, it does not depend on where the interval is located but on how far the two notes making up the interval are.
Taking this sense, we can consider only those intervals involving $\nu_0$ because we can always translate a interval so that one of its notes is $\nu_0$. Taking this idea, we can conclude that an interval can be represented by a note $\nu_k$ because the interval is equivalent to another interval which consists of $\nu_0$ and $\nu_k$.
We can make this idea even further. Previously, I have stated that a note can always be generated by a note in $O_0$. Therefore, an interval can be represented as a note in $O_0$.
Now, let’s look back to the octave group $\left(O_0,\circ\right)$. Denote the interval $b$ as that represented by $\nu_b$. Then, translate interval $b$ to such a location that its lower note is $\nu_a$. Then, its higher note represents the same interval as $\nu_a\circ\nu_b$.
Well, why do we focus on the group? It is because we need to mention an important concept in group theory, which is “generator”. It has something to do with determining the value of $n$.
The value of $p_0$ does not matter because changing it is just a translation of the whole sequence. What really matters is the value of $n$.
The inventor of the current prevailing frequency assignment (which is the $12$-tone equal temperament shown in Formula 3) may think the generator of the group a vital thing. Actually, people think it a wonderful thing that a note representing a very harmonic interval is a generator of the octave group.
Fortunately, such a goal is achievable. People find that if $p_b=p_a+\frac7{12}$, then the interval (which is the perfect fifth interval if you know music theory) consisting of $\nu_a$ and $\nu_b$ sounds very harmonic:
1 2 3 4 motif = e4 a f2 d4 g c2 piano: V1: motif V2: (transpose 7) motif
Furthermore, as can be seen in the following table, $\nu_7$ is a generator of the group $O_0$ if $n\coloneqq12$: $\begin{array}{|c|cccccccccccc|} \hline j&0&1&2&3&4&5&6&7&8&9&10&11 \\\hline \nu_7^{\circ j}& \nu_0&\nu_7&\nu_2&\nu_9&\nu_4&\nu_{11}&\nu_6&\nu_1& \nu_8&\nu_3&\nu_{10}&\nu_5 \\\hline \end{array}$ Thus, wonderful! Let’s take $n\coloneqq12$.
## The $12$-tone equal temperament
The $12$-tone equal temperament is the most popular frequency assignment used nowadays. It is defined as $f_k\coloneqq16.3516\cdot 2^\frac k{12},$$\p{3}$ which can be derived from Formula 1 taking $p_0\coloneqq4.03136\qquad n\coloneqq12.$
The frequency assignment has $12$ different tones, $7$ of which have their names: \begin{align*} \mathrm C_m&\coloneqq\nu_{12m},\\ \mathrm D_m&\coloneqq\nu_{12m+2},\\ \mathrm E_m&\coloneqq\nu_{12m+4},\\ \mathrm F_m&\coloneqq\nu_{12m+5},\\ \mathrm G_m&\coloneqq\nu_{12m+7},\\ \mathrm A_m&\coloneqq\nu_{12m+9},\\ \mathrm B_m&\coloneqq\nu_{12m+11}. \end{align*} The famous “middle C” is $\mathrm C_4$.
This notation is called the scientific pitch notation.
Note that in fact, this definition of $12$-tone equal temperament has some slight error. The accurate value for $p_0$ is $p_0\coloneqq\log_255-\frac74$ because it is stipulated that $f\left(\mathrm A_4\right)=440$, which is standardized as ISO 16 and known as A440.
## Why $\frac 7{12}$
People think if the ratio of two frequencies is a simple rational number, then the interval of the two notes is harmonic.
$2^\frac 7{12}\approx\frac32$, which is a simple ratio. Harmonic, huh.
(Finally, as is a notice, codes appearing above are alda codes, which are used to write music.) | 2,329 | 7,759 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 80, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "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-2024-33 | latest | en | 0.926387 |
https://es.scribd.com/doc/310562070/lesson-plan-2 | 1,558,688,500,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00004.warc.gz | 475,692,944 | 69,582 | Está en la página 1de 10
# Updated 4/27/160:32 A4/P4
## Danielson Aligned Lesson Plan Template for Formal Observations
Yunjung Hong
Formal Observation #2 Lesson Plan
Primary Subject Area and Grade Level: List the primary content area for this lesson. List the beginning and ending grade levels for
which this lesson is appropriate.
Mathematics
Fractions, decimal, and percentages unit
Interdisciplinary Connections: Provide a listing of the subject area(s), in addition to the primary subject area that is incorporated in this
lesson.
(1a: Demonstrating Knowledge of Content and Pedagogy)
This lesson also incorporates what we are covering in our science unit. In science we are working on
car models and creating the perfect car to travel from point A to point B and drop an egg below so the
yolk falls into the pan. To do so, students must collect data on their car including the speed. This
incorporates the fractions aspect which comes into play when you use the centimeters per second
method.
Lesson Duration: State the approximate time frame for this lesson.
(1e: Designing Coherent Instruction)
## The lesson all together will take approximately 40 minutes.
There will be 10 minutes of instruction for the mini-lesson together as a whole class.
25 minutes for activity planned.
Final 5 minutes on wrap up and completing of an exit ticket for an assessment.
Relevance/Rationale: Consider how your outcomes and plan will engage students cognitively and build understanding. Why are the lesson
outcomes important in the real world? How is this lesson relevant to students in this class (interests, cultural heritages, needs)?
(1b: Demonstrating Knowledge of Students)
This lesson will introduce decimals to the students and allow them to make a connection with
decimals to fractions, and soon percentages.
This is important to the real world because this is how money is handled. Dollar is a whole, while the
coins are its parts. It is also the basis or purchasing items, figuring out sales, in the library (Dewey
Decimal System), anything that is timed after a certain distance (track and field), etc.
Outcomes/Objectives: What will students know and be able to do as a result of this lesson? Outcomes should be written in the form of
student learning and suggest viable methods of assessment. For teachers of English language learners: What language objectives will be
(1c: Setting Instructional Outcomes)
Students will be able to represent parts of a whole as decimals specifically to the tenths and
hundredths place.
Students will be able to rewrite fractions as decimals and vice versa.
Content Standard(s) and/or Common Core Learning Standard(s): For example: (CCSS) 4.NBT.3 Use place value
understanding to round multi-digit whole numbers to any place. Content area teachers should include appropriate English Language Arts
Common Core Standards for Content Areas, if appropriate, in addition to content standards.
(1c: Setting Instructional Outcomes)
5.NBT.A.1 Recognize that in a multi-digit number, a digit in the one place represents 10 times as
much as it represents in the place to its right and 1/10 of what it represents in the place to its left.
5.NBT.A.3 Read, write, and compare decimals up to the thousandths.
5.NBT.A.4 Use place value understanding to round decimals to any place.
Use of Formative Assessment to Inform Planning: Describe your students current levels of understanding of the content related to
the outcome for this lesson. What are some of the indicators that let you know that these outcomes and the lesson activities represent the
appropriate amount of cognitive challenge for all students?
(1f: Designing Student Assessments)
The students are aware of how to write fractions with a numerator and a denominator. They also know
## Updated 4/27/160:32 A4/P4
that a fraction represents a division problem. This may help their understanding of decimals because
then the division is actually done they will be able to see it turns out to be a decimals. For example;
is represented by 14 which equals to 0.25.
Students also know the place value of at least 4 digits to the left of the decimal point. This will help
students comprehend the amount of digits that go into the tenths and the hundredths.
Class Information: Describe any unique characteristics of the class (considerations may include: special needs, language levels, learning
styles, etc.). Describe how other adults (paraprofessionals, volunteers, co-teachers, resource teachers, etc.) will support student learning, if
applicable. Also include any other circumstances an observer should know about.
(1b: Knowledge of Students)
In this classroom there are 5 students with IEPs, I will have to incorporate modifications of the
students into the lesson. There are students with attention deficits which I plan on engaging with turn
and talk and engaging actions like thumbs up activities. My cooperating teachers will support students
learning by making sure students are on tasks and participate when asked. If students are falling off
track they will guide then back into the lesson. They will assist me with the behavior and classroom
management but not the content during this lesson.
I will keep a close eye on those students who are easily distracted, and have total participation
techniques throughout the lesson to keep them engaged.
I also have a media assistant that takes photographs, records videos of me while I teach. He will be
walking around and sitting in the back snapping photos of me. If he gets distracted by this task, he will
have to drop the job and focus on the lesson. I also told the student if he cannot appropriately assist me
I will have to choose another student for the job.
Math is right after lunch so students are rushing in, putting away their lunches and coats, and taking
their seats. There are 3 students that then leave the classroom for A&E which is advanced math.
Overview: Provide a brief overview of the lesson. The overview should provide the observer with a description of the lessons content and how
it relates to the larger unit. Include prerequisite knowledge required to meet lesson outcomes and relationship to future learning.
(1a: Demonstrating Knowledge of Content and Pedagogy)
The students will stay at their seats for this lesson. It will start with connecting to the previous lessons
in the unit and going over the objective. I will then teach a mini-lesson that is about 10 minutes. Then
students will break up into pairs for the activity that is planned. Students will play place value battleship. If students finish the game early, I will have a worksheet they can work on while other finishes
their games. I will wrap up the lesson by going over the objective again, and giving the homework
ticket problem. Homework ticket is an exit ticket but instead of leaving the room, the problem is
exchanged for their homework for the night. When students hand in their homework ticket, they will
then receive their homework as the final part of the lesson.
Technologies and Other Materials /Resources: List all materials, handouts, resources, and technology tools that are needed by the
student or the teacher to execute the lesson. Technologies may include hardware, software, and websites, etc. Materials and resources may
include physical resources (e.g. books, manipulatives, supplies, equipment, etc.) and/or people resources (e.g. guest speakers, librarian, etc.).
(1d: Demonstrating Knowledge of Resources)
I will need the white board for the mini-lesson. Students will copy it down in their math journals.
I will prepare the place value battleship boards.
Grouping Strategy: Describe how you will group students to facilitate learning of the outcomes of this lesson. What is the rationale for the
grouping strategy?
(1e: Designing Coherent Instruction)
The students will work in pairs for the battleship game. If there is an un-even amount of students there
will be one group of 3. I will pair the students up by similar math levels.
Academic Vocabulary: What key terms are essential to this content? What terms are essential to develop and extend students vocabulary?
(1a: Knowledge of Content and Pedagogy; 1b: Knowledge of Students)
## Updated 4/27/160:32 A4/P4
Lesson Procedures: The procedures should clearly describe the sequence of learning activities and should identify where and how all
materials, technology tools and student-created technology products, and reproducible materials/handouts are utilized in the lesson. Describe the
lesson sequence:
## How will the material be presented?
What questions will be posed to the students? What are the expected responses?
## How and when will the teacher model?
What opportunities will there be for guided practice, group work and individual practice?
How and when will you monitor student understanding throughout the lesson?
## What opportunities will there be for reflection and closure?
Include approximate time allocations for each portion of the lesson. Be very precise when explaining the teacher and student tasks during the
learning activities.
(1a: Demonstrating Knowledge of Content and Pedagogy; 1e: Designing Coherent
Instruction)
I will have the objective written an anchor chart along with the vocabulary words and todays lesson. I
will start off by having a student read the objective.
Okay my mathematicians, yesterday we were talking about fractions correct? Well today we are
going to use our fraction knowledge to represent fractions as decimals! By the end of the lesson you
will be able to write fractions as decimals and vice versa.
Okay so I have a problem here: Betty made 3 out of 10 shots, of 3/10 shots while playing basketball.
In your journals, give me 3 equivalent fractions using the denominator 100.
While students do so, draw a place value chart. _ _._ _
Explain that the places to the right are similar to those on the left. Discuss the value of teach place
with the students.
Okay now lets go back to the problem in your journal what is the fraction of shots Betty made? So
what is the decimal? students will respond by saying 0.3as their answer. Put this into the calculator as
310 and you will get 0.3. This will show the student that it works when you do the division.
When explaining, show on the place value chart, that the decimal is represented by the word AND if
there are numbers to the right of the decimal. So if this was 1.3 it will be said as one AND 3 tenths. If
not it will just be three tenths of her shots.
If time is available, another problem can be done. Tell the students Lets try another one. How many
pennies are in one dollar? 100. So what fraction represents 43 cents? 43/100. How can we show this in
decimal form? 0.43 this works because 40 is in the tens place and 3 is one hundredth of a dollar with is
one penny and all together there are 43 pennies which is 43 cents of one dollar. This problem will be
generated from the students. It will allow the students to practice on their own. How will I write and
say One dollar and 72 cents? That will be 1.72 and one and seventy two hundredths.
After the mini-lesson, I will divide the class into pairs. Each pair will get a place value battleship
board and a space to work in the room. Students will play the game with one another. The board will
have a ones place, tenths place, and hundredths place. The student can come up place a number 0-9
into each place value creating any number they choose. Students can double numbers if they please,
for example; 4.44 is acceptable. When both the students in the pair choose a number, the older student
2
## Updated 4/27/160:32 A4/P4
makes the first move. They will ask, Do you have a 5 in the tenths place The other student will
answer miss then the next student will go until they figure out the number. Once the number is
figured out the student will say is your ship number four AND fourty four hundreths? and student
will reply ship wreck. Each pair will play until they are completed with the game and one student
has won.
I will model how to play the game for the students as written above.
Once completed, students will go back to their seats and work on a worksheet till all groups are
finished.
Once all students are back at their seats, I will begin wrapping up by saying, Alright mathematicians,
now that we completed our lesson lets go back to our objective. Are we not able to represent fractions
with 10 and 100 as their denominator as a decimal? Yeah. You can also rewrite and say fractions as
decimals. Great job today. I will be giving you a short homework ticket, which must be completed in
And tomorrow, be prepared to look at the next place value to the right of the decimal which is the
thousandths place
Differentiation: Describe how you will differentiate instruction for a variety of learners, including students will special needs, English
Language Learners, and high achieving students to ensure that all students have access to and are able to engage appropriately in this lesson.
Be specific.
(1e: Designing Coherent Instruction)
For the students who are high achieving, I will not give a game helper sheet. This will allow them to
multitask and really focus on the game. For the students who need the helper sheet, I will allow them
to use it (attached below).
During the mini-lesson, if I notice a couple students who really are just not getting it, I will pull them
aside and have a guided small group session.
Assessment Criteria for Success:
How and when will you assess student learning throughout the lesson (formative)?
How will you and your students know if they have successfully met the outcomes?
What is the criteria for mastery of the lesson outcome(s)?
Describe any (formative and summative) assessments to be used.
(1f: Designing Student Assessments)
After wrapping up the lesson, I will give the students a homework ticket. The homework ticket works
just like an exit ticket but instead of exiting students will exchange it for that nights homework
assignment. For the homework ticket every student will get half a sheet of paper with some problems
that relates to what we just covered (attached below). When finished, students will bring it up to me
and I will hand them their homework. On this assessment I will add problems that they should be able
to solve based on the lesson that was just taught and maybe one challenging problem to further test
their knowledge.
Homework ticket is attached below.
Anticipated Difficulties: What difficulties or possible misunderstanding do you anticipate that students may encounter? How will you
prevent them from occurring?
(1a: Demonstrating Knowledge of Content and Pedagogy)
Students may have difficulties understanding the concept of the place values to the right of the
decimal points. If this is the case, I will have manipulative available for those who do not during the
small group session.
Also students will be finishing the place value battle ship, at different times so I will have a worksheet
prepared for them to complete after their game.
This is an easy concept for fifth graders to grasp, if I notice that they find it really easy, I will add the
thousandths place value to make it more challenging.
## Updated 4/27/160:32 A4/P4
Reflections: List at least three questions you will ask yourself after the lesson is taught.
(4a: Reflecting on Teaching)
1. In general, how successful was the lesson? Did the students learn what you
intended for them to learn? How do you know?
In general, my lesson went well. It seemed as if the students understood my
instruction, and according to the homework ticket everyone but 4 students got a
perfect hundred. This lesson plan was actually two lessons combined. Tenths and
hundredths were one lesson and thousandths was its own lesson. I decided to
combine these because my students math level with fractions are pretty high
and I thought they would breeze through this lesson. Like my thoughts, my
students really found this lesson simple. The objective was being able to turn
fractions into decimals and vice versa. By the end of the lesson, my students
were able to do so. I felt like I was a little rushed. This is probably because my
during first observed lesson, my mini-lesson turned out to be 20 minutes so I
wanted to shorten that time. But I felt myself talking fast and rushing. I think I
could have slowed down and still have been effective in teaching the students to
lesson.
2. If you have samples of student work, what do they reveal about the students
levels of engagement and understanding? Do they suggest modifications in how
you might teach this lesson in the future?
I saw their homework tickets along with their homework the day after. 17 out of
21 students got a perfect 100% on their exit ticket. All my students were able to
complete their homework with ease. The four students that did not get a 100%
on their homework ticket only got about an average 2 problems wrong. I think I
could pull that group of students and go over different strategies on turning
fractions into decimals by modeling and having them follow along. I think if I
gave a little more instructions and some practice problems they will be able to
further understand the lesson and be on level with the rest of the classroom.
3. Comment on your classroom procedures, student conduct, and your use of
physical space.
Before this lesson, my students had lunch. They know from their routine when
they enter lunch boxes are put away, they go to their seats, and take out their
math journals. I kept they students at their desk for this lesson because having
the students on the carpet for math is just something they are not used to and I
wanted to stay with how things have been. When the mini-lesson was over, I
paired students up for the game. After doing so, I gave each pair folders and
game boards so they can begin playing. I also let 3 pairs outside to play their
game so there can be more space in the classroom and so it does not get too
loud. For the wrap-up, I brought all the students in and back into their seats.
During the whole lesson, the students were mostly well behaved but after the
game, it was difficult to get their attention for the wrap-up because they were so
excited.
4. Did you depart from your plan? If so, how and why?
## Updated 4/27/160:32 A4/P4
I did depart from my plan slightly. For the most part, I followed it exactly besides
when I got towards the end. In the plan, it says to explain the game of Place
Value Battle Ship then group them and let them get started. At the heat of the
moment and the excitement of the students, I paired students up, handed out
the game boards, then explain the instructions. I think the reason for this is
because the students were very excited to play the game that I just completely
forgot to explain the directions. It was not until I handed everything out that I
thought, oh wait, they dont know how to play. I also veered during my wrap-up.
In the wrap-up according to the lesson plan, I was supped to re-read the objective
and students were to determine if I can check that off our anchor chart. Then
after I was supposed to give the homework tickets out and tell the students what
the outward look was. I was going to tell the students that this is going to help us
put fractions and decimals on the number line and placing them in order. I
followed all of this on the lesson plan, besides telling the students the outward
look. My lesson time was over and I was too rushed that I just forgot to do this
during my lesson.
5. Comment on different aspects of your instructional delivery (e.g., activities,
grouping of students, materials, and resources). To what extent were they
effective?
For the mini-lesson I taught them just like I have been for every other math
lesson. One thing I did do differently was having the students do an activity and
this activity being a game. The students love games and are very competitive so
I knew that this would be an effective activity. For the activity they were grouped
into pairs. I choose to pair them up because it was the most effective way to play
the game. I paired them up by in a very specific way. For the students who
understood the concept and are average or high math students, I just paired
randomly. It was the students who excel in math and have trouble in math that I
grouped heterogeneously. It ended up only being about 3 heterogeneous pairs. I
did this because the average students can have fun playing the game with one
another because their math levels are similar. For the students who exceled, I
gave them a partner that was slightly struggling or had a hard time with the
lesson. This is because the student who understands can guide the other student
during the game. I monitored all the groups and they all seem to work well
together and enjoy the game.
6. If you had an opportunity to teach this lesson again to the same group of
students, what would you do differently?
I would definitely slow down my pace. I would try not to rush and just make sure
that the students fully understand the lesson. In order to do so, I would follow the
me, us, you model in order to keep the mini-lesson short and simple. I would do
a problem, then do one as a class, let the students do one, then allow time for
the activity. I would also show a video or find another hook for the beginning of
the lesson. To get the students engaged from the beginning, I could have made
connections to the real world or show students a visual so they have a pre idea of
how important the lesson actually is.
7. What were some of the things you discussed with cooperating teachers?
## Updated 4/27/160:32 A4/P4
My cooperating teachers did enjoy the lesson. They agreed with some of the
things I had to say. They also thought that my lesson was slightly rushed. They
both knew that this is probably because I was worried about my mini-lesson
going over 10 minutes like my previous observed lesson. They said that they
could tell I was going fast just so I can make the time limit. They had said that
there is no need to rush because it makes my students and I both very flustered.
They also said that I could have made the game board a little more challenging
by having the students fill in the tenths, hundredths, and thousandths place and
not labeling it for them. Other than these two words of advice they liked the
lesson and was impressed on how well the students did with it.
## Name __________________ ** this is a copy of the homework ticket**
1.831
In the number above, 8 is in the ______________ place.
In the number above, 3 is in the ______________ place.
In the number above, 1 is in the ____________________ place.
Write out the number above in words.
____________________________________________________________
Write each fraction in decimal form.
## Updated 4/27/160:32 A4/P4
a.
5
= _______
10
98
b. 100 =_______
20
c. 100 =_______
987
d. 1000
=_______
Write each decimal as a fraction.
a. .35= _______
b. .7=_______
c. 1.4=_______
d. .298=________
Name __________________
1.83
In the number above, 8 is in the ______________ place.
In the number above, 3 is in the ______________ place.
In the number above, 1 is in the ____________________ place.
## Write out the number above in words.
____________________________________________________________
Write each fraction in decimal form.
b.
5
= _______
10
98
b. 100 =_______
20
c. 100 =_______
987
d. 1000
=_________
Write each decimal as a fraction.
b. .35= _______
b. .7=_______
c. 1.4=_______
d. .298=________
## Place value battle ship!
My number:
ONES
HUNDRETH
S
Opponents number:
TENTHS
Thousandth
s
## Updated 4/27/160:32 A4/P4
Thousandths Hundredths
Tenths
Ones
1 2 34567890
1 2 34567890
1 2 34567890
1 2 34567890
2 | 5,290 | 23,789 | {"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-2019-22 | latest | en | 0.895704 |
https://drjeanandfriends.blogspot.com/2013/08/?m=0 | 1,726,518,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00899.warc.gz | 190,528,709 | 38,040 | ## Saturday, August 31, 2013
### DOING THE DOTS!
Subitizing is the ability to recognize numbers without counting. Dot cards can contribute to children’s understanding of number concepts, counting, composing and decomposing numbers, and a variety of standards.
Hint! Start with dot patterns up to 6 and then extend it to 10 when they are ready.
*Make sure to download dot cards on card stock or heavy paper.
One to One
Children match up pompoms, beans, erasers, pebbles, and other small objects with dots on cards.
Dot Flash
The teacher quickly holds up a dot card and then places it face down. The children hold up that number of fingers on their chest. Ask, “How did you know it was that number?”
Clip It
Children use the appropriate number of paper clips or clothespins to attach to the dot cards.
Copy Cat
The teacher holds up a dot card. The children try to reproduce the pattern with their own counters.
Sorting
Sort dot cards by amount.
Sort odd and even cards.
Line Up
Students line up the dot cards in numerical order from largest to smallest or smallest to largest.
Matching
Match dot cards with dots on dice.
Match dot cards with ten frames with the same amount.
Make puzzle games where children match dots with numerals or words.
Partner Count
Cards are placed face down on the table. One card is turned over and the first child to say the number gets to keep the card. The partner must count the dots to verify it’s correct.
*To make the game more challenging, ask them to say one more than the quantity of dots, one less, two more, etc.
Memory
Run off two sets of dot cards. Place them face down on the floor. One child at a time turns over two cards. If the cards match they make keep them and take another turn. If the cards don’t match, they are turned back over and the next child takes a turn.
Top It
You will need several sets of dot cards for this game. Shuffle the cards and lay them face down in a pile. One child at a time chooses a card. The child with the largest number of dots wins both cards. If they turn over the same amount they continue to draw cards until one player has a higher number.
In Between
Shuffle dot cards and place them face down on the table. Each child chooses two cards and lays them in front of them. Another card is selected. If it fits between their two cards they get a point. (Use tally marks to keep score.) Place the cards on the bottom of the pile and continue the game.
Two children have a set of cards and face each other. They each turn over a card and add up the amount. The first child to correctly say the answer gets to keep the cards.
*Tally to keep score.
Paper Plates
How about making some dot plates?
Whew! I think you’ve got the idea!!!
## Friday, August 30, 2013
### MATH TOOLS AND MUST HAVES
There are three other math tools that will build a foundation for the Common Core Math Standards. Today I’ll share ideas for ten frames, tomorrow we’ll explore dot cards, and then it's rekenrek time.
www.k-5mathteachingresources.com
You’ll find games children can play on these sites:
Check out these activities for using tens frames with a Smart Board:
The Georgia Department of Education has some good activities for both ten frames and dot cards:
And here are a few videos that demonstrate ten frames:
An understanding of “ten” is essential for working with larger numbers. It helps students visualize the numbers and gives them an understanding of the relationship of numbers to ten.
Note! Many researchers suggest starting with a five frame before presenting a ten frame.
Building Sets
Give children counters (bears, buttons, small erasers, dry snacks, seasonal objects, etc.). Call out a number and demonstrate how to place the counters in the frame starting on the left. Remind them to always start with the first frame on the left.
*Turn the frame vertically to make sets.
*Can you place the counters in a different way?
*Have children count forwards and backwards on their frames.
Rock and Roll
Children take one or two dice, roll them, and build that amount on their frame.
Flash
Show a numeral or dot card. Challenge children to make the set on their frame.
Have students fill up their frame with counters. Call out a number. Can they remove counters or add counters to build the new number?
Place counters for the first addend on the top row and the second addend in the bottom row. How many altogether?
*Take away counters for subtraction problems.
Double Frames
Extend to a double ten frame for building numbers to 20.
*Use a file folder to make a single or double ten frame for each child.
Giant Ten Frame
Draw a large ten frame on the sidewalk with chalk. Let children stand in the frames to build sets, add, subtract, etc.
*Make a giant ten frame with tape on the floor in your classroom or draw one on a shower curtain with a permanent marker.
## Thursday, August 29, 2013
### NUMBER VESTS
I can’t think of a good reason why everyone who teaches young children shouldn’t
have these number vests in their classroom. You can write them yourself or download them free at makinglearningfun.com. (Better yet, ask a parent to download them for you!) Put them in clear sheet protectors, punch holes at the top, tie on string, and you’ve got a math tool to use all year long!
Hint! You’ll have to make your own =, +, -, and < and > signs.
Counting - Have children get in numerical order according to the number they are wearing.
Songs - Wear number vests as you sing “Five Little Monkeys,” “I Know An Old Lady Who Swallowed a One,” and other songs.
Writing – Children can trace over the numerals with a dry erase marker and then erase. They could also drive over the numerals with a little car or roll play dough and put it on top of the numerals.
Number Words – Write number words on the back of the vests. Use them interchangeably in rhymes and counting.
Inequalities - Put up two numbers and have children choose “<” or “>” to go between them.
Addition and Subtraction - Have children make number sentences using the numbers and signs on the vests.
Fact Families – Move numbers around to demonstrate different fact families.
Word Problems – Use number vests to engage children in solving word problems.
Dot to Dot - Make a giant pencil by covering a paper towel roll with yellow paper. Wrap orange paper around the bottom for the “eraser” and insert a black cone in the other end for the “point.” Pass out numbers and have children scatter around the room. One child takes the pencil and goes from “0” to “10” by “connecting the dots.”
## Wednesday, August 28, 2013
### COUNT ON
High Five
Make a “High Five” book with children’s fingers. Trace around each child’s hand and let them decorate it. Attach pages with tape to make an accordion book. Number pages 5, 10, 15, 20…etc.
*Make a “Piggie Book” by tracing around children’s feet. Practice counting by ten’s with this book.
Body Counting
Use different body parts for counting to 100. Touch head as you count 1-
10. Touch shoulders as you count from 11-20. Touch knees as you count
from 21-30, and so forth.
Skip Counting
Patty cake or cross and tap as you skip count.
Odd and Even
Slap thighs on odd numbers and clap hands on even numbers.
Numbo
Draw the face of a worm (Numbo) on a circle or paper plate. Cut 10-15 circles out of construction paper and number 1-10 or 1-25. Pass the circles out to the children. Place Numbo’s head on the floor and ask the children to help him grow. The child with “1” puts her circle down, followed by “2,” “3,” etc. Ask questions, such as: “What number comes between 7 and 9? What comes before 13? What is 2 more than 4?
*Cut circles out of different colors of construction paper. Start a pattern on Numbo and see if the children can extend it.
Count On
There are many variations of this game. Children stand in a circle and each child says a number in order. When you reach ten that child sits down. At the beginning of the year start again with one to reinforce counting to ten.
*As children progress, count higher and have children sit down every time you reach a ten. For example, 10, 20, 30, etc. would sit down.
*Adapt this game for counting by 5's or other multiples.
*The teacher randomly picks a starting number and stopping number. For example: 17 and 32.
## Tuesday, August 27, 2013
### MORE ADVENTURES IN NUMBER LAND
What better way to develop counting and cardinality with a song or a rhyme!
Over in the meadow in the sand and the sun
Lived an old mother froggie and her little froggie one. (Hold up 1 finger.)
“Hop,” said the mother. “I hop,” said the one. (Pretend to hop finger around.)
So they hopped and were glad in the sand and the sun.
Over in the meadow where the stream runs blue
Lived an old mother fishie and her little fishies two. (Hold up 2 fingers.)
“Swim,” said the mother. “We swim,” said the two. (Pretend to swim fingers.)
So they swam and were glad where the stream runs blue.
Over in the meadow in the nest in the tree
Lived an old mother birdie and her little birdies three. (Hold up 3 fingers.)
“Fly,” said the mother. “We fly,” said the three. (Fly fingers over your head.)
So they flew and were glad in the nest in the tree.
Over in the meadow by the old apple core
Lived an old mother wormie and her little wormies four. (Hold up 4 fingers.)
“Squirm,” said the other. “We squirm,” said the four. (Wiggle fingers.)
So they squirmed and were glad by the old apple core.
Over in the meadow by the big beehive
Lived an old mother bee and her baby bees five. (Hold up 5 fingers.)
“Buzz,” said the mother. “We buzz, “ said the five. (Fly fingers in front of you.)
So they buzzed and were glad by the big beehive.
Activities: Assign children to be the different animals in the song. Have them move hop, swim, fly, wiggle, and buzz around the room when their verse is sung.
Fold 3 sheets of paper in half. Staple. Let children illustrate a number book to go with the song.
Make up additional verses for numerals 6-10. For example, “Over in the meadow in a nest made of sticks lived an old mother beaver and her little beavers six…”
Number March
(“The Ants Go Marching” - Totally Math CD)
The ants go marching one by one, hurrah, hurrah. (Hold up 1 finger.)
The ants go marching one by one, hurrah, hurrah. (Put fist in the air and cheer.)
They all were red and the first one said:
“You’d better catch up, I’m way ahead.”
And they all went marching one by one by one.
The spiders go crawling two by two hurrah, hurrah… (Crawl 2 fingers.)
They were side by side and the second one cried,
“I wish I had someone to give me a ride.”
And they all went crawling two by two by two.
The birds go flying three by three hurrah, hurrah… (Hold up 3 fingers and
Their feet were bare as they flew through the air and pretend to fly.)
The third one said, “I’d like shoes to wear.”
And they all went flying three by three by three.
The rabbits go hopping four by four hurrah, hurrah… (Pretend to hop 4 fingers.)
They hipped and hopped and bounced and bopped—
The fourth one got tired and down she plopped.
And they all went hopping four by four by four.
The horses go galloping five by five hurrah, hurrah… (Gallop 5 fingers
The fifth in line said “I feel fine; tapping them on your thigh.)
“I love to gallop all the time.”
And they all went galloping five by five by five.
The fish go swimming six by six hurrah, hurrah… (Pretend to swim 6 fingers.)
Their tails went swish and the sixth one wished
He wouldn’t end up as a tasty dish.
And they all went swimming six by six by six.
The mice go creeping seven by seven hurrah, hurrah… (Creep 7 fingers up the
The seventh was meek, he let out a squeak: front of your body.)
“I can’t see a thing; I’m afraid to peek!”
And they all went creeping seven by seven by seven.
The worms go wiggling eight by eight hurrah, hurrah…(Wiggle 8 fingers.)
The eighth one thought, “It’s awfully hot—
I’d like to rest in a shady spot.”
But they all kept wiggling eight by eight by eight.
The monkeys go swinging nine by nine hurrah, hurrah… (Swing 9 fingers.)
The ninth one called to one and all,
“I hope you’ll catch me if I fall!”
And they all went swinging nine by nine by nine.
The kids go walking ten by ten hurrah, hurrah… (Hold up 10 fingers and
The tenth one knew they were so cool, pretend to walk.)
‘Cause they were on their way to school.
And they all went walking ten by ten by ten.
You can download a free book that goes with this song at drjean.org June. Click “Free Activities” and scroll down to Number March.
Activities: Let children take different verses and illustrate them. Put their pictures together to make a class book.
Five Little Hot Dogs (“Five Little Ducks” - Just for Fun CD)
Five little hotdogs frying in the pan. (Hold up five fingers.)
The grease got hot, and one went BAM! (Clap.)
Four little hotdogs… (Four fingers.)
Three… (Three fingers.)
Two… (Two fingers.)
One… (One finger.)
No little hotdogs frying in the pan. (Hold up fist.)
The pan got hot and it went BAM!
Cut hotdogs out of paper and glue them to spring clothespins. Draw a pan similar to the one shown on a file folder and make a slit along the middle of the pan. Attach the hotdogs and remove one at a time as you sing the song. When the pan goes “Bam!” close the file folder.
*Change the words to: “Five little kernels sizzling in the pot. When the oil got hot one went ‘POP’!”
Five Little Monkeys (Totally Math CD)
Five little monkeys jumping on the bed. (Hold up 5 fingers.)
Momma called the doctor, and the doctor said, (Hold hand to ear.)
“That’s what you get for jumping on the bed.” (Point finger.)
Four…three…two…one…
No little monkeys jumping on the bed. (Shake head “no.”)
They’re all sick with broken heads! (Hold palms up in the air.)
Activities: Choose five children to act out the rhyme.
Ten in the Bed and Ten in the Sled
There were ten in the bed and the little one said,
“Roll over! Roll over!”
So they all rolled over and one fell out.
There were nine in the bed and the little one said…
Activities: Choose ten children and let them “roll” out of the bed one at a time as you sing the song.
*In the winter you can do “Ten in the Sled.”
One Small Noodle
One small noodle on my noodle plate. (Hold up one finger.)
Salt and pepper, tastes just great. (Pretend to shake salt.)
Mother’s going to the store.
Mother, mother, get some more.
Two…Three…Four…
Five small noodles on my noodle plate.
Salt and pepper, tastes just great.
Mother, mother, I am stuffed.
I think that I have had enough!
Activities: Take a paper plate and cut out five holes as shown. Stick fingers in the holes to match the noodles in the rhyme.
## Monday, August 26, 2013
### NUMBER LAND
Enough with letters! What about us numbers?
Well, guess what? You can use almost all of the activities from the past week to help children learn numerals and shapes. Here’s a numeral song you can sing to the tune of “Skip to My Lou.”
The Numeral Song (“Sing to Learn” CD)
Come right down and that is all. (Use index finger and middle finger
Come right down and that is all. to do “invisible” writing in the air.)
Come right down and that is all
To make the numeral one. (Hold up 1 finger.)
2 – Curve around and slide to the right…
3 – Curve in and around again…
4 – Down, over, down some more…
5 – Down, around, put on a hat…
6 – Curve in and around again…
7 – Slide to the right and slant it down…
8 – Make an “s” then close the gate…
9 – Circle around then come right down…
10 – Come right down, then make a zero…
We can sing the “Numeral Song”…
And make numerals all day long!
*Give children a strip of tissue paper and let them make the numerals in the air.
*Make dots with water soluble markers and then put drops of glue on top. Children can trace over the dots as they say the number chant.
*Put a squirt of shaving cream on each child’s desk so they can practice writing the numerals as they sing.
*Hide magnetic numerals and shapes in the sand for children to find.
*Have children lay on the floor to make shapes and numerals with their bodies.
*Glue magnetic numerals and shapes to craft sticks and challenge children to match them up with numerals or similar shapes in the classroom.
Come back tomorrow for another exciting adventure in Number Land!
## Sunday, August 25, 2013
### WE'RE GREAT!
We’re Great!
This alphabet book will build your classroom community and introduce vocabulary words.
We’re great, but no one knows it.
No one knows it so far.
Some day they’ll realize how wonderful we are!
They’ll look at us, and point at us,
and then they’ll shout, “Hurray!”
Let’s cheer how we’re wonderful beginning with A.
A- We’re awesome.
B- We’re brave.
C- We’re creative.
D- We’re dynamic
E- We’re enthusiastic
F- We’re fantastic.
H- We’re honest
I- We’re imaginative
J- We’re joyful.
K- We’re kind.
L- We’re lovable
M- We’re magnificent.
N- We’re nice.
O- We’re outgoing
P- We’re polite.
Q- We’re quick.
R- We’re responsible
S- We’re special.
T- We’re terrific.
U- We’re unique.
V- We’re valuable.
W- We’re wonderful.
X- We’re excellent.
Y- We’re youthful
Z- We’re zany!
Note! This song is actually on my “Just for Fun” CD, but I can’t think of a familiar
tune??? It will work if you just say the words.
*On the front of the notebook or pocket folder write “We’re Great!” Write a
different letter of the alphabet on each page. Show children the dictionary and
ask if they know what it is. Explain that words are written in the book in
alphabetical order (ABC). When people want to know how to spell a word or
want to know what a word means they look it up in the dictionary. Tell the
children that you want them to help you make a special dictionary with
WONDERFUL words that describe special people just like them! Use the words
from the above chant to start your dictionary. Add words that children suggest.
*Let children make an acrostic poem by writing the letters in their name vertically down the left side on a piece of paper. Can they write an adjective that describes them for each letter?
*How about a vocabulary parade? Have each child choose a "wonderful" word. Write it on a sentence strip and let them decorate it. Pin it to them and let them walk around the room parading their words. | 4,648 | 18,850 | {"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} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.92316 |
https://www.extramarks.com/studymaterials/formulas/cotangent-formula/ | 1,721,098,652,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00646.warc.gz | 658,183,684 | 75,124 | # Cotangent Formula
## Cotangent Formula
Mathematics is one of the most important disciplines that are taught to students. Mathematics is a subject that involves numbers, forms, facts, measurements, and logical processes. Every aspect of our humanity, including medicine, engineering, finance, the natural sciences, economics, etc., is greatly affected by it. The use of Mathematics can be seen in daily life.
Mathematical ideas, theories, and formulas have a wide range of practical applications.
Students must master the formulae and principles in order to solve diverse problems. To comprehend its many uses and relevance, it is crucial to understand this subject.
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What is Cotangent?
The study of the correlation between a right-angled triangle’s sides and angles is the focus of one of the most significant branches of Mathematics in history: Trigonometry.
One of the most significant areas of Mathematics, trigonometry has a wide range of applications. The study of the relationship between the sides and angles of the right-angle triangle is essentially the focus of the field of Mathematics known as “Trigonometry.”
Therefore, employing trigonometric formulas, functions, or trigonometric identities can be helpful in determining the missing or unknown angles or sides of a right triangle. Angles in Trigonometry can be expressed as either degrees or radians. 0°, 30°, 45°, 60°, and 90° are some of the Trigonometry angles that are most frequently employed in computations.
The Cotangent Formula is a topic of Trigonometry and therefore one of the many challenging topics of the subject. At Extramarks the students get all the help they require in order to understand the topic of Cotangent Formula and other Trigonometry related topics. The Cotangent Formula is just one of the many topics that are covered in Trigonometry.
There are two other sub-branch classifications for Trigonometry. The two varieties of Trigonometry are as follows:
• Trigonometry in Planes
• Trigonometry using spheres
### Properties of Cotangent
The Trigonometry functions are another name for the triangle’s trigonometric ratios. Three crucial trigonometric functions—sine, cosine, and tangent—are represented by the abbreviations sin, cos, and tan. How these ratios or functions are assessed when a right-angled triangle is present-
In a right-angled triangle, its longest side is referred to as the hypotenuse, and its adjacent and opposite sides are its adjacent opposite sides. One topic of Trigonometry is the Cotangent Formula
### Cotangent in Terms of Cos and Sin
Trigonometry is the branch of Mathematics that studies the relationship between the side ratios and angles of right triangles. Explore these relationships using trigonometric ratios such as sine, cosine, tangent, cotangent, secant, and cosecane. Trigonometry is one of the most important concepts in Mathematics. It plays an important role in almost every field, including aviation, physics, criminology, and the military. Trigonometry is used to find the angles or sides of a triangle. Among the many topics of Trigonometry, one is the Cotangent Formula
Basic Trigonometry Angle measurements and angle problems are discussed in Basic Trigonometry. One of the topics in it is the Cotangent Formula. Trigonometry has three basic functions: sine, cosine, and tangent. Other important trigonometric functions can be derived from these three fundamental relationships. These functions form the basis for all basic Trigonometry topics. In summary, the six main features of angles commonly used in trigonometry are
• sine (sin),
• cosine (cos),
• tangent (tan),
• cotangent (cot),
• secant (sec), and
• cosecant (csc).
All these have different formulas like the Cotangent Formula, etc. At Extramarks, students get the required help regarding all these topics, including the Cotangent Formula and more.
### Cotangent in Terms of Tan
Throughout history, Trigonometry has been employed in a wide range of disciplines, including surveying, theoretical physics, and architecture. It can be applied to many different fields, such as oceanography, seismology, meteorology, physical sciences, astronomy, acoustics, navigation, electronics, and many others. It can also be used to calculate the length of long rivers, the height of mountains, and other things. Spherical trigonometry has been used to calculate the locations of the sun, moon, and stars.
The Cotangent Formula is one of the many topics that are covered under the vast theme of Trigonometry. The Cotangent Formula can be a challenging topic for some, but with the right tools, students can master it. The Extramarks website, and the mobile application for the same, have several tools that can help students with the topic of the Cotangent Formula and other related topics.
Students have the notion that topics like the Cotangent Formula and Trigonometry, in general, are not relevant topics when it comes to everyday life. This is a misconception. Trigonometry and the Cotangent Formula, just like most topics of Mathematics, are closely related to the real world. Due to this misconception, the students develop a negative relationship with Mathematics. This is bad for the students in the long run as they end up losing interest in the subject and that leads to them getting overwhelmed by the subject too.
To avoid such situations, it is important for students to have access to the right type of tools and resources that can help them understand the importance of Mathematics in the real world. As well as, tools that can make understanding and preparation for Mathematics interesting and fun.
The experts at Extramarks understand this, and hence they come up with resources that can be of great help in this context. The students can access different types of preparation tools available on the Extramarks website for topics like the Cotangent Formula and many more. The Cotangent Formula is just one of the topics that are covered in the resources available at Extramarks.
### Cotangent in Terms of Cosec
In Trigonometry, several examples from everyday life are commonly used. To learn Trigonometry better, studying through examples is really helpful. The tools prepared by the experts of Mathematics at Extramarks have several examples that help the students not only understand the topics like the Cotangent Formula better but also make the process interesting for them.
The Cotangent Formula can be understood and mastered with the help of the tools prepared by the experts of Mathematics at Extramarks.
### Cotangent Law
Trigonometry and its topics like the Cotangent Formula are generally introduced to the students when they are in Class 10. Class 10 is a very crucial class for students. Majority of boards of education conduct board examinations in Class 10. This means there is added pressure to perform well in the Class 10 examinations. It also means that the syllabus is comparatively a little more difficult than what they were studying in earlier classes.
Students of Class 10 also have to particularly perform well in this class because it’s based on these marks they get to choose the stream they’ll continue their senior secondary education. All these reasons make Class 10 to be really crucial in a student’s school life.
### Sign of Cotangent
The experts at Extramarks understand the importance of Class 10 and the amount of performance-related pressure students have to deal with in this particular class. Therefore, experts in different subjects come up with helpful resources to make the preparation of examinations easier for the students of Class 10.
### Period of Cotangent
The tools prepared by the experts of Mathematics at Extramarks for topics like the Cotangent Formula and other topics as well, are not only great for preparations of examinations but can also be of great help to the students when they need help with their homework and assignments, etc. Students can simply look for the topic they need help with, and they can find various types of study material and preparation tools made specifically for that particular topic.
### Cotangent of Negative Angle
Candidates can demonstrate their familiarity and level of skill with a particular topic or problem during exams, which serves as a formal evaluation process. They are typically granted by hiring or assessing authority for a variety of reasons. For several reasons, exams are crucial to a student’s academic career. To prepare well for the examinations, students are recommended to make use of the many tools available on the Extramarks website. These include tools for topics such as the Cotangent Formula and many more.
Exams improve knowledge acquisition and the capacity to absorb new information. As the student’s brain adapts to new information over time through trial and error, their memory for facts and figures gets better. Being a good learner is surely a wonderful trait. Students can prepare well for their exams by using tools that are designed to make exam preparations easy. All these tools are created by experts in different fields who have a clear understanding of the needs and requirements of the students of different classes. The Cotangent Formula can be studied efficiently with the help of study tools available on the Extramarks website.
### Cotangent on Unit Circle
Exams are an excellent way to determine how much knowledge a student has about a certain subject. The exam reveals the lessons each student recalled and found most interesting. Teachers can watch students’ interactions and use their own judgment as they work, which heightens the excitement of the exam environment.
When studying for an exam, using the Extramarks, experts-created resources for different topics like the Cotangent Formula can be of great help. Exams are a good way to identify students’ skills and limitations.
### Domain, Range, and Graph of Cotangent
A wonderful strategy to prepare for the final exam is to use mock exams. This gives both teachers and students the opportunity to recognise their weaknesses before official assessments. This will provide students with the tools they need in the classroom to realise their full potential and reap the rewards in the long run. To realise their full potential, students are recommended to take advantage of tools available on the Extramarks website and the mobile application for the same. These include tools for topics like the Cotangent Formula and several more.
As students become older, school gets more difficult. Students develop personally, and their classes get harder. Exams assist universities in determining whether a candidate can fulfil the demands of the profession. Therefore, the importance of examinations and good performances increases as the students move to higher classes. To help students consistently perform well, the experts have developed many helpful resources that the students can very easily access by visiting the Extramarks website. The topics are vast and everything that is taught in the classroom can be studied with the help of these tools. For example, topics like the Cotangent Formula
### Domain and Range of Cotangent
Students can find useful resources on Extramarks to assist with topics like the Cotangent Formula. Along with the resources for the Cotangent Formula, students may also find the answers to all the other topics of Trigonometry.
With the intention of assisting students, specialists create helpful tools for topics like the Cotangent Formula and many more. These resources, created for the Cotangent Formula, have step-by-step solutions to many questions. These Cotangent Formula based resources help students understand how to respond to specific questions that are based on the topic of the Cotangent Formula.
### Graph of Cotangent
Starting the revision process early is essential for the students to score well in their examinations. Students must allow themselves enough time to thoroughly analyse and comprehend all they have learned (or to read around the subject or ask for help if they are struggling). Cramming at the last minute is far less effective.
The revision will be a lot easier if students review each subject as they go and make sure they thoroughly understand it. The best advice is to study diligently and thoroughly, and the best way to do this is to begin early.
The use of tools available at Extramarks can help the students follow this preparation tip.
### Derivative and Integral of Cotangent
There’s a good chance that some topics will be simpler for the students than others. Additionally, they might discover that some subjects require more revision than others. It is also beneficial to think about the time and duration of the daily study sessions.
Planning the revision so that they may make the most of their time is an essential skill to prepare well for the exams. Which part of the day—morning, afternoon, or evening—do they prefer most? Can they read more frequently at certain times? This will assist students in making a general plan for what they intend to do, but they should always leave room for flexibility in case circumstances change.
### Derivative of Cotangent
Exams can be stressful for most students and this is a result of the students not having a positive experience with exams in their initial years of school education. The use of the resources available at Extramarks can help ensure that students get to prepare well for their examinations.
When students are well-prepared, they tend to not fear, but rather enjoy the process of examination. Therefore, it is recommended by the experts at Extramarks to make the most of all the helpful resources available on Extramarks. These tools include resources for topics like the Cotangent Formula. The Cotangent Formula is part of the broad topic of Trigonometry. Apart from the Cotangent Formula, there are several other tools available for the topics of Trigonometry, on the Extramarks website that students can take help with.
### Integral of Cotangent
Students can easily access all the tools available on the Extramarks website and the mobile application. The experts understand that the students of younger classes also require help with their homework, assignments and examinations and hence these tools are made to be extremely user-friendly as well as easy to access.
### Examples on Cotangent
The experts at Extramarks are highly educated and masters in their respective fields. This makes the resources available on Extramarks to be highly reliable and therefore the students can easily depend on these tools.
In today’s day and age when there are endless options available for students to refer to, it is important for students to know that not all the available resources on the internet can be trusted. Hence, a lot of times, students end up wasting their crucial time before the examinations looking up the right sources of information and preparation tools.
The students are advised to keep their reference tools limited and only use reliable sources like Extramarks. This helps avoid confusion and saves a lot of crucial preparation time for students.
### Qs on Cotangent Formula
There is a lot of strain on many students, especially before exams. The Extramarks website, which is recommended for students, has resources for topics like the Cotangent Formula. This is what Extramark specialists advise using to lessen exam stress. Students can also consult the question answers sections if students still have queries after looking over the other resources related to the Cotangent Formula. | 3,179 | 16,367 | {"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-2024-30 | latest | en | 0.948882 |
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# Practice Test: Coordinate Geometry - Class 10 MCQ
Test Description
## 25 Questions MCQ Test Mathematics (Maths) Class 10 - Practice Test: Coordinate Geometry
Practice Test: Coordinate Geometry for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The Practice Test: Coordinate Geometry questions and answers have been prepared according to the Class 10 exam syllabus.The Practice Test: Coordinate Geometry MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Coordinate Geometry below.
Solutions of Practice Test: Coordinate Geometry questions in English are available as part of our Mathematics (Maths) Class 10 for Class 10 & Practice Test: Coordinate Geometry solutions in Hindi for Mathematics (Maths) Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Practice Test: Coordinate Geometry | 25 questions in 25 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Mathematics (Maths) Class 10 for Class 10 Exam | Download free PDF with solutions
Practice Test: Coordinate Geometry - Question 1
### The distance between the points (a, b) and (– a, – b) is :
Detailed Solution for Practice Test: Coordinate Geometry - Question 1
We have distance formula as d =
Where x1=a,y1=b,x2=-a,y2=-b
Practice Test: Coordinate Geometry - Question 2
### The distance between points (a + b, b + c) and (a – b, c – b) is :
Detailed Solution for Practice Test: Coordinate Geometry - Question 2
Distance between two point is
root[{(a+b)-(a-b)}^2] + [{(b+c)-(c-b)}^2]
=root [{(a+b-a+b)^2}] + [((b+c-c+b)}^2]
=root[(2b)^2 + (2b)^2]
=root[4b^2 +4b^2]
=root[8b^2]
=2.root(2)b
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Practice Test: Coordinate Geometry - Question 3
### The distance between A (1, 3) and B (x, 7) is 5. The value of x > 0 is :
Practice Test: Coordinate Geometry - Question 4
The distance between the points (a cos 20° + b sin 20°, 0) and (0, a sin 20° – b cos 20°) is :
Practice Test: Coordinate Geometry - Question 5
Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is :
Practice Test: Coordinate Geometry - Question 6
The co-ordinates of the points which divides the join of (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is :
Practice Test: Coordinate Geometry - Question 7
The co-ordinates of the point on x-axis which is equidistant from the points (5, 4) and (– 2, 3) are :
Detailed Solution for Practice Test: Coordinate Geometry - Question 7
Let the point on x-axis be (a,0)
Since PA = PB PA2 = PB2
(a + 2)2 + (0 - 5)2 = (a - 2)2 + (0 + 3)2
(a + 2)2 - (a - 2)2 = 9 - 25 = -16
8a = -16 a = -2.
The required point is (-2,0).
Practice Test: Coordinate Geometry - Question 8
The co-ordinates of the point on y-axis which is equidistant from the points (3, 1) and (1, 5) are:
Practice Test: Coordinate Geometry - Question 9
The coordinates of the centre of a circle are (– 6, 1.5). If the ends of a diameter are (– 3, y) and (x, – 2) then:
Practice Test: Coordinate Geometry - Question 10
The points (– 2, 2), (8, – 2) and (– 4, – 3) are the vertices of a :
Practice Test: Coordinate Geometry - Question 11
The points (1, 7), (4, 2), (– 1, 1) and (– 4, 4) are the vertices of a :
Detailed Solution for Practice Test: Coordinate Geometry - Question 11
Step-by-step explanation:
Given points are (1,7),(4,2),(-1,-1),(-4,4)
Let the points are A,B,C,D.
Distance formula = √(x₂-x₁)²+(y₂-y₁)²
AB = √(4-1)²+(2-7)²
= √9+25
= √34
BC = √(-1-4)²+(-1-2)²
= √25+9
= √34
CD = √(-4-(-1))²+(4-(-1))²
= √9+25
= √34
DA = √(1-(-4))²+(7-4)²
= √25+9
= √34
also diagonals AC and BD
AC = √(1+1)² + (7+1)² = √68
BD = √(4+4)² + (2-4)² = √68
We know that the all sides and both diagonals of the square are equal
So,from this we can say that these points are the vertices of a square.
Practice Test: Coordinate Geometry - Question 12
The line segment joining (2, – 3) and (5, 6) is divided by x-axis in the ratio:
Detailed Solution for Practice Test: Coordinate Geometry - Question 12
Hence, the ratio is 1:2 and the division is internal.
Practice Test: Coordinate Geometry - Question 13
The line segment joining the points (3, 5) and (– 4, 2) is divided by y-axis in the ratio:
Detailed Solution for Practice Test: Coordinate Geometry - Question 13
Practice Test: Coordinate Geometry - Question 14
If (3, 2), (4, k) and (5, 3) are collinear then k is equal to :
Practice Test: Coordinate Geometry - Question 15
If the points (p, 0), (0, q) and (1, 1) are collinear then 1/p+1/q is equal to :
Practice Test: Coordinate Geometry - Question 16
Two vertices of a triangle are (–2, – 3) and (4, –1) and centroid is at the origin. The coordinates of the third vertex of the triangle are :
Practice Test: Coordinate Geometry - Question 17
A (5, 1), B(1, 5) and C (–3, –1) are the vertices of ?ABC. The length of its median AD is :
Practice Test: Coordinate Geometry - Question 18
Three consecutive vertices of a parallelogram are (1, –2), (3, 6) and (5, 10). The coordinates of the fourth vertex are :
Practice Test: Coordinate Geometry - Question 19
The vertices of a parallelogram are (3, –2), (4, 0), (6, –3) and (5, –5). The diagonals intersect at the point M. The coordinates of the point M are :
Detailed Solution for Practice Test: Coordinate Geometry - Question 19
Practice Test: Coordinate Geometry - Question 20
If two vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5), then the other two vertices are :
Practice Test: Coordinate Geometry - Question 21
The circumcentre of the triangle formed by the lines xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is :
Practice Test: Coordinate Geometry - Question 22
The vertices of a triangle are (a, b – c), (b, c – a) and (c, a – b), then it's centroid lies on :
Practice Test: Coordinate Geometry - Question 23
The points (1, 2), (3, 8) and (x, 20) are collinear if x =
Detailed Solution for Practice Test: Coordinate Geometry - Question 23
Practice Test: Coordinate Geometry - Question 24
For the triangle whose sides are along the lines x = 0, y = 0 and x/6+y/8 = 1, the incentre is :
Practice Test: Coordinate Geometry - Question 25
For the triangle whose sides are along the lines y = 15, 3x – 4y = 0, 5x + 12y = 0, the incentre is :
Detailed Solution for Practice Test: Coordinate Geometry - Question 25
Given equations:
3x – 4y = 0 …(1)
5x+12y = 0 …(2)
Y-15 = 0 …(3)
From the given equations, (1), (2) and (3) represent the sides AB, BC and CA respectively.
Solving (1) and (2), we get
x= 0, and y= 0
Therefore, the side AB and BC intersect at the point B (0, 0)
Solving (1) and (3), we get
x= 20, y= 15
Hence, the side AB and CA intersect at the point A (20, 15)
Solving (2) and (3), we get
x= -36, y = 15
Thus, the side BC and CA intersect at the point C (-36, 15)
Now,
BC = a = 39
CA = b = 56
AB = c = 25
Similarly, (x1, y1) = A(20, 15)
(x2, y2) = B(0, 0)
(x3, y3) = C(-36, 15)
Therefore, incentre is
## Mathematics (Maths) Class 10
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## Mathematics (Maths) Class 10
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Thus, π A is a left inverse of ι b and ι b is a right inverse of π A. Indeed, the frame inequality (5.2) guarantees that Φf = 0 implies f = 0. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Example. Notice that f … The left inverse g is not necessarily an inverse of f, because the composition in the other order, f ∘ g, may differ from the identity on Y. The left inverse g is not necessarily an inverse of f, because the composition in the other order, f ∘ g, may differ from the identity on Y. (* im_dec is automatically derivable for functions with finite domain. apply f_equal with (f := g) in eq. It is easy to show that the function $$f$$ is injective. then f is injective. Does an injective group homomorphism between countable abelian groups that splits over every finitely generated subgroup, necessarily split? For each function f, determine if it is injective. Solution. (a) Show that if has a left inverse, is injective; and if has a right inverse, is surjective. A frame operator Φ is injective (one to one). Let f : A ----> B be a function. That is, given f : X → Y, if there is a function g : Y → X such that, for every x ∈ X. g(f(x)) = x (f can be undone by g). (b) Give an example of a function that has a left inverse but no right inverse. The calculator will find the inverse of the given function, with steps shown. Then we plug into the definition of left inverse and we see that and , so that is indeed a left inverse. assumption. The inverse of a function with range is a function if and only if is injective, so that every element in the range is mapped from a distinct element in the domain. The type of restrict f isn’t right. De nition. One of its left inverses is … The equation Ax = b either has exactly one solution x or is not solvable. i) ). if r = n. In this case the nullspace of A contains just the zero vector. We wish to show that f has a left inverse, i.e., there exists a map h: B → A such that h f =1 A. We write it -: → and call it the inverse of . We will show f is surjective. De nition 1. Ask Question Asked 10 years, 4 months ago. Functions with left inverses are always injections. Suppose f is injective. Let $f \colon X \longrightarrow Y$ be a function. If f : A → B and g : B → A are two functions such that g f = 1A then f is injective and g is surjective. an element b b b is a left inverse for a a a if b ... Then t t t has many left inverses but no right inverses (because t t t is injective but not surjective). (But don't get that confused with the term "One-to-One" used to mean injective). A function f from a set X to a set Y is injective (also called one-to-one) if distinct inputs map to distinct outputs, that is, if f(x 1) = f(x 2) implies x 1 = x 2 for any x 1;x 2 2X. [Ke] J.L. Then is injective iff ∀ ⊆, − (()) = is surjective ... For the converse, if is injective, it has a left inverse ′. Proof. If the function is one-to-one, there will be a unique inverse. One to One and Onto or Bijective Function. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Any function that is injective but not surjective su ces: e.g., f: f1g!f1;2g de ned by f(1) = 1. The matrix AT )A is an invertible n by n symmetric matrix, so (AT A −1 AT =A I. Often the inverse of a function is denoted by . unfold injective, left_inverse. Let A and B be non-empty sets and f : A !B a function. The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism. When does an injective group homomorphism have an inverse? In other words, an injective function can be "reversed" by a left inverse, but is not necessarily invertible , which requires that the function is bijective . Since g(x) = b+x is also injective, the above is an infinite family of right inverses. Suppose f has a right inverse g, then f g = 1 B. A bijective group homomorphism $\phi:G \to H$ is called isomorphism. So there is a perfect "one-to-one correspondence" between the members of the sets. (a) f:R + R2 defined by f(x) = (x,x). Let A be an m n matrix. It has right inverse iff is surjective: Advanced Algebra: Aug 18, 2017: Left and right inverse: Calculus: May 13, 2014: right and left inverse: Calculus: May 10, 2014: May I have a question about left and right inverse? (exists g, left_inverse f g) -> injective f. Proof. Show Instructions. IP Logged "I always wondered about the meaning of life. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. 2. Its restriction to Im Φ is thus invertible, which means that Φ admits a left inverse. Bijective means both Injective and Surjective together. Note that this wouldn't work if $f$ was not injective . Proof: Left as an exercise. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. i)Function f has a right inverse i f is surjective. If h is a right inverse for f, f h = id B, so f is surjective by problem 4(e). In order for a function to have a left inverse it must be injective. Kolmogorov, S.V. Since $\phi$ is injective, it yields that $\psi(ab)=\psi(a)\psi(b),$ and thus $\psi:H\to G$ is a group homomorphism. 9. g is an inverse so it must be bijective and so there exists another function g^(-1) such that g^(-1)*g(f(x))=f(x). For each b ∈ f (A), let h (b) = f-1 ({b}). Proposition: Consider a function : →. Left inverse Recall that A has full column rank if its columns are independent; i.e. By definition of left inverse we have then x = (h f)(x) = (h f)(y) = y. LEFT/RIGHT INVERTIBLE MATRICES MINSEON SHIN (Last edited February 6, 2014 at 6:27pm.) g(f(x))=x for all x in A. A, which is injective, so f is injective by problem 4(c). For instance, if A is the set of non-negative real numbers, the inverse map of f: A → A, x → x 2 is called the square root map. Then we say that f is a right inverse for g and equivalently that g is a left inverse for f. The following is fundamental: Theorem 1.9. left inverse (plural left inverses) (mathematics) A related function that, given the output of the original function returns the input that produced that output. If yes, find a left-inverse of f, which is a function g such that go f is the identity. Liang-Ting wrote: How could every restrict f be injective ? ⇐. (b) Given an example of a function that has a left inverse but no right inverse. Kelley, "General topology" , v. Nostrand (1955) [KF] A.N. Left inverse ⇔ Injective Theorem: A function is injective (one-to-one) iff it has a left inverse Proof (⇒): Assume f: A → B is injective – Pick any a 0 in A, and define g as a if f(a) = b a 0 otherwise – This is a well-defined function: since f is injective, there can be at most a single a such that f(a) = b iii)Function f has a inverse i f is bijective. In other words, an injective function can be "reversed" by a left inverse, but is not necessarily invertible, which requires that the … My proof goes like this: If f has a left inverse then . This example shows that a left or a right inverse does not have to be unique Many examples of inverse maps are studied in calculus. Injections can be undone. *) Qed. Relating invertibility to being onto (surjective) and one-to-one (injective) If you're seeing this message, it means we're having trouble loading external resources on our website. We say that A is left invertible if there exists an n m matrix C such that CA = I n. (We call C a left inverse of A.1) We say that A is right invertible if there exists an n m matrix D such that AD = I m. Note that the does not indicate an exponent. Function has left inverse iff is injective. Tags: group homomorphism group of integers group theory homomorphism injective homomorphism. Hence, f is injective. We define h: B → A as follows. repeat rewrite H in eq. intros A B f [g H] a1 a2 eq. ii)Function f has a left inverse i f is injective. If there exists v,w in A then g(f(v))=v and g(f(w))=w by def so if g(f(v))=g(f(w)) then v=w. Active 2 years ago. An injective homomorphism is called monomorphism. For example, For example, in our example above, is both a right and left inverse to on the real numbers. Question 3 Which of the following would we use to prove that if f:S + T is injective then f has a left inverse Question 4 Which of the following would we use to prove that if f:S → T is bijective then f has a right inverse Owe can define g:T + S unambiguously by … A function that is both injective and surjective is called bijective (or, if domain and range coincide, in some contexts, a permutation). What’s an Isomorphism? For example, the image of a constant function f must be a one-pointed set, and restrict f : ℕ → {0} obviously shouldn’t be a injective function.. require is the notion of an injective function. (a) Prove that f has a left inverse iff f is injective. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. Injective mappings that are compatible with the underlying structure are often called embeddings. Calculus: Apr 24, 2014 (c) Give an example of a function that has a right inverse but no left inverse. Think of it as a "perfect pairing" between the sets: every one has a partner and no one is left out. The function f: R !R given by f(x) = x2 is not injective … The function f is called as one to one and onto or a bijective function, if f is both a one to one and an onto function More clearly, f maps distinct elements of A into distinct images in B and every element in B is an image of some element in A. But do n't get that confused with the term one-to-one correspondence '' between members. Often the inverse map of an isomorphism is again a homomorphism, and hence isomorphism that has! Group of integers group theory homomorphism injective homomorphism b be a unique inverse calculator find! Matrices MINSEON SHIN ( Last edited February 6, 2014 at 6:27pm. splits over every finitely subgroup. Map of left inverse injective injective function '' used to mean injective ) work if [ math ] f /math... At 6:27pm. note that this would n't work if [ math ] f x! * ) bijective means both injective and surjective together [ g h a1. No left inverse i f is injective this would n't work if [ math ] f \colon x \longrightarrow [! One-To-One '' used to mean injective ) with ( f ( x ) ( b. That f … require is the identity defined by f ( x ) = ( x ) a web,... * ) bijective means both injective and surjective together thus invertible, which a. Have a left inverse but no left inverse of are independent ; i.e in eq February,! B+X is also injective, so 5x is automatically derivable for functions finite! Homomorphism group of integers group theory homomorphism injective homomorphism a inverse i f injective... If r = n. in this case the nullspace of a function Last edited February,. Inverse it must be injective filter, please make sure that the function (., there will be a function that has a right inverse but right... 5.2 ) guarantees that Φf = 0 implies f = 0 a1 eq... Injective function 4 ( c ) Give an example of a function is by... Tags: group homomorphism have an inverse of life ] f [ /math ] was not injective r R2... A2 eq looked it up in the dictionary under ' L ' and there it was -... Injective mappings that are compatible with the term one-to-one correspondence '' between the members of the sets: one... ) is injective ; and if has a partner and no one is left out How could every f! Note that this would n't work if [ math ] f \colon x \longrightarrow Y [ /math ] was injective! = 0 b and ι left inverse injective is a left inverse iff f is identity... By f ( x ) ( c ) Give an example of a function that has a left,. Nullspace of a function to have a left inverse then x ) let [ math ] f \colon \longrightarrow... Column rank if its columns are independent ; i.e proof goes like this: if f a! To mean injective ) 4 months ago iii ) function f has a left inverse it must be injective How! 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Perfect one-to-one correspondence '' between the members of the sets: every has. ) is injective so 5x is equivalent to 5 * x of its inverses. Is left out Question Asked 10 years, 4 months ago x.. The multiplication sign, so f is the notion of an injective group homomorphism \$ \phi: g h! Multiplication sign, so 5x is equivalent to 5 * x.! This would n't work if [ math ] f [ g h ] a2... If it is easy to Show that if has a left inverse but no right inverse, injective... A b f [ /math ] was not injective the zero vector term one-to-one... A ) f: r + R2 defined by f ( a ) Prove that f … is... Do n't get that confused with the term one-to-one correspondence '' between the sets i f is identity... Intros a b f [ /math ] was not injective f \colon x \longrightarrow Y [ /math ] not. | 4,610 | 17,552 | {"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": 3, "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} | 3.5625 | 4 | CC-MAIN-2022-49 | latest | en | 0.907734 |
https://nrich.maths.org/public/topic.php?code=-99&cl=2&cldcmpid=1981 | 1,582,459,400,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00303.warc.gz | 494,263,861 | 9,470 | # Resources tagged with: Working systematically
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### There are 339 results
Broad Topics > Mathematical Thinking > Working systematically
### Wonky Watches
##### Age 7 to 11 Challenge Level:
Stuart's watch loses two minutes every hour. Adam's watch gains one minute every hour. Use the information to work out what time (the real time) they arrived at the airport.
### 5 on the Clock
##### Age 7 to 11 Challenge Level:
On a digital clock showing 24 hour time, over a whole day, how many times does a 5 appear? Is it the same number for a 12 hour clock over a whole day?
### Calendar Sorting
##### Age 7 to 11 Challenge Level:
The pages of my calendar have got mixed up. Can you sort them out?
### A Child Is Full of ...
##### Age 7 to 11 Short Challenge Level:
My cousin was 24 years old on Friday April 5th in 1974. On what day of the week was she born?
### Palindromic Date
##### Age 7 to 11 Challenge Level:
What is the date in February 2002 where the 8 digits are palindromic if the date is written in the British way?
### How Many Times?
##### Age 7 to 11 Challenge Level:
On a digital 24 hour clock, at certain times, all the digits are consecutive. How many times like this are there between midnight and 7 a.m.?
### Crack the Code
##### Age 7 to 11 Challenge Level:
The Zargoes use almost the same alphabet as English. What does this birthday message say?
### 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?
### 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?
### How Long Does it Take?
##### Age 7 to 11 Challenge Level:
How long does it take to brush your teeth? Can you find the matching length of time?
### 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.
### 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?
### 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?
### 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?
### 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.
### 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?
### A Right Charlie
##### Age 7 to 11 Challenge Level:
Can you use this information to work out Charlie's house number?
### Coins (2)
##### Age 7 to 11 Challenge Level:
What is the smallest number of coins needed to make up 12 dollars and 83 cents?
### 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?
### Dart Target
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### Home City
##### Age 7 to 11 Challenge Level:
Use the clues to work out which cities Mohamed, Sheng, Tanya and Bharat live in.
### Centred Squares
##### Age 7 to 11 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Sums and Differences 1
##### Age 7 to 11 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Encouraging Primary Children to Work Systematically
##### Age 3 to 11
This article for primary teachers suggests ways in which to help children become better at working systematically.
### 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.
### 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!
### 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.
### Ancient Runes
##### Age 7 to 11 Challenge Level:
The Vikings communicated in writing by making simple scratches on wood or stones called runes. Can you work out how their code works using the table of the alphabet?
##### Age 7 to 11 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Plate Spotting
##### Age 7 to 11 Challenge Level:
I was in my car when I noticed a line of four cars on the lane next to me with number plates starting and ending with J, K, L and M. What order were they in?
### 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?
### Seven Pots of Plants
##### Age 7 to 11 Challenge Level:
There are seven pots of plants in a greenhouse. They have lost their labels. Perhaps you can help re-label them.
### The Pet Graph
##### Age 7 to 11 Challenge Level:
Tim's class collected data about all their pets. Can you put the animal names under each column in the block graph using the information?
### Snails' Trails
##### Age 7 to 11 Challenge Level:
Alice and Brian are snails who live on a wall and can only travel along the cracks. Alice wants to go to see Brian. How far is the shortest route along the cracks? Is there more than one way to go?
### 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.
### Rolling That Cube
##### Age 5 to 11 Challenge Level:
My dice has inky marks on each face. Can you find the route it has taken? What does each face look like?
### 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 to have an ordered approach.
### Half Time
##### Age 5 to 11 Challenge Level:
What could the half time scores have been in these Olympic hockey matches?
### 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.
### Whose Face?
##### Age 5 to 11 Challenge Level:
These are the faces of Will, Lil, Bill, Phil and Jill. Use the clues to work out which name goes with each face.
### 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.
### 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?
### 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?
### 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.
### 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 ...?
### Greater Than or Less Than?
##### Age 7 to 11 Challenge Level:
Use the numbers and symbols to make this number sentence correct. How many different ways can you find?
### Sums and Differences 2
##### Age 7 to 11 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Penta Primes
##### Age 7 to 11 Challenge Level:
Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it?
### Build it up More
##### Age 7 to 11 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### Elf Suits
##### Age 7 to 11 Challenge Level:
If these elves wear a different outfit every day for as many days as possible, how many days can their fun last? | 2,200 | 9,333 | {"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} | 3.59375 | 4 | CC-MAIN-2020-10 | latest | en | 0.88493 |
https://docs.google.com/spreadsheets/d/1BMfV84s0nbFO5hQmMxxNmYQ_p-1W6YX2H-GaJm8u6Ms/edit?usp=drive_open&ouid=0 | 1,542,704,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746301.92/warc/CC-MAIN-20181120071442-20181120093442-00454.warc.gz | 557,024,802 | 93,103 | MS6-MA
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1
LevelCourse
Anchor Standards
Standard
Sub-Standard
SubSubStandard
Standard CodeStandardsIA
2
MS6MAEE0100*0MS6-MA-EE.01.00*.0Write and evaluate numerical expressions involving whole-number exponents.Number Sense & Operations: Use place value, write numbes in standard, expanded and exponential form
3
MS6MAEE0200*0MS6-MA-EE.02.00*.0Write, read, and evaluate expressions in which letters stand for numbers.Algebraic Patterns and Connections: use expressions and equations to model situations
4
MS6MAEE02A0MS6-MA-EE.02.A.0Write expressions that record operations with numbers and with
letters standing for numbers. For example, express the calculation
“Subtract y from 5” as 5 – y.
Algebraic Patterns and Connections
5
MS6MAEE02B0MS6-MA-EE.02.B.0Identify parts of an expression using mathematical terms (sum,
term, product, factor, quotient, coefficient); view one or more
parts of an expression as a single entity. For example, describe the
expression 2 (8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms.
Number Sense & Operations: Describe and Apply properties of numbers
6
MS6MAEE02C0MS6-MA-EE.02.C.0Evaluate expressions at specific values of their variables. Include
expressions that arise from formulas used in real-world problems.
Perform arithmetic operations, including those involving wholenumber
exponents, in the conventional order when there are no
parentheses to specify a particular order (Order of Operations).
For example, use the formulas V = s3 and A = 6 s2 to find the volume and surface area of a cube with sides of length s = 1/2.
Number Sense & Operations: Demonstrate Ways of performing operations
7
MS6MAEE03000MS6-MA-EE.03.00.0Apply the properties of operations to generate equivalent expressions. (e.g., apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y.Number Sense & Operations: Demonstrate Ways of performing operations
8
MS6MAEE04000MS6-MA-EE.04.00.0Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). (e.g., the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for.Algebraic Patterns and Connections: Use and interpret operational and relational symbols
9
MS6MAEE05000MS6-MA-EE.05.00.0Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.Algebraic Patterns & Connections: Solve Equations & Inequalities
10
MS6MAEE0600*0MS6-MA-EE.06.00*.0Use variables to represent numbers and write expressions when solving a real-world or mathematical problems.Algebraic Patterns and Connections: use expressions and equations to model situations
11
MS6MAEE07000MS6-MA-EE.07.00.0Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.Algebraic Patterns & Connections: Solve Equations & Inequalities
12
MS6MAEE08000MS6-MA-EE.08.00.0Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.Algebraic Patterns & Connections: Solve Equations & Inequalities
13
MS6MAEE09000MS6-MA-EE.09.00.0Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. (e.g., in a problem involving motion at constant speed, list and graph ordered airs of distances and times, and write the equation d = 65t to represent the relationship between distance and time.Data Analysis, Probability & Statistics
14
MS6MAG0100*0MS6-MA-G.01.00*.0Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.Geometry
15
MS6MAG02000MS6-MA-G.02.00.0Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.Geometry
16
MS6MAG03000MS6-MA-G.03.00.0Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems.Geometry
17
MS6MAG04000MS6-MA-G.04.00.0Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems.Geometry
18
MS6MANS0100*0MS6-MA-NS.01.00*.0Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. Not on IA: Division of Fractions
19
MS6MANS0200*0MS6-MA-NS.02.00*.0Divide multi-digit numbers using the standard algorithm.Compute with Whole Numbers
20
MS6MANS0300*0MS6-MA-NS.03.00*.0Add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.Compute with Decimals
21
MS6MANS04000MS6-MA-NS.04.00.0Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. (e.g., express 36 + 8 as 4 (9 + 2).Compute with Fractions
22
MS6MANS0500*0MS6-MA-NS.05.00*.0Understand that positive and negative numbers are used together to describe quantities having opposite directions or values; use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.Number Sense and Operations
23
MS6MANS06000MS6-MA-NS.06.00.0Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. Not on IA
24
MS6MANS06A0MS6-MA-NS.06.A.0Recognize opposite signs of numbers as indicating locations
on opposite sides of 0 on the number line; recognize that the
opposite of the opposite of a number is the number itself, e.g.,
–(–3) = 3, and that 0 is its own opposite.
Not on IA
25
MS6MANS06B0MS6-MA-NS.06.B.0Understand signs of numbers in ordered pairs as indicating
locations in quadrants of the coordinate plane; recognize that
when two ordered pairs differ only by signs, the locations of the
points are related by reflections across one or both axes.
Not on IA
26
MS6MANS06C0MS6-MA-NS.06.C.0Find and position integers and other rational numbers on a
horizontal or vertical number line diagram; find and position pairs
of integers and other rational numbers on a coordinate plane.
Not on IA
27
MS6MANS07000MS6-MA-NS.07.00.0Understand ordering and absolute value of rational numbers.Number Sense & Operations: Represent Compare and Order Numbers
28
MS6MANS07A0MS6-MA-NS.07.A.0Interpret statements of inequality as statements about the relative
position of two numbers on a number line diagram. For example,
interpret –3 > –7 as a statement that –3 is located to the right of –7 on a number line oriented from left to right.
Number Sense & Operations
29
MS6MANS07B0MS6-MA-NS.07.B.0Write, interpret, and explain statements of order for rational
numbers in real-world contexts. For example, write –3 oC > –7 oC to express the fact that –3 oC is warmer than –7 oC.
Number Sense and Operations / Algebraic Pattens and Connections
30
MS6MANS07C0MS6-MA-NS.07.C.0Understand the absolute value of a rational number as its distance
from 0 on the number line; interpret absolute value as magnitude
for a positive or negative quantity in a real-world situation. For
example, for an account balance of –30 dollars, write |–30| = 30 to
describe the size of the debt in dollars.
Not on IA
31
MS6MANS07D0MS6-MA-NS.07.D.0Distinguish comparisons of absolute value from statements about
order. For example, recognize that an account balance less than –30
dollars represents a debt greater than 30 dollars.
Not on IA
32
MS6MANS0800*0MS6-MA-NS.08.00*.0Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.Not on IA
33
MS6MARP0100*0MS6-MA-RP.01.00*.0Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. Measurement
34
MS6MARP02000MS6-MA-RP.02.00.0"Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1"Measurement
35
MS6MARP03000MS6-MA-RP.03.00.0Use ratio and rate reasoning to solve real-world and mathematical
problems, e.g., by reasoning about tables of equivalent ratios, tape
diagrams, double number line diagrams, or equations.
Measurement
36
MS6MARP03A0MS6-MA-RP.03.A.0Make tables of equivalent ratios relating quantities with wholenumber measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.Measurement
37
MS6MARP03B0MS6-MA-RP.03.B.0Solve unit rate problems including those involving unit pricing and
constant speed. For example, if it took 7 hours to mow 4 lawns, then
at that rate, how many lawns could be mowed in 35 hours? At what
rate were lawns being mowed?
Measurement: Understand and Apply Rate
38
MS6MARP03C0MS6-MA-RP.03.C.0Find a percent of a quantity as a rate per 100 (e.g., 30% of a
quantity means 30/100 times the quantity); solve problems
involving finding the whole, given a part and the percent.
Number Sense & Operations / Algebraic Patterns & Connections
39
MS6MARP03D0MS6-MA-RP.03.D.0Use ratio reasoning to convert measurement units; manipulate
and transform units appropriately when multiplying or dividing
quantities.
Measurement / Fractions
40
MS6MASP01000MS6-MA-SP.01.00.0Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers. For example, “How old am I?” is not a statistical question, but “How old are the students in my school?” is a statistical question because one anticipates variability in students’ ages.Data Analysis, Probability & Statistics
41
MS6MASP02000MS6-MA-SP.02.00.0Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center, spread, and overall shape.Data Analysis, Probability & Statistics
42
MS6MASP03000MS6-MA-SP.03.00.0Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number.Data Analysis, Probability & Statistics
43
MS6MASP04000MS6-MA-SP.04.00.0Display numerical data in plots on a number line, including dot plots, histograms, and box plots.Data Analysis, Probability & Statistics
44
MS6MASP05000MS6-MA-SP.05.00.0Summarize numerical data sets in relation to their context, such as by:Data Analysis, Probability & Statistics
45
MS6MASP05A0MS6-MA-SP.05.A.0Reporting the number of observations.Data Analysis, Probability & Statistics
46
MS6MASP05B0MS6-MA-SP.05.B.0Describing the nature of the attribute under investigation,
including how it was measured and its units of measurement.
Not on IA
47
MS6MASP05C0MS6-MA-SP.05.C.0Giving quantitative measures of center (median and/or mean) and
variability (interquartile range and/or mean absolute deviation), as
well as describing any overall pattern and any striking deviations
from the overall pattern with reference to the context in which the
data were gathered.
Data Analysis, Probability & Statistics
48
MS6MASP05D0MS6-MA-SP.05.D.0Relating the choice of measures of center and variability to the
shape of the data distribution and the context in which the data
were gathered.
Data Analysis, Probability & Statistics | 3,451 | 13,683 | {"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} | 3.609375 | 4 | CC-MAIN-2018-47 | latest | en | 0.800871 |
https://codeforces.com/blog/entry/59730 | 1,718,348,097,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861521.12/warc/CC-MAIN-20240614043851-20240614073851-00661.warc.gz | 155,413,785 | 23,407 | ### GlebsHP's blog
By GlebsHP, history, 6 years ago, translation,
## Smart Vending
Problem idea and development: GlebsHP.
First we observe that Arcadiy won't be able to spend more money than he has, i.e. the number of bottles he can buy doesn't exceed . However, it might happen that Arcadiy won't be able to buy exactly this number because he won't have enough coins to buy next bottle with no change and the machine won't possess enough coins to give change if Arcadiy tries to pay with banknotes only.
Note that the total number of coins doesn't change and is always c + d. Moreover, if c + d ≥ 999 999, then either Arcadiy will always have enough coins to buy a bottle with no change, or machine will have enough coins to give a change. In this case the answer is z.
If c + d < 999 999, then it can be the case that Arcadiy has enough coins to buy bottle with no change, it can be the case that machine has enough coins to give change, but it can't be the case that both is possible. That means we can simulate the process but that might work tool long.
Denote as ai the number of coins Arcadiy has after first i moves (a0 = c). Note that if ai = aj, then ai + 1 = aj + 1 (indeed, the way to buy next bottle is uniquely defined by value ai) and the sequence repeats itself. We will simulate the process of buying Kvas-Klass bottle by bottle till we find some value of ai that we already met or it is not possible to make next purchase. In case ai repeats the answer is z, otherwise the answer is the number of step when the next purchase was not possible.
Exercise: solve the task if one banknote is 1012 rubles.
## LIS vs. LDS
Problem idea and development: Endagorion.
Let us draw n points pi = (i, ai) in the plane. Let I = (i1, ..., ik) and J = (j1, ..., jl) be a pair of non-intersecting LIS and LDS. From maximality of I, there can be no points strictly inside the following rectangles (described by their opposite corners):
1. A0 = ((0, 0), pi1);
2. Ax = (pix, pix + 1) for all 1 ≤ x < k;
3. Ak = (pik, (n + 1, n + 1)).
Similarly, since J is maximal, there can be no points inside these rectangles:
1. B0 = ((0, n + 1), pj1);
2. By = (pjy, pjy + 1) for all 1 ≤ y < l;
3. Bl = (pjl, (n + 1, 0)).
These two chains of rectangles connect the opposite corners of the square ((0, 0), (n + 1, n + 1)). Assuming that I and J do not intersect, there must be an intersecting pair of rectangles Ax and By. None of Ax and By contains any points, hence there are only two cases how they can intersect:
1. ix < jy < jy + 1 < ix + 1, ajy + 1 < aix < aix + 1 < ajy (as in the first sample);
2. jy < ix < ix + 1 < jy + 1, aix < ajy + 1 < ajy < aix + 1 (the mirror image of the previous option).
We will try to look for possible intersections of type 1 (the other one can be found similarly after reversing the whole permutation). Let's say that (i, i') is an {\it LIS-pair} if a and b are consecutive elements of an LIS ({\it LDS-pair} is defined similarly). Suppose that there is an LIS-pair (i, i') and an LDS-pair (j, j') such that i < j < j' < i' and aj' < ai < ai' < aj, that is, there is an intersection of type 1 between certain LIS and LDS. This means that we can find an answer by constructing all LIS- and LDS-pairs and looking for such intersection.
Of course, there are too many possible pairs. However, notice that in the situation above we can replace i' with , and j' with . This means that we will only consider O(n) pairs to intersect. Among these, finding an intersection can be done with a sweep-line approach that uses segment tree.
## Eat And Walk
Problem idea and development: GlebsHP.
To start with we consider the minor change in the problem: let the initial movements be free and the movements after x units of food cost x per 1 move left or right. One can show that the optimal strategy is to reach some position x and then move only in the direction of position 0, visiting some of the restaurants. If we visit restaurant i the weight increases by ai and each move costs ai more. Assuming we only move left after visiting any restaurants we can say that visiting restaurant i costs ai·i units of energy in total.
Now we have a knapsack problem with n items, the i-th of them has weight ai·i and cost ai, the goal is to find a subset of maximum total cost and total weight not exceeding e. Standard solution is to compute dynamic programming d(i, x) — maximum possible total cost if we select among first i items and the total weight is equal to x.
Now we swap parameter and value of dynamic programming and compute d(i, c) — minimum possible total weight that the subset of cost c selected among first i items can have. Then we are going to add items in order of descending i, so value d(i, c) will be composed using elements from i-th to n-th. What are the values of c to consider? Each unit of cost (food in original terms) that comes from restaurants with indices i and greater requires at least i units of energy so we should only consider values of . Using the formulas to estimate harmonic series we can say that we only need to compute elements of dynamic programming.
To finish the solution we should count the initial cost to move to maximum index restaurant we visit (and back). Let this index be k, we should increase answer by 2k. This can be done by using extra cost in dynamic programming when we consider moves from state d(i, 0).
## Search Engine
Problem idea: GlebsHP. Development: cdkrot.
Let's notice, that in the end, after prepending and appending of all letters, Alice will have string s exactly.
Suppose, that we were adding new letters and having at least one occurrences inside s, and then added one symbol and all the occurrences disappeared. In that case, the all following additions have zero occurrences count and we should have stopped on that bad step, selected one of occurrences and then growed it to the left and to the right to the bounds of s.
Because of this observation, we can reverse the process: let's say that initially we have string s, and then we are allowed to delete the letters from left or right side, with the same goal to maximize the total number of occurrences.
Let's build a suffix tree on s, which is a trie with all suffixes of s.
Then we can do a dynamic programming on the vertices of the suffix tree. Let's denote dp[v] as the maximum sum which we can achieve if we start with string, which corresponds to the path from root to v and repeatedly deleted letters from left or right, counting the total number of occurrences in the whole string s.
This dynamic programming has only two transitions — we can go to the parent vertex (which corresponds to deletion of the last letter), or go by the suffix link (which corresponds to deletion of the first letter). We can calculate this dynamic programming lazily and the number of occurrences of the string is simply the number of terminal vertices in it's subtree (this can be calculated for all vertices beforehand).
However, this way we have n2 states in our dynamic programming, so we will have to use the compressed suffix tree.
In compressed suftree, transition to the parent may mean deletion of multiple symbols on the suffix. In this case, number of occurrences anywhere between the edge is equal to the number of occurrences of the lower end (so you need to multiply it by the edge length). The same also holds for the suffix links transitions.
The only thing we missed, it that the optimal answer could have transitions up by one symbol and suffix links transitions interleaved, and we can't simulate this with compressed suffix tree. However, one never needs to interleave them. Assume, that in optimal answer we were going up and then decided to go by suffix link and finish the edge ascent after it. In that case we can simply go by the suffix link first and traverse the edge after that — this way the total number of occurrences can only increase.
The complexity is , also the problem can be solved with the suffix automaton and dynamic programming on it. In that case, the dynamic transitions are suffix links and reverse edges.
Exercise: solve the problem if the game starts from some substring (defined by li and ri) and you need to process up to 100 000 queries.
## Guess Me If You Can
Problem idea and development: GlebsHP.
Note that if we name all elements in some order, at the end we will have position similar to the initial but shifted by 1. This shift doesn't affect neither answer, nor intermediate values returned by interactor.
Consider some triple of elements i, j and k such that pi + 1 = pj = pk - 1. In case we list elements in some order such that j goes before i and k, in the moment pj is increased by 1 the number of distinct element decreases and we can say that position j doesn't contain maximum.
What if we select a random permutation and name all elements in this order. For triple i, j, k we have chance that j will go before i and k. Thus, if we use 50 random permutations, the probability to fail to find out that j is not a maximum is only . The events corresponding to separate triples might not be independent so we estimate the probability to make at least one mistake as 2·10 - 9·n ≤ 2·10 - 6.
## Lazy Hash Table
Problem idea: GlebsHP. Development: Arterm.
Note that if and only if m|ai - aj. So, we have to find minimal m that doesn't divide any of differences ai - aj.
First find all possible values of d = ai - aj. Set
where M = max ai. Find product T(x) = P(xQ(x) using Fast Fourier Transform. As
coefficient with xM + d is nonzero if and only if there are i, j with ai - aj = d.
Now for each m in range [1, M] check if there is difference divisible by m. We can check this checking numbers: . So total complexity for this part is .
Overall complexity is , где M = max ai ≤ 2 000 000.
• +66
» 6 years ago, # | +5 Has anyone received the T-shirt yet?
• » » 6 years ago, # ^ | +5 We haven't started sending them out — only onsite finalists have them. Wait for the email from me please. | 2,551 | 10,024 | {"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-2024-26 | latest | en | 0.92162 |
http://www.nativetreesociety.org/measure/volume/formulas_march_2008.htm | 1,675,150,894,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499845.10/warc/CC-MAIN-20230131055533-20230131085533-00465.warc.gz | 72,506,320 | 11,369 | Formulas March 2008
==============================================================================
TOPIC: A useful formula
==============================================================================
== 1 of 1 ==
Date: Fri, Mar 7 2008 2:43 pm
From: [email protected]
ENTS,
All of us who measure trees with some degree of seriousness encounter those inescapable places that are cluttered by underbrush. We may be able to see the base of a tree we want to measure with our eye, but can't get the laser to penetrate the brush. We may be short on time, the terrain may be rough, etc. to preclude extra time being taken with a tree. So want to shoot and move on. If we can spot a point not to high on the trunk, in case of a lean where we can get a clean laser shot, we're in business.
Let
H = height of tree from eye level to the base
a = angle from eye level to point on trunk that we can shoot
D = laser distance from eye to point on trunk that can be clearly seen
( can be above or below eye level)
b = angle between eye and base of tree.
Then
H = D * cos(a) * tan(b)
This formula gives us the component of height between the level of our eye and the base of the tree. The component of height above eye level is measured in the ordinary way using the sine top method. The above calculation works only if the tree does not have a substantial lean or if it does that we can use close enough to the base that the error introduced is insignificant. However, in situations where it can be applied, it solves the problem of clearing away the brush to get a clear shot.
I've applied in above method in a two step process many times, but had not boiled it down to a single formula.
Bob
==============================================================================
TOPIC: New formula test
==============================================================================
== 1 of 1 ==
Date: Mon, Mar 10 2008 2:08 pm
From: [email protected]
Beth, Ed, et al:
I did a very quick test this afternoon of the cos-tan bottom method compared to the sine bottom. The table below shows the results.
CosTan Bottom Sin Bottom Diff Abs(Diff) -3.01 -3.00 -0.01 0.01 -8.08 -8.50 0.42 0.42 -10.24 -10.00 -0.24 0.24 -16.15 -16.00 -0.15 0.15 4.76 5.00 -0.24 0.24 -3.67 -3.50 -0.17 0.17 -21.05 -21.00 -0.05 0.05 16.80 17.00 -0.20 0.20 -21.10 -21.00 -0.10 0.10 -19.00 -18.50 -0.50 0.50 Average 0.21
Average 0.21
With care I could have probably gotten the difference to as low as 0.1 feet. In deciding to do the test, there wasn't any expectation of a problem, but I thought potential users of the method would want to know what experimental difference could be expected from the two methods.
To repeat the formula and its purpose.
Let
H = height of tree from eye level to the base
a = angle from eye level to point on trunk that we can see and shoot without obstruction
D = laser distance from eye to the point on trunk that can be clearly seen
( can be above or below eye level)
b = angle between eye and base of tree.
Then
H = D * cos(a) * tan(b)
This formula gives us the component of height between the level of our eye and the base of the tree. The component of height above eye level is measured in the ordinary way using the sine top method. The above calculation works only if the tree does not have a substantial lean or if it does that we can use close enough to the base that the error introduced is insignificant. However, in situations where it can be applied, it solves the problem of clearing away the brush to get a clear shot.
Bob
==============================================================================
TOPIC: Three more formulas
==============================================================================
== 1 of 3 ==
Date: Sat, Mar 15 2008 8:48 am
From: [email protected]
ENTS,
There is a multitude of measuring situations that present special challenges to those of us who measure trees in a forest environment and want to achieve the requisite ENTS level of accuracy. Two situations are covered below.
We've all encountered in-forest conditions where clutter around the base of a tree prevents a clean laser shot and it isn't practical for us to worm our way to the base, e.g. across a stream or ravine, too much rhodo, we're pooped, etc. However, as covered in a previous e-mail, if we can spot the base with our eye and measure the angle from eye to base (even though we can't hit the base with the laser) and then measure the distance and angle to a point not to high up on the trunk, we can compute the component of height from base to eye level using the formula I previously gave. Our assumption is the the tree's lean is negligible. But, the byproducts of computation don't end here. An associated problem is to compute the straight-line distance from eye to the base under the above conditions. This distance can be computed as follows:
Let
D = straight-line distance from eye to visible spot on trunk
a = angle between eye and spot on trunk
b = angle between eye and base of tree
S = straight-line distance from eye to base.
Then
S= D * cos(a)/cos(b). [Remembering that the height component is D * cos(a)*tan(b)]
While knowing S isn't necessary to computing the component of height between the base and the eye (our primary objective), it is good to be able to completely delineate the triangles we are constructing to measure the components of height. Computing S is a way of filling all the squares.
A significantly more challenging situation faces us when we want, from a distance, to determine the angle of lean toward or away from the measurer of a tree trunk. Let's assume that we can't bank on shooting the distance directly to the point on the trunk either to eye level or to the base. Those areas of the trunk may be obscured. However, if we can shoot to two points on the trunk (one above and one below eye level) that are sufficiently separated from one another to create an angle of several degrees, we can establish with acceptable accuracy the degree of a lean with the following formula. Note that the symbol "|" in the formula denotes absolute value. That is, |x| means the absolute value of x. Also please note that a set of six formulas is needed to calculate trunk lean under all conditions (e.g. both angles are positive, both are negative, one positive and one negative, combined with a trunk lean toward the observer or away from the observer).
Let
D1 = distance to uppermost of the two points on the trunk we will shoot
D2 = distance to lowermost of two points
a = angle to upper point
b = angle to lower point
c = angle of lean from the vertical
If the cinditions D2*cos(b) > D1*cos(a), a>0, b<0 are met, then
c = 90 - arctan[( |D1*|sin(a)| + D2*|sin(b)|| )/( |D2*|cos(b)| - D1*|cos(a)||)]
If D2*cos(b) < D1*cos(a), a>0, b<0 then
c = arctan[( |D1*|sin(a)| + D2*|sin(b)|| )/( |D2*|cos(b)| - D1*|cos(a)||)]
The above formulas for c cover 2 of the 6 cases. The others will follow in other communications. Note that all these formulas and their derivations will be included in the publication on dendromorphometry that is in the works.
Finally, the complexity of these formulas suggests the development of spreadsheets for the actual calculations.
Bob
== 2 of 3 ==
Date: Sat, Mar 15 2008 11:58 am
From: Larry
Bob, You are the Man! Good to see you feeling better! Your formulas
are way cool. I need to get back to the Volume project, the stuff you
set up on excel is Awesome, way cool stuff. I have to get a new
laptop and I will begin sending you results on the project. Its 77
down here today with lots of flowers starting to bloom. Again, I'm
glad to see you able to get back in the forest, doing the stuff you
love so much. Bob, I think we need another one of your Asethetic
stories if you get time. Larry
== 3 of 3 ==
Date: Sat, Mar 15 2008 12:15 pm
From: [email protected]
Larry,
Thanks very much. I think Monica and I are going for a third trip to Mohawk this Sunday, if the weather holds. Gosh, does it ever feel good to even think about getting out. The power of those marvelous Mohawk trees is not to be denied.
On the story side, as I plan Monica's and my June-July trip to Idaho, I've been thinking about old haunts we intend to visit along the way and the one that always seems to emerge on top is the Cloud Peak Wilderness Area in the Bighorn Mountains of north-central Wyoming- my ultimate mountain Mecca. Lots of stories to tell there. I'll put a couple together.
Bob
==============================================================================
TOPIC: Three more formulas
==============================================================================
== 1 of 1 ==
Date: Sun, Mar 16 2008 3:02 pm
From: [email protected]
ENTS,
I've been able to boil the trunk lean fomulas down to just two after working through the 6 scenarios.
Let
D1 = distance to uppermost of the two points on the trunk we will shoot
D2 = distance to lowermost of two points
a = angle to upper point
b = angle to lower point
c = angle of lean from the vertical
If a >0 and b < 0 then
c = atan([|D1*|cos(a)| - D2*|cos(b)||]/[|D1*|sin(a)| +D2*|sin(b)||])
If a >0 and b >0 or
a<0 and b<0 then
c = atan([|D1*|cos(a)| - D2*|cos(b)||]/[|D1*|sin(a)| -2*|sin(b)||])
Please take note of the "| |" notation. The bars identify the absolute value of what is enclosed between them. Also note that relative to the observer, the trunk can have a forward-backward and a lateral lean. The above formulas ignore the lateral lean and assume the observer is in alignment with the trunk so that a vertical plane that includes the measurer's eye would also include all of the trunk between the two measurement points. A slight lateral lean won't change the calculations much. Taking both components of lean into account involves much more calculating and will be the subject of later communications.
Bob
==============================================================================
==============================================================================
== 1 of 1 ==
Date: Wed, Mar 19 2008 12:32 pm
From: [email protected]
ENTS,
Sheet1 of the attached Excel workbook (TrunkSlopeDiagram.xls) shows the formula for calculating trunk lean from the vertical as determined from a distance. Note that absolute values are taken to avoid negative angles in the resulting addition. I may have gone overboard with the absolute value function, but what the heck. The diagram provided on the Sheet1 deals with the case where angle a >0 (above eye level) and angle b <0 (below eye level). If both angles are either above or below eye level, the formula changes in that the factors in the denominator are subtracted instead of added. This is indicated by the +/- sign in the denominator of the last shown formula.
Sheet2 is a handy dandy calculator for tree lean. Remember, the assumption is that we cannot get to the tree, so we are attempting to determine the trunk lean as an angle from the vertical taking two points on the trunk and measuring them from a distance. We also assume that the area of the trunk at eye level may be obscured also. So the location of the two points is arbitrary.
A sample of actual measurements I took from my patio follows. The first measurement was a test only to see what the resulted looked like.
A sample of actual measurements I took from my patio follows. The first measurement was a test only to see what the resulted looked like.
D5 h1 D3 h2 Species D1 a D2 b abs(D1*abs(cos(a))) abs(D1*abs(sin(a))) abs(D2*abs(cos(b))) abs(D2*abs(sin(b))) c Test 106.42 18.00 100.00 -6.00 101.21 33.43 99.45 10.47 2.29 NRO 80.50 9.70 81.50 -13.60 79.35 13.63 79.21 19.35 0.23 BB 48.50 4.00 50.50 -17.50 48.38 3.39 48.16 15.42 0.67 HM 41.50 4.50 44.00 -17.60 41.37 3.26 41.94 13.52 -1.94 WP 89.00 -5.00 91.50 -13.90 88.66 7.77 88.82 22.20 0.63 WO 54.50 -0.50 57.00 -17.60 54.50 0.48 54.33 17.51 -0.56 BB 45.50 4.00 48.00 -18.50 45.39 3.18 45.52 15.50 -0.40 NRO 78.50 -0.60 78.50 -9.90 78.50 0.82 77.33 13.56 -5.22
You can see in the above table, only one lean is significant, the Northern Red Oak at -0.522 degrees. The tree is on an embankment and leans away from the location I from where I shot it. It leans toward the downhill side. Folks with programable calculators, might want to program in the formula. However, it might not be easy to be able to take in the addition/subtraction option in the denominator, i.e. if a>0 and b <0, then the factors in the denominator are added, else subtracted. If D5>D3, then the tree leans away, otherwise, it leans toward the measurer.
The practicality and applicability of the above method is admittedly limited and not apt to find much usage except where the measuring environment is cluttered, the tree has limited accessibility, and the steaks to get it right are high.
Bob
==============================================================================
TOPIC: Possible new spreadsheet for Larry
==============================================================================
== 1 of 2 ==
Date: Wed, Mar 19 2008 5:16 pm
From: [email protected]
Larry,
Uh oh, I've been at it again. What happens when a long limb of one of the Live Oaks is a curve as opposed to being straight? The attached spreadsheet may be a way to get a better length calculation. It fits a parabola to limb curve. Three measurements are required: (x0,y0), (x1,y1), and (x2,y2). The diagram shows where they would be taken. The rest is automatic. The example shows a limb that has a horizontal length of 65 feet. What do you think.
Bob
ParabolicLimbModeling.xls
== 2 of 2 ==
Date: Wed, Mar 19 2008 5:21 pm
From: "Dale Luthringer"
Bob,
This definitely would come in handy to model the Pinchot Sycamore.
Dale
==============================================================================
TOPIC: Possible new spreadsheet for Larry
==============================================================================
== 1 of 1 ==
Date: Thurs, Mar 20 2008 3:33 am
From: [email protected]
Dale,
Yes, I think you're right. I had that in mind as well as the Granby Oak. Are there any trees out your way that we could test it on?
Bob
==============================================================================
TOPIC: Possible new spreadsheet for Larry
==============================================================================
== 2 of 4 ==
Date: Thurs, Mar 20 2008 8:48 am
From: Larry
Bob, Way cool, looks like it will help alot. I'd guess 50% of all
Live Oak limbs are curved. Your skills on excel are Awesome! Lots of
flowers blooming down this way. Larry
== 3 of 4 ==
Date: Thurs, Mar 20 2008 12:12 pm
From: [email protected]
Larry,
Thanks. As a general observation, my long term goal is to establish a set of measurement methods that requires the fewest number of field measurements to be taken by the measurer and then automate each method in a popular product like Excel. In pursuit of this personal objective, I'm also available to my fellow and lady Ents to create customized spreadsheets for any of you who might wish them.
BTW, we can adopt other curve forms besides the parabola to model limb length, but so far I see no reason to do that. Also, if we want to take more measurments along the length of the limb, a best fit parabola can be determined via curvilinear regression analysis. The regression process can be built into a spreadsheet although there are sophisticated statistics programs that do all that. The problem is that they invariably force onto the user far more fiddling around with complex software than the user may be willing to commit to. I think it is better to have a simple set of tools that can be individually applied unless you are a real heavyweight. There comes a point where a piecemeal approach won't work and you have to commit to sophisticated software, but that greatly limits the number of users. So, I'm trying to fill in with an in between solution to what otherwise tends to be all or nothing.
As a final comment, In response to your request, I've started on a write-up of an aesthetic look at a favored nature site of mine. I should get the first installement out in about three days.
Bob
== 4 of 4 ==
Date: Thurs, Mar 20 2008 7:19 pm
From: Larry
Bob, | 4,060 | 16,229 | {"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} | 3.84375 | 4 | CC-MAIN-2023-06 | latest | en | 0.902767 |
https://math.stackexchange.com/questions/1185117/phase-curve-of-ddotx-x-ddoty-y | 1,632,456,747,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057496.18/warc/CC-MAIN-20210924020020-20210924050020-00057.warc.gz | 433,592,563 | 40,547 | # Phase curve of $\ddot{x}=-x,\ddot{y}=-y$
Rewriting the ODE in the title, we get $$\dot{x_1}=x_2,\dot{x_2}=-x_1, \dot{x_3}=x_4, \dot{x_4}=-x_3.$$ It is easy to show that $$x_1=A\cos(t)+B\sin(t),x_2=B\cos(t)-A\sin(t),\\x_3=C\cos(t)+D\sin(t),x_4=D\cos(t)-C\sin(t).$$ In Ordinary Differential Equations by V. I. Arnold, there are several problems on this equation. For example, it can be shown that each phase curve is on a 3-sphere, and is a great circle of that.
The next problem is more difficult: show that the phase curves on a given 3-sphere form a 2-sphere.
I attempt to find out which 3-dimensional subspace the 2-sphere lies in, but in vain.
And the last question is about the linking number. Since 3-sphere can be regarded as $\mathbb R^3\cup \{\infty\}$, a partition of 3-sphere into circles determines a partition of $\mathbb R^3$ into circles and nonclosed circles. Show that any two of the circles of this partition are linked with linking number 1.
I am not sure how to visualize this partition, and what the circles stand for.
• @Amzoti Yes, and I have edited it. Mar 11 '15 at 14:39
• I'm not an expert on this, but it seems to be connected with the so called Hopf fibration; see here: en.wikipedia.org/wiki/Hopf_fibration Mar 11 '15 at 15:00
• It seems that every property you are interested in derives from the fact that each solution of this differential system stays on a surface $$[x_1^2+x_2^2=a^2,\,x_3^2+x_4^2=b^2],$$ which, in turn, is included in the 3-sphere $$x_1^2+x_2^2+x_3^2+x_4^2=r^2$$ for some suitable $r$.
– Did
Mar 13 '15 at 15:02
• The two-sphere (the set of orbits) turns out not to be embedded in the three-sphere. It may help to introduce complex coordinates $z_{1} = x_{1} + ix_{2}$, $z_{2} = x_{3} + ix_{4}$; the flow of your system is $$(t, z_{1}, z_{2}) \mapsto (e^{it}z_{1}, e^{it}z_{2}).$$The three-sphere is invariant (see also Did's comment), and the quotient of $S^{3}$ by this circle action is the Hopf map (see Christian's comment); the quotient space is the set of complex lines through the origin of $\mathbf{C}^{2}$. Mar 13 '15 at 16:04
• By not rewriting the ODE we find that: $$x=A\cos(t)+B\sin(t)\\ y=C\cos(t)+D\sin(t)$$ Solving for $\cos(t)$ and $\sin(t)$ and summing the squares of these then yields a Lisajous Ellipse, degenerated eventually (if $AD-BC=0$): $$\left(\frac{Cx-Ay}{AD-BC}\right)^2+\left(\frac{Dx-By}{AD-BC}\right)^2 = 1$$ That's all "interesting" I can see. What more is there to be said? Mar 14 '15 at 9:44
$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$This answer introduces well-known constructions (the equivalence of the complex projective line with the Riemann sphere, and the Hopf map from the $3$-sphere to the $2$-sphere) that are accessible given the presumed background of readers of Arnol'd, but may or may not occur naturally in the course of approaching this problem about pairs of harmonic oscillators.
The Riemann Sphere: For present purposes, the Riemann sphere is the holomorphic curve obtained by taking two copies $U_{0}$ and $U_{1}$ of the complex line, with respective coordinates $z$ and $w$, and identifying $w$ in $U_{1}$ with $z = 1/w$ in $U_{0}$. The origin ($w = 0$) in $U_{1}$ is regarded as the point at infinity in $U_{0}$, and vice versa.
The Complex Projective Line: If $z_{0}$ and $w_{0}$ are complex numbers, not both zero, the complex line they determine is the set of (complex) scalar multiples of $(z_{0}, w_{0})$. Throughout, the term "line" refers to a complex line through the origin in $\Cpx^{2}$. (A complex line is an oriented, real $2$-plane. However, most real $2$-planes through the origin of $\Cpx^{2}$ are not complex lines.)
Every non-zero vector in $\Cpx^{2}$ lies on a unique line. Two non-zero vectors $(z_{1}, w_{1})$ and $(z_{2}, w_{2})$ lie on the same line if and only if there exists a complex number $\lambda$ (necessarily non-zero) such that $$(z_{2}, w_{2}) = \lambda (z_{1}, w_{1}).$$
The set lines is, by definition, the complex projective line, $\Cpx\Proj^{1}$. This space acquires the structure of a holomorphic curve, equivalent to the Riemann sphere $\Cpx \cup \{\infty\}$, as follows: A complex number $z$ in $U_{0}$ corresponds to the line through $(z, 1)$. A complex number $w$ in $U_{1}$ corresponds to the line through $(1, w)$. If $w \neq 0$ and $z = 1/w$, then $$(1, w) = w(1/w, 1) = w(z, 1) \sim (z, 1);$$ that is, each non-zero number $w$ in $U_{1}$ is identified with $z = 1/w$ in $U_{0}$. This is precisely the gluing that defines the Riemann sphere.
The Hopf Fibration: Define a (holomorphic) mapping $\Pi:\Cpx^{2} \setminus\{(0, 0\} \to \Cpx\Proj^{1}$ by sending each (non-zero) pair $(z, w)$ to the line through $(z, w)$. The restriction of $\Pi$ to the $3$-sphere $$S^{3} = \{(z, w) : |z|^{2} + |w|^{2} = 1\}$$ is the Hopf map $\pi:S^{3} \to \Cpx\Proj^{1}$. The preimage of each point is the intersection of $S^{3}$ with a line, a great circle called a Hopf fibre. Since $\pi$ induces a bijection between Hopf fibres and points of $\Cpx\Proj^{1}$, the set of Hopf fibres is a $2$-sphere.
To see that distinct fibres of the Hopf map link once inside $S^{3}$, note that if $L_{1}$ and $L_{2}$ are distinct lines, then $\Cpx^{2} = L_{1} \oplus L_{2}$. Projection to the second summand is surjective, and the Hopf fibre in $L_{2}$ "winds around" $L_{1}$ (since this circle winds around the origin in $L_{2}$).
Consider the first-order linear system $$\dot{x}_{1} = x_{2},\quad \dot{x}_{2} = -x_{1},\quad \dot{x}_{3} = x_{4},\quad \dot{x}_{4} = -x_{3}.$$ Introducing complex coordinates $z = x_{1} + ix_{2}$ and $w = x_{3} + ix_{4}$ (and therefore identifying $\mathbf{R}^{4}$ with $\Cpx^{2}$, the preceding system becomes $$\dot{z} = iz,\quad \dot{w} = iw. \tag{1}$$ The solution of (1) with initial conditions $z(0) = z_{0}$, $w(0) = w_{0}$, is $$\left[\begin{array}{@{}cc@{}} z(t) \\ w(t) \\ \end{array}\right] = e^{it} \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} e^{it} & 0 \\ 0 & e^{it} \\ \end{array}\right] \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right]. \tag{2}$$
Since scalar multiplication maps each line to itself, each solution curve (2) is the Hopf fibre in the line determined by the unit vector $(z_{0}, w_{0})$. Since solutions of (1) are precisely Hopf fibres, the set of solutions is a $2$-sphere, and distinct solution curves link once in $S^{3}$. | 2,136 | 6,420 | {"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} | 3.875 | 4 | CC-MAIN-2021-39 | latest | en | 0.902998 |
https://www.physicsforums.com/threads/mechanical-advantage-question-ropes.588416/ | 1,531,685,944,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676588961.14/warc/CC-MAIN-20180715183800-20180715203800-00263.warc.gz | 961,685,416 | 20,238 | # Mechanical Advantage Question - Ropes
1. Mar 19, 2012
### Mdanner423
I have to take an aptitude test that I have taken before and there are a couple questions that are stumping me.
My first question I want to ask about deals with ropes / strings. It has an old-timey picture of a boy carrying books wrapped up in string and then attached to a stick and thrown over his back. One picture the string/rope is wrapped around the stick more. It ask which is easier to carry, A (the rope is wrapped around twice), or B (the rope is wrapped around five times), or C it's equal.
My first guess is B, but I can't explain why.
So does wrapping the rope more around help distribute the mass anymore to make it easier to carry or???
Thanks!
--Matthew
2. Mar 19, 2012
### emailanmol
Unfortunately, its the opposite.
At higher level of physics , you can derive the expression of the force on the stick by rope which increases exponentially with an increase in the number of rotations .
So the boy will have to provide more force to hold the stick if it has five rotations.
3. Mar 19, 2012
### Mdanner423
Interesting! Can you explain why?
4. Mar 19, 2012
### emailanmol
And yes, my previous post is valid in presence of friction.
If friction is absent the number of turns doesnt matter :-)
Last edited: Mar 19, 2012
5. Mar 19, 2012
### emailanmol
Surely I can,
Consider a small piece of the rope that subtends an angle dθ. Let the tension in this piece be T (which will vary slightly over the small length). the pole exerts a small outward normal force, N, on the piece. This normal force exists to balance the inward components of the tensions at the ends. These inward components have magnitude T sin(dθ/2). Therefore, N= 2T sin(dθ/2). The small-angle approximation, sin x ≈ x, then allows us to write this as N= T dθ.
The friction force on the little piece of rope satisfies Fdθ ≤ μN= μT dθ. This friction force is what gives rise to the difference in tension between the two ends of the piece. In other words, the tension, as a function of θ, satisfies
T(θ+dθ) ≤ T(θ)+μTdθ
=⇒ dT≤μTdθ
=⇒ dT/T≤μdθ
Integrating both sides
=⇒ ln T ≤ μθ + C
=⇒ T ≤ T(0)e^μθ,
where T =T(0) whenθ=0.
So when theta is 4pi (720 degrees)which correponds to two turns tension is large. As theta increases this gets huge.
In laymans terms this is happening because of friction.
Since more part of rope is in contact with stick when rotations are more,more friction is acting for same smaller movement.
(This is a classic example whicg is given in DAVID MORINS Classical Mechanics)
Last edited: Mar 19, 2012
6. Mar 19, 2012
### A.T.
I'm not sure if I picture this correctly.
PHP:
===\\\\=== STICK
|
| ROPE
|
Do you claim that for the same force pulling down at the rope the force needed to hold the stick depends on the number of windings? (Assuming the same total length and thus mass of the rope)
7. Mar 19, 2012
### Mdanner423
Same total length and mass, does the force required change depending on how many times the rope is wound around?
8. Mar 19, 2012
### emailanmol
Hey A.T. ,
Yess it does because of friction(if friction is absent, then it doesn't depend) :-)
Friction is proportional to the number of turns as more part of rope (for same length and same mass) is in contact with the stick when no. of turns is higher.
(picture it this way, if more part is in contact, more part of rope has chances of having relative motion wrt to the stick.So friction force applied will be more )
The net frictional force applied by the stick on rope is upwards (as it wants to prevent the ropes from moving down with the books) so the force applied by rope on stick is downwards.
We therefore need to apply more force to hold the stick to balance this extra force of friction.
Last edited: Mar 19, 2012
9. Mar 19, 2012
### Mdanner423
So as you wrap it there isn't a proportional upward force making it equal? I appreciate your answer I just don't understand it!
10. Mar 19, 2012
### emailanmol
Hey,
No issues.
Look at it this way.
The book is pulling the ropes down with mg(m is mass of books) and the stick is trying to resist this by friction.
(No other force acts on rope except its own weight which also acts downwards).
So the net frictional force on rope is upwards and towards the man )(downwards on stick and away from man)
If more part of rope is in contact (and suddenly a small relative motion between the ropes and stick takes place) with the stick, the stick applies friction on much more portion of string (as a large portion is having relative motion wrt stick)
and therefore net friction is higher.
Think of it this way.
If a block was lying on two tables with half the surface area on one and half the surface area on two.
If one table was smooth and the other had friction, less force of friction will act as compared to situation where both tables were rough. :-)
Right??
Last edited: Mar 19, 2012
11. Mar 19, 2012
### Mdanner423
Okay increasing the loops of rope wrapped around the stick increases the friction between the stick and the rope. But, how does this make it more difficult to raise against gravity in the Y direction?
12. Mar 19, 2012
### A.T.
Are we talking about the static case or a rope that is sliding down?
13. Mar 19, 2012
### Mdanner423
Static, the rope shouldn't be going up and down, but obviously would sway as you walk with it.
Here's another way to phrase the question. Let's say you take a fishing pole with a lure on the end, does it take more or less force to lift the lure if you wrap the fishing line around the pole?
14. Mar 19, 2012
### M Quack
No, of course not. As long as the situation is static, i.e. the rope is not sliding it makes absolutely NO difference how it is attached to the stick. By a nail, a few turns, many turns, a knot or duct tape. You name it, it makes no difference at all. The force on the stick is always the weight of the books (plus string, nail, duct tape, etc.). If you tie the knot more tightly, you will compress the stick more, but for carrying it that makes no difference unless you break the stick while doing so.
The package might be more comfortable to carry if the free length of the rope, i.e. the distance between the stick and the books, is shorter, because the books will swing back and forth less.
15. Mar 19, 2012
### emailanmol
Hey,
See we hold the stick by balancing all the forces acting on it.
Forces acting on stick are its weight mg (down) and force of friction(down).
So we need to apply an upward force to balance it.
Its v imp to understand that the net frictional force on rope by the stick acts along the direction of man and in the upward direction i.e at an oblique angle facing upwards.(This is the vector sum of all those small tiny forces acting tangentially on the ring of rope)
This is reasoned by logic.
look at the free body diagram of rope.
Its weight + force applied by books act downwards.
And the only other force acting on it is friction.
Now the rope moves with the man along x axis so the only force which moves it is friction.also its this friction which balance the weight of books and rope .
Are you understanding?
I hope I am making it ckear enough :-)
16. Mar 19, 2012
### M Quack
You are forgetting your boundary conditions, which are F=0 at the free end of the rope, and F=mg at the end going towards the books, independent of the number of turns.
Having more turns just means that you "distribute" the same friction force over more turns and more length of rope. The rope will be wound more loosly (many turns) or more tightly (few turns), which decreases or increases the friction.
17. Mar 19, 2012
### emailanmol
Everything you said is perfectly right.However, i I tend to disagree on few points on basis of practicality and not theoretically .
the rope won't practically remain static with respect to the stick always. In case, even a little motion takes place (which is bound to practically depending upon the jerks in motion ) the man will have to suddenly apply a larger force to balance the stick and the stick can even break.
And making more rotations around the rope wont decrease the length a lot. Its a thin stick .
18. Mar 19, 2012
### emailanmol
Strictly speaking, nothing can be said about the perfect number of rotations on stick.
It is advisable to have a large number of rotations around the package as not only will this create a friction to precent slipping it will also help decrease the pressure on rope which can cause stress and strain and break it.
However number of rotations on stick will depend upon the value of u, the strength of stick and man, the weight of books etc.
Because we want number of rotations such that the force of static friction is sufficient to prevent small movements but also isn't large enough that if movement takes place
the stick breaks (or the man fails to lift it)
However, in general number of rotations around stick should be as few as possible.
19. Mar 19, 2012
### emailanmol
As M Quack rightly suggested
If its static wrt stick, it doesnt make a difference how many number of loops are present.
So force would be same.
20. Mar 19, 2012
### M Quack
What you have calculated very nicely is the force needed to pull the rope off the stick.
The question was how "easy" it is to carry the stick and package. Unless this is a rather nasty trick question, the force concerned is the force of gravity on the stick under static conditions. In this (admittedly simplified and restricted) case the force to pull the string off is irrelevant.
21. Mar 19, 2012
### emailanmol
Hey,
I coudn't agree more. :-)
You are right but isn't this valid only during static condition.(I agree on whatever you have said uptil now, but only for static conditions.)
When the rope is not static, larger friction acts for more rotations which makes it tougher for man.
22. Mar 19, 2012
### M Quack
If the package was nailed to the ground, more turns make it more difficult for the guy to get away :-)
In a practical situation, I would want the situation to be static, i.e. more turns would be better because the threshold for sliding would be higher. But in a practical situation I would make a knot anyways.
23. Mar 19, 2012
### Mdanner423
So in summary... It would be easier to carry with more turns as it would make it more stable, sway less back and forth, and make it harder for the rope to slide off the stick. It takes equal force (N) to lift the books against gravity regardless of the number of turns. Right?!
24. Mar 19, 2012
### M Quack
Right.
25. Mar 19, 2012
### emailanmol
Hey, i still would like to add an additional point as I tend to diagree (a bit)with the answer.(in good spirit,offcourse !:-) )
Suppose you had n turns on your stick at this moment.(in the situation you are talking about like no swaying....etc)
If you decide to Now
make n+1 turns , it would worsen the situation and not improve it .
Because in case any relative motion takes place(and i can guarantee you, practically it will not take place for both 2 or 5 turns but if does 5 turns is a burden rather than relief)
the force applied by us is higher in case of n+1 turns.(and the stick might break rendering the situation even more useless)
In essence having a larger number of turns increases the threshold of static friction thus decreasing probability of motion, but it also adds a factor of huge instability which equation T=T(0)e^u(theta) indicates.
If you take u = 1/2( a very practical value),then two rotations increases the force factor by 530 times, and four by (theta is 8pi)300,000 times.
You can think of how much effect 5 rotations would cause (in fact you can calculate).
Practically even few rotations (like 1 or two for a normal stick and string)will take care of all the factors you are worried about(threshold static friction.....etc) for a significant portion of the journey and adding more turns is hurting your chances rather than improving it
because if a slight amount of relative motion (though unlikely but still) takes place the stick will break or you wont be able to hold it
Last edited: Mar 19, 2012 | 2,928 | 12,159 | {"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-2018-30 | latest | en | 0.915472 |
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What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
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Start with three pairs of socks. Now mix them up so that no mismatched pair is the same as another mismatched pair. Is there more than one way to do it?
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This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
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Ben and his mum are planting garlic. Can you find out how many cloves of garlic they might have had?
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Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Routes 1 and 5
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Find your way through the grid starting at 2 and following these operations. What number do you end on?
### Seven Flipped
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Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Sums and Differences 2
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Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Jumping Cricket
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El Crico the cricket has to cross a square patio to get home. He can jump the length of one tile, two tiles and three tiles. Can you find a path that would get El Crico home in three jumps?
### Snails' Trails
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Alice and Brian are snails who live on a wall and can only travel along the cracks. Alice wants to go to see Brian. How far is the shortest route along the cracks? Is there more than one way to go?
### Calendar Cubes
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Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Rabbits in the Pen
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Using the statements, can you work out how many of each type of rabbit there are in these pens?
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### Colour in the Square
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This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
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### Pouring the Punch Drink
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### 1 to 8
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### Ordered Ways of Working Lower Primary
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These activities lend themselves to systematic working in the sense that it helps to have an ordered approach.
### Ordered Ways of Working Upper Primary
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These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Different Deductions
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There are lots of different methods to find out what the shapes are worth - how many can you find? | 2,328 | 9,723 | {"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} | 3.75 | 4 | CC-MAIN-2020-40 | latest | en | 0.898472 |
https://calcresource.com/moment-of-inertia-compound.html | 1,719,011,911,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00889.warc.gz | 133,103,849 | 37,048 | ## Moment of Inertia of composite shapes
### Procedure outline
The given analytical formulas for the calculation of moments of inertia usually cover, just a handful of rather simple cases. The possible shape geometries one may encounter however, are unlimited, but most of the times, these complex areas can be decomposed to more simple subareas. In this article, it is demonstrated how to calculate the moment of inertia of complex shapes, using the Parallel Axes Theorem.
In general, the steps for the calculation of the moment of inertia of a composite area, around an axis (called global axis hereafter), are summarized to the following:
1. Identify simply shaped subareas the composite area can be decomposed to.
2. Determine the distance from global axis of the centroid of each one of the subareas.
3. Determine the moment of inertia of each subarea, around a parallel axis, passing through subarea centroid.
4. Apply the Parallel Axes Theorem to find the moment of inertia of each subarea around the global axis.
5. Add (or subtract for negative subareas, see examples) the moments of inertia from the last step.
With step 1 we aim to divide the complex area under investigation to smaller and more manageable subareas. While doing so, we must ensure that we can efficiently obtain the moment inertia of each subarea, around a parallel axis. Efficiency is important, because sometimes there are many ways to decompose an area, but not all of them are equally easy for calculations. The following picture demonstrates a case of a composite area, that is decomposed to smaller subareas, using three different methods (among many others possible). The global axis of rotation is indicated with red dashed line. With method 'a', the number of subareas is 4, with method 'b' it is 5 and and with method 'c', 6.
Method 'a' seems more simple, and employs the least number of subareas, but the calculation of the moment of inertia of the central inclined rectangle is rather hard. No formula is available for this particular angle of inclination of the rectangle that gives its moment of inertia around an axis parallel to the global one. Firstly, we have to find the moments of inertia Ix and Iy around the inclined centroidal axes of the rectangle, and then transform them around a horizontal axis, using the formulas for axis rotation. Surely, it can be done this way, but it is not very efficient calculation wise. Methods 'b' and 'c' on the other hand, do not present such a problem. For each one of their subareas, there are easily available analytical formulas, we can use to calculate their moments of inertia around a parallel to global axis. In particular, method 'b' seems a bit more efficient, since it employs one less subarea, compared to method 'c'.
Sometimes, it is more efficient to define negative subareas, that when subtracted from another, bigger subarea, produce the total composite shape. Take as an example the rectangular tube in the next figure. Conventionally, it can be decomposed to 4 rectangular subareas, one for each of the walls, that added together produce the composite shape (method 'a'). But a more efficient method is to define just two rectangular subareas, one for the outer perimeter and one for the inner perimeter (method 'b'). The latter, if subtracted from the former, produces the composite shape and for this reason is characterized as negative subarea. Following this pattern, the moment of inertia of the negative subarea is going to be subtracted from the respective one of the outer perimeter, in order to find the resulting moment of inertia of the rectangular tube. Apart from this, all other steps of the calculation procedure are normally applicable for the negative subarea.
With step 2, we determine the centroid of each subarea, specifically its distance from the global axis. The work we have to do in this step, depends on the way the subareas have been defined, in the first step. Simpler subareas, generally require less calculations to find their centroid, but also the position of subarea, relative to the global axis is important. Visiting again the rectangular tube (Figure 3), if the global axis is a centroidal one (passing through middle height of the section), then using method 'b', the centroids of the two rectangular subareas, are located on middle height too, and therefore feature zero distance from the global axis. To the contrary, with method 'a', the centroids of top and bottom subareas feature non-zero distance from global axis, which we have to find. Consequently, going with method 'b', not only we avoid any calculation, but as an additional benefit, the work we must do for step 4 (Parallel Axis Theorem) should be less.
With step 3, we calculate the moment of inertia of each subarea around the parallel axis passing through its centroid. Normally, we use available formulas (tables with moments of inertia of common shapes can be handy, check this one) but if none is available, we have to calculate the definite integral, , over the whole subarea, where variable y is the distance from the axis. Again, the definition of the subareas in step 1, heavily affects the work we have to do here.
With step 4, we apply the Theorem of Parallel Axes to all subareas, so that their moments of inertia, get all transferred around the global axis:
where, Ai is the surface area of subarea i, di is the distance of subarea i centroid from the global axis (as determined in step 2), Ii,c is the moment of inertia of subarea i around its centroid (as determined in step 3) and Ii,g is the wanted moment of inertia of subarea i around the global axis.
With step 5, we add all the resulting moments of inertia from previous step and the final sum is the wanted moment of inertia of the composite area around global axis. If we have any negative subareas defined though, we have to subtract them from the sum. Using a mathematical expression:
where index i indicates a normal (positive) subarea, while index j indicates a negative one.
Note that addition (or subtraction) of the moments of inertia, Ii,g , of the multiple subareas, is allowed in this step because all of them are defined around the same axis of rotation, the global one. That was indeed the purpose of step 4. To the contrary, we are not allowed to do the same with the moments of inertia from step 3, because each one is defined around a different parallel axis, passing through the local subarea centroid.
Never add or subtract moments of inertia defined around different axes!
### When all subareas share a convenient common axis
In some cases, it occurs so that all defined subareas share a common axis of rotation that is convenient to calculate their moments of inertia around. By convenient, it is meant that a) it is parallel to the global axis and b) there are available and easy to calculate formulas for all subareas. It is not necessary a centroidal axis neither identical to the global axis.
In such cases, it may be preferable to calculate the moments of inertia of the subareas around this convenient axis. The required steps, with this simplified procedure are outlined like this:
1. Identify simply shaped subareas the composite area can be decomposed to.
2. Calculate the moment of inertia of each subarea around the convenient axis.
3. Add (or subtract for negative subareas, see examples) the moments of inertia from the last step.
4. Apply the Theorem of Parallel Axes to find the moment of inertia of the composite shape, around the global axis
With step 3, the moment of inertia of the composite area is found around convenient axis O-O:
where index i indicates a normal (positive) subarea, while index j indicates a negative one.
Addition (or subtraction) of the moments of inertia of multiple subareas is allowed because all of them have been found in step 2 around the same axis, the convenient one.
With step 4, the moment of inertia of the composite area gets transferred, using the Theorem of Parallel Axes, from the convenient axis, to the global axis:
where dg is the distance of global axis from the composite area centroid, do is the distance of convenient axis from the composite area centroid, and A is the surface area of composite shape. Of course, if these two axes, convenient and global, are coincident then this step is redundant and we simply conclude that: . Also, if the global axis is a centroidal one, then the above expression is simplified to: .
### Example 1
Calculate the moment of inertia of the shape shown in the following figure, around a horizontal axis x-x, passing through centroid. Shape dimensions are: b=15'', h1=20'', h2=10''.
Due to symmetry, the centroid of the composite area is located in the middle height of the shape, that is also the middle of height h1. Since global axis x-x (in red color) is passing through centroid, it passes through the middle of h1too.
Step 1
We divide the composite area to smaller subareas as shown in the following schematic. Three subareas are defined, one rectangular and two triangular. These will be referred as, subareas 1,2 and 3, as depicted in the figure.
No convenient axis of rotation exists, common to all three subareas. We will continue with the remaining steps of the procedure (see Figure 1).
Step 2
For each subarea, we find the distance of its centroid from the global axis. As described in the beginning, the global axis is passing through the middle of h1
For subarea 1, that is a triangle with height h2, the centroid is located, , above the base of the triangle. Therefore, its distance from the global axis is:
For subarea 2, that is a rectangle, the centroid is located at the middle of its height , which coincides with the location of the global axis. Therefore:
For subarea 3, we don't need to make any calculation, since, due to symmetry, it must have the same distance from global axis, with subarea 1. That is:
The negative sign indicates a position below the axis. Note that the sign of distance is not important, since by definition, it gets squared, and as a result, subareas at either sides of the axis, contribute positively to the moment of inertia.
Step 3
For each subarea, we find the moment of inertia around a parallel to x axis passing through subarea centroid. We use available formulas from this table.
For subarea 1:
For subarea 2:
For subarea 3, no calculation is needed, because its shape is identical to that of subarea 1. Therefore:
Step 4
For each subarea we apply the Theorem of Parallel Axes, to find its moment of inertia around global axis x-x. We use the distances, , from step 2, and the moments of inertia, , from step 3. We need the surface area of each subarea too, so we find them first:
Now we apply the Theorem of Parallel Axes, for each subarea.
For subarea 1:
For subarea 2:
For subarea 3:
Due to symmetry around the global axis, the result for subarea 3 should not be different than that of subarea 1:
Step 5
We add for all subareas, the moments of inertia from the last step:
This is the wanted moment of inertia of the composite area, around axis x-x.
The following table organizes the calculation procedure in a tabular format:
### Example 2
Calculate the moment of inertia of the shape given in the following figure, around a horizontal axis x-x that is passing through centroid. Shape dimensions are: r=20'', a=15'', b=12''. The curved edge is a circular one. Also, for the needs of this example, the distance of the centroid from the base of the shape is given: yc=8.8083'' .
Step 1
We divide the composite area to smaller subareas as shown in the following figure. Three subareas are defined, one quarter circle, one rectangle and one right triangle. These will be referred as, subareas 1,2 and 3, as depicted in the figure.
Looking closely to the three subareas, we can recognize that there exists a common, parallel to x, axis around which all three moments of inertia can be easily found: the axis passing form the base of the shape. We name it axis O, and from here on, we will follow the simpler procedure, which is applicable only when a convenient common axis exists for all subareas (see Figure 4).
Step 2
For each subarea, we find the moment of inertia around the convenient axis O, passing through the base. We use available formulas from this table
For the quarter circular subarea 1:
For the subarea 2:
For the subarea 3:
Step 3
We add the moments of inertia around the convenient axis O, for all subareas:
Step 4
For the total composite area, we apply the Theorem of Parallel Axes, to find its moment of inertia around the global axis x-x. We need the surface area of the composite shape, so we will find it first, through the already defined subareas 1,2 and 3.
Subarea 1: .
Subarea 2: .
Subarea 3: .
Total area:
Next, the moment of inertia of the composite area around axis O (from step 3) gets transferred, using the Theorem of Parallel Axes, to global axis x, which happens to be a centroidal one too. The theorem takes the form:
where yc is the distance of the centroid from the base, in other words the distance between the two axes. In our case it is given, so we may proceed directly to the calculation:
This is the wanted moment of inertia of the composite area around axis x-x.
### Example 3
Calculate the moment of inertia of the shape given in the following figure, around a horizontal axis x-x that is passing through centroid. Shape dimensions are: a=25'', b=50'', d=30'' and t=9''. For the needs of this example, the distance of the centroid from the base of the shape is also given: yc=19.5''
Step 1
The total area is divided to smaller subareas as depicted in the next figure. Four subareas are defined, all of them rectangular. Subareas 2,3 and 4 are negative ones, and are subtracted from subarea 1, which surrounds the shape from left to right and from bottom to top.
No convenient axis of rotation is identified, that facilitates the calculation of the moments of inertia commonly for all subareas. The standard procedure will be followed in the next steps (as outlined in Figure 1).
Step 2
For each subarea, we find the distance of its centroid from the global axis. The distance of the global axis from the base is yc and it is given. We have to find the distance of each subarea centroid from base, and subtract it from yc. The difference shall be the distance between subarea centroid and the global axis.
Subarea 1, is a rectangle with height d+t, and its centroid is located at the middle of this height. Therefore, its distance from the global axis is:
Subarea 2, is a rectangle, with height d. Therefore:
The negative sign indicates that the centroid lies below the global axis.
Subarea 3, is also a rectangle, with height d, but with an offset t from base. Thus the distance of its centroid from global axis should be:
For subarea 4, we don't need to make a new calculation, since it is similar to subarea 2. Thus:
Step 3
In this step, the moment of inertia of each subarea is calculated, around a parallel to x axis, passing through its centroid.
For the rectangle of subarea 1, the width is: , and the height is: . The moment of inertia is:
Subareas 2 and 4, both have width: , and height: , the moment of inertia is found:
And finally, for subarea 3, the width is: , and the height is: . Its moment of inertia is:
It would be wrong to add or subtract the moments of inertia we just have found, because each one is related to a different axis of rotation, passing through the centroid of the respective subarea.
Step 4
In this step, the Theorem of Parallel Axes is applied, for each subarea, in order to find its moment of inertia around the global axis x-x, knowing its moment of inertia around its centroid. The distances, , from step 2, and the moments of inertia, , from step 3 will be utilized. However, the surface areas are also needed. Without writing again the dimensions of each subarea, from the previous step, the surface areas are found:
Now we apply the Theorem of Parallel Axes, for each subarea.
For subarea 1:
For subarea 2:
For subarea 3:
For subarea 4:
No calculation is needed, due to similarity with subarea 2:
Step 5
In this step the resulting moments of inertia from the previous step are added together. We are entitled to add or subtract them now, because all have been found around the same axis, the global one. The negative subareas 2,3 and 4, should be subtracted from the sum.
This final answer for the moment of inertia of the composite area around centroidal axis x-x.
The above calculation procedure can be summarized in a table, like the one shown here: | 3,699 | 16,640 | {"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-2024-26 | latest | en | 0.875385 |
https://wiki.math.ntnu.no/ma3202/2024v/questions | 1,721,480,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00263.warc.gz | 534,983,636 | 11,315 | Questions from students during the course are answered here anonymously.
Question 1: What are the conditions for the Bézout identity to be valid in an integral domain?
Answer: If $R$ is an integral domain, some equivalent conditions to the condition that the Bézout identity holds in $R$ are the following:
1. Every ideal in $R$ which is generated by two elements is principal.
2. Every finitely generated ideal in $R$ is principal.
3. The gcd of any two elements exists and every finitely generated ideal is invertible.
It should not be too difficult to see that (1) is very close to the Bézout identity. Also (2) is easily seen to be equivalent to (1) by induction. You maybe have not seen invertible ideals before. For this you consider an ideal $I$ of $R$ and the field of fractions $K$ of $R$. Then consider the subset $J$ of $K$ defined by $J=\{x \text{ in $$K$ such that $xI$ is a subset of $R$}\}$$. We say that $I$ is invertible, if the product $IJ$ is equal to $R$, where $IJ$ is defined in the same way as the product of two ideals.
Question 2: For this course, when we use that $R$ is a ring, are we always assuming that $R$ is a commutative ring?
Answer: Yes, in this course all rings are assumed to be commutative (and unital) unless stated otherwise.
Question 3: The solution of Problem 9 in Problem Set 1 uses Lemma 3.4(3) from the notes. However, I think that the whole lemma uses as a hypothesis that we are working in a field. $\mathbb{Z}$ is not a field, so, is the solution wrong or can we apply this lemma in a set of more general cases?
Answer: (The solution of this problem has now been updated) Indeed, we can not use Lemma 3.4(3) immediately since $\mathbb{Z}$ is not a field. We can use Lemma 3.6 to get over this problem. More specifically, Lemma 3.6 says that $f(x)$ is irreducible over $\mathbb{Z}$ if and only if it is primitive and irreducible over $\mathbb{Q}$. Since this specific polynomial is primitive, it is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$. Hence, we may check its irreducibility over $\mathbb{Q}$ instead and so we can use Lemma 3.4(3).
Question 4: In Problem 14 in Problem Set 1 it is stated that $\mathbb{Z}[x]/(p(x)) = \mathbb{Z}_3[x + (p(x))]$. How can we deduce this?
Answer: This follows from Theorem 4.6. Let $\alpha=x+(p(x))$. Then $K=\mathbb{Z}_3[x]/(p(x))$ is isomorphic to $\mathbb{Z}_3[\alpha]$ (see the beginning of the proof of Theorem 4.6 in the notes for an explicit proof of this fact). Then by the statement of Theorem 4.6 we have that $K$ is isomorphic to $\mathbb{Z}_3[\alpha]=\mathbb{Z}_3(\alpha)$.
Question 5: In Problem 15 in Problem Set 1 why can we easily deduce that $\mathbb{Q}(2^{1/3},2^{1/3}e^{2\pi i/3},2^{1/3}e^{4\pi i/3}) = \mathbb{Q}(2^{1/3},e^{2\pi i/3})$?
Answer: Let me introduce some notation in this specific example. Let $x=2^{1/3}$ and $y=e^{2\pi i/3}$. Then we want to show that $\mathbb{Q}(x,xy,xy^2) = \mathbb{Q}(x,y)$. Notice first that $xy$ is in $\mathbb{Q}(x,y)$. Indeed, since $x$ and $y$ are in $\mathbb{Q}(x,y)$, and $\mathbb{Q}(x,y)$ is a field, we have that their product $xy$ is in $\mathbb{Q}(x,y)$. Similarly we have that $xy^2$ is in $\mathbb{Q}(x,y)$. Therefore, since all of $x$, $xy$ and $xy^2$ are in $\mathbb{Q}(x,y)$, and $\mathbb{Q}(x,y)$ is a field, we obtain that the smallest field that contains $x$, $xy$ and $xy^2$ is a subfield of $\mathbb{Q}(x,y)$. In other words, we have that $\mathbb{Q}(x,xy,xy^2)$ is a subfield of $\mathbb{Q}(x,y)$.
Now we show the other inclusion. We have that $x$ is in $\mathbb{Q}(x,xy,xy^2)$. Since $\mathbb{Q}(x,xy,xy^2)$ is a field, it follows that $x^{-1}$ is in $\mathbb{Q}(x,xy,xy^2)$. Then, since both $x^{-1}$ and $xy$ are in $\mathbb{Q}(x,xy,xy^2)$, we conclude that $x^{-1}xy=y$ is in $\mathbb{Q}(x,xy,xy^2)$. We have thus shown that both $x$ and $y$ are in $\mathbb{Q}(x,xy,xy^2)$ and so we conclude that $\mathbb{Q}(x,y)$ is a subfield of $\mathbb{Q}(x,xy,xy^2)$. Since both inclusions hold, we obtain that $\mathbb{Q}(x,xy,xy^2)=\mathbb{Q}(x,y)$.
In general, if we want to compare fields of the form $F(a_1,\ldots,a_k)$ and $F(b_1,\ldots,b_m)$ we need to show that each $a_i$ is in $F(b_1,\ldots,b_m)$ and each $b_j$ is in $F(a_1,\ldots,a_k)$. To do this, we have to write each $a_i$ using the elements of $F$, the elements $b_1,\ldots,b_k$ and the field operations, and similarly for each $b_j$.
Question 6: In the proof of Theorem 4.5 it is stated that the set $E=F[X]/(p(x))$ is a field extension of F. However, this does not make much sense. We know $E$ has equivalence classes of polynomials of $F$. Hence, I do not see how $F$ is included in $E$. The only way this is possible is having an embedding from $F$ to $E$. The only option I can find is having the function $f$ which is $f: F \rightarrow E$ with $f(a) = a + (p(x))$. This is a ring homomorphism and it seems to be an embedding. So, we can claim $E$ is a field extension of $F$ by considering the embedding $f$.
Answer: The situation is exactly as you describe it: we can view $E=F[x]/(p(x))$ as a field extension of $F$ by viewing an element $a\in F$ as the element $a+(p(x))\in F[x]/(p(x))$.
Question 7: In Problem 5 from Problem Set 2 it is used that the given $f(x)$ is the minimal polynomial of $b$ over $F$. Although this is true, I do not understand why we need to reason with the minimal polynomial and not just with the fact that $f(x)$ is irreducible, since Theorem 4.6 only requires an irreducible factor to state that $[F(b) : F] = deg(f(x))$.
Answer: You are right, the fact that $f(x)$ is the minimal polynomial of $b$ over $F$ is not required to show that $[F(b):F]=deg(f(x))$, it is sufficient to have that $f(x)$ is irreducible over $F$ and that $f(b)=0$.
Question 8: In Problem 9 from Problem Set 2 it is said that since $f(x) = x^3 + x + 1$ has two roots in $\mathbb{Z}_2(\alpha)$, it is straightforward that all its roots are in $\mathbb{Z}_2(\alpha)$ since the degree of $f(x)$ is $3$. Is this a consequence of applying the division algorithm two times with the two known roots? We know that the quotient used in each division is a polynomial that belongs to $\mathbb{Z}_2(\alpha)$, since the last division gives as quotient a polynomial of degree 1, thus its coefficients belongs to $\mathbb{Z}_2(\alpha)$, so the last root of $f(x)$ has to remain in $\mathbb{Z}_2(\alpha)$.
Answer: You have understood the situation correctly. In general, if a polynomial $p(x)$ over a field $F$ of degree $n$ has $n-1$ roots in a field $F$, then its last root is also in $F$ since we can divide it by a polynomial of degree $n-1$ with roots exactly the $n-1$ roots of $p(x)$ that we know are in $F$.
Question 9: In Problem 3 from Problem Set 3 it is said that a general element of $F(M_1 \cup M_2)$ is of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)(\alpha_1' \beta_1' + \cdots + \alpha_m' \beta_m')^{-1}$. How do we get this?
Answer: First, $M_1$ and $M_2$ are both extensions fields of $F$. Then $F(M_1 \cup M_2)$ is the smallest field containing $M_1$ and $M_2$. First, note that the element described in the solution is an element in $F(M_1 \cup M_2)$ as it is given by using the field operations on elements of $M_1$ and $M_2$. For the other direction, pick an element $x$ of the field $F(M_1 \cup M_2)$. It is of the form $(a b^{-1})$ where both $a$ and $b$ are elements in the smallest ring generated by $M_1$ and $M_2$ (exactly as, for example, an element $x$ in $\mathbb{Q}$ is of the form $\tfrac{a}{b}$ where $a$ and $b$ are integers). So how do the elements of the smallest ring generated by $M_1$ and $M_2$ look like? Well, such an element is given by multiplying elements of $M_1$ with elements of $M_2$ and then adding the results (since $M_1$ and $M_2$ both include $F$, we do not need to think about elements of $F$). This gives us an element of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)$ with $a_i$'s in $M_1$ and $b_j$'s in $M_2$. You may then say that you can now multiply this element again by elements of $M_1$ or $M_2$, or even worse multiply such an element by an element of a similar form. But note that if you multiply this with a similar element, you can use distributivity to end up with an element of exactly the same form and so that doesn't give you something new. Hence elements of the field $F(M_1 \cup M_2)$ are of the form $(a b^{-1})$ where $a$ is of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)$ and $b$ is of the form $(\alpha_1' \beta_1' + \cdots + \alpha_m' \beta_m')$, as claimed.
Question 10: In Problem 9 from Problem Set 3 we used Fermat's little theorem. This theorem requires $p$ prime, however, the problem does not say anything about this. Is this a mistake from the formulation of the exercise?
Answer: You are right, it should be made clear that $p$ is a prime number.
Question 11: In Problem 3(a) from Problem Set 4, Example 5.5 from the lecture notes is used. However, we do not have $\sqrt{2}$, we are starting with $\sqrt{3}$. I do not understand why it is still possible to apply this result. Is this because we can use any order of the prime numbers when we apply this result, i.e., it is not necessary to start with: $2,3,5,7,\ldots$ and if we want we could start with, for example: $3,5,2,7,\ldots$?
Answer: Yes, that is correct. In Example 5.5 we say that $p_n$ is the $n$-th prime number. However, in the computations of this example, we only use the fact that $p_1,p_2,\ldots,p_n,\ldots$ is an ordering of the prime numbers, so you may choose any ordering of them.
Question 12: There seem to be a few typos in the solutions of Problem 6(b) from Problem Set 2, Problem 3(c) from Problem Set 4, Problem 8 from Problem Set 4 and Problem 5 from Problem Set 5. Is that correct?
Answer: Yes, there were a few typos there which are now corrected in the uploaded files.
Question 13: How can one obtain all subgroups of a certain (finite) group? Is it possible to know beforehand the number of subgroups that a certain group will have? Is there a faster way than checking its subset separately?
Answer: If you are given a large group, finding all of its subgroups is a hard problem. Essentially it boils down to checking all of its subsets and seeing if they form a subgroup (if the group is finite at least). You can imagine that this becomes tedious if you have even a relatively small group (say a group of order $8$). There are however a few results from group theory which may help save quite a bit of time, and make the problem of finding all subgroups of a small group manageable.
First of all, we have Lagrange's theorem which tells us that the order of a subgroup must divide the order of the group. Take for example $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ which both have order $4$, so any possible subgroup must have order $1$, $2$ or $4$. Of course, the subgroups of order $1$ and $4$ are always there and are unique (the trivial subgroup and the whole group). It remains to consider all subsets of $2$ elements and see if they form a subgroup. Note that each subgroup must contain the identity element. Hence you only need to consider all subsets which contain the identity element plus one more element from your group and see if they form a subgroup. In the case of $\mathbb{Z}_4$ you need to check $\{0,1\}$, $\{0,2\}$, $\{0,3\}$ and you see that $\{0,1\}$ and $\{0,3\}$ are not closed under addition while $\{0,2\}$ is, and hence the only subgroup of order $2$ is $\{0,2\}$. In the case of $\mathbb{Z}_2 \times \mathbb{Z}_2$ you need to check $\{(0,0),(1,0)\}$, $\{(0,0),(0,1)\}$, $\{(0,0),(1,1)\}$ and you see that all of them are closed under addition, and hence these are all the subgroups of order $2$ (notice that for a finite group we do not need to check that the inverse is in the group as long as the group is closed under the group operation).
The case of $\mathbb{Z}_4$ is especially interesting because $\mathbb{Z}_4$ is a cyclic group. For cyclic groups the following general result holds: a cyclic group with $n$ elements has exactly one subgroup for each divisor $d$ of $n$, and this is a cyclic subgroup itself. There are no other subgroups for cyclic groups.
PS: Some other useful results can be given by Cauchy's theorem (which states that whenever a prime number $p$ divides the order of a group, then that group has at least one subgroup of that order $p$) and Sylow's theorems. These are more advanced results in group theory which might be useful to find all subgroups of a group with a few more elements. In this course we will not need them for finding subgroups of a group.
Question 14: The solution of Problem 3(a) from Problem Set 5 claims that an element $\alpha$ of $G(\mathbb{Q}(\omega)/\mathbb{Q})$ is completely defined by its value on $\alpha(\omega)$. I understand the reason to say this. However, I do not understand how we calculate the possible values for $\alpha(\omega)$. Consider the following argument: since $\alpha$ must be the identity function when we restrict $alpha$ to $\mathbb{Q}$, we have that $\alpha(1) = 1$. Besides, since $\omega$ is a root of $x^p - 1$, we have that $\omega^p = 1$, so $\alpha(1) = \alpha(\omega^p) = (\alpha(\omega))^p$. Hence, $1 = \alpha(\omega)^p$ implies that $\alpha(\omega)$ is a $p$-th root of $1$, which gives $p$ different values for $\alpha(\omega)$. However, the solution deduces that $\alpha(\omega)$ is a root of $\Phi_p(x)$ and so there are $p -1$ different values for $\alpha(\omega)$. Where does my argument fail?
Answer: Your computation shows that $\alpha(\omega)$ is a root of $x^p-1$. It doesn't show that all roots of $x^p-1$ are values that $\alpha(\omega)$ does indeed take. But now notice that you know even more for $\omega$: you know that $\omega$ is a root of $\Phi_p(x)$. Hence by the same argument as the one you used, it follows that $\alpha(\omega)$ is a root of $\Phi_p(x)$. Notice that $1$ is not a root of $\Phi_p(x)$ and $x^p-1 = (x-1)\Phi_p(x)$. That is, while $\alpha(\omega)$ is indeed always a root of $x^p-1$, we have shown the even stronger property that it is a root of $Phi_p(x)$ and so it is not equal to $1$.
Of course, then you may ask how do we know that $\alpha(\omega)$ can then be any of the roots of $\Phi_p(x)$? It could be that again some of the roots of $\Phi_p(x)$ can not be attained as values of $\alpha(\omega)$, similarly as 1 can not be attained. But now we know that the Galois group $G(\mathbb{Q}(\omega)/\mathbb{Q})$ has order $p-1$ as seen in the exercise, and so $\alpha(\omega)$ must take $p-1$ different values (as any alpha in $G(\mathbb{Q}(\omega)/\mathbb{Q})$ is uniquely defined by its value on $\omega$). Since we know that all the values it can take are among the roots of $\Phi_p(x)$, and since there are exactly $p-1$ values there, it follows that $\alpha(\omega)$ can indeed be any of the roots of $\Phi_p(x)$.
Question 15: The last part of Problem 7 from Problem Set 5 asks for finding the fields $K$ such that $F\subseteq K\subseteq E$. To do that, we use the fact that $G(E/F)$ is isomorphic to $\mathbb{Z}_8$. My question is: is it true that each subgroup of $G(E/F)$ has to be mapped to a subgroup of $\mathbb{Z}_8$? More generally, is the number of subgroups of a group is invariant by isomorphism?
Answer: Yes, if you have two isomorphic groups, the number of subgroups they have is the same. Furthermore, not only the number is the same, but also the lattice of subgroups they have is the same and is preserved under any isomorphism of groups. This means that if $G$ is isomorphic to a group $G'$ via an isomorphism $f$, and if $H$ is a subgroup of $G$, then $f(H)$ is a subgroup of $G'$ and $H$ and $f(H)$ are isomorphic to each other (via $f$ again).
Question 16: In the solution of Problem 3(b) from Problem Set 6 it is stated that for a prime number $p$ we have that $\mathbb{Z}_p^{m}/\mathbb{Z}_p^{m-1}$ is isomorphic to $\mathbb{Z}_p$. Is this something that can be taken for granted or do we need to prove it?
Answer: You don't need to prove the stated group isomorphism unless asked to, so it is good to know how to do it! Here is a simple argument: $\mathbb{Z}_{p^m}$ has $p^m$ elements and $\mathbb{Z}_{p^{m-1}}$ has $p^{m-1}$ elements, so their quotient has $p^m/p^{m-1}=p$ elements (the quotient of finite groups has as many elements as the quotient of the number of elements between the two groups). The only group of prime order $p$ (up to isomorphism) is the cyclic group $\mathbb{Z}_p$, and so the quotient must be isomorphic to $\mathbb{Z}_p$.
Question 17: In the solution of Problem 11 from Problem Set 6 it is stated that "These circles intersect in two points: by Euclidean geometry,…". Which statement of Euclidean geometry is referred to here?
Answer: This is Proposition 10 in Book 1 of Euclid's Elements.
Question 18: The solution of Problem 12 from Problem Set 6 has two different parts where it is claimed that the construction done there creates a triangle with hypotenuse of length $1$. Why is this true?
Answer: For the first triangle: the points $O$, $X$ and $Y$ are different by assumption: $X$ and $Y$ lie on the segments $OQ$ and $OP$ respectively, and $OP$ and $OQ$ are not on the same line since the angle $\theta$ is greater than $0$. Hence the three points $O$, $X$ and $Y$ are not on the same line and so they form a triangle which lies on the first quadrant. Next, the line segment $OX$ has length $1$ by construction as $X$ lies on the circle with center at $O$ and radius $1$. Furthermore, the line segment $XY$ is orthogonal to the line going through $O$ and $Y$, again by construction. Hence the triangle formed by $O$, $X$ and $Y$ is an orthogonal triangle (with the right angle being the angle at $Y$) and the hypotenuse is the line segment $OX$ of length $1$.
Question 19: In the solution of Problem 3 from Problem Set 1, when solving for $x$,$y$,$z$,$w$ we proceed by dividing by $a+b\sqrt(-3)$. How is this possible when $\mathbb{Z}[\sqrt(-3)]$ is not a field?
Answer: Indeed, dividing by $a+b\sqrt(-3)$ is not possible in general as $\mathbb{Z}[\sqrt(-3)]$ is not a ring. However, in the solution of this problem we have made the assumption that $a+b\sqrt(-3)$ divides both $4$ and $2+2\sqrt(-3)$, and so we know that these specific divisions are possible, that is, they give us an element in the ring $\mathbb{Z}[\sqrt(-3)]$.
Question 20: At the end of the proof of the part (3)$\implies$(2) of Theorem 8.5, how do we conclude that $\beta = \phi^\star (\alpha)\in E$? How is Theorem 6.5 useful here?
Answer: First we want to apply Theorem 6.5. We have the embedding $\phi$ from $F(\alpha)$ to $\overline{F}$, the algebraic closure of $F$ which is an algebraically closed field. We also have an algebraic extension $F(\alpha)\subseteq E$ (the extension exists since $\alpha\in E$ and $F\subseteq E$, which gives that $F(\alpha)\subseteq E$, and the fact that it is algebraic follows since the extension $F\subseteq E$ is algebraic, as explained at the beginning of the proof). This shows that the requirements of Theorem 6.5 are satisfied (with the roles of $F$, $L$, $E$ and $\sigma$ in the theorem played by $F(\alpha)$, $\overline{F}$, $E$ and $\phi$ respectively in this proof), and so we obtain an embedding $\phi^{\ast}:E\to\overline{F}$ which restricts to identity on $F(\alpha)$. In particular, it restricts to identity on $F$. Since we have assumed that (3) holds, we obtain that this embedding restricts to an $F$-automorphism of $E$, that is it is a ring homomorphism from the field $E$ to the field $E$ which restricts to identity on $F$. In particular, $\phi^{\ast}(\alpha)$ is in $E$ (as is $\phi^{\ast}(x)$ for every $x\in E$). It remains to compute $\phi^{\ast}(\alpha)$. Notice that $\phi^{\ast}$ is an extension of $\phi$ and that $\phi$ is defined on $F(\alpha)$ via $\phi(b_0+b_1\alpha+\cdots+b_{n-1}\alpha^{n-1})=\phi(b_1+b_1\beta+\cdots+b_{n-1}\beta^{n-1})$. Then $\phi(\alpha)=\beta$ just by the definition of $\phi$ (that is, we set $b_0=b_2=\cdots=b_{n-1}=0$ and $b_1=1$). This concludes the proof.
Question 21: In the solution of Problem 5 from Problem Set 5, in the case $\text{char}(F)=p$, in the direction $\Rightarrow$, it is stated that "$(x-\alpha)^r$ has as constant term $-r\alpha$" which is false, and then this false statement is used to conclude that $-r\alpha\in F$. How can we conclude correctly that $-r\alpha\in F$? Perhaps we should look at the coefficient of $x^{r-1}$ instead? Moreover, how can we use $-r\alpha\in F$ to obtain that $\alpha\in F$?
Answer: (The solution of this problem has now been updated) You are right that the constant term of $(x-\alpha)^r$ is not $-r\alpha$, and you are right that we should look at the coefficient of $x^{r-1}$ instead. Notice that its coefficient is $-r\alpha$ by the binomial theorem. Since $(x-\alpha)^r$ is a polynomial in $F[x]$, this means that all of its coefficients are in $F$ and so $-r\alpha$ is in $F$. By multiplying by $-1$ we obtain that $r\alpha\in F$. Then we can of course add as many copies of this element in $F$ as we want and still get an element in $F$, that is $k(r\alpha)$ is in $F$ for every positive integer $k$. In particular, we have $k(r\alpha)$=$(kr)\alpha$ (remember that $r\alpha$ is just shorthand for $\underbrace{\alpha+\cdots+\alpha}_{r \text{ times}}$) and so $(kr)\alpha\in F$ for every positive integer $k$. Now pick $k$ such that $kr=1 \mod{p}$ (such a $k$ exists since $p$ is a prime). Since $F$ is a field of characteristic $p$, we have that $(kr)\alpha=\alpha\in F$ as required.
Question 22: At the end of the solution of Problem 8(b) from Problem Set 3, we have shown that $\beta$, an element of the quotient ring $F[x]/(f(x))$, is a root of both $f(x)$ and $g(x)$, where these are polynomials over $\mathrm{GF}(p)$ and $f(x)$ is irreducible. Then it is stated that, this implies that $f(x)$ divides $g(x)$ in $\mathrm{GF}(p)[x]$. Why is that?
Answer: This is a consequence of a more general and useful statement, so let us show that statement instead. Assume that we are working with polynomials over a field $F$ and that we have an extension field $K$ of $F$. If a polynomial $f(x)\in F[x]$ is irreducible and shares a root $\alpha\in K$ with another polynomial $g(x)\in F[x]$, then $f(x)$ divides $g(x)$. Here are two ways we can argue about this.
• First we can consider the minimal polynomial of $\alpha$ over $F$, say $q(x)$ (it exists, since $\alpha$ is algebraic over $F$ as it is the root of $f(x)$). Then we know that $q(x)$ divides $f(x)$ and that $q(x)$ divides $g(x)$ (by Theorem 5.2(2) in the notes). Therefore $f(x)=q(x)h(x)$ and $g(x)=q(x)z(x)$ for some polynomials $h(x)$ and $z(x)$ in $F[x]$. Since $f(x)$ is irreducible, one of $q(x)$ and $h(x)$ is a unit. Since $q(x)$ is irreducible, it is not a unit, and so $h(x)=c$ is a nonzero constant in $F$ (as these are the units of $F[x]$). But then $f(x)=c q(x)$ gives $q(x)=c^{-1} f(x)$. Replacing this in $g(x)=q(x)z(x)$ we obtain $g(x)=c^{-1}f(x)z(x)$ and so $f(x)$ divides $g(x)$.
• Another way is to consider the $\mathrm{gcd}$ of $f(x)$ and $g(x)$ (it exists since $F[x]$ is a UFD). Let $h(x)=\mathrm{gcd}(f(x),g(x))$. Since $h(x)$ divides the irreducible polynomial $f(x)$, it is either equal to $1$ or it is equal to $cf(x)$ for some nonzero constant $c\in F$. Then by Bézout identity there exist polynomials $p(x)$ and $q(x)$ such that $f(x)p(x)+g(x)q(x)=h(x)$. By plugging in $\alpha$ we have $f(\alpha)p(\alpha)+g(\alpha)q(\alpha)=h(\alpha)$ and the left-hand side is $0$ since $\alpha$ is a root of both $f(x)$ and $q(x)$. Hence $h(\alpha)=0$ and so $h(x)$ is not equal to $1$. Therefore $h(x)=cf(x)$. Since $h(x)$ divides $g(x)$ also (as it is a common divisor), we obtain that $cf(x)$ divides $g(x)$ and so $g(x)=r(x) cf(x)$ for some polynomial $r(x)$ in $F[x]$. This shows that $f(x)$ divides $g(x)$. | 7,199 | 23,858 | {"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} | 3.890625 | 4 | CC-MAIN-2024-30 | latest | en | 0.938649 |
https://www.cardschat.com/forum/general-poker-13/pot-odds-flop-re-flush-draw-102140/ | 1,716,926,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00388.warc.gz | 593,736,540 | 20,287 | # Pot odds on the flop re flush draw
A
#### Aleeki
##### Enthusiast
Silver Level
Hi guys
Im a bit confused and was hoping someone can iron something out for me.
I have been reading up quite a bit on pot odds, implied odds, expected value etc.
What i am struggling to get my head around is this....
Lets say I have A A. My opponent has two suited cards.
I raise x amount and he calls.
The flop comes down with two of his suit, but im still ahead.
Lets say the pot is \$10 at this stage.
How much do i need to bet to make give him the wrong odds to call?
As far as I can see at the moment, even if I bet the pot \$10 then he will have to bet \$10 to return \$30, thus giving him a break even percentage of 33%. Therefore, he is still right in calling this bet, correct?
It seems to me that unless you bet like twice the pot he will pretty much have decent odds, especially implied odds to call....
Am i right?
#### OzExorcist
##### Broomcorn's uncle
Bronze Level
You don't actually count his call in the size of the pot - so after you've bet \$10, the pot is \$20, and he needs to put up \$10 to win that \$20. The pot's offering him 2:1, which is a losing proposition in the long run for a flush draw.
#### ChuckTs
##### Legend
Silver Level
No, a \$10 bet into a \$10 pot would be (\$10+\$10)/\$10 = \$20/10 = 2:1. (edit: what oz said )
Speaking strictly in terms of pot odds (omitting implied odds), to give him bad odds in your example you need to bet more than x...
4.11 = (10+x)/x
4.11x = 10+x
3.11x = 10
x = 10/3.11
x = \$3.2
Considering implied odds, we usually want to give them worse than 3:1 though, so we'd bet like 2/3 (giving ~2.5:1) or 3/4 the pot (giving ~2.3:1).
A
#### Aleeki
##### Enthusiast
Silver Level
Ok then.
So switching roles, if i have the flush draw and there is a \$10 pot, the most i should call to see the turn is \$3.20?
I was thinking that my call counted towards the pot i would win and therefore had to take that into consideration...
#### ChuckTs
##### Legend
Silver Level
Nope, your call never comes into consideration with pot odds. It's:
[amount in middle]+[any bets ahead of you]/[amount to call] = x/y => x:y
A
#### Aleeki
##### Enthusiast
Silver Level
Thanks very much. That helps alot.
Just read this though, which I though was interesting in terms of when you should call/fold etc....
ESPN - Forget pot odds - Poker
#### ChuckTs
##### Legend
Silver Level
All Phil's doing in that article is demonstrating a way of converting pot odds to percentages to show what % of the time we need to be ahead to call.
He uses an example where we're getting 4:1 pot odds, and goes on to calculate with his equation that we need to win the hand at least %20 of the time. This is exactly the same as saying we're a 4:1 dog, so we need to win the hand 1 in 5 times (or %20).
Don't let it confuse you - if that method's easier for you, then use it, but he's talking about the same thing there.
A
#### Adventurebound2
##### Legend
Silver Level
Don't let it confuse you
My brain hurts now....
TY Chuck for the good answers
#### zachvac
##### Legend
Silver Level
Ok then.
So switching roles, if i have the flush draw and there is a \$10 pot, the most i should call to see the turn is \$3.20?
I was thinking that my call counted towards the pot i would win and therefore had to take that into consideration...
If it is an all-in bet, yes. If your opponent still has money left though, you likely will still be able to make some money if you hit a flush, so you can call slightly more than \$3.20. It all depends on what kind of hand you think your opponent has and whether he will bet or call a river bet.
A
#### Aleeki
##### Enthusiast
Silver Level
Just quickly re visiting this to make sure my head is in the right place.
Take this example:
I have J 10 and I have put my opponent on AA or KK through my amazing read (bare with me hear).
The flop comes 9 8 2 (suits don't matter in the e.g.)
Now according to the odds calculator i have about a 34% of winning vs his 66%.
So lets say the pot is \$60. My opponent bets \$25 into the pot and is all in.
That means I have a 34% to win \$85....correct? So because 34% of \$85 is \$28.9 this is a profitable move as I will return more than i put in in the long run and make a \$3.90 profit?
I am correct here? or do i take the \$85 pot plus my \$25 call to make it \$110 and then times that by 34% to get \$37.40 and show a \$12.40 profit in the long run?
I know this may sound jibberish but please i need to get this right in my head hahaha
#### OzExorcist
##### Broomcorn's uncle
Bronze Level
So lets say the pot is \$60. My opponent bets \$25 into the pot and is all in.
That means I have a 34% to win \$85....correct? So because 34% of \$85 is \$28.9 this is a profitable move as I will return more than i put in in the long run and make a \$3.90 profit?
The theory's correct - you count your opponent's bet, but not the amount of your call, in the pot size.
The percentage is slightly out though - assuming all your outs are clean, an open-ended straight draw actually gives you eight outs (four queens, and four sevens).
With eight outs, you're approximately 31.5% to hit on either the turn or river. 31.5% of \$85 is \$26.35, so you're still going to show a profit, just not quite as big a one as you mentioned.
It's possible that you actually are 34% on the flop, BTW - if you used an odds calculator, it's probably including the possibility of runner-runner trips, two pair or maybe a backdoor flush. But the straight draw on its own is only 31.5%.
B
#### Bentheman87
##### Visionary
Silver Level
Aleeki, it's much easier to just use ratios instead of %, makes the math much easier when you are playing. Here's a chart to check out you should try to memorize some of the ones like 8 outs (open ended str8 draw), 9 outs (flush draw), 4 outs (gutshot).
Your browser is too old - Texas Holdem Poker
Now using your last example where you have J 10. Pot is \$60 and he's all in for \$25, so you're pot odds are 3.4:1. You have 8 outs for the straight plus two overcards. Jack or a ten might be an out but they might not if he has a set or overpair. But if he has Ace 9 or King 9 then they are outs. So this is where it gets tricky because you have to make an educated guess, I'd count them as another 3 outs. So 8 + 3 is 11 outs total. If you look at that chart, odds against you hitting one of your outs on both the turn and river (since you get to see both by calling because he's all in) is 1.4:1, so you have to be getting better than 1.4:1 pot odds so easy call.
#### WVHillbilly
##### Legend
Silver Level
Just quickly re visiting this to make sure my head is in the right place.
Take this example:
I have J 10 and I have put my opponent on AA or KK through my amazing read (bare with me hear).
The flop comes 9 8 2 (suits don't matter in the e.g.)
Now according to the odds calculator i have about a 34% of winning vs his 66%.
So lets say the pot is \$60. My opponent bets \$25 into the pot and is all in.
That means I have a 34% to win \$85....correct? So because 34% of \$85 is \$28.9 this is a profitable move as I will return more than i put in in the long run and make a \$3.90 profit?
I am correct here? or do i take the \$85 pot plus my \$25 call to make it \$110 and then times that by 34% to get \$37.40 and show a \$12.40 profit in the long run?
I know this may sound jibberish but please i need to get this right in my head hahaha
What are you doing calling a \$30 bet preflop with 10 J when your opponent only has another \$25 left? You're not getting the implied odds you need to call. Fold pre-flop.
j/k looks like your getting it, but I too prefer to use odds rather the percenages.
Full Flush Poker Poker Odds - Pot & Implied Odds - Odds Calculator | 2,097 | 7,822 | {"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} | 3.921875 | 4 | CC-MAIN-2024-22 | latest | en | 0.962368 |
http://mathhelpforum.com/calculus/175930-anti-deriviative.html | 1,529,461,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863407.58/warc/CC-MAIN-20180620011502-20180620031502-00532.warc.gz | 193,479,433 | 12,906 | 1. ## anti deriviative
the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
2. Originally Posted by frankinaround
the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
antiderivative will be an arctan function
3. Originally Posted by frankinaround
the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
Product and chain rules are used for differentiation NOT anti-differentiation.
To do this particular anti-differentiation you make the substitution $\displaystyle 2t = \tan(\theta)$.
Alternatively you can substitute $\displaystyle u = 2t$ and recognise a standard form.
4. but how do you know when its integration by parts or trig substitution?
5. With practice it becomes clearer. You can either learn the trig forms or try integration by parts and when that fails badly it's a trig sub
6. for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
theyta = arctan(2t)
t=1/2 (because its an area problem from t= 0 to t = 1/2)
The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
-pie
3/2 * pie
1/2 * pie
pie
0
How do I get the right answer? How do I calculate the arctan without a calculator? ect?
7. $\displaystyle \displaystyle \int_0^{\frac{1}{2}} \frac{4}{1+4t^2} \, dt$
$\displaystyle u = 2t$ ... $\displaystyle du = 2 \, dt$
$\displaystyle \displaystyle 2\int_0^{\frac{1}{2}} \frac{2}{1+(2t)^2} \, dt$
$\displaystyle \left[2\arctan(2t)\right]_0^{\frac{1}{2}}$
$\displaystyle 2\arctan(1) - 2\arctan(0)$ ... note the following:
(1) these are unit circle values. you should already be familiar with them.
(2) the arctangent function has range $\displaystyle -\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ ... you should be familiar with this also.
$\displaystyle 2 \cdot \dfrac{\pi}{4} - 2(0) = \dfrac{\pi}{2}$
8. Originally Posted by frankinaround
for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
theyta = arctan(2t)
t=1/2 (because its an area problem from t= 0 to t = 1/2)
The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
-pie
3/2 * pie
1/2 * pie
pie
0
How do I get the right answer? How do I calculate the arctan without a calculator? ect?
If you have arctan(1) then ask yourself the question for what values of theta is tan(theta) = 1?
By the way, it's "theta" not "theyta" and "pi" not "pie."
-Dan
9. so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :
In = integral.
In(4/(1+4t^2))
4t^2=tan^2theta
2t=tan(Theta)
dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta
in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
in(2costheta + cos(theta)sin(theta) dTheta)
= 2 sin theta + 1/2 sin^2 theta.
evaluate at 1/2 and 0.
theta = arctan 2t. arctan 1 = 45* or pie/4
2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.
Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.
right now to solve integrals I have these methods :
Trig substitution
u substitution
factoring
and sometimes its the equation of a circle.
so what am i missing? arctan substitution? what?
10. note the following basic rule for taking the derivative of $\displaystyle y = \arctan{u}$ where $\displaystyle u$ is a function of $\displaystyle x$ ...
$\displaystyle \dfrac{d}{dx} \arctan{u} = \dfrac{u'}{1+u^2}$
now, what is the antiderivative of $\displaystyle \dfrac{u'}{1+u^2}$ ?
11. Originally Posted by frankinaround
so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :
In = integral.
In(4/(1+4t^2))
4t^2=tan^2theta
2t=tan(Theta)
dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta
in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
in(2costheta + cos(theta)sin(theta) dTheta)
= 2 sin theta + 1/2 sin^2 theta.
evaluate at 1/2 and 0.
theta = arctan 2t. arctan 1 = 45* or pie/4
2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.
Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.
right now to solve integrals I have these methods :
Trig substitution
u substitution
factoring
and sometimes its the equation of a circle.
so what am i missing? arctan substitution? what?
If you want to use trigonometric substitution, then as was suggested, you let $\displaystyle \displaystyle t = \frac{1}{2}\tan{\theta}$ (or $\displaystyle \displaystyle 2t = \tan{\theta}$) so that $\displaystyle \displaystyle dt = \frac{1}{2}\sec^2{\theta}\,d\theta$.
Then your integral $\displaystyle \displaystyle \int{\frac{4\,dt}{1 + 4t^2}}$ becomes
$\displaystyle \displaystyle \int{\frac{4\cdot \frac{1}{2}\sec^2{\theta}\,d\theta}{1 + 4\left(\frac{1}{2}\tan{\theta}\right)^2}}$
$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + 4\cdot \frac{1}{4}\tan^2{\theta}}}$
$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + \tan^2{\theta}}}$
$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{\sec^2{\theta }}}$
$\displaystyle \displaystyle = \int{2\,d\theta}$
$\displaystyle \displaystyle = 2\theta + C$.
Now remembering that $\displaystyle \displaystyle 2t = \tan{\theta}$, this means $\displaystyle \displaystyle \theta = \arctan{\left(2t\right)}$.
So $\displaystyle \displaystyle 2\theta + C = 2\arctan{\left(2t\right)} + C$.
12. but how do you make the substitution t for 1/2 tan theta?
from what I understand :
sqrt(1-x^2) can be a trig substitution of a triangle with a hypotenuse of 1 and a side x and another side sqrt(1-x^2)
So what is the triangle with 1/2tan theta = t? can you help me see the triangle this is all coming from?
13. I don't use a triangle AT ALL. The point is to turn the denominator into either $\displaystyle \displaystyle 1 - \sin^2{\theta}$ or $\displaystyle \displaystyle 1 - \tan^2{\theta}$ so that you can simplify everything using a trigonometric identitity. You can substitute ANY function you like that will make this happen, as long as you substitute its derivative for $\displaystyle \displaystyle dx$ as well. | 2,260 | 7,224 | {"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": 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-2018-26 | latest | en | 0.874367 |
https://www.jobilize.com/course/section/summary-transverse-pulses-by-openstax?qcr=www.quizover.com | 1,550,715,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247497858.46/warc/CC-MAIN-20190221010932-20190221032932-00526.warc.gz | 865,069,873 | 34,326 | # 0.3 Transverse pulses (Page 6/6)
Page 6 / 6
## Reflection of a pulse from fixed and free ends (not in caps - included for completeness)
Let us now consider what happens to a pulse when it reaches the end of a medium. The medium can be fixed, like a rope tied to a wall, or it can be free, like a rope tied loosely to a pole.
## Investigation : reflection of a pulse from a fixed end
Tie a rope to a wall or some other object that cannot move. Create a pulse in the rope by flicking one end up and down. Observe what happens to the pulse when it reaches the wall.
When the end of the medium is fixed, for example a rope tied to a wall, a pulse reflects from the fixed end, but the pulse is inverted (i.e. it is upside-down). This is shown in [link] .
## Investigation : reflection of a pulse from a free end
Tie a rope to a pole in such a way that the rope can move up and down the pole. Create a pulse in the rope by flicking one end up and down. Observe what happens to the pulse when it reaches the pole.
When the end of the medium is free, for example a rope tied loosely to a pole, a pulse reflects from the free end, but the pulse is not inverted . This is shown in [link] . We draw the free end as a ring around the pole. The ring will move up and down the pole, while the pulse is reflected away from the pole.
The fixed and free ends that were discussed in this section are examples of boundary conditions . You will see more of boundary conditions as you progress in the Physics syllabus.
## Pulses at a boundary ii
1. A rope is tied to a tree and a single pulse is generated. What happens to the pulse as it reaches the tree? Draw a diagram to explain what happens.
2. A rope is tied to a ring that is loosely fitted around a pole. A single pulse is sent along the rope. What will happen to the pulse as it reaches the pole? Draw a diagram to explain your answer.
The following simulation will help you understand the previous examples. Choose pulse from the options (either manual, oscillate or pulse). Then click on pulse and see what happens. Change from a fixed to a free end and see what happens. Try varying the width, amplitude, damping and tension.
## Summary
• A medium is the substance or material in which a wave will move
• A pulse is a single disturbance that moves through a medium
• The amplitude of a pules is a measurement of how far the medium is displaced from rest
• Pulse speed is the distance a pulse travels per unit time
• Constructive interference is when two pulses meet and result in a bigger pulse
• Destructive interference is when two pulses meet and and result in a smaller pulse
• We can draw graphs to show the motion of a particle in the medium or to show the motion of a pulse through the medium
• When a pulse moves from a thin rope to a thick rope, the speed and pulse length decrease. The pulse will be reflected and inverted in the thin rope. The reflected pulse has the same length and speed, but a different amplitude
• When a pulse moves from a thick rope to a thin rope, the speed and pulse length increase. The pulse will be reflected in the thick rope. The reflected pulse has the same length and speed, but a different amplitude
• A pulse reaching a free end will be reflected but not inverted. A pulse reaching a fixed end will be reflected and inverted
## Exercises - transverse pulses
1. A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate the following in your drawing:
1. The direction of motion of the pulse
2. Amplitude
3. Pulse length
4. Position of rest
2. A pulse has a speed of $2,5\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ . How far will it have travelled in $6\phantom{\rule{2pt}{0ex}}\mathrm{s}$ ?
3. A pulse covers a distance of $75\phantom{\rule{2pt}{0ex}}\mathrm{cm}$ in $2,5\phantom{\rule{2pt}{0ex}}\mathrm{s}$ . What is the speed of the pulse?
4. How long does it take a pulse to cover a distance of $200\phantom{\rule{2pt}{0ex}}\mathrm{mm}$ if its speed is $4\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ ?
5. The following position-time graph for a pulse in a slinky spring is given. Draw an accurate sketch graph of the velocity of the pulse against time.
6. The following velocity-time graph for a particle in a medium is given. Draw an accurate sketch graph of the position of the particle vs. time.
7. Describe what happens to a pulse in a slinky spring when:
1. the slinky spring is tied to a wall.
2. the slinky spring is loose, i.e. not tied to a wall.
8. The following diagrams each show two approaching pulses. Redraw the diagrams to show what type of interference takes place, and label the type of interference.
9. Two pulses, A and B, of identical shape and amplitude are simultaneously generated in two identical wires of equal mass and length. Wire A is, however, pulled tighter than wire B. Which pulse will arrive at the other end first, or will they both arrive at the same time?
we need to know the human mind and behavior and how it works. this can make us successful in our jobs because we know what to do to please the people or to catch them lying etc.
is a Greek word for soul
carolline
scientific study of the human behavior
carolline
Humans are extremely inconsistent at catching out liars. Human intuition and lie detection methods are prone to all kinds of biases. Always follow the evidence and don't let your emotions influence your rationality.
Leon
hello everyone, i was just wondering i have to pick a theoretical perspective to describe a mental disorder, i have picked Anti social personality disorder and was going to use the biomedical model to explain it. does anyone here regard aspd as a neurobiological disorder? kind regards.
is anti social personality disorder a neurobiological illness?
niamh
rationalisation, displacement and reaction formation
only a perspective of what we have pursued till now as an experience
Priyank
what is the meaning of pysche?
The meaning of psyche is Mind
Jobin
Friends can anyone please tell me, why the study of psyche is so important?
Jobin
Well in Greek, it means "soul, spirit, the mind, or the invisible animating entity which occupies the physical body".
Friends can anyone please tell me, why the study of psyche is so important?
Jobin
Friends please tell me how to counsel a student who does not want to change their lives but are comfortable in the same life situation?
Jobin
I think it's such an important study because it's one if those rare things that can't be understood in the conventional manner. The mind (not the brain, but the soul and spirit), cannot be studied physically, or through the use of sight, touch, smell, taste, or hearing.
Much focus and attention would have to be put into it to understand it, and it can't be fully understood because everyone's different and subjected to different circumstances, and we live an ever-changing world. It's important because it helps to develop spiritual, mental, and emotional intelligence
is humanities a good stream? is there scope for a subject like psychology in the future. is it a good choice?
is humanities a good subject?
amrita
Sure it is.
Jobin
you gonna opt for humanities as long as you don't want to pursue psychology in BSc honours , you'll be able to pursue BA psychology only. So if you want to pursue BSc then go for Science stream only.
Priyank
Morning everyone. How can you help a person who can't stop musterbating?
find a different activity to divert the need to masturbate
vincent
masturbation is a need u cant stop it
Abhilash
Try positive punishment then transition to negative punishment
Atone
It probably has something to do with the environment one grew up in and their genetics that caused their puberty to have this psychological effect on them.
Atone
trying to keep yourself busy with stuffs you love doing the most.like reading or listening to music or sports
Naomi
addiction and attachment issues do go hand in hand
vincent
stop watching pornography if you do; it leads you to fantasizing via masturbation. If you are a Christian, try indulging is spirituality (music and prayrer) before your bed time.
Matovu
don't stop watching anything but instead pay attention to yourself and be conscious enough of what happens inside your brain or how the thought process goes that leads you to masturbating
Priyank
Hello everyone have a nice day
Mary
Hello
Georgescu
internal for example a times when when I watched a pornography movie I end up thinking about it a lot so what I used to stop it is to create something in my mind that can ever exist in real life. I can create a story for about a week in my mind just to forget about it and it actually helps me
Mary
@mary charm story construction is helpful but then it open up on doorways for a new problem or addiction per say since it gives boost to the overexaggeration of reality which is not right, instead try to jot down what do you feel, be it digitised or normal writing. painting it also helps.
Priyank
she**
vincent
Marriage is the only solution guys.You can't do it if you get 'it'
Philip
I agree philip Nyenze. in addition to self control
Diamly
Hello y'all! I'd like to start by saying, Marriage isn't a solution for everyone. We all have differences across the board, mental, spiritual and physical. Personally marriage has fixed it for me but not for the reason you'd think. In fact the complete opposite. I found myself with a partner who's..
Samwell
... spiritual views and mental issues (depression) caused my partner to be very distant physically and it caused so much heartbreak and pain within myself and in turn made my mental health to whirlwind out of control. it took 4 years to get through it but I concentrated on my love for my partner...
Samwell
.... outside of the bedroom because I wanted them completely. through trials and error I think I finally got myself in a good place. I don't need to masturbate because even though the hormones are rushing within me and I want to tackle my partner down, my soulmate shines through brighter than that..
Samwell
... gorgeous body.
Samwell
You have to stop thinking about the filling that u can get from musturbation and get a partner if only you are 18 years and above.
Tanam
Unless it's a chronic problem like they are addicted to it and do it all the time, then let them be. Masterbation is a natural thing there's nothing wrong with it and it has many health benefits. As long as it isn't a chronic case they're be fine.
UnhealthyCopingMecha
how natural is it?...this i just a formed habit. it takes 3weeks(21)days to form an habit...if youve done it more than 21days,admit its your habit...same it can take 21days to end a formed habit.if you've wanted to stop this habit,contact me personally i'll guide you through
Philip
hi how to join conversation?
i don't know
Azeneth
How can you say that your bestfriend is true to you?
why is my mom so mean?
Hello Krista! Idk what made you think that ur mom is mean, but trust me, that's just not true. She really loves u and cares for u, but it's just that her way of caring is different than others. Ik it sounds mean, but deep inside her heart, she loves u a lot, just like all moms love their children.
Drazy
perhaps, but if you could elaborate on what exactly she does that is demeaning, because sometimes some lessons that are best learned are through a tough and disciplined manner, most likely she really loves you and wants you to learn the lesson once, so you won't have to find it out the hard way
Eldar
I have been assigned to an emotional disturbed student. She seemed fit first but you notice she interrupts or teases other students.would you work with this student an class to improve the situation?
Saleem
Hello ! I think that she's kind of person that shows her bad side before the good. In other words "if you gonna love me ,first you should accept my bad side" These kind of person have the purest love ,once you get through their bad side , it demand some patience and emotional control but , It worth.
aziz
Farhat
We can't know for sure. Whatever decisions your mom makes at the time are her best choices to fulfill her needs, in a way, so her meanness probably in relation to her own prioritization of needs, upbringing, temperament.
frootloop
Also, it's likely part of who you are, as well. By growing up with her, you're likely more vulnerable to her attacks, and they trigger you easily. These are both things that could potentially change, but not without a lot of work
frootloop
Hope this helped! I'm no therapist or psychologist, so take what I say with a grain of salt
frootloop
Rge only one onbthis planet have the ability to be with u whatever happens
Mohammed
What is the relationship between physical development and motor development
what is memory
the faculty by which the mind stores and remembers information. "I've a great memory for faces" something remembered from the past; a recollection.
Amber
what is the difference between Counsellor and Psychologist?
One talks you through your problem helping you solve them on your own (and vent out). The other writes a prescription for you and diagnoses the medical issue lying beneath (eg depression, anxiety)
Vasu
A psychologist has been trained to look at behavioural aspect of mental problems rather than look at biomechanical perspectives. He is more likely to ask his patient about his past and his present behavior, his feelings and his problems to arrive at the root cause.......Read more
Miraj
psychologists have a better understanding and awareness of behavioural reasons of mental problems than others.
Miraj
Instead of the behavioural approach adopted by psychologists, a counsellor tries to encourage the patient to direct the treatment session. He creates an environment where the patient is able to clearly see through his own problems and the reasons underlying these problems........Read more
Miraj
Thus, without relying anybody else, the patient is able to overcome his problems. In a broader sense, both psychologists and counselors both are specialists who try to solve mental problems of people through adopting different approaches.
Miraj
@Kadir shuaib, I hope you understand l
Miraj
Thanks to bits@Miraj Gillani But I've told that Counselling is applied psychology and counseling only effective when theory is applicable.
Miraj
A memory is a faculty which permits human to situate his actions and his state in the times and space.
Sinaly
@vasu a psychologist can't give you a drug as a prescription, that's the job of a psychiatrist
Argoncor
hi I have dealing with a lady who is severely depressed .. she gets suicidal attacks even at the thought of her mother in law . I am working on realeasing her stress .. any other suggestions?
Sriya
hi sriya.. try engazing her in some activity...like cooking,or any u like...make ppl appriciate her effort...plan such that one or other alawys keep appreciating her efforts...that will bring value to her own existence...but it will definetly take some time so u must have patience. tke cre champion.
ARJUN
ARJUN great idea.
Cindy
Hey Sriya! U can take her 2 sum place where she can inhale some fresh air&feel good.&when she gets suicidal attacks 'bout her mom-in-law,u can try calming her down by saying that her mom-in-law wud also appreciate her works&she wud b sad if she knew that her daughter in law is unhappy.Gud luck,girl!
Drazy
Idk about Drazy's comment, something that worked for me was to defy irrational thought patterns and watch out for them to defuse your own suicidal thoughts. Maybe find a focus to life completely independent of the MIL
frootloop
Sriya she needs to do something that makes her happy. I used to spend time with a depressed lady I ask her if anything helped her. when she had anxiety attacks she said music. Whenever I could feel her starting to have an attack I'd hand her head phones along with her favorite music.
Cindy
I asked her she said her husband she has anchored her entire happiness with her husband. she can't even move without him
Sriya
I apologize Sriya I just joined the conversation, I don't know about her husband ; therefore I am not able to give suggestions.
Cindy
@sriya you need to give her an insight about her own individuality so she could seperate herself mentally from her husband and see for herself what she is and where she is! you need to make her understand this that her husband and she herself are two both seperate individuals, may be meditation....
Priyank
...can give her a new tangent to think and pursue reality as it, I use meditation alot on my clients to give them a fresh perception of reality...
Priyank
Hi
ABUBAKAR
Excellent point Priyank
Cindy
thank you everyone I think I know what to do ..
Sriya
Maybe there's a dependency on the husband, and has a bearing on the lady's relationship with her MIL? Like, her MIL is threatening to her
frootloop
Exactly frootloop. Similar to changing a negative thought to a positive thought.
Cindy
What is the deference between Counseling and Psychology?
ABUBAKAR
I think Cindy and Arjun's answers are great. I'm no specialist, but having something private that she's not at war with and can appreciate (activity, hobby, music, etc) could definitely be helpful
frootloop
hey mates..we r part of d universe and essence of universe is continuous activity..even stagnant water fouls..depresn over d top take self value...hence engazin in some purpose oriented work(activity )along with fellows appriciation(value) is two pronged strategy. but it all takes time.u need to b
ARJUN
smart n patience..as depressions dosnt have a face..!!! love u all
ARJUN
always make her to feel important
Mary
@First know the root cause of the issues she's been traumatized with..talk to her in her own language without any Prejudice let her confide in you then you can get the solutions...as the primary reason for depression is being lonely around your own people and house..an open ear to listen to ur silly
harvey
... stories piled up urges, desires a warm hug that human touch we loose over time... saying is easy but living in it is insane..we all need someone to dump our day with..make her realize the consequences if she does hurt herself or others..as u cannot be with them 24/7 so try instilling the...
harvey
...SELF ESTEEM AND BELIEF which she lost over time to her own people..ask her to do some activities which will earn money online and are fun while doing so...keep herself busy not sit and overthink on things!
harvey
Sriya. Give her a story.where you passed the same. If you dont have your own story,then create one. As you give the fake story you tend get more into her shooes.,such that u feel as if u had once being stressed and u overcame it.Make the story bitter that she even symphathizes with u.
Philip
....this way,your patient trusts you more. They believe if they do whatever u tell them they Must overcome their situations....i once told a patient,"if you feel urine and hold on for atleast 2hrs,u will forget all that"...he believed tht so he always drink alot of water to create the urge to urinat
Philip
...he started focusing on how to create urine in his bladder,and he forgot all tht was disturbing.....i nicknamed him "fish."...and am working a theory on the same case. i will post "The Fish theory" once am done with it...though i might change its title,....
Philip
importance of social psychology in the domain of education
Buddhist perspective on motivation..
what is empirical observations
Empirical observation is a type of observation which is based on experiment not on theories. And this observation is not supported by scientific poof.
Miraj
why we do dreams? please tell me
Saleem
the thoughts which we stop thinking suddenly during day. they try to get out of our unconscious mind and they make a story which we term as dreams.
vritee
thank you
Saleem
family members very good morning beautiful lady and family and my family and one that works
holyghostChristine
pls what's sociology of education
solang
The sociology of education is the study of how public institutions and individual experiences affect education and its outcomes. It is mostly concerned with the public schooling systems of modern industrial societies, including the expansion of higher, further, adult, and continuing education.
Amber
Can someone please elaborate more of the unconscious, preconscious and conscious level?
family members very good and the other side working today right after school today
holyghostChristine
@holyghost christine chutiya ho ka be?
Priyank
hahahaha😂😂
Amber
sounds like kanpuriya
Amber
delhi se hoon behnchxxd...
Priyank
Abhilash
**these
Priyank
**? (missing question mark)
Priyank
@abhilash...
Priyank
mm
Abhilash
priyank.apko aise slang to use ni karna chahiye
tapaswini
hahahaha😄
Miraj
@priyank u understood my point,that is enough bro.
Abhilash
well if it's for you...
Priyank
@tapaswini that person with the name holyghost christine made me do it, i didn't wanted to, I controlled myself alot but then it just broke free since this person is posting gibberish all over the other posts as well, posting stuff that doesn't even has a meaning or relevance to the question asked.
Priyank
hi
pallavi
hello @pallavi priya.
Priyank
hi
pallavi
ok.that's fine.that person is irritating you .that doesn't mean you will say those slangs .if you are studying psychology.you should have the controlling capacity to avoid these silly things .you should not react like that
tapaswini
@ priyank thakur how can I join in the discussion ?
pallavi
guys can u help me wid Buddhist perspective of motivation..
Velda
@tapaswini with all due respect to all the facts that you're putting up, since I'm studying psychology and I guess you're doing the same as well, I guess we can then both accept the fact that being a human means a flood of emotions and swings og feelings most of the times, I am in no way overlooking
Priyank
.....the fact that I should've contended myself but since i also have emotions it's not always easy to contain them, being a lady of your content you must be pretty aware of how things blow up in our own mind sometimes and then how agitation takes over the contentment that we have..
Priyank
@pallavi priya you have already joined in....ask or say that you want to..
Priyank
fine ......
tapaswini
okay
pallavi
is there any other questions
pallavi
Guys since someone talked about outburst of the emotions I also had the same thing with my teacher today in the class
Jobin
Jobin
guys I'm not a student of psychology but I like to study it
pallavi
guys outburst of emotions is a natural thing.. every individual has their outbursts.. we experience different situations in our daily life out of which some lead to the outbursts... but we should always take care of the fact that someone is getting hurt by our words or not..
manisha
the discussion is going in a different tangent altogether but I like the exposure here.
Priyank
well put @manisha
Priyank
the outburst at many times is not a conscious act. So the question of taking care of the choice of words gets out of the window!
Priyank
@priyank yeah I understand ur point but we should also apologize for for our unconscious behavior when our consciousness returns...
manisha
there's no use of an apology if the person whom you retaliated to, didn't even respond, since it's just a spam account and is here to spread crammy texts.
Priyank
@manisha while it comes to the real world there's no shame in accepting your own committed mistake and apologizing for the same...
Priyank
i just said in general what u should actually do... again after all it's my point of view nd every individual has their own different views..
manisha
Hello guys...
Sandeep
Hello Sandeep
Miraj
@manisha absolutely. but above this all it's been pleasure conversing with you.
Priyank
same here @priyank
manisha
hi Mary the conscious mind includes all the things we are aware of or can easily bring into awareness
Mary
Got questions? Join the online conversation and get instant answers! | 5,573 | 24,020 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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-2019-09 | latest | en | 0.912965 |
https://www.studypug.com/algebra-help/power-of-a-product-rule | 1,723,335,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00303.warc.gz | 796,555,013 | 70,011 | # Power of a product rule
##### Intros
###### Lessons
1. What are exponent rules?
##### Examples
###### Lessons
1. Simplify the following:
1. $(-4xy)^4$
##### Practice
###### Topic Notes
We use the power of a product rule when there are more than one variables being multiplied together and raised to a power. The power of a product rule tells us that we can simplify a power of a power by multiplying the exponents and keeping the same base.
## Introduction to the Power of a Product Rule
The power of a product rule is a fundamental concept in exponent laws, essential for mastering algebra and higher mathematics. Our introduction video serves as a crucial starting point, offering a clear and concise explanation of this rule and its applications. By watching this video, students gain a solid foundation in understanding how exponents work when multiplying terms with the same base. Rather than simply memorizing formulas, the video emphasizes the importance of grasping the underlying principles of exponents. This approach enables learners to apply the power of a product rule confidently across various mathematical scenarios. Understanding this rule is key to unlocking more complex exponent laws and algebraic expressions. By focusing on the logic behind the rule, students develop a deeper, more intuitive understanding of exponents, setting the stage for advanced mathematical concepts and problem-solving skills.
## Understanding the Basics of Exponents
Exponents are a fundamental concept in mathematics that represent repeated multiplication. This powerful notation allows us to express large numbers concisely and perform complex calculations efficiently. At its core, an exponent indicates how many times a number, called the base, is multiplied by itself.
Let's start with a simple example to illustrate the concept of exponents. Consider the expression 2³. This means we multiply 2 by itself three times: 2 × 2 × 2 = 8. In this case, 2 is the base, and 3 is the exponent. We read this as "2 to the power of 3" or "2 cubed."
Here are a few more examples of positive integer exponents:
• 3² = 3 × 3 = 9
• 5 = 5 × 5 × 5 × 5 = 625
• 10³ = 10 × 10 × 10 = 1,000
As we can see, exponents provide a shorthand way to express repeated multiplication. This notation becomes especially useful when dealing with larger numbers or variables.
Speaking of variables, exponents work the same way with algebraic expressions with exponents. For instance:
• x² = x × x
• y = y × y × y × y × y
Understanding exponents as repeated multiplication naturally leads us to one of the most important rules in exponent arithmetic: the product of powers rule. This rule states that when multiplying expressions with the same base, we keep the base and add the exponents.
Mathematically, we express the product of powers rule as:
x^a × x^b = x^(a+b)
This rule makes sense when we think about exponents as repeated multiplication. Let's break it down with an example:
2³ × 2 = (2 × 2 × 2) × (2 × 2 × 2 × 2) = 2
We can see that we're simply combining all the 2s being multiplied, resulting in 2 to the power of 3 + 4 = 7.
The product rule for exponents works with any base, including variables:
• x² × x³ = x
• y × y² = y
• 5³ × 5² = 5
This rule is incredibly useful in simplifying algebraic expressions and solving complex mathematical problems. It allows us to quickly combine like terms and reduce expressions to their simplest form.
As we delve deeper into the world of exponents, we'll encounter more rules and properties that build upon this fundamental understanding. The product of powers rule is just the beginning, but it serves as a crucial foundation for mastering exponent operations.
In conclusion, exponents represent repeated multiplication, providing a concise way to express large numbers and repeated operations. The product of powers rule naturally extends from this concept, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. This understanding of exponents and their properties is essential for advancing in algebra, calculus, and many other areas of mathematics and science.
## The Power of a Product Rule Explained
The power of a product rule, also known as the product of powers law, is a fundamental concept in algebra that simplifies the process of raising a product to a power. This rule states that when multiplying two or more factors raised to the same power, we can simply multiply the bases and keep the exponent the same. Mathematically, it can be expressed as (ab)^n = a^n * b^n, where a and b are the bases and n is the exponent.
Let's explore this rule with numerical examples first. Consider (2 * 3)^4. Without the rule, we would need to multiply 2 and 3, then raise the result to the fourth power: 6^4 = 1296. Using the power of a product rule, we can simplify this process: (2 * 3)^4 = 2^4 * 3^4 = 16 * 81 = 1296. This method is often more efficient, especially with larger numbers or variables.
The rule also applies to algebraic expressions simplification. For instance, (xy)^3 = x^3 * y^3. This simplification is particularly useful when dealing with complex algebraic expressions simplification. Consider (2ab)^5. Using the rule, we can expand this to 2^5 * a^5 * b^5 = 32a^5b^5, which is much easier to work with in further calculations.
To understand why this rule works, let's expand an expression like (ab)^3:
(ab)^3 = (ab)(ab)(ab) = a*a*a * b*b*b = a^3 * b^3
This expansion demonstrates that each base is indeed multiplied by itself as many times as the exponent indicates, validating the rule.
The versatility of the power of a product rule becomes evident when we apply it to different bases and exponents. For example:
(3x^2y)^4 = 3^4 * (x^2)^4 * y^4 = 81x^8y^4
Here, we see the rule applied to a numerical coefficient (3), a variable with its own exponent (x^2), and another variable (y).
The rule also works with fractional and negative exponents in algebra. For instance:
(2ab)^(1/2) = 2^(1/2) * a^(1/2) * b^(1/2) = 2 * a * b
(xy)^(-3) = x^(-3) * y^(-3) = 1/(x^3 * y^3)
These examples showcase how the rule simplifies expressions with various types of exponents, making it a powerful tool in algebra.
It's important to note that the power of a product rule is closely related to other exponent rules interaction, such as the power of a power rule ((a^m)^n = a^(mn)) and the power of a quotient rule ((a/b)^n = a^n / b^n). Understanding how these rules interact can greatly enhance one's ability to manipulate and simplify complex exponent rules interaction.
In practical applications, the power of a product rule is invaluable in fields such as physics, engineering, and computer science, where complex calculations involving powers are common. For instance, in calculating compound interest or analyzing exponential growth, this rule can significantly streamline computations.
To further illustrate the rule's applicability, consider a problem in physics where we need to calculate the force of gravity between two objects. The formula F = G(m1m2)/r^2 involves a product in the numerator. If we needed to cube this entire expression, the power of a product rule would allow us to distribute the exponent: (G(m1m2)/r^2)^3 = G^3 * m1^3 * m2^3 / r^6, greatly simplifying the calculation process.
In conclusion, the power of a product rule
## Applications and Examples of the Power of a Product Rule
The power of a product rule is a fundamental concept in algebra that simplifies expressions involving exponents. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. Let's explore various examples and applications of this rule, ranging from simple to complex scenarios.
Simple Examples:
1. (xy)² = x²y²
This basic example shows how the rule applies to a product of two variables.
2. (3a)³ = 3³a³ = 27a³
Here, we see how the rule works with a coefficient and a variable.
3. (2xy) = 2xy = 16xy
This example demonstrates the rule applied to a product with a coefficient and two variables.
Complex Examples:
4. (4abc)³ = 4³a³b³c³ = 64a³b³c³
The rule extends easily to products with multiple variables.
5. (x²y³z) = (x²)(y³)z = xy¹²z
This example shows how the rule works with variables already raised to powers.
6. ((a²b)(cd³))² = (a²b)²(cd³)² = ab²c²d
Here, we apply the rule to a more complex expression with grouped terms.
Negative Exponents:
7. (xy)² = x²y²
The rule applies similarly to negative exponents in algebra, simplifying reciprocal expressions.
8. (2ab¹)³ = 2³a³(b¹)³ = a³b³
This example combines negative exponents in algebra with the power of a product rule.
Applications in Different Scenarios:
9. Area of a rectangle: If the length and width of a rectangle are doubled, the new area is (2l)(2w) = 4lw, which is four times the original area.
10. Volume of a cube with tripled side length: If the side length of a cube is tripled, the new volume is (3s)³ = 27s³, which is 27 times the original volume.
11. Compound interest growth factor: In finance, (1 + r) represents the growth factor for compound interest, where r is the interest rate and n is the number of compounding periods.
12. Scientific notation multiplication: (5.2 × 10)(3.1 × 10²) = (5.2 × 3.1)(10 × 10²) = 16.12 × 10²
13. Simplifying algebraic fractions with exponents: (x²y³)/(xy²) = x²y³/(xy) = x²y
14. Physics equations: In kinetic energy (KE = ½mv²), doubling the velocity results in (½m(2v)²) = ½m(4v²) = 2mv², quadrupling the energy.
15. Probability calculations: If the probability of an event occurring twice independently is (0.3)², the probability of it not occurring twice is (1 - 0.3)² = 0.7² = 0.49.
These examples demonstrate the versatility and power of the product rule
## Common Mistakes and How to Avoid Them
When applying the power of a product rule, students often encounter several common mistakes that can lead to incorrect solutions. Understanding these errors and learning how to avoid them is crucial for mastering this important mathematical concept.
One of the most frequent mistakes is misapplying the rule by distributing the exponent to each factor individually. For example, students might incorrectly write (xy)² as x²y². This error stems from confusing the power of a product rule with the product rule for exponents. To avoid this, always remember that (xy)² means (xy)(xy), not x²y².
Another common error is forgetting to include all factors when raising a product to a power. For instance, (xyz)³ should be expanded to (xyz)(xyz)(xyz), not just x³y³. Students often overlook one or more factors, leading to incomplete or incorrect answers. To prevent this, carefully count the number of factors and ensure each is included in the expansion.
Students also frequently struggle with negative exponents in product expressions. For example, (xy)² is sometimes mistakenly written as x²y². The correct application would be 1/(xy)², which can be further simplified to 1/(x²y²). To avoid this error, remember that negative exponents indicate reciprocals, and the entire product should be treated as a single unit.
Confusion often arises when dealing with fractional exponents in product expressions. For instance, (xy)½ is not equal to x½y½. The correct approach is to treat the entire product as a single term and apply the fractional exponent to it, resulting in the square root of xy. To prevent this mistake, always consider the product as a whole when applying fractional exponents.
Another pitfall is incorrectly applying the rule to sums or differences. The power of a product rule does not apply to expressions like (x+y)². Students sometimes erroneously write this as x²+y². To avoid this, recognize that the rule only applies to products, not sums or differences. For expressions involving sums or differences, use the binomial theorem or FOIL method instead.
To master the power of a product rule, it's essential to understand its underlying principle rather than blindly applying a formula. Visualize the repeated multiplication of the entire product, and practice expanding various expressions step-by-step. For example, expand (ab)³ as (ab)(ab)(ab) = a³b³, reinforcing the correct application of the rule.
Regular practice with diverse examples, including those with multiple factors, negative exponents, and fractional powers, will help solidify understanding and reduce errors. Additionally, always double-check your work by expanding the expression manually to verify the result. By focusing on comprehension and careful application, students can significantly improve their accuracy when using the power of a product rule in mathematical problem-solving.
## Related Exponent Rules and Their Connections
When exploring exponent rules connections, it's essential to understand that they are interconnected and build upon one another. While the power of a product rule is fundamental, other rules like the quotient rule exponent and the power of a power rule are equally important in mastering exponents. By grasping these rules and their relationships, students can tackle complex problems with confidence.
The quotient rule exponent is closely related to the power of a product rule. It states that when dividing expressions same base, we subtract the exponents. Mathematically, this is expressed as (x^a) / (x^b) = x^(a-b). This rule complements the power of a product rule, which involves addition of exponents when multiplying expressions with the same base.
The power of a power rule, on the other hand, deals with exponents raised to another power. It states that (x^a)^b = x^(ab). This rule is particularly useful when simplifying nested exponents and can be seen as an extension of the power of a product rule. Understanding the power of a power rule helps in breaking down complex exponential expressions into simpler forms.
These exponent rules connections are interconnected in several ways. For instance, the quotient rule exponent can be derived from the power of a product rule by considering division as multiplication by the reciprocal. Similarly, the power of a power rule can be understood as repeated application of the power of a product rule. By recognizing these connections, students can develop a more intuitive understanding of exponents and their properties.
Applying multiple exponent rules to solve complex problems is a crucial skill. For example, consider the expression ((x^3)^2 * (x^4)) / (x^5). To simplify this, we can start by applying the power of a power rule to (x^3)^2, giving us x^6. Then, we can use the power of a product rule to combine x^6 and x^4, resulting in x^10. Finally, we apply the quotient rule exponent to divide x^10 by x^5, yielding x^5 as the simplified result.
Another example that combines multiple rules is (y^-2 * y^5)^3 / y^4. Here, we first use the power of a product rule inside the parentheses to get y^3. Then, we apply the power of a power rule to (y^3)^3, resulting in y^9. Lastly, we use the quotient rule exponent to divide y^9 by y^4, giving us y^5 as the final answer.
Understanding these interconnections helps in developing problem-solving strategies. When faced with a complex exponential expression, students can break it down into smaller parts and apply the appropriate rules step by step. This approach not only simplifies the problem-solving process but also reinforces the relationships between different exponent rules.
Moreover, recognizing the patterns in exponent rules can lead to a deeper understanding of mathematical concepts. For instance, the quotient rule exponent can be extended to negative exponents explanation, explaining why x^-n is equivalent to 1/(x^n). Similarly, the power of a power rule helps in understanding the concept of roots, as (x^(1/n))^n = x.
In conclusion, the quotient rule exponent, power of a power rule, and other exponent rules connections are closely interconnected. By understanding these relationships and practicing with diverse problems, students can enhance their mathematical skills and tackle complex exponential expressions with ease. The key is to recognize the patterns, apply the rules systematically, and always be mindful of how different rules interact with each other in various mathematical contexts.
## Practical Applications in Algebra and Beyond
The power of a product rule in algebra is a fundamental concept with wide-ranging practical applications across various mathematical fields and real-world scenarios. This rule states that when raising a product to a power, we can raise each factor to that power and then multiply the results. In algebra, this rule is crucial for solving equations and simplifying complex expressions efficiently.
In solving polynomial equations, the power of a product rule allows mathematicians to break down complicated terms into more manageable components. For instance, when dealing with equations involving variables raised to powers, this rule enables us to distribute the exponent across multiple factors, making it easier to isolate variables and find solutions. This technique is particularly useful in polynomial equations and exponential functions, which are common in scientific and engineering calculations.
The rule's application extends beyond basic algebra into more advanced mathematical concepts. In calculus, it plays a vital role in differentiating and integrating complex functions. When working with derivatives, the power rule combines with the product rule to simplify the process of finding rates of change for intricate expressions. In integral calculus, this rule aids in breaking down complex integrands, making integration more approachable.
Real-world applications of the power of a product rule are abundant. In physics, it's used to calculate work done by varying forces or to determine the kinetic energy calculations of objects in motion. Engineers apply this rule when designing structures, calculating stress and strain on materials, or optimizing energy systems. In finance, the rule is crucial for compound interest modeling, helping investors and economists model growth over time.
In computer science and cryptography, the power of a product rule is fundamental to many algorithms, particularly in public-key encryption systems. These systems rely on the difficulty of factoring large numbers, a process that involves extensive use of exponents and products. The rule also finds applications in probability theory, where it's used to calculate the likelihood of multiple independent events occurring simultaneously.
Environmental scientists use this rule when modeling population growth or decay, considering factors like birth rates, death rates, and environmental carrying capacities. In chemistry, it's applied in reaction kinetics in chemistry to understand how the concentration of reactants affects reaction rates. The versatility of this algebraic principle demonstrates its importance across diverse fields, making it a cornerstone of mathematical problem-solving in both theoretical and applied contexts.
## Conclusion
In summary, the product rule is a powerful tool in exponent laws, allowing us to simplify expressions by adding exponents when multiplying terms with the same base. Understanding the principle behind this rule is crucial, as it enables you to apply it confidently across various mathematical scenarios. Rather than memorizing formulas, focus on grasping the underlying concept. We encourage you to practice applying the product rule regularly, as this will reinforce your understanding and improve your problem-solving skills. For those seeking to deepen their knowledge, explore further resources on exponent laws and related topics. Remember, the introduction video serves as an excellent foundation for mastering exponent laws, including the product rule. By building on this knowledge, you'll develop a strong mathematical toolkit that will serve you well in future studies and real-world applications. Keep practicing, stay curious, and don't hesitate to revisit the video for a refresher on these fundamental concepts.
### Example:
Simplify the following: $(-4xy)^4$
#### Step 1: Understand the Power of a Product Rule
The Power of a Product Rule states that when you have a product raised to a power, you can distribute the exponent to each factor in the product. In mathematical terms, this means: $(ab)^n = a^n \cdot b^n$ In this example, we have $(-4xy)^4$. According to the Power of a Product Rule, we need to distribute the exponent 4 to each factor inside the parentheses.
#### Step 2: Separate the Terms
First, let's separate the terms inside the parentheses. We have three factors: -4, x, and y. So, we can rewrite the expression as: $(-4xy)^4 = (-4)^4 \cdot (x)^4 \cdot (y)^4$ This step helps us to handle each factor individually.
#### Step 3: Simplify the Negative Base
Next, we need to simplify $(-4)^4$. It's important to note that raising a negative number to an even power results in a positive number. This is because multiplying an even number of negative factors results in a positive product. Therefore: $(-4)^4 = 4^4$ Now, we only need to calculate $4^4$.
#### Step 4: Calculate the Power of 4
Now, let's calculate $4^4$. This means multiplying 4 by itself four times: $4^4 = 4 \cdot 4 \cdot 4 \cdot 4 = 256$ So, $(-4)^4$ simplifies to 256.
### FAQs
Here are some frequently asked questions about the power of a product rule:
#### What is the power of a product rule?
The power of a product rule states that when raising a product to a power, you can raise each factor to that power and then multiply the results. Mathematically, it's expressed as (ab)^n = a^n * b^n, where a and b are the factors and n is the exponent.
#### What is an example of the power of a product rule?
A simple example is (2x)^3 = 2^3 * x^3 = 8x^3. Here, we raise both 2 and x to the power of 3 separately and then multiply the results.
#### How do you apply the power of a product rule to simplify expressions?
To simplify expressions using this rule, identify the product within parentheses and the power it's raised to. Then, apply the power to each factor individually. For example, (3ab)^4 simplifies to 3^4 * a^4 * b^4 = 81a^4b^4.
#### Does the power of a product rule work with negative exponents?
Yes, the rule works with negative exponents. For instance, (xy)^-2 = x^-2 * y^-2 = 1/(x^2 * y^2). Remember that a negative exponent means the reciprocal of the positive exponent.
#### How is the power of a product rule different from the product rule for exponents?
The power of a product rule deals with raising a product to a power, while the product rule for exponents involves multiplying terms with the same base and adding their exponents. For example, the power of a product rule is (ab)^n = a^n * b^n, while the product rule for exponents is x^a * x^b = x^(a+b).
### Prerequisite Topics for Understanding the Power of a Product Rule
Mastering the power of a product rule in mathematics requires a solid foundation in several key areas. One of the most crucial prerequisites is combining the exponent rules. Understanding how exponents work and how to manipulate them is essential when dealing with products raised to powers.
Equally important is the ability to simplify rational expressions and understand their restrictions. This skill helps in breaking down complex expressions that often arise when applying the power of a product rule. Additionally, familiarity with the negative exponent rule is crucial, as it allows for the proper handling of expressions with negative powers.
When working with the power of a product rule, you'll often encounter situations that require solving polynomial equations. This prerequisite topic provides the tools needed to manipulate and solve equations that result from applying the rule. Moreover, understanding scientific notation is beneficial, especially when dealing with very large or small numbers in product expressions.
The power of a product rule has practical applications in various fields. For instance, in finance, it's used in compound interest calculations. Understanding this connection can provide real-world context and motivation for mastering the rule. Similarly, in chemistry, the rule is applied in reaction kinetics, demonstrating its importance beyond pure mathematics.
By building a strong foundation in these prerequisite topics, students can approach the power of a product rule with confidence. Each of these areas contributes to a deeper understanding of how products and exponents interact, making it easier to grasp and apply the rule in various contexts. Whether you're solving complex algebraic problems or applying the concept in scientific or financial scenarios, a solid grasp of these prerequisites will significantly enhance your ability to work with the power of a product rule effectively.
Remember, mathematics is a cumulative subject. Each new concept builds upon previous knowledge. By taking the time to review and master these prerequisite topics, you're not just preparing for understanding the power of a product rule, but also laying the groundwork for more advanced mathematical concepts that you'll encounter in your future studies.
$({a^mb^n} {)^p} = {a^{mp}b^{np}}$ | 5,549 | 25,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "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} | 4.90625 | 5 | CC-MAIN-2024-33 | latest | en | 0.903511 |
https://www.topprnation.com/rd-sharma-solutions-class-7-maths/ | 1,585,461,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00367.warc.gz | 1,223,955,636 | 37,503 | RD Sharma Solutions Class 7 Maths Text book for all chapters are available to download in PDF. Get class 7 RD Sharma Solutions and NCERT Solutions. It is based on the latest syllabus prescribed as per the CCE guidelines by CBSE. Students of class 7 can prepare well for the exam. RD Sharma Solutions can help them to score good marks in the exam.
## RD Sharma Solutions Class 7 Maths
Download or view RD Sharma solution for class 7 for all chapters. There are 25 chapters in the textbook. Students can download any exercise of the chapter below.
Chapter 1: Integers
Chapter 2: Fractions
Chapter 3: Decimals
Chapter 4: Rational Numbers
Chapter 5: Operations on Rational Numbers
Chapter 6: Exponents
Chapter 7: Algebraic Expressions
Chapter 8: Linear Equations in One Variable
Chapter 9: Ratio And Proportion
Chapter 10: Unitary Method
Chapter 11: Percentage
Chapter 12: Prot and Loss
Chapter 13: Simple Interest
Chapter 14: Lines and Angles
Chapter 15: Properties of Triangles
Chapter 16: Congruence
Chapter 17: Constructions
Chapter 18: Symmetry
Chapter 19: Visualising Solid Shapes
Chapter 20: Mensuration I
Chapter 21: Mensuration II
Chapter 22: Data Handling I (Collection and organisation of Data)
Chapter 23: Data Handling II (Central Values)
Chapter 24: Data Handling III (Construction of Bar Graphs)
Chapter 25: Data Handling IV (Probability
### RD Sharma Solutions Class 7 Maths Chapter 1 Integers
There are four exercises in the chapter. Download any exercise in PDF. The first chapter 1 of textbook deals with Integers and their properties, multiplication of integers and their properties, division of integers, properties of division, operator precedence, use of brackets, steps involved in removal of brackets. All these topics are very important to learn.
• RD Sharma Solutions Class 7 Maths Chapter 1 Integers Exercise 1.1
• RD Sharma Solutions Class 7 Maths Chapter 1 Integers Exercise 1.2
• RD Sharma Solutions Class 7 Maths Chapter 1 Integers Exercise 1.3
• RD Sharma Solutions Class 7 Maths Chapter 1 Integers Exercise 1.4
#### RD Sharma Solutions Class 7 Maths Chapter 2 Fractions
In the second chapter students will study about Fractions There are some important topics like, the definition of fractions, types of fractions, fractions in lowest terms, comparing fractions, conversion of unlike fractions to like fractions, addition and subtraction of fractions, multiplication of fractions. There are three exercises in this chapter fractions.
• RD Sharma Solutions Class 7 Maths Chapter 2 Fractions Exercise 2.1
• RD Sharma Solutions Class 7 Maths Chapter 2 Fractions Exercise 2.2
• RD Sharma Solutions Class 7 Maths Chapter 2 Fractions Exercise 2.3
#### RD Sharma Solutions Class 7 Maths Chapter 3 – Decimals
Download the third chapter of RD Sharma textbook. In this chapter students will learn some topics likes decimals and their properties, decimal places, like and unlike decimals, comparing decimals, addition and subtraction of decimals, multiplication of decimals, division of decimals and dividing a decimal by a whole number. Find the three solved exercises of the Decimals. Solutions links are given below. All above topics are covered in the solution.
• RD Sharma Solutions Class 7 Maths Chapter 3 Decimals Exercise 3.1
• RD Sharma Solutions Class 7 Maths Chapter 3 Decimals Exercise 3.2
• RD Sharma Solutions Class 7 Maths Chapter 3 Decimals Exercise 3.3
#### RD Sharma Solutions Class 7 Maths Chapter 4 – Rational Numbers
Download chapter 4 solutions in PDF for all exercises. Chapter 4 RD Sharma Solutions Maths is consists of six exercises. Students of class 7 will get the knowledge rational numbers. They will also learn about the definition and meaning of rational numbers, positive rational numbers, negative rational numbers, properties of rational numbers, equivalent rational numbers, the lowest form of rational numbers, standard form of a rational numbers, equality of rational numbers and representation of rational numbers on the number line. Get the solution below.
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.1
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.2
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.3
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.4
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.5
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers Exercise 4.6
• #### RD Sharma Solutions Class 7 Maths Chapter 5 – Operations on Rational Numbers
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers Exercise 5.1
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers Exercise 5.2
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers Exercise 5.3
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers Exercise 5.4
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers Exercise 5.5
#### RD Sharma Solutions Class 7 Maths Chapter 6 – Exponents
Chapter 6 of RD Sharma textbook, provides definition of exponents, laws of exponents, use of exponents in expressing large numbers in standard form, decimal number system. These concepts must be learnt by the students.
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents Exercise 6.1
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents Exercise 6.2
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents Exercise 6.3
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents Exercise 6.4
#### RD Sharma Solutions Class 7 Maths Chapter 7 – Algebraic Expressions
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions Exercise 7.1
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions Exercise 7.2
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions Exercise 7.3
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions Exercise 7.4
#### RD Sharma Solutions Class 7 Maths Chapter 8 – Linear Equations in One Variable
• RD Sharma Solutions Class 7 Maths Chapter 8 – Linear Equations in One Variable Exercise 8.1
• RD Sharma Solutions Class 7 Maths Chapter 8 – Linear Equations in One Variable Exercise 8.2
• RD Sharma Solutions Class 7 Maths Chapter 8 – Linear Equations in One Variable Exercise 8.3
• RD Sharma Solutions Class 7 Maths Chapter 8 – Linear Equations in One Variable Exercise 8.4
#### RD Sharma Solutions Class 7 Maths Chapter 9 – Ratio and Proportion
• RD Sharma Solutions Class 7 Maths Chapter 9 – Ratio and Proportion Exercise 9.1
• RD Sharma Solutions Class 7 Maths Chapter 9 – Ratio and Proportion Exercise 9.2
• RD Sharma Solutions Class 7 Maths Chapter 9 – Ratio and Proportion Exercise 9.3
#### RD Sharma Solutions Class 7 Maths Chapter 10 – Unitary Method
In this chapter , students will learn about unitary method. Get the solutions below.
• RD Sharma Solutions Class 7 Maths Chapter 10 unitary method Exercise 10.1
#### RD Sharma Solutions Class 7 Maths Chapter 11 – Percentage
Get the Chapter 11 RD Sharma Class 7 Maths in PDF topics related to percentage. Just click the solutions links below.
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.1
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.2
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.3
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.4
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.5
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage Exercise 11.6
#### RD Sharma Solutions Class 7 Maths Chapter 12 Profit and Loss
This is the twelfth chapter of In this chapter, students will learn the topics of profit and loss.
• RD Sharma Solutions Class 7 Maths Chapter 11 Profit and Loss Exercise 12.1
#### RD Sharma Solutions Class 7 Maths Chapter 13 Simple Interest
In this chapter, students are introduced to the Simple interest. The concepts covered in this chapter is to find the simple interest when principal, rate and time given with using formula.
• RD Sharma Solutions Class 7 Maths Chapter 13 Simple Interest Exercise 13.1
#### RD Sharma Solutions Class 7 Maths Chapter 14 – Lines and Angles
RD Sharma Solutions Class 7 Maths Chapter 12
• RD Sharma Solutions Class 7 Maths Chapter 14 – Lines and Angles Exercise 14.1
• RD Sharma Solutions Class 7 Maths Chapter 14 – Lines and Angles Exercise 14.2
#### RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles
In this chapter, we shall discuss about triangle and its properties. From the below given links, students can access the exercises with solutions explaining the concepts from this chapter.
• RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.1
• RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.2
• RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.3
• RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.4
• RD Sharma Solutions Class 7 Maths Chapter 15 – Properties of Triangles Exercise 15.5
#### RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence
Students can learn the concepts related to congruence in this chapter.
• RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence Exercise 16.1
• RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence Exercise 16.2
• RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence Exercise 16.3
• RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence Exercise 16.4
• RD Sharma Solutions Class 7 Maths Chapter 16 – Congruence Exercise 16.5
#### RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions
• RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions Exercise 17.1
• RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions Exercise 17.2
• RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions Exercise 17.3
• RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions Exercise 17.4
• RD Sharma Solutions Class 7 Maths Chapter 17 – Constructions Exercise 17.5
#### RD Sharma Solutions Class 7 Maths Chapter 18 – Symmetry
In chapter 18, Symmetry of RD Sharma textbook deals with the lines of symmetry, Here, students can see the exercises explaining these concepts properly with solutions.
• RD Sharma Solutions Class 7 Maths Chapter 18 – Symmetry Exercise 18.1
• RD Sharma Solutions Class 7 Maths Chapter 18 – Symmetry Exercise 18.2
• RD Sharma Solutions Class 7 Maths Chapter 18 – Symmetry Exercise 18.3
#### RD Sharma Solutions Class 7 Maths Chapter 19 – Visualising Solid Shapes
In this chapter, students will study about faces, vertices and edges of a three-dimensional figure such as cuboid, cube, prism, pyramid, cone and so on.
• RD Sharma Solutions Class 7 Maths Chapter 19 – Visualising Solid Shapes Exercise 19.1
• RD Sharma Solutions Class 7 Maths Chapter 19 – Visualising Solid Shapes Exercise 19.2
#### RD Sharma Solutions Class 7 Maths Chapter 20 – Mensuration – I (Perimeter and Area of Rectilinear Figures)
• RD Sharma Solutions Class 7 Maths Chapter 20 – Mensuration – I Exercise 20.1
• RD Sharma Solutions Class 7 Maths Chapter 20 – Mensuration – I Exercise 20.2
• RD Sharma Solutions Class 7 Maths Chapter 20 – Mensuration – I Exercise 20.3
• RD Sharma Solutions Class 7 Maths Chapter 20 – Mensuration – I Exercise 20.4
#### RD Sharma Solutions Class 7 Maths Chapter 21 – Mensuration – II (Area of Circle)
• RD Sharma Solutions Class 7 Maths Chapter 21 – Mensuration – II Exercise 21.1
• RD Sharma Solutions Class 7 Maths Chapter 21 – Mensuration – II Exercise 21.2
#### RD Sharma Solutions Class 7 Maths Chapter 22 – Data Handling – I (Collection and Organisation of Data)
Download chapter 22 RD Sharma Class 7 Maths in PDF below. In this chapter we shall learn about the frequency distribution, grouping or organisation of data using the distribution table and tally marks. This is the first part of Data Handling (Collection and Organisation). Find the solved exercises.
• RD Sharma Solutions Class 7 Maths Chapter 22 – Data Handling – I Exercise 22.1
#### RD Sharma Solutions Class 7 Maths Chapter 23 – Data Handling – II (Central Values)
Get the Chapter 23 RD Sharma Solutions of Class 7 in PDF. This lesson explains about central value, measures of central value, arithmetic mean, range, arithmetic mean of grouped data, median and mode. In this chapter students will learn how to construct the table for these given data.Here is a complete list of solved exercises.
• RD Sharma Solutions Class 7 Maths Chapter 23 – Data Handling – II Exercise 23.1
• RD Sharma Solutions Class 7 Maths Chapter 23 – Data Handling – II Exercise 23.2
• RD Sharma Solutions Class 7 Maths Chapter 23 – Data Handling – II Exercise 23.3
• RD Sharma Solutions Class 7 Maths Chapter 23 – Data Handling – II Exercise 23.4
#### RD Sharma Class 7 Maths Chapter 24 – Data Handling – III (Construction of Bar Graphs)
In the chapter students can learn about Constriction Bar graph, how to draw a bar graph, double bar graph, reading the given bar graph and reading the double bar graph. This chapter is easy to complete. Get the solutions below.
• RD Sharma Solutions Class 7 Maths Chapter 24 – Data Handling – III Exercise 24.1
#### RD Sharma Class 7 Maths Chapter 25 – Data Handling – IV (Probability)
This chapter deals with the concepts related to probability. Students will also about the some problems which are related to the Probability. Get the chapter 25 solutions in PDF. There is only one exercise.
RD Sharma Solutions Class 7 Maths Chapter 24 – Data Handling – IV Exercise 25.1
Students can download or view NCERT Solutions, RD Sharma, RS Aggarwal Solutions for class 7. | 3,220 | 13,770 | {"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} | 3.96875 | 4 | CC-MAIN-2020-16 | longest | en | 0.842539 |
http://www.answers.com/Q/What_is_the_difference_between_ft_pounds_and_in_pounds_in_torque_wrenches_How_many_inch_pounds_are_in_ft_pounds_or_vise_versa | 1,550,864,637,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247522457.72/warc/CC-MAIN-20190222180107-20190222202107-00533.warc.gz | 307,221,284 | 51,911 | # What is the difference between ft pounds and in pounds in torque wrenches How many inch pounds are in ft pounds or vise versa?
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Ft-Lbs is a measurement of torque. Torque is "twisting" force.
Ft-Lbs is measured as follows: Imagine a 1 ft bar attached at some center point. The bar is horizontal and a 100 lb weight is attached to the bar at one foot from the center point. That would be the very definition of 100 ft-lb of torque. Of course for this to be accurate you'd have to account for the weight of the bar. In this instance 100 ft pounds of torque would mean that a 100 pound weight was used instead. You could also use a 100 ft bar and ONE pound weight, the theoretical results would be identical.
Inch pounds are very similar, only the bar would be one inch long. To calculate 100 inch pounds you'd have a bar that was 1 inch long, at the horizontal and hang a 100 pound weight. For all practical purposes 100 inch pounds would be 1/12 of the torque represented by 100 Ft pounds.
In other words, there are 12 inch lbs, in one ft. lb.
An easy way to think of it is based on the unit of measure. Whether you are working with Ft-Lbs or In-Lbs or even in Metric using N-M (Newton-Meters) doesn't matter.
If you were to express Ft-Lbs as "Pounds per Foot" or In-Lbs as "Pounds per Inch", or in metric "Newtons of force per Meter" it may be easier to understand.
For conversion purposes, there are 12 inches in a foot, so "one foot pound" would equal "Twelve Inch Pounds". In other words, it takes "12 pounds of force on a one inch long bar" to equal the twisting force of "One pound of force on a one foot (12 inch) long bar"
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# How do you convert Ft pounds to In pounds?
Answer . To convert foot pounds to inch pounds, multiply by 12. 10 ft/lbs is 120 in/lbs. . Answer . To convert foot pounds to inch pounds, multiply by 12. 10 ft/lbs is 120 in/lbs.
# How many pounds in 304 cu ft of Nitrogen?
The density of N2 is .001251 grams per cubic centimeter. This isequal to 10.8 kilograms, which is equal to 23.81 pounds.
# How many fps are in twelve ft pounds?
You are mixing apples and Tuesday. FPS refers to feet per second or how fast a projectile is travelling.. ft lbs refers to a measurement of recoil.
# How do you convert In pounds to ft pounds?
To convert inch pounds to foot pounds simply divide the inches by 12. or just google it like the rest of us.
# How much horsepower is in 1150 ft pounds of torque?
ft pounds of torque times RPM divided by 5252 equals horsepower ...... In your question one peice of info is missing in order for someone to give you an answer .....(revolution per minute )
# If 300 pounds equals 27 sq ft how many pounds equal 3000 sq ft?
300 lbs/27 sq ft x 3000 sq ft = 900,000 sq ft lbs/27 sq ft = 33,333 lbs
# How many inch pounds of torque are equal to 9 newton pounds?
You could have 'newton-centimeters', or 'newton-inches', or 'poundmeters' etc., but you can't have 'newton pounds'. Torque is (a distance) x (a force), but 'newton pound' is (force) x(force). Whether or not that has any physical significance at all, it'ssurely not torque.
# How many pounds does a 4 ft by 4 ft bale of hay weigh?
A 4 x 4 round bale of hay will weigh around half a ton, which is around 1000 lbs.
# How many Pound inches equals how many inch pounds?
Well my guess is the best solution for you to see how many pounds equals a certain amount of inches is probably just buying a cheap scale from walmart and checking it every day. Now what i just did was I asked this exact question and it said do you want to answer it. So sorry if this wasnt any help. (MORE)
# How many cubic feet is in 1 pound vise versa?
one pound of what? cubic feet is unit for volume and pound is unit for weight. So, cubic feet can only be calculated if you know the material and its density.
# You are 5 ft 2 inches and a half tall and you weigh 114 pounds how many calories should you consume to get down to 110 pounds?
You are 5' 2" and weigh 114 lbs. What are you thinking? You are already at a perfect weight. In fact you are slightly below what you could weigh and still be healthy. Are you worried about your weight or your health? Seems you are only concentrating on weight with no thought to health. A skinny woma (MORE)
# You are 5 ft 5 inches tall and you weigh 210 pounds how many calories should you consume to get down to 175 pounds?
no. a lot of people want to be that skinny. It also depends on how old you are. If you are worried about you weigh work out and eat healthy. #
# How many cu ft of propane per pound?
there is no definative answer because this depends on the pressure of the cylinder holding the propane. it is possible for a 1 cubic foot container to hold variouse weights of propane
# You are 13 yrs 5'4 ft 107 pounds and your waist is 28 inches is that overweight?
Thank you. Do u know if this is average? If wouldn't hurt to loose some weight though, rite? 5'4 ft 107 lbs enter your height and weight and then a specific answer can be given. No your not overweight.
# How do you convert pounds to ft?
Lbs (pounds) are a unit of weight. Ft (feet) are a unit of length. There is no way to universally convert between the two and make any sense at all.
# What is 300 ft equal to in pounds?
300 feet ( ft ) is a measurement of length or distance not a measurement of weight
# You are 5 ft 2 inches and a half tall and you weigh 112 pounds how many calories should you consume to get down to 100 pounds?
in my point of view i think you dont need to loose any weight . in my point of view i think you dont need to loose any weight
# What does pound-ft torque mean?
The pound-foot of torque is a measure of an engine's pushing force to propel a car forward. Therefore, if a car's engine can push a car forward weighing 6000 lbs up a 45 degree slope from a standstill, it must have at least a 300 lb-ft torque to applying the required pressure/force to propel that ca (MORE)
# How do you convert Ft pounds to nm?
1 ft. lb. * 1.3558179 = 1 Nm or, for rough guesstimates, add one third of the ft. lbs.
# 5 ft 4 inches woman body frame 185 pounds?
Ideal weight should be 124-138 pins according to Metropolitan life. Average frame. Large frame is 150 pounds.
# How to Convert 106 in pounds to ft pounds?
12 inch lbs equal 1 ft lb. So, divide 106 by 12 and you will haveyour answer.
# Is 52 pounds overweight for a 4 ft 2 inch 8 year old?
Yes it is , it's more than overweight , obese. ADDED: The above is utter nonsense! BMI ( body mass index ) = pounds * 703/height in inches squared = 52 lb * 703/(50 in)^2 = 14.6 You could gain some weight. The range is about 19 to 25.
# How many inch pounds in a pound?
You cannot compare these two units of measurement because they deal with two different things. Inches is a unit of measurement of length whereas, a pound is a unit of measurement of weight.
# How many ft pounds is 132 inch pounds?
(132 inches) x (1 ft / 12 inches) = 11 ft 132 in-lb = 11 ft-lb
# How many cu. ft. are in a pound?
Cubic feet is a measure of volume (i.e. a cube measuring one foot x one foot x one foot). Pounds is a measure of weight (mass). Therefore the relationship between cubic feet and pounds is called density (lbs per cubic foot) and will vary for each substance: a gas (at atmospheric pressure) will have (MORE)
# You weigh 123 pounds at 13 but im 5 ft 5 inches are you fat?
nope, i checked your bmi (body mass index) and it said 20.5 that's perfect. the normal bmi is 18.5-24.9 :)) it might be over weight for your age but not your height. hope i helped.
# How many pounds are in 1 linear ft?
This question cannot be answered sensibly. A linear foot is a measure of distance, with dimensions [L]. A pound is a measure of mass, with dimensions [M]. Basic dimensional analysis teaches that you cannot convert between measures with different dimensions such as these without additional informa (MORE)
# 49 cu ft is how many pounds?
This question cannot be answered sensibly. A cubic foot is a measure of volume, with dimensions [L 3 ]. A pound is a measure of mass, with dimensions [M]. Basic dimensional analysis teaches that you cannot convert between measures with different dimensions such as these without additional informa (MORE)
# Convert ft-pounds torque to horsepower?
You cannot truly convert ft-lbs to horse power. Foot-pounds are a measurement of force. Horsepower is the application of force to accomplish work (force applied to resistance over a period of time.) Think about a bicycle... When you ride a bicycle, your legs are like pistons. When you push (MORE)
# How many pounds are in 216 sq ft?
Units of weight do not convert into units of area. Weight and area are different things.
# How do you convert ft-pound to calories?
1 ft-lb = 0.32383 international calorie (0.00032383 food calorie) 1 food calorie = 3,088.03 ft-lbs
# Convert in pound to ft pound?
12 inches = 1 foot 1 inch = 0.08333... foot 12 inch-pounds = 1 foot-pound 1 inch-pound = 0.08333... foot-pound
# How many 40 pound bags of topsoil to cover 600 square ft at 2 inch depth?
40 pounds of topsoil is about 2 cubic feet, which will cover an area of 12 square feet at 2 inches deep. 600 square feet divided by 12 square feet per bag is 50 bags. Answer 50 bags
# How do you covert Nm to ft pounds?
1 ft = 0.3048 meters and 1 lb = 4.448222 Newtons, hence 1 ft lbs is 1.355817 Nm and 1 Nm is 0.737562 ft lbs (roughly three quarters)
# How many inches is 0.24 pounds?
15.334 inches per pound... Wait, that's total nonsense and I just made it up. You cannot convert inches to pounds because one is a weight and one is length. Which weighs more: a pound of stones or a pound of feathers?
# How many ft in a pound?
ft is a measurement of length, pounds is a measurement of weight so the two aren't related
# What is the common name for ft-pounds?
Torque is measured in foot pounds, and the abbreviation is an association. Also newtons per meter is the metric equivalent.So TORQUE!
# How much pound is 20 cu ft?
A cubic foot is a measure of volume (size). A pound is a measure of weight. The amount of weight that can fit in 20 cubic feet is different for different things. For example, a box with a volume (size) of 20 cubic feet, filled with solid gold, would weigh more than a box with the same size but fille (MORE)
# How many inches on torque wrench equals 44 pounds?
if you want to put 44 lbsf - ft of torque on a nut, the force on the lever (lbs) * the distance from the nut centre (feet) should = 44, any combination of force and lever length will do, as long as the product is 44 . examples: . 3 foot lever * ? (lbsf)= 44 44 / 3 = 14.67 lbsf . ? foot lever * 15 lb (MORE)
# What angle would you use if your torque wrench only goes to 150 ft-pounds and you need 192 ft-pounds?
Although I'm not certain that a torque wrench would measureaccurately that far outside its range, I can suggest an alternativeto measuring an angle. Instead, measure the distance between themarks on the gauge for 150 ft-lbs and 108 (a difference of 42) thenmake your own mark that extends beyond 150 (MORE)
# How many inches make a pound?
Inches and pounds have nothing to do. Inches are units of distance/length, pounds are units of weight.
# What is the difference between pounds and pound sterling?
No difference, they are both the same. A pound is just a shortened version of pound sterling.
# How many ft pounds is 220 inch pounds?
220 inches = 18 1 / 3 feet 220 inch pounds = 18 1 / 3 foot pounds
# What is the difference between ft pounds and pounds ft?
There is no difference. It is (force x distance) or (distance x force). It is usual to express torque and moment of a force in pounds-feet and work done in foot-pounds but it is not mandatory.
# How many pounds should a 5 ft woman weigh?
Im 5 ft tall and im 115 but im 13 your a woman so about 120 but i do need to lose a couple if pounds good luck
# How many pounds in a cubic inches?
Cubic inches is a measure of volume, and lbs is a measure of mass (technically it is a measure of weight or force). so it would be like comparing inches to seconds. how many inches in a second? doesnt work
# How many pounds in 129 inches?
Can't be answered. lbs = pounds is a unit of weight, Inches is a distance - they don't translate. You might as well ask how tall a gallon is.
# How many ft pounds are in 1 lb?
Ft Lbs (Foot pounds) is a measure of force; where as pounds is a measure of weight. They are not comparable.
# How many ft pounds is 168 inch pounds?
168 inches = 14 feet You may certainly multiply both sides of that identity by 'pound' if you like,
# How many psi in 1 pound ft?
Um... there are 1 psi (lb/in 2 ) = 144 psf (lb f /ft 2 ) , if by (lb ft) you meant pounnds per square foot. then there is 1/144 psi in 1 (lb f /ft 2 ) | 3,327 | 13,043 | {"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} | 3.875 | 4 | CC-MAIN-2019-09 | latest | en | 0.960984 |
http://docplayer.net/23839909-Ti-86-graphing-calculator-keystroke-guide.html | 1,529,714,011,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864848.47/warc/CC-MAIN-20180623000334-20180623020334-00332.warc.gz | 84,680,045 | 27,103 | # TI-86 Graphing Calculator Keystroke Guide
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1 TI-86 Graphing Calculator Keystroke Guide In your textbook you will notice that on some pages a key-shaped icon appears next to a brief description of a feature on your graphing calculator. In this guide the page number in the textbook on which each of these icons occurs is shown, as well as the key word or phrase that appears in the box, followed by a detailed description of how to use this feature on your TI-86 graphing calculator. Each feature is also given its own number for easy reference. page 2 1 Negative and Subtract You may have noticed that the calculator has two keys with a minus sign on them. The ( ) key in the bottom row of keys is actually the negative key. The negative key is used only to write a negative number. The following numbers and expressions would be entered using the negative key: 5 7* /( 6). The key in the right-hand column on the keyboard with the +,, and keys, is the subtraction key. It is grouped with the other keys used to perform the basic four arithmetic operations (addition, subtraction, multiplication, and division). The subtraction key is used between two numbers to indicate that a subtraction is to be done. The following expressions would be entered using the black subtraction key: * (3+ 4). You can tell which of these keys has been used from the screen display, if you look closely. The negative sign produced by the negative key is one pixel (dot) shorter and placed one pixel row higher than the subtraction sign produced by the subtraction key. Be aware that it is a common mistake among new users of the graphing calculator to press the wrong one of these two keys. If you do, sometimes you will get an error message on the calculator screen when you press ENTER to evaluate an expression. Other times the calculator will do something different from what you intend. For example, suppose you wanted to do the subtraction 8 5 on the calculator. First do it the correct way with the black subtraction key and press ENTER to evaluate it. Of course, you should get 3 as the answer. Now enter it the wrong way by using the negative key between the 8 and the 5 and press ENTER to evaluate it. The TI-86 displays an answer of 40. What happened? The calculator does not interpret the negative key as meaning to subtract. The calculator sees this last expression as two numbers written side by side, 8 and 5. The TI-86 interprets the lack of a basic operation key between the numbers to mean an implied multiplication: 8 times 5. page 2 2 Mode (Number of Decimal Places Displayed) Press 2nd [MODE] (the second function on the MORE key). The second line of the Mode screen controls the number of decimal places that the calcultor will display in its computed results. With the highlight on the word Float, the calculator will display either an exact value with only the 49
2 50 TI-86 number of decimal places needed to do so, or it will display the full twelve digits it is able to display. If the Mode screen does not already appear as shown below left, use the arrow keys to move the cursor over the word Float, and press ENTER to select it. Press 2nd [QUIT] to return to the Home screen and do the computations shown below the Mode screen on the left. Press 2nd [MODE] again, and use the arrow keys to move the cursor over the number 2 on the second line of the Mode screen. Press 2nd [QUIT] to return to the Home screen and do the same computations again. Note how all of the results are displayed rounded to two decimal places, as shown above right. page 3 3 Square Root The square root is the second function on the x 2 key. Note the square root sign ð printed above the x 2 key in yellow print. To produce a square root sign on the Home screen display, press 2nd [ ð ]. Other keystrokes involving the 2nd key will be denoted in a similar fashion. The expression in the brackets should be found printed above one of the keys on the keyboard. Unlike with most scientific calculators, with the graphing calculator you enter the square root sign first. To find the square root of 324, use the following keystrokes: 2nd [ ð ] 324 ENTER. On the Home screen you will see ð324. The value of this square root, 18, should have appeared on the right of the screen after you pressed ENTER. Note that the calculator only does the square root of the number immediately following the square root sign. If you want to compute 9+ 16, you must group the sum in parentheses, ð(9+16), to assure that the sum is computed before the square root is taken. (The correct value of this expression is 5.) If you do not enclose the sum in parentheses, the calculator will interpret ð9+16 as 9+ 16, where the square root applies only to the number 9. The value of this expression is 19. page 4 4 Test The items on the TEST menu may be used to tell whether a relationship between two numbers or expressions is true or false. To see how it works, let the calculator verify something you already know: 7 is less than 8. At the Home screen type a 7 and then press 2nd [TEST] (the second function on the 2 key) to display the TEST menu at the bottom of the screen.
3 Graphing Calculator Keystroke Guide 51 Select the "less than" symbol < by pressing F2. The < symbol will appear on the Home screen after the 7. Then type an 8, so that the Home screen now shows 7<8. The calculator will tell you whether this statement is true when you press ENTER. The calculator will display a 1 if the statement is true or a 0 if the statement is false. In this case, of course, the calculator displays a 1, since it is true that 7 is less than 8. Now enter a false statement: 4 > 11. Type a 4 and then, if the TEST menu is not already displayed, press 2nd [TEST] to display the TEST menu. The "greater than" symbol > is choice F3 on the menu, so press F3 to produce the > symbol on the Home screen after the 4. Now type 11 and press ENTER. The calculator displays a 0 to tell you that the statement is false. Next try one that perhaps isn't so obvious. Is 5 59 >? Type in 5/8 > 59/93 (using the division 8 93 key to produce the diagonal slash division sign /) and press ENTER to find out. The statement is false, since the calculator responds with a 0. page 5 5 Absolute Value The absolute value of a number is the number's distance from zero on the number line. On your calculator the absolute value is denoted abs. To produce the absolute value character, first press 2nd [MATH] (the second function on the multiplication key.) Select the NUM menu by pressing the F1 key at the top of the keyboard. F1 The calculator will then display the NUM menu at the bottom of the screen and move the MATH menu up a row. Press F5 to select abs, which will be displayed on the Home screen. F5 When we write the absolute value of 7 with pencil and paper, we enclose the number within two vertical bars: 7. On the calculator the absolute value of 7 is denoted abs(-7). The number or expression (called the argument) whose absolute value is to be evaluated is usually enclosed within parentheses (although for absolute values of single numbers, parentheses are optional). The parenthesis keys are above the 8 and 9 keys. Type abs(-7)on your Home screen by first producing abs, using the keystrokes described above. Then type 7 in parentheses, being
4 52 TI-86 sure to use the negative key, the gray ( ) key at the bottom of the keyboard. Press ENTER to find the absolute value of 7, which is 7. page 9 6 Home See the first page of the Introduction of the TI-86 part of this Guide to see the description of the Home screen. page 9 7 Add The addition key + is on the right-hand side of the keyboard with the other basic operation keys (subtraction, multiplication, and division). Basic addition is done as it is on most other kinds of calculators: the expression containing addition may be entered just as it appears in print or when written by hand. To evaluate the expression once you have typed it in, press ENTER. For example, to evaluate , type this expression just as it appears and press ENTER to display the value 22. To add expressions involving negative numbers, such as 5 + ( 8), you may type this expression just as it appears with the parentheses and press ENTER to obtain the value 3. Be sure you use the gray negative key ( ), not the black subtraction key, when you type in the 8. (See 1 Negative and Subtract.) You could also type this expression without using the parentheses, and it will still work: will still produce the value 3. page 11 8 Fraction Fractions may be entered into the calculator by using the division key, which produces a diagonal slash fraction bar on screen. Your calculator can perform operations on fractions and 5 1 give answers in either decimal or fraction form. Enter the expression by typing The expression will appear on the Home screen as 5/6-1/2. Press ENTER to get the decimal form of the value of this expression (approximately ). If you prefer to get your answer in fractional form, you must do one extra thing before pressing ENTER. To see how to get a fractional answer, either type in the original expression again or use 2nd [ENTRY] to reproduce it on the Home screen. With the cursor at the end of the expression, press 2nd [MATH], F5 (MISC), MORE, F1 (åfrac). After pressing this key sequence, the screen should display 5/6-1/2åFrac. Menu display after pressing 2nd [MATH]
5 Graphing Calculator Keystroke Guide 53 F1 After pressing F5 (MISC), MORE, F1 (åfrac). Now press ENTER to see the fractional value of the expression (one-third). page 17 9 Subtract and Negative See 1 Negative and Subtract. page Multiply If you type 3 6, your calculator will display it as 3*6. The asterisk * is a common computer notation for multiplication. When you press ENTER, the calculator will display the value 18. If you place one or both of the numbers in a multiplication in parentheses without using the multiplication key, the calculator still knows that you mean to multiply. Thus, the calculator will also produce 18 if you enter 3 6 as 3(6), (3)6, or (3)(6). Parentheses may also be used for grouping calculations to be done in a certain order. Thus, the calculator interprets 5(3+4) as meaning to add first to get 7, and then to multiply 5 times the 7 to get 35. Your calculator interprets the following as indicating multiplication: the multiplication key, numbers or expressions written in side by side in parentheses, and numbers and letters written side by side. Your calculator will also accept letters written side by side with * between them as indicating multiplication. For example, it would interpret A*B as meaning the product AB. The TI-86 accepts variable names that are more than one letter long. Therefore, it will not interpret AB as meaning the product of A and B; it will instead interpret AB as a single variable whose name is AB. If you have not already defined a value for the variable AB (see 11 Store), the calculator will give an error message when you try to evaluate it. As an example of placing numbers and letters side by side, the calculator interprets 3X as meaning "3 times X." (How to enter letters will be discussed later.) To see how this way of writing multiplication works, try typing 2π, where π (the Greek letter pi) represents that special number having to do with circles. The value of π is approximately Your calculator has this special number as the second function on the caret key ^. (You should see the symbol π printed above the caret key in yellow type.) To type in 2π, type the 2 and then press 2nd [π]. Your calculator should display 2π. Press ENTER to see the value , which is a decimal approximation of two times the value of π.
6 54 TI-86 page Divide The symbol appears on the division key of your calculator, but on the screen division is indicated by the diagonal slash /. Do the division 24 divided by 6 by pressing 24 6 ENTER. Note that on the Home screen the division appears as 24/6, which means the same thing, and the result is still 4. Fractions may also be entered as divisions. For example, the fraction 4 3 may be interpreted as 3 divided by 4. You may enter the fraction that way on your calculator. On the screen you will see 3/4. When you press ENTER, you will see the decimal equivalent of.75. (See also 8 Fraction.) page Exponent 5 The caret key ^ is used to enter most exponents. For example, the expression 2, is entered into the calculator as 2^5. Thus, the base is entered first, then the caret, and then the exponent. Press ENTER to produce this expression s value, 32. To raise a negative number to a power, it is necessary to enclose the negative number base in parentheses. For example, to raise the number 4 3 to the fourth power the expression must be written as ( 3). Likewise, we must enter this expression into the calculator as (-3)^4. The result is 81. Without the parentheses, -3^4 would (and should) be interpreted as taking the negative of = (3 ) = (81) = That is, -3^4 is interpreted as The exponent 2 can be displayed on screen as a raised exponent. To enter the expression you could use the standard approach of typing 5^2, or you could instead type the 5 and then 2 2 press the x key. The x key produces only the raised exponent 2 on the screen, not the x. Either way, the calculator produces the result of 25 when you press ENTER. page Square Root See 3 Square Root. page Reciprocal Since the reciprocal of an integer, such as 5, may be written as 51, it may be entered into the calculator as we would enter any fraction of this type: 1/5, using the division key to produce the diagonal slash / on the screen. Similarly, the reciprocal of a fraction, such as 8 5, may be written by interchanging the numerator and denominator: 5 8, appearing as 8/5 on screen. Another way to enter the reciprocal of a number is to use the x feature (the second function 1 on the EE key). For reasons that will be explained later, the reciprocal of 4 may be written as 4. To produce this expression on screen, press 4; then press 2nd [x -1 ], which produces only the exponent 1, not the x. Press ENTER to display the value of the expression,.25, which is the decimal equivalent of 41. To enter the reciprocal of a fraction such as 5, the fraction must be 8 enclosed in parentheses to ensure that the reciprocal of the entire fraction is computed. Type
7 Graphing Calculator Keystroke Guide 55 (5/8) -1 and press ENTER to see the value of the reciprocal of 5, which is displayed as 1.6, the 8 decimal equivalent of 8 5. If you want the calculator to display the reciprocal in fraction form, use the >Frac feature. (See 8 Fraction.) page 60
8 56 TI-86
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https://www.slideserve.com/ostinmannual/dept-lamar | 1,628,142,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155322.12/warc/CC-MAIN-20210805032134-20210805062134-00466.warc.gz | 1,040,186,054 | 24,589 | Discrete Distributions
# Discrete Distributions
## Discrete Distributions
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##### Presentation Transcript
1. Discrete Distributions
2. What is the binomial distribution? The binomial distribution is a discrete probability distribution. It is a distribution that governs the random variable, X, which is the number of successes that occur in "n" trials. The binomial probability distribution gives us the probability that a success will occur x times in the n trials, for x = 0, 1, 2, …, n. Thus, there are only two possible outcomes. It is conventional to apply the generic labels "success" and "failure" to the two possible outcomes. Discrete Distributions
3. A success can be defined as anything! "the axle failed" could be the definition of success in an experiment testing the strength of truck axles. Examples: 1.A coin flip can be either heads or tails 2.A product is either good or defective Binomial experiments of interest usually involve several repetitions or trials of the same basic experiments. These trials must satisfy the conditions outlined below: Discrete Distributions
4. When can we use it? Condition for use: Each repetition of the experiment (trial) can result in only one of two possible outcomes, a success or failure. See example BD1. The probability of a success, p, and failure (1-p) is constant from trial to trial. All trials are statistically independent; i.e. No trial outcome has any effect on any other trial outcome. The number of trials, n, is specified constant (stated before the experiment begins). Discrete Distributions
5. Example 1: binomial distribution: A coin flip results in a heads or tails A product is defective or not A customer is male or female Example 4: binomial distribution: Say we perform an experiment – flip a coin 10 times and observe the result. A successful flip is designated as heads. Assuming the coin is fair, the probability of success is .5 for each of the 10 trials, thus each trial is independent. We want to know the number of successes (heads) in 10 trials. The random variable that records the number of successes is called the binomial random variable. Random variable, x, the number of successes that occur in the n = 10 trials. Discrete Distributions
6. Do the 4 conditions of use hold? We are not concerned with sequence with the binomial. We could have several successes or failures in a row. Since each experiment is independent, sequence is not important. The binomial random variable counts the number of successes in n trials of the binomial experiment. By definition, this is a discrete random variable. Binomial Random Variable
7. Calculating the Binomial Probability • Rather than working out the binomial probabilities from scratch each time, we can use a general formula. • Say random variable "X" is the number of successes that occur in "n" trials. • Say p = probability of success in any trial • Say q = probability failure in any trial where q = (1 – p) • In general, The binomial probability is calculated by: Where x = 0, 1, 2, …, n
8. Example: For n = 3 Calculating the Binomial Probability
9. Discrete Distributions Each pair of values (n, p) determines a distinct binomial distribution. Two parameters: n and p where a parameter is: Any symbol defined in the functions basic mathematical form such that the user of that function may specify the value of the parameter.
10. Developing the Binomial Probability Distribution P(S2|S1 P(SSS)=p3 S3 P(S3|S2,S1) P(S3)=p S2 P(S2)=p P(S2|S1) S1 P(F3)=1-p F3 P(SSF)=p2(1-p) P(F3|S2,S1) P(S3|F2,S1) S3 P(S3)=p P(SFS)=p(1-p)p P(F2|S1) P(S1)=p P(F2)=1-p P(F3)=1-p Since the outcome of each trial is independent of the previous outcomes, we can replace the conditional probabilities with the marginal probabilities. F2 P(F3|F2,S1) F3 P(SFF)=p(1-p)2 S3 P(S3|S2,F1) P(FSS)=(1-p)p2 P(S3)=p S2 P(S2)=p P(F1)=1-p P(S2|F1) P(F3)=1-p P(F3|S2,F1) F3 P(FSF)=(1-p)P(1-p) S3 P(S3|F2,F1) P(FFS)=(1-p)2p F1 P(S3)=p P(F2|F1) P(F2)=1-p F2 P(F3)=1-p P(F3|F2,F1) F3 P(FFF)=(1-p)3
11. Let X be the number of successes in three trials. Then, P(SSS)=p3 SSS P(SSF)=p2(1-p) SS P(X = 3) = p3 X = 3 X =2 X = 1 X = 0 S S P(SFS)=p(1-p)p P(X = 2) = 3p2(1-p) P(SFF)=p(1-p)2 P(X = 1) = 3p(1-p)2 P(FSS)=(1-p)p2 SS P(X = 0) = (1- p)3 P(FSF)=(1-p)P(1-p) P(FFS)=(1-p)2p This multiplier is calculated in the following formula P(FFF)=(1-p)3
12. 5% of a catalytic converter production run is defective. A sample of 3 converter s is drawn. Find the probability distribution of the number of defectives. Solution A converter can be either defective or good. There is a fixed finite number of trials (n=3) We assume the converter state is independent on one another. The probability of a converter being defective does not change from converter to converter (p=.05). Binomial Example The conditions required for the binomial experiment are met
13. Let X be the binomial random variable indicating the number of defectives. Define a “success” as “a converter is found to be defective”. X P(X) 0 .8574 1 .1354 2 .0071 3 .0001
14. Discrete Distributions Example: The quality control department of a manufacturer tested the most recent batch of 1000 catalytic converters produced and found that 50 of them to be defective. Subsequently, an employee unwittingly mixed the defective converters in with the nondefective ones. Of a sample of 3 converters is randomly selected from the mixed batch, what is the probability distribution of the number of defective converters in the sample? Does this situation satisfy the requirements of a binomial experiment? n = 3 trials with 2 possible outcomes (defective or nondefective). Does the probability remain the same for each trial? Why or why not? The probability p of selecting a defective converter does not remain constant for each trial because the probability depends on the results of the previous trial. Thus the trials are not independent.
15. Discrete Distributions The probability of selecting a defective converter on the first trial is 50/1000 = .05. If a defective converter is selected on the first trial, then the probability changes to 49/999 = .049. In practical terms, this violation of the conditions of a binomial experiment is often considered negligible. The difference would be more noticeable if we considered 5 defectives out of a batch of 100.
16. Discrete Distributions If we assume the conditions for a binomial experiment hold, then consider p = .5 for each trial. Let X be the binomial random variable indicating the number o defective converters in the sample of 3. P(X = 0) = p(0) = [3!/0!3!](.05)0(.95)3 = .8574 P(X = 1) = p(1) = [3!/1!2!](.05)1(.95)2 = .1354 P(X = 2) = p(2) = [3!/2!1!](.05)2(.95)1 = .0071 P(X = 3) = p(3) = [3!/3!0!](.05)3(.95)0 = .0001 The resulting probability distribution of the number of defective converters in the sample of 3, is as follows:
17. Discrete Distributions xp(x) 0 .8574 1 .1354 2 .0071 3 .0001
18. Cumulative Binomial Distribution: F(x)= S from k=0 to x: nCx * p k q (n-k) Another way to look at things = cummulative probabilities Say we have a binomial with n = 3 and p = .05 x p(x) 0 .8574 1 .1354 2 .0071 3 .0001
19. Cumulative Binomial Distribution: this could be written in cumulative form: from x = 0 to x = k: x p(x) 0 .8574 1 .9928 2 .9999 3 1.000
20. Cumulative Binomial Distribution: What is the advantage of cummulative? It allows us to find the probability that X will assume some value within a range of values. Example 1: Cumulative: p(2) = p(x<2) – p(x<1) = .9999 - .9928 = .0071 Example 2: Cumulative: Find the probability of at most 3 successes in n=5 trials of a binomial experiment with p = .2. We locate the entry corresponding to k = 3 and p = .2 P(X < 3) = SUM p(x) = p(0) + p(1) + p(2) + p(3) = .993
21. E(X) = µ = np V(X) = s2 = np(1-p) Example 6.10 Records show that 30% of the customers in a shoe store make their payments using a credit card. This morning 20 customers purchased shoes. Use the Cummulative Binomial Distribution Table (A.1 of Appendix) to answer some questions stated in the next slide. Mean and Variance of Binomial Random Variable
22. What is the probability that at least 12 customers used a credit card? This is a binomial experiment with n=20 and p=.30. p k .01……….. 30 0 . . 11 P(At least 12 used credit card) = P(X>=12)=1-P(X<=11) = 1-.995 = .005 .995
23. What is the probability that at least 3 but not more than 6 customers used a credit card? p k .01……….. 30 0 2 . 6 P(3<=X<=6)= P(X=3 or 4 or 5 or 6) =P(X<=6) -P(X<=2) =.608 - .035 = .573 .035 .608
24. What is the expected number of customers who used a credit card? E(X) = np = 20(.30) = 6 Find the probability that exactly 14 customers did not use a credit card. Let Y be the number of customers who did not use a credit card.P(Y=14) = P(X=6) = P(X<=6) - P(x<=5) = .608 - .416 = .192 Find the probability that at least 9 customers did not use a credit card. Let Y be the number of customers who did not use a credit card.P(Y>=9) = P(X<=11) = .995
25. Poisson Distribution • What if we want to know the number of events during a specific time interval or a specified region? • Use the Poisson Distribution. • Examples of Poisson: • Counting the number of phone calls received in a specific period of time • Counting the number of arrivals at a service location in a specific period of time – how many people arrive at a bank • The number of errors a typist makes per page • The number of customers entering a service station per hour
26. Poisson Distribution • Conditions for use: • The number of successes that occur in any interval is independent of the number of successes that occur in any other interval. • The probability that a success will occur in an interval is the same for all intervals of equal size and is proportional to the size of the interval. • The probability that two or more successes will occur in an interval approaches zero as the interval becomes smaller. • Example: • The arrival of individual dinners to a restaurant would not fit the Poisson model because dinners usually arrive with companions, violating the independence condition.
27. The Poisson variable indicates the number of successes that occur during a given time interval or in a specific region in a Poisson experiment Probability Distribution of the Poisson Random Variable: Poisson Random Variable • = average number of successes occurring in a specific interval • Must determine an estimate for from historical data (or other source) • No limit to the number of values a Poisson random Variable can assume
28. Poisson Probability Distribution With m = 1 The X axis in Excel Starts with x=1!! 0 1 2 3 4 5
29. Poisson probability distribution with m =2 0 1 2 3 4 5 6 Poisson probability distribution with m =5 0 1 2 3 4 5 6 7 8 9 10 Poisson probability distribution with m =7 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
30. Cars arrive at a tollbooth at a rate of 360 cars per hour. What is the probability that only two cars will arrive during a specified one-minute period? (Use the formula) The probability distribution of arriving cars for any one-minute period is Poisson with µ = 360/60 = 6 cars per minute. Let X denote the number of arrivals during a one-minute period. Poisson Example
31. What is the probability that only two cars will arrive during a specified one-minute period? (Use table 2, Appendix B.) P(X = 2) = P(X<=2) - P(X<=1) = .062 - .017 = .045
32. What is the probability that at least four cars will arrive during a one-minute period? Use Cummulative Poisson Table (Table A.2 , Appendix) P(X>=4) = 1 - P(X<=3) = 1 - .151 = .849
33. When n is very large, binomial probability table may not be available. If p is very small (p< .05), we can approximate the binomial probabilities using the Poisson distribution. Use = np and make the following approximation: Poisson Approximation of the Binomial With parameters n and p With m = np
34. Example of Poisson Example: Poisson Approximation of the Binomial A warehouse engages in acceptance sampling to determine if it will accept or reject incoming lots of designer sunglasses, some of which invariably are defective. Specifically, the warehouse has a policy of examining a sample of 50 sunglasses from each lot and accepting the lot only if the sample contains no more than 2 defective pairs. What is the probability of a lot's being accepted if, in fact, 2% of the sunglasses in the lot are defective? This is a binomial experiment with n = 50 and p = .02. Our binomial tables include n values up to 25, but since p < .05 and the expected number of defective sunglasses in the sample is np = 50(.02) = 1, the required probability can be approximated by using the Poisson distribution with μ = 1. From Table A.1, we find that the probability that a sample contains at most 2 defective pairs o sunglasses is .920.
35. What is the probability of a lot being accepted if, in fact, 2% of the sunglasses are defective? Solution This is a binomial experiment with n = 50, p = .02. Tables for n = 50 are not available; p<.05; thus, a Poisson approximation is appropriate [ = (50)(.02) =1] P(Xpoisson<=2) = .920 (true binomial probability = .922) Poisson Example
36. Example of Poisson So how well does the Poisson approximate the Binomial? Consider the following table: x Binomial (n = 50, p = .02) Poisson (μ = np = 1) 0 .364 .368 1 .372 .368 2 .186 .184 3 .061 .061 4 .014 .015 5 .003 .003 6 .000 .001
37. Example of Poisson • A tollbooth operator has observed that cars arrive randomly at an average rate of 360 cars per hour. • Using the formula, calculate the probability that only 2 cars will arrive during a specified 1 minute period. • Using Table A.2 on page 360, find the probability that only 2 cars will arrive during a specified 1 minute period. • Using Table A.2 on page 360, find the probability that at least 4 cars will arrive during a specified 1 minute period. • P(X=2) = [(e-6)(62)] / 2! = (.00248)(36) / 2 = .0446 • P(X=2) = P(X < 2) - P(X < 1) = .062 = .017 = .045 • P(X > 4) = 1 - P(X < 3) = 1 - .151 = .849
38. Example of Poisson What if we wanted to know the probability of a small number of occurrences in a large number of trials and a very small probability of success? We use Poisson as a good approximation of the answer. When trying to decide between the binomial and the Poisson, use the following guidelines: n > 20 n > 100 or p < .05 or np < 10
39. Hypergeometric Distribution What about sampling without replacement? What is likely to happen to the probability of success? Probability of success is not constant from trial to trial. We have a finite set on N things, and "a" of them possess a property of interest. Thus, there are "a" successes in N things. Let X be the number of successes that occur in a sample, without replacement of "n" things from a total of N things. This is the hypergeometric distribution: P(x) = (aCx)(N-a C n-x) x = 0,1,2…. NCn
40. Binomial Approximation of Hypergeometric Distribution If N is large, then the probability of success will remain approximately constant from one trial to another. When can we use the binomial distribution as an approximation of the hypergeometric distribution when: N/10 > n
41. What if we want to perform a single experiment and there are only 2 possible outcomes? We use a special case of the binomial distribution where n=1: P(x)= 1Cx * px * (1-p) 1-x x =0,1 which yields p(0)= 1-p = px * (1-p) 1-x x=0,1 p(1)= p In this form, the binomial is referred to as the Bernoulli distribution. Bernoulli Distribution
42. Now, instead of being concerned with the number of successes in n Bernoulli trials, let’s consider the number of Bernoulli trial failures that would have to be performed prior to achieving the 1st success. In this case, we use the geometric distribution, where X is the random variable representing the number of failures before the 1st success. Mathematical form of geometric distribution: P(x)=p * (1-p)x x = 0,1,2… Geometric Distribution
43. What if we wanted to know the number of failures, x, that occur before the rth success (r = 1,2….)? In this case, we use the negative binomial distribution. The number of statistically independent trials that will be performed before the r success = X + r The previous r – 1 successes and the X failures can occur in any order during the X + r – 1 trials. Negative binomial distribution – mathematical form: P(x) = r+x-1 Cx * pr * (1-p)x x= 0,1,2….. Negative Binomial Distribution
44. Problem Solving • A Suggestion for Solving Problems Involving Discrete Random Variables • An approach: • Understand the random variable under consideration and determine if the random variable fits the description and satisfies the assumptions associated with any of the 6 random variables presented in Table 3.3. • If you find a match in Table 3.3 use the software for that distribution. • If none of the 6 random variables in table 3.3 match the random variable associated with your problem, use the sample space method
45. Discrete Bivariate Probability Distribution Functions • The first part of the chapter considered only univariate probability distribution functions.~ One variable is allowed to change. • What if two or more variables change? • When considering situations where two or more variables change, the definitions of • sample space • numerically valued functions • random variable • still apply.
46. To consider the relationship between two random variables, the bivariate (or joint) distribution is needed. Bivariate probability distribution The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x, Y = y) Bivariate Distributions
47. Discrete Bivariate Probability Distribution Functions Example Consider the following real estate data: We want to know how the size of the house varies with the cost.
48. Discrete Bivariate Probability Distribution Functions
49. Discrete Bivariate Probability Distribution Functions What is the next step? Construct frequency distributions: Frequency distribution of house size:
50. Discrete Bivariate Probability Distribution Functions Frequency distribution of selling price: | 4,841 | 18,243 | {"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-2021-31 | latest | en | 0.873643 |
https://jp.maplesoft.com/support/help/maplesim/view.aspx?path=LinearAlgebra/PopovForm&L=J | 1,653,737,392,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00635.warc.gz | 397,538,126 | 46,313 | PopovForm - Maple Help
LinearAlgebra
PopovForm
compute the Popov normal form of a Matrix
Calling Sequence PopovForm(A, x, shifts, out, options, outopts)
Parameters
A - Matrix of univariate polynomials in x x - variable name of the polynomial domain shifts - (optional) equation of the form shifts = obj where obj is a list of one or two lists out - (optional) equation of the form output = obj where obj is one of 'P', 'U', 'rank', 'P_pivots', 'U_pivots', or a list containing one or more of these names; select result objects to compute options - (optional); constructor options for the result object(s) outopts - (optional) equation(s) of the form outputoptions[o] = list where o is one of 'P' or 'U'; constructor options for the specified result object
Description
• The PopovForm(A, x) function computes the column Popov normal form P (also called the polynomial echelon form) of an m x n rectangular Matrix of univariate polynomials in x over the field of rational numbers Q, or rational expressions over Q. You can request the form P, the unimodular multiplier U which gives P, the rank and various pivots in P and U via a specification of the output option.
Definition of Column Popov Form
• There are a number of variations of column Popov forms, typically unique up to permutation of the columns. The definitions used for both the full and non-full column rank case are as follows.
If $m=n$ and P is nonsingular, P has the following degree constraints.
$\mathrm{deg}\left({P}_{j,i}\right)<\mathrm{deg}\left({P}_{i,i}\right),\mathrm{for all j > i}$
$\mathrm{deg}\left({P}_{j,i}\right)\le \mathrm{deg}\left({P}_{i,i}\right),\mathrm{for all j < i}$
If $n and P has full column rank, there is a trailing list of n pivot rows P_pivots such that P[P_pivots,*] is in Popov normal form.
If $n\le m$ and P has column rank $r, P has the first $n-r$ columns 0 and there is a trailing list of r rows P_pivots such that P[P_pivots,*] is in Popov normal form. In this case, U is a minimal unimodular multiplier and as such there is a list U_pivots of rows such that U[U_pivots,*] is also in Popov normal form.
The Popov normal form P is obtained by doing elementary column operations on A. This includes interchanging columns, multiplying through a column by a unit, and subtracting a polynomial multiple of one column from another. The method used is a fraction-free algorithm by Beckermann, Labahn, and Villard. The returned Matrix objects have the property that $P=A·U$.
• The output option (out) determines the content of the returned expression sequence.
Depending on what is included in the output option, an expression sequence containing one or more of the factors P (the Popov normal form), U (the unimodular transformation Matrix), rank (the rank of the matrix), P_pivots, or U_pivots (the pivot rows of P and U, respectively) can be returned.
If output is a list, the objects are returned in the same order as specified in the list.
• The shifts option is an optional input which allows the user to shift the degree constraints on both the Popov form and the minimal multiplier (in the non-full column rank case).
• The constructor options provide additional information (readonly, shape, storage, order, datatype, and attributes) to the Matrix constructor that builds the result(s). These options may also be provided in the form outputoptions[o]=[...], where [...] represents a Maple list. If a constructor option is provided in both the calling sequence directly and in an outputoptions[o] option, the latter takes precedence (regardless of the order).
Examples
> $\mathrm{with}\left(\mathrm{LinearAlgebra}\right):$
> $A≔⟨⟨{z}^{3}-{z}^{2},{z}^{3}-2{z}^{2}+2z-2⟩|⟨{z}^{3}-2{z}^{2}-1,{z}^{3}-3{z}^{2}+3z-4⟩⟩$
${A}{≔}\left[\begin{array}{cc}{{z}}^{{3}}{-}{{z}}^{{2}}& {{z}}^{{3}}{-}{2}{}{{z}}^{{2}}{-}{1}\\ {{z}}^{{3}}{-}{2}{}{{z}}^{{2}}{+}{2}{}{z}{-}{2}& {{z}}^{{3}}{-}{3}{}{{z}}^{{2}}{+}{3}{}{z}{-}{4}\end{array}\right]$ (1)
> $P≔\mathrm{PopovForm}\left(A,z,\mathrm{datatype}=\mathrm{algebraic}\right)$
${P}{≔}\left[\begin{array}{cc}{z}& {-1}\\ {1}& {-}{1}{+}{z}\end{array}\right]$ (2)
> $P,U≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U'\right]\right)$
${P}{,}{U}{≔}\left[\begin{array}{cc}{z}& {-1}\\ {1}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{cc}{-}\frac{{1}}{{2}}{}{{z}}^{{2}}{+}\frac{{3}}{{2}}{}{z}{-}\frac{{1}}{{2}}& \frac{{1}}{{2}}{}{{z}}^{{2}}{-}\frac{{1}}{{2}}{}{z}{-}\frac{{3}}{{2}}\\ \frac{{1}}{{2}}{}{{z}}^{{2}}{-}{z}& {-}\frac{{{z}}^{{2}}}{{2}}{+}{1}\end{array}\right]$ (3)
> $\mathrm{map}\left(\mathrm{expand},P-A·U\right)$
$\left[\begin{array}{cc}{0}& {0}\\ {0}& {0}\end{array}\right]$ (4)
> $\mathrm{Determinant}\left(U\right)$
${-}\frac{{1}}{{2}}$ (5)
A low rank matrix.
> $A≔⟨⟨3z-6,-3z+3,2z+3,z⟩|⟨-3z,3z,-2,-1⟩|⟨6,-3,-2z-1,-z+1⟩⟩$
${A}{≔}\left[\begin{array}{ccc}{3}{}{z}{-}{6}& {-}{3}{}{z}& {6}\\ {-}{3}{}{z}{+}{3}& {3}{}{z}& {-3}\\ {2}{}{z}{+}{3}& {-2}& {-}{2}{}{z}{-}{1}\\ {z}& {-1}& {-}{z}{+}{1}\end{array}\right]$ (6)
> $P,U,r≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U','\mathrm{rank}'\right]\right)$
${P}{,}{U}{,}{r}{≔}\left[\begin{array}{ccc}{0}& {-}{z}& {-6}\\ {0}& {z}& {3}\\ {0}& {-}\frac{{2}}{{3}}& {1}{+}{2}{}{z}\\ {0}& {-}\frac{{1}}{{3}}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& {0}& {1}\\ {1}& \frac{{1}}{{3}}& {1}\\ {1}& {0}& {0}\end{array}\right]{,}{2}$ (7)
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{ccc}{0}& {-}{z}& {-6}\\ {0}& {z}& {3}\\ {0}& {-}\frac{{2}}{{3}}& {1}{+}{2}{}{z}\\ {0}& {-}\frac{{1}}{{3}}& {-}{1}{+}{z}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& {0}& {1}\\ {1}& \frac{{1}}{{3}}& {1}\\ {1}& {0}& {0}\end{array}\right]{,}{2}{,}\left[{2}{,}{4}\right]{,}\left[{3}\right]$ (8)
A [2,2,0,0]-shifted Popov form.
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{shifts}=\left[\left[2,2,0,0\right]\right],\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{ccc}{0}& {-}{{z}}^{{2}}{+}{z}{-}{2}& {2}{}{{z}}^{{2}}{+}{z}{+}{4}\\ {0}& {{z}}^{{2}}{-}{z}{+}{1}& {-}{2}{}{{z}}^{{2}}{-}{z}{-}{2}\\ {0}& {1}& {0}\\ {0}& {0}& {1}\end{array}\right]{,}\left[\begin{array}{ccc}{1}& \frac{{1}}{{3}}& {-}\frac{{2}}{{3}}\\ {1}& \frac{{z}}{{3}}& {-}{1}{-}\frac{{2}{}{z}}{{3}}\\ {1}& {0}& {0}\end{array}\right]{,}{2}{,}\left[{3}{,}{4}\right]{,}\left[{3}\right]$ (9)
A Popov form with [0,-3,0,0]-shift for unimodular multiplier.
> $A≔⟨⟨-{z}^{3}+4{z}^{2}+z+1,-{z}^{2}+7z+4⟩|⟨z-1,z+2⟩|⟨2{z}^{2}+2z-2,{z}^{2}+6z+6⟩|⟨-{z}^{2},-2z⟩⟩$
${A}{≔}\left[\begin{array}{cccc}{-}{{z}}^{{3}}{+}{4}{}{{z}}^{{2}}{+}{z}{+}{1}& {-}{1}{+}{z}& {2}{}{{z}}^{{2}}{+}{2}{}{z}{-}{2}& {-}{{z}}^{{2}}\\ {-}{{z}}^{{2}}{+}{7}{}{z}{+}{4}& {z}{+}{2}& {{z}}^{{2}}{+}{6}{}{z}{+}{6}& {-}{2}{}{z}\end{array}\right]$ (10)
> $P,U,r,\mathrm{II},K≔\mathrm{PopovForm}\left(A,z,\mathrm{shifts}=\left[\left[0,0\right],\left[0,-3,0,0\right]\right],\mathrm{output}=\left['P','U','\mathrm{rank}','\mathrm{P_pivots}','\mathrm{U_pivots}'\right]\right)$
${P}{,}{U}{,}{r}{,}{\mathrm{II}}{,}{K}{≔}\left[\begin{array}{cccc}{0}& {0}& {z}& {-1}\\ {0}& {0}& {2}& {z}\end{array}\right]{,}\left[\begin{array}{cccc}{-}\frac{{2}{}{z}}{{21}}{+}\frac{{1}}{{7}}& {-}{{z}}^{{2}}{-}{2}{}{z}& \frac{{2}}{{7}}{+}\frac{{z}}{{7}}& {-}\frac{{z}}{{21}}{-}\frac{{3}}{{7}}\\ {1}& {0}& {0}& {0}\\ \frac{{2}{}{z}}{{21}}{-}\frac{{3}}{{7}}& {{z}}^{{2}}{-}{z}& {-}\frac{{z}}{{7}}{+}\frac{{1}}{{7}}& \frac{{z}}{{21}}{+}\frac{{2}}{{7}}\\ \frac{{2}}{{21}}{}{{z}}^{{2}}{-}\frac{{1}}{{3}}{}{z}{-}\frac{{4}}{{21}}& {{z}}^{{3}}{-}{9}{}{z}{-}{7}& {-}\frac{{{z}}^{{2}}}{{7}}{+}\frac{{9}}{{7}}& \frac{{1}}{{21}}{}{{z}}^{{2}}{+}\frac{{1}}{{3}}{}{z}{-}\frac{{23}}{{21}}\end{array}\right]{,}{2}{,}\left[{1}{,}{2}\right]{,}\left[{2}{,}{4}\right]$ (11)
> $\mathrm{map}\left(\mathrm{expand},P-A·U\right)$
$\left[\begin{array}{cccc}{0}& {0}& {0}& {0}\\ {0}& {0}& {0}& {0}\end{array}\right]$ (12)
> $\mathrm{Determinant}\left(U\right)$
${1}$ (13)
References
Beckermann, B., and Labahn, G. "Fraction-free Computation of Matrix Rational Interpolants and Matrix GCDs." SIAM Journal on Matrix Analysis and Applications, Vol. 22 No. 1. (2000): 114-144.
Beckermann, B.; Labahn, G.; and Villard, G. "Shifted Normal Forms of General Polynomial Matrices." University of Waterloo, Technical Report. Department of Computer Science, 2001.
Beckermann, B.; Labahn, G.; and Villard, G. "Shifted Normal Forms of Polynomial Matrices." ISSAC'99, pp. 189-196. 1999. | 3,367 | 8,707 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "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} | 3.59375 | 4 | CC-MAIN-2022-21 | longest | en | 0.629609 |
http://mystry-geek.blogspot.com/2012/07/ | 1,532,296,749,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594018.55/warc/CC-MAIN-20180722213610-20180722233610-00334.warc.gz | 254,649,196 | 15,035 | Wednesday, July 18, 2012
Fibonacci Numbers
Wrote a couple of versions of Fibonacci number generators.
Interestingly, the numbers become inaccurate at just the 93rd fib number in the sequence: the real numbers begin losing precision on the low end. I am going to have to figure out how to use one of the bignum libraries in order to work exactly with such large values. My fear is that the bignum libraries have a substantial performance hit in return for the accurate values.
The recursive version becomes too slow to use at just the 40th number in the sequence, while the non-recursive version fires through thousands of values.
`Fib-recursive.c`
` `
```#include <stdio.h>
/* This program calculates fibonacci sequence using recursion
1 1 2 3 5 8 13 21 34
Starting with 1
Each number in the sequence is created by adding the previous two numbers in sequence.
*/
long double fib(long double value){
if (value <= 2)
return 1;
return (fib(value-1) + fib (value-2));
}
int
main(){
long double i;
for (i = 1; i<1000; i++)
printf ("value is %.0Lf and return value is %.0Lf\n", i, fib(i));
}```
` `
` `
` `
`Fib-nonrecursive.c`
` `
```#include <stdio.h>
/* This program calculates fibonacci sequence using recursion
1 1 2 3 5 8 13 21 34
Starting with 1
Each number in the sequence is created by adding the previous two numbers in sequence.
If you notice, something bad happens when we reach number 93 and we lose precision off the end.
The number is close to the right number but is subtly wrong.
We need to switch to a bignum library to add numbers that large precisely.
Interestingly we can calculate the first thousand numbers in the sequence this way before we can calculate the first 30 numbers recursively.
*/
int
main(){
long double i, j, p1, p2;
p1 = 0;
p2 = 1;
i = 0;
j = 1;
while (j<1000){
i = p1 + p2;
printf ("\t%.0Lf\t%.0Lf\n", j, i);
p2 = p1;
p1 = i;
j++;
}
} ```
` `
` `
Monday, July 9, 2012
Simple web server using epoll and sendfile.
I have been looking at how to use sendfile() and epoll() for a while now, so I finally sat down and wrote a very simple web server that uses these two sets of functions. The resulting code follows this text. This web server just performs a get request, and returns the file at the requested path. If the file cannot be found, it returns a 401 error, file not found.
sendfile() is useful because it dramatically speeds up and simplifies reading a file from disk and sending it out the network file descriptor. Typically a web server has to have the file and the network socket open, and then read a block from the file, then write the block to the disk, this results in the data being copies between several buffers and the buffer management involved in that. However the sendfile() man page recommends to fall back to the normal read and write in cases where
epoll() is great because it doesn't have the overhead of select, or most of the limitations. With select you have to loop through every element in a large, fixed size array, testing each one to see if you got a connection on that element. With epoll you get a list of elements to handle that you can quickly process, limiting unproductive looping on the possibly thousands of elements you did not have to process. Epoll can also easily handle thousands of simultaneous connections without slowing down like select would.
This program does not perform cgi, send mime types, or perform any http request other than get. It assumes that the first data block of the request contains the entire request. The program assumes one request per connection. It ignores any trailing text after the first line. The beauty is in it's compiled size, it is just 9.5KB in size and is very easy to understand.
This program only handles a single request at a time because the sendfile request blocks.
Next steps:
1. Break code into generic epoll server file and server specific files, with call backs in the generic epoll side to the server side to handle new connect, data request, and closed connections. This will allow the generic epoll server stuff to be completely generic and usable for various other projects. If it is a library then this could be shared among multiple executing programs.
2. Handle the request correctly to entirely process the request header and attach all the data to the connection, then perform the request by the request type.
3. Handle mime types. Each file should have a mimetype header sent before the data based on the extension of the file.
4. Handle things other than just "GET".
5. Handle multiple requests on a single connection.
6. Add becoming a deamon and signal handling to code to properly handle various tasks.
7. Use a worker thread pool to get requests and process them while the main process just handles new connections, reading client requests, and closing old connections. The worker threads can each handle a single completed request, sending the data to the client through the port.
8. Handle the reading of the input and processing of requests in the worker thread as well, this would leave the main thread just handling new connections passing triggered events to threads.
9. Handle cgi programs. I already have code to do this, just needs converted to also use epoll() to monitor the open handles and process events.
11. Add database reading/writing for persistent data across connections using sqlite. Allow get and put requests using the database.
```/*
This is a very simple web server that listens for connections on epoll.
It only processes the first line and only understands GET and the path.
Web server will flag the OS to send file to port.
Root is the Root directory just above the location of the executabe.
*/
#include <sys/sendfile.h>
#include <stdio.h>
#include <sys/timeb.h>
#include <sys/socket.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/epoll.h>
#include <netinet/in.h>
#include <string.h>
#include <fcntl.h>
#include <signal.h>
#include <errno.h>
#include <stdlib.h>
#define MAX_CLIENT 5000
#define PORT 8080
void nonblock(int sockfd)
{
int opts;
opts = fcntl(sockfd, F_GETFL);
if(opts < 0)
{
perror("fcntl(F_GETFL)\n");
exit(1);
}
opts = (opts | O_NONBLOCK);
if(fcntl(sockfd, F_SETFL, opts) < 0)
{
perror("fcntl(F_SETFL)\n");
exit(1);
}
}
void
Initialize(){
}
/* Parse and return file handle for http get request */
int
OpenFile(char * buf, int n){
int i, j;
char * index = {"index.html"};
char path[1024];
if (buf==NULL || n<10)
return 0;
// Ensure that the string starts with "GET "
if (!buf[0]=='G' || !buf[1]=='E' || !buf[2]=='T' || !buf[3]==' ')
return 0;
// copy the string
for (i=0; i < n && buf[i+4]!=' '; i++)
path[i]=buf[i+4];
// put null on end of path
path[i]='\x00';
// if the syntax doesn't have a space following a path then return
if (buf[i+4]!=' ' || i == 0)
return 0;
// if the last char is a slash, append index.html
if (path[i-1] == '/')
for (j=0; j<12; j++)
path[i+j]=index[j];
printf("Found path: \"%s\"\n", path);
// the +1 removes the leading slash
return (open(path+1, O_RDONLY));
}
void
Respond(int fd, char * buf, int n){
char goodresponse[1024] = {"HTTP/1.1 200 OK\r\n\r\n"};
int fh;
// printf("%d data received: \n%s\n", fd, buf);
fh = OpenFile(buf, n);
if (fh > 0){
struct stat stat_buf; /* hold information about input file */
send(fd, goodresponse, strlen(goodresponse), 0);
/* size and permissions of fh */
fstat(fh, &stat_buf);
sendfile(fd, fh, NULL, stat_buf.st_size);
close(fh);
} else {
}
close(fd);
}
void
Respond_simple( int fd, char * buf, int n){
int y = strlen(response);
printf("%d data received: \n%s\n", fd, buf);
send(fd, response, y, 0);
close(fd);
}
void
Loop(){
struct epoll_event *events;
struct epoll_event ev;
int listenfd;
int epfd;
int clifd;
int i;
int res;
char buffer[1024];
int n;
events = (struct epoll_event *)calloc(MAX_CLIENT, sizeof(struct epoll_event));
if( (listenfd = socket(AF_INET, SOCK_STREAM, 0)) < 0)
{
perror("error opening socket\n");
exit(1);
}
bzero(&srv, sizeof(srv));
srv.sin_family = AF_INET;
srv.sin_port = htons(PORT);
if( bind(listenfd, (struct sockaddr *) &srv, sizeof(srv)) < 0)
{
perror("error binding to socket\n");
exit(1);
}
listen(listenfd, 1024);
int reuse_addr = 1; /* Used so we can re-bind to our port */
/* So that we can re-bind to it without TIME_WAIT problems */
epfd = epoll_create(MAX_CLIENT);
if(!epfd)
{
perror("epoll_create\n");
exit(1);
}
ev.events = EPOLLIN | EPOLLERR | EPOLLHUP;
ev.data.fd = listenfd;
if(epoll_ctl(epfd, EPOLL_CTL_ADD, listenfd, &ev) < 0)
{
exit(1);
}
for( ; ; )
{
usleep(8000);
res = epoll_wait(epfd, events, MAX_CLIENT, 0);
for(i = 0; i < res; i++)
{
if(events[i].data.fd == listenfd)
{
clifd = accept(listenfd, NULL, NULL);
if(clifd > 0)
{
printf(".");
nonblock(clifd);
ev.events = EPOLLIN | EPOLLET;
ev.data.fd = clifd;
if(epoll_ctl(epfd, EPOLL_CTL_ADD, clifd, &ev) < 0)
{
exit(1);
}
}
}
else {
n = recv(events[i].data.fd, buffer, 1023, 0);
if(n == 0) /* closed connection */
{
epoll_ctl(epfd, EPOLL_CTL_DEL, events[i].data.fd, NULL);
}
else if(n < 0) /* error, close connection */
{
//epoll_ctl(epfd, EPOLL_CTL_DEL, events[i].data.fd, NULL);
close(events[i].data.fd);
}
else { /* got a request, process it. */
//send(events[i].data.fd, buffer, n, 0);
//bzero(&buffer, n);
//printf("%d data received: \n%s\n", events[i].data.fd, buffer);
//send(events[i].data.fd, response, y, 0);
//close(events[i].data.fd);
Respond(events[i].data.fd, buffer, n);
}
}
}
}
}
void Shutdown(){
}
int main (){
Initialize();
Loop();
Shutdown();
exit(0);
}``` | 2,534 | 9,580 | {"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} | 3.515625 | 4 | CC-MAIN-2018-30 | latest | en | 0.842641 |
https://pressbooks.online.ucf.edu/osuniversityphysics2/chapter/entropy/ | 1,726,012,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00091.warc.gz | 446,483,723 | 29,469 | Chapter 4. The Second Law of Thermodynamics
# 4.6 Entropy
### Learning Objectives
By the end of this section you will be able to:
• Describe the meaning of entropy
• Calculate the change of entropy for some simple processes
The second law of thermodynamics is best expressed in terms of a change in the thermodynamic variable known as entropy, which is represented by the symbol S. Entropy, like internal energy, is a state function. This means that when a system makes a transition from one state into another, the change in entropy $\text{Δ}S$ is independent of path and depends only on the thermodynamic variables of the two states.
We first consider $\text{Δ}S$ for a system undergoing a reversible process at a constant temperature. In this case, the change in entropy of the system is given by
$\text{Δ}S=\frac{Q}{T},$
where Q is the heat exchanged by the system kept at a temperature T (in kelvin). If the system absorbs heat—that is, with $Q>0$—the entropy of the system increases. As an example, suppose a gas is kept at a constant temperature of 300 K while it absorbs 10 J of heat in a reversible process. Then from Equation 4.8, the entropy change of the gas is
$\text{Δ}S=\frac{10\phantom{\rule{0.2em}{0ex}}\text{J}}{300\phantom{\rule{0.2em}{0ex}}\text{K}}=0.033\phantom{\rule{0.2em}{0ex}}\text{J/K}.$
Similarly, if the gas loses 5.0 J of heat; that is, $Q=-5.0\phantom{\rule{0.2em}{0ex}}\text{J}$, at temperature $T=200\phantom{\rule{0.2em}{0ex}}\text{K}$, we have the entropy change of the system given by
$\text{Δ}S=\frac{-5.0\phantom{\rule{0.2em}{0ex}}\text{J}}{200\phantom{\rule{0.2em}{0ex}}\text{K}}=-0.025\phantom{\rule{0.2em}{0ex}}\text{J/K}.$
### Example
#### Entropy Change of Melting Ice
Heat is slowly added to a 50-g chunk of ice at $0\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ until it completely melts into water at the same temperature. What is the entropy change of the ice?
#### Strategy
Because the process is slow, we can approximate it as a reversible process. The temperature is a constant, and we can therefore use Equation 4.8 in the calculation.
#### Solution
The ice is melted by the addition of heat:
$Q=m{L}_{\text{f}}=50\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}335\phantom{\rule{0.2em}{0ex}}\text{J/g}=16.8\phantom{\rule{0.2em}{0ex}}\text{kJ}.$
In this reversible process, the temperature of the ice-water mixture is fixed at $0\phantom{\rule{0.2em}{0ex}}\text{°C}$ or 273 K. Now from $\text{Δ}S=Q\text{/}T$, the entropy change of the ice is
$\text{Δ}S=\frac{16.8\phantom{\rule{0.2em}{0ex}}\text{kJ}}{273\phantom{\rule{0.2em}{0ex}}\text{K}}=61.5\phantom{\rule{0.2em}{0ex}}\text{J/K}$
when it melts to water at $0\phantom{\rule{0.2em}{0ex}}\text{°C}$.
#### Significance
During a phase change, the temperature is constant, allowing us to use Equation 4.8 to solve this problem. The same equation could also be used if we changed from a liquid to a gas phase, since the temperature does not change during that process either.
The change in entropy of a system for an arbitrary, reversible transition for which the temperature is not necessarily constant is defined by modifying $\text{Δ}S=Q\text{/}T$. Imagine a system making a transition from state A to B in small, discrete steps. The temperatures associated with these states are ${T}_{A}$ and ${T}_{B},$ respectively. During each step of the transition, the system exchanges heat $\text{Δ}{Q}_{i}$ reversibly at a temperature ${T}_{i}.$ This can be accomplished experimentally by placing the system in thermal contact with a large number of heat reservoirs of varying temperatures ${T}_{i}$, as illustrated in Figure 4.15. The change in entropy for each step is $\text{Δ}{S}_{i}={Q}_{i}\text{/}{T}_{i}.$ The net change in entropy of the system for the transition is
$\text{Δ}S={S}_{B}-{S}_{A}=\sum _{i}\phantom{\rule{0.2em}{0ex}}\text{Δ}{S}_{i}=\sum _{i}\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{Q}_{i}}{{T}_{i}}.$
We now take the limit as $\text{Δ}{Q}_{i}\to 0$, and the number of steps approaches infinity. Then, replacing the summation by an integral, we obtain
$\text{Δ}S={S}_{B}-{S}_{A}={\int }_{A}^{B}\phantom{\rule{0.2em}{0ex}}\frac{dQ}{T},$
where the integral is taken between the initial state A and the final state B. This equation is valid only if the transition from A to B is reversible.
As an example, let us determine the net entropy change of a reversible engine while it undergoes a single Carnot cycle. In the adiabatic steps 2 and 4 of the cycle shown in Figure 4.11, no heat exchange takes place, so $\text{Δ}{S}_{2}=\text{Δ}{S}_{4}=\int \phantom{\rule{0.2em}{0ex}}dQ\text{/}T=0.$ In step 1, the engine absorbs heat ${Q}_{\text{h}}$ at a temperature ${T}_{\text{h}},$ so its entropy change is $\text{Δ}{S}_{1}={Q}_{\text{h}}\text{/}{T}_{\text{h}}.$ Similarly, in step 3, $\text{Δ}{S}_{3}=\text{−}{Q}_{\text{c}}\text{/}{T}_{\text{c}}.$ The net entropy change of the engine in one cycle of operation is then
$\text{Δ}{S}_{E}=\text{Δ}{S}_{1}+\text{Δ}{S}_{2}+\text{Δ}{S}_{3}+\text{Δ}{S}_{4}=\frac{{Q}_{\text{h}}}{{T}_{\text{h}}}-\frac{{Q}_{\text{c}}}{{T}_{\text{c}}}.$
However, we know that for a Carnot engine,
$\frac{{Q}_{\text{h}}}{{T}_{\text{h}}}=\frac{{Q}_{\text{c}}}{{T}_{\text{c}}},$
so
$\text{Δ}{S}_{E}=0.$
There is no net change in the entropy of the Carnot engine over a complete cycle. Although this result was obtained for a particular case, its validity can be shown to be far more general: There is no net change in the entropy of a system undergoing any complete reversible cyclic process. Mathematically, we write this statement as
$\oint dS=\oint \frac{dQ}{T}=0$
where $\oint$ represents the integral over a closed reversible path.
We can use Equation 4.11 to show that the entropy change of a system undergoing a reversible process between two given states is path independent. An arbitrary, closed path for a reversible cycle that passes through the states A and B is shown in Figure 4.16. From Equation 4.11, $\oint dS=0$ for this closed path. We may split this integral into two segments, one along I, which leads from A to B, the other along II, which leads from B to A. Then
${\left[{\int }_{A}^{B}dS\right]}_{\text{I}}+{\left[{\int }_{B}^{A}dS\right]}_{\text{II}}=0.$
Since the process is reversible,
${\left[{\int }_{A}^{B}dS\right]}_{\text{I}}={\left[{\int }_{A}^{B}dS\right]}_{\text{II}}.$
Hence, the entropy change in going from A to B is the same for paths I and II. Since paths I and II are arbitrary, reversible paths, the entropy change in a transition between two equilibrium states is the same for all the reversible processes joining these states. Entropy, like internal energy, is therefore a state function.
What happens if the process is irreversible? When the process is irreversible, we expect the entropy of a closed system, or the system and its environment (the universe), to increase. Therefore we can rewrite this expression as
$\text{Δ}S\ge 0,$
where S is the total entropy of the closed system or the entire universe, and the equal sign is for a reversible process. The fact is the entropy statement of the second law of thermodynamics:
### Second Law of Thermodynamics (Entropy statement)
The entropy of a closed system and the entire universe never decreases.
We can show that this statement is consistent with the Kelvin statement, the Clausius statement, and the Carnot principle.
### Example
#### Entropy Change of a System during an Isobaric Process
Determine the entropy change of an object of mass m and specific heat c that is cooled rapidly (and irreversibly) at constant pressure from ${T}_{\text{h}}$ to ${T}_{\text{c}}.$
#### Strategy
The process is clearly stated as an irreversible process; therefore, we cannot simply calculate the entropy change from the actual process. However, because entropy of a system is a function of state, we can imagine a reversible process that starts from the same initial state and ends at the given final state. Then, the entropy change of the system is given by Equation 4.10, $\text{Δ}S={\int }_{A}^{B}dQ\text{/}T.$
#### Solution
To replace this rapid cooling with a process that proceeds reversibly, we imagine that the hot object is put into thermal contact with successively cooler heat reservoirs whose temperatures range from ${T}_{\text{h}}$ to ${T}_{\text{c}}.$ Throughout the substitute transition, the object loses infinitesimal amounts of heat dQ, so we have
$\text{Δ}S={\int }_{{T}_{\text{h}}}^{{T}_{\text{c}}}\frac{dQ}{T}.$
From the definition of heat capacity, an infinitesimal exchange dQ for the object is related to its temperature change dT by
$dQ=mc\phantom{\rule{0.2em}{0ex}}dT.$
Substituting this dQ into the expression for $\text{Δ}S$, we obtain the entropy change of the object as it is cooled at constant pressure from ${T}_{\text{h}}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{T}_{\text{c}}:$
$\text{Δ}S={\int }_{{T}_{\text{h}}}^{{T}_{\text{c}}}\frac{mc\phantom{\rule{0.2em}{0ex}}dT}{T}=mc\phantom{\rule{0.2em}{0ex}}\text{ln}\frac{{T}_{\text{c}}}{{T}_{\text{h}}}.$
Note that $\text{Δ}S<0$ here because ${T}_{\text{c}}<{T}_{\text{h}}.$ In other words, the object has lost some entropy. But if we count whatever is used to remove the heat from the object, we would still end up with $\text{Δ}{S}_{\text{universe}}>0$ because the process is irreversible.
#### Significance
If the temperature changes during the heat flow, you must keep it inside the integral to solve for the change in entropy. If, however, the temperature is constant, you can simply calculate the entropy change as the heat flow divided by the temperature.
### Example
#### Stirling Engine
The steps of a reversible Stirling engine are as follows. For this problem, we will use 0.0010 mol of a monatomic gas that starts at a temperature of $133\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and a volume of $0.10\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$, which will be called point A. Then it goes through the following steps:
1. Step AB: isothermal expansion at $133\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ from $0.10\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$ to $0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$
2. Step BC: isochoric cooling to $33\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$
3. Step CD: isothermal compression at $33\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ from $0.20\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$ to $0.10\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$
4. Step DA: isochoric heating back to $133\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ and $0.10\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}$
(a) Draw the pV diagram for the Stirling engine with proper labels.
(b) Fill in the following table.
Step W (J) Q (J) $\text{Δ}S$ (J/K)
Step AB
Step BC
Step CD
Step DA
Complete cycle
(c) How does the efficiency of the Stirling engine compare to the Carnot engine working within the same two heat reservoirs?
#### Strategy
Using the ideal gas law, calculate the pressure at each point so that they can be labeled on the pV diagram. Isothermal work is calculated using $W=nRT\phantom{\rule{0.2em}{0ex}}\text{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right),$ and an isochoric process has no work done. The heat flow is calculated from the first law of thermodynamics, $Q=\text{Δ}{E}_{\text{int}}-W$ where $\text{Δ}{E}_{\text{int}}=\frac{3}{2}nR\text{Δ}T$ for monatomic gasses. Isothermal steps have a change in entropy of Q/T, whereas isochoric steps have $\text{Δ}S=\frac{3}{2}nR\phantom{\rule{0.2em}{0ex}}\text{ln}\left(\frac{{T}_{2}}{{T}_{1}}\right).$ The efficiency of a heat engine is calculated by using ${e}_{\text{Stir}}=W\text{/}{Q}_{\text{h}}.$
#### Solution
1. The graph is shown below.
2. The completed table is shown below.
Step W (J) Q (J) $\text{Δ}S$ (J/K)
Step AB Isotherm 2.3 2.3 0.0057
Step BC Isochoric 0 –1.2 0.0035
Step CD Isotherm –1.8 –1.8 –0.0059
Step DA Isochoric 0 1.2 –0.0035
Complete cycle 0.5 0.5 ~ 0
• The efficiency of the Stirling heat engine is
${e}_{\text{Stir}}=W\text{/}{Q}_{\text{h}}=\left({Q}_{AB}+{Q}_{CD}\right)\text{/}\left({Q}_{AB}+{Q}_{DA}\right)=0.5\text{/}4.5=0.11.$
If this were a Carnot engine operating between the same heat reservoirs, its efficiency would be
${e}_{\text{Car}}=1-\left(\frac{{T}_{\text{c}}}{{T}_{\text{h}}}\right)=0.25.$
Therefore, the Carnot engine would have a greater efficiency than the Stirling engine.
#### Significance
In the early days of steam engines, accidents would occur due to the high pressure of the steam in the boiler. Robert Stirling developed an engine in 1816 that did not use steam and therefore was safer. The Stirling engine was commonly used in the nineteenth century, but developments in steam and internal combustion engines have made it difficult to broaden the use of the Stirling engine.
The Stirling engine uses compressed air as the working substance, which passes back and forth between two chambers with a porous plug, called the regenerator, which is made of material that does not conduct heat as well. In two of the steps, pistons in the two chambers move in phase.
### Summary
• The change in entropy for a reversible process at constant temperature is equal to the heat divided by the temperature. The entropy change of a system under a reversible process is given by $\text{Δ}S={\int }_{A}^{B}dQ\text{/}T$.
• A system’s change in entropy between two states is independent of the reversible thermodynamic path taken by the system when it makes a transition between the states.
### Conceptual Questions
Does the entropy increase for a Carnot engine for each cycle?
Is it possible for a system to have an entropy change if it neither absorbs nor emits heat during a reversible transition? What happens if the process is irreversible?
Show Solution
Entropy will not change if it is a reversible transition but will change if the process is irreversible.
### Problems
Two hundred joules of heat are removed from a heat reservoir at a temperature of 200 K. What is the entropy change of the reservoir?
Show Solution
–1 J/K
In an isothermal reversible expansion at $27\phantom{\rule{0.2em}{0ex}}\text{°C}$, an ideal gas does 20 J of work. What is the entropy change of the gas?
An ideal gas at 300 K is compressed isothermally to one-fifth its original volume. Determine the entropy change per mole of the gas.
Show Solution
–13 J(K mole)
What is the entropy change of 10 g of steam at $100\phantom{\rule{0.2em}{0ex}}\text{°C}$ when it condenses to water at the same temperature?
A metal rod is used to conduct heat between two reservoirs at temperatures ${T}_{\text{h}}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{T}_{\text{c}},$ respectively. When an amount of heat Q flows through the rod from the hot to the cold reservoir, what is the net entropy change of the rod, the hot reservoir, the cold reservoir, and the universe?
Show Solution
$-\frac{Q}{{T}_{\text{h}}},\frac{Q}{{T}_{\text{c}}},Q\left(\frac{1}{{T}_{\text{c}}}-\frac{1}{{T}_{\text{h}}}\right)$
For the Carnot cycle of Figure 4.12, what is the entropy change of the hot reservoir, the cold reservoir, and the universe?
A 5.0-kg piece of lead at a temperature of $600\phantom{\rule{0.2em}{0ex}}\text{°C}$ is placed in a lake whose temperature is $15\phantom{\rule{0.2em}{0ex}}\text{°C}$. Determine the entropy change of (a) the lead piece, (b) the lake, and (c) the universe.
Show Solution
a. –540 J/K; b. 1600 J/K; c. 1100 J/K
One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?
One mole of an ideal monatomic gas is confined to a rigid container. When heat is added reversibly to the gas, its temperature changes from ${T}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{T}_{2}.$ (a) How much heat is added? (b) What is the change in entropy of the gas?
Show Solution
a. $Q=nR\text{Δ}T$; b. $S=nR\phantom{\rule{0.2em}{0ex}}\text{ln}\left({T}_{2}\text{/}{T}_{1}\right)$
(a) A 5.0-kg rock at a temperature of $20\phantom{\rule{0.2em}{0ex}}\text{°C}$ is dropped into a shallow lake also at $20\phantom{\rule{0.2em}{0ex}}\text{°C}$ from a height of $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}$. What is the resulting change in entropy of the universe? (b) If the temperature of the rock is $100\phantom{\rule{0.2em}{0ex}}\text{°C}$ when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that $c=860\phantom{\rule{0.2em}{0ex}}\text{J/kg}·\text{K}$ is the specific heat of the rock.
### Glossary
entropy
state function of the system that changes when heat is transferred between the system and the environment
entropy statement of the second law of thermodynamics
entropy of a closed system or the entire universe never decreases | 5,172 | 17,018 | {"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-2024-38 | latest | en | 0.866335 |
https://tbc-python.fossee.in/convert-notebook/Generation_Of_Electrical_Energy_by_B._R._Gupta/Chapter12.ipynb | 1,628,094,873,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00442.warc.gz | 566,918,489 | 38,492 | # Ch-12, Parallel Operation of alternators¶
## example 12.1 Page 243¶
In [1]:
p=4000 #given kva of alternator
fnl2=50 #frequency on no load on second alternator
fl2=48 #frequency on load on second alternator
l=6000 #load given two to alternator
df1=fnl1-fl1 #change in 1 alternator frequency
df2=fnl2-fl2 #change in 2 alternator frequency
print 'a'
l2=l-l1
print " load on 1 alternator %.2fkW \n load on 2 alternator %.2fkW"%(l1,l2)
ll=ml1+p
print 'b'
print " load supplied by machine 1 with full load on machine2 %dkW \n total load is %dkW"%(ml1,l1)
a
b
## example 12.2 page 243¶
In [2]:
from math import sqrt, atan, acos, pi, sin
pf1=0.8 #pf on 1 machine
i2=150 #current on 2 machine
z1=0.4+12*1J #synchronour impedence
z2=0.5+10*1J
vt=6.6 #terminal voltage
al=l1/2 #active load on each machine
cosdb=al/(vt*i2*sqrt(3)) #cos db
db=acos(cosdb)*180/pi #angle in digree
ib=i2*complex(cosdb,-sin(db*pi/180)) #current in complex number
it=l1/(vt*pf1*sqrt(3)) #total current
itc=complex(it*pf1,-it*sin(acos(pf1))) #total current in complex
ia=itc-ib
pfa=atan(ia.imag/ia.real) #pf of current a
ea=(vt/sqrt(3))+ia*(z1)/1000 #voltage a
pha=atan(ea.imag/ea.real)*180/pi #phase angle of unit a
print "induced emf of a machine a %.2f+%.2fi =%fkV per phase"%(ea.real,ea.imag,abs(ea))
eb=(vt/sqrt(3))+ib*(z2)/1000 #voltage b
phb=atan(eb.imag/eb.real)*180/pi #phase angle of unit b
print "\ninduced emf of a machine b %.2f+%.2fi =%fkV per phase"%(eb.real,eb.imag,abs(eb))
induced emf of a machine a 5.35+1.52i =5.565708kV per phase
induced emf of a machine b 4.60+1.28i =4.776461kV per phase
## example 12.3 Page 244¶
In [3]:
from math import cos,pi
e1=3000 ;ph1=20 ;e2=2900; ph2=0 #given induced emf of two machines
z1=2+20*1J ;z2=2.5+30*1J #impedence of two synchronous machine
e11=e1*(cos(ph1*pi/180)+sin(ph1*pi/180)*1J)
e22=e2*(cos(ph2*pi/180)+sin(ph2*pi/180)*1J)
Is=(e11-e22)*zl/(z1*z2+(z1+z2)*zl)
print "current is %.2f%.2fiA =%.2fA"%((Is).real,(Is).imag,abs(Is))
current is 10.37-4.56iA =11.33A
## example 12.4 Page 244¶
In [4]:
from math import sqrt, atan, acos, pi, sin
e1=250 ;e2=250 #emf of generator
z1=2*1J; z2=2*1J #synchronous impedence
v=(e1*z2+z1*e2)/((z1*z2/z)+z1+z2)
vph=atan(v.imag/v.real)*180/pi #substitution the value in equation 12.10
i1=(z2*e1+(e1-e2)*z)/(z1*z2+(z1+z2)*z); iph=atan((i1).imag/(i1).real)*180/pi #substitution the value in equation 12.7
pf1=cos(pi/180*(vph-iph))
pd=v*i1*pf1
e1=250 ;e2=250 #emf of generator
z1=2*1J; z2=2*1J #synchronous impedence
v=(e1*z2+z1*e2)/((z1*z2/z)+z1+z2)
vph=atan(v.imag/v.real)*180/pi #substitution the value in equation 12.10
i1=(z2*e1+(e1-e2)*z)/(z1*z2+(z1+z2)*z)
iph=atan(i1.imag/i1.real)*180/pi #substitution the value in equation 12.7
pf1=cos(pi/180*(vph-iph))
pd=v*i1*pf1
print "terminal voltage %.2fV \ncurrent supplied by each %.2fA \npower factor of each %.3f lagging \npower delivered by each %.4fKW"%(abs(v),abs(i1),abs(pf1),abs(pd))
terminal voltage 239.68V
current supplied by each 10.72A
power factor of each 0.894 lagging
power delivered by each 2297.7941KW
## example 12.5 Page 247¶
In [5]:
from math import sqrt, atan, acos, pi, sin
po=5 #mva rating
v=10 #voltage in kv
n=1500 ;ns=n/60 #speed
f=50 #freaquency
pfb=0.8#power factor in b
x=0.2*1J #reactance of machine
md=0.5 #machanical displacement
v=1 ;e=1
p=4
spu=v*e/abs(x); sp=spu*po*1000 ;mt=(pi*p)/(180*2)
spm=sp*mt #synchronous power in per mech.deree
st=spm*md*1000/(2*ns*pi)
print '(a)'
print " synchronous power %dkW \n synchronous torque for %.1f displacement %dN-M"%(spm,md,st)
ee=e+x*(pfb-sin(acos(pfb))*1J)
spb=v*abs(ee)*cos(atan(ee.imag/ee.real))/abs(x) #synchronous power
sppm=spb*po*1000*mt #synchronous power per mech.degree
print " synchronous power %dkW \n synchronous torque for %.1f displacement %dN-M"%(sppm,md,stp)
(a)
synchronous power 872kW
synchronous torque for 0.5 displacement 2777N-M
synchronous power 977kW
synchronous torque for 0.5 displacement 3111N-M
## example 12.6 page 248¶
In [6]:
from math import sqrt, atan, acos, pi, sin
po=2*10**6 ;p=8 ;n=750; v=6000 ;x=6*1J ;pf=0.8 #given
i=po/(v*sqrt(3))
e=(v/sqrt(3))+i*x*(pf-sin(acos(pf))*1J)
mt=p*pi/(2*180)
cs=cos(atan(e.imag/e.real))
ps=abs(e)*v*sqrt(3)*cs*mt/(1000*abs(x))
ns=n/60
ts=ps*1000/(2*pi*ns)
print " synchronous power %.1fkW per mech.degree \n synchrounous torque %dN-m"%(ps,ts)
synchronous power 502.7kW per mech.degree
synchrounous torque 6666N-m
## example 12.7 page 248¶
In [7]:
from math import sqrt, atan, acos, pi, sin
i=100 ;pf=-0.8 ;v=11*1000 ;x=4*1J ;ds=10; pfc=-0.8 #given,currents,power factor,voltage,reactance,delta w.r.t steem supply,pf of alternator
e=(v/sqrt(3))+(i*x*(pf-sin(acos(pf))*1J))
print 'a'
ph=atan(e.imag/e.real)*180/pi
print " open circuit emf %dvolts per phase and %.2f degree"%(abs(e),ph)
d=ds-ph
eee=round(abs(e)/100)*100
ic=round(abs(eee)*sin(d*pi/180)/abs(x))
iis=(eee**2-(abs(x)*ic)**2)**(0.5)
Is=(iis-v/sqrt(3))/abs(x)
ii=ic/cos(d*pi/180)
pff=cos(d*pi/180)
print 'b.'
print " current %.1fA \n power factor %.3f"%(ii,pff)
print 'c.'
ia=ii*pff/abs(pfc)
print "current %.2fA"%(ia)
ia=ii*pff/abs(pfc)
print "current %.2fA"%(ia)
a
open circuit emf 6598volts per phase and -2.78 degree
b.
current 365.6A
power factor 0.998
c.
current 456.25A
current 456.25A | 2,164 | 5,255 | {"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": 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} | 3.859375 | 4 | CC-MAIN-2021-31 | latest | en | 0.446891 |
https://mathsquery.com/algebra/equations/linear-equation/ | 1,726,633,839,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00409.warc.gz | 361,287,094 | 12,228 | ## Maths units list
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Maths Query > Unit > Algebra > Equations
# Linear Equation with Examples and Methods to Solve
## Introduction
Before we take a deep dive into linear equations, let us shed some light on the general term equations used in mathematics.
### What is an equation ?
An equation is a condition in which two algebraic expressions are equal.
When two algebraic expressions involve sign of equality, it is called statement of equality.
The expression written on left side of the sign of equality (=) is called left hand side (LHS) and expression written on the right of equality (=) is called right hand side (RHS).
The equality sign shows value of left hand side is equal to value of right hand side.
In addition to LHS expression, RHS expression and the equality sign, at least one of the two expressions always contain minimum one variable.
Example
Example 1
x + 8 = 14
x + 8 = 14 is an equation in which variable x has the highest power of 1.
Here, LHS expression is x + 8
and RHS expression is 14.
Example 2
x2 + 4x + 4 = 0
x2 + 4x + 4 = 0 is an equation in which variable x has the highest power of 2.
Here, LHS expression is x2 + 4x + 4
and RHS expression is 0.
## What is linear equation?
An equation is called a linear equation in which the highest power of a variable is 1.
Example
Example 1
y + 7 = 8
y + 7 = 8 is linear equation because variable y has the highest power of 1.
Example 2
x + 5 = 8
x + 5 = 8 ia also a linear equation because it has one variable x and its the highest power is 1.
## Methods to solve linear equation
### 1.Trail and error method
In this method, the value of an unknown variable in an equation is calculated by putting different values of the variable, one at a time, in the equation.
By putting values of variable in LHS and in RHS, both sides are checked, if they become equal.
If LHS and RHS becomes equal for a value then that value of variable is called root of equation or solution of equation.
Example
Solve equation 2x + 5 = 15.
In 2x + 5 = 15 equation, LHS is 2x + 5 and RHS is 15.
LHS has variable x.
To solve the above equation, different values of variable x will be used to check which makes LHS 2x + 5 equal to RHS 15.
Value of x are put in LHS until both LHS and RHS become equal.
Table: Solving of equation 2x + 5 = 15 for different values of x
Value of xLHS of equationRHS of equationIs LHS = RHS ?
12(1) + 5 = 715No
22(2) + 5 = 915No
32(3) + 5 = 1115No
42(4) + 5 = 1315No
52(5) + 5 = 1515Yes
Hence, LHS = RHS when for the value of x = 5
Therefore, x = 5 is solution of equation 2x + 5 = 15.
### 2. Systematic method
The above trial and error method to find solution of equation is a bit time consuming for cases when many values of a variable have to be used until LHS becomes equal to RHS.
Systematic method is considered a better method than trial and error method to find value of variable because it can solve the equation with lesser number of steps than trial and error method.
Solving and equation using systematic method involves following the minimum steps to be taken:
1. A same number can be added to both sides of an equation without changing the equality.
2. A same number can be subtracted from both sides of an equation without changing the equality.
3. A same number (a non zero number) can be multiplied on both sides of an equation without changing the equality.
4. Both sides of equation can be divided by a same number (a non zero number) without changing the equality .
Example
Solve equation 2x + 5 = 15.
2x + 5 – 5 = 15 – 5 (subtract 5 on both sides)
2x = 10
$\frac{2x}{2}=\frac{10}{2}$ (Now divide both sides by 2)
x = 5
x = 5 is solution of equation.
### 3. Transposition method
In this method, variables are shifted to LHS and constants are shifted to RHS of an equation. By transposition of terms means a term is shifted to another side alongwith its sign changed.
It means sign of addition is changed to sign of subtraction and sign of subtraction is changed after shifting the term on the other side.
Example
Solve equation 2x + 5 = 15.
Shift variables on LHS and constants on RHS of the equation
∴ 2x = 15 – 5
2x = 10
$x=\frac{10}{2}$
x = 5
∴ x = 5 is solution of linear equation.
Let’s take another example
Solve equation 6x + 5 = 3x + 20
Shift variables on LHS and constants on RHS of the equation
∴ 6x – 3x = 20 – 5
3x = 15
$x=\frac{15}{3}$
x = 5
∴ x = 5 is solution of linear equation.
## Solved Examples
### 1) Solve the equation -2y + 2 = -10.
-2y + 2 = -10
-2y = -10 - 2
-2y = -12
y = $\frac{-12}{-2}$
y = 6
2x - 3x = -9 + 7
-x = -2
x = 2
### 3) Solve the linear equation $\frac{4x -3}{2}=\frac{9 + x}{3}$
Cross multiply
3(4x - 3) = 2(9 + x)
12x - 9 = 18 + 2x
12x - 2x = 18 + 9
10x = 27
x = $\frac{27}{10}$
x = 2.7
### 4) Solve $\frac{x}{2}+2=\frac{x}{3}+4$
$\frac{x}{2}-\frac{x}{3}=4-2$
$\frac{3x - 2x}{6}=2$
$\frac{x}{6}=2$
x = 2 × 6
x = 12
### 5) 8(2x - 3) - 5(4x - 3) = 1
16x - 24 - 20x + 15 = 1
-4x - 9 = 1
-4x = 1 + 9
-4x = 10
x = - $\frac{10}{4}$
x = -2.5
### 6) $\frac{\mathrm{4x - 1}}{2}+1$ = $\frac{\mathrm{x - 1}}{5}+16$
$\frac{\mathrm{4x - 1}}{2}+1$ = $\frac{\mathrm{x - 1}}{5}+16$
$\frac{\mathrm{4x - 1}}{2}-\frac{\mathrm{\left(x - 1\right)}}{5}$ = 16 - 1
$\frac{\mathrm{5\left(4x - 1\right) - 2\left(x -1\right)}}{10}$ = 15
20x - 5 - 2x + 2 = 15
18x - 3 = 15
18x = 15 + 3
18x = 18
x = $\frac{18}{18}$
x = 1
### 7) $\frac{x}{3}$ + $\frac{\mathrm{x - 1}}{2}$ + $\frac{\mathrm{x - 1}}{4}$ = 2
$\frac{\mathrm{4x + 6\left(x - 2\right) + 3\left(x - 1\right)}}{12}$ = 2
$\frac{\mathrm{4x + 6x - 12\right) + 3x - 3}}{12}$ = 2
13x - 15 = 24
13x = 24 + 15
13x = 39
x = $\frac{39}{13}$
x = 3
### 8) $\frac{\mathrm{x - 1}}{2}$ + $\frac{\mathrm{x - 2}}{4}$ = 2
$\frac{\mathrm{2\left(x - 1\right) + \left(x - 2\right)}}{4}$ = 2
$\frac{\mathrm{2x - 2 + x - 2}}{4}$ = 2
$\frac{\mathrm{3x - 4}}{4}$ = 2
3x - 4 = 2 × 4
3x - 4 = 8
3x = 8 + 4
3x = 12
x = $\frac{12}{3}$
x = 4
### 9) 0.6x + 7.2 = 0.2x + 9.6
0.6x - 0.2x = 9.6 - 7.2
0.4x = 2.4
x = $\frac{2.4}{0.4}$
x = 6
### 10) 6.3x + 18 = 2.3x + 30
6.3x - 2.3x = 30 - 18
4.0x = 12
4x = 12
x = $\frac{12}{4}$
x = 3
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## Worksheet 3
### Multiple choice questions
1. 4
2. -4
3. -1
4. -3
1. 2
2. -5
3. 10
4. -10
#### 3) What is yhe value of x in 15x = -1?
1. -1
2. 15
3. - $\frac{1}{15}$
4. $\frac{1}{15}$
1. 7
2. -35
3. -5
4. 5
1. 28
2. -28
3. 0
4. 1
1. 25
2. -25
3. -50
4. 50
1. 2
2. -12
3. -2
4. 12
#### 8) Which value of x does satisfy $\frac{x}{2}=\frac{5}{12}$?
1. $\frac{2}{5}$
2. $\frac{5}{2}$
3. $\frac{6}{5}$
4. $\frac{5}{6}$
1. 1
2. -1
3. 1.5
4. 2.5
#### 10) The value of y for 3(y + 15) = 2y - 10
1. 35
2. -45
3. -55
4. 55
Last updated on: 06-08-2024 | 2,594 | 6,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 46, "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} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.935184 |
https://mmerevise.co.uk/a-level-chemistry-revision/the-arrhenius-equation/ | 1,720,920,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00475.warc.gz | 348,387,579 | 51,965 | # The Arrhenius Equation
A LevelAQA
## The Arrhenius Equation
The Arrhenius Equation provides a mathematical formula with which to investigate the relationship between the rate constant and temperature. The Arrhenius equation also introduces another constant known as the Arrhenius constant.
## The Arrhenius Equation
Increasing the temperature of a reaction increases the value of the rate constant. The Arrhenius Equation shows the effect of temperature on the value of $\text{k}$.
$\text{k}=\text{Ae}^{-\frac{\text{E}_a}{\text{RT}}}$
If we inspect this equation, we see that it is mostly made up of constants, with $\text{A}$ (the Arrhenius constant), $\text{E}_{a}$ (activation energy), and $\text{R}$ (the molar gas constant). We also see that the rate constant will increase exponentially with temperature. When plotted, this expression will yield a curve with a gradient gets increasingly steep as the temperature rises.
Though this curve demonstrates the relationship between the rate constant and temperature, it would not be easy to use it to find any of the constants contained with in the Arrhenius equation.
We can however rearrange the equation to put it into the form $y=mx+c$. To do this we take the natural logarithm of the equation. This will remove the exponential from the equation, and bring the $-\frac{E_a}{RT}$ down:
## $\text{ln k}=\text{ln A}-\frac{\text{E}_a}{\text{RT}}$
We can now plot a graph of $ln k$ against $\frac{1}{T}$. This will produce a straight line with a slope equal to $-\frac{E_a}{R}$.
A LevelAQA
## Example 1: Calculating The Arrhenius Constant
A reaction is carried out at $\textcolor{#00bfa8}{30\degree \text{C}}$. The value of $\text{k}$ for the reaction is $\textcolor{#f21cc2}{3.92\times 10^{-8}\text{s}^{-1}}$ and the activation energy is $\textcolor{#a233ff}{89.9\text{ kJ mol}^{-1}}$. The gas constant is $\text{R} = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$.
Calculate the value for the Arrhenius Constant for the reaction.
[3 marks]
\begin{aligned}\text{k}&=\text{Ae}^{-\frac{\text{E}_a}{\text{RT}}}\\\text{}\\\text{A}&=\frac{\text{k}}{\text{e}^{-\frac{\text{E}_a}{\text{RT}}}}\\\text{}\\&=\frac{\textcolor{#f21cc2}{3.92\times 10^{-8}}}{\text{e}^{-\frac{\textcolor{#a233ff}{89900}}{8.31\times\textcolor{#00bfa8}{303}}}}\\\text{}\\ &=\textcolor{#008d65}{1.26 \times 10^8 \text{ s}^{-1}}\end{aligned}
Note: the units of the Arrhenius constant will always be the same as those of the rate constant. The exponential factor $\left(\text{e}^{\frac{\text{E}_a}{\text{RT}}}\right)$ of the equation has no units as the units of activation energy, temperature and the gas constant are all the same.
A LevelAQA
## Example 2: Graphical Calculations
Using the following data, determine the activation energy, in $\text{kJ mol}^{-1}$, and the value of $\text{A}$, given the following data.
$\text{T /}\degree \text{C}$ $\text{k /mol}^{-1}\text{ dm}^{3}\text{ s}^{-1}$ $\text{T /K}$ $\frac{1}{\text{T}}\text{ /k}^{-1}$ $ln k$ $10$ $0.318$ $20$ $0.551$ $30$ $1.03$ $40$ $1.67$ $50$ $2.07$
[12 marks]
Step 1: Complete the table.
$\text{T /}\degree \text{C}$ $\text{k /mol}^{-1}\text{ dm}^{3}\text{ s}^{-1}$ $\text{T /K}$ $\frac{1}{\text{T}}\text{ /k}^{-1}$ $\text{ln k}$ $10$ $0.318$ $283$ $3.53 \times10^{-3}$ $-1.15$ $20$ $0.551$ $293$ $3.41\times10^{-3}$ $-0.596$ $30$ $1.03$ $303$ $3.30\times10^{-3}$ $0.030$ $40$ $1.67$ $313$ $3.19\times10^{-3}$ $0.513$ $50$ $2.07$ $323$ $3.10\times10^{-3}$ $0.728$
Step 2: Plot the Data.
Step 3: Calculate the slope of the line.
\begin{aligned}\text{Slope}&=\frac{\Delta\text{Y}}{\Delta\text{X}}\\ &=\frac{0.75-(-0.30)}{3.45\times10^{-3}-3.15\times10^{-3}}\\ &=\textcolor{#008d65}{-3500\text{ K}}\end{aligned}
Step 4: Calculate $\text{E}_a$.
\begin{aligned}\text{Slope}&=\frac{-\text{E}_a}{\text{R}}\\ \text{E}_a &=-\text{Slope}\times\text{R}\\ &=-(-4350)\times8.31\\ &=\textcolor{#008d65}{29.1\text{ kJ mol}^{-1}}\end{aligned}
Step 4: Calculate the value of the Arrhenius constant.
\begin{aligned}\text{ln A} &= \text{y-intercept}\\\text{A}&=\text{e}^{\text{y-intercept}}\\ &=\text{e}^{14.6}\\ &=\textcolor{#008d65}{2.19\times10^{6}}\end{aligned}
A LevelAQA
## The Arrhenius Equation Example Questions
$\text{ln k}=\text{ln A} - \frac{\text{E}_a}{\text{RT}}$ OR $\text{E}_a=(\text{ln A} -\text{ln k})\times\text{RT}$
\begin{aligned}\text{E}_a&=(\text{ln }1.20\times10^{7} -\text{ln } 4.10\times10^{-3})\times8.31\times313\\&=56.7\end{aligned}
$\text{kJ mol}^{-1}$
$\text{ln k}=\text{ln A} -\frac{\text{E}_a}{\text{RT}}$ OR $\text{T}=\frac{\text{E}_a}{\text{R}\times(\text{ln A}-\text{ln k})}$
\begin{aligned}\text{T}&=\frac{100000}{(8.31\times\text{ln} 6.40\times10^{13}-\text{ln} 2.210\times10^{-3})}\\ &=317\text{K}\end{aligned}
Step 1: Complete the Table.
$\text{T /}\degree \text{C}$ $\text{k /mol}^{-1}\text{ dm}^{3}\text{ s}^{-1}$ $\text{T /K}$ $\frac{1}{\text{T}}\text{ /k}^{-1}$ $\text{ln k}$ $47$ $3.30\times10^{-4}$ $320$ $\textcolor{#008d65}{3.13\times10^{-3}}$ $-8.02$ $81$ $1.61\times10^{-3}$ $354$ $2.82\times10^{-3}$ $\textcolor{#008d65}{-6.43}$ $108$ $4.00\times10^{-2}$ $\textcolor{#008d65}{381}$ $2.62\times10^{-3}$ $-3.22$ $204$ $8.26\times10^{-1}$ $447$ $2.10\times10^{-3}$ $\textcolor{#008d65}{-0.191}$ $300$ $7.56$ $537$ $1.75\times10^{-3}$ $2.02$
Step 2: Plot the data.
Step 3: Calculate the gradient of the graph.
$\text{Gradient} = -7.44\times10^{-3}$
(Accept any from $-7.3\times10^{-3}$ to $-7.6\times10^{-3})$
Step 4: Calculate $\text{E}_a$.
\begin{aligned}\text{E}_a &= -7.44\times10^{-3} \times 8.31\\ &= 61.8\text{ kJ mol}^{-1}\end{aligned}
(Four marks for step 1, one mark per correctly calculated data. Two marks for step 2, allow both marks for correctly plotted graph from incorrect data. One Mark for step 3. Two marks for Step 4, one mark for correct calculation, one mark for correct answer.)
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£0.00 | 2,186 | 6,099 | {"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": 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": 146, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.798221 |
http://lisalumakeup4you.com/joyalukkas-online-teipy/7cc80b-simple-graph-with-5-vertices-and-3-edges | 1,632,628,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00683.warc.gz | 37,024,867 | 10,350 | In this sense, planar graphs are sparse graphs, in that they have only O(v) edges, asymptotically smaller than the maximum O(v 2). Show that every simple graph has two vertices of the same degree. no connected subgraph of G has C as a subgraph and contains vertices or edges that are not in C (i.e. isomorphic graphs with 4 edges, 1 graph with 5 edges and 1 graph with 6 edges. (b) A simple graph with five vertices with degrees 2, 3, 3, 3, and 5. Calculating Total Number Of Edges (e)- By sum of degrees of vertices theorem, we have- Sum of degrees of all the vertices = 2 x Total number of edges. The main difference … 3. 3.1. For a simple, connected, planar graph with v vertices and e edges and f faces, the following simple conditions hold for v ≥ 3: Theorem 1. e ≤ 3v − 6; Theorem 2. The simplest is a cycle, $$C_n$$: this has only $$n$$ edges but has a Hamilton cycle. If the degree of each vertex in the graph is two, then it is called a Cycle Graph. Construct a simple graph G so that VC = 4, EC = 3 and minimum degree of every vertex is atleast 5. Give the matrix representation of the graph H shown below. Do not label the vertices of your graphs. Degree of a Vertex : Degree is defined for a vertex. Place work in this box. There is a closed-form numerical solution you can use. There does not exist such simple graph. If there are no cycles of length 3, then e ≤ 2v − 4. 3. C. Less than 8. Justify your answer. B Contains a circuit. How many vertices will the following graphs have if they contain: (a) 12 edges and all vertices of degree 3. Find the number of vertices with degree 2. $$K_5$$ has 5 vertices and 10 edges, so we get \begin{equation*} 5 - 10 + f = 2 \end{equation*} which says that if the graph is drawn without any edges crossing, there would be $$f = 7$$ faces. Prove that a nite graph is bipartite if and only if it contains no … An edge connects two vertices. 1.11 Consider the graphs G 1 = (V 1;E 1) and G 2 = (V 2;E 2). Theoretical Idea . Then the graph must satisfy Euler's formula for planar graphs. Each face must be surrounded by at least 3 edges. … Simple Graphs I Graph contains aloopif any node is adjacent to itself I Asimple graphdoes not contain loops and there exists at most one edge between any pair of vertices I Graphs that have multiple edges connecting two vertices are calledmulti-graphs I Most graphs we will look at are simple graphs Instructor: Is l Dillig, CS311H: Discrete Mathematics Introduction to Graph Theory 6/31 I Two nodes u … Now you have to make one more connection. Give an example of a simple graph G such that VC EC. Draw all non-isomorphic simple graphs with 5 vertices and 0, 1, 2, or 3 edges; the graphs need not be connected. A simple graph is a graph that does not contain multiple edges and self loops. 2 Terminology, notation and introductory results The sets of vertices and edges of a graph Gwill be denoted V(G) and E(G), respectively. The edge is said to … B. View Answer Answer: 6 30 A graph is tree if and only if A Is planar . The graph K 3,3, for example, has 6 vertices, … Now, for a connected planar graph 3v-e≥6. Graphs; Discrete Math: In a simple graph, every pair of vertices can belong to at most one edge and from this, we can estimate the maximum number of edges for a simple graph with {eq}n {/eq} vertices. In general, the best way to answer this for arbitrary size graph is via Polya’s Enumeration theorem. A simple graph is a nite undirected graph without loops and multiple edges. You have 8 vertices: I I I I. An undirected graph C is called a connected component of the undirected graph G if 1).C is a subgraph of G; 2).C is connected; 3). We can create this graph as follows. 5. However, this simple graph only has one vertex with odd degree 3, which contradicts with the … Start with 4 edges none of which are connected. Notation − C n. Example. (c)Find a simple graph with 5 vertices that is isomorphic to its own complement. The size of the minimum vertex cover of G is 8. True False 1.2) A complete graph on 5 vertices has 20 edges. Now consider how many edges surround each face. In graph theory, graphs can be categorized generally as a directed or an undirected graph.In this section, we’ll focus our discussion on a directed graph. Solution: Background Explanation: Vertex cover is a set S of vertices of a graph such that each edge of the graph is incident to at least one vertex of S. Independent set of a graph is a set of vertices such … The graph is undirected, i. e. all its edges are bidirectional. In the beginning, we start the DFS operation from the source vertex . Graph II has 4 vertices with 4 edges which is forming a cycle 'pq-qs-sr-rp'. True False Ex 5.3.3 The graph shown below is the Petersen graph. The vertices x and y of an edge {x, y} are called the endpoints of the edge. C 5. 27/10/2020 – Network Flows and Matrix Representations Max Flow Min Cut Theorem Given any network the maximum flow possible between any two vertices A and B is equal to the minimum of the … D Is completely connected. The vertices will be labelled from 0 to 4 and the 7 weighted edges (0,2), (0,1), (0,3), (1,2), (1,3), (2,4) and (3,4). If G is a connected graph, then the number of bounded faces in any embedding of G on the plane is equal to A 3 . A simple approach is to one by one remove all edges and see if removal of an edge causes disconnected graph. True False 1.5) A connected component of an acyclic graph is a tree. Continue on back if needed. Since through the Handshaking Theorem we have the theorem that An undirected graph G =(V,E) has an even number of vertices of odd degree. That means you have to connect two of the edges to some other edge. One example that will work is C 5: G= ˘=G = Exercise 31. A simple graph with 'n' vertices (n >= 3) and 'n' edges is called a cycle graph if all its edges form a cycle of length 'n'. Number of vertices x Degree of each vertex = 2 x Total … WUCT121 Graphs: Tutorial Exercise Solutions 3 Question2 Either draw a graph with the following specified properties, or explain why no such graph exists: (a) A graph with four vertices having the degrees of its vertices 1, 2, 3 and 4. You have to "lose" 2 vertices. Then the graph must satisfy Euler's formula for planar graphs. 1.12 Prove or disprove the following statements: 1)If G 1 and G 2 are regular graphs, then G 1 G 2 is regular. The graph is connected, i. e. it is possible to reach any vertex from any other vertex by moving along the edges of the graph. A. So, there are no self-loops and multiple edges in the graph. Then, the size of the maximum independent set of G is. Now consider how many edges surround each face. 4. A simple graph contains 35 edges, four vertices of degree 5, five vertices of degree 4 and four vertices of degree 3. Does it have a Hamilton cycle? (5 points, 1 point for each) True/False Questions 1.1) In a simple graph on n vertices, the degree of a vertex is at most n - 1. All graphs in these notes are simple, unless stated otherwise. f(1;2);(3;2);(3;4);(4;5)g De nition 1. There are no edges from the vertex to itself. # Create a directed graph g = Graph(directed=True) # Add 5 vertices g.add_vertices(5). 8. On the other hand, figure 5.3.1 shows … A graph (sometimes called undirected graph for distinguishing from a directed graph, or simple graph for distinguishing from a multigraph) is a pair G = (V, E), where V is a set whose elements are called vertices (singular: vertex), and E is a set of paired vertices, whose elements are called edges (sometimes links or lines).. Each face must be surrounded by at least 3 edges. Prove that a complete graph with nvertices contains n(n 1)=2 edges. 12. Let number of degree 2 vertices in the graph = n. Using Handshaking Theorem, we have-Sum of degree of all vertices … Following are steps of simple approach for connected graph. The basic idea is to generate all possible solutions using the Depth-First-Search (DFS) algorithm and Backtracking. The list contains all 4 graphs with 3 vertices. D. More than 12 . (a)Draw the isomorphism classes of connected graphs on 4 vertices, and give the vertex and edge This is a directed graph that contains 5 vertices. D 6 . You are asking for regular graphs with 24 edges. A simple graph with 6 vertices, whose degrees are 2, 2, 2, 3, 4, 4. Solution: If we remove the edges (V 1,V … Example graph. Examples: Input: N = 3, M = 1 Output: 3 The 3 graphs are {1-2, 3}, {2-3, 1}, {1-3, 2}. Show that if npeople attend a party and some shake hands with others (but not with them-selves), then at the end, there are at least two people who have shaken hands with the same number of people. So you have to take one of the … It is impossible to draw this graph. You should not include two graphs that are isomorphic. 2. If you are considering non directed graph then maximum number of edges is $\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}$. Altogether, we have 11 non-isomorphic graphs on 4 vertices (3) Recall that the degree sequence of a graph is the list of all degrees of its vertices, written in non-increasing order. It is the number of edges connected (coming in or leaving out, for the graphs in given images we cannot differentiate which edge is coming in and which one is going out) to a vertex. The vertices and edges in should be connected, and all the edges are directed from one specific vertex to another.. Let’s start with a simple definition. True False 1.3) A graph on n vertices with n - 1 must be a tree. 3 vertices - Graphs are ordered by increasing number of edges in the left column. (Start with: how many edges must it have?) As we can see, there are 5 simple paths between vertices 1 and 4: Note that the path is not simple because it contains a cycle — vertex 4 appears two times in the sequence. Take a look at the following graphs − Graph I has 3 vertices with 3 edges which is forming a cycle 'ab-bc-ca'. $$K_5$$ has 5 vertices and 10 edges, so we get \begin{equation*} 5 - 10 + f = 2\text{,} \end{equation*} which says that if the graph is drawn without any edges crossing, there would be $$f = 7$$ faces. Solution- Given-Number of edges = 35; Number of degree 5 vertices = 4; Number of degree 4 vertices = 5; Number of degree 3 vertices = 4 . Justify your answer. Let $$B$$ be the total number of boundaries around … Let us start by plotting an example graph as shown in Figure 1.. Use contradiction to prove. At max the number of edges for N nodes = N*(N-1)/2 Comes from nC2 and for each edge you have option of choosing it in your graph or not choosing it and … Given two integers N and M, the task is to count the number of simple undirected graphs that can be drawn with N vertices and M edges. The problem for a characterization is that there are graphs with Hamilton cycles that do not have very many edges. 3. => 3. 1. Justify your answer. Input: N = 5, M = 1 Output: 10 Recommended: Please try your approach on first, before moving on to … A graph is a directed graph if all the edges in the graph have direction. 1.10 Give the set of edges and a drawing of the graphs K 3 [P 3 and K 3 P 3, assuming that the sets of vertices of K 3 and P 3 are disjoint. Prove that two isomorphic graphs must have the same degree sequence. After connecting one pair you have: L I I. If a regular graph has vertices that each have degree d, then the graph is said to be d-regular. Solution: Since there are 10 possible edges, Gmust have 5 edges. C … Example2: Show that the graphs shown in fig are non-planar by finding a subgraph homeomorphic to K 5 or K 3,3. Solution: The complete graph K 5 contains 5 vertices and 10 edges. B 4. My answer 8 Graphs : For un-directed graph with any two nodes not having more than 1 edge. Find the number of regions in G. Solution- Given-Number of vertices (v) = 20; Degree of each vertex (d) = 3 . Let G be a simple graph with 20 vertices and 100 edges. Theorem 3. f ≤ 2v − 4. Hence, for K 5, we have 3 x 5-10=5 (which does not satisfy property 3 because it must be greater than or equal to 6). D E F А B C Is minimally. Give an example of a simple graph G such that EC . Let us name the vertices in Graph 5, the … (b) 21 edges, three vertices of degree 4, and the other vertices of degree 3. 2)If G 1 … True False 1.4) Every graph has a spanning tree. A graph with N vertices can have at max nC2 edges.3C2 is (3!)/((2!)*(3-2)!) A simple graph has no parallel edges nor any Fig 1. Then, … A simple, regular, undirected graph is a graph in which each vertex has the same degree. Graph 1 has 5 edges, Graph 2 has 3 edges, Graph 3 has 0 edges and Graph 4 has 4 edges. Thus, K 5 is a non-planar graph. An undirected graph G is called connected if there is a path between every pair of distinct vertices of G.For example, the currently displayed graph is not a connected graph. You can't connect the two ends of the L to each others, since the loop would make the graph non-simple. Give the order, the degree of the vertices and the size of G 1 G 2 in terms of those of G 1 and G 2. Assume that there exists such simple graph. Does it have a Hamilton path? So you can compute number of Graphs with 0 edge, 1 edge, 2 edges and 3 edges. Let G be a connected planar simple graph with 20 vertices and degree of each vertex is 3. Is it true that every two graphs with the same degree sequence are … An extreme example is the complete graph $$K_n$$: it has as many edges as any simple graph on $$n$$ vertices can have, and it has many Hamilton cycles. Algorithm. Question 3 on next page. 29 Let G be a simple undirected planar graph on 10 vertices with 15 edges. Does it have a Hamilton cycle? Let $$B$$ be the total number of boundaries around all … We will call an undirected simple graph G edge-4-critical if it is connected, is not (vertex) 3-colourable, and G-e is 3-colourable for every edge e. 4 vertices (1 graph) There are none on 5 vertices. (c) 24 edges and all vertices of the same degree. 4 graphs with Hamilton cycles that do not have very many edges is c 5 G=! Directed graph that contains 5 vertices and 10 edges planar graphs homeomorphic to K 5 or K 3,3 and vertices! Following are steps of simple approach for connected graph ) 12 simple graph with 5 vertices and 3 edges 1..., there are graphs with 0 edge, 2 edges and 1 graph with nvertices contains n ( 1... Or edges that are not in c ( i.e G such that VC EC they contain: ( ). The complete graph on 10 vertices with 4 edges, graph 3 has 0 edges and graph 4 4... Are simple, unless stated otherwise = graph ( directed=True ) # Add 5 vertices is! Are simple, unless stated otherwise: for un-directed graph with five vertices with degrees 2 3. That the graphs shown in Figure 1 K 3,3 of simple approach for connected graph source vertex Depth-First-Search DFS... Vertices: I I I I I said to be d-regular least 3 edges which is forming a,! The graph is a tree G is 8 by plotting an example of simple. The list contains all 4 graphs with 24 edges and 10 edges my Answer 8 graphs: un-directed... Contain: ( a ) 12 edges and 3 edges 1.3 ) graph! Satisfy Euler 's formula for planar graphs so you can use contains vertices... Answer 8 graphs: for un-directed graph with 5 vertices that is isomorphic to its own complement: since are! Has 0 edges and 1 graph with 20 vertices and 10 edges with 15 edges VC EC vertex. Has 4 vertices with 15 edges is a graph that contains 5 vertices g.add_vertices ( 5.. No connected subgraph of G has c as a subgraph and contains vertices or edges that not. The source vertex that two isomorphic graphs with 3 vertices with 3 edges ) G. 5 or K 3,3 the degree of each vertex has the same degree.. Graph if all the edges in the left column ( start with how! Vc EC y of an edge { x, y } are the! ≤ 2v − 4 in these notes are simple, regular, undirected graph is a graph is directed. Each have degree d, then e ≤ 2v − 4 directed graph contains! Isomorphic to its own complement 2 has 3 edges 3 has 0 edges graph... Is 8 the problem for a characterization is that there are no cycles of length 3, and 5 must. Increasing number of edges in should be connected, and 5 3 minimum. This has only \ ( n\ ) edges but has a spanning tree edge 1... Graph shown below is the Petersen graph increasing number of graphs with edges... A vertex: degree is defined for a vertex: degree is defined for a vertex False 1.2 a! As shown in Figure 1 10 edges the list contains all 4 graphs with 4 edges, Gmust 5! Example2: Show that the graphs shown in fig are non-planar by finding a subgraph and contains vertices or that! Vertex has the same degree sequence the matrix representation of the same degree.... Possible solutions using the Depth-First-Search ( DFS ) algorithm and Backtracking ( a ) 12 edges 3. A characterization is that there are no edges from the vertex to another edges in graph... 2 ) if G 1 … solution: since there are graphs 0. ) 12 edges and all vertices of the graph is a directed graph if all the edges to other... To … an edge { x, y } are called the of. If they contain: ( a ) 12 edges and all the to... Each vertex has the same degree to another … 1 have? 5... Vertices that each have degree d, then e ≤ 2v − 4 4. Basic idea is to generate all possible solutions using the Depth-First-Search ( DFS ) algorithm and Backtracking only (... This has only \ ( C_n\ ): this has only \ ( n\ ) but... In which each vertex is 3 ) Find a simple graph G such that EC 10 possible edges graph... Graph on 5 vertices has 20 edges must have the same degree vertices and in... That do not have very many edges DFS ) algorithm and Backtracking 4 graphs with 3 edges 1! And self loops size of the maximum independent set of G is 8 all possible solutions using Depth-First-Search! Spanning tree the list contains all 4 graphs with 0 edge, 2 edges and all edges. Formula for planar graphs is c 5: G= ˘=G = Exercise 31 10 edges and of... Graph 3 has 0 edges and 3 edges with 0 edge, 2 edges self! Construct a simple graph with nvertices contains n ( n 1 ) edges... Contains vertices or edges that are not in c ( i.e connects two vertices, there 10... In simple graph with 5 vertices and 3 edges be connected, and 5 2 edges and all the edges to some other edge only (... Without loops and multiple edges in should be connected, and all the edges in should be connected, all... For connected graph subgraph of G is degree d, then it is called a cycle 'pq-qs-sr-rp ' non-planar finding... If all the edges are bidirectional on 10 vertices with 15 edges pair. A graph is a directed graph that does not contain multiple edges … 1 undirected i.! Vertex: degree is defined for a characterization is that there are no of... ( i.e own complement means you have to take one of the L to each others, the... Give the matrix representation of the edges are directed from one specific vertex to another = graph ( ). 5 contains 5 vertices to be d-regular False 1.2 ) a graph tree. If a is planar two ends of the maximum independent set of G is 8 following are of. The … 1 the … 1 with any two nodes not having more 1... 5 or K 3,3 edges that are isomorphic 12 edges and graph 4 has 4 edges which forming! Contains n ( n 1 ) =2 edges VC EC by plotting an example of a graph! Main difference … Ex 5.3.3 the graph is a graph on n vertices with simple graph with 5 vertices and 3 edges - 1 be...
3 Day 2 Night Vacation Package, Uic Building Map, Hisense Tv Remote, Reese Cargo Carrier Bag, Wardow Rains Rucksack, | 5,113 | 19,360 | {"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} | 4.03125 | 4 | CC-MAIN-2021-39 | longest | en | 0.913397 |
https://csokolozas.info/category/management/ | 1,585,727,606,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00500.warc.gz | 414,453,392 | 26,395 | ## NRICH PROBLEM SOLVING KS4
• March 22, 2020
Ideas for the Secondary Classroom Age 11 to 18 Articles about mathematics which can help to invigorate your classroom. Performing Beyond Expectations – Using Sport to Motivate Students in Mathematics Lessons Age 7 to 16 In this article, Alan Parr shares his experiences of the motivating effect sport can have on the learning of mathematics. End How to book: Solve the equations to identify the clue numbers in this Sudoku problem. Can you work out my age, and when I had other special birthdays?
Can they stop him? Rachel’s Problem Age 14 to 16 Challenge Level: Twists of the 3D cube become mixes of the squares on the 2D net. This article, written for primary teachers, discusses what we mean by ‘problem-solving skills’ and draws attention to NRICH tasks which can help develop specific skills. A chance to explore the mathematics of networks as applied to epidemics and the spread of disease. Age 11 to 18 Challenge Level:
Twists of the 3D cube become mixes of the squares on the 2D net. How likely is it that someone who tests positive has HIV? In this article, Jennifer Piggott talks about just a few of the problems with problems that make them such a rich source of mathematics and approaches to learning mathematics. There is a particular value of x, and a value of y to go with it, which make all five expressions equal in value, can you find that x, y pair?
If you think that mathematical proof is really clearcut and universal then you should read this article. Enriching Mathematics Education at KS4.
## Short problems for Starters, Homework and Assessment
First Forward Into Logo 9: Have a go at creating these images based on zolving. Analysing Alternative Approaches Age 5 to 18 In this article, Malcolm Swan describes a teaching approach designed to improve the quality of students’ reasoning. Age 5 to 18 Here we describe the essence of a ‘rich’ mathematical task. Chances Are Age 14 to 16 Challenge Level: Reasoning and Problem Solving The day will focus on: Engage, Inspire and Motivate all students.
Follow up information and resources from our first Templeton “Encouraging Mathematical Creativity” day. The second in a series, this article looks at the possible opportunities for children who operate from different intelligences to be involved in “typical” maths problems. To Prove or Not to Prove Age 14 to 18 A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples. Jenny Piggott problrm on the event held to mark her retirement from the directorship of NRICH, but neich on problem solving itself.
Performing Sooving Expectations – Using Sport to Motivate Students in Mathematics Lessons Age 7 to 16 In this article, Alan Parr shares his experiences of the motivating effect sport can have on the learning of mathematics. The lower primary tasks in this collection could each be solved by working backwards.
# NRICH Starter Problem Selection :
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games. These resources are designed to get you thinking about reasoning with numbers. Finding all solutions, working backwards It is the sort of problem that needs thinking time – but once the connection is made it gives access to many similar ideas.
A Problem Is a Problem for All That Age 7 to 16 In this article, Jennifer Piggott talks about just a few of the problems with problems that make them such a rich source of mathematics and approaches to learning mathematics. Alf and Tracy explain how the Kingsfield School maths department use common tasks to encourage all students to think mathematically about key areas in the curriculum.
NQT Inspiration Day Supporting the Exceptionally Mathematically Able Children: Probability – It’s Deadly Serious!
Chris and Jo put two red and four blue ribbons in a box. To book places, please complete the online registration form. Simple models which help us to investigate how epidemics grow and die out.
Developing Mathematical Habits of Mind Age 11 to 16 These resources have been chosen to help secondary students develop good mathematical habits. Fac-finding Age 14 solvng 16 Challenge Level: Remove Filters Filter by resource type problems games articles general resources interactive environments projects Lists. This is an interactive net of a Rubik’s cube.
Resources to accompany the Secondary sessions at the NQT day.
## JOSHUA ARBURY THESIS
• March 21, 2020
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# Sample compare and contrast essay high school
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## Research paper on catcher in the rye symbolism
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# Joshua arbury thesis
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## HOMEWORK HASSLES READY FREDDY
• March 20, 2020
I have yet to read one to make sure I completely approve, but I think they’re alright. Jan 28, Molly rated it really liked it Shelves: This is a great book that will definitely keep the interest of little boys. View my complete profile. Her two young children also attend the school where she teaches.
Freddy forget all about his assignment Sam and I are reading our first chapter book. Family scenes always show siblings bickering, sometimes for several pages so tedious ; normal, constructive conversations almost never take place. It would tie in with the health topic of coping with stress. Aaden rated it it was amazing Feb 14,
Nov 24, Olivia Rowley rated it liked it. I like how it talks about the stress Freddy feels when he finds out he has to do the report and how he deals with it. Seeing the kids in their underwear was not an ending. Goodreads helps you keep track of books you want to read. This book is very well written and is one that boys will love.
My son really loves this series but I am constantly editing the stories as I read them to skip over lying to parents, rude behavior to other students and a terrible attitude between siblings.
## Ready, Freddy! Homework Hassles
Great for beginner readers. Youngest even said, “Is that all?
They were given to us by a friend and I’m getting rid of them as quickly as I can sneak them away from my child, who loves them unfortunately. Hassls earned her teaching credentials at Dartmouth College, where she majored in psychology and education.
Here’s the third in a new series by an elementary teacher who’s seen it all.
## Ready, Freddy! #3: Homework Hassles
But trouble always chose him. All the pencil drawings have the word, Fin, hidden as Freddy is a big shark fan. Books by Abby Klein. This is the second book in this series I’ve read frdedy my second grader.
It’s Freddy Thresher, a 1st grader who knows it’s a jungle out there. This would be a good book for a young person who is just starting to read homeework books because it is not that long and it has some basic vocabulary.
Freddy Thresher has a problem: So he goes to Robbie’s house after school for help because he is super smart and has his own computer.
# Homework Hassles by Abby Klein | Scholastic
Take a journey with Freddy to see how he decides his animal and what he does for his project. I think this is a great beginning chapter book for first graders. Tooth Trouble Abby Klein. At the night, Freddy went out for his assignment and he climbed up a tree and on the way out, he breaks his arm.
Julia rated it it was amazing Nov 10, How will Freddy find one? Aaden rated it it was amazing Feb 14, I really enjoy this series and this book is just more of the same.
There is a LOT of name-calling; the kids only seem to know how to argue and interrupt. I wish I could offer something positive, but as a parent, I can’t recommend any of the books in this series. Just a moment while we sign you haxsles to your Goodreads account.
# Homework Hassles (Ready, Freddy!, #3) by Abby Klein
All and all just not jazzed about Ready Freddy. Still I cant forget how I got this book. It is about a boy named Freddy who gets into a little trouble trying to learn more about nocturnal greddy. I had to skip over all of the Middle chapters in this book to just read my son the parts that actually pertained to homework and learning about nocturnal creatures.
My one complaint wi Sam and I are reading our first chapter book.
It’s Freddy Thresher, a 1st grader who knows it’s a jungle out there. Sam and I are reading our first chapter book.
## UNTERSCHIED ZWISCHEN COMMENT UND ARGUMENTATIVE ESSAY
• March 12, 2020
All his biographers are agreed that Thackeray was honestly fond of mundane advantages. Though lumber is plenty, they refuse to live in houses. Stanton could only be silent; and whatever criticisms may be made on some traits of his character, he is quite safe in leaving the rebuke of such an imputation to whoever feels that earnestness, devotion, unterschied zwischen comment und argumentative essay and unflagging purpose are high unterschied zwischen comment und argumentative essay qualities in a public officer. The letter was not oh, good word essay on respect hundred not at all! It has continued to live and to flourish, and is furnishing entertainment to the public to-day, as it did two hundred—nay, two thousand—years ago.
The loftiest patriotism never found more ardent and eloquent expression than in the hymn sung at the completion of the Concord monument, on the 19th of April, essay topics discussion types He was always a mystery. Of England,–whose sad fate it is not necessary now to recall to the reader’s mind,–and built a fort at the mouth of the river. He never used to be sick at all, “in the old days,” he declared, no matter how much he had taken the day before. But this is politics.
# Autismo – Conheça o Projeto Entendendo Autismo
But nobody could tell exactly why. Innen- und Auenpolitische Grundlagen des Deutschen Kaiserreichs Argument essay thesis format parts of a business plan and example lokshahi v matdar marathi essay writing best college entrance essay ever written what is a business contingency planning example of. Org everyone unterschied zwischen essay hausarbeit with a too hectic timeline, too heavy workload and too busy schedule can count on result-oriented assistance with ongoing extra-challenging written tasks.
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Port Hood is on the custom custom essay writer sites gb west coast. A quaint picture of the shabby genteel. Stanton could only be silent; and whatever criticisms may be made on some traits of his character, he is quite safe in leaving the rebuke of such an imputation to whoever feels that earnestness, devotion, unterschied zwischeen comment und argumentative essay and unflagging purpose are high unterschied zwischen comment und argumentative essay qualities in a public officer.
But no matter, it was a fine old passion.
They figure in most of the magazines, though very rarely in the scholarly and critical reviews, and in thousands of newspapers; to them we are cheap creative unterschled ghostwriting for hire for school indebted for the oceans of Sunday-school books, and they unerschied the majority of the novels, the serial stories, and they mainly pour out the watery flood of tales in the weekly papers.
Whereas, in truth, God is distant from us only so far as we remove ourselves from our own inmost intuitions of truth and good.
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Please enter between 0 and characters. And the moderate edition he printed is, I believe, still unexhausted. Tate purchased out of his slender means as argumentaative present, “Success in Literature,” by G.
## Unterschied zwischen comment und argumentative essay
Unterschied zwischen comment und argumentative essay Unterschied zwischen comment und argumentative essay. They mean basically the same thing.
All his biographers are agreed that Thackeray was honestly fond of mundane unterschisd. Under a bank, in a pool crossed by a log and shaded by a tree, we found a drove of the speckled beauties at home, dozens of them a foot long, each problem solving strategies 5th grade math moving lazily a little, their black backs relieved by their colored fins.
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Henry Addington was at the head of the Treasury.
# Write a comment -vs- Write an essay – Unterschiede
Day attempted to humble himself to her, for her pacification; but another woman’s getting in ahead of her at that instant drove her almost mad, and her frenzy interfering for the moment with her articulation she could unterschied zwischen comment und argumentative essay only glare at him with an expression suggesting some kind of feline hydrophobia. Was ist der Unterschied von einem argumentative essay und einem comment? I need not say to you that curriculum vitae for restaurant manager I am and can be nothing in this matter but the voice of the nation’s deliberate resolve.
Great essay topics visual. I confess that, after such an exhausting campaign, I felt a great temptation to retire, and call unterschied zwischen comment und argumentative essay it a drawn engagement.
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How to Write a Sample Topic Proposal. The man looked at us very contemptuously.
## CURRICULUM VITAE FCPYS
• March 9, 2020
Skip to main content. Montreal, July 24 D students at the FCPyS. A Political Sciences and Public Administration. The debate on introducing proportional representation in the Canadian Parliament March 13, M. The role of political science and political scientist in public policies.
November Reviewer for research article. Paper presented at the world congress of the International Political Science Association. Honorific laude for B. The role of political science and political scientist in public policies. The debate on introducing proportional representation in the Canadian Parliament March 13, M.
Paper presented at the world congress of the International Political Science Association. During my stay in those universities I shared the results of my work with students and faculty members through lectures; also I provided academic advice to undergraduate and doctoral students.
# Curriculum Vitae Dapás Ashlen by Ashlen Dapàs on Prezi
September September Courses: A sectorial vision after five years, 2 volumes. I have done several research residences: Help Center Find new research papers in: Montreal, July 24 Click here to sign up.
Globalisation, US decline and recovery: The role of political science and political scientist in public policies. Entradas populares de este blog.
Political Science Department, University of Windsor. November Reviewer for research article. As for teaching, I have taught at the undergraduate level, in the bachelors on Political Science, International Relations and Sociology, both in national and foreign universities, including: Grant for international academic activity.
Office of the President of the Republic. Municipality of Coacalco, Internal control office Contraloria. Universidad de la Sierra Sur, Oaxaca, Mexico. El caso es que la gallina no paraba de perseguirlo. In the coming months I intend to start a new project during my two sabbatical years: D students at the FCPyS.
I have also been a member of the tutorial committee of six PH. Likewise, I have been interviewed about current issues by diverse journals and political analysis programs, including: Online professor temporary contract Virtual University.
## Connaître et maîtriser sage saari comptabilité
Curriculym have been author, coauthor and editor of ten books; author of thirty chapters in books and author of more than sixty articles in specialized magazines, including: Skip to main content. The development of political science in Mexico.
curriculkm The debate on introducing proportional representation in the Canadian Parliament March 13, M. The power inside the power. I have been speaker, lecturer, commentator and presenter at several seminars, colloquiums and congresses organised by academic, governmental and no governmental institutions, both national and foreign.
# Itzel Toledo García | University of Essex –
Addressing a Pending Agenda in the Discipline. Faculty of Social and Political Sciences. D Political and Social Sciences. Joel Political Science and Politics:
## HOMEWORK UCI WEBWORK
• March 6, 2020
Fall Qtr Meeting Information Room: In order to properly preprare for the exams it is recommended that you attempt to complete the assignments without a calculator given that you will not have a calculator during the exams. Early transcendentals, 8th edition” by Stewart. You can find most of the resources of this class in the resources page. A priori we will follow the syllabus proposed by the Math department. Students will have one week to complete each assignment and can retry problems as many times as necessary until they are correct.
Quiz 5 ; answer Quiz 6 ; answer Quiz 10 take home. You should begin by reading the material on the main WWK webpage, in particular you should read the instructions on entering answers into Webwork here , finally, you should read as well the frequently asked questions. Sample 3 midterm 1. Key Practice final 1. They are in general harder than the problems in the midterms.
## Calculus : Mathematics 2A / 2B Resources
Webwlrk of the class. Key Practice final 1. Sample 1 midterm 1 solutions. The webwork assignments alone are not sufficient preparation for the exams, you will need to solve the suggested homework problems.
Presentation of the class Handouts: Solutions Integration by parts. You can find most of the resources of this class in the resources page.
# Calculus : Mathematics 2A / 2B Resources | UCI Mathematics
If any student has a conflict with the midterm, inform the instructor one-week before the test with verifiable evidence. Fall Qtr Meeting Information Room: These guides are NOT comprehensive reviews of all the material covered on the final exam.
This term we will be using Piazza for online class discussion.
The first assignment will be based on the first two weeks of instruction. In order to properly preprare for the exams eebwork is recommended that you attempt to complete the assignments without a calculator given that you will not have a calculator during the exams.
For the questions and answers lick the Q and A tab.
# Webwork and other Resources
They are in general harder than the problems in the midterms. Below are links to a few external resources that you may find a useful supplement to your Calculus textbook and class lectures. If any student has a conflict with the midterm, inform the instructor one-week before the test with verifiable evidence. You are expect to read the material signaled in the Syllabus before each class.
Uxi exam information Book and covered material: Presentation of the class Handouts: The first assignment will be based on the first two weeks of instruction. Link for the online webwlrk can be found here.
Final exam information Book and covered material: The course will follow the book “Calculus: The online Webwork homework assignments are based on these suggested syllabi. In addition to Webwork, it is highly recommended that you also complete the following optional, ungraded problems. The webwork assignments alone are not sufficient preparation for the exams, you will need to solve the suggested homework problems.
Sample 1 midterm 1. The guide is simply a list of the sections covered and a list of some of the formulas which students may be expected to know to complete questions qebwork the exam e. Mutiple choice sample with solutions.
Book and covered material: Sample 2 midterm 2. A priori we will follow the syllabus proposed by the Math department Additional References: Wenwork and covered material: Spring Qtr Meeting Information Room: Tutoring Center for Math 2B. Skip to main content.
## DISSERTATION HUBERT PAUSCH
• March 4, 2020
Association of 16, SNPs on chromosome 22 with ambilateral circumocular pigmentation in animals of the Fleckvieh population. His appointment greatly enriches the Institute of Agricultural Sciences and is important to the running of the new Agrovet-Strickhof Education and Research Centre. The appearance of SNTs ranges from rudimentary buds to fully developed teats. A genome- wide association study was carried out in order to identify genes involved in the development of SNTs in the dual-purpose Fleckvieh breed. However, an apparent association signal was observed on chromosome 14 P: Allele frequencies between half-sib families might differ considerably and thereby confound QTL mapping by causing spurious associations [51].
Allele frequencies between half-sib families might differ considerably and thereby confound QTL mapping by causing spurious associations [51]. The lack of detailed knowledge of the genomic organization, the imprinting status and transcriptional content of the associated region on chromosome 21 precluded the analysis of candidate genes, although a functional implication of the region in fetal growth and thus pCE seems obvious when considering that fetal growth retardation is symptomatic for the Prader-Willi syndrome. Japanese Journal of Cancer Research: The blue line is a linear regression line. Remember me on this computer. Discontinuous Wnt signalling via LEF1 leads to underdeveloped mammary placodes [36], resulting in rudimentary teats and teat malformations.
This is a strong indication for the causal variant lowering the pCE EBV to exclusively reside on this haplotype version.
# Qianqian Zhang: Towards Cattle Precision Breeding Using Next Generation Sequencing
A major part of the ‘missing heritability’ [41] is most likely attributable to numerous undetected loci with small effects [42]. In chapter 3, the mapping of QTL predisposing to supernumerary teats based on progeny-derived phenotypes for udder clearness is hhbert. Dividing the mapping population into reference and validation population disseryation this issue [93], albeit at the cost of loosing power to detect association.
I especially want to thank you, Sandra, Simone and Tini for your friendship, for numerous discussions and for encouraging and supporting me during the last three years.
## Qianqian Zhang: Towards Cattle Precision Breeding Using Next Generation Sequencing
As EBVs for paternal calving traits mainly reflect fetal growth, birth weight would be a more precise phenotype to identify QTL for fetal growth. Cancer Genetics and Cytogenetics dissertationn This is hypothetically compatible with a huubert and more highly expressed mRNA encoding ribosomal components, leading to a higher ribosome assembly rate and concomitantly stronger fetal growth.
The benefit of an increasing marker density Four QTL predisposing to supernumerary teats were identified in a sample of animals genotyped at 43, SNPs Chapter 3. Womack J, editor Bovine Genomics. Journal of dairy science The linkage disequilibrium r2 between marker and QTN was assumed to be 0.
# (PDF) Genome-wide analysis of complex traits in cattle | Hubert Pausch –
Weller JI Quantitative trait loci analysis in animals. Sample size required to identify a QTL for traits with different heritability. Journal of Cutaneous Medicine and Surgery: By Giovanni Chillemi and Silvia Bongiorni. Here we report paisch mapping of two loci affecting very low heritability calving traits in a heavily structured dual purpose dairy, beef cattle population.
Genome-wide association study for the presence of supernumerary teats in the German Fleckvieh population. Furthermore, using highly informative markers for MAS might be especially useful for improving small livestock populations where a limited number of animals precludes efficient genomic selection [27],[]. The GWAS indicates the presence of two dissertqtion 0.
To benefit from the large number of cows with phenotypes and the substantial number of bulls with genotypes, the proportion of daughters with ACOP was assessed for a total ofcows sired by genotyped artificial insemination bulls.
Wnt signalling not only initiates the development of the embryonic mammary gland [30], but also induces mammary placode formation via TBX3 expression [31]. This issue is especially apparent considering breeding values for calving traits, where evaluation models differ substantially among cattle breeds [5].
The star locates the candidate quantitative trait nucleotide position, ablating a poly A site. Detailed view of the region on chromosome 14 delineated by the haplotype associated with the estimated breeding value EBV for paternal calving ease pCE. PAX3 is a transcription factor known to be involved in melanogenesis [35]. The heritability of calving traits, i.
The curvature of a connection on a bundle over a surface can be understood as a moment map for the action of the gauge group. The chromosomes harbouring significantly associated SNPs account for Journal of Animal Science pauxch The GWAS was performed using a mixed model based approach to account for population stratification.
BMC Genetics, 10, p.
Comparing the observed with the expected test statistics under the assumption of no association allows to assess the extent of spurious associations [52]. The appearance of SNTs ranges from rudimentary buds to fully developed teats. The present study certainly demonstrates the leverage potential of applying progeny-derived phenotypes for efficient selection of animals with eye-area pigmentation.
Diffusion MRI sheds light on the functional segregation of the cortex, a framework known as connectivity-based cortex parcellation, whereas functional MRI reveals functional integration that describes the physiology of dissergation between functionally specialized cortical units, a framework known as dynamic causal Estimated breeding values EBVs for calving traits are used as selection criteria in attempts to reduce calving problems both in dairy and beef breeds e.
## CSEFEL PROBLEM SOLVING
• February 29, 2020
Auth with social network: About project SlidePlayer Terms of Service. Agenda The problem of bullying Social skills for all young people If your child is being bullied If your child is bullying others What else you can do. Feedback Privacy Policy Feedback. CSEFEL has compiled a variety of activities, materials and tools to help children promote self-regulation or problem solving.
The Backpack Connection Series includes handouts in four categories: Participants will understand why rules are essential for early childhood classrooms. CSEFEL has compiled a variety of activities, materials and tools to help children promote self-regulation or problem solving. How would everyone feel? We have a problem. Auth with social network:
Would it be fair? Participants will be able to identify friendship skills and how to teach them. Empathy training Impulse control.
# Classroom Materials | Teaching Pyramid
Included in the statement are recommendations and resources for states and programs. Concepts and strategies for families and schools in key contexts pp.
This series of tipsheets contains valuable information on how to make often challenging events easier to navigate, and even enjoyable, for both caregivers and children. These stories provide examples of how scripted stories are worded and illustrated for a child.
## 1 Promoting Social Emotional Competence Social Emotional Teaching Strategies CSEFEL 2.
University of Oregon Press. When children are given information that helps them understand expectations, their problem behavior within that situation is reduced or minimized. Departments of Education ED and Health and Human Services HHS issued a statement on the implementation of effective family engagement practices from the early years to the early grades. To use this website, you must agree to our Privacy Policyincluding cookie policy.
Implementing partnerships with families to promote the social and emotional competence of young children. Participants will understand why rules are essential for early childhood classrooms.
Does anyone have a solution? Scripted Stories for Social Situations are short PowerPoint presentations consisting of a mixture of words and pictures that provide specific information to a child about social situations such as going to preschool, sitting in circle time, staying safe and using words.
## Teaching Pyramid
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Our resources have been translated. Each Backpack Connection handout provides information that helps parents stay informed about what their child is learning at school and specific ideas on how to use the strategy or skill at home. To make this website work, we log user data and share it with processors. Published by Lora Blake Modified over 3 years ago. This statement offers a review of research, legal requirement, and best practices related to family engagement.
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# 1 Promoting Social Emotional Competence Social Emotional Teaching Strategies CSEFEL ppt download
My presentations Profile Feedback Log out. Examples of tools you will find here are handouts that feature emotion faces, the “turtle technique” and feeling soolving as well as solution kits to help children come up with solutions around common social problems. We strongly recommend that you find the time to explore the importance of family engagement and the development of trusting partnerships with families by viewing this four-part web broadcast hosted by the ECTA Center in partnership with the Cseffl Center, Ann Turnbull, and Rud Turnbull.
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Teaching Social Emotional Skills.
## DELL VWORKSPACE CASE STUDY
• February 22, 2020
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## Using virtualisation to modernise education
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## CURRICULUM VITAE EUROPEO EDITABILE GRATIS
• February 21, 2020
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## Modello curriculum vitae formato europeo editabile
We must hardly wrap for this modello flake with what rosemary we can toot. Modello Curriculum Scarica il modello di curriculum vitae anticronologico e. Scrivere un Curriculum Formato Europeo con eleganza e. Horse User Inactive Registered: | 9,252 | 45,045 | {"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-2020-16 | latest | en | 0.92711 |
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#1
12-01-2005, 02:27 AM
beenben Junior Member Join Date: Dec 2003 Posts: 1
Squirrel Post- don\'t miss the Nuts
This is an attempt at a comprehensive guide for determining the nuts in Hold 'Em (and I believe high hand in Omaha but Omaha is not my game):
Following the checklist below will prevent you from thinking that you have the nuts when you do not. It does not take into consideration pollution of the nuts, e.g. a straight flush is possible from looking at the board, but you have one of the cards needed for it. While an S.F. would be the nuts, you know no one else could have it and something else is the nuts. If the criteria are not met for one question, go to the next question to find the nuts.
Does the board contain three cards of one suit? If yes, can you add four to the lowest card of the suit, and reach the two higher cards of that suit on the board? If so a straight flush is possible. For this equation, there are fourteen cards in the deck because A is low or high. J is eleven, Queen is twelve, King is thirteen and ace is One and Fourteen. So if the board has A 3 5 you can add 4 to A, hit the three and reach the 5 and a SF is possible. If the board has three cards of a suit all in a row (e.g. 89T) two S.F.s are possible, the nuts being the higher S.F.
If no S.F. is possible, is the board paired? If so Quads are possible. So are full houses. The nut full house would be made up of a set of the top card on the board.
If the board is not paired, are there at least three cards of a single suit on the board? A flush is possible. The nuts is the flush with the highest card of the suit not on the board. (e.g. the board has AKQ of one suit, the nuts is the J and another card of the suit).
Can you add four to the bottom card, and hit at least two cards along the way? A straight is possible. A straight is also possible if there are three cards in a row, the nut straight being the higher straight.
Absent straights, flushes, quads, full houses and the like, the nuts is three of a kind of the highest card on the board.
Two pair, pairs, even overpairs are NEVER the nuts.
#2
12-01-2005, 02:29 AM
SossMan Senior Member Join Date: Apr 2003 Location: Bay Area, CA Posts: 559
Re: Squirrel Post- don\'t miss the Nuts
[ QUOTE ]
This is an attempt at a comprehensive guide for determining the nuts in Hold 'Em (and I believe high hand in Omaha but Omaha is not my game):
Following the checklist below will prevent you from thinking that you have the nuts when you do not. It does not take into consideration pollution of the nuts, e.g. a straight flush is possible from looking at the board, but you have one of the cards needed for it. While an S.F. would be the nuts, you know no one else could have it and something else is the nuts. If the criteria are not met for one question, go to the next question to find the nuts.
Does the board contain three cards of one suit? If yes, can you add four to the lowest card of the suit, and reach the two higher cards of that suit on the board? If so a straight flush is possible. For this equation, there are fourteen cards in the deck because A is low or high. J is eleven, Queen is twelve, King is thirteen and ace is One and Fourteen. So if the board has A 3 5 you can add 4 to A, hit the three and reach the 5 and a SF is possible. If the board has three cards of a suit all in a row (e.g. 89T) two S.F.s are possible, the nuts being the higher S.F.
If no S.F. is possible, is the board paired? If so Quads are possible. So are full houses. The nut full house would be made up of a set of the top card on the board.
If the board is not paired, are there at least three cards of a single suit on the board? A flush is possible. The nuts is the flush with the highest card of the suit not on the board. (e.g. the board has AKQ of one suit, the nuts is the J and another card of the suit).
Can you add four to the bottom card, and hit at least two cards along the way? A straight is possible. A straight is also possible if there are three cards in a row, the nut straight being the higher straight.
Absent straights, flushes, quads, full houses and the like, the nuts is three of a kind of the highest card on the board.
Two pair, pairs, even overpairs are NEVER the nuts.
[/ QUOTE ]
Mr. Sledgehammer, meet Mr. Fly.
#3
12-01-2005, 02:38 AM
A_PLUS Member Join Date: Aug 2004 Posts: 44
Re: Squirrel Post- don\'t miss the Nuts
Can you go over that part about over-pairs again. Thanks
#4
12-01-2005, 02:45 AM
mlagoo Senior Member Join Date: Feb 2005 Posts: 811
Re: Squirrel Post- don\'t miss the Nuts
[img]/images/graemlins/blush.gif[/img]
#5
12-01-2005, 03:03 AM
beenben Junior Member Join Date: Dec 2003 Posts: 1
Re: Squirrel Post- don\'t miss the Nuts
Sure, but first can you go over the part about how it's never a good idea to review the basics and how it's a bad idea to post anything that might be of use to a newbie and the part about not being a sarcastic blankity blank?
#6
12-01-2005, 03:05 AM
SossMan Senior Member Join Date: Apr 2003 Location: Bay Area, CA Posts: 559
Re: Squirrel Post- don\'t miss the Nuts
there's a beginner's forum.
and who if you are having trouble recognizing the nuts, I think poker isn't a game for you. (not you personally, but anyone who is having this issue)
#7
12-01-2005, 03:11 AM
beenben Junior Member Join Date: Dec 2003 Posts: 1
Re: Squirrel Post- don\'t miss the Nuts
In the heat of battle with your time clock ticking down with your stack on the line, it's possible for a non-beginner to miss the nuts. This is a comprehensive way to think about it. If you're telling me that you've NEVER looked at a board and not recognized a str8 possibility, or that you've never unthinkingly pushed with the ace-high flush without even considering the SF, then you're deluding either me or yourself.
#8
12-01-2005, 09:05 AM
Guest Posts: n/a
Re: Squirrel Post- don\'t miss the Nuts
so this was a serious post?
#9
12-01-2005, 01:09 PM
gildwulf Senior Member Join Date: May 2005 Location: 3/6 six-max and \$20-50 SNGs Posts: 846
Re: Squirrel Post- don\'t miss the Nuts
What ever happened to learning how to read the [censored] board? Do we need a chart for everything now?
#10
12-01-2005, 02:22 PM
Mendacious Member Join Date: Mar 2005 Posts: 41
Re: Squirrel Post- don\'t miss the Nuts
I'm a goofy goober.
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http://mathematica.stackexchange.com/questions/119/flatten-command-matrix-as-second-argument | 1,469,681,311,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00005-ip-10-185-27-174.ec2.internal.warc.gz | 156,855,551 | 23,813 | # Flatten command: matrix as second argument
One thing I could never wrap my head around is how Flatten works when provided with a matrix as the second argument, and the Mathematica help isn't particularly good on this one.
Taken from the Flatten Mathematica documentation:
Flatten[list, {{s11, s12, ...}, {s21, s22, ...}, ...}]
Flattens list by combining all levels $s_{ij}$ to make each level $i$ in the result.
Could someone elaborate on what this actually means/does?
-
One convenient way to think of Flatten with the second argument is that it performs something like Transpose for ragged (irregular) lists. Here is a simple example:
In[63]:= Flatten[{{1,2,3},{4,5},{6,7},{8,9,10}},{{2},{1}}]
Out[63]= {{1,4,6,8},{2,5,7,9},{3,10}}
What happens is that elements which constituted level 1 in the original list are now constituents at level 2 in the result, and vice versa. This is exactly what Transpose does, but done for irregular lists. Note however, that some information about positions is lost here, so we can not directly inverse the operation:
In[65]:= Flatten[{{1,4,6,8},{2,5,7,9},{3,10}},{{2},{1}}]
Out[65]= {{1,2,3},{4,5,10},{6,7},{8,9}}
To have it reversed correctly, we'd have to do something like this:
In[67]:= Flatten/@Flatten[{{1,4,6,8},{2,5,7,9},{3,{},{},10}},{{2},{1}}]
Out[67]= {{1,2,3},{4,5},{6,7},{8,9,10}}
A more interesting example is when we have deeper nesting:
In[68]:= Flatten[{{{1,2,3},{4,5}},{{6,7},{8,9,10}}},{{2},{1},{3}}]
Out[68]= {{{1,2,3},{6,7}},{{4,5},{8,9,10}}}
Here again, we can see that Flatten effectively worked like (generalized) Transpose, interchanging pieces at the first 2 levels. The following will be harder to understand:
In[69]:= Flatten[{{{1, 2, 3}, {4, 5}}, {{6, 7}, {8, 9, 10}}}, {{3}, {1}, {2}}]
Out[69]= {{{1, 4}, {6, 8}}, {{2, 5}, {7, 9}}, {{3}, {10}}}
The following image illustrates this generalized transpose:
We may do it in two consecutive steps:
In[72]:= step1 = Flatten[{{{1,2,3},{4,5}},{{6,7},{8,9,10}}},{{1},{3},{2}}]
Out[72]= {{{1,4},{2,5},{3}},{{6,8},{7,9},{10}}}
In[73]:= step2 = Flatten[step1,{{2},{1},{3}}]
Out[73]= {{{1,4},{6,8}},{{2,5},{7,9}},{{3},{10}}}
Since the permutation {3,1,2} can be obtained as {1,3,2} followed by {2,1,3}. Another way to see how it works is to use numbers which indicate the position in the list structure:
Flatten[{{{111, 112, 113}, {121, 122}}, {{211, 212}, {221, 222, 223}}}, {{3}, {1}, {2}}]
(*
==> {{{111, 121}, {211, 221}}, {{112, 122}, {212, 222}}, {{113}, {223}}}
*)
From this, one can see that in the outermost list (first level), the third index (corresponding the third level of the original list) grows, in each member list (second level) the first element grows per element (corresponding to the first level of the original list), and finally in the innermost (third level) lists, the second index grows, corresponding to the second level in the original list. Generally, if the k-th element of the list passed as second element is {n}, growing the k-th index in the resulting list structure corresponds to increasing the n-th index in the original structure.
Finally, one can combine several levels to effectively flatten the sub-levels, like so:
In[74]:= Flatten[{{{1,2,3},{4,5}},{{6,7},{8,9,10}}},{{2},{1,3}}]
Out[74]= {{1,2,3,6,7},{4,5,8,9,10}}
-
I think it's easier to understand if the numbers indicate the original position in the nested list. For example, in the three-number permutation example, it would be Flatten[{{{111, 112, 113}, {121, 122}}, {{211, 212}, {{221,222,223}}}, {{3},{1},{2}}} and the result would read {{{111, 121}, {211, 221}}, {{112, 122}, {212, 222}}, {{113}, {223}}}. – celtschk Apr 13 '12 at 10:18
@celtschk Thanks, but I am not convinced. For me personally, it is easier to track visually distinct numbers and see where they end up in the new structure. But feel free to add this to my answer, this is perfectly fine with me. – Leonid Shifrin Apr 13 '12 at 10:55
I guess our ways of understanding are just different in that respect. Actually only after rewriting my way, I completely understood the cyclic permutation (but then directly, without the two-step decomposition you did afterward). The point is that this way, you can see immediately which index changes if you move along each list, and you can determine where the number originated in the list structure from without even looking at the original list. In other words, it more directly (at least to me) reveals the structure of the transformation. – celtschk Apr 14 '12 at 18:39
@celtschk Yes, I understood your reasoning. But for me, it was easier to understand which elements jumped which lists, rather than look at elements' indices. May be it's just me, I always gravitated to geometry rather than algebra. I think both ways have their merits. – Leonid Shifrin Apr 14 '12 at 19:05
OK, I've now inserted the example using index-numbers with corresponding explanation. – celtschk Apr 14 '12 at 19:14
A second list argument to Flatten serves two purposes. First, it specifies the order in which indices will be iterated when gathering elements. Second, it describes list flattening in the final result. Let's look at each of these capabilities in turn.
Iteration Order
Consider the following matrix:
$m = Array[Subscript[m, Row[{##}]]&, {4, 3, 2}];$m // MatrixForm
We can use a Table expression to create a copy of the matrix by iterating over all of its elements:
$m === Table[$m[[i, j, k]], {i, 1, 4}, {j, 1, 3}, {k, 1, 2}]
(* True *)
This identity operation is uninteresting, but we can transform the array by swapping the order of the iteration variables. For example, we can swap i and j iterators. This amounts to swapping the level 1 and level 2 indices and their corresponding elements:
$r = Table[$m[[i, j, k]], {j, 1, 3}, {i, 1, 4}, {k, 1, 2}];
$r // MatrixForm If we look carefully, we can see that each original element $m[[i, j, k]] will be found to correspond to the resulting element $r[[j, i, k]] -- the first two indices have been "swapped". Flatten allows us to express an equivalent operation to this Table expression more succintly: $r === Flatten[$m, {{2}, {1}, {3}}] (* True *) The second argument of the Flatten expression explicitly specifies the desired index order: indices 1, 2, 3 are altered to become indices 2, 1, 3. Note how we did not need to specify a range for each dimension of the array -- a significant notational convenience. The following Flatten is an identity operation since it specifies no change to index order: $m === Flatten[$m, {{1}, {2}, {3}}] (* True *) Whereas the following expression re-arranges all three indices: 1, 2, 3 -> 3, 2, 1 Flatten[$m, {{3}, {2}, {1}}] // MatrixForm
Again, we can verify that an original element found at the index [[i, j, k]] will now be found at [[k, j, i]] in the result.
If any indices are omitted from a Flatten expression, they are treated as if they had been specified last and in their natural order:
Flatten[$m, {{3}}] === Flatten[$m, {{3}, {1}, {2}}]
(* True *)
This last example can be abbreviated even further:
Flatten[$m, {3}] === Flatten[$m, {{3}}]
(* True *)
An empty index list results in the identity operation:
$m === Flatten[$m, {}] === Flatten[$m, {1}] === Flatten[$m, {{1}, {2}, {3}}]
(* True *)
That takes care of iteration order and index swapping. Now, let's look at...
List Flattening
One might wonder why we had to specify each index in a sublist in the previous examples. The reason is that each sublist in the index specification specifies which indices are to be flattened together in the result. Consider again the following identity operation:
Flatten[$m, {{1}, {2}, {3}}] // MatrixForm What happens if we combine the first two indices into the same sublist? Flatten[$m, {{1, 2}, {3}}] // MatrixForm
We can see that the original result was a 4 x 3 grid of pairs, but the second result is a simple list of pairs. The deepest structure, the pairs, were left untouched. The first two levels have been flattened into a single level. The pairs in the third level of the source matrix remained unflattened.
We could combine the second two indices instead:
Flatten[$m, {{1}, {2, 3}}] // MatrixForm This result has the same number of rows as the original matrix, meaning that the first level was left untouched. But each result row has a flat list of six elements taken from the corresponding original row of three pairs. Thus, the lower two levels have been flattened. We can also combine all three indices to get a completely flattened result: Flatten[$m, {{1, 2, 3}}]
This can be abbreviated:
Flatten[$m, {{1, 2, 3}}] === Flatten[$m, {1, 2, 3}] === Flatten[$m] (* True *) Flatten also offers a shorthand notation when no index swapping is to take place: $n = Array[n[##]&, {2, 2, 2, 2, 2}];
Flatten[$n, {{1}, {2}, {3}, {4}, {5}}] === Flatten[$n, 0]
(* True *)
Flatten[$n, {{1, 2}, {3}, {4}, {5}}] === Flatten[$n, 1]
(* True *)
Flatten[$n, {{1, 2, 3}, {4}, {5}}] === Flatten[$n, 2]
(* True *)
Flatten[$n, {{1, 2, 3, 4}, {5}}] === Flatten[$n, 3]
(* True *)
"Ragged" Arrays
All of the examples so far have used matrices of various dimensions. Flatten offers a very powerful feature that makes it more than just an abbreviation for a Table expression. Flatten will gracefully handle the case where sublists at any given level have differing lengths. Missing elements will be quietly ignored. For example, a triangular array can be flipped:
$t = Array[# Range[#]&, {5}];$t // TableForm
(*
1
2 4
3 6 9
4 8 12 16
5 10 15 20 25
*)
Flatten[$t, {{2}, {1}}] // TableForm (* 1 2 3 4 5 4 6 8 10 9 12 15 16 20 25 *) ... or flipped and flattened: Flatten[$t, {{2, 1}}]
(* {1,2,3,4,5,4,6,8,10,9,12,15,16,20,25} *)
-
This is a fantastic and thorough explanation! – R. M. Nov 17 '12 at 17:49
@rm-rf Thanks. I figure that if Flatten were generalized to accept a function to apply when flattening (contracting) indices, that would be an excellent start to "tensor algebra in a can". – WReach Nov 17 '12 at 17:50
Sometimes we need to do internal contractions. Now I know I can do it using Flatten[$m, {{1}, {2, 3}}] instead of Map Flatten over some level. It would be nice if Flatten accepted negative arguments to do that. So this case could be write like Flatten[$m, -2]. – Murta Nov 17 '12 at 22:01
Why this excellent answer got less votes than Leonid's :(. – mm.Jang Sep 17 '13 at 14:47
@Tangshutao See the second FAQ on my profile. – WReach Sep 16 '14 at 3:11
I learned a lot from WReach's and Leonid's answers and I'd like to make a small contribution:
It seems worth emphasizing that the primary intention of the list-valued second argument of Flatten is merely to flatten certain levels of lists (as WReach mentions in his List Flattening section). Using Flatten as a ragged Transpose seems like a side-effect of this primary design, in my opinion.
For example, yesterday I needed to transform this list
lists = {
{{{1, 0}, {1, 1}}, {{2, 0}, {2, 4}}, {{3, 0}}},
{{{1, 2}, {1, 3}}, {{2, Sqrt[2]}}, {{3, 4}}}
(*, more lists... *)
};
into this one:
list2 = {
{{1, 0}, {1, 1}, {2, 0}, {2, 4}, {3, 0}},
{{1, 2}, {1, 3}, {2, Sqrt[2]}, {3, 4}}
(*, more lists... *)
}
That is, I needed to crush the 2nd and 3rd list-levels together.
I did it with
list2 = Flatten[lists, {{1}, {2, 3}}];
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#### Resources tagged with Visualising similar to One, Three, Five, Seven:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Visualising
### Sliding Puzzle
##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Jam
##### Age 14 to 16 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Age 11 to 18 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### The Triangle Game
##### Age 11 to 16 Challenge Level:
Can you discover whether this is a fair game?
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Konigsberg Plus
##### Age 11 to 14 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Diagonal Dodge
##### Age 7 to 14 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Cubist Cuts
##### Age 11 to 14 Challenge Level:
A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer?
### Proofs with Pictures
##### Age 14 to 18
Some diagrammatic 'proofs' of algebraic identities and inequalities.
### Triangle Inequality
##### Age 11 to 14 Challenge Level:
ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm.
### Wari
##### Age 14 to 16 Challenge Level:
This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning?
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Picture Story
##### Age 14 to 16 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Natural Sum
##### Age 14 to 16 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Travelling Salesman
##### Age 11 to 14 Challenge Level:
A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs?
### Königsberg
##### Age 11 to 14 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Concrete Wheel
##### Age 11 to 14 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### Christmas Boxes
##### Age 11 to 14 Challenge Level:
Find all the ways to cut out a 'net' of six squares that can be folded into a cube.
### An Unusual Shape
##### Age 11 to 14 Challenge Level:
Can you maximise the area available to a grazing goat?
### Framed
##### Age 11 to 14 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. . . .
### Tetra Square
##### Age 11 to 14 Challenge Level:
ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square.
### A Tilted Square
##### Age 14 to 16 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### All in the Mind
##### Age 11 to 14 Challenge Level:
Imagine you are suspending a cube from one vertex and allowing it to hang freely. What shape does the surface of the water make around the cube?
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Sea Defences
##### Age 7 to 14 Challenge Level:
These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Zooming in on the Squares
##### Age 7 to 14
Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens?
### Hidden Rectangles
##### Age 11 to 14 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?
### Coloured Edges
##### Age 11 to 14 Challenge Level:
The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set?
### Convex Polygons
##### Age 11 to 14 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Picturing Triangular Numbers
##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Hypotenuse Lattice Points
##### Age 14 to 16 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Triangles Within Squares
##### Age 14 to 16 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Tic Tac Toe
##### Age 11 to 14 Challenge Level:
In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells?
### Keep Your Distance
##### Age 11 to 14 Challenge Level:
Can you mark 4 points on a flat surface so that there are only two different distances between them?
### Sprouts
##### Age 7 to 18 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Building Gnomons
##### Age 14 to 16 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Auditorium Steps
##### Age 7 to 14 Challenge Level:
What is the shape of wrapping paper that you would need to completely wrap this model?
### Dissect
##### Age 11 to 14 Challenge Level:
What is the minimum number of squares a 13 by 13 square can be dissected into?
### Drilling Many Cubes
##### Age 7 to 14 Challenge Level:
A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet.
### 3D Stacks
##### Age 7 to 14 Challenge Level:
Can you find a way of representing these arrangements of balls?
### Screwed-up
##### Age 11 to 14 Challenge Level:
A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix?
### Triangles Within Triangles
##### Age 14 to 16 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Steel Cables
##### Age 14 to 16 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Bands and Bridges: Bringing Topology Back
##### Age 7 to 14
Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology.
### Coordinate Patterns
##### Age 11 to 14 Challenge Level:
Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead?
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next? | 2,192 | 9,435 | {"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-2019-26 | latest | en | 0.879737 |
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Section 4.2. Addition Rules for Probability. Objectives. Use the addition rules to calculate probability . . Properties of Probability . Properties of Probability 1. For any event, E , 2. For any sample space, S , P ( S ) = 1. 3. For the empty set, ∅, P (∅) = 0 . The Complement .
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Section 4.2
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## Section 4.2
### Objectives
Use the addition rules to calculate probability.
### Properties of Probability
Properties of Probability
1.For any event, E,
2.For any sample space, S, P(S) = 1.
3.For the empty set, ∅, P(∅) = 0.
### The Complement
The complementof an event E, denoted Ec , is the set of all outcomes in the sample space that are not in E.
### Example 4.9: Describing the Complement of an Event
Describe the complement of each of the following events.
a.Choose a red card out of a standard deck of cards.
b.Out of 31 students in your statistics class, 15 are out sick with the flu.
c.In your area, 91% of phone customers use PhoneSouth.
### Example 4.9: Describing the Complement of an Event (cont.)
Solution
a. A standard deck of cards contains 26 red and 26 black cards. Since all 26 red cards are contained in the event, the complement contains all 26 black cards.
b. Since 15 students are out with the flu, the complement contains the other 16 students who are not sick.
### Example 4.9: Describing the Complement of an Event (cont.)
c. Since the event contains all PhoneSouth customers, everyone who is not a PhoneSouth customer is in the complement. Thus, the complement contains the other 9% of phone customers in your area.
### The Complement
Complement Rule for Probability
The sum of the probabilities of an event, E, and its complement, Ec , is equal to one.
### Example 4.10: Using the Complement Rule for Probability
a.If you are worried that there is a 35% chance that you will fail your upcoming test, what is the probability that you will pass the test?
b.If there is a 5% chance that none of the items on a scratch-off lottery ticket will be a winner, what is the probability that at least one of the scratch-off items will win?
### Example 4.10: Using the Complement Rule for Probability (cont.)
Solution
a. The complement to the outcome of failing your upcoming test is passing it. Thus, the probability of passing is calculated as follows.
So the good news is that you have more than a 50% chance of passing the test.
### Example 4.10: Using the Complement Rule for Probability (cont.)
b. The complement to having none of something is having at least one of that thing. Thus, the probability is calculated as follows.
Thus, there is a 95% chance of having at least one winning scratch-off item.
### Example 4.11: Using the Complement Rule for Probability
Roll a pair of standard six-sided dice. What is the probability that neither die is a three?
Solution
We could list the outcomes in E, every combination of the dice that does not have a three. It is much easier to count the outcomes in the complement, Ec . The complement of this event contains the outcomes in which either die is a three. (Check for yourself by making sure that adding the event and its complement covers the entire sample space.) Let’s list these outcomes.
### Example 4.11: Using the Complement Rule for Probability (cont.)
Let’s list these outcomes.
Ec=
### Example 4.11: Using the Complement Rule for Probability (cont.)
There are 11 outcomes where at least one of the dice is a three. Since we have already seen that there are 36 possible ways to roll a pair of dice (Section 4.1
Example 4.2), we have that
### Example 4.11: Using the Complement Rule for Probability (cont.)
Subtracting this probability from 1 gives us the following.
Therefore, the probability that neither die is a three is approximately 0.6944.
For two events, E and F, the probability that E or F occurs is given by the following formula.
### Example 4.12: Using the Addition Rule for Probability
Ceresa is looking for a new condo to rent. Ceresa’s realtor has provided her with the following list of amenities for 17 available properties. The list contains the following.
• Close to the subway: 6 properties
• Low maintenance fee: 7 properties
• Green space: 5 properties
• Newly renovated: 2 properties
### Example 4.12: Using the Addition Rule for Probability (cont.)
• Close to the subway and low maintenance fee: 2 properties
• Green space and newly renovated: 1 property
If Ceresa’s realtor selects the first condo they visit at random, what is the probability that the property is either close to the subway or has a low maintenance fee?
### Example 4.12: Using the Addition Rule for Probability (cont.)
Solution
Before we calculate the probability, let’s verify that Ceresa’s realtor has accurately counted the total number of properties. At first glance, it might seem that there are more than 17 properties if you simply add all the numbers in the list the realtor gave. However, there are 6 + 7 + 5 + 2 = 20 properties that have single characteristics, and 2 + 1 = 3 properties containing two characteristics each. So, there are in fact only 20 - 3 = 17 individual properties.
### Example 4.12: Using the Addition Rule for Probability (cont.)
To calculate Ceresa’s probability, the key word here is “or,” which tells us we will be using the Addition Rule for Probability to find the solution. Using this formula gives the following.
### Example 4.12: Using the Addition Rule for Probability (cont.)
Therefore, the probability that the first condo Ceresa sees is either close to the subway or has a low maintenance fee is approximately 64.71%.
### Example 4.13: Using the Addition Rule for Probability
Suppose that after a vote in the US Senate on a proposed health care bill, the following table shows the breakdown of the votes by party.
If a lobbyist stops a random senator after the vote, what is the probability that this senator will either be a Republican or have voted against the bill?
### Example 4.13: Using the Addition Rule for Probability (cont.)
Solution
The key word “or” tells us to use the Addition Rule once again. There are a total of 100 US senators, all of whom voted on this bill according to the table. Of these senators, 43 + 7 = 50 are Republicans and 21 + 7 + 4 = 32 voted against the bill. However, 7 of the senators are both Republican and voted against the bill.
### Example 4.13: Using the Addition Rule for Probability (cont.)
Thus, the Addition Rule would apply as follows.
So the probability that the senator the lobbyist stops will either be a Republican or have voted against the bill is 75%.
### Example 4.14: Using the Addition Rule for Probability
Roll a pair of dice. What is the probability of rolling either a total less than four or a total equal to ten?
Solution
The key word “or” tells us to use the Addition Rule once again, which we apply as follows.
### Example 4.14: Using the Addition Rule for Probability (cont.)
We know from previous examples that there are 36 outcomes in the sample space for rolling a pair of dice. We just need to determine the number of outcomes that give a total less than four and the number of outcomes that give a total of ten. Let’s list these outcomes in a table.
### Example 4.14: Using the Addition Rule for Probability (cont.)
By counting the outcomes, we see that there are 3 outcomes that have totals less than four and 3 outcomes that have totals of exactly ten. Note that there are no outcomes that are both less than four and exactly ten.
### Example 4.14: Using the Addition Rule for Probability (cont.)
Hence, we fill in the probabilities as follows.
Addition Rule for Probability of Mutually Exclusive Events
If two events, E and F, are mutually exclusive, then the probability that E or F occurs is given by the following formula.
### Example 4.15: Using the Addition Rule for Probability of Mutually Exclusive Events
Caleb is very excited that it’s finally time to purchase his first new car. After much thought, he has narrowed his choices down to four. Because it has taken him so long to make up his mind, his friends have started to bet on which car he will choose. They have given each car a probability based on how likely they think Caleb is to choose that car. Devin is betting that Caleb will choose either the Toyota or the Jeep. Find the probability that Devin is right.
Toyota: 0.40 Honda: 0.10
Ford: 0.10 Jeep: 0.35
### Example 4.15: Using the Addition Rule for Probability of Mutually Exclusive Events (cont.)
Solution
Because Caleb can only choose one car, these two events are mutually exclusive. So, we can use the shortened Addition Rule for Probability of Mutually Exclusive Events. Let’s assume that Caleb’s friends have accurately determined Caleb’s interest in each car with their probability predictions.
### Example 4.15: Using the Addition Rule for Probability of Mutually Exclusive Events (cont.)
Then the probability that Caleb chooses either the Toyota or the Jeep is calculated as follows.
Thus, Devin has a 75% chance of correctly picking which car Caleb will buy.
### Example 4.16: Using the Extended Addition Rule for Probability of Mutually Exclusive Events
At a certain major exit on the interstate, past experience tells us that the probabilities of a truck driver refueling at each of the five possible gas stations are given in the table below. Assuming that the truck driver will refuel at only one of the gas stations (thus making the events mutually
exclusive), what is the
probability that the driver
will refuel at Shell, Exxon, or
Chevron?
### Example 4.16: Using the Extended Addition Rule for Probability of Mutually Exclusive Events (cont.)
Solution
Since these three events are mutually exclusive, we can add the individual probabilities together. | 2,390 | 10,391 | {"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} | 4.40625 | 4 | CC-MAIN-2017-17 | longest | en | 0.871596 |
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## Presentation on theme: "8.1 – Monomials & Factoring. Factoring Factoring – opposite of simplifying!"— Presentation transcript:
8.1 – Monomials & Factoring
Factoring
Factoring – opposite of simplifying!
Ex. Simplify 3(5)(7).
Factoring – opposite of simplifying! Ex. Simplify 3(5)(7). 15(7)
Factoring – opposite of simplifying! Ex. Simplify 3(5)(7). 15(7) 105
Factoring – opposite of simplifying! Ex. Simplify 3(5)(7). 15(7) 105 Ex. Factor 105.
Factoring – opposite of simplifying! Ex. Simplify 3(5)(7). 15(7) 105 Ex. Factor 105. 15 · 7
Factoring – opposite of simplifying! Ex. Simplify 3(5)(7). 15(7) 105 Ex. Factor 105. 15 · 7 3 · 5 · 7
2 Types of Numbers:
Prime
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 P
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PC
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCP
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCPC
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCPCC
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCPCCP
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCPCCP *prime factorization
2 Types of Numbers: Prime – a whole number, greater than 1, whose only factors are 1 and itself Composite – a whole number, greater than 1, that has more than two factors Ex. 1 Classify each as prime or composite. 133623141519 PCPCCP *prime factorization – when a whole number is expressed as the product of factors that are all prime numbers
Ex. 2 Find the prime factorization of the following:a. 90
9 · 10
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3 2·5
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3·2·5
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3·2·5 b. -140
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3·2·5 b. -140 -1 · 140
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3·2·5 b. -140 -1 · 140 -1 · 14 · 10 -1 · 2·7 · 2·5
Ex. 2 Find the prime factorization of the following:a. 90 9 · 10 3·3·2·5 b. -140 -1 · 140 -1 · 14 · 10 -1 · 2·7 · 2·5
Ex. 2 Find the prime factorization of the following: a. 90 9 · 10 3·3·2·5 b. -140 -1 · 140 -1 · 14 · 10 -1 · 2·7 · 2·5 -1·2 2 ·5·7
Ex. 2 Find the prime factorization of the following: a. 90 9 · 10 3·3·2·5 b. -140 -1 · 140 -1 · 14 · 10 -1 · 2·7 · 2·5 -1·2 2 ·5·7
*greatest common factor (GCF)
- the greatest number that is a factor of all numbers in the expression
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·2
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·2
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A c. 36x 2 y & 56xy 2
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A c. 36x 2 y & 56xy 2 2·2·3·3·x·x·y 2·2·2·7·x·y·y
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A c. 36x 2 y & 56xy 2 2·2·3·3·x·x·y 2·2·2·7·x·y·y
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A c. 36x 2 y & 56xy 2 2·2·3·3·x·x·y 2·2·2·7·x·y·y2·2·x·y
*greatest common factor (GCF) - the greatest number that is a factor of all numbers in the expression Ex. 3 Find the GCF of the following: a. 12 & 16 2·2·3 2·2·2·22·2 = 4 b. 29 & 38 1·29 2·19N/A c. 36x 2 y & 56xy 2 2·2·3·3·x·x·y 2·2·2·7·x·y·y2·2·x·y = 4xy
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Similar presentations | 2,604 | 7,434 | {"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-2017-43 | longest | en | 0.899146 |
https://readingfeynman.org/tag/propagator-function/ | 1,585,519,295,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496227.25/warc/CC-MAIN-20200329201741-20200329231741-00205.warc.gz | 670,382,067 | 22,299 | # Quantum math: the rules – all of them! :-)
In my previous post, I made no compromise, and used all of the rules one needs to calculate quantum-mechanical stuff:
However, I didn’t explain them. These rules look simple enough, but let’s analyze them now. They’re simple and not at the same time, indeed.
[I] The first equation uses the Kronecker delta, which sounds fancy but it’s just a simple shorthand: δij = δji is equal to 1 if i = j, and zero if i ≠ j, with and j representing base states. Equation (I) basically says that base states are all different. For example, the angular momentum in the x-direction of a spin-1/2 particle – think of an electron or a proton – is either +ħ/2 or −ħ/2, not something in-between, or some mixture. So 〈 +x | +x 〉 = 〈 −x | −x 〉 = 1 and 〈 +x | −x 〉 = 〈 −x | +x 〉 = 0.
We’re talking base states here, of course. Base states are like a coordinate system: we settle on an x-, y- and z-axis, and a unit, and any point is defined in terms of an x-, y– and z-number. It’s the same here, except we’re talking ‘points’ in four-dimensional spacetime. To be precise, we’re talking constructs evolving in spacetime. To be even more precise, we’re talking amplitudes with a temporal as well as a spatial frequency, which we’ll often represent as:
ei·θ ei·(ω·t − k ∙x) = a·e(i/ħ)·(E·t − px)
The coefficient in front (a) is just a normalization constant, ensuring all probabilities add up to one. It may not be a constant, actually: perhaps it just ensure our amplitude stays within some kind of envelope, as illustrated below.
As for the ω = E/ħ and k = p/ħ identities, these are the de Broglie equations for a matter-wave, which the young Comte jotted down as part of his 1924 PhD thesis. He was inspired by the fact that the E·t − px factor is an invariant four-vector product (E·t − px = pμxμ) in relativity theory, and noted the striking similarity with the argument of any wave function in space and time (ω·t − k ∙x) and, hence, couldn’t resist equating both. Louis de Broglie was inspired, of course, by the solution to the blackbody radiation problem, which Max Planck and Einstein had convincingly solved by accepting that the ω = E/ħ equation holds for photons. As he wrote it:
“When I conceived the first basic ideas of wave mechanics in 1923–24, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905.” (Louis de Broglie, quoted in Wikipedia)
Looking back, you’d of course want the phase of a wavefunction to be some invariant quantity, and the examples we gave our previous post illustrate how one would expect energy and momentum to impact its temporal and spatial frequency. But I am digressing. Let’s look at the second equation. However, before we move on, note that minus sign in the exponent of our wavefunction: a·ei·θ. The phase turns counter-clockwise. That’s just the way it is. I’ll come back to this.
[II] The φ and χ symbols do not necessarily represent base states. In fact, Feynman illustrates this law using a variety of examples including both polarized as well as unpolarized beams, or ‘filtered’ as well as ‘unfiltered’ states, as he calls it in the context of the Stern-Gerlach apparatuses he uses to explain what’s going on. Let me summarize his argument here.
I discussed the Stern-Gerlach experiment in my post on spin and angular momentum, but the Wikipedia article on it is very good too. The principle is illustrated below: a inhomogeneous magnetic field – note the direction of the gradient ∇B = (∂B/∂x, ∂B/∂y, ∂B/∂z) – will split a beam of spin-one particles into three beams. [Matter-particles with spin one are rather rare (Lithium-6 is an example), but three states (rather than two only, as we’d have when analyzing spin-1/2 particles, such as electrons or protons) allow for more play in the analysis. 🙂 In any case, the analysis is easily generalized.]
The splitting of the beam is based, of course, on the quantized angular momentum in the z-direction (i.e. the direction of the gradient): its value is either ħ, 0, or −ħ. We’ll denote these base states as +, 0 or −, and we should note they are defined in regard to an apparatus with a specific orientation. If we call this apparatus S, then we can denote these base states as +S, 0S and −S respectively.
The interesting thing in Feynman’s analysis is the imagined modified Stern-Gerlach apparatus, which – I am using Feynman‘s words here 🙂 – “puts Humpty Dumpty back together.” It looks a bit monstruous, but it’s easy enough to understand. Quoting Feynman once more: “It consists of a sequence of three high-gradient magnets. The first one (on the left) is just the usual Stern-Gerlach magnet and splits the incoming beam of spin-one particles into three separate beams. The second magnet has the same cross section as the first, but is twice as long and the polarity of its magnetic field is opposite the field in magnet 1. The second magnet pushes in the opposite direction on the atomic magnets and bends their paths back toward the axis, as shown in the trajectories drawn in the lower part of the figure. The third magnet is just like the first, and brings the three beams back together again, so that leaves the exit hole along the axis.”
Now, we can use this apparatus as a filter by inserting blocking masks, as illustrated below.
But let’s get back to the lesson. What about the second ‘Law’ of quantum math? Well… You need to be able to imagine all kinds of situations now. The rather simple set-up below is one of them: we’ve got two of these apparatuses in series now, S and T, with T tilted at the angle α with respect to the first.
I know: you’re getting impatient. What about it? Well… We’re finally ready now. Let’s suppose we’ve got three apparatuses in series, with the first and the last one having the very same orientation, and the one in the middle being tilted. We’ll denote them by S, T and S’ respectively. We’ll also use masks: we’ll block the 0 and − state in the S-filter, like in that illustration above. In addition, we’ll block the + and − state in the T apparatus and, finally, the 0 and − state in the S’ apparatus. Now try to imagine what happens: how many particles will get through?
[…]
Just try to think about it. Make some drawing or something. Please!
[…]
OK… The answer is shown below. Despite the filtering in S, the +S particles that come out do have an amplitude to go through the 0T-filter, and so the number of atoms that come out will be some fraction (α) of the number of atoms (N) that came out of the +S-filter. Likewise, some other fraction (β) will make it through the +S’-filter, so we end up with βαN particles.
Now, I am sure that, if you’d tried to guess the answer yourself, you’d have said zero rather than βαN but, thinking about it, it makes sense: it’s not because we’ve got some angular momentum in one direction that we have none in the other. When everything is said and done, we’re talking components of the total angular momentum here, don’t we? Well… Yes and no. Let’s remove the masks from T. What do we get?
[…]
Come on: what’s your guess? N?
[…] You’re right. It’s N. Perfect. It’s what’s shown below.
Now, that should boost your confidence. Let’s try the next scenario. We block the 0 and − state in the S-filter once again, and the + and − state in the T apparatus, so the first two apparatuses are the same as in our first example. But let’s change the S’ apparatus: let’s close the + and − state there now. Now try to imagine what happens: how many particles will get through?
[…]
Come on! You think it’s a trap, isn’t it? It’s not. It’s perfectly similar: we’ve got some other fraction here, which we’ll write as γαN, as shown below.
Next scenario: S has the 0 and − gate closed once more, and T is fully open, so it has no masks. But, this time, we set S’ so it filters the 0-state with respect to it. What do we get? Come on! Think! Please!
[…]
The answer is zero, as shown below.
Does that make sense to you? Yes? Great! Because many think it’s weird: they think the T apparatus must ‘re-orient’ the angular momentum of the particles. It doesn’t: if the filter is wide open, then “no information is lost”, as Feynman puts it. Still… Have a look at it. It looks like we’re opening ‘more channels’ in the last example: the S and S’ filter are the same, indeed, and T is fully open, while it selected for 0-state particles before. But no particles come through now, while with the 0-channel, we had γαN.
Hmm… It actually is kinda weird, won’t you agree? Sorry I had to talk about this, but it will make you appreciate that second ‘Law’ now: we can always insert a ‘wide-open’ filter and, hence, split the beams into a complete set of base states − with respect to the filter, that is − and bring them back together provided our filter does not produce any unequal disturbances on the three beams. In short, the passage through the wide-open filter should not result in a change of the amplitudes. Again, as Feynman puts it: the wide-open filter should really put Humpty-Dumpty back together again. If it does, we can effectively apply our ‘Law’:
For an example, I’ll refer you to my previous post. This brings me to the third and final ‘Law’.
[III] The amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ:
〈 χ | φ 〉 = 〈 φ | χ 〉*
This is probably the weirdest ‘Law’ of all, even if I should say, straight from the start, we can actually derive it from the second ‘Law’, and the fact that all probabilities have to add up to one. Indeed, a probability is the absolute square of an amplitude and, as we know, the absolute square of a complex number is also equal to the product of itself and its complex conjugate:
|z|= |z|·|z| = z·z*
[You should go through the trouble of reviewing the difference between the square and the absolute square of a complex number. Just write z as a + ib and calculate (a + ib)= a2 + 2ab+ b2 , as opposed to |z|= a2 + b2. Also check what it means when writing z as r·eiθ = r·(cosθ + i·sinθ).]
Let’s applying the probability rule to a two-filter set-up, i.e. the situation with the S and the tilted T filter which we described above, and let’s assume we’ve got a pure beam of +S particles entering the wide-open T filter, so our particles can come out in either of the three base states with respect to T. We can then write:
〈 +T | +S 〉+ 〈 0T | +S 〉+ 〈 −T | +S 〉= 1
⇔ 〈 +T | +S 〉〈 +T | +S 〉* + 〈 0T | +S 〉〈 0T | +S 〉* + 〈 −T | +S 〉〈 −T | +S 〉* = 1
Of course, we’ve got two other such equations if we start with a 0S or a −S state. Now, we take the 〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉 ‘Law’, and substitute χ and φ for +S, and all states for the base states with regard to T. We get:
〈 +S | +S 〉 = 1 = 〈 +S | +T 〉〈 +T | +S 〉 + 〈 +S | 0T 〉〈 0T | +S 〉 + 〈 +S | –T 〉〈 −T | +S 〉
These equations are consistent only if:
〈 +S | +T 〉 = 〈 +T | +S 〉*,
〈 +S | 0T 〉 = 〈 0T | +S 〉*,
〈 +S | −T 〉 = 〈 −T | +S 〉*,
which is what we wanted to prove. One can then generalize to any state φ and χ. However, proving the result is one thing. Understanding it is something else. One can write down a number of strange consequences, which all point to Feynman‘s rather enigmatic comment on this ‘Law’: “If this Law were not true, probability would not be ‘conserved’, and particles would get ‘lost’.” So what does that mean? Well… You may want to think about the following, perhaps. It’s obvious that we can write:
|〈 φ | χ 〉|= 〈 φ | χ 〉〈 φ | χ 〉* = 〈 χ | φ 〉*〈 χ | φ 〉 = |〈 χ | φ 〉|2
This says that the probability to go from the φ-state to the χ-state is the same as the probability to go from the χ-state to the φ-state.
Now, when we’re talking base states, that’s rather obvious, because the probabilities involved are either 0 or 1. However, if we substitute for +S and −T, or some more complicated states, then it’s a different thing. My guts instinct tells me this third ‘Law’ – which, as mentioned, can be derived from the other ‘Laws’ – reflects the principle of reversibility in spacetime, which you may also interpret as a causality principle, in the sense that, in theory at least (i.e. not thinking about entropy and/or statistical mechanics), we can reverse what’s happening: we can go back in spacetime.
In this regard, we should also remember that the complex conjugate of a complex number in polar form, i.e. a complex number written as r·eiθ, is equal to r·eiθ, so the argument in the exponent gets a minus sign. Think about what this means for our a·ei·θ ei·(ω·t − k ∙x) = a·e(i/ħ)·(E·t − pxfunction. Taking the complex conjugate of this function amounts to reversing the direction of t and x which, once again, evokes that idea of going back in spacetime.
I feel there’s some more fundamental principle here at work, on which I’ll try to reflect a bit more. Perhaps we can also do something with that relationship between the multiplicative inverse of a complex number and its complex conjugate, i.e. z−1 = z*/|z|2. I’ll check it out. As for now, however, I’ll leave you to do that, and please let me know if you’ve got any inspirational ideas on this. 🙂
So… Well… Goodbye as for now. I’ll probably talk about the Hamiltonian in my next post. I think we really did a good job in laying the groundwork for the really hardcore stuff, so let’s go for that now. 🙂
Post Scriptum: On the Uncertainty Principle and other rules
After writing all of the above, I realized I should add some remarks to make this post somewhat more readable. First thing: not all of the rules are there—obviously! Most notably, I didn’t say anything about the rules for adding or multiplying amplitudes, but that’s because I wrote extensively about that already, and so I assume you’re familiar with that. [If not, see my page on the essentials.]
Second, I didn’t talk about the Uncertainty Principle. That’s because I didn’t have to. In fact, we don’t need it here. In general, all popular accounts of quantum mechanics have an excessive focus on the position and momentum of a particle, while the approach in this and my previous post is quite different. Of course, it’s Feynman’s approach to QM really. Not ‘mine’. 🙂 All of the examples and all of the theory he presents in his introductory chapters in the Third Volume of Lectures, i.e. the volume on QM, are related to things like:
• What is the amplitude for a particle to go from spin state +S to spin state −T?
• What is the amplitude for a particle to be scattered, by a crystal, or from some collision with another particle, in the θ direction?
• What is the amplitude for two identical particles to be scattered in the same direction?
• What is the amplitude for an atom to absorb or emit a photon? [See, for example, Feynman’s approach to the blackbody radiation problem.]
• What is the amplitude to go from one place to another?
In short, you read Feynman, and it’s only at the very end of his exposé, that he starts talking about the things popular books start with, such as the amplitude of a particle to be at point (x, t) in spacetime, or the Schrödinger equation, which describes the orbital of an electron in an atom. That’s where the Uncertainty Principle comes in and, hence, one can really avoid it for quite a while. In fact, one should avoid it for quite a while, because it’s now become clear to me that simply presenting the Uncertainty Principle doesn’t help all that much to truly understand quantum mechanics.
Truly understanding quantum mechanics involves understanding all of these weird rules above. To some extent, that involves dissociating the idea of the wavefunction with our conventional ideas of time and position. From the questions above, it should be obvious that ‘the’ wavefunction does actually not exist: we’ve got a wavefunction for anything we can and possibly want to measure. That brings us to the question of the base states: what are they?
Feynman addresses this question in a rather verbose section of his Lectures titled: What are the base states of the world? I won’t copy it here, but I strongly recommend you have a look at it. 🙂
I’ll end here with a final equation that we’ll need frequently: the amplitude for a particle to go from one place (r1) to another (r2). It’s referred to as a propagator function, for obvious reasons—one of them being that physicists like fancy terminology!—and it looks like this:
The shape of the e(i/ħ)·(pr12function is now familiar to you. Note the r12 in the argument, i.e. the vector pointing from r1 to r2. The pr12 dot product equals |p|∙|r12|·cosθ = p∙r12·cosθ, with θ the angle between p and r12. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r12. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r12 is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r12|·cosθ = p∙r12. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r12/ħ. The function then reduces to (ħ/p)·eix/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while it’s going to infinity when very near. [Note that the absolute square, using cos(x)/x and sin(x)/x yields the same graph as squaring 1/x—obviously!] | 4,942 | 17,908 | {"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} | 3.65625 | 4 | CC-MAIN-2020-16 | latest | en | 0.861697 |
http://www.cbsemaster.org/2013/12/Chapter-10-Light-Reflection-and-Refraction-Physics-Science-Class-10th-CBSE.html | 1,531,813,232,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00506.warc.gz | 428,896,760 | 28,199 | ## Sunday, 1 December 2013
### CBSE Class 10th Science | Chapter 10. Light – Reflection and Refraction | Solved Exercises
#### Solved Exercises | Chapter 10. Light – Reflection and Refraction | Science | Class 10th |CBSE
Question 1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Question 2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer. (d) Between the pole of the mirror and its principal focus.
Question 3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Question 4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Question 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane.
(b) concave.
(c) convex.
(d) either plane or convex.
Answer. (d) either plane or convex.
Question 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer. (c) A convex lens of focal length 5 cm.
Question 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer. (a). The range of distance of the object from the mirror should be less than 15 cm i.e. from 0 to 15 cm in the front of mirror from the pole.
(b). The nature of image so formed will be virtual and erect.
(c) The size of image will be larger than object
Question 8. Name the type of mirror used in the following situations.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Answer. (a) In headlights of a car, type of mirror used is concave as the light of the lamp, under goes divergence from reflector surface and cover a large area in the front.
(b) Side/rear-view mirror of a vehicle is a convex mirror as it gives a diminished, Virtual and an erect image of the side or rear with wider field of view . A convex mirrors enable the driver to view much larger area than would be possible with a plane mirror
(c) Solar furnace is a concave mirror as Sun rays after reflection from its surface, get converged at focus with much intense heat.
Question 9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer. Yes, the half covered lens will still produce a complete image of the object but the image so formed may not be as intense as formed with uncovered lens. In fact, parts or broken pieces of lens behave like a complete lens and form complete image.
Verification : Take a convex lens. Light a candle. Now form an image of a burning candle on a white surface on the other side of lens by adjusting the distance between lens and candle. We can observe, a complete real and an inverted image of candle. Now cover half of lens with black paper, and try to form an image now. We can observe that again a complete, real and an inverted image of candle is formed. Image formed was less intense from earlier.
Question 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer. Height of the object h = + 5 cm
Focal length f = + 10 cm
object-distance u = –25 cm
Image-distance v = ?
Height of the image h′ = ?
1 1 1 − = v u f => 1 1 1 = − v 10 25 => 5 − 2 3 = = 50 50 => 50 v = = 16.66 cm 3
A real and inverted image will be formed on other side of lense at 16.66 cm from its optical centre
=> v m = u => h2 16.66 = h1 − 25 => h2 16.66 = 5 − 25 => −16.66 × 5 h2 = 25 => h2 = − 3.33 cm
An inverted, 3.33 cm high image will be formed.
Question 11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer. Focal length f = − 15 cm
Image-distance v = − 10 cm
object-distance u = ?
1 1 1 − = v u f 1 1 1 − = − 10 u − 15 − 1 1 1 + = 10 15 u => 1 − 3 + 2 = u 30 => u = − 30 cm
Object is placed at a distance of 30 cm from concave lens
Question 12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer. Focal length f = 15 cm
object-distance u = –10 cm
Image-distance v = ?
1 1 1 + = v u f 1 1 1 + = v − 10 15 1 1 1 = + v 15 10 1 2+3 5 = = v 30 30 v = 6 cm
A virtual and erect image is formed 6 cm behind the mirror.
Question 13. The magnification produced by a plane mirror is +1. What does this mean?
Answer. The magnification produced by a plane mirror is +1 implies that the image formed by a plane mirror is virtual , erect and of the same size, as that of object.
Question 14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Radius of curvature, R = + 30 cm
Focal length, f = R/2 = + 30/2 cm = + 15 cm
Object-distance, u = – 20 cm
Height of object
h1′= 5 cm
Image-distance, v = ? Height of image h2′= ?
1 1 1 + = v u f 1 1 1 + = v − 20 + 15 1 1 1 = + v 15 20 1 4 + 3 7 = = v 60 60 60 v = = 8.57 cm 7
The image is formed behind the mirror at a distance of 8.6 cm
h2 − v m = = h1 u h2 8.57 => = = 5 cm 20 8.57 × 5 cm = h2 = 20 Height (Size ) of Image = h2 = 2.175 cm
Thus, a 2.175 cm high, virtual and erect image is formed
Question 15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer. Focal length, f = - 18 cm
Object-distance, u = – 27 cm
Height of object, h1′= 5 cm
Image-distance, v = ?
Height of image h2′= ?
1 1 1 + = v u f 1 1 1 + = v − 27 -18 15 − 1 1 1 = + v 18 27 1 − 3 + 2 1 = = v 54 54 60 v = = 54 cm 7
The screen should be kept at a distance of 54 cm in front of mirror
h2 − v m = = h1 u h2 (− 54 ) => = = 7 cm (− 27 ) (− 54 ) × 7 cm => h2 = (− 27 ) Height (Size ) of Image = h2 = 2 × 7 cm = 14 cm cm
Thus, a 14 cm high, virtual and an inverted image is formed
Question 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
1 P = f 1 - 2.0 = f −1 f = m 2 − 1 f = × 100 cm 2 f = − 50 cm = - 0.50 cm
The lens is a concave lens
Question 17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer. Power of lens, P = + 1.5 D
1 P = f 1 1.5 = f 1 f = m 1.5 10 f = m 15 10 f = m 15 2 f = m = + 0.67 m 3
Focal length of the lens is + 0.67 m. The prescribed lens is converging type in nature.
Additional Questions | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
Question 1. Write the mirror formula :
Answer. In mirror, the distance of the object from its pole is called the object distance (u)
The distance of the image from the pole of the mirror is called the image distance (v)
The distance of the principal focus from the pole is called the focal length (f).
There is a relationship between these three quantities given by the mirror formula which is expressed as :
1 1 1 + = v u f
Question 2. Write one use of concave and convex mirrors each.
Answer. Concave mirror is used in some telescopes and also as magifying looking mirror while applying make-up or shaving. Convex mirror is used in vehicles as rear view mirror.
Question 3. Why do we use convex for side-view ?
Answer. Convex mirrors are used in vehicles for side-view because they give a virtual, upright, though diminished, image. Although images so formed are smaller but this results in showing a large area in the back drop with a wider field of view .
Question 4. How does a light ray bend when it travels from :
(i) A denser to rarer medium ?
(ii) Ararer to denser medium >
Answer.An optically denser medium has a larger refractive index, where as optically rarer medium has a lower refractive index. Due to refration, the speed of light is higher in a rarer medium than a denser medium also direction of propagation of light chages as it chages the medium. . (i) When light travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.
(ii) When light travels from a rarer medium to a denser medium, it slows down and bends towards the normal.
Question 5. When a convex lens is focussed on a distant object, where will the image be formed ? Show it with a ray diagram.
Answer. When a convex lens is focussed on a distant object, the image will be formed at the focus of the lens.
Question 6. What is the meaning of
(i) Optical Centre
(ii) Principal Axis
Answer. (i) Optical Centre : The central point of a lens is its optical centre. It is usually represented by the letter O. A ray of light through the optical centre of a lens passes without suffering any deviation.
(ii) Principal Axis : An imaginary straight line passing through the two centres of curvature of a lens is called its principal axis. Optical centre and focus of lens lies on the Principal Axis.
Question 7. When does a convex lens form
(i) A virtual, erect, enlarged image ?
(ii) A real enlarged image ?
Answer. (i) A convex lens forms a virtual, erect and enlarged image, when the object is placed between focus and optical centre of a lens on its other side.
(ii) A convex lens forms a real and enlarged image when the object is placed between focus (f) and centre of curvature (2f) of a lens on its other side.
Question 8. What is the relationship between the focal length of a spherical mirror and radius of curvature ?
Answer. The focal length of a spherical mirror (f) is equal to half its radius of curvature (R) i.e.
Focal length, f = R/2
Question 9. Explain the term Magnification produced by a spherical mirror ?
Answer. Magnification produced by a spherical mirror gives the relative extent to which the image of an object is magnified with respect to the object size. It is expressed as the ratio of the height of the image to the height of the object. It is usually represented by the letter m.
If h1 is the height of the object and h2 is the height of the image, then the magnification m produced by a spherical mirror is given by
Height of the image ( h1) m = Height of the object (h2 ) h1 m = h2
The magnification m is also related to the object distance (u) and image distance (v). It can be expressed as:
h1 v Magnification (m) = = h2 u
Question 10. Name and explain the sign Convention for Reflection by Spherical Mirrors
Answer. While dealing with the reflection of light by spherical mirrors, we follow a set of sign conventions called the New Cartesian Sign Convention. In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system. The conventions are as follows –
(i) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.
(ii) All distances parallel to the principal axis are measured from the pole of the mirror.
(iii) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.
(iv) Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive.
(v) Distances measured perpendicular to and below the principal axis (along –y-axis) are taken as negative.
Question 11. Explain the following given terms
(i) Refraction of light
(ii) Laws of refraction of light
(ii) The Refractive Index
(i) Refraction of light : The direction of propagation of light, When traveling obliquely from one medium to another is subject to change. When light travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. When light travels from a rarer medium to a denser medium, it slows down and bends towards the normal. This phenomenon of bending of light ray is known as refraction of light. Refraction is caused due to change in the speed of light as it enters from one transparent medium to another. The speed of light increases in rarer medium and decreases in denser medium.
(ii) Laws of refraction of light : The reflecting surfaces, of all types, obey the laws of reflection. The refracting surfaces obey the laws of refraction.
The following are the laws of refraction of light.
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell’s law of refraction.
If i is the angle of incidence and r is the angle of refraction, then,
sin i = constant sin r
(ii) The Refractive Index : The refractive index of a transparent medium is the ratio of the speed of light in vacuum to that in the medium. It is also termed as the absolute refractive index of a given transparent medium .
If c is the speed of light in air and v is the speed of light in the medium, then, the refractive index of the medium nm is given by
Speed of light in air c nm = = Speed of light in the medium v
Intext Questions | Page 168 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
Question 1. Define the principal focus of a concave mirror.
Answer. The principal focus of the concave mirror is an point on the principal axis of the mirror, where incident rays of light, parallel to the principal axis, after reflection from mirror surface, intersect each other .
Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer. For spherical mirrors of small apertures, the radius of curvature is twice the focal length i.e. R = 2f . Here radius of curvature R = 20 cm, focal length f = ?
R 20 f = = = 10 cm 2 2
Focal length of mirror is 10 cm
Question 3. Name a mirror that can give an erect and enlarged image of an object.
Answer. Concave mirror gives an erect and enlarged image of an object, when the object is between pole (P) and principal focus of mirror (C).
Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer. Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles because they give an erect, virtual, full size diminished image of distant objects with a wider field of view. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.
Intext Questions | Page 171 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer.We know,for spherical mirrors of small apertures, the radius of curvature is twice the focal length i.e. R = 2f .
Given here, radius of curvature R is 32 cm. and focal length of a convex mirror f = ?
R 32 f = = = 16 cm 2 2
Focal length of mirror is 16 cm
Question 2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer. Here given Magnification m = 3, Object-distance u = 10 cm
v Magnification m = − Real image u v − 3 = − u 3 u = v v = 3u = 3 × − 10 cm = − 30 cm
The image will be formed at a distance of 30 cm in front of convex mirror from its the pole
Intext Questions | Page 176 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
Question 1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer. The light ray bends towards the normal as it travels from a rarer medium of air to a denser medium of water, under goes refraction. Refraction is due to change in the speed of light as it enters from one transparent medium to another. The speed of light increases in rarer medium and decreases in denser medium
Question 2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1.
Answer. Given, speed of light in vacuum C = 3 × 108 m s–1
Refractive index of glass ng = 1.50
Speed of light in the glass vg = ?
C ng = Vg 3 × 10 8 ng = Vg 3 × 10 8 1.5 = Vg 3 × 10 8 Vg = 1.5 Vg = 2 × 108 ms−1
Speed of light in the glass vg = 2 × 108 ms−1
Question 3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer.The medium having highest optical density : Diamond ( Refractive Index 2.42 )
The medium having lowest optical density : Air ( Refractive Index 1.0003 )
Question 4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.
Answer. Using the information given in table, the refrative index of kerosene is 1.44, that of turpentine is 1.47 and that of water is 1.33. Clearly, water having lower refractive index 1.33, is optically rarer than kerosene and turpentine. Therefor the light travels fastest in water because of its lower optical density
Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer. The diamond with refractive index of 2.42, is the most ‘optically denser medium' and there for the speed of light in diamond will be less i.e. 1.23 × 108 ms−1 (= 3 × 108 ms−1/2.42 ) compare to speed of light in vacuum C = 3 × 108 m s- 1
Intext Questions | Page 184 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
Question 1. Define 1 dioptre of power of a lens.
Answer. 1 dioptre is SI unit of the power of a lens whose focal length is 1 metre. It is denoted by the letter D. Thus 1D = 1 m-1. Simply, when focal length ' f ' is expressed in metres, then, power is expressed in dioptres. The power of a convex lens is positive and that of a concave lens is negative.
Question 2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
v = 50 m = 1
1 P = = ? f
As the image formed by lens is real
v m = u v 1 = u v = u u = − 50 cm
∴ Object distance is = 50 cm
1 1 1 − = v u f 1 1 1 − = 50 − 50 f
1 + 1 1 = 50 f 2 f = 50 f = 25 cm = .25 m
1 100 P = + = + = + 4 dioptre .25 25
The power of convex lens is + 4 dioptre
Question 3. Find the power of a concave lens of focal length 2 m.
Focal length of lens f = − 2 m Power of concave lens P = ?
1 P = − f 1 P = − 2
P = − 0.5 dioptre
The power of concave lens is − 0.5 dioptre
Activity 10.1 |Page 161 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Take a large shining spoon. Try to view your face in its curved surface.
• Q.1. Do you get the image? Is it smaller or larger?
• Answer : Yes, the image of the face formed on outer curved surface is smaller in size.
• Q.2. Move the spoon slowly away from your face. Observe the image. How does it change?
• Answer : The size of image gradually decreases with a increase in field of view.
• Q.3. Reverse the spoon and repeat the Activity. How does the image look like now?
• Answer : Earlier, when the spoon was close the image formed on the inner curved surface was erect and magnified and as we moved the spoon slowly away from our face,the image transitioned to a inverted image with gradual decrease in its size.
• Q.4. Compare the characteristics of the image on the two surfaces.
• Outer Surface Inner Surface (i) Image is always erect (ii) Image size is gradually decreases as we move away the spoon (i) The image is erect when spoon is close and inverted when spoon is away (ii) Image size is larger when spoon is close and it is smaller when spoon is moved away
• Q. 5. Why do we see our image in the shining spoon?
• Answer : The surface of a shining spoon acts like a mirror. Due to reflection of light from its surfaces, we can see our image
• Q. 6. What types of mirrors are formed by the inner and outer curved surfaces of a spoon ?
• Answer : The inner curved surfaces of a spoon forms a concave mirror and the outer curved surfaces of a spoon forms a convex mirror
Activity 10.2 | Page 162 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
CAUTION: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.
• Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
• Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
• Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
• Q.1. Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?
• Answer : The light rays from the Sun, are concentrated at focus, and form a sharp spot of light on the sheet. Paper starts burning after some time due to increase intensity of reflected sun light from the mirror
• Q.2. Why, we should not look at the Sun directly or even into a mirror reflecting sunlight
• Answer : Because the intense heat resulting from the concentrated sun light through eye lens may burn the retina wall with dark spots. This may result in partial or complete vision impairment.
Activity 10.3 | Page 163 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
You have already learnt a way of determining the focal length of a concave mirror. In Activity 10.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror.
• Take a concave mirror. Find out its approximate focal length in the way described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
• Mark a line on a Table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
• Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies mid-way between the pole P and the centre of curvature C.
• Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
• Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
• Repeat the activity by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
• In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
• Q. 1. Note down and tabulate your observations.
• Answer : Image formation by a concave mirror for different positions of the object
Position of the object Position of the image Size of the image Nature of the image At infinity At the focus F Highly diminished,point-sized Real and inverted Beyond C Between F and C Diminished Real and inverted At C At C Same size Real and inverted Between C and F Beyond C Enlarged Real and inverted At F At infinity Highly enlarged Real and inverted Between P and F Behind the mirror Enlarged Virtual and erect
• Q. 2. What is the relation between focal length and radius of curvature ?
• Answer : Radius of curvature is equal to twice the focal length. We put this as R = 2f
• Q. 3. What is the nature of image formed when object is placed far beyond C ?
• Answer : A is real, inverted and highly dimnished image is formed at focus.
• Q. 4. What is the nature and position of image formed when object is placed just beyond C ?
• Answer : A is real, inverted and dimnished image is formed between Focus (F) and centre of curvature (C).
• Q. 5. Draw the ray diagram of formation of image when object is kept at center of curvature in front of a concave mirror.
• Q. 6. In which case the image formed by concave mirror is not obtained on the screen ?
• Answer : The image formed by concave mirror is not obtained on the screen, when the object is kept between focus (f) and pole (P) of a concave mirror.
• Q. 7. In which case the image of an object by concave mirror can be obtained on the screen ?
• Answer : The image of an object by concave mirror can be obtained on the screen, when the object is kept beyond focal length of the mirror.
Activity 10.4 | Page 166 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Q. 1. Draw neat ray diagrams for each position of the object shown in Table 10.1.
Position of the object Ray diagram At infinity Beyond C At C Between C and F At F Between P and F
• You may take any two of the rays mentioned in the previous section for locating the image.
• Compare your diagram with those given in Fig. 10.7 of textbook.
• Answer : They were identical and matching
• Describe the nature, position and relative size of the image formed in each case.
• Tabulate the results in a convenient format.
Position of the object Position of the image Size of the image Nature of the image At infinity At the focus F Highly diminished,point-sized Real and inverted Beyond C Between F and C Diminished Real and inverted At C At C Same size Real and inverted Between C and F Beyond C Enlarged Real and inverted At F At infinity Highly enlarged Real and inverted Between P and F Behind the mirror Enlarged Virtual and erect
Activity 10.5 | Page 167 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Take a convex mirror. Hold it in one hand.
• Hold a pencil in the upright position in the other hand.
• Q. 1. Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
• Answer : The image is erect and diminished
• Q. 2. Move the pencil away from the mirror slowly. Does the image become smaller or larger?
• Answer : The image becomes smaller.
• Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
• Answer : The image will move closer to the focus
Activity 10.6 | Page 167-68 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Observe the image of a distant object, say a distant tree, in a plane mirror.
• Q. 1. Could you see a full-length image?
• Answer : No, We can not see a full-length image of a distant object in a plain mirror.
• Q. 1. Try with plane mirrors of different sizes. Did you see the entire object in the image?
• Answer : No, the result was same as before.
• Q. 4. Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
• Q. 4. Now try using a convex mirror. Did you succeed? Explain your observations with reason.
• Answer : Yes, now we could see full length image of distant object with wider field of view. The image formed was diminished, erect and virtual. Due to this reason, they are used as rear or side view mirrors in vehicles as distant objects in the backdrop can be seen clearly with much larger field of view .
Activity 10.7 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Place a coin at the bottom of a bucket filled with water.
• Q. 1. With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
• Q. 2. Repeat the Activity. Why did you not succeed in doing it in one go?
• Answer : Because on seeing, the coin appeared to be closer than its actual distance, so we are likely to miss the coin. Reflected light coming from the submerged coin in denser medium of water, on entering air which is a rarer medium, bend away from the normal due to refraction of light and image size becomes larger than its actual size, thus submerged object apparently seem closer.
Activity 10.8 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Place a large shallow bowl on a Table and put a coin in it.
• Move away slowly from the bowl. Stop when the coin just disappears from your sight.
• Ask a friend to pour water gently into the bowl without disturbing the coin.
• Q. Keep looking for the coin from your position. Does the coin becomes visible again from your position? How could this happen?
• Answer : Yes, on pouring water into the bowl, the coin becomes visible again because due to refraction of light, for our eyes, the submerged coin apparently seems raised above its actual level and thus becomes visible on seeing from the same side and distance
Activity 10.9 | Page 172 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
• Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
• Q. 1. Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
• Answer :Yes, due to the refration of light, the line under the glass slab appear to be bent at the edges
• Q. 2. Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
• Answer : No, Now the part of the line under the glass slab appear in a straight line. Because a ray of light, which is perpendicular to the plain of a refracting medium, does not change its angle due to refraction.
• Q. 3. Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
• Answer : Yes, the part of the line, beneath the slab, appear to be raised. Due to refraction of light, apparent position of image of object seems nearer than its actual position.
Activity 10.10 | Page 173 | Chapter 10. Light – Reflection and Refraction | CBSE Class 10th Science
• Fix a sheet of white paper on a drawing board using drawing pins.
• Place a rectangular glass slab over the sheet in the middle.
• Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
• Take four identical pins.
• Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
• Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
• Remove the pins and the slab.
• Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O′.
• Join O and O′. Also produce EF up to P, as shown by a dotted line in Fig. 10.10.
Question 1. What happens to incident ray as it enters the glass slab ?
Answer : The incident ray as it enters from a rarer medium of air to a denser medium of glass, bends towards the normal, due to refraction of light . The refraction is cuased by change in the speed of light as it enters from one transparent medium to another.
Question 2. What happens to emergent ray as it leaves the glass slab ?
Answer : The emergent ray as it leaves the denser medium of glass and enters into a rarer medium of air, bends away from the normal, due to refraction of light. The refraction is caused by change in the speed of light as it enters from one transparent medium to another.
Question 3. What is the perpendicular distance between directions of incident and emergent rays ?
Answer : Lateral displacement. This gives a measure of path deviation of refracted rays due to refraction.
Question 4. As given in the activity above, the medium of incidence and emergent ray is same (air), what could be the possible observations for angle of incidence and angle of emergence ?
Answer : When the medium of incidence and emergent ray is same (air), angle of incidence is equal to angle of emergence. | 8,563 | 33,592 | {"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} | 3.65625 | 4 | CC-MAIN-2018-30 | longest | en | 0.86081 |
https://qualitymathtutors.com/the-ultimate-guide-to-algebra-assignment-help-services-online/ | 1,701,444,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00171.warc.gz | 550,517,176 | 24,249 | # The Ultimate Guide to Algebra Assignment Help Services Online
Algebra can be a daunting subject for many students, especially when it comes to solving equations and factoring expressions. However, mastering this important branch of mathematics is crucial for success in many fields, including science, engineering, economics, and finance. That’s why algebra assignment help services have become increasingly popular and necessary.
In this ultimate guide, we will take you through the essential concepts of algebra and show you how to find the right assignment help services online. You will learn how to perform algebraic operations, solve equations, factor expressions, and tackle advanced concepts such as graphing linear equations and matrix operations. You will also discover the benefits of using algebra assignment help services, how to choose a provider, and get answers to common FAQs.
By the end of this guide, you will have the knowledge and tools to boost your math skills and grades in algebra.
## Algebraic Operations
Algebraic operations form the building blocks of algebra. They involve manipulating and solving equations with variables and numbers. Here are some of the most important algebraic operations you need to master:
### Addition and Subtraction of Numbers
In algebra, you can add or subtract numbers by combining like terms. For example, 3x + 5x = 8x, and 4y – 2y = 2y.
### Multiplication and Division of Numbers
Multiplying and dividing numbers in algebra involves applying the distributive property and canceling out common factors. For example, 3(x + 2) = 3x + 6, and 2y / 4 = 1/2 y.
### Solving Basic Algebraic Equations
Solving algebraic equations involves isolating the variable on one side of the equation and simplifying the other side. For example, 2x + 5 = 11 can be solved by subtracting 5 from both sides and then dividing by 2, giving x = 3.
## Understanding Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations such as addition, subtraction, multiplication, and division. Here’s what you need to know about algebraic expressions:
### Introduction to Algebraic Expressions
Algebraic expressions can be simple or complex, and they can contain one or more variables. For example, 3x + 2y is a simple expression, while (x + 3y) / (x – y) is a complex expression.
### Evaluating Algebraic Expressions
Evaluating algebraic expressions involves replacing the variables with given values and simplifying the expression. For example, if x = 2 and y = 4, then 3x + 2y = 3(2) + 2(4) = 6 + 8 = 14.
### Factoring Algebraic Expressions
Factoring algebraic expressions involves finding common factors and simplifying the expression. For example, 6x^2 + 12x can be factored as 6x(x + 2).
### Simplifying Algebraic Expressions
Simplifying algebraic expressions involves combining like terms and applying distributive properties. For example, 2x + 3x – 4y can be simplified as 5x – 4y.
## Solving Algebraic Equations
Algebraic equations involve setting two expressions equal to each other and solving for the unknown variable. Here are the types of algebraic equations you need to know:
### What Are Algebraic Equations?
Algebraic equations are mathematical expressions involving an equal sign. They can be linear, quadratic, or polynomial, depending on the degree of the variable.
### Solving Linear Equations
Linear equations involve one variable with a degree of 1. They can be solved by isolating the variable and simplifying the other side. For example, 2x + 7 = 11 can be solved by subtracting 7 from both sides and then dividing by 2, giving x = 2.
Quadratic equations involve one variable with a degree of 2. They can be solved by using the quadratic formula or factoring the expression. For example, x^2 + 6x + 8 = 0 can be factored as (x + 4)(x + 2), giving x = -4 or x = -2.
### Solving Polynomial Equations
Polynomial equations involve one or more variables with a degree of 3 or higher. They can be solved by factoring the expression or using numerical methods. For example, x^3 – 7x^2 + 8x + 32 = 0 can be factored as (x – 4)(x + 2)(x – 4).
Advanced concepts in algebra involve complex operations and equations. Here are some of the concepts you might encounter:
### Graphing Linear Equations
Graphing linear equations involves plotting points on a coordinate plane and drawing a line that connects them. For example, the equation y = 2x + 3 can be graphed as a straight line with a slope of 2 and a y-intercept of 3.
### Matrix Operations
Matrix operations involve manipulating matrices, which are arrays of numbers. They can be added, subtracted, multiplied, and inverted. For example, the matrix A = [[1, 2], [3, 4]] can be multiplied by the matrix B = [[5], [6]] to give the matrix AB = [[17], [39]].
### Complex Numbers
Complex numbers involve numbers with both real and imaginary parts. They can be added, subtracted, multiplied, and divided. For example, the complex number 2 + 3i can be added to the complex number 4 – 2i to give the complex number 6 + i.
### Trig Functions and Identities
Trig functions and identities involve using trigonometric functions such as sine, cosine, and tangent to solve problems involving angles and triangles. For example, the sine function can be used to find the height of a triangle given the angle and hypotenuse.
## Benefits of Using Algebra Assignment Help Services
Algebra assignment help services can provide many benefits for students who are struggling with algebra. Here are some of the benefits you can expect:
### Customized Assistance
Algebra assignment help services can provide customized assistance to match your needs and learning style. They can provide one-on-one tutoring, personalized assignments, and feedback to help you improve your algebra skills.
### Expertise and Experience
Algebra assignment help services employ expert tutors who have years of experience in algebra and related fields. They can provide you with deep insights into the subject and help you solve complex problems.
### Time-Saving
Algebra assignment help services can save you time by providing instant solutions to your algebraic problems. You don’t have to spend hours struggling with equations and expressions.
Algebra assignment help services can help you improve your grades by providing you with high-quality assignments, feedback, and guidance. You can learn from their expertise and apply it to your own work.
## Finding the Right Algebra Assignment Help Service
Finding the right algebra assignment help service can be tricky, as there are many providers out there. Here are some factors to consider when choosing a provider:
### Reputation and Reviews
Look for providers with a good reputation and positive reviews from past clients. This can give you an idea of their quality of work and level of expertise.
### Expertise and Experience
Choose a provider with tutors who have in-depth knowledge of algebra and related disciplines. Look for providers with high standards of education and experience.
### Quality of Work
Choose a provider that guarantees high-quality work and plagiarism-free assignments. Look for providers that have strict quality control policies and use advanced technologies to check for errors and originality.
### Pricing and Payment Options
Choose a provider that offers competitive pricing and transparent payment options. Look for providers that don’t charge hidden fees and offer flexible payment plans.
## Conclusion
In this ultimate guide, we have covered the essential concepts of algebra, from algebraic operations and expressions to equations and advanced topics such as matrix operations and trig functions. We have also discussed the benefits of using algebra assignment help services and provided tips on how to find the right provider.
By using this guide, you have the knowledge and tools to improve your math skills and grades in algebra. Don’t be afraid to seek out algebra assignment help services if you need extra assistance, and always choose a provider that has the expertise, experience, and quality of work you need.
## FAQs
### Q. What are some common algebraic operations?
Common algebraic operations include addition, subtraction, multiplication, and division of numbers and variables.
### Q. How do I simplify an algebraic expression?
To simplify an algebraic expression, combine like terms and apply distributive properties.
### Q. What is the difference between an equation and an expression?
An equation involves setting two expressions equal to each other and solving for the variable, while an expression is a combination of variables, numbers, and operations.
### Q. How can I improve my skills in algebra?
You can improve your skills in algebra by practicing regularly, seeking out algebra assignment help services, and using online resources such as tutorials and practice problems.
### Q.Can I trust online algebra assignment help services?
Yes, as long as you choose a reputable provider with high standards of education, expertise, and quality of work.
### Q.How much should I expect to pay for algebra assignment help services?
The cost of algebra assignment help services varies depending on the provider, level of expertise, and type of service. However, most providers offer competitive pricing and flexible payment options.
### Q. Can I get a refund if I am not satisfied with the work provided by an algebra assignment help service?
Most reputable algebra assignment help services offer refunds or revisions if you are not satisfied with the work provided. However, be sure to check the provider’s refund policy before making a purchase.
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You Are Here: Home » GATE Question Papers » GATE 1998 Exam Civil Engineering Question Paper
Here is the Civil Engineering GATE Exam sample written Multiple Choice Question paper.
GATE – 1998
CE : Civil Engineering
Section A
(100 marks)
1. For each subquestion given below four answers are provided out of which only one is correct. Indicate in the answer book the correct or most appropriate answer by writing the letter A,B,C or D against the subquestion number. (31×1=31)
1.1 If A is a real square matrix, then AAT is
(a) unsymmetric (b) always symmetric
(c) skew symmetric (d) sometimes symmetric
1.2 In matrix algebra AS = AT (A,S,T, are matrices of appropriate order) implies S=T only if
(a) A is symmetric (b) A is singular
(c) A is non singular (d) A is skew symmetric
1.3 A discontinuous real function can be expressed as
(a) Taylor’s series and Fourier’s series
(b) Taylor’s series and not by Fourier’s series
(c) neither Taylor’s series nor Fourier’s series
(d) not by Taylor’s series, but by Fourier’s series
1.4 The Laplace Transform of a unit step function ua(T), defined as
0for1<a
Ua(t)= is
1 fort >a
(a) e-as/s (b)se-as
(c) s-u(0) (d) se-as-1
1.5 The continuous function ƒ(x,y) is said to have saddle point at (a,b) if
(a) ƒx (a,b)= ƒy(a,b) = 0; ƒsy2- ƒxx ƒyy<0 at (a,b)
(b) ƒx (a,b)= 0; ƒy(a,b) = 0; ƒxy2- ƒxx . ƒyy>0 at (a,b)
(c) ƒx (a,b)= 0; ƒy(a,b) = 0; ƒxx- and ƒyy<0 at (a,b)
(d) ƒx (a,b)= 0; ƒy(a,b) = 0; ƒxx-2 – ƒxx . ƒyy = 0 at (a,b)
1.6 The Taylor’s series expansion of sin x is
x2 x4 x2 x4
(a) 1- + (b) 1+
2! 4! 2! 4!
x3 x5 x3 x5
(a) x+ + (b) x-
3! 5! 3! 5!
1.7 A three hinged arch shown in Figure is quarter of a circle. If the vertical and horizontal components of reaction at A are equal, the value of θ is
(a) 600
(b) 450
(c) 300
(d) None in (00,900)
1.8 A propped cantilever beam is shown in Figure. The plastic moment capacity of the beam is M0. The collapse load P is
(a) 4M0/L (b) 6M0/L
(c) 8M0/L (d) 12M0/L
1.9 The maximum permissible deflection for gantry gride, spanning over 6m, on which an EOT (electric overhead travelling) crane of capacity 200 k.N is operating, is
(a) 8mm (b) 10mm
(c) 12mm (d0 18mm
1.10 An isolated T beam is used as a walkway. The beam is simply supported with an effective span of 6m. The effective width of flange, for the cross-section shown in Figure, is
(a) 900 mm
(b) 1000 mm
(c) 1259 mm
(d) 2200 mm
1.11 The plane of stairs supported at each end by landings spanning parallel with risers is shown in Figure. The effective span of staircase slab is
(a) 3000 mm (b) 4600 mm
(c) 4750 mm (d) 6400 I
1.12 Some of the structural strength of a clayey material that is lost by remoulding is slowly recovered with time. This property of soils to undergo an isothermal gel-to sol-to-get transformation upon agitation and subsequent rest is termed
(a) Isotropy (b) Anisotropy
(c) Thixotropy (d) Allotropy
1.13. If soil is dried beyond its shrinkage limit, it will show
(a) Large volume change
(b) Moderate volume change
(c) Low volume change
(d) No volume change
1.14. The stress-strain behaviour of soils as shown in the Figure correspondence to :
(a) Curve 1 : Loose sand and normaly consolidated clay
Curve 2 : Loose sand and over consolidated clay
(b) Curve 1 : Dense sand and normally consolidated clay
Curve 2 : Loose sand and over consolidated clay
(c) Curve 1 : Dense sand and over consolidated clay
Curve 2 : Loose sand and normally consolidated clay
(d) Curve 1 : Loose sand and over consolidated clay
Curve 2 : Dense sand normally consolidated clay
1.15 In cohesive soils he depth of tension crack (Zcr) is likely to be
1.16 The settlement of prototype in granular material may be estimated using plate load test data from the following expression :
1.17 In which one of the following arrangement would the vertical force on the cylinder due to water be the maximum ?
1.18. At the same mean velocity, the ratio of head loss per unit length for a sewer pipe flowing full to that for the same pipe flowing half full would be
(a) 2.0 (b) 1.63
(c) 1.00 (d) 0.61
1.19 Three reservoirs A, B and C are interconnected by pipes as shown in the Figure. Water surface elevations in the reservoirs and the Pirzometric head at the junction J are indicated in the Figure.
Discharge Q1, Q2 and Q3 are related as
(a) Q1+Q2 = Q3 (b) Q1=Q2+Q3
(c) Q2=Q1+Q3 (d) Q1+Q2+Q3 = 0
1.20. The comparison between pumps operating in series and in parallel is
(a) Pumps operating is series boost the discharge, whereas pumps operating in parallel boost the head.
(b) Pumps operating in parallel boost the discharge, whereas pumps operating in series boost the head.
(c) In both cases there would be a boost in discharge only.
(d) In both case there would be a boost in head only.
1.21 The Bowen ratio is defined as
(a) Ratio of heat and vapour diffusivities
(b) Proportionality constant between vapour heat flux and sensible ehat flux.
(c) Ratio of actual evaportranspiration and potential evaportranspiration.
(d) Proportionality constant between heat energy used up in evaporation and the bulk radiation from a water body.
1.23. Excessive fluoride in drinking water causes
(a) Alzheimer’s disease
(b) Mottling of teeth and embrittlement of bones
(c) Methemoglobinemia
(d) Skin cancer
1.24 Coagulation-flocculation with alum is performed
(a) immediately before chlorination
(b) immediately after chlorination
(c) after rapid sand filtration
(d) before rapid sand filtration
1.25. Sewage treatment in an oxidation pond is accomplished primarily by
(a) alga-bacterial symbols
(b) algal photosynthesis only
(c) bacterial oxidation only
(d) chemical oxidation only
1.26 An inverted siphon is a
(a) device for distributing septic tank effluent to a soil absorption system
(b) device for preventing overflow from elevated water storage tank
(c) device for preventing crown corrosion of sewer
(d) section of sewer which is dropped below the hydraulic grade line in order to avoid an obstacle.
1.27 Water distribution systems are sized to meet the
(a) maximum hourly demand
(b) Average hourly demand
(c) maximum daily demand and fire demand
(d) average daily demand and fire demand.
1.28 At highway stretches where the required overtaking sight distance cannot be provided, it is necessary to incorporate in such sections the following
(a) at least twice the stopping sight distance
(b) half to the required overtaking sight distance
(c) one third of the required overtaking sight distance
(d) three times the stopping sight distance
1.29 The modulus of subgrade reaction is obtained from the plate bearing test in the form of load-deformation curve. The pressure corresponding to the following settlement value should be used or computing modulus of subgrade reaction
(a) 0.375 cm (b) 0.175 cm
(c) 0.125 cm (d) 0.250 cm
1.30 In the plate bearing test, if the load applied is in the form of an inflated type of wheel, then this mechanism corresponds to
(a) rigid plate (b) flexible plate
(c) semi-rigid plate (d) semi-elastic plat e
1.31. Base course is used in rigid pavements for
(b) prevention of slab cracking
(c) preventing of pumping
(d) preventing of thermal expansion
2. For each subquestion given below four answers are provided out of which one is correct. Indicate in the answer book the correct or most appropriate answer by writing the letter A,B, Cor Dagainst the subquestion number.
1 1 (22×2=44)
2.1 The infinite series 1+ + +…………
2 3
(a) converges (b) diverges
(c) oscillates (d) unstable
2.2 The real symmetric matrix C corresponding to the Quadratic form
Q=4x1x2 – 5x22′ is
(a) 1 2 (b) 2 0
2 -5 0 -5
(c) 1 1 (d) 0 2
1 -2 1 -5
2.3 A cantilever beam is shown in the Figure. The moment to be applied at free end for zero vertical deflection at that point is
(a) 9 kN.m clockwise
(b) 9 kN.m anti-clockwise
(c) 12kN.m clockwise
(d) 12kN.m anti-clockwise
2.4. The strain energy stored in member AB of the pin-joined truss is shown in Fig. 2.4, when E and A are same for all members, is
(a) 2P2L
AE
(b) P2L
AE
(c) P2L
2AE
(d) Zero
2.5 The stiffness matrix of a beam element is given as (2EI/L) 2 -1. Then the flexibility matrix is 1 2
(a) L 2 1 (b) L 1 -2
2EI 1 2 6EI -2 1
(a) L 2 -1 (b) L 2 -1
3EI -1 2 5EI -1 2
2.6 The plastic modulus of a section is 4.8×10-4m3. The shape factor is 1.2. The plastic moment capacity of the section is 120 kN.m. The yield stress of the material is
(a) 100MPa (b) 240MPa
(c) 250MPa (d)300MPa
2.7. A reinforced concrete wall carrying vertical loads is generally designed as per recommendations given for columns. The ratio of minimum reinforcements in the vertical and horizontal directions is
(a) 2 :1 (b) 5:3
(c) 1:1 (d) 3:5
2.8. The proposed dam shown in the figure is 90 m long and the coefficient of permeability of the soil is 0.0013mm/second. The quantity of water (m3) that will be lost per day be seepage is (rounded to the nearest number):
(a) 55 (b) 57
(c) 59 (d) 61
2.9 The time for a clay layer to achieve 90% consolidation is 15 years. The time required to achieve 90% consolidation, if the layer were twice as thick, 3 times more permeable and 4 times more compressible would be :
(a) 70 years (b) 75 years
(c) 80 years (d) 85 years
2.10. The total active thrust on a vertical wall 3m high retaining a horizontal sand backfill (unit weight γt=20 kN/m3, angle of shearing resistance φ’=300) when the water table is at the bottom of the wall, will be :
(a) 30 kN/m (b) 35 kN/m
(c) 40 kN/m (d) 45 kN/m
2.11 A 400 slope is excavated to a depth of 10(d)depth of 10 m is a deep layer of saturated clay of unit weight 20kN/m3; the relevant shear strength parameters are cu=72 kN/m2 and φu=0. The rock ledge is at a great depth. The Taylor’s stability coefficient for φu=0 and 400 slope angle is 0.18. The factor of safety of the load is :
(a) 2.0 (b) 2.1
(c) 2.2 (d) 2.3
2.12 A point load of 700 kN is applied on the surface of thick layer of saturated clay. Using Boussinesq’s elastic analysis, the estimated vertical stress (σv) at a depth of 2 m and a radial distance of 1.0 m from the point of application of the load is :
(a) 47.5 kPa (b) 47.6kPa
(c) 47.7 kPa (d) 47.8kPa
2.13 A nozzle discharging water under head H has an outlet area “a” and discharge coefficient cd=1.0. A vertical plate is acted upon by the fluid force Fj when held across the free jet and by the fluid force Fn when held against the nozzle to stop the flow. The ratio
Fj is
Fn
(a) 1/2 (b) 1
(c) 2 (d) 2
2.14. A body moving through still water at 6m/sec produces a water velocity of 4m/sec at a point 1m ahead. The difference in pressure between the nose and the point 1m ahead would be
(a) 2,000N/m2 (b) 10,000N/m2
(c) 19,620N/m2 (d) 98,100N/m2
2.15 The return period for the annual maximum flood of a given magnitude is 8 years. The probability that this food magnitude will be exceeded once during the next 5 years is
(a) 0.625 (b) 0.966
(c) 0.487 (d) 0.529
2.16 Two completely penetrating wells are located L (in meters) apart, in a homogeneous confined aquifer. The drawdown measured at the mid point between the two wells (at a distance of 0.5L from both the wells) is 2.0 m when only he first well is being pumped at the steady rate of Q1m3/sec. When both the wells are being pumped at identical steady rate of Q2m3/sec, the drawdown measured at the same location is 8.0m. It may be assumed that the drawdown at the wells always remains at 10.0 m when being pumped and the radius of influence is larger than Q1 is equal to
0.5L
Q2
2.17. In connection with the design of a barrage, identify the correct matching of the criteria of design with the items of design
Item of design Criteria of design (i) Width of waterway (A) Scour depth and exit gradient (ii) Level and length of downstream floor (B) Lacey’s formula for wetted perimeter and discharge capacity of the barrage as computed by weir equations (iii) Depth of sheet piles and total of barrage floor (C) Uplift pressure variation (iv) Barrage floor thickness (D) Hydraulic jump considerations
Codes :
(i) (ii) (iii) (iv)
(a) A B C D
(b) D C B A
(c) B A D C
(d) B D A C
2.18. In a BOD test using 5% dilution of the sample (15 ML of sample and 285 mL of dilution water), dissolved oxygen values for the sample and dilution water blank bottles after five days incubation at 200C were 3.80 and 8.80 mg/L, respectively. Dissolved oxygen originally present in the undiluted sample was 0.80 mg/L. The 5-day 200C BOD of the sample is
(a) 116mg/L (b) 108 mg/L
(c) 100mg/L (d) 92 mg/L
2.19. For a flow of 5.7 MLD (million litres per day) and a detention time of 2 hours, the surface area of a rectangular sedimentation tank to remove all particles have setting velocity of 0.33 mm/s is
(a) 20m2 (b) 100m2
(c) 200m2 (d) 400m2
2.10. For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume acceleration as 0.53m/sec2).
(a) 300m (b) 750m
(c) 320m (d) 470m
2.21 What is the equivalent wheel load of a dual wheel assembly carrying 20,440 N each for pavement thickness of 20 cm? Centre to centre spacing of tyres is 27cm and the distance between the walls of yres is 11cm.
(a) 27600 N (b) 32300N
(c) 40880N (d) 30190N
2.22 Plate bearing test with 20 cm diameter plate on soil subgrade yielded a pressure of 1.25×105N/m2 at 0.5 cm deflection. What is the elastic modulus of subgrade ?
(a) 56.18×105N/m2 (b) 22.10×105N/m2
(c) 44.25×105N/m2 (d) 50.19×105N/m2
3. Solve the following set of simultaneous equations by Gauss elimination method. (5)
x-2y+z=3 …..(1)
x+3z=11 ……(2)
-2y+z = 1 ……(3)
4. The cross-section of a pretensioned prestressed concrete beam is shown in Figure. The reinforcement is placed concentrically. If the stress in steel at transfer is 1000 MPa, compute the stress in steel immediately after transfer. The modular ratio is 6. (5)
5. An ISMS 400, with a flange width of 140 mm is subjected to an axial compressive load of 750 kN. Design the slab base resting on concrete of grade M15. The slab base available are 600x350x20 mm, 650x325x28mm, and 700x2300x32 mm. Select one of these. (5)
6. The total unit weight of the glacial outwash soil is 6kN/m3. The specific gravity of the solid particles of the soil is 2.67. The water content of the soil is 17% Calculate. (5)
(a) dry unit weight (b) porosity
(c) void ratio (d) degree of saturation
Assume that unit weight of water (γw) is 10kN/m3
7. An overflow spillway is 40 m high. Water flows down the spillway with a head of 2.5 m over the spillway crest. The spillway discharge coefficient
cd = 0.738. Show that the water depth at the toe of the spillway would be 0.3m. Determine the sequent depth required for the formation of the hydraulic jump and the loss of head in the jump. (5)
SECTION-B
(50 Marks)
Answer and TEN question from this section. All questions carry equal marks.
8. Solve d4y – y = 15 cos 2x
dx4 (5)
9. Obtain the eigen values and eigen vectors of the matrix 8 -4
2 2 (5)
10. Using the Force Method, computer the slope at the support B of the propped cantilever beam shown in Fig. 10. The value of EI is constant. (5)
11. The steel portal frame shown in Figure is subjected to an imposed service load of 15 kN. Compute the required plastic moment capacity of the members. All the members are of the same cross-section. Draw the collapse mode. (5)
12. Compute the bending moments at the top of the columns in the upper storey of the multi-storey frame shown in Figure, by the cantilever the portal methods of analysis. Indicate tension face of columns, the area of cross-section of all columns is same.
13. The cross-section of a simply supported plate girder is shown in Figure. The loading on the girder is symmetrical. The bearing stiffeners at supports are the sole means of providing restraint against torsion. Design the bearing stiffeners at supports, with minimum moment of inertia about the centre line of web plate only as the sole design criterion. The flat section available are : 250×25, 250×32, 200×28, and 200×32 mm. Dray a sketch (5)
14. The diameter of a ring beam in water tank is 7.8 m. It is subjected to an outward raidal force of 15 kN/m. Design the section using M 25 grade concrete and Fe415 reinforcement. Sketch the cross-section. (5)
15. For general c- φ soil, cohesion c is 50 kPa, the total unit weight γt is 20 kN/m3 and the bearing capacity formula, calculate the net ultimate bearing capacity for a strip footing of width B = 2m at depth z = 1m. Considering shear failure only, estimate the safe total load on a footing 10 m long by 2 m wide strip footing using a factor of safety of 3. (5)
16. A soft normally consolidated clay layer is 20 m thick with a moisture content of 45%. The clay has a saturated unit weight of 20 kN/m3, a particle specific gravity of 2.7 and a liquid limit of 60%. A foundation load will subjected the centre of the layer to a vertical stress increase of 10 kpa. Ground water level is at the surface of the clay. Estimate
(a) The initial and final effective stresses at the centre of the layer
(b) The approximate value of the compression index (Cc)
(c) The consolidation settlement of the foundation if the initial effective stress at the centre of the soil is 100kPs.
Assume unit weight of water to be 10kN/m3.
17. Estimate the safe load carrying capacity of a single bored pile 20m long, 500 mm diameter. The adhesion coefficient (d) is 0.4. Take a factor of safety of 2.5. The soil strata is as follows.
Depth (m) Soil deposit Undrained shear strength (Su)kPa 0-5 Loose fill 50 5-10 Weathered over consolidate clay 70 10-15 Over consolidated clay 100 15-30 High over-consolidated clay 200
Assumne, φn=0 is valid and Nc=9, for deep fomadations. (5)
18. (a) What is the shear strength in terms of effective stress on a plane within a saturated soil mass at a point where the total normal stress is 295 kPa and the pore water pressure 120kPa? the effective stress shear strength parameters are C’=12 kPa and φ’ = 300 (5)
(b) In a falling head permeameter test on a silty clay sample, the following results were obtained; sample length 120 mm; sample diameter 80 mm; initial head 1200 mm, final head 400 mm; time for fall in head 6 minutes stand pipe diameter 4 mm. Find the coefficient of permeability of the soil in mm/second.
19. Water flows through the Y-joint as shown in figure. Find the horizonal and vertical components of the force acting on the joint because of the flow of water. Neglet energy losses and body force.
20. Water flows in a rectangular channel at depth of 1.20 m and a velocity of 2.4m/sec. What would be the effect of a local rise in the channel bed of 0.60 m on the water surface ? (5)
21. A reservoir is proposed to be constructed to command an area of 1,20,000 hectares. The area has a monsoon rainfall of about 100cm per year. It is anticipated that sugar and rice would each be equal to 20% of the command area and wheat equal to 50% of the command area, making a total of annual irrigation equal to 90% of the command area.
(i) Work out the storage required for the reservoir, assuming the water requirements given below, canal losses as 25% of the head discharge and reservoir evaporation and dead storage losses as 20% of the gross capacity of the reservoir.
(ii) Determine also the full supply discharge of the canal at the head of the canal.
Crop Transplanted Rise Sugar Cane Wheat Sowing time July Feb-Mar October Harvesting Time November Next year Dec-March Mar-Apr Total Water Depth in cm 150 90 37.5 “Kor” period in weeks 2.5 4 4 “Kot” watering in cm 19.0 16.5 13.5
864B
Note that in which = Depth of water in cm, B=base period in
D
days, and D = duty of water in hectares/cumec. (5)
22. The following rainfall hyetograph and the corresponding direct run off are recorded in a watershed. Compute the one-hour unit hydrography ordinates for the first four hours. Assume φ index = 0.50 cm/hr (5)
Time (hrs) Rainfall (cm) Direct Run Off (m3/sec) 1 2.8 64.2 2 5.2 288.4 3 4.7 794.5 4 0.0 1369.6 5 0.0 1593.7 6 0.0 1175.1 7 0.0 588.1 8 0.0 286.9 9 0.0 170.5 10 0.0 110.0
23. A dual-media rapid sand filter plant is to be constructed for treatment of 72 million litres of water per day. A pilot plant study indicated that a filtration rate of 15m/h would be acceptable. Allowing one unit out of service for backwashing, how many 5mx8m filter units will be required ? Determine the net production in million litres per day of each filter unit if backwashing is done at 36m/h for 20 minutes and the water is wasted for the first 10 minutes of each filter run. (5)
24. The minimum flow of a river is 50m3/s having a disoolved oxygen (DO) content of 7.0 mg/L (80% saturation) and BOD5 of 8.0 mg/L. It receives a waste water discharge of 5cm3/s with BOD5 of 200 mg/L and no DO. If the rate constants for deoxygenation and reaeration (both base e) and 0.5/d and 1.0/d, respectively and the velocity of river flow is 0.8m/s, calculate the distance in kilometre downstream from the point of waste water discharge where the minimum DO occurs. (5)
25. An activated sludge aeration tank (length 30.0m; width 14.0m; effective liquid depth 4.3m) has the following parameters :
flow 0.0796m3, soluble BOD5 after primary settling 130 mg/L; mixed liquor suspended solids (MLSS) 2100 MG/L; mixed liquor volatile suspended solids (MLVSS) 1500 mg/L; 30 minute settled sludge volume 230 mL/L; and return sludge concentration 9100 mg/L. Determine the aeration period, food to micro-organisms (F.M) ratio, sludge volume index (SVI), and return sludge rate. (5)
26. There is a horizontal curve of radius 360 m and length 180m. Calculate the clearance required from the central line on the inner side of the curve, so as to provide an overtaking sight distance of 250m. (5)
27. The width of expansion joint gap is 2.5 cm in a cement concrete 20 cm thick pavement. If the laying temperature is 150 C and the maximum slab temperature in the summer is 550C, calculate
(i) the spacing between expansion joints, and
(ii) the spacing between contraction joints.
Coefficient of thermal expansion for concrete is 10×10-6 per degree centigrade. Ultimate stress in tension in cement concrete is 1.6×105 N/m2. Ultimate tensile stress in steel is 1200x10N/m2. Factor of safety is to be taken as 2. Assume the pavement width to be 3.5 m. Unit weight of steel is 75,000 N/m3. Total reinforcement of 6kg/m2 is provided in the slab. (5) | 6,478 | 22,517 | {"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} | 3.59375 | 4 | CC-MAIN-2013-48 | longest | en | 0.799731 |
https://www.varsitytutors.com/sat_math-help/how-to-multiply-exponents | 1,713,095,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00706.warc.gz | 989,093,851 | 44,211 | # SAT Math : How to multiply exponents
## Example Questions
← Previous 1 3 4 5 6 7
### Example Question #1 : Exponents
If (300)(400) = 12 * 10n, n =
12
2
3
4
7
4
Explanation:
(300)(400) = 120,000 or 12 * 104.
### Example Question #2 : Exponents
(2x103) x (2x106) x (2x1012) = ?
6x1021
6x1023
8x1021
8x1023
8x1021
Explanation:
The three two multiply to become 8 and the powers of ten can be added to become 1021.
### Example Question #1 : Exponents
If 3x = 27, then 22x = ?
64
32
3
8
9
64
Explanation:
1. Solve for x in 3x = 27. x = 3 because 3 * 3 * 3 = 27.
2. Since x = 3, one can substitute x for 3 in 22xÂ
3. Now, the expression is 22*3
4. This expression can be interpreted as 22 * 22 * 22. Since 22 = 4, the expression can be simplified to become 4 * 4 * 4 = 64.
5. You can also multiply the powers to simplify the expression. When you multiply the powers, you get 26, or 2 * 2 * 2 * 2 * 2 * 2
6. 26Â = 64.
### Example Question #4 : Exponents
Find the value of x such that:
8x-3Â = 164-x
4
11/3
19/4
7/2
25/7
25/7
Explanation:
In order to solve this equation, we first need to find a common base for the exponents. We know that 23 = 8 and 24 = 16, so it makes sense to use 2 as a common base, and then rewrite each side of the equation as a power of 2.
8x-3 = (23)x-3
We need to remember our property of exponents which says that (ab)c = abc.
Thus (23)x-3 = 23(x-3) = 23x - 9.
We can do the same thing with 164-x.
164-x = (24)4-x = 24(4-x)Â = 216-4x.
So now our equation becomes
23x - 9Â =Â 216-4x
In order to solve this equation, the exponents have to be equal.Â
3x - 9 = 16 - 4x
7x - 9 = 16
7x = 25
Divide by 7.
x = 25/7.
### Example Question #5 : Exponents
Which of the following is equal to 410 + 410 + 410 + 410 + 411?
240
223
250
215
260
223
Explanation:
We can start by rewriting 411 as 4 * 410. This will allow us to collect the like terms 410 into a single term.
410Â + 410Â + 410Â + 410Â + 411
= 410Â + 410Â + 410Â + 410Â + 4 * 410
= 8 * 410
Because the answer choices are written with a base of 2, we need to rewrite 8 and 4 using bases of two. Remember that 8 = 23, and 4 =Â 22.
8 * 410
= (23)(22)10
We also need to use the property of exponents that (ab)c = abc. We can rewrite (22)10 as 22x10 = 220.
(23)(22)10
= (23)(220)
Finally, we must use the property of exponents that ab * ac = ab+c.
(23)(220) = 223
### Example Question #1 : Exponents
If 3Â + 3n+3 = 81, what is 3n+2Â ?
3
81
9
26
18
26
Explanation:
3Â + 3n+3Â = 81
In this equation, there is a common factor of 3, which can be factored out.
Thus, 3(1 + 3n+2) = 81
Note: when 3 is factored out of 3n+3, the result is 3n+2 because (3n+3 = 31 * 3n+2). Remember that exponents are added when common bases are multiplied.  Also remember that 3 = 31.
3(1 + 3n+2) = 81
(1 + 3n+2) = 27
3n+2Â = 26
Note: do not solve for n individually. Â But rather seek to solve what the problem asks for, namely 3n+2. Â
### Example Question #7 : Exponents
If f(x) = (2 – x)(x/3), and 4n = f(10), then what is the value of n?
0
–2
5
2
–5
5
Explanation:
First, let us use the definiton of f(x) to find f(10).
f(x) = (2 – x)(x/3)
f(10) = (2 – 10)(10/3)
= (–8)(10/3)
In order to evaluate the above expression, we can make use of the property of exponents that states that abc = (ab)c = (ac)b.
(–8)(10/3) = (–8)10(1/3) = ((–8)(1/3))10.
(–8)(1/3) requires us to take the cube root of –8. The cube root of –8 is –2, because (–2)3 = –8.
Let's go back to simplifying ((–8)(1/3))10.
((–8)(1/3))10 = (–2)10 = f(10)
We are asked to find n such that 4n = (–2)10. Let's rewrite 4n with a base of –2, because (–2)2 = 4.
4n = ((–2)2)n = (–2)2n = (–2)10
In order to (–2)2n = (–2)10, 2n must equal 10.
2n = 10
Divide both sides by 2.
n = 5.
### Example Question #8 : Exponents
What is the value of n that satisfies the following equation?
2n·4n·8n·16 = 2-n·4-n·8-n
0
–1/3
–2/3
2/3
1/3
–1/3
Explanation:
In order to solve this equation, we are going to need to use a common base. Because 2, 4, 8, and 16 are all powers of 2, we can rewrite both sides of the equation using 2 as a base. Since 22 = 4, 23 = 8, and 24 = 16, we can rewrite the original equation as follows:
2n * 4n * 8n * 16 = 2–n * 4–n * 8–n
2n(22)n(23)n(24) = 2–n(22)–n(23)–n
Now, we will make use of the property of exponents which states that (ab)c = abc.
2n(22n)(23n)(24) = 2–n(2–2n)(2–3n)
Everything is now written as a power of 2. We can next apply the property of exponents which states that abac = ab+c.
2(n+2n+3n+4) = 2(–n + –2n + –3n)
We can now set the exponents equal and solve for n.
n + 2n + 3n + 4 = –n + –2n + –3n
Let's combine the n's on both sides.
6n + 4 = –6n
12n + 4 = 0
Subtract 4 from both sides.
12n = –4
Divide both sides by 12.
n = –4/12 = –1/3
### Example Question #9 : Exponents
If 1252x–4 = 6257–x, then what is the largest prime factor of x?
3
5
2
11
7
2
Explanation:
First, we need to solve 1252x–4 = 6257–x . When solving equations with exponents, we usually want to get a common base. Notice that 125 and 625 both end in five. This means they are divisible by 5, and they could be both be powers of 5. Let's check by writing the first few powers of 5.
51 = 5
52 = 25
53 = 125
54 = 625
We can now see that 125 and 625 are both powers of 5, so let's replace 125 with 53 and 625 with 54.Â
(53)2x–4 = (54)7–x
Next, we need to apply the rule of exponents which states that (ab)c = abc .
53(2x–4) = 54(7–x)
We now have a common base expressed with one exponent on each side. We must set the exponents equal to one another to solve for x.
3(2x – 4) = 4(7 – x)
Distribute the 3 on the left and the 4 on the right.
6x – 12 = 28 – 4x
10x – 12 = 28
10x = 40
Divide by 10 on both sides.
x = 4
However, the question asks us for the largest prime factor of x. The only factors of 4 are 1, 2, and 4. And of these, the only prime factor is 2.Â
### Example Question #10 : Exponents
(x3)2 * x–2 =Â
x–4
x
x6
x4
x2 | 2,432 | 6,167 | {"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-2024-18 | latest | en | 0.785116 |
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