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With concerns over moving to a clean energy platform worldwide with electric vehicles and renewables, the energy we consume, or should we say do NOT consume, is as important as the green energy we produce. Thus, solid state lighting, more efficient than incandescent and fluorescent bulbs, based upon light emitting diodes (LED) is touted as the solution. However, LEDs struggle to deliver high brightness for the shorter-wavelength end of lighting needs. And emitted short wavelengths facilitate white light through known phosphor downconverters.
In the October 24, 2017 issue of Light: Science & Applications, currently online, the Ohio State University researchers, with scientists at Wright State University and Naval Research Laboratory, describe a promising new semiconductor LED made with GaN-based materials that could boost wallsocket efficiency by reducing energy losses and self-heating.
If this new technology can be harnessed for large light output, the breakthrough could enhance LED solid state lighting without a significant change to the existing LED manufacturing facility.
The new LEDs could provide more light with less voltage and resistance than in conventional GaN LEDs, thereby boosting the overall lumens per watt output and avoiding the efficiency droop that plagues high brightness LEDs.
One way the team overcomes this problem is by completely removing all p-type doping in gallium nitride, which historically is hard to dope and leads to a high series resistance.
The key to the team's discovery is the ability to create "holes" for radiative recombination with electrons by quantum-mechanical tunneling, not by p doping. The tunneling occurs by the Zener mechanism, delivering the holes to the zone of recombination, mitigating the need for clumsy p-type ohmic contacts and resistive p-type semiconductor injectors.
The team includes Paul R. Berger and Tyler A. Growden at Ohio State University; Elliott R. Brown and Weidong Zhang at Wright State University; and David F. Storm and David J. Meyer at the Naval Research Laboratory.
Their discovery was made while advancing resonant tunneling diodes (RTD) in the gallium nitride system for the Office of Naval Research under program manager Dr. Paul Maki. As reported in the August 2016 issue of Applied Physics Letters, their effort also established a stable GaN-based RTDplatform for high microwave power generation and potentially terahertz sources.
The fundamental science behind this advancement is the utilization of the extremely highelectric fields induced by the polarization effects within wurtzite GaN based heterostructures. These high fields allow the new device to not only inject electrons across a classic RTD double-barrier structure in the conduction band, but alsosimultaneously inject holes by Zener tunneling across the GaN band gap into the valence band. Thus, the new LED uses only n-type doping, but includes bipolar tunneling charges to create the new LED light source.
To be useful for commercialization, the team is working to balance the injected electron and hole ratio to create and therefore deliver up to one emitted photon for each injected electron.<|endoftext|>
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Folic acid, also known as folate, is one of the B complex vitamins. It received its name from the Latin word “folium,” alluding to its presence in dark green foliage, such as spinach, kale, chard and beet greens. Folate is a delicate vitamin, and it is easily lost during the processing of foods or by exposure to heat, light or acid. Even prolonged storage at room temperature decreases the potency of folate. Since it is critical for normal fetal development, pregnant women or women who are contemplating pregnancy should take extra folate.
Folate is essential for the synthesis of DNA and RNA. Hence, any process that demands extra cellular production, such as pregnancy, increases your need for folate. Folate is instrumental in the formation of hemoglobin, which is the oxygen-carrying protein in your red blood cells, and in the metabolism and conversion of proteins and amino acids. Folic acid plays a major role in the production of neurotransmitters in your brain, and it is needed for the normal development of the fetal nervous system.
The typical Western diet, which is rich in processed foods, meats, fats and sugars, is a poor source of folic acid. Nutritionist Elson Haas, M.D., believes that folate deficiency is one of the most common nutrient deficiencies in America. Folic acid deficiency leads to fatigue, loss of appetite, irritability, mental disorders and megaloblastic anemia, which is characterized by abnormally large and poorly functioning red blood cells. During pregnancy, folate deficiency inhibits the development of an infant’s brain and spinal cord, leading to malformations called neural tube defects.
Sources and Recommendations
The best source of folate is fresh, dark green leafy vegetables. Bean sprouts and wheat germ are good sources, as are liver, kidney and brewer’s yeast. Many foods are fortified with folate, but storage reduces its potency. In 1998 the National Academy of Sciences established recommended daily allowances for folate, ranging from 65mcg daily for infants to 600mcg daily for pregnant women. People with increased needs for folate, such as pregnant women, may need to take supplements.
Demands of Pregnancy
Normal fetal brain development is dependent on adequate maternal levels of folate. According to the Linus Pauling Institute, an infant’s brain and spinal cord undergo critical development during the fourth week of gestation, when many women do not even realize they are pregnant. Therefore, the U.S. Public Health Service recommends that all women who are capable of becoming pregnant take 400mcg of folate daily. The Institute of Medicine recommends that pregnant women take 600mcg of folate daily. This dosage ensures adequate levels for fetal development as well as the expectant mother’s need for additional red blood cell production.
Folate's benefits during pregnancy are well documented, and prenatal vitamins are usually formulated to meet the increased requirement. However, prenatal vitamins are not all the same. Some preparations may require two or more capsules or pills to furnish an adequate daily intake of folate, iron or other nutrients. Your prenatal vitamin should supply at least 600mcg of folic acid in each day’s dose. Read the label carefully to determine what that dosage is, and consult your doctor if you are not certain that your prenatal vitamin meets your needs.<|endoftext|>
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# Finding a matrix and its transpose from a given matrix
Show 40 post(s) from this thread on one page
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• Aug 5th 2007, 12:23 AM
beberenne
Finding a matrix and its transpose from a given matrix
Hi all, I'm new here. I'm currently doing a thesis on image compression and embedding. However, I'm stuck at a final part of my calculations.
Lets say you have a square matrix A. Taking the transpose of the matrix A you will get another square matrix A'.
Multiplying A by A' you get another square matrix B.
Example: A X A' = B
Now assuming you only have the matrix B given to you. How do you find the value of matrix A?
Hope you guys can help in this. Thanks alot.:)
• Aug 5th 2007, 01:04 AM
tukeywilliams
Given that $A \times A^{T} = B$, and we are given the square matrix $B$, the problem is to find $A$ and $A^{T}$.
So lets say $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$.
Then $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 11 \\ 11 & 25 \end{bmatrix}$ which is a symmetric matrix.
So if $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} a & c \\ b & d \end{bmatrix}$ then
$a^2 + b^2 = 5$
$ac+bd = 11$
$c^2+d^2 = 25$
• Aug 5th 2007, 01:42 AM
tukeywilliams
There also are probably some efficient algorithms (similar to LU-decomposition) that allow you to decompose a given matrix to a product of a matrix and its transpose (numerical linear algebra).
• Aug 7th 2007, 12:24 AM
beberenne
i have checked out the LU-decomposition method and i found it didn't help much in obtaining the product of a matrix and its transpose. Writing the long equation in MATLAB will be a bit difficult as i'm dealing with a 512 X 512 image which will be having many unknowns. Any more suggestions will me most welcomed. By the way, many thanks to tukeywilliams. :)
• Aug 7th 2007, 03:21 AM
JakeD
Since if $AA^T = B$ implies $B$ is symmetric, there exists an orthogonal matrix $P$ such that $P^TBP = \Lambda,$ and $\Lambda$ is a diagonal matrix with the eigenvalues of $B$ along the diagonal and the columns of $P$ are the eigenvectors of $B$. $P^TBP = \Lambda$ is called a unitary transformation. Since $P$ is orthogonal, $PP^T = I$ and thus $B = PP^T B PP^T = P\Lambda P^T.$
Further, since $x^TBx = x^TAA^Tx = (A^Tx)^T A^Tx > 0$ when $B$ is nonsingular, $B$ is positive definite and thus has positive eigenvalues. Thus $\Lambda = D D^T$ where $D$ has the square roots of the eigenvalues of $B$ on the diagonal.
Then $B = P\Lambda P^T = PD D^TP^T = A A^T$ where $A = PD$. So the problem is reduced to finding the eigenvalues and eigenvectors of a symmetric matrix $B.$ There are efficient numerical methods for this.
• Aug 7th 2007, 04:27 AM
beberenne
Correct me if I'm wrong.
Assuming that I extracted the eigenvectors and eigenvalues of B using MATLAB. So i now have a matrix of the eigenvectors of B (eig_vec) and a matrix of the eigenvalues of B (eig_val).
[eig_vec eig_val] = eig(B);
Then D has the square roots of the eigenvalues of B on the diagonal. So i take the square root of the eigenvalues of B.
D = sqrtm(B);
The columns of P are the eigenvectors of B, which means P is equal to the eigenvectors of B.
P = eig_vec;
To find A,
A = PD;
• Aug 7th 2007, 05:12 AM
JakeD
Quote:
Originally Posted by beberenne
Correct me if I'm wrong.
Assuming that I extracted the eigenvectors and eigenvalues of B using MATLAB. So i now have a matrix of the eigenvectors of B (eig_vec) and a matrix of the eigenvalues of B (eig_val).
[eig_vec eig_val] = eig(B);
The eigenvectors must be orthonormal, which means the matrix $P$ must satisfy $P^TP = PP^T = I.$ I'm not familiar with MATLAB, so I don't know if this is the case. The normalized eigenvectors corresponding to distinct eigenvalues will be orthonormal. And multiple eigenvectors corresponding to a single eigenvalue can be made orthonormal through Gram-Schmidt orthogonalization. So the eigenvectors can be made orthonormal. But I don't know if MATLAB does that automatically.
Quote:
Then D has the square roots of the eigenvalues of B on the diagonal. So i take the square root of the eigenvalues of B.
D = sqrtm(B);
Is this the square root of B or the eigenvalues of B?
Quote:
The columns of P are the eigenvectors of B, which means P is equal to the eigenvectors of B.
P = eig_vec;
To find A,
A = PD;
Yes. And then you must check that you actually get $AA^T = B.$
• Aug 7th 2007, 07:58 AM
beberenne
The eigenvectors are non orthonormal. For the square root, the function squares the whole matrix.
• Aug 7th 2007, 05:29 PM
beberenne
I have experimented with a 4X4 image. The matrix obtain from the image is B.
$
B = \begin{bmatrix} 1 & -3.8858e-016 & 1.0408e-016 & -2.7756e-017 \\ -3.8858e-016 & 1 & -2.3592e-016 & -2.7756e-01 \\ 1.0408e-016 & -2.3592e-016 & 1 & -1.8041e-016 \\ -2.7756e-017 & -2.7756e-016 & -1.8041e-016 & 1\end{bmatrix}
$
After having extracted the eigenvalues and eigenvectors, P of B;
$
eigenvalues = \begin{bmatrix} -1.1102e-016 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}
$
$
eigenvectors = \begin{bmatrix} -0.5 & 0-0.81917 & 0.14281 & 0.24199 \\ -0.5 & 0.026681 & -0.15422 & -0.85177 \\ -0.5 & 0.31193 & -0.68558 & 0.42741 \\ -0.5 & 0.48056 & 0.69699 & 0.18236\end{bmatrix}
$
The transpose of the eigenvectors is
$
eigenvectors transposed = \begin{bmatrix} -0.5 & -0.5 & -0.5 & -0.5 \\ -0.81917 & 0.026681 & 0.31193 & 0.48056 \\ 0.14281 & -0.15422 & -0.68558 & 0.69699 \\ 0.24199 & -0.85177 & 0.42741 & 0.18236\end{bmatrix}
$
The inverse of the eigenvectors is
$
eigenvectors inversed = \begin{bmatrix} -0.5 & -0.5 & -0.5 & -0.5 \\ -0.81917 & 0.026681 & 0.31193 & 0.48056 \\ 0.14281 & -0.15422 & -0.68558 & 0.69699 \\ 0.24199 & -0.85177 & 0.42741 & 0.18236\end{bmatrix}
$
The product of $
PP^T
$
is not equal to $
P^TP
$
The square root of the eigenvalues of B consist of a complex number
$
eigenvalues squared = \begin{bmatrix} 0 +1.0537e-008i & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}
$
Hope JakeD can comment on this. Thanks
• Aug 7th 2007, 08:02 PM
JakeD
Quote:
Originally Posted by beberenne
I have experimented with a 4X4 image. The matrix obtain from the image is B.
$
B = \begin{bmatrix} 1 & -3.8858e-016 & 1.0408e-016 & -2.7756e-017 \\ -3.8858e-016 & 1 & -2.3592e-016 & -2.7756e-01 \\ 1.0408e-016 & -2.3592e-016 & 1 & -1.8041e-016 \\ -2.7756e-017 & -2.7756e-016 & -1.8041e-016 & 1\end{bmatrix}
$
[snipped]
First, the matrix is not symmetric as is. The entry 2,4 (row 2, column 4) is not equal to the entry 4,2. It looks like a typo and entry 2,4 is missing a 6 at the end. Then the matrix is symmetric.
Second, on any typical computer, numbers like -3.8858e-016 are effectively 0 except for rounding error because numbers like this are at the limits of machine accuracy. But all the entries not on the diagonal are like that. Setting all those entries to zero means the matrix is effectively the identity matrix and the problem is the trivial one B = I and A = A^T = I.
Where did this matrix come from?
• Aug 7th 2007, 10:03 PM
beberenne
Sorry about the typo error. The matrix is symmetrical. The matrix is obtained through a test matrix A multiplied by A transposed. The matrix is generated from a test image of 4 X4. It seems that any image i use, i still get a symmetrical matrix with ones in the diagonal.
The matrix B comes from the product of A and A transposed.
$A= \begin{bmatrix} 0.12452 & 0.65243 & 0.56703 & 0.48714 \\ 0.32858 & 0.34674 & 0.25193 & -0.84163 \\ 0.49466 & -0.65422 & 0.56457 & 0.092583 \\ 0.79489 & 0.16158 & -0.5443 & 0.21398\end{bmatrix}
$
$A'= \begin{bmatrix} 0.12452 & 0.32858 & 0.49466 & 0.79489 \\ 0.65243 & 0.34674 & -0.65422 & -0.16158 \\ 0.56703 & 0.25193 & 0.56457 & -0.5443 \\ 0.48714 & -0.84163 & 0.092583 & 0.21398\end{bmatrix}
$
• Aug 9th 2007, 11:12 PM
beberenne
JakeD, if i let the roundings to be zero, then i would not get the correct answer. As you can see, matrix B is the product of A and A transpose as shown above. By letting the roundings to be zero, then i would not get back A.
• Aug 10th 2007, 12:10 AM
JakeD
Quote:
Originally Posted by beberenne
I have experimented with a 4X4 image. The matrix obtain from the image is B.
$
B = \begin{bmatrix} 1 & -3.8858e-016 & 1.0408e-016 & -2.7756e-017 \\ -3.8858e-016 & 1 & -2.3592e-016 & -2.7756e-01 \\ 1.0408e-016 & -2.3592e-016 & 1 & -1.8041e-016 \\ -2.7756e-017 & -2.7756e-016 & -1.8041e-016 & 1\end{bmatrix}
$
Quote:
Originally Posted by beberenne
Sorry about the typo error. The matrix is symmetrical. The matrix is obtained through a test matrix A multiplied by A transposed. The matrix is generated from a test image of 4 X4. It seems that any image i use, i still get a symmetrical matrix with ones in the diagonal.
The matrix B comes from the product of A and A transposed.
$A= \begin{bmatrix} 0.12452 & 0.65243 & 0.56703 & 0.48714 \\ 0.32858 & 0.34674 & 0.25193 & -0.84163 \\ 0.49466 & -0.65422 & 0.56457 & 0.092583 \\ 0.79489 & 0.16158 & -0.5443 & 0.21398\end{bmatrix}
$
$A'= \begin{bmatrix} 0.12452 & 0.32858 & 0.49466 & 0.79489 \\ 0.65243 & 0.34674 & -0.65422 & -0.16158 \\ 0.56703 & 0.25193 & 0.56457 & -0.5443 \\ 0.48714 & -0.84163 & 0.092583 & 0.21398\end{bmatrix}
$
Quote:
Originally Posted by beberenne
JakeD, if i let the roundings to be zero, then i would not get the correct answer. As you can see, matrix B is the product of A and A transpose as shown above. By letting the roundings to be zero, then i would not get back A.
Multiplying AA' yields
Code:
1.0000e+00 -1.4105e-06 -8.6817e-06 3.1304e-06
-1.4105e-06 1.0000e+00 2.6298e-06 -6.2810e-06
-8.6817e-06 2.6298e-06 1.0000e+00 6.8791e-06
3.1304e-06 -6.2810e-06 6.8791e-06 1.0000e+00
which is not the matrix B you gave above. Is this a different matrix, or is it different because the A matrix wasn't printed with enough precision? Anyway, the B matrix still looks strange. Why is it so close to the identity matrix?
• Aug 10th 2007, 12:18 AM
beberenne
JakeD, i also noticed this problem and did some manual calculations on it. I think is the rounding error of the MATLAB. Mentioning earlier about the eigenvectors of B are not orthogonal, actually they are. At first they seem not as there are rounding errors. But when i manually check, they are orthogonal.
Now the problem is to find D. I'm having problems with finding D as my eigenvalues obtained trough MATLAB are all ones in the diagonal except for [1,1] which is a negative small value.
On why the B similar to an identity matrix, this i do not know. :confused:
• Aug 10th 2007, 12:35 AM
JakeD
Quote:
Originally Posted by beberenne
JakeD, i also noticed this problem and did some manual calculations on it. I think is the rounding error of the MATLAB. Mentioning earlier about the eigenvectors of B are not orthogonal, actually they are. At first they seem not as there are rounding errors. But when i manually check, they are orthogonal.
Now the problem is to find D. I'm having problems with finding D as my eigenvalues obtained trough MATLAB are all ones in the diagonal except for [1,1] which is a negative small value.
From what you showed before, that eigenvalue is effectively zero except for rounding error. In the absence of rounding error, the eigenvalues have to be positive because B is positive definite as I argued before.
Quote:
On why the B similar to an identity matrix, this i do not know. :confused:
I think this is the main issue. Resolve this and things should work.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last<|endoftext|>
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Sickle Cell Disease
(SCD; Sickle Cell Anemia)
by Editorial Staff and Contributors
Sickle cell disease (SCD) is a blood disorder. The shape of some red blood cells (RBCs) change to a sickle (crescent moon) shape. The sickle RBCs can block blood flow in blood vessels. This will slow or stop the flow of blood and oxygen to tissues and organs.
The body will also destroy these RBCs because they are abnormal. This means you have fewer overall RBCs. Low levels of RBCs is a condition called anemia. This makes it harder for the body to get the oxygen it needs.
A defect in the genes causes SDC. This gene causes a problem with a part of the RBC called hemoglobin.
Parents pass the defect to their children. SCD will only develop if you get the defective gene from both of your parents.
If you only get one defective gene, you will not have SCD. But, you can pass this gene on to your children.
Sickle cell disease is most common in people who are Black or have with African ancestry.
Other backgrounds with high risk include Hispanic, Asian Indian, Middle Eastern, and southern European.
Symptoms may be noticed as early as 6 months old. They are different in everyone depending on where problems are in the body. The most common include:
Signs of acute chest syndrome (ACS):
SDC crisis may cause:
SCD may cause other serious problems such as a stroke, or painful erections lasting 4 hours or more. These problems need immediate care.
All newborns have a screening test at birth. Blood tests will show the sickle shaped RBCs and other problems that indicate SCD. You may also have:
The goal of care is to lower the chances of health problems. It will include both medical care and lifestyle habits:
Young children may be given penicillin daily. It can help to lower the risk of certain infections. It’s given until the age 5 years. Penicillin is also used for people who had their spleen removed.
People with SCD have a higher risk of infections. Infections can also trigger other health problems. The doctor may advise:
Hydroxyurea is a pill that can help lower the risk of or prevent health problems. It increases the production of healthy hemoglobin.
Hydroxyurea lowers the chances of sickle cell crisis. It’s not right for everyone. It’s also unsafe for pregnant women.
Blood transfusions raise the number of healthy RBCs. Transfusions treat anemia. They also help with liver problems or stroke.
Hematopoietic Stem Cell Transplantation (HSCT)
HSCT is currently the only cure. It's not right for everyone. A donor with matching blood proteins will be needed. This is most often found in a brother or sister.
There are also a number of medical risks involved. For one, the immune system will need to be suppressed for the rest of your life. This can increase the risk for other illnesses. HSCT may be considered in children who have had severe complications.
Regular medical care is important. Keep in touch with your healthcare team. Early care of health problems may keep them from getting worse.
Lifestyle steps that may help prevent problems:
There is no way to prevent sickle cell disease. If you plan on having a family, talk to your doctor about genetic testing if you're at high risk. If you're pregnant, a prenatal test called amniocentesis can detect SCD before birth.
American Sickle Cell Anemia Association
Sickle Cell Information Center
About Kids Health—The Hospital for Sick Children
Sickle Cell Foundation of Alberta
Sickle cell disease. Family Doctor—American Academy of Family Physicians website. Available at: https://familydoctor.org/condition/sickle-cell-disease. Updated September 8, 2017. Accessed July 18, 2018.
Sickle cell disease. Kids Health—Nemours Foundation website. Available at:
...(Click grey area to select URL)
Updated May 2018. Accessed July 18, 2018.
Sickle cell disease. National Heart, Lung, and Blood Institute website. Available at: https://www.nhlbi.nih.gov/health-topics/sickle-cell-disease. Accessed July 18, 2018.
Sickle cell disease (SCD). Centers for Disease Control and Prevention website. Available at: https://www.cdc.gov/ncbddd/sicklecell/index.html. Updated June 18, 2018. Accessed July 18, 2018.
Sickle cell disease in adults and adolescents. EBSCO DynaMed Plus website. Available at: http://www.dynamed... . Updated January 19, 2018. Accessed July 18, 2018.
Sickle cell disease in infants and children. EBSCO DynaMed Plus website. Available at: http://www.dynamed... . Updated May 7, 2018. Accessed July 18, 2018.
1/4/2011 DynaMed's Systematic Literature Surveillance http://www.dynamed... : Vichinsky EP, Neumayr LD, Gold JI, et al. Neuropsychological dysfunction and neuroimaging abnormalities in neurologically intact adults with sickle cell anemia. JAMA. 2010;303(18):1823-1831.
Last reviewed May 2018 by EBSCO Medical Review Board Michael Fucci, DO, FACC
Last Updated: 7/18/2018
EBSCO Information Services is fully accredited by URAC. URAC is an independent, nonprofit health care accrediting organization dedicated to promoting health care quality through accreditation, certification and commendation.
This content is reviewed regularly and is updated when new and relevant evidence is made available. This information is neither intended nor implied to be a substitute for professional medical advice. Always seek the advice of your physician or other qualified health provider prior to starting any new treatment or with questions regarding a medical condition.
To send comments or feedback to our Editorial Team regarding the content please email us at [email protected]. Our Health Library Support team will respond to your email request within 2 business days.<|endoftext|>
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# From Paper to App: Nonogram Tips and Tricks
Nonogram or picture cross games have traveled a long way. From just being paper puzzles to now being fully computer-generated, even their rules have evolved. These are logic puzzles where each square in the grid must be painted or left blank according to the numerical clues outside, revealing a pixel image in the end. It may seem complicated, but with basic elimination techniques and using a logical bend of mind, you’ll soon have a hang of it.
But, before getting onto the tips and tricks of solving a picture cross puzzle, here’s an essential piece of information. Every time you paint a box or cross it, remember you’re doing it for both that particular column and row. So, keep in mind that you’ve to scan through both sides (top and left) carefully. Now let’s get right into the details …
Nonogram Tricks
Trick #1 – The trick is to catch hold of the larger numbers first to clear your vision and simplify the task. So, for instance, if it’s a 15×15 grid, and you have the digit 15 on either side, paint those boxes first i.e. the ones you’re sure of.
Trick #2 – Rows or columns with multiple clues are slightly difficult to crack. But the rule is to have a single space between those numbers and paint the remaining boxes accordingly. For example, if in a 15×15 grid, you get 2, 3, and 8 as clues on either side, these are to be direct inputs since there’ll be no room on the grid for any other possibilities. So, 2 + (1 space) + 3 + (1 space) + 8 equals to 15, completing the line.
Trick #3 – When you apply the above tricks, you’ll already see an image developing on the grid. What you can do next is, target the smaller numbers – since you’d already have come halfway or are just there. These smaller digits will certainly help you paint and block more cells.
Trick #4 – If there’s a bigger number, it guarantees a cell to be crossed since it’ll create an overlap. So, assuming we have a 15×15 grid, and there’s a 9, you could either paint it from the top left to the right or from the very right towards the left. But, in both cases, some cells have to be marked either way leaving us with six guaranteed cells. This method works in most cases where it’s a single number, and it’s also greater than half of the line.
Trick #5 – Complex puzzles require deeply searching for contradictions with advanced reasoning. So, when all methods exhaust, deep recursion comes to rescue. There can be several possibilities to explore in the puzzle; hence this trick is practically possible on automated picture cross puzzles only.
Now that you’re aware of some basic tricks of solving a picture cross puzzle, try them out. Where? On the Easybrain Nonogram.com mobile app, that’ll excellently challenge your skills. You can put these tips and tricks to the test without ever resorting to guesswork. And, in the end, you’ll be presented with a reward in the shape of a pretty picture. But of course, the real reward is in applying your logical and systematic thinking.<|endoftext|>
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Primary sources enable the researcher to get as close as possible to the truth of what actually happened during an event or time period. Primary sources are the evidence left behind by participants or observers. The following are generally considered primary sources:
If you are attempting to find evidence documenting the mentality or psychology of a time, or of a group (evidence of a world view, a set of attitudes, or the popular understanding of an event or condition), the most obvious source is public opinion polls taken at the time. Since these are generally very limited in availability and in what they reveal, however, it is also possible to make use of ideas and images conveyed in the mass media, and even in literature, film, popular fiction, self-help literature, textbooks, etc. Again the point is to use these sources, written or produced at the time, as evidence of how people were thinking.
Elihu Burritt Library
(access via Harold Lewis Drive)
Central Connecticut State University, 1615 Stanley Street, New Britain, CT 06050<|endoftext|>
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Rectilinear Motion of Particles
Motion is one of the most common phenomena we come across in our daily lives. For example, a moving car, a kid running on the road or a fly moving in the air are all said to be in motion. So, in general terms, a body is said to be in motion if it changes its position with respect to a reference point and time. Depending upon the path taken by the particle the motion can be of different types like projectile motion, rectilinear motion, rotational motion, etc. For now, we will only focus on the rectilinear motion which is also known as linear motion.
What is Rectilinear Motion?
When we require only one co-ordinate axis along with time to describe the motion of a particle it is said to be in linear motion or rectilinear motion. Some examples of linear motion are a parade of soldiers, a train moving along a straight line, and many more.
Distance and Displacement
So now that we have learned about linear motion we will discuss two terms related to change in position. These are called – ‘Distance’ and ‘Displacement’.
Distance is defined as,
The total path length covered during a journey
While displacement is defined as,
The path length from final position of the particle to the origin O.
Consider the following figure:
We have an origin O, measurements to the right of O are taken as a positive while to the left are taken as negative. Suppose a person who starts from origin O reaches point A,
Distance = OA
Displacement = OA
Now he turns and reaches point B,
Distance = OA + AB
Displacement = -OB
As we can see, displacement is negative since it is measured to the left of the origin. From the above example, we can infer that distance is always positive while displacement can either be positive or negative.
Speed and Velocity
These terms are used to describe the rate of change of position. Speed is the rate of change of distance while velocity is the rate of change of displacement. Comparing from above as distance can never be negative so the speed is never negative while velocity can be both positive and negative. In mathematical terms, these are defined as follows:
Speed = $\frac{Distance~ Travelled}{Time ~Taken}$
Velocity =$\frac{(Final~ position – Initial~ position)}{Time~ Taken}$
Examples for Rectilinear Motion
Following are the rectilinear motion examples:
• The use of elevators in public places is an example of rectilinear motion.
• Gravitational forces acting on objects resulting in free fall is an example of rectilinear motion.
• Kids sliding down from a slide is a rectilinear motion.
• The motion of planes in the sky is a rectilinear motion.
Types of Rectilinear Motion
There are three types of rectilinear motion and they are:
• Uniform rectilinear motion: When an object travels at a constant speed with zero acceleration it is known as uniform rectilinear motion.
• Uniformly accelerated rectilinear motion: When an object travels with constant acceleration it is known as uniformly accelerated rectilinear motion.
• Rectilinear movement with non-uniform acceleration: When an object travels at an irregular speed and acceleration it is known as a rectilinear movement with non-uniform acceleration.
What is the difference between linear and rectilinear motion?
Linear motion Rectilinear motion An object is said to be in a linear motion when the object moves in a straight line An object is said to be in a rectilinear motion when two objects are moving in a straight line and are parallel to each other There are two types of linear motion, and they are uniform motion and non- uniform motion There are three types of rectilinear motion, and they are uniform rectilinear motion, uniformly accelerated rectilinear motion and rectilinear movement with non-uniform acceleration Athlete running 200 m along the straight track Elevator movement is an example of rectilinear motion
1. 2) How does the velocity change with time in uniform rectilinear motion of an object?
3) A body is moving with a velocity of 10 m/s. if the motion is uniform what will be the velocity
after 10 s?
4) Can a body have constant speed but variable velocity?
5) Differentiate between uniform and non-uniform velocity.
6) Express average velocity when the velocity of the body changes at a uniform and a non-
uniform rate.
7) An object travels along a circular path of radius 70 m. After travelling half a revolution in 20s,
find the average speed and average velocity of the object.
1. 2) When the object changes its direction, the velocity changes.
3) Uniform velocity implies zero acceleration. Hence, v = u + at becomes v = u. The velocity of the body at 10 seconds will be the same as the velocity of the body at 0 seconds, in this case, it will be 10 m/s.
4) Yes, the body can have a constant speed and variable velocity. Whenever the direction changes, the velocity changes.
5) A body is said to have uniform velocity when it covers equal distances in equal intervals of time. A body is said to have non-uniform velocity when it covers unequal distances in equal intervals of time.
6) When the velocity of a body changes at a uniform rate then the average velocity is the mean of initial velocity & final velocity for a specific period of time. Mathematically, it is written as follows:
Average Velocity = (Initial Velocity + Final velocity)/2
When the velocity of a body changes at a non-uniform rate then its average velocity is net displacement covered by the total time taken and is given as,
Average Velocity = Net Displacement/ Total Time Taken
7) The average velocity is calculated as follows:
Average Velocity = Net Displacement/ Time
Average Velocity = (70+70)/20 = 7 m/s
The average speed is calculated as follows:
Average Speed = Total Distance/ Total Time
Ï€r/20 = 220/20 (220 is the approximated value)
=11 m/s
2. Express average velocity when the velocity of a body changes at a non – uniform rate and a
uniform rate.<|endoftext|>
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People of the Neolithic age around 6,000 years ago were closely connected both in life and death. This became evident in a detailed archaeological and anthropological of a collective grave containing 50 bodies near Burgos, northern Spain. In the pioneering study, researchers used a whole array of modern methods to examine the way of life in the region at that time. Published in the academic journal PLOS ONE, the research was conducted by anthropologists at the University of Basel and archaeologists from the University of Valladolid.
The collective graves of the Neolithic period were made mostly of stone and were large enough to hold many bodies in a communal space. The megalithic tomb in Alto de Reinoso, Burgos differs from this in only one respect: the burial chamber had originally been made from wood over which a stone mound was erected afterwards. Radiocarbon dating suggests that the tomb was used by a community living between 3700 and 3600 AD and spanning about three to four generations. At least 47 bodies have been found. "While the lower layer has been relatively well preserved, numerous disturbances have been observed in the upper layers such as missing skulls, which could be due to a certain kind of ancestral worship," says Prof. Manuel Rojo Guerra, from the University of Valladolid, Spain.
Way of life revealed
In order to examine the Neolithic community's way of life, researchers used modern methods to collect individual data such as age, sex, body height, disease, stress markers and signs of violence. This data was also supplemented by information about diet, area of origin, mobility and familial relationships. "This is the first study that presents a detailed picture of how Neolithic people were connected in life and death," says lead author Prof. Kurt W. Alt, visiting professor at the University of Basel.
Almost half of the deceased were adults; the other half were children and adolescents. The average body height was 159 cm ± 2 cm for men and 150 cm ± 2 cm for women. The adults showed skeletal stress markers and various stages of degenerative diseases of the spine and joints, healed fractures, head injuries and dental diseases such as caries.
Molecular genetic studies also revealed the relationships between individuals in the group, especially on the maternal side. Furthermore, there is evidence that bodies buried close to each other were in some cases also closely related genetically. Apart from three individuals, all of the deceased grew up in the area around the collective grave. A reconstruction of the diet further demonstrates the homogeneous structure of the farming community: the staple food for everyone was cereals (wheat and barley) and animal proteins (sheep, goat and pig in particular).<|endoftext|>
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Do some LEGO® math! Here’s a fun way to teach kids about probability and graphing with LEGO® bricks. Draw bricks out of a bag at random and record which colors are chosen. Then graph the results!
How to Play
We used a baseplate and some tan bricks to build a graph. Then, I told my 8 year old that we were going to put some LEGO bricks in a bag and choose one out one at a time.
We chose to put 5 bricks of each color in our paper bag. So 25 total bricks. Then Owen drew one brick out at a time and graphed his results.
In order to keep the math accurate, we used a separate pile of bricks for filling in the graph. Otherwise, we would have changed the odds every time we removed a brick from the bag and then didn’t return it! So we put each brick back each time.
Here are our results after drawing a brick out of the bag 10 times:
Teach the Concepts of Probability and Actual Results
Then we discussed our results. The odds of drawing each of the five colors were equal at a 1 in 5 chance, but we discussed the fact that the actual outcome of an event does not always perfectly reflect the odds. The fact that we did not draw any red bricks was slightly unlikely, but not impossible. On the other hand, it would be technically possible to draw only one color of brick, but not at all likely.
It would be fun to do this project with a class, because the more times that this activity is repeated, the closer the actual results will reflect the odds. We didn’t draw any red bricks, but the more we repeated this activity, the average of the results would get closer and closer to 2 bricks of each color when 10 were drawn.
Then we changed the odds! We swapped out our bricks so that we had 20 red bricks and 25 of other colors.
Our results after drawing 10 bricks reflected the increase in red bricks!
It would also be fun to try a bag with 30 blue bricks, and only 1 red brick. How many times of drawing a brick would it take before drawing the red brick? It would be possible to draw the red one on the first try, but it would be far more likely for it to take 10 or more attempts.
See, math is actually fun after all! I really think that probability and statistics is a very helpful and practical math topic to study. It shows up all over the place in real life!
More LEGO® Math Ideas:
LEGO® is a registered trademark of the LEGO® group of companies which does not sponsor, authorize, or endorse this site.<|endoftext|>
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question
A bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by a Styroform sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used.
Thank you very much.
2. I'm not sure if I'm correct but here it goes...
Instead of calling the radius R, call it x as it's probably something we're both familiar with.
Lets only imagine the top left corner of the box and cut it out so we have box of sides of length of the radius x. Also in the box is top left quarter of our circle of radius x so it's not touching the top left corner.
Now we should find the length of the diagonal that goes from the centre of the big circle to the top left corner of the box:
$\displaystyle \sqrt x^2 + x^2=\sqrt 2x^2 = x\sqrt 2$
How do we get the radius of the smaller circle?
Well we get the length from the edge of the big circle to the top left corner leaving just the smaller circle within it. To do this it is just the diagonal length minus the radius of the big circle:
$\displaystyle x\sqrt{2} - x$
Which can not be canceled down further.
HOWEVER, this is the diagonal of the line from where the big circle touches the smaller one so really we have the diameter of the circle plus the bit in the corner. So in plain terms, we have the diameter plus top left corner minus the bottom right corner (PS. to visualise this, draw it all out with the diameter always drawn as the diagonal through the box).
So since the corners cancel out, we actually had the diameter all along! And in order to get the radius, just halve it. Which gives:
$\displaystyle \frac{x\sqrt{2} - x}{2}$
And since you want it in terms of R....
$\displaystyle SmallRadius=\frac{R\sqrt{2} - R}{2}$
I hope this is alright for you to understand! I was really working it out as I was going along so you have my thoughts here too and this is my second helping post I think too!
I think a good tip would be to draw out each stage as you read it!
Also you should note that you can keep working out the radiuses of each smaller circle like this an infinite number of times as it is impossible to reach the corner with circles!
Anthmoo
3. Originally Posted by anthmoo
I'm not sure if I'm correct but here it goes...
Instead of calling the radius R, call it x as it's probably something we're both familiar with.
Lets only imagine the top left corner of the box and cut it out so we have box of sides of length of the radius x. Also in the box is top left quarter of our circle of radius x so it's not touching the top left corner.
I'm having difficulty following your explanation, but to me it looks like you
are solving the 2-D analog of this problem rather than the 3-D problem
RonL
4. Originally Posted by Jenny20
question
A bowling ball of radius R is placed inside a box just large enough to hold it, and it is secured for shipping by a Styroform sphere into each corner of the box. Find the radius of the largest Styrofoam sphere that can be used.
Thank you very much.
Hello Jenny,
I've attached a sketch of the described situation.
By Pythagorian rule you get:
$\displaystyle r^2+r^2=(r+x)^2$. Thus $\displaystyle r+x=r \cdot \sqrt{2}$
You get the propoertion:
$\displaystyle \frac{R}{r}=\frac{R+2r+x}{r+x}$
$\displaystyle \frac{R}{r}=\frac{R+r+r+x}{r+x}=\frac{R+r}{r+x}+1$
$\displaystyle \frac{R-r}{r}=\frac{R+r}{r+x}$. Now plug in the term for (r+x)
$\displaystyle \frac{R-r}{r}=\frac{R+r}{r \cdot \sqrt{2}}$. Now solve for r. You'll get 2 solutions. The neagtive solution isn't very realistic with your problem. I've got:
$\displaystyle r=R\cdot \left(\frac{\sqrt{2}-1}{2} \right)$
EB
5. Originally Posted by earboth
Hello Jenny,
I've attached a sketch of the described situation.
By Pythagorian rule you get:
$\displaystyle r^2+r^2=(r+x)^2$. Thus $\displaystyle r+x=r \cdot \sqrt{2}$
You get the propoertion:
$\displaystyle \frac{R}{r}=\frac{R+2r+x}{r+x}$
$\displaystyle \frac{R}{r}=\frac{R+r+r+x}{r+x}=\frac{R+r}{r+x}+1$
$\displaystyle \frac{R-r}{r}=\frac{R+r}{r+x}$. Now plug in the term for (r+x)
$\displaystyle \frac{R-r}{r}=\frac{R+r}{r \cdot \sqrt{2}}$. Now solve for r. You'll get 2 solutions. The neagtive solution isn't very realistic with your problem. I've got:
$\displaystyle r=R\cdot \left(\frac{\sqrt{2}-1}{2} \right)$
EB
2-D analog of the problem, not what was asked, which is for the maximum
sizes sphere that can be fitted into the corner of a box in the form of a
cube of side 2R containing a sphere of radius R.
(Probably wrong for the 2-D analog as well)
RonL
6. Hello, Jenny!
A different approach . . . (and a different answer)
A bowling ball of radius $\displaystyle R$ is placed inside a box just large enough to hold it,
and it is secured for shipping by a Styroform sphere into each corner of the box.
Find the radius $\displaystyle r$ of the largest Styrofoam sphere that can be used.
Code:
A R B
*-------*-*-*-------*
| * : * / |
| * : / * |
|* R: / *|R
| : / |
* : / *
* * - - - - *C
* O R *
| |
|* *|
| * * |
| * * |
*-------*-*-*-------*
In the upper-right, we have an R-by-R square: $\displaystyle OABC.$
The diagonal $\displaystyle OB$ has length $\displaystyle R\sqrt{2}.$
Diagonal $\displaystyle OB$ intersects the circle at $\displaystyle D$ (not shown).
Then: .$\displaystyle DB \:=\:R\sqrt{2} - R$
This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: .$\displaystyle \frac{R\sqrt{2} - R}{\sqrt{2}}$
The radius of the inscribed circle is: .$\displaystyle \frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$
Therefore: .$\displaystyle r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$
7. Originally Posted by CaptainBlack
I'm having difficulty following your explanation, but to me it looks like you
are solving the 2-D analog of this problem rather than the 3-D problem
RonL
I know what you mean - would it be good enough to treat the situation as a 2-D one if we take the diagonal cross-section of the box?
8. Originally Posted by anthmoo
I know what you mean - would it be good enough to treat the situation as a 2-D one if we take the diagonal cross-section of the box?
No because the smaller sphere would not touch one side of the crossection.
RonL
9. Originally Posted by Soroban
Hello, Jenny!
A different approach . . . (and a different answer)
Code:
A R B
*-------*-*-*-------*
| * : * / |
| * : / * |
|* R: / *|R
| : / |
* : / *
* * - - - - *C
* O R *
| |
|* *|
| * * |
| * * |
*-------*-*-*-------*
In the upper-right, we have an R-by-R square: $\displaystyle OABC.$
The diagonal $\displaystyle OB$ has length $\displaystyle R\sqrt{2}.$
Diagonal $\displaystyle OB$ intersects the circle at $\displaystyle D$ (not shown).
Then: .$\displaystyle DB \:=\:R\sqrt{2} - R$
This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: .$\displaystyle \frac{R\sqrt{2} - R}{\sqrt{2}}$
The radius of the inscribed circle is: .$\displaystyle \frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$
Therefore: .$\displaystyle r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$
Ho-Hum...
RonL - Too lazy to type the same thing a third time
10. Originally Posted by CaptainBlack
2-D analog of the problem, not what was asked, which is for the maximum
sizes sphere that can be fitted into the corner of a box in the form of a
cube of side 2R containing a sphere of radius R.
RonL
I tried to do it the diagonal way quickly and got something which I think is very incorrect...
$\displaystyle \frac{\sqrt{4x^2 + x^2\sqrt{8}} - x}{2}$
11. Originally Posted by Soroban
Hello, Jenny!
A different approach . . . (and a different answer)
Code:
A R B
*-------*-*-*-------*
| * : * / |
| * : / * |
|* R: / *|R
| : / |
* : / *
* * - - - - *C
* O R *
| |
|* *|
| * * |
| * * |
*-------*-*-*-------*
In the upper-right, we have an R-by-R square: $\displaystyle OABC.$
The diagonal $\displaystyle OB$ has length $\displaystyle R\sqrt{2}.$
Diagonal $\displaystyle OB$ intersects the circle at $\displaystyle D$ (not shown).
Then: .$\displaystyle DB \:=\:R\sqrt{2} - R$
This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: .$\displaystyle \frac{R\sqrt{2} - R}{\sqrt{2}}$
The radius of the inscribed circle is: .$\displaystyle \frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$
Therefore: .$\displaystyle r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$
Well I guess I'm wrong..
Worth a try though!
12. Hi all,
r =(2-sqrt(3))R
13. And also the hint:
Take the origin of a Cartesian coordinate system at a corner of the box with the coordinate axes along the edges.
14. Consider one corner of the box, the ball of radius $\displaystyle R$, and a smaller sphere
which is the largest that will fit between the large sphere (the ball) and the corner.
Let the radius of the small sphere be $\displaystyle \rho$
Take one corner of the box as origin with axes along the edges of the box.
Then the diagonal of the box is the line $\displaystyle \lambda (1,\ 1,\ 1)$ and the
centres of the small sphere in the corner choosen as origin, and the
large sphere both lie on this diagonal, and so does the kissing point
between the two spheres.
The equation of the small sphere is:
$\displaystyle (x-\rho)^2+(y-\rho)^2+(z-\rho)^2=\rho^2$
and the points at which the diagonal meets this satisfy:
$\displaystyle (\lambda-\rho)^2+(\lambda-\rho)^2+(\lambda-\rho)^2=\rho^2$,
or: $\displaystyle \lambda=\rho(1 \pm 1/\sqrt{3})$
Now doing the same for the large sphere gives:
$\displaystyle \lambda=R(1 \pm 1/\sqrt{3})$,
and as the spheres are kissing the larger of the roots for the small sphere
is equal to the smaller for the large sphere, so:
$\displaystyle \rho(1 + 1/\sqrt{3})=R(1 - 1/\sqrt{3})$
Hence:
$\displaystyle \rho=R\ \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}}$
Which simplifies down to:
$\displaystyle \rho=R\ (2-\sqrt{3})$
RonL
15. Originally Posted by Soroban
Hello, Jenny!
A different approach . . . (and a different answer)
Code:
A R B
*-------*-*-*-------*
| * : * / |
| * : / * |
|* R: / *|R
| : / |
* : / *
* * - - - - *C
* O R *
| |
|* *|
| * * |
| * * |
*-------*-*-*-------*
In the upper-right, we have an R-by-R square: $\displaystyle OABC.$
The diagonal $\displaystyle OB$ has length $\displaystyle R\sqrt{2}.$
Diagonal $\displaystyle OB$ intersects the circle at $\displaystyle D$ (not shown).
Then: .$\displaystyle DB \:=\:R\sqrt{2} - R$
This is the diagonal of a smaller square in the upper-right corner.
. . Its side is: .$\displaystyle \frac{R\sqrt{2} - R}{\sqrt{2}}$
The radius of the inscribed circle is: .$\displaystyle \frac{1}{2} \times \frac{R\sqrt{2} - R}{\sqrt{2}}$
Therefore: .$\displaystyle r \;= \;R\left(\frac{\sqrt{2} - 1}{2\sqrt{2}}\right)$
The 2-D analog of the full 3-D solution is:
$\displaystyle R(3-2\sqrt{2})\approx0.172\,R$,
while the above solution gives:
$\displaystyle R(\sqrt{2}-1)/(2\sqrt{2})\approx 0.146\,R$,
and construction (drawing a scal diagram and measuring) gives a factor more
like $\displaystyle \approx 0.175$ which allowing for errors in construction is consistent with the
first but not the second of these candidates.
RonL
Page 1 of 2 12 Last<|endoftext|>
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# ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures
## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 13 Rectilinear Figures
Exercise 13.1
Question 1.
If two angles of a quadrilateral are 40° and 110° and the other two are in the ratio 3 : 4, find these angles.
Solution:
Question 2.
If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.
Solution:
Question 3.
If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram.
Solution:
Question 4.
(a) In figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.
(b) In figure (2) given below, ABCD is a parallelogram. Find the angles of the AAOD.
(c) In figure (3) given below, ABCD is a rhombus. Find the value of x.
Solution:
Question 5.
(a) In figure (1) given below, ABCD is a parallelogram with perimeter 40. Find the values of x and y.
(b) In figure (2) given below. ABCD is a parallelogram. Find the values of x and y.
(c) In figure (3) given below. ABCD is a rhombus. Find x and y.
Solution:
Question 6.
The diagonals AC and BD of a rectangle > ABCD intersect each other at P. If ∠ABD = 50°, find ∠DPC.
Solution:
Question 7.
(a) In figure (1) given below, equilateral triangle EBC surmounts square ABCD. Find angle BED represented by x.
(b) In figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of the ∆ AOB.
(c) In figure (3) given below, ABCD is rhombus and diagonals intersect at O. If ∠OAB : ∠OBA = 3:2, find the angles of the ∆ AOD.
Solution:
Question 8.
(a) In figure (1) given below, ABCD is a trapezium. Find the values of x and y.
(b) In figure (2) given below, ABCD is an isosceles trapezium. Find the values of x and.y.
(c) In figure (3) given below, ABCD is a kite and diagonals intersect at O. If ∠DAB = 112° and ∠DCB = 64°, find ∠ODC and ∠OBA.
Solution:
Question 9.
(i) Prove that each angle of a rectangle is 90°.
(ii) If the angle of a quadrilateral are equal, prove that it is a rectangle.
(iii) If the diagonals of a rhombus are equal, prove that it is a square.
(iv) Prove that every diagonal of a rhombus bisects the angles at the vertices.
Solution:
Question 10.
ABCD is a parallelogram. If the diagonal AC bisects ∠A, then prove that:
(i) AC bisects ∠C
(ii) ABCD is a rhombus
(iii) AC ⊥ BD.
Solution:
Question 11.
(i) Prove that bisectors of any two adjacent angles of a parallelogram are at right angles.
(ii) Prove that bisectors of any two opposite angles of a parallelogram are parallel.
(iii) If the diagonals of a quadrilateral are equal and bisect each other at right angles, then prove that it is a square.
Solution:
Question 12.
(i) If ABCD is a rectangle in which the diagonal BD bisect ∠B, then show that ABCD is a square.
(ii) Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Question 13.
P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.
Solution:
Question 14.
(a) In figure (1) given below, ABCD is a parallelogram and X is mid-point of BC. The line AX produced meets DC produced at Q. The parallelogram ABPQ is completed. Prove that:
(i) the triangles ABX and QCX are congruent;
(ii)DC = CQ = QP
(b) In figure (2) given below, points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other.
Solution:
Question 15.
ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP=DQ, prove that AP and DQ are perpendicular to each other.
Solution:
Question 16.
If P and Q are points of trisection of the diagonal BD of a parallelogram ABCD, prove that CQ || AP.
Solution:
Question 17.
A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B ; the four bisectors form a quadrilateral ABCD. Prove that
(i) ABCD is a rectangle.
(ii) CD is parallel to the original parallel lines.
Solution:
Question 18.
In a parallelogram ABCD, the bisector of ∠A meets DC in E and AB = 2 AD. Prove that
(i) BE bisects ∠B
(ii) ∠AEB = a right angle.
Solution:
Question 19.
ABCD is a parallelogram, bisectors of angles A and B meet at E which lie on DC. Prove that AB
Solution:
Question 20.
ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO =AP, prove that 3 ∠POB = ∠AOP.
Solution:
Question 21.
ABCD is a square. E, F, G and H are points on the sides AB, BC, CD and DA respectively such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:
Question 22.
(a) In the Figure (1) given below, ABCD and ABEF are parallelograms. Prove that
(i) CDFE is a parallelogram
(ii) FD = EC
(iii) Δ AFD = ΔBEC.
(b) In the figure (2) given below, ABCD is a parallelogram, ADEF and AGHB are two squares. Prove that FG = AC
Solution:
Question 23.
ABCD is a rhombus in which ∠A = 60°. Find the ratio AC : BD.
Solution:
Exercise 13.2
Question 1.
Using ruler and compasses only, construct the quadrilateral ABCD in which ∠ BAD = 45°, AD = AB = 6cm, BC = 3.6cm, CD = 5cm. Measure ∠ BCD.
Solution:
Question 2.
Draw a quadrilateral ABCD with AB = 6cm, BC = 4cm, CD = 4 cm and ∠ ABC = ∠ BCD = 90°
Solution:
Question 3.
Using ruler and compasses only, construct the quadrilateral ABCD given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm, ∠BAD = 90° and the diagonal AC = 5.5 cm.
Solution:
Question 4.
Construct a quadrilateral ABCD in which AB = 3.3 cm, BC = 4.9 cm, CD = 5.8 cm, DA = 4 cm and BD = 5.3 cm.
Solution:
Question 5.
Construct a trapezium ABCD in which AD || BC, AB = CD = 3 cm, BC = 5.2cm and AD = 4 cm
Solution:
Question 6.
Construct a trapezium ABCD in which AD || BC, ∠B= 60°, AB = 5 cm. BC = 6.2 cm and CD = 4.8 cm.
Solution:
Question 7.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Solution:
Question 8.
Using ruler and compasses only, construct a parallelogram ABCD in which AB = 4.6 cm, BC = 3.2 cm and AC = 6.1 cm.
Solution:
Question 9.
Using ruler and compasses, construct a parallelogram ABCD give that AB = 4 cm, AC = 10 cm, BD = 6 cm. Measure BC.
Solution:
Question 10.
Using ruler and compasses only, construct a parallelogram ABCD such that BC = 4 cm, diagonal AC = 8.6 cm and diagonal BD = 4.4 cm. Measure the side AB.
Solution:
Question 11.
Use ruler and compasses to construct a parallelogram with diagonals 6 cm and 8 cm in length having given the acute angle between them is 60°. Measure one of the longer sides.
Solution:
Question 12.
Using ruler and compasses only, draw a parallelogram whose diagonals are 4 cm and 6 cm long and contain an angle of 75°. Measure and write down the length of one of the shorter sides of the parallelogram.
Solution:
Question 13.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 6 cm, altitude = 3.5 cm and side BC = 4 cm. Measure the acute angles of the parallelogram.
Solution:
Question 14.
The perpendicular distances between the pairs of opposite sides of a parallelogram ABCD are 3 cm and 4 cm and one of its angles measures 60°. Using ruler and compasses only, construct ABCD.
Solution:
Question 15.
Using ruler and compasses, construct a rectangle ABCD with AB = 5cm and AD = 3 cm.
Solution:
Question 16.
Using ruler and compasses only, construct a rectangle each of whose diagonals measures 6cm and the diagonals intersect at an angle of 45°.
Solution:
Question 17.
Using ruler and compasses only, construct a square having a diagonal of length 5cm. Measure its sides correct to the nearest millimeter.
Solution:
Question 18.
Using ruler and compasses only construct A rhombus ABCD given that AB 5cm, AC = 6cm measure ∠BAD.
Solution:
Question 19.
Using ruler and compasses only, construct rhombus ABCD with sides of length 4cm and diagonal AC of length 5 cm. Measure ∠ABC.
Solution:
Question 20.
Construct a rhombus PQRS whose diagonals PR and QS are 8cip and 6cm respectively.
Solution:
Question 21.
Construct a rhombus ABCD of side 4.6 cm and ∠BCD = 135°, by using ruler and compasses only.
Solution:
Question 22.
Construct a trapezium in which AB || CD, AB = 4.6 cm, ∠ ABC = 90°, ∠ DAB = 120° and the distance between parallel sides is 2.9 cm.
Solution:
Question 23.
Construct a trapezium ABCD when one of parallel sides AB = 4.8 cm, height = 2.6cm, BC = 3.1 cm and AD = 3.6 cm.
Solution:
Question 24.
Construct a regular hexagon of side 2.5 cm.
Solution:
Multiple Choice Questions
Choose the correct answer from the given four options (1 to 12):
Question 1.
Three angles of a quadrilateral are 75°, 90° and 75°. The fourth angle is
(a) 90°
(b) 95°
(c) 105°
(d) 120°
Solution:
Sum of 4 angles of a quadrilateral = 360° Sum of three angles = 75° + 90° + 75° = 240° Fourth angle = 360° – 240° = 120° (d)
Question 2.
A quadrilateral ABCD is a trapezium if
(a) AB = DC
(c) ∠A + ∠C = 180°
(d) ∠B + ∠C = 180°
Solution:
A quadrilateral ABCD is a trapezium if ∠B + ∠C= 180°
(Sum of co-interior angles) (d)
Question 3.
If PQRS is a parallelogram, then ∠Q – ∠S is equal to
(a) 90°
(b) 120°
(c) 0°
(d) 180°
Solution:
PQRS is a parallelogram ∠Q – ∠S = 0
(∵ Opposite angles of a parallelogram, are equal) (c)
Question 4.
A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
(a) 55°
(b) 50°
(c) 40°
(d) 25°
Solution:
In a rectangle a diagonal is inclined to one side of the rectangle is 25°
Question 5.
ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is
(a) 40°
(b) 45°
(c) 50°
(d) 60°
Solution:
Question 6.
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠D AC = 32° and ∠AOB = 70°, then ∠DBC is equal to
(a) 24°
(b) 86°
(c) 38°
(d) 32°
Solution:
Question 7.
If the diagonals of a square ABCD intersect each other at O, then ∆OAB is
(a) an equilateral triangle
(b) a right angled but not an isosceles triangle
(c) an isosceles but not right angled triangle
(d) an isosceles right angled triangle
Solution:
Question 8.
If the diagonals of a quadrilateral PQRS bisect each other, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Solution:
Diagonals of a quadrilateral PQRS bisect each other, then quadrilateral must be a parallelogram.
(∵ A rhombus, rectangle and square are also parallelogram) (a)
Question 9.
If the diagonals of a quadrilateral PQRS bisect each other at right angles, then the quadrilateral PQRS must be a
(a) parallelogram
(b) rectangle
(c) rhombus
(d) square
Solution:
Diagonals of quadrilateral PQRS bisect each other at right angles, then quadrilateral PQRS [ must be a rhombus.
(∵ Square is also a rhombus with each angle equal to 90°) (c)
Question 10.
Which of the following statement is true for a parallelogram?
(a) Its diagonals are equal.
(b) Its diagonals are perpendicular to each other.
(c) The diagonals divide the parallelogram into four congruent triangles.
(d) The diagonals bisect each other.
Solution:
For a parallelogram an the statement ‘The diagoanls bisect each other’ is true. (d)
Question 11.
Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) opposite angles are bisected by the diagonals
(d) diagonals bisect each other
Solution:
The statement that in a parallelogram, .the opposite angles are bisected by the diagonals, is not true in each case. (c)
Question 12.
A quadrilateral in which the diagonals are equal and bisect each other at right angles is a
(a) rectangle which is not a square
(b) rhombus which is not a square
(c) kite which is not a square
(d) square
Solution:
In a quadrilateral, if diagonals are equal and bisect each other at right angles, is a square. (d)
Chapter Test
Question P.Q.
The interior angles of a polygon add upto 4320°. How many sides does the polygon have ?
Solution:
Question P.Q.
If the ratio of an interior angle to the exterior angle of a regular polygon is 5:1, find the number of sides.
Solution:
Question P.Q.
In a pentagon ABCDE, BC || ED and ∠B: ∠A : ∠E =3:4:5. Find ∠A.
Solution:
Question 1.
In the given figure, ABCD is a parallelogram. CB is produced to E such that BE=BC. Prove that AEBD is a parallelogram.
Solution:
Question 2.
In the given figure, ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD || BA. Show that
(i) ∠DAC=∠BCA
(ii) ABCD is a parallelogram.
Solution:
Question 3.
Prove that the quadrilateral obtained by joining the mid-points of an isosceles trapezium is a rhombus.
Solution:
Question 4.
Find the size of each lettered angle in the Following Figures.
Solution:
Question 5.
Find the size of each lettered angle in the following figures :
Solution:
Question 6.
In the adjoining figure, ABCD is a rhombus and DCFE is a square. If ∠ABC = 56°, find
(i) ∠DAG
(ii) ∠FEG
(iii) ∠GAC
(iv) ∠AGC.
Solution:
Question 7.
If one angle of a rhombus is 60° and the length of a side is 8 cm, find the lengths of its diagonals.
Solution:
Question 8.
Using ruler and compasses only, construct a parallelogram ABCD with AB = 5 cm, AD = 2.5 cm and ∠BAD = 45°. If the bisector of ∠BAD meets DC at E, prove that ∠AEB is a right angle.
Solution:<|endoftext|>
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# Are Vertical Angles Supplementary
by -2 views
The two angles form a linear angle such that if one angle is x then the other the angle is 180 x. When the sum of two angles is 180 then the angles are known as supplementary angles.
Adjacent And Vertical Angles Notes Vertical Angles Geometry Lessons Junior High Math
### We can see this if we start at the top left and work our way clockwise around the figure.
Are vertical angles supplementary. A pair of vertically opposite angles are always equal to each other. Two Angles are said to be Supplementary when they add up to 180 degrees. Identify each pair of angles as vertical supplementary complementary or none of the above.
3 a1 and a3 are supplementary to the same angle so a1 is congruent to a3 by the Congruent Supplements Theorem. In the image above angles A and B are supplementary so add up to 180. Students find unknown angles using supplementary adjacent and vertical angle relationships.
Supplementary angles are pairs of angles that add up to 180. A 133 degree angle is supplementary to the complement of a 43 degree angle. Complementary angles are two angles with a sum of 90º.
Also a vertical angle and its adjacent angle are supplementary angles ie they add up to 180 degrees. A B 180. Because all straight lines are 180 we know Q and S are supplementary adding to 180.
Complementary angles add up to 90º. Vertical angles are the angles which are opposite of each other at the intersection of two lines. When 2 lines intersect they make vertical angles.
Supplementary angles are two angles with a sum of 180º. For example if two lines intersect and make an angle say X45 then its opposite angle is also equal to 45. These angles are NOT adjacent.
Those two adjacent angles will always add to 180. Slide 6 Slide 7 Slide 8 Supplementary angles add up to 180º. Vertical Angles Theorem This is a type of proof regarding angles being equal when they are vertically opposite.
In one particular case when vertical angles are right angles that is lines that form this pair of vertical angles are perpendicular their sum is equal to a straight angles and therefore they are supplemental. Vertical angles are angles formed when two lines intersect. A pair meaning only two angles of angles that add up to exactly 90 degrees.
Both supplementary angles and complementary angles are much broader – they do not even have to be touching or near each other but they could be. 1 a1 and a2 are a linear pair so a1 and a2 are supplementary. Identifying supplementary complementary and vertical angles This is the currently selected item.
Vertical angles are very specific – you have two have two intersecting lines to form two sets of vertical angles which are across from each other and congruent. Complementary and supplementary angles visual. So if the two lines are perpendicular then the vertical angles will sum to 180.
Supplementary angles are two angles that sum to 180. Vertical angles are two angles whose sides form two pairs of opposite rays. Take any two adjacent angles from among the four angles created by two intersecting lines.
We can think of these as opposite angles formed by an X. While vertical angles are not always supplementary adjacent angles are always supplementary. See the picture below that represents angles formed by perpendicular lines.
They do not need to be adjacent they can be separate. If they are placed adjacent they will make a straight angle. In other words if two angles add up to form a straight angle then those angles are referred to as supplementary angles.
Supplementary angles are not limited to just transversals. Supplementary angles need not have to be placed adjacent to each other. 2 a2 and a3 are a linear pair so a2 and a3 are supplementary.
The following steps show why the Vertical Angles Theorem is true. This calculator will calculate the difference of the given angle with 180 degree. These are called supplementary angles.
Another property of vertical angles is that angles that are next to each other are supplementary angles meaning that they add up to 180 degrees. Prior to completing this lesson students should understand that angles are a measurement of rotation know that one full rotation measures 360 and recognize angle measure as additive. So if one angle measures 35 degrees then the.
What Is Supplementary Vertical And Complementary And Adjacent Angles Adjacent Angles Complementary Angles Vertical Angles
Angle Recognition 7th Grade Math Anchor Chart Angles Worksheet Angles Math Graphing Linear Equations
Pin On High School Math Posters And Anchor Charts
This Math Shorts Video Reviews Different Types Of Angles Complementary Supplementary Adja Angle Activities Interior And Exterior Angles Complementary Angles
Image Result For Angles Adjacent Vertical Complementary Supplementary Adjacent Angles Supplementary Angles Vertical Angles
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Special Angle Pairs Mini Posters Sol 2016 8 5 Teaching Geometry Math Geometry Angle Pairs
Geometry Angles Frayer Model Vocabulary Vocabulary They Will Like And Use Geometry Angles Vertical Angles Vocabulary
Angles Complementary Supplementary Vertical Adjacent Common Core 7 G Everyday Math Teaching Middle School Maths Teaching Geometry
Math Rap Complementary And Supplementary Angles Youtube Angles Worksheet Middle School Math Homeschool Math
Complementary Supplementary And Vertical Angle Practice With These Fun Stations Includes Card Sort Angle Relationships Vertical Angles Supplementary Angles
Vertical Supplementary Adjacent Angles Using A City Map Students Use A Map Of Streets In A City That Creates Ver Teaching Geometry Math Teacher Math Tutor
Angles Relationships Complementary Adjacent Supplementary Vertical Congruent Math Anchor Charts Math Methods Teaching Math
Complementary And Supplementary Angles Lindsay Bowden Supplementary Angles Junior High Math Middle School Math
Adjacent And Vertical Angles Notes Vertical Angles Angle Activities Angles Worksheet
Angle Relationships Graphic Organizer Angle Relationships Graphic Organizers Interactive Math Journals
Angle Relationships Foldable Angle Relationships Angle Relationships Foldable Math Interactive Notebook
Pin On Math Class
Needed A Real World Application For Angles In This Activity Students Use Vertical Angles Complement Free Math Lessons Elementary School Math Education Math<|endoftext|>
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Meningitis is an infection and consequent inflammation of the fluid and membranes surrounding your spinal cord and brain and is sometimes referred to as spinal meningitis. Most often, it is caused by a bacteria or viruses. Bacterial meningitis is typically more severe than its viral counterpart. Over two-thirds of the approximately 10,000 cases that occur each year are in children under five years of age. Those under two years of age have the highest incidence due to their immature immune systems. 10% of those afflicted with the condition die as a result.
Contact a medical malpractice attorney with experience handling failure to diagnose meningitis lawsuits.
If meningitis is not diagnosed in a timely manner and treated properly, the consequence can manifest itself in the form of deafness, brain damage, and seizure disorders. Diagnosis may be difficult because no single symptom is necessarily a positive indicator, and many of the symptoms are subtle or confusing. It is a combination of symptoms and the lack of alternate grounds for the infection that should raise suspicion.
At the onset of meningitis, symptoms may not be unlike those of a flu. A change in the usual pattern of the child’s behavior, however, quickly ensues, and may include lethargy and drowsiness alternating with irritability, and a reluctance to eat. Those afflicted with meningitis often complain of nausea, fever, vomiting, confusion, and sensitivity to light. As the condition progresses, another indicator is bruises which form under the skin and spread quickly.
Bacterial meningitis can be diagnosed with a lumbar puncture or spinal tap. But because the initial symptoms are so similar to those of influenza, a spinal tap may be foregone. By the time the symptoms worsen, it is often too late to prevent brain damage or even death.
Treatment often involves the administration of intravenous antibiotics and cortico-steroids to prevent hearing loss. In cases with the bacterial form of the disease, the right antibiotic must be introduced to kill the specific bacteria causing the meningitis.
Following are issues that continue to contribute to misdiagnosis and mistreatment of the disease:
- Failure to see the patient when requested
- Failure to act with the necessary degree of urgency
- Failure to examine the patient appropriately
- Sending the patient home instead of admitting him or her to the hospital
- Failure to consult with more experienced doctors or specialists
- Failure to arrange adequate follow-up
- Communication breakdown between the doctor and the patient or patient's guardians
If you suspect that your child’s meningitis could have been prevented or moderated with a timely diagnosis, you may wish to sue to recover damages. Be sure, however, that the attorney you retain is an experienced failure to diagnose meningitis lawyer. He or she may need to be able to garner expert testimony to demonstrate that the misdiagnosis was a breach of standard of care resulting in exacerbated injuries to your child.
Contact our Personal Injury Lawyers and Attorneys today to find an experienced lawyer specializing in failure to diagnose meningitis cases.<|endoftext|>
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In symmetric key cryptography, both parties must possess a secret key which they must exchange prior to using any encryption. Distribution of secret keys has been problematic until recently, because it involved face-to-face meeting, use of a trusted courier, or sending the key through an existing encryption channel. The first two are often impractical and always unsafe, while the third depends on the security of a previous key exchange.
In public key cryptography, the key distribution of public keys is done through public key servers. When a person creates a key-pair, they keep one key private and the other, known as the public-key, is uploaded to a server where it can be accessed by anyone to send the user a private, encrypted, message.
Secure Sockets Layer (SSL) uses Diffie–Hellman key exchange if the client does not have a public-private key pair and a published certificate in the public key infrastructure, and Public Key Cryptography if the user does have both the keys and the credential.
Key distribution is an important issue in wireless sensor network (WSN) design. There are many key distribution schemes in the literature that are designed to maintain an easy and at the same time secure communication among sensor nodes. The most accepted method of key distribution in WSNs is key predistribution, where secret keys are placed in sensor nodes before deployment. When the nodes are deployed over the target area, the secret keys are used to create the network. For more info see: key distribution in wireless sensor networks.
Storage of keys in the cloud
Key distribution and key storage are more problematic in the cloud due to the transitory nature of the agents on it. Secret sharing can be used to store keys at many different servers on the cloud. In secret sharing, a secret is used as a seed to generate a number of distinct secrets, and the pieces are distributed so that some subset of the recipients can jointly authenticate themselves and use the secret information without learning what it is. But rather than store files on different servers, the key is parceled out and its secret shares stored at multiple locations in a manner that a subset of the shares can regenerate the key.
Secret sharing is used in cases where one wishes to distribute a secret among N shares so that M < N of them (M of N) can regenerate the original secret, but no smaller group up to M − 1 can do so.
- Du, Wenliang Kevin; Deng, Jing; Han, Yunghsiang S.; and Varshney, Pramod K., "A Pairwise Key Pre-Distribution Scheme for Wireless Sensor Networks" (2000). Electrical Engineering and Computer Science. Paper 36. http://surface.syr.edu/eecs/36
- Yu, S. et al. Achieving Secure, Scalable, and Fine-grained Data Access Control in Cloud Computing. IEEE INFOCOM 2010.
- Parakh, A. and Kak, S. Online data storage using implicit security. Information Sciences, vol. 179, pp. 3323-3331, 2009.
- Shamir, Adi (1979). "How to share a secret". Communications of the ACM 22 (11): 612–613.
- Blakley, G. R. (1979). "Safeguarding cryptographic keys". Proceedings of the National Computer Conference 48: 313–317.<|endoftext|>
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The Planck telescope team has released its first full-sky picture – the clearest yet taken, showing the glow of the first light in the universe.
Pieced together over the last six months, the microwave images show the plane of our galaxy, the Milky Way, as a glowing band across the middle.
The red and yellow mottling in the high-latitude regions of the map shows the subtle temperature variations that derive from the irregular distribution of matter in the early Universe, 380 million years ago – the focus of Planck’s mission.
To the right of the main image, below the plane of the Milky Way, is a large cloud of gas in our galaxy. The arc of light surrounding it is Barnard’s Loop – the expanding bubble of an exploded star.
The €600 million telescope reveals other galaxies too. The great spiral galaxy of Andromeda, 2.2 million light-years from Earth, shows up as a sliver of microwave light, released by the coldest dust in its giant body.
Other, more distant, galaxies with supermassive black holes appear as single points of microwaves dotting the image.
Learning more about the early Universe will requre the removal of these elements with sophisticated image processing software.
This will give scientists a clearer focus on the primordial microwaves – the Cosmic Microwave Background Radiation – represented by the red and yellow portions of the image. ESA hopes to use this data to establish the age of the universe and its composition more precisely than ever before.
The Planck telescope was launched in May last year. The new image is derived from data collected between August 2009 and June 2010.<|endoftext|>
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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2012 AMC 12A Problems/Problem 24"
## Problem
Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general,
$$a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}$$
Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$
$\textbf{(A)}\ 671\qquad\textbf{(B)}\ 1006\qquad\textbf{(C)}\ 1341\qquad\textbf{(D)}\ 2011\qquad\textbf{(E)}\ 2012$
## Solution
First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$(decreasing exponential function), and $g(x) = x^k$ for $k > 0$(increasing exponential function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant.
We will now examine the first few terms.
Comparing $a_1$ and $a_2$, $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$.
Therefore, $0 < a_1 < a_2 < 1$.
Comparing $a_2$ and $a_3$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$.
Comparing $a_1$ and $a_3$, $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$.
Therefore, $0 < a_1 < a_3 < a_2 < 1$.
Comparing $a_3$ and $a_4$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$.
Comparing $a_2$ and $a_4$, $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$.
Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$.
Continuing in this manner, it is easy to see a pattern.
We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$.
We will now use induction to prove this statement. (Note that this is not necessary on the AMC):
Base Case: We have already shown the base case above, where $0 < a_1 < a_2 < 1$.
Inductive Step:
Rearranging in decreasing order gives
$1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$.
Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{\textbf{(C)} 1341}$.<|endoftext|>
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Activity 1: Story: Barn School, Free School
Activity time: 10 minutes
Materials for Activity
- A copy of the story "Barn School, Free School"
- Optional: "Barn School, Free School" coloring sheet, and crayons
Preparation for Activity
- Read the story several times; it has a lot of conversation. Practice using differing voices for each character.
- Optional: Download, print, and copy the coloring sheet (PDF). Place coloring sheets and crayons where children can use them when invited but will not be distracted beforehand.
Description of Activity
Participants will hear the story of an imagined result of Dorothea Dix's real life choice to begin a free school for the poor.
Gather the children to hear a story. Say, in your own words:
Dorothea Dix exemplified the UU Principles in her life choices. Born into a poor family, Dorothea took responsibility for her own life. She left home at the age of 12 and went to live with her grandparents in Boson. She chose to go to school and learn. Dorothea never forgot the lack of hope in her life when she was poor. She offered to teach a free school in the barn of her Grandmother's house. After a period of exhaustion, Dorothea was exposed to the plight of the mentally ill. Again Dorothea chose to take that experience and make the world a better place. For the rest of her life, Dorothea was an active reformer for the mentally ill. Dorothea Dix was a Unitarian woman who made choices to make the world fair for others.
Lead a discussion with these process questions:
- Why did Dorothea choose to go to school and learn?
- How does going to school and learning help all of us learn to make good choices?
Say, "Now let's hear the story about Dorothea Dix."
Read/tell the story enthusiastically. Use these questions to process after the story.
- What do you think "free school" means?
- How did Dorothea use what she had to help others?
- If you could choose to go to school or not to go to school, what would you choose? Why?
- How does going to school and getting an education help us follow the sixth Principle?<|endoftext|>
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Reading for children at this developmental stage is focused on reading for multiple viewpoints. This is also a developmental stage during which reading is often replaced by alternate activities and interests. Motivation and engagement are key factors for children becoming and remaining capable confident readers. During this stage parents need to continue to support their child’s development of reading habits.
The main areas of development are Vocabulary, Comprehension and Reading Processes.
Children of this age are able to draw on their knowledge of word origins to work out the meaning of unknown or new words. They use these new words for familiar concepts such as blissful for happy.
- 11-12 year old children develop opinions through inference skills
Older primary kids are able to justify their personal interpretations and can re-examine a text or section of text for evidence to support their opinion or argument.
- 11-12 year olds can compare how texts are presented
11-12 year olds are able to read and compare different texts such as multi modal form, imaginative, informative or persuasive, and understand how this format alters the way ideas are presented
- 11-12 year olds can comprehend layered text
Children are able to identify multiple purposes within a text. As texts become increasingly complex they often contain more than one purpose.
- 11-12 year olds can apply their knowledge of visual literacy
Kids can apply their visial literacy to evaluate how written information that is read differs to the meaning shaped by visual images on the same topic.
- 11-12 year olds can understand how others can read texts
Children at this developmental stage understand how a person’s personal experience – as the audience – can alter their understanding and inference of the text. They can also interpret a text to create an accurate, unbiased summary which represents alternate perspectives on a central idea for example the discussion genre.
- Children at this age are reading for a sustained period of time (more than 30 minutes) using effectively numerous strategies for maintaining meaning of what they have read.
- Children read more demanding texts that have an increased level of technical terms and abstract concepts.
- They are increasingly independent in their reading and viewing of a wide range of literal, factual and multi modal texts.
- Children at this stage should also be capable of confidently engaging with a wide range of authentic texts for example newspapers, TV documentaries, letters and websites.
Tips for parents
- Prioritise reading in your house and encourage them to continue reading and choosing texts which interest them.
- Link their interest in technology and multi modal texts to written texts by checking out online book reviews, author’s websites or reading forums.
- Help your child choose the correct text for them. Continue to expose them to a wide range of genres pitched at the appropriate level of difficulty.
- Focus on engagement. Use humour when dealing with a reluctant reader. The book “The Day my Bum went Psycho” by Andy Griffiths is a fantastic story about rebel bums who attempt to overthrow the world. For readers in this age group, particularly boys, the topic of bums is irresistible.
- Continue to read aloud to your kids and have them read aloud to you – they are never too old to share a book with you.
Engaged readers seek to understand; they enjoy learning and the believe in their reading abilities (Guthrie, 2001). As parents we should all be aiming to raise engaged readers.
This article was written by Michelle Barrington for Kidspot, New Zealand’s leading education resource for parents. Michelle is a teacher and mother and blogs at Gee, You’re Brave.<|endoftext|>
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by TeachThought Staff
A decent working definition of intrinsic motivation is “motivation that stems directly from an action rather than a reward.” Dr. Richard Ryan and Edward Deci explain in their Intrinsic and Extrinsic Motivations: Classic Definitions & New Directions.
“Intrinsic motivation is defined as the doing of an activity for its inherent satisfaction rather than for some separable consequence.
When intrinsically motivated, a person is moved to act for the fun or challenge entailed rather than because of external products, pressures, or rewards….In Self-Determination Theory, we distinguish between different types of motivation based on the different reason or goals that give rise to an action. The most basic distinction is between intrinsic motivation, which refers to doing something because it is inherently interesting or enjoyable, and extrinsic motivation, which refers to doing something because it leads to a separable outcome. Extrinsic motivation thus contrasts with intrinsic motivation, which refers to doing an activity simply for the enjoyment of the activity itself, rather than its instrumental value.”
Put another way, if a student studies for a test to make a qualifying grade to play for the basketball team, that would be an example of extrinsic motivation. Another example? Studying to “get good grades.” And as you probably know by now, its polar opposite, intrinsic motivation, is the more powerful of the two, though not necessarily more common.
In the following video, Daniel Pink explores the incredible impact of intrinsic motivation on performance, innovation, and the way we learn. While he frames the idea around “business,” he is clearly discussing learning and performance, which is ground zero for educators.
The Definition Of Intrinsic Motivation; image attribution flickr user bengrey<|endoftext|>
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Show that the Signum Function $$f : R \to R$$, given by $f(x) = \left\{ \begin{array}{l l} 1, & if \; x \; > 0 \\ 0, & if\; x\;= 0 \text{ is neither one-one nor onto } \\ -1, & if\; x \;< 0 \end{array} \right.$
Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f: R \to R$
$f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$
Step1: Injective or One-One function:
Consider x=1 and y=2, since both 1 and 2 > 0
$f(1)=f(2)=1$
but $1 \neq 2$
$x\neq y$
$f: R \to R$ $f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$ is not one one
Step 2: Surjective or On-to function:
consider the element -2
$-2 \in R$ there does not exists any element x in R such that $f(x)=-2$
f(x) takes only three values {1,0,-1}
$f: R \to R$ $f(x)= \left\{ \begin{array}{1 1} 1 & \quad x>0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$ is not onto
Solution:Hence singum function is neither one-one nor onto
edited Mar 20, 2013 by meena.p<|endoftext|>
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It’s one of the most iconic images of the modern Space Age. In 1995, the Hubble Space Telescope team released an image of towering columns of gas and dust that contained newborn stars in the midst of formation. Dubbed the “Pillars of Creation,” these light-years long tendrils captivated the public imagination and now grace everything from screensavers to coffee mugs. This is a cosmic portrait of our possible past, and the essence of the universe giving birth to new stars and worlds in action.
Now, a study out on Thursday from the 2014 National Astronomy Meeting of the Royal Astronomical Society has shed new light on just how these pillars may have formed. The announcement comes out of Cardiff University, where astronomer Scott Balfour has run computer simulations that closely model the evolution and the outcome of what’s been observed by the Hubble Space Telescope.
The ‘Pillars’ lie in the Eagle Nebula, also known as Messier 16 (M16), which is situated in the constellation Serpens about 7,000 light years distant. The pillars themselves have formed as intense radiation from young massive stars just beginning to shine erode and sculpt the immense columns.
But as is often the case in early stellar evolution, having massive siblings nearby is bad news for fledgling stars. Such large stars are of the O-type variety, and are more than 16 times as massive as our own Sun. Alnitak in Orion’s belt and the stars of the Trapezium in the Orion Nebula are examples of large O-type stars that can be found in the night sky. But such stars have a “burn fast and die young” credo when it comes to their take on nuclear fusion, spending mere millions of years along the Main Sequence of the Hertzsprung Russell diagram before promptly going supernova. Contrast this with a main sequence life expectancy of 10 billion years for our Sun, and life spans measured in the trillions of years — longer than the current age of the universe — for tiny red dwarf stars. The larger a star you are, the shorter your life span.
Such O-Type stars also have surface temperatures at a scorching 30,000 degrees Celsius, contrasted with a relatively ‘chilly’ 5,500 degree Celsius surface temperature for our Sun.
This also results in a prodigious output in energetic ultraviolet radiation by O-type stars, along with a blustery solar wind. This carves out massive bubbles in a typical stellar nursery, and while it may be bad news for planets and stars attempting to form nearby any such tempestuous stars, this wind can also compress and energize colder regions of gas and dust farther out and serve to trigger another round of star formation. Ironically, such stars are thus “cradle robbers” when it comes to potential stellar and planetary formation AND promoters of new star birth.
In his study, Scott looked at the way gas and dust would form in a typical proto-solar nebula over the span of 1.6 million years. Running the simulation over the span of several weeks, the model started with a massive O-type star that formed out of an initial collapsing smooth cloud of gas.
That’s not bad, a simulation where 1 week equals a few hundred million years…
As expected, said massive star did indeed carve out a spherical bubble given the initial conditions. But Scott also found something special: the interactions of the stellar winds with the local gas was much more complex than anticipated, with three basic results: either the bubble continued to expand unimpeded, the front would expand, contract slightly and then become a stationary barrier, or finally, it would expand and then eventually collapse back in on itself back to the source.
The study was notable because it’s only in the second circumstance that the situation is favorable for a new round of star formation that is seen in the Pillars of Creation.
“If I’m right, it means that O-type and other massive stars play a much more complex role than we previously thought in nursing a new generation of stellar siblings to life,” Scott said in a recent press release. “The model neatly produces exactly the same kind of structures seen by astronomers in the classic 1995 image, vindicating the idea that giant O-type stars have a major effect in sculpting their surroundings.”
Such visions as the Pillars of Creation give us a snapshot of a specific stage in stellar evolution and give us a chance to study what we may have looked like, just over four billion years ago. And as simulations such as those announced in this week’s study become more refined, we’ll be able to use them as a predictor and offer a prognosis for a prospective stellar nebula and gain further insight into the secret early lives of stars.<|endoftext|>
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a theorem
That day a friend asked me about alternate angle theorem, im going to post it here.
The theorem is called
Alternate Segment Theorem
The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
In simple words: a = b (refer to the diagram above).
So my friend asked me: how to prove that a=b
First, we draw an isosceles triangles inside the bigger triangle with a point touching the center. Then we split the isosceles triangle into two parts, each with angle “y”.
As we all know, the line from the tangent to the center of the circle is 90º, then a+x = 90º.
On the other hand, y+x = 90º
From that, we can deduce y+x = a+x
Hence, cancle off the x, we get a=y.
b=y, this is because the angle at the center is two times the angle at the circumference, with the same arc. since the angle at the center is 2y, then b = 1/2(2y) = y.
a=y, b=y. Conclusion a=b.<|endoftext|>
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# Anderson Cooper Under The Microscope (10/16/2019)
How will Anderson Cooper do on 10/16/2019 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is not at all guaranteed – do not take this too seriously. I will first find the destiny number for Anderson Cooper, and then something similar to the life path number, which we will calculate for today (10/16/2019). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 10/16/2019: We will take the month (10), the day (16) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 10 and add the digits together: 1 + 0 = 1 (super simple). Then do the day: from 16 we do 1 + 6 = 7. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 1 + 7 + 12 = 20. This still isn’t a single-digit number, so we will add its digits together again: 2 + 0 = 2. Now we have a single-digit number: 2 is the path number for 10/16/2019.
DESTINY NUMBER FOR Anderson Cooper: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Anderson Cooper we have the letters A (1), n (5), d (4), e (5), r (9), s (1), o (6), n (5), C (3), o (6), o (6), p (7), e (5) and r (9). Adding all of that up (yes, this can get tiring) gives 72. This still isn’t a single-digit number, so we will add its digits together again: 7 + 2 = 9. Now we have a single-digit number: 9 is the destiny number for Anderson Cooper.
CONCLUSION: The difference between the path number for today (2) and destiny number for Anderson Cooper (9) is 7. That is larger than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t fear! As mentioned earlier, this is not at all guaranteed. If you want really means something, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.<|endoftext|>
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Rhetoric and Composition/Memoirs
A memoir is a form of first-person nonfiction writing that portrays a person's experiences during periods of his or her life. Influential people, such as former U. S. Presidents Bill Clinton and Ronald Reagan, often write lengthy memoirs depicting the many critical events of their time in office. Some celebrities write memoirs, or they have ghostwriters help them (or write them entirely). However, fame is not a prerequisite for writing a memoir; any writer who thoughtfully reflects on events in his or her life can write a memoir that will interest others.
Memoirs differ from autobiographies in that they do not cover the person’s entire life; they retell, in a narrative format, significant events or experiences from it. For instance, someone might write about her daughter’s illness or traveling the world – both could make interesting subjects of a memoir. Not all memoirs have to center on uncommon or fascinating events, though; even average experiences can be made into exceptional memoirs if the writer brings them to life with skillful storytelling, and if she communicates a message worth hearing.
If you’d like to write a memoir but are not sure how, it might be a good idea to start small. Think of a significant event in your life and consider how it impacted you, and keep that impact in mind while you’re writing. To begin writing, simply start recording the event and everything you remember seeing, thinking, smelling, feeling, etc. From this recollection of the experience, you can choose which details matter for the story. The most important ones will likely stand out, but even some of the periphery details, like the mustard stain you noticed on your father’s shirt as he yelled at you for coming home late, can bring a narrative to life and tie into the memoir, even if you initially thought they had nothing to do with the story. You may be surprised what else you have to learn about an experience once you’ve begun to write about it.
It can be hard to recall little details from a past event, especially if it happened a long time ago. If you’re struggling to write a recollection, it might help to attempt to partially relive the event. You can do this rather easily by looking at photos taken at the event itself or during that period of your life. If possible, it could be really helpful to visit the place where it happened. If not, looking at and holding whatever material items you still have from that period of your life may help some details resurface. If you’re willing to spend a little time re-immersing yourself in that time period, you could try listening to the music you loved then or rereading a book that had an impact on you. Essentially, you want to attempt to put yourself back in the mindset you were in then so you can more fully reconstruct what you may have experienced during that event. Furthermore, if you can talk to other people who were there, they may be able to fill in some of the details you’re missing.
Although memoirs are supposed to be factual, no one’s memory is perfect, so you may have to invent some details. Unless you’re fabricating the story entirely or painting very inaccurate or unfair depictions of people, there’s no need to feel bad about making up some details in order to tell the story and communicate the message. Some details even get in the way of your story and can be cut out without compromising the message. What matters most is that you feel your memoir tells the truth. Also, remember that even if another person’s recollection of the event doesn’t match up perfectly with yours, you are telling your story; their experiences might help you recall and reflect on your own, but they don’t need to be accounted for in your memoir.<|endoftext|>
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# Learn To Find The Volume Of Prisms And Cylinders.
2y ago
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9-2 Volume of Prisms and CylindersLearn to find the volume of prisms andcylinders.Course 2
9-2 VolumeInsert LessonTitleHereof PrismsandCylindersVocabularyvolumeCourse 2
9-2 Volume of Prisms and CylindersAny solid figure can be filled completely withcongruent cubes and parts of cubes. The volumeof a solid is the number of cubes it can hold. Eachcube represents a unit of measure called a cubicunit.Course 2
9-2 Volume of Prisms and CylindersAdditional Example 1: Using Cubes to Find theVolume of a Rectangular PrismFind how many cubes the prism holds. Thengive the prism’s volume.You can find the volume ofthis prism by counting howmany cubes tall, long, andwide the prism is and thenmultiplying.1 · 4 · 3 12There are 12 cubes in the prism, so the volume is 12cubic units.Course 2
9-2 InsertTitleandHereVolumeLessonof PrismsCylindersTry This: Example 1Find how many cubes the prism holds. Thengive the prism’s volume.You can find the volume ofthis prism by counting howmany cubes tall, long, andwide the prism is and thenmultiplying.2 · 4 · 3 24There are 24 cubes in the prism, so the volume is 24cubic units.Course 2
9-2 Volume of Prisms and CylindersA cube that measures onecentimeter on each siderepresents one cubic centimeterof volume. Suppose the cubes inthe prism in Additional Example 1measure one centimeter on eachside. The volume of the prismwould be 12 cm3.Volume 1 cm31 cm1 cm1 cmThis volume is found by multiplying the prism’s lengthtimes its width times its height.Reading MathAny unit of measurement with an exponent of 3 is a cubicunit. For example, cm3 means “cubic centimeter” and in3means “cubic inch.”Course 2
9-2 Volume of Prisms and Cylinders34cm·3cm·1cm 12cm1 cmlength · width · height volume3 cm4 cmarea ofbase· height volumeNotice that for the rectangular prism, thevolume is found by multiplying the area of itsbase times its height. This method can beused for finding the volume of any prism.VOLUME OF A PRISMThe volume V of a prism is the area of its base Btimes its height h.V BhCourse 2
9-2 Volume of Prisms and CylindersAdditional Example 2: Using a Formula to Find theVolume of a PrismFind the volume of the prism to the nearesttenth.4.1 ft4.1 ft12 ftV BhUse the formula.The bases are rectangles.The area of each rectangular base is 12 · 4.1 49.2V 49.2 · 4.1 Substitute for B and h.Multiply.V 201.72The volume to the nearest tenth is 201.7 ft3.Course 2
9-2 Volume of Prisms and CylindersTry This: Example 2Find the volume of the prism to the nearesttenth.6.3 ft6.3 ft8 ftV BhUse the formula.The bases are rectangles.The area of each rectangular base is 8 · 6.3 50.4V 50.4 · 6.3 Substitute for B and h.Multiply.V 317.52The volume to the nearest tenth is 317.5 ft3.Course 2
9-2 Volume of Prisms and CylindersFinding the volume of a cylinder is similar to findingthe volume of a prism.VOLUME OF A CYLINDERThe volume V of a cylinder is the area of its base, r2, times its height h.V r2hCourse 2
9-2 Volume of Prisms and CylindersAdditional Example 3: Using a Formula to Find theVolume of a CylinderFind the volume of a cylinder to the nearesttenth. Use 3.14 for .V r2hUse the formula.The radius of the cylinder is 5 m, andthe height is 4.2 mV 3.14 · 52 · 4.2 Substitute for r and h.V 329.7Multiply.The volume is about 329.7 m3.Course 2
9-2 InsertTitleandHereVolumeLessonof PrismsCylindersTry This: Example 3Find the volume of a cylinder to the nearesttenth. Use 3.14 for .V r2h7m3.8 mUse the formula.The radius of the cylinder is 7 m, andthe height is 3.8 mV 3.14 · 72 · 3.8 Substitute for r and h.V 584.668Multiply.The volume is about 584.7 m3.Course 2
9-2 VolumeInsert Lessonof PrismsTitleandHereCylindersLesson QuizFind the volume of each solid to the nearesttenth. Use 3.14 for .1.4,069.4 m32.861.8 cm33. triangular prism: base area 24 ft2, height 13 ft312 ft3Course 2
Homework9-2 Worksheet
9-2 Volume of Prisms and Cylinders Find how many cubes the prism holds. Then give the prism’s volume. You can find the volume of this prism by counting how many cubes tall, long, and wide the prism is and then multiplying. 2 · 4 · 3 24 There are 24 cubes in the prism, so the volume is 24 cubic units.
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Le genou de Lucy. Odile Jacob. 1999. Coppens Y. Pré-textes. L’homme préhistorique en morceaux. Eds Odile Jacob. 2011. Costentin J., Delaveau P. Café, thé, chocolat, les bons effets sur le cerveau et pour le corps. Editions Odile Jacob. 2010. 3 Crawford M., Marsh D. The driving force : food in human evolution and the future.<|endoftext|>
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# What Is A Factor Of 108?
## What is the factor of 21?
21 is a composite number.
21 = 1 x 21 or 3 x 7.
Factors of 21: 1, 3, 7, 21.
Prime factorization: 21 = 3 x 7..
## IS 392 a perfect cube?
If not, find the smallest natural number by which 392 must be multiplied so that the product is a perfect cube. No, this is not a perfect cube. new number=392*7=2744.
## IS 256 a perfect cube?
After grouping the factors of equal triples, 2 x 2 is left. So, 256 is not a perfect-cube.
## What is the factor tree of 108?
The number 108 is a composite number because 108 can be divided by 1, by itself and at least by 2 and 3. So, it is possible to draw its prime tree. The prime factorization of 108 = 22•33.
## What are the factors of 111?
111 is a composite number. 111 = 1 x 111 or 3 x 37. Factors of 111: 1, 3, 37, 111. Prime Factorization: 111 = 3 x 37.
## What are the common factors of 72 and 108?
Example 1: What is the HCF of 72 and 108? 72 : 1, 2, 3, 4, 6, 9, 12, 18, 36, 72. 108 : 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108.
## What is a prime factor of 100?
Answer and Explanation: The prime factorization of 100 is 2 * 2 * 5 * 5. Using exponents, this number can also be written as 2^2 * 5^2.
## What is the highest common factor of 108?
Greatest common factor (GCF) of 108 and 123 is 3. We will now calculate the prime factors of 108 and 123, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 108 and 123.
## What is the LCM of 108?
Least Common Multiple – 108NumbersLCM108 and 9108108 and 10540108 and 111,188108 and 12108243 more rows
## What is the highest common factor of 108 and 24?
Finding the Greatest Common FactorTwo NumbersThinking …Greatest Common Factor24 and 1082 × 2 × 2 × 3 = 24, and 2 × 2 × 3 × 3 × 3 = 1082 × 2 × 3 = 12
## What are factors of 72?
1 Answer. The factors are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.
## IS 108 is a perfect cube?
The value of cube root of one is 108. The nearest previous perfect cube is 64 and the nearest next perfect cube is 125 . Cube root of 108 can be represented as 3√108. … The nearest previous perfect cube is 64 and the nearest next perfect cube is 125 .
## What is the prime factorization of 108?
prime factorization calculator of 108 Positive Integer factors of 108 = 2, 4, 3, 12, 36, 108 divided by 2, 2, 3, 3, 3, gives no remainder. They are integers and prime numbers of 108, they are also called composite number.
## What is the factor of 81?
Factors of 81: 1, 3, 9, 27, 81. Prime factorization: 81 = 3 x 3 x 3 x 3 which can also be written 3⁴.
## What is the exponent of 108?
If we put all of it together we have the factors 2 x 2 x 3 x 3 x 3 = 108. It can also be written in exponential form as 22 x 33.
## What are the multiples of 108?
Answer : 108,216,324,432,540,648,756,864,972,1080,1188,1296,1404,1512,1620,1728,1836,1944,2052,2160,2268,2376,2484,2592,2700,2808,2916,3024,3132,3240,3348,3456,3564,3672,3780,3888,3996,4104,4212,4320,4428,4536,4644,4752,4860,4968,5076,5184,5292, Related Links : What are the factors of 108?
## Is 108 prime or composite?
For 108, the answer is: No, 108 is not a prime number. The list of all positive divisors (i.e., the list of all integers that divide 108) is as follows: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108. For 108 to be a prime number, it would have been required that 108 has only two divisors, i.e., itself and 1.
## How do you factor 108?
Answer and Explanation: The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54 and 108. These are the factors of 108 because: 1 x 108 = 108. 2 x 54 = 108.
## What is the factor tree of 100?
Factor Tree Calculator The number 100 is a composite number because 100 can be divided by 1, by itself and at least by 2 and 5. So, it is possible to draw its prime tree. The prime factorization of 100 = 22•52.
## What is the prime factorization of 111?
111 = 1 x 111 or 3 x 37. Factors of 111: 1, 3, 37, 111. Prime Factorization: 111 = 3 x 37.
## IS 512 a perfect cube?
Since 512 is a whole number, it is a perfect cube. The nearest previous perfect cube is 343 and the nearest next perfect cube is 729 . 512 is said to be a perfect cube because 8 x 8 x 8 is equal to 512. Since 512 is a whole number, it is a perfect cube.
## What is the factor of 25?
Table of Factors and MultiplesFactorsMultiples1, 5, 2525501, 2, 13, 2626521, 3, 9, 2727541, 2, 4, 7, 14, 28285641 more rows
## What are the factors of 109?
109 and Level 2109 is a prime number.Prime factorization: 109 is prime.The exponent of prime number 109 is 1. Adding 1 to that exponent we get (1 + 1) = 2. … Factors of 109: 1, 109.Factor pairs: 109 = 1 x 109.109 has no square factors that allow its square root to be simplified. √109 ≈ 10.4403.
## IS 144 a perfect square?
A number is a perfect square (or asquare number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, thesquare root of 144 is 12. Therefore, thesquare root of 144 is an integer, and as a consequence 144 is a perfect square.
## What are the factors of 110?
Factors of 110: 1, 2, 5, 10, 11, 22, 55, 110. Prime factorization: 110 = 2 x 5 x 11.
## What is the perfect square of 108?
The square root of 108 is 10 and 8/21 or 10.381. We know that the square root of 108 falls between the two perfect squares of the square root of 100…<|endoftext|>
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# Using Percents Part 2 - PowerPoint PPT Presentation
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Using Percents Part 2. Rate of interest is the percent charged or earned. Simple Interest. Time that the money is borrowed or invested (in years). Principal is the amount of money borrowed or invested. What is interest?.
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Using Percents Part 2
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## Using PercentsPart 2
Rate of interestis the percent charged or earned
Simple Interest
Time that the money is borrowed or invested (in years)
Principal is the amount of money borrowed or invested
### What is interest?
When you borrow money from a bank, you pay interest for the use of the bank’s money. When you deposit money into a savings account, you are paid interest. Simple interest is one type of fee paid for the use of money.
I= Prt
### Calculate Simple Interest
To buy a car, Jessica borrowed \$15,000 for 3 years at an annual simple interest rate of 9%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay?
First, find the interest she will pay.
I = PrtUse the formula.
I = 15,000 0.09 3 Substitute. Use 0.09 for 9%.
• I = 4050Solve for I.
### Calculate Simple Interest…
Jessica will pay \$4050 in interest.
You can find the total amount A to be repaid on a loan by adding the principal P to the interest I.
P+ I = Aprincipal + interest = amount
15,000 + 4050 = ASubstitute.
Jessica will repay a total of \$19,050 on her loan.
### Calculate Simple Interest
• Nancy’s grandmother opened a savings account for her and put in \$1,250. The account pays 3% simple interest.
• How much interest will the account earn in 5 years?
• How much will be in the account after 5 years?
First, find the interest she will earn.
I = PrtUse the formula.
I = 1,250 0.03 5Substitute. Use 0.09 for 9%.
• I = \$187.50Solve for I.
### Calculate Simple Interest
Nancy will earn \$187.50 in interest.
You can find the total amount Ain her account by adding the principal P to the interest I.
P+ I = Aprincipal + interest = amount
1,250 + 187.50= ASubstitute.
Nancy will have a total of \$1,437.50 in her savings account.
### Practice
• A bank is offering 2.5% simple interest on a savings account. If you deposit \$5000, how much interest will you earn in one year?
\$125<|endoftext|>
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Young children act a lot like scientists when they learn through their senses. Sensory play includes any activity that stimulates young children’s senses: smell, touch, sight, hearing and taste. Sensory activities facilitate exploration and encourage children to use scientific processes while they play, create, and explore. Spending time stimulating their senses benefits children on multiple levels.
You will need: sorting tray (I purchased the above at a local Dollar Store); various pasta, beans, marshmallows, etc. to sort.
1. There are multiple ways to introduce this sorting activity to your child(ren). For my 4 year old, I set-up the tray with the pastas, beans, marshmallows, etc. all mixed together in the various slots of the tray.
2. I then explained that she was to sort the objects anyway she wanted to (by color, size, shape, texture, etc.).
3. Sort accordingly.
4. Discuss how your child(ren) sorted the objects. Discuss why they sorted the way they did.
5. Sort again with different criteria (instead of shape, sort by color).<|endoftext|>
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Effects of Pollution
Polluted storm water runoff can have many adverse effects on plants, fish, animals and people.
Sediment can cloud the water and make it difficult for or impossible for aquatic plants to grow. Sediment can also destroy aquatic habitats.
Excess nutrients can cause algae blooms. When algae die, they sink to the bottom and decompose in a process that removes oxygen from the water. Fish and other aquatic organisms can't exist water with low dissolved oxygen levels.
Bacteria and other pathogens can wash into swimming areas and create health hazards, often making beach closures necessary.
Debris - plastic bags, six-pack rings, bottles, and cigarette butts - washed into water bodies can choke, suffocate, or disable aquatic life like ducks, fish, turtles, and birds.
Household hazardous wastes like insecticides, pesticides, paint, solvents, used motor oil, and other auto fluids can poison aquatic life. Land animals and people can become sick or die from eating diseased fish and shellfish or ingesting polluted water.
Polluted storm water often affects drinking water sources. This, in turn, can affect human health and increase drinking water treatment costs.<|endoftext|>
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How is it that melting an asteroid can "reset" its atomic clock? The recent news article on a meteorite that originated on the moon was able to list the major impact events it experienced while on the moon, and to date them. It stated that this is possible because each event melted part of it, and thus the atomic clock of that part was reset. So how does melting a stone cause decayed radioactive elements to return to their original form? When we melt lead it doesn't revert to uranium. The same question applies to dating the solar system. Shouldn't the radioactive elements have been decaying for 8 billion, 12 billion years; ever since they were blasted into existence in ancient supernovae?
In the solid state, everything is pretty well locked into place, so that relatively few particles can escape. In the liquid or gaseous state, however, particles can move around and boil off. Because of radioactive decay, the material starts out with a larger amount of daughter isotope than is chemically favorable, so some of it will escape to bring the material to equilibrium. Think of air bubbles frozen in an ice cube--the air is prevented from escaping by the solid ice, but will escape when the cube melts because the energy of the system is lower without the bubbles. (Further separation could occur during solidification, due to different crystallizations of the parent and daughter isotopes.) This means that the information about how much decay took place prior to melting is lost.
Therefore we can only figure out the time since melting if we can determine the amount of daughter isotope present at solidification. One way to do this is to find another isotope which doesn't participate in the radioactive decay (call it isotope B, for boring), find a non-radioactive rock (or one that was recently liquid), and measure the ratio of the daughter isotope to isotope B. This ratio will be the same for any material in equilibrium, so it was the ratio in the sample when it solidified. You would then measure the amount of isotope B in your meteorite (or whatever), and multiply it by the equilibrium ratio, yielding the amount of daughter isotope at solidification. If you subtract this from the total amount of daughter isotope that you measure in the sample, you have the amount of daughter isotope due to radioactive decay since solidification, which tells you the time since solidification.
This page was last updated on June 27, 2015.<|endoftext|>
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# Divisibility Rules For numbers 10 – 19
###### Divisibility Rules For numbers 10 – 19
The rules for numbers greater than 10 especially numbers between 10 and 20 work in this way – Given a number y = 10T + U, where U is the unit digit number and T is the number formed by the rest digits.
A number y is divisible by N (where N is a number greater than 10) if (N-10)T – U is divisible by N. That is, find the difference between the number N and 10
Note: This is true of all numbers greater than 10 but is not efficient in some cases
The rule for 11 using the above:
A number y is divisible by 11 if and only if (11 – 10)T – U is divisible by 11
The difference between 11 and 10 is 1
Example 1: Test whether 132 is divisible by 11,
Multiply 13 by 1 and subtract 2 from the product. That is (1×13) – 2 = 13 -2 = 11, a number divisible by 11.
Example 2: Test whether 1023 is divisible by 11,
Multiply 102 by 1 and subtract 3 from the product. That is (1×102) – 3 = 102 – 3 = 99 a number divisible by 11.
The rule for 12
Old: A number is divisible by 12 if and only if the number is divisible by 3 and 4.
Using the above: A number y is divisible by 12 if and only if (12 -10)T – U (that is 2T – U ) is divisible by 12
The difference between 12 and 10 is 2
Example 1: Test whether 156 is divisible by 12,
Multiply 15 by 2 and subtract 6 from the product. That is (2 x 15) – 6 = 30 – 6 = 24, gives a number divisible by 12.
Example 2: Test whether 576 is divisible by 12,
Multiply 57 by 2 and subtract 6 from the product. That is (2 x 57) – 6 = 114 – 6 = 108, a number divisible by 12.
The rule for 13
Using the above: A number y is divisible by 13 if and only if (13 -10)T – U (that is 3T – U ) is divisible by 13.
The difference between 13 and 10 is 3
Example 1: Test whether 169 is divisible by 13,
Multiply 16 by 3 and subtract 9 from the product. That is (3 x 16) – 9 = 48 – 9 = 39, a number divisible by 13.
Example 2: Test whether 312 is divisible by 13,
Multiply 31 by 3 and subtract 2 from the product. That is (3 x 31 ) – 2 = 93 – 2 = 91, a number divisible by 13
###### Rules for 14
(a) A number is divisible by 14 if it is divisible by 2 and 7
(b) A number y is divisible by 14 if and only if (14 -10)T – U (that is 4T – U ) is divisible by 14.
(c) A number is divisible by 14 if 6T + 2U is divisible by 14
Using rule (b)
The difference between 14 and 10 is 4
Example 1: Test whether 98 is divisible by 14,
Multiply 9 by 4 and subtract 8 from the product. That is (4 x 9) – 8 = 36 – 8 = 28, a number divisible by 14.
Using rule (c)
Example : Test whether 182 is divisible by 14,
182: (6 x 18 ) +( 2 x 2) = 108 + 4 = 112,
112: (6 x 11) + 2 x 2 = 70 = 14 x 5, a number divisible by 14.
The rule for 15
(a): A number is divisible by 15 if and only if the number is divisible by 3 and 5.
(b) : A number y is divisible by 15 if and only if (15 -10)T – U (that is 5T – U ) is divisible by 15.
(c) : A number y is divisible by 15 if and only if 5T + 2U ) is divisible by 15.
Using rule (b)
The difference between 15 and 10 is 5
Example 1: Test whether 135 is divisible by 15,
135: (5 x 13) – 5 = 65 – 5 = 60 = 15 x 4, a number divisible by 15.
Using rule (c)
Example : Test whether 225 is divisible by 15,
225: is (5 x 22 ) + (2 x 5) = 110 + 10 = 120 = 15 x 8, a number divisible by 15.
The rule for 16
(a) A number y is divisible by 16 if and only if (16 -10)T – U (that is 6T – U ) is divisible by 16.
(b) A number is divisible by 16 if 4T + 2U is divisible by
The difference between 16 and 10 is 6
Example 1: Test whether 128 is divisible by 16,
Multiply 12 by 6 and subtract 8 from the product. That is (12 x 6) – 8 = 72 – 8 = 64, a number divisible by 16.
Example 2: Test whether 256 is divisible by 16,
Multiply 25 by 6 and subtract 6 from the product. That is (6 x 25 ) – 6 = 150 – 6 = 144, a number divisible by 16.
The rule for 17
(a): A number y is divisible by 17 if and only if (17 -10)T – U (that is 7T – U ) is divisible by 17.
(b) A number is divisible by 17 if 3T + 2U is divisible by 17
Using rule (a)
The difference between 17 and 10 is 7
Example : Test whether 68 is divisible by 17,
68: (7 x 6) – 8 = 42 – 8 = 34 = 17 x 2, a number divisible by 17.
Using rule (b)
Example 2: Test whether 204 is divisible by 17,
204: (7 x 20) – 4 = 140 – 4 = 136, a number divisible by 17.
If one is not sure that 136 is divisible by 17, this number can be tested recursively. That is, (7 x 13) – 6 = 91 – 6 = 85= 17 x 5, a number divisible by 17.
The rule for 18
(a): A number is divisible by 18 if and only if the number is even and divisible by 9.
(b): A number y is divisible by 18 if and only if (18 -10)T – U (that is 8T – U ) is divisible by 18.
(c): A number is divisible by 18 if 2T + 2U (that is, 2(T + U) is divisible by 18
Using rule (b)
The difference between 18 and 10 is 8
Example : Test whether 72 is divisible by 18,
72: (8 x 7) – 2 = 56 – 2 = 54 = 18 x 3, a number divisible by 18.
Using rule (c) that is testing for 2 and (T + U)
Example : Test whether 126 is divisible by 18,
Since 126 is even and 126: 12 + 6 = 18, 126 is divisible by 18.
The rule for 19
Using the above: A number y is divisible by 19 if and only if (19 -10)T – U (that is 9T – U ) is divisible by 19.
The difference between 19 and 10 is 9
Example 1: Test whether 76 is divisible by 19,
Multiply 7 by 9 and subtract 6 from the product. That is (9 x 7) – 6 = 63 – 6 = 57, a number divisible by 19.
Example 2: Test whether 209 is divisible by 19,
Multiply 20 by 9 and subtract 9 from the product. That is (9 x 20) – 9 = 180 – 9 = 171, a number divisible by 19.
If one is not sure that 171 is divisible by 19, this number can be tested recursively. That is, (9 x 17) – 1 = 153 – 1 = 152, which is a number divisible by 19.
A simpler Rule for 19:
A number y is divisible by 19 if and only if T + 2U is divisible by 19.
Example 1: Test whether 209 is divisible by 19,
Using T + 2U, 20 + (2 x 9) = 38, a number divisible by 19.
Example 2: Test whether 437 is divisible by 19,
Using T + 2U, 43 + (2 x 7) = 43 + 14 = 57, a number divisible by 19.
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The Iban trace their origins to the Kapuas Lake region of Kalimantan. With a growing population creating pressures on limited amounts of productive land, the Iban fought members of other tribes aggressively, practicing headhunting and slavery. Enslavement of captives contributed to the necessity to move into new areas. By the middle of the nineteenth century, they were well established in the First and Second Divisions, and a few had pioneered the vast Rejang River valley. Reacting to the establishment of the Brooke Raj in Sarawak in 1841, thousands of Iban migrated to the middle and upper regions of the Rejang, and by the last quarter of the century had entered all remaining divisions. The most dramatic changes in the past three decades have been abandonment of longhouses and permanent settlement in Sarawak's towns and cities. Iban have lived near other ethnic groups with whom they have interacted. The most important of these societies have been the Malays, Chinese, Kayan and, during the Brooke Raj and the period of British colonialism, Europeans. The dynamic relations between Iban and these societies have produced profound changes in Iban society and culture.<|endoftext|>
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NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4
NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 1 Integers Ex 1.4.
Ex 1.4 Class 7 Maths Question 1.
Evaluate each of the following:
(a) (-30) ÷ 10
(b) 50 ÷ (-5)
(c) (-36) ÷ (-9)
(d) (-49) ÷ (49)
(e) 13 ÷ [(-2) + 1]
(f) 0 ÷ (-12)
(g) (-31) ÷ [(-30) + (-1)]
(h) [(-36) ÷ 12] ÷ 3
(i) [(-6) + 5] ÷ [(-2) + 1]
Solution:
(a) (-30) ÷ 10 = 30/10 = -3
(b) 50 ÷ (-5) = 50/5 = -10
(c) (-36) ÷ (-9) = 36/9 = 4
(d) (-49) ÷ (49) = 49/49 = -1
(e) 13 ÷ [(-2) + 1] = 13 ÷ -1 = 13/1 = -13
(f) 0 ÷ (-12) = 0/12 = 0
(g) (-31) ÷ [(-30) + (-1)] = (-31) ÷ (-31) = 31/31 = 1
(h) [(-36) ÷ 12] ÷ 3 = [−36/12] ÷ 3 = -3 ÷ 3 = 3/3 = -1
(i) [(-6) + 5] ÷ [(-2) + 1] = (-1) ÷ (-1) = 1/1 = 1
Ex 1.4 Class 7 Maths Question 2.
Verify that a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c) for each of the following values of a, b and c.
(а) a = 12, b = – 4, c = 2
(b) a = (-10), b = 1, c = 1
Solution:
(a) a = 12, 6 = – 4, c = 2
a ÷ (b + c) = 12 ÷ [(-4) + 2]
= 12 ÷ (-2) = 12/2 = -6
(a ÷ b) + (a ÷ c) = [12 ÷ (-4)] + [12 ÷ 2]
= 12/4 + 12/2 = −3 + 6 = 3
Since, (-6) ≠ 3
Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
(b) a = (-10), b = 1, c = 1
a ÷ (b + c) = (-10) ÷ (1 + 1)
=(-10) ÷ 2 = 10/2 = -5
(a ÷ b) + (a ÷ c)
=[(-10) ÷ 1] + [(-10) ÷ 1]
=(−10)/1 + (−10)/1
= (-10) + (-10) = -20
Since (-5) ≠ (-20)
Hence, a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)
Ex 1.4 Class 7 Maths Question 3.
Fill in the blanks:
(a) 369 ÷ _____ = 369
(b) (-75) ÷ _____ = -1
(c) (-206) ÷ _____ = 1
(d) -87 ÷ _____ = 87
(e) _____ ÷ 1 = -87
(f) _____ ÷ 48 = -1
(g) 20 ÷ _____ = -2
(h) _____ ÷ (4) = -3
Solution:
(a) 369 ÷ _____ = 369 Hence, 369 ÷ 1 = 369
(b) (-75) ÷ _____ = -1 Hence, (-75) ÷ 75 = -1
(c) (-206) ÷ _____ = 1 Hence, (-206) ÷ (-206) = 1
(d) -87 ÷ _____ = 87 Hence, -87 ÷ (-1) = 87
(e) _____ ÷ 1 = -87 Hence, -87 ÷ 1 = -87
(f) _____ ÷ 48 = -1 Hence, (-48) ÷ 48 = -1
(g) 20 + _____ = -2 Hence, 20 ÷ (-10) = -2
(h) _____ + (4) = -3 Hence, (-12) ÷ (4) = -3
Ex 1.4 Class 7 Maths Question 4.
Write five pairs of integers (a, b) such that a ÷ b = -3. One such pair is (6, -2) because 6 ÷ (-2) = -3.
Solution:
(i) (9, -3) because 9 ÷ (-3) = -3
(ii) (-12, 4) because (-12) ÷ 4 = -3
(iii) (15, -5) because 15 ÷ (-5) = -3
(iv) (21, -7) because 21 ÷ (-7) = -3
(v) (30, -10) because 30 ÷ (-10) = -3
Ex 1.4 Class 7 Maths Question 5.
The temperature at 12 noon was 10°C above zero. If it decreases at the rate of 2°C per hour until midnight, at what time would the temperature be 8°C below zero? What would be the temperature at midnight?
Solution:
The temperature at 12 noon was 10°C above zero, i.e., +10°C
Rate of decrease in temperature per hour = 2°C
Number of hours from 12 noon to midnight = 12
Change in temperature in 12 hours
= 12 × (-2°C) = -24°C
Temperature at midnight = +10°C + (-24°C) = -14°C
Hence, the temperature at midnight = -14°C
Difference in temperature between +10°C and -8°C
= +10°C – (-8°C) = +10°C + 8°C = 18°C
Number of hours required = 18°C
/2°C = 9 hours
Time after 9 hours from 12 noon = 9 pm.
Ex 1.4 Class 7 Maths Question 6.
In a class test (+3) marks are given for every correct answer and (-2) marks are given for every incorrect answer and no marks for not attempting any question:
(i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly?
(ii) Mohini scores -5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
Solution:
It is given that +3 marks are given for each correct answer and (-2) marks are given for each incorrect answer. Zero marks are given for not attempting questions.
(i) Marks obtained by Radhika for 12 correct answers = (+3) × 12 = 36
Total marks obtained by Radhika = 20
Marks obtained by Radhika for incorrect answers = 20 – 36 = -16
Number of incorrect answers = (−16) ÷ (−2) = (−16)/(−2) = 8
Hence, the required number of incorrect answers is 8.
(ii) Marks scored by Mohini = -5
Number of correct answers by Mohini = 7
Marks obtained by Mohini for 7 correct answers = 7 × (+3) = 21
Marks obtained for incorrect answers = -5 – 21 = (-26)
Number of incorrect answers = (-26) ÷ (-2) = 13
Hence, the required number of incorrect answers is 13.
Ex 1.4 Class 7 Maths Question 7.
An elevator descends into a nine shaft at the rate of 6 m/min. If the descent starts from 10 m above the ground level, how long will it take to reach -350 m.
Solution:
The current position of the elevator is at 10 m above the ground level.
Distance moved by the elevator below the ground level = 350 m
Total distance moved by the elevator = 350 m + 10 m = 360 m
Rate of descent = 6 m/min.
Total time taken by the elevator = 360/6 min
= 60 minutes = 1 hour
Hence, the required time is 1 hour.
You can also like these:
NCERT Solutions for Maths Class 8
NCERT Solutions for Maths Class 9
NCERT Solutions for Maths Class 10
NCERT Solutions for Maths Class 11
NCERT Solutions for Maths Class 12
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Up: The Shape of Space Curriculum Materials: Symmetry and Tiling
# Symmetry and 3-Dimensional Space
## Objective
Determine different symmetrical tilings of space using the cube as the fundamental domain.
## Materials
• 3-D Tiling Puzzle
• Clear Tape Or Glue
• Scissors
• 3/4" Diameter Color Coded Circular Labels
## Introduction
From previous activities with 2-dimensional surfaces you learned that gluing opposite edges together of a square in various combinations created surfaces which could be possible universes for a Flatlander. In addition, each gluing creates a symmetrical tiling that demonstrates what one would see as an observer on that surface. In other words, the gluing, the symmetry, and the surface all determine one another. This is also true of 3-dimensional space.
Euclidean 3-space is space that is unbounded and infinite. Can you think of a space that is unbounded and finite? Suppose our universe is a closed space in the same way a Flatland's universe is a closed surface. We can form a closed 3-dimensional universe by gluing opposite faces of a cube together in various combinations. This creates a symmetrical tiling of our motif in the images of it that would be seen from an intrinsic observer's point of view.
## Activity 1
Using an overhead transparency with the drawing from the 3-D Tiling Puzzle burned into it, assemble a cube with the figure suspended in the cube's interior. Placing color coded circular labels on the six faces of the cube enhance identifying the front, back, top, bottom, left, and right faces. Repeat the process to assemble eight cubes total. The cube will be the shape we will use as a fundamental domain for the tiling of 3-dimensional Euclidean space and the space ship will be the motif in space.
## Activity 2
Create the following closed 3-dimensional universe and describe what an observer would see concerning the motif and its moves. Use the eight fundamental domains created in Activity 1 to help you visualize. Create different 3-D tilings with each the following rules:
1. Attach the front faces to back faces.
2. Attach the left faces to right faces.
3. Attach the top faces to bottom faces.
4. Do both 1 and 2.
5. Do both 1 and 3.
6. Do both 2 and 3.
7. Do all of 1, 2, and 3. (This is called a 3-torus closed universe.)
(To save time, just do 1, 4, and 7 above.)
## Activity 3
Create the following closed 3-dimensional universe by adding a "half twist" before gluing as shown in the diagram below. Describe what the an observer would see concerning the motif and its images.
Take the cube's top and bottom, rotate one of them a quarter turn, rotate it another a quarter turn, and glue.
1. Glue front faces to back faces with a half twist.
2. Glue left faces to right faces with a half twist.
3. Glue top faces to bottom faces with a half twist.
4. Do both 1 and 2.
5. Do both 1 and 3.
6. Do both 2 and 3.
7. Do both 1, 2, and 3.
(To save time, just do 1, 4, and 7 above.)
## Activity 4
Create the following closed 3-dimensional universe by adding a quarter turn clockwise, from the point of view of the figurine, before gluing as shown in the diagram below. Describe what an observer in this universe would see concerning the motif and its image. Which do not tile space symmetrically?
Take the cube's top and bottom, rotate one of them a quarter turn, and glue.
1. Glue front faces to back face with a quarter turn.
2. Glue left faces to right face with a quarter turn.
3. Glue top faces to bottom face with a quarter turn.
4. Combine 1 and 2.
5. Combine 1 and 3.
6. Combine 2 and 3.
7. Combine 1, 2, and 3.
(To save time, just do 1, 4, and 7 above.)
## Questions
1. Does gluing two opposite faces with a three quarter turn clockwise, from the point of view of the figurine, create a closed 3-dimensional universe that is the same as one created from a quarter turn?
2. What are other ways of gluing opposite faces together?
## Extensions
As a classroom activity, choose any combination of gluing faces and place the cubes from the entire class to tile space in a single stack. Each student takes a turn placing a cube in the stack with the correct orientation. Each student must verify whether the placing of the cube is consistent with the tiling in all directions. What other regular polyhedra will tessellate 3-dimensional space?
Up: The Shape of Space Curriculum Materials: Symmetry and Tiling<|endoftext|>
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Introduction Recall that the imaginary unit i is equal to. A fraction with i in the denominator does not have a rational denominator, since is not a rational.
Presentation on theme: "Introduction Recall that the imaginary unit i is equal to. A fraction with i in the denominator does not have a rational denominator, since is not a rational."— Presentation transcript:
Introduction Recall that the imaginary unit i is equal to. A fraction with i in the denominator does not have a rational denominator, since is not a rational number. Similar to rationalizing a fraction with an irrational square root in the denominator, fractions with i in the denominator can also have the denominator rationalized. 1 4.3.4: Dividing Complex Numbers
Key Concepts Any powers of i should be simplified before dividing complex numbers. After simplifying any powers of i, rewrite the division of two complex numbers in the form a + bi as a fraction. To divide two complex numbers of the form a + bi and c + di, where a, b, c and d are real numbers, rewrite the quotient as a fraction. 2 4.3.4: Dividing Complex Numbers
Key Concepts, continued Rationalize the denominator of a complex fraction by using multiplication to remove the imaginary unit i from the denominator. The product of a complex number and its conjugate is a real number, which does not contain i. Multiply both the numerator and denominator of the fraction by the complex number in the denominator. Simplify the rationalized fraction to find the result of the division. 3 4.3.4: Dividing Complex Numbers
Key Concepts, continued In the following equation, let a, b, c, and d be real numbers. 4 4.3.4: Dividing Complex Numbers
Common Errors/Misconceptions multiplying only the denominator by the complex conjugate incorrectly determining the complex conjugate of the denominator 5 4.3.4: Dividing Complex Numbers
Guided Practice Example 2 Find the result of (10 + 6i ) ÷ (2 – i ). 6 4.3.4: Dividing Complex Numbers
Guided Practice: Example 2, continued 1.Rewrite the expression as a fraction. 7 4.3.4: Dividing Complex Numbers
Guided Practice: Example 2, continued 2.Find the complex conjugate of the denominator. The complex conjugate of a – bi is a + bi, so the complex conjugate of 2 – i is 2 + i. 8 4.3.4: Dividing Complex Numbers
Guided Practice: Example 2, continued 3.Rationalize the fraction by multiplying both the numerator and denominator by the complex conjugate of the denominator. 9 4.3.4: Dividing Complex Numbers
Guided Practice: Example 2, continued 4.If possible, simplify the fraction. The answer can be left as a fraction, or simplified by dividing both terms in the numerator by the quantity in the denominator. 10 4.3.4: Dividing Complex Numbers
Guided Practice: Example 2, continued 11 4.3.4: Dividing Complex Numbers
Guided Practice Example 3 Find the result of (4 – 4i) ÷ (3 – 4i 3 ). 12 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 1.Simplify any powers of i. i 3 = –i 13 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 2.Simplify any expressions containing a power of i. 3 – 4i 3 = 3 – 4(–i) = 3 + 4i 14 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 3.Rewrite the expression as a fraction, using the simplified expression. Both numbers should be in the form a + bi. 15 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 4.Find the complex conjugate of the denominator. The complex conjugate of a + bi is a – bi, so the complex conjugate of 3 + 4i is 3 – 4i. 16 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 5.Rationalize the fraction by multiplying both the numerator and denominator by the complex conjugate of the denominator. 17 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 6.If possible, simplify the fraction. The answer can be left as a fraction, or simplified by dividing both terms in the numerator by the quantity in the denominator. 18 4.3.4: Dividing Complex Numbers
Guided Practice: Example 3, continued 19 4.3.4: Dividing Complex Numbers
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# Converting Between Standard Form and Vertex Form
Hello! Welcome to this video on converting between standard form and vertex form of a quadratic equation. Before we start converting, let’s review what these forms look like.
Standard form of a quadratic equation is:
$$y=ax^{2}+bx+c$$
Where $$a$$, $$b$$, and $$c$$ are real numbers, and $$a\neq 0$$.
Vertex form of a quadratic equation is:
$$y=a(x-h)^{2}+k$$
Where $$a$$, $$h$$, and $$k$$ are real numbers, $$a\neq 0$$, and $$(h,k)$$ is the vertex of the parabola.
Now that we’ve reviewed these forms, let’s start by converting a standard form equation to vertex form.
Convert the standard form equation $$y=x^{2}+14x-9$$ to vertex form.
In order to do this, we need to complete the square. To complete the square, $$a$$ must equal 1. In this equation, $$a$$ is already 1, so we can move on to the next step.
Find the $$b$$-value, divide it by 2, and square it.
$$(\frac{b}{2})^{2}=(\frac{14}{2})^{2}=(7)^{2}=49$$
Now, add and subtract this value after the $$x$$-term.
$$y=x^{2}+14x+49-49-9$$
I’m going to add parentheses around these first three terms because this will be our perfect square trinomial.
$$y=(x^{2}+14x+49)-49-9$$
This trinomial factors to the perfect square $$(x+7)^{2}$$.
$$y=(x+7)^{2}-49-9$$
Remember, the point of completing the square is to get it to factor into something that looks like this. So now, all we have to do is combine our like terms right here.
$$y=(x+7)^{2}-58$$
Now, notice that we went through this completing the square process fairly quickly. If you want some more help on this, see one of our other videos where we go more in depth over how to do this process.
Let’s try another problem where we convert a standard form equation to vertex form.
$$y=-2x^{2}-4x+8$$
Remember, in order to complete the square, the value of a must be 1. Factor a –2 out of the right side of the equation.
$$y=-2(x^{2}+2x-4)$$
Remember, we factor a –2 out of each of the terms. Now, identify the $$b$$-value, halve it, and square it. So, we want the $$b$$-value of this trinomial where a is 1 $$(x^{2}+2x-4)$$, so our $$b$$-value is 2.
$$(\frac{b}{2})^{2}=(\frac{2}{2})^{2}=(1)^{2}=1$$
Add and subtract this value after the $$x$$-term.
$$y=-2(x^{2}+2x+1-1-4)$$
Put parentheses around the first three terms to identify the perfect square trinomial.
$$y=-2((x^{2}+2x+1)-1-4)$$
Factor the perfect square trinomial.
$$y=-2((x+1)^{2}-1-4)$$
Combine like terms inside the parentheses.
$$y=-2((x+1)^{2}-5)$$
Now, the final step in simplifying this just a little bit more, is distributing the –2 inside this larger set of parentheses, so to the $$(x+1)^{2}$$ and to the –5.
$$y=-2(x+1)^{2}+10$$
Now let’s move on to converting a vertex form equation to standard form. This conversion process is much simpler.
Convert the equation $$y=3(x-4)^{2}+7$$ to standard form.
The first thing to do is expand the squared term.
$$y=3(x-4)(x-4)+7$$
And then we can FOIL to simplify this part.
$$y=3(x^{2}-4x-4x+16)+7$$
Now, we can combine like terms.
$$y=3(x^{2}-8x+16)+7$$
Now, distribute 3 to the trinomial.
$$y=3x^{2}-24x+48+7$$
Finally, combine like terms.
$$y=3x^{2}-24x+55$$
And there’s our equation in standard form.
Let’s try one more example before we go.
Convert the equation $$y=-11(x+8)2-9$$ to standard form.
Start by expanding the squared binomial.
$$y=-11(x+8)(x+8)-9$$
And again, we can FOIL these two terms.
$$y=-11(x^{2}+8x+8x+64)-9$$
We can simplify by combining like terms.
$$y=-11(x^{2}+16x+64)-9$$
Then, distribute –11 to the trinomial.
$$y=-11x^{2}-176x-704-9$$
Finally, combine like terms.
$$y=-11x^{2}-176x-713$$
And there you have it!
I hope this video on converting between standard form and vertex form of a quadratic equation was helpful. Thanks for watching, and happy studying!
Return to Algebra I Videos
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Posts Tagged: "word problems"
Two-Step Equations Containing Integers
- -
Oftentimes, when we are confronted with a problem, we translate it first into an equation then use it to solve for the missing quantity. These equations may involve one, two or more steps before arriving at its solutions. Equations such as 2x + 3 = 5 and 3x - 5 = 7, require two–step solutions. These equations are called two–step equations.
These are equations that require two steps or two operations before arriving at its solutions. In solving two–step equations, there is no definite rule “which” operation to undo first. However, one can apply the rule for order of operations or whichever will result to a simpler equation. Again, the purpose in solving two-step equation is to combine the constants in one side of the equation and see to it that the variable must only have 1 as its numerical coefficient.
Before illustrating the steps on how to solve two-step equations, listed below are examples of two–step equations containing integers.
1. $2x-7=5$
2. $2n+ \left(-3\right ) =-1$
3. $5a-1=4$
4. $2a-\left(-2\right)=16$
5. $\displaystyle \frac{a}{3}+2=\left(-5\right)$
Solving Two-Step Equations Containing Integers
To solve two–step equations, do the inverse of the operations involved one at a time. Since, we are dealing with equations, it must be noted that whatever operations are done on one side of the equation must as well be done on the other side. Also, we apply the rules for performing operations with integers. To illustrate, consider the following examples:
Solve for the unknown variable:
1. $2x-7=5$
Solution:
If we divide both sides by 2, we are going to have fractions. It is more preferable to undo subtraction first. Since twice the variable $x$ is subtracted by 7 and the opposite of addition is subtraction, then we add 7 to both sides of the equation. Then divide the resulting equation by 2.
2. $2x+(-3)=-1$
Solution:
Again, division of 2 in the equation will lead to fractions. We undo the addition first. Add the opposite of , which is 3 to both sides of the equation then divide both sides by 2.
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &2x+\left ( -3 \right )&=&\left ( -1 \right ) \\ &2n+\left ( -3 \right )+3&=&\left ( -1 \right )+3 \\&2n&=&2 \\&\displaystyle \frac{2n}{2}&=&\displaystyle \frac{2}{2} \\ &\therefore n&=&1 \\ \end{array}$
3. $5a-1=4$
Solution:
Since the variable term $5a$ is subtracted by 1, then we add both sides by 1 then divide both sides both sides by 5.
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &5a-1&=&4 \\ &5a-1+1&=&4+1 \\&5a&=&5 \\&\displaystyle \frac{5a}{5}&=&\displaystyle \frac{5}{5} \\ &\therefore a&=&1 \\ \end{array}$
4. $\displaystyle \frac{a}{3}+2=\left(-5\right)$
Solution:
Subtract 2 from both sides of the equation then multiply both sides by the inverse of $\frac {1}{3}$, which is 3.
$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &\displaystyle \frac{a}{3}+2&=&\left ( -5 \right ) \\ &\displaystyle \frac{a}{3}+2-2&=&\left ( -5 \right )-2 \\&\displaystyle \frac{a}{3}&=&-7 \\&\left ( 3 \right )\displaystyle \frac{a}{3}&=&-7\left ( 3 \right ) \\ &\therefore a&=&-21 \\ \end{array}$
Practice Exercises:
Solve for the unknown variable.
1. $2x-4x=-3$
2. $3w+\left(-7\right)=-19$
3. $-3y-3=12$
4. $\displaystyle \frac{z}{-3}+2=9$
5. $3q+3=-36$
One-Step Equation Word Problems
- -
One–step equation word problem is a simple word problem involving only addition or subtraction of a constant to an unknown value and either multiplying or dividing a certain number to an unknown quantity.
Similar Figures Word Problems
- - Pre-Algebra
Similar figures and scale drawings are very useful topics in mathematics and in many other subjects. In social sciences, for instance, students are provided with maps of diverse countries and cities all over the world. The students must be able to interpret the maps in order to comprehend them. Proportions provide students an idea of what a scale is. Scales are obtained from a scale factor. A figure may be tightened in any direction by a certain scale factor, which then makes the resulting figure similar to the original.
Proportion Word Problems
- - Pre-Algebra
After learning proportions, let us now try to solve word problems involving proportions. To do so, we form the proportion between the ratios given in the problem then solve for the missing value by applying the property between its means and extremes.<|endoftext|>
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A warm-up is a short activity at the start of a lesson to jump-start students' thinking. Warm-ups have many purposes including:
- to introduce a new topic
- to reinforce skills
- to motivate students
- to prepare students to learn
- to assess how much students know
Warm-up time is also a great time to promote higher order thinking. For example, students who have mastered the math facts may be bored with the following warm-up:
Fact masters need the "benefits of a proper warm-up" such as,
Needless to say, the second warm-up:
- is more challenging.
- reinforces relationships between facts.
- requires higher-order thinking!<|endoftext|>
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## Bullets Per Minute
This was a two part question, the first part I was able to calculate.
Question Part 1:
A machine gun fires a stream of bullets into a block that is free to move on a horizontal frictionless tabletop. Each bullet has mass 66 grams; their speed is 930 m/sec, and the block a mass of 7.36 kg. After 15 bullets, the speed of the block is?
Calculation Part 1:
$$\frac{(0.066 kg * 15 bullets)}{(0.066 kg * 15 bullets) + 7.36 kg}=110.263 m/sec$$
Question Part 2:
If a man in the previous statement can exert an average force of 180 N against the gun, determine the maximum number of bullets he can fire per minute.
Attempt to Calculate:
I am unsure of how I would go about this. I attempted to think of it as a reverse of the previous question, but what didn't work out was getting bullets per minute without figuring out a basic equation then using sample bullets/minute numbers to see where the breaking point of 0 m/sec is. Also, I kept in mind the need to change the velocity of m/sec to m/min if needed.
The equation I attempted to use was:
$$V_{gun+man}=\frac{m_{bullet}*numberofbullets}{(m_{bullet}*numberofbulle ts)+M_{man}}v_{bullet}$$
Any help with where to start?
it is basically just a simple problem of conservation of momentum.
You have the "force equation": F*t=m*v the total momentum during a minute is m*v*k, where k is the number of bullets per minute.
## Bullets Per Minute
Quote by Kurret You have the "force equation": F*t=m*v the total momentum during a minute is m*v*k, where k is the number of bullets per minute.
So I get...
$$F \times t=m \times v$$
$$F \times t=m \times v \times k$$
$$\frac{F \times t}{m \times v}=k; k=bullets/min$$
$$\frac{180 N \times 60 sec}{0.066 kg \times 930 m/sec}=175.953 bullets/sec$$
Thank you both.
Quote by blue5t1053 So I get... $$F \times t=m \times v$$ $$F \times t=m \times v \times k$$ $$\frac{F \times t}{m \times v}=k; k=bullets/min$$ $$\frac{180 N \times 60 sec}{0.066 kg \times 930 m/sec}=175.953 bullets/sec$$ Thank you both.
.953????
Never knew Bullets went to the target in fractions.!!!!!
ever heard of splinters/shotguns/shells
Recognitions: Homework Help Science Advisor Part 1 is correct, although your expression is not correct. You need the muzzle velocity in there: $$m_{bullet}v_{muzzle} N_{bullets} = \Delta P = (M_{block} + m_{bullet}N_{bullets}) v_{block/bullets}$$ The analysis for Part 2 is not clear.Since the force applied by the gun to the shooter is the time rate of change of momentum of the gun/bullet system F = dp/dt: $$F = \frac{dm}{dt}v_{muzzle}$$ $$\frac{dm}{dt} = F/v = 180/930 = .194 \text{kg/sec}$$ Therefore, the number of bullets per second is .194/.066 = 2.94 or 176 bullets/minute, but this is only if you want to provide an average force of 180 N. The actual peak force that must be applied will be greater than this, so actually firing this number of bullets will cause the force on the shooter to exceed 180N. AM<|endoftext|>
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DESCRIPTIVE STATISTICS
# DESCRIPTIVE STATISTICS
Télécharger la présentation
## DESCRIPTIVE STATISTICS
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##### Presentation Transcript
1. DESCRIPTIVE STATISTICS UNIT 4: Measures of Position TouchText • Z-Scores • Percentile Rankings Problems and Exercises Next
2. Position of a Single Observation Within a Distribution If a student were told that he/she had received an 83 on an exam, the questions that student would most likely (and rightly) ask are: Dictionary • 83 out of how many possible? • Is an 83 good or bad? • What was the highest score in the class? • What was the lowest score in the class? • What was the average score in the class? • Etc. These are questions of position – of one observation (in this case, a single test score) relative to the rest of the population or sample. Take Notes Back Next
3. Z-Score: Position of a Single Observation Standardized, Relative to the Mean and Standard Deviation One way of describing the position of a single variable relative to a distribution is to report its distance from the mean. To make this independent of units or scale, we measure this distance in units of standard deviation. Dictionary An observation’s Z-score measures how many standard deviations it is away from the mean: Take Notes Back Next
4. Calculating the Z-Score To calculate the Z-score, it is first necessary to calculate the distribution’s mean and standard deviation; then, calculate a specific observation’s Z-score as above. Dictionary *Because the deviations sum to zero, so too will the z-scores. Take Notes Back Next
5. Observations Measured in Standard Deviations Away From the Mean In the following exercises, calculate how many standard deviations away from the mean is each observation. Dictionary Example: Take Notes Back Next
6. Observations Measured in Standard Deviations Away From the Mean In the following exercises, calculate how many standard deviations away from the mean is each observation. Dictionary Solutions: Zi = (30 – 40)/10 = -10/10 = -1 s below the mean. Zi = (100 – 100)/6 = 0/6 = 0 s exactly at the mean. Zi = (82.5 – 80)/5 = 2.5/5 = 0.5 s above the mean. Zi = (295 – 300)/25 = -5/25 = -0.2 s below the mean. Zi – (760 – 1000)/120 = -240/120 = -2 s below the mean. Take Notes Back
7. Z-Score Within a Distribution How good or bad was the score of 83 in this distribution? Dictionary Score = 83 high score = 94 (2 students) low score = 56 mean = 75.3 Take Notes Back Next
8. Z-score: Example In the previous example, the Z-score for 83 is: Dictionary Score = 83 high score = 94 (2 students) low score = 56 mean = 75.3 Z83 = 1.033 So, we can see that a positive Z-score of 1.033 is above average. What, if anything, more precisely can we say from the Z-score alone? Take Notes Back Next
9. What Does and Doesn’t the Z-score Say? It is generally not possible to translate an observation’s Z-score into a specific ranking or position in an ordered data set unless we know the exact distribution of that data. However, we can make the following statements: Dictionary • Z-score = 0: Exactly at the mean (average) • Z-score > 0: Above the mean (average). Also, the greater the Z-score, the greater (larger, higher, etc.) and rarer is that score. • Z-score < 0: Below the mean (average). Also, the more negative the Z-score, the lesser (smaller, lower, etc.) and rarer is that score. * In a subsequent unit, we introduce the standard normal distribution, in which Z-scores can be translated exactly into a percentile ranking. Take Notes Back Next
10. Distributions of Z-Scores Not only can a single observation be translated into a Z-score, but the entire data set can be re-characterized in terms of Z-scores. Dictionary The following example (task icon, below) was introduced in the previous unit. It is a sample of tourist expenditures of passengers leaving Bangkok Airport. On completion of the task(s), one will be able to say, in this example, how many of the observations are within one, two or three standard deviations of the mean (Z-score < 1, Z-score < 2, Z-score < 3). The ability to do this provides more information about the relative ranking and rarity of the observation. Take Notes Back Next
11. Example: Calculating Observations as Z-Scores Example: Expenditures of passengers leaving Bangkok Airport On your previous MS Excel table (linked below), you should already have a column which calculates deviations (Xi – X). From this, create another column which calculates deviations per unit of standard deviation ((Xi – X)/s). Use the Data Analysis > Histogram feature to create a frequency distribution. Use bins of 0.5 between -3 and +3 (i.e. -3.0, -2.5, -2, -1.5,…., 0, 0.5, ….2.5, 3.0). Once the histogram is created, change the distribution to a relative distribution, so that you can calculate the percentage of scores within each group of data (bin). Dictionary Take Notes Back Next
12. Calculating Observations as Deviations From the Mean This is what your table should look like (formulas below). Dictionary Cells E6, E7, etc. Cell H\$100 is the standard deviation Cell E\$100 is the mean Take Notes Back
13. Calculating Observations as Deviations From the Mean * If you did the exercise properly, you should have gotten a histogram similar to the one depicted at right. Dictionary Relatively how many scores are within 1 standard deviation of the mean? Two standard deviations? Three standard deviations? Does this distribution appear to be symmetrical or skewed? Can you confirm this using statistics (mean and median)? Take Notes Back Next
14. Calculating Observations as Deviations From the Mean The distribution results are shown below right. This distribution appears to be positively skewed. Dictionary Positive skewness can be confirmed by generating descriptive statistics with the Data Analysis add-in, and noting that the mean (5,085 baht) is greater than the median (4,377 baht). Positive kurtosis in the distribution confirms this. Take Notes Back
15. Using Z-Scores Backwards: Retrieving the Raw Data From the Z-Score As we’ve seen, one can calculate a Z-score from an observation’s value, knowing the mean and standard deviation of the distribution. Thus, it is also possible to retrieve the observation’s raw value, knowing its Z-score and the distribution’s mean and standard deviation. Dictionary Example: In a sample distribution with mean X = 200 and standard deviation s = 25, a Z-score of Zi = -0.2 translates into a raw score of Xi = -0.2(25) + 200 = 195. Take Notes Back Next
16. Percentiles Observations in quantitative data sets can be ordered from smallest/lowest to largest/highest, and then assigned percentiles (hundredths). An observation’s percentile indicates the percentage of observations that are below or lower than that particular observation. Dictionary For example, in an ordered database (smallest to largest values), the 64th percentile P64 would appear as follows: Take Notes Back Next
17. Calculating Percentiles Percentiles are calculated as follows: Dictionary • Adding 0.5 to the numerator of the calculation essentially splits Xi in half, placing one half of Xi in the “below Xi” category, and the other half in the “above Xi”category. This way, the two categories sum to 100%. • When there are two or more identical scores of Xi, the percentile (below) calculation only measures scores strictly below Xi, essentially treating the observation as the lowest of all identical scores. The percentile ranking is strictly below the score, not “at or below”. Take Notes Back Next
18. Calculating Percentiles: Example Dictionary Example: Using the same data sate introduced earlier, calculate the percentile of the score of 88. % above Answer: Rank the scores, revealing nine scored of 14 below X = 88. The percentile is therefore (9 + 0.5)/14 = 67.86%. % below Take Notes Back Next
19. Calculating Percentiles: Further Examples Examples: Using the same data sate as in the preceding example, calculate the percentile rankings of the following scores: 97, 81, 68 and 47. Dictionary Take Notes Back Next
20. Calculating Percentiles: Further Examples Examples: Using the same data sate as in the preceding example, calculate the percentile rankings of the following scores: 97, 81, 68 and 47. Dictionary Answers: P97 = {(13.5)/14}x100% = 96.43%. P81= {(7.5)/14}x100% = 53.57%. P68= {(3.5)/14}x100% = 25.00%. P47= {(0.5)/14}x100% = 3.57%. * Because (only) half of each score is included in the “below…” category, the highest score doesn’t have a 100% percentile ranking, and the lowest score does not have a 0% percentile ranking. Take Notes Back
21. Using MS Excel for Ranks and Percentiles MS Excel’s Data Analysis Add-In (Data > Data Analysis > Rank and Percentile) also calculates rank and percentile from a set of data. However, Excel (a) ranks the data from highest (#1) to lowest (#n) and (b) includes the whole score (rather than half) in the numerator of their calculation. Dictionary • When there is more than one instance of the same value, MS Excel’s formula also calculates the percentile relative to scores strictly below the observation, plus the (one only) observation itself. • The two alternate formulas converge in value for large n. Take Notes Back
22. Raw Data From Percentiles: Example If one wants to know a raw score associated with a particular percentile, use the percentile formula backwards: Dictionary However, instead of simply adding 0.5, make the following adjustment, (a) if the Raw Score Rank is not a whole number, simply round up to the next whole number; or, (b) if the Raw Score Rank is a whole number, get the average if it a and the next highest ordered whole number in the ordered set. Take Notes Back Next
23. Raw Data From Percentiles: Example Dictionary Example: Using the same data sate introduced earlier, calculate the raw data associated with a percentile ranking of 67.86. % above Answer: The unadjusted Raw Score is (67.86 x 14)/100 = 9.5. Rounding up gives a rank of 10. The 10th highest score is X = 88; an exact reverse as previously. % below Take Notes Back Next
24. Calculating Raw Scores From Percentiles: Further Examples Examples: Using the same data sate as in the preceding examples, calculate the raw data score associated with the following percentiles: 94%, 80%, 50%, 39%. Dictionary Take Notes Back Next
25. Calculating Raw Scores From Percentiles: Further Examples Examples: Using the same data sate as in the preceding examples, calculate the raw data score associated with the following percentiles: 94, 80, 50, 39. Dictionary Rank and Raw Score Round Up: Rank = 14th (top score of 97). Round Up: Rank = 12th (score of 89). Average Up: Rank = Average 7th and 8th.(score of (78+81)/2 = 79.5). Round Up: Rank = 6th (score of 76). Take Notes Back
26. Special Percentiles: The Median Recall that the median is one of three measures of central tendency. It is the score with an equal number of scores below it as above it. In percentile terms, the median is simply the 50th percentile. Dictionary Median: If n is an odd number, rounding up would be to the median score. If n is an even number, averaging it with the next highest score would be taking the average of the two most middle scores. These steps are just as the median rule suggested. Previous example: n = 14 (even), so the median is the average of the middle most 7th and 8th ranked scores: (78 + 81)/2) = 79.5, just as above. Take Notes Back Next
27. Special Percentiles: The Median and the Interquartile Range The Median, the middle ranked score with exactly one half scores higher and one half scores lower, can also be described as: (a) the 50th percentile, (b) the 5thdecile, or (c) the 2nd quartile. Dictionary Median The Interquartile Range (IQR) is the range of the middle two quartiles (Q3 – Q1). IQR Take Notes Back Next
28. Identifying “Outliers”: Using the Interquartile Range An Outlieris an extreme score either far below or far above the other scores, and is viewed as non-representative of the data set. Outliers can result in statistics mis-representing the data set, and must be dealt with. Many times, outliers are dealt with by focusing on only the interquartile range of the data set. Dictionary Outliersare often identified using some rule comparing raw scores to the Interquartile Range. For example: (low) Scores less than: Q1 – 0.5 x (Q3 – Q1) and (high) Scores more than: Q3 + 0.5 x (Q3 – Q1) Can be considered outliers. IQR Outliers Outliers Take Notes Back Next
29. Special Percentiles: Deciles and Quartiles Decilesdivide a data set into ten ordered and ranked subsets of 10% each of the data set. There are nine deciles markers needed to divide the whole data set into ten parts. Quartilesdivide a data set into four ordered and ranked quarters. Three quartile markers are needed to divide the whole data set into four quarters. Dictionary To calculate decile and quartile scores, just use the formula introduced above. (A whole number) So the score is the average of the 55th and 56th ranked scores. Example: In a data set of n = 220, the first quartile would be: Take Notes Back Next
30. Ignoring Outliers: The Interquartile Range Decilesdivide a data set into ten ordered and ranked subsets of 10% each of the data set. There are nine deciles markers needed to divide the whole data set into ten parts. Quartilesdivide a data set into four ordered and ranked quarters. Three quartile markers are needed to divide the whole data set into four quarters. Dictionary To calculate decile and quartile scores, just use the formula introduced above. Example: In a data set of n = 220, the first quartile would be: In this example, because it is a whole number, the score is the average of the 55th and 56th ranked scores (from lowest). Take Notes Back Next
31. Box Plots A Box Plot illustrates the interquartile range, the median, and the high and low values (excluding outliers) of a data set. Outliers are separately identified. Dictionary Example: low score = 46; Q1 (median) = 60; Q3 = 71 high score (excluding outlier) = 85; outlier = 98. Because the median is towards the left of the (yellow) interquartile box, this data set would appear to be positively skewed. * Box plots provide a useful visual aid comparing two or more data sets. However, they are not part of the MS Excel chart menu, and are not utilized here. Take Notes Back Next
32. End of Unit 4 Questions and Problems The following problems require the calculation of various statistics using MS Excel. The problems are linked to actual Excel spreadsheets, where students should do their work. xxxx Dictionary Take Notes Back End<|endoftext|>
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Do you listen well? How would you know? Really that is a judgement made by others. Listening is an act of generosity. This is a generosity that is not always evident.
What does it mean to listen well?
There are many well-known factors that contribute to effective listening. Forbes magazine suggests these:
- Maintain eye contact
- Be attentive but relaxed
- Keep an open mind
- Listen to the words and try to picture what the speaker is saying
- Don’t interrupt and don’t impose your solutions
- Wait for the speaker to pause, to pose clarifying questions
- Ask questions only to ensure understanding
- Try to feel what the speaker is feeling
- Give the speaker regular feedback
- Give attention to the unspoken, non-verbal cues
Listening is an act of generosity
Listening well requires a lot of generosity on the part of the listener. How so? It requires the listener to suspend their preferences and judgements and be open to the ideas being presented, without bias.
Listening is one of the hardest skills to master and requires uncovering deeper barriers within oneself
(Su & Wilkins, 2014).
These barriers can be uncovered by increasing your awareness of your own behaviour. The following are some considerations to take on board.
- Ignore your inner critic
When engaged in group situations like meetings or presentations, it is important not to get caught up in your own nerves or let your inner critic take precedence. This preoccupation will prevent your focus on other speakers, the questions they may pose and answers that may follow.Tip: Find a way to be fully focused on the speaker and their perspective. Be aware when this focus waivers.
- Consider your role as a listener
You need to know when to stop talking and start listing. Sometimes we are intent on providing solutions before someone has had the opportunity to fully express their situation. Generosity lies in allowing that person to express themselves without interruption. It is also about suspending judgement until all ideas have been presented.
Tip: When listening, close your mouth. You have two ears and one mouth. Try and listen twice as much as you speak.
- Be aware of your own triggers
Listening requires being “present” to the conversation you are part of. That means attending to what is being said and being aware and alert to things in the conversation that you react to. It is about staying with the conversation rather than anticipating what might happen before it does. This might happen in conversations with people that have a pattern to them, triggering habitual responses. You anticipate the way the conversation may go and rush to respond.Listening is especially important in difficult conversations. Multiple interests and agendas may be at play, leading to anticipated confrontation and a desire to avoid conflict. Defensiveness often kicks in and listening shuts down. When sensitive issues are the topics of conversation, full attention is needed. Often the person’s need to be heard defuses a situation, whereas cutting them off or offering solutions may inflame it.Tip: Take stock of your reactions in difficult conversations. Are your buttons being pressed? Be aware of what triggered this. Try and detach from the situation.
- Listen with confidence
Enter conversations with the confidence that solutions will be revealed. Consider that all parties can contribute. No one needs to try hard to convince the others of their confidence or ability to solve issues at hand. In so doing, not all perspectives are given an airing or due consideration. Providing a space where everyone can speak and be heard establishes confidence for all participants. Tip: Establish ground rules about each participant having air time. In meetings this may be timed so that contributions are fair and perspectives heard.
How’s your listening?
Do people say you are a good listener? Could you be a better listener? Perhaps try some of these ideas and see if they change anything for you.<|endoftext|>
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# Illustrative Mathematics Grade 8, Unit 4, Lesson 7: All, Some, or No Solutions
Learning Targets:
• I can determine whether an equation has no solutions, one solution, or infinitely many solutions.
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Illustrative Math
Grade 8
#### Lesson 7: All, Some, or No Solutions
Let’s think about how many solutions an equation can have.
Illustrative Math Unit 8.4, Lesson 7 (printable worksheets)
#### Lesson 7 Summary
An equation is a statement that two expressions have an equal value. The equation
2x = 6
is a true statement if x is 3: 2 · 3 = 6.
It is a false statement if x is 4: 2 · 4 ≠ 6.
The equation 2x = 6 has one and only one solution, because there is only one number (3) that you can double to get 6.
Some equations are true no matter what the value of the variable is. For example:
2x = x + x
is always true, because if you double a number, that will always be the same as adding the number to itself. Equations like 2x = x + x have an infinite number of solutions. We say it is true for all values of x.
Some equations have no solutions. For example:
x = x + 1
has no solutions, because no matter what the value of x is, it can’t equal one more than itself. When we solve an equation, we are looking for the values of the variable that make the equation true. When we try to solve the equation, we make allowable moves assuming it has a solution. Sometimes we make allowable moves and get an equation like this:
8 = 7
This statement is false, so it must be that the original equation had no solution at all.
#### Lesson 7.1 Which One Doesn’t Belong: Equations
Which one doesn’t belong?
1. 5 + 7 = 7 + 5
2. 5 · 7 = 7 · 5
3. 2 = 7 - 5
4. 5 - 7 = 7 - 5
#### Lesson 7.2 Thinking About Solutions
Without solving, identify whether these equations have a solution that is positive, negative, or zero.
1. Sort these equations into the two types: true for all values and true for no values.
2. Write the other side of this equation so that this equation is true for all values of u.
6(u - 2) + 2 =
3. Write the other side of this equation so that this equation is true for no values of u.
6(u - 2) + 2 =
#### Are you ready for more?
Consecutive numbers follow one right after the other. An example of three consecutive numbers is 17, 18, and 19. Another example is -100, -99, -98.
How many sets of two or more consecutive positive integers can be added to obtain a sum of 100?
• Show Answers
Are there sets with two numbers?
We can try two numbers grouped around 100/2 = 50.
49 + 50 ≠ 100 and 50 + 51 ≠ 100
And so, we cannot obtain two consecutive positive integers that will add to 100.
Are there sets with three numbers?
We can try three numbers grouped around 100/3 = 33.
Again, we cannot obtain three consecutive positive integers that will add to 100.
Similarly, we can try with four numbers.
We can try four numbers grouped around 100/4 = 25.
Again, we cannot obtain four consecutive positive integers that will add to 100.
We can try with five numbers.
We can try five numbers grouped around 100/5 = 20.
Yes, 18 + 19 + 20 + 21 + 22 = 100.
Six? No
Seven? No
Eight? Yes, 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 100.
Nine? No. ….
We can stop at 14, where the numbers would be grouped around 7.
At 15, the numbers would be grouped around 6 and the smallest number would be non-positive.
So there are two sets of consecutive positive integers whose sum is 100: one of five numbers (18 + 19 + 20 + 21 + 22), the other of eight (9 + 10 + 11 + 12 + 13 + 14 + 15 + 16).
#### Lesson 7.3 What’s the Equation?
1. Complete each equation so that it is true for all values of x.
2. Complete each equation so that it is true for no values of x.
3. Describe how you know whether an equation will be true for all values of or true for no values of x.
#### Lesson 7 Practice Problems
1. For each equation, decide if it is always true or never true.
2. Mai says that the equation 2x + 2 = x + 1 has no solution because the left hand side is double the right hand side. Do you agree with Mai? Explain your reasoning. 3 a. Write the other side of this equation so it’s true for all values of x: 1/2(6x - 10) - x =
b. Write the other side of this equation so it’s true for no values of x: 1/2(6x - 10) - x =
3. Here is an equation that is true for all values of x: 5(x + 2) = 5x + 10. Elena saw this equation and says she can tell 20(x + 2) + 31 = 4(5x + 10) + 31 is also true for any value of x. How can she tell? Explain your reasoning.
4. Elena and Lin are trying to solve 1/2 x + 3 = 7/2 x + 5. Describe the change they each make to each side of the equation.
a. Elena’s first step is to write 3 = 7/2 x - 1/2 x + 5.
b. Lin’s first step is to write x + 6 = 7x + 10.
5. Solve each equation and check your solution.
6. The point (-3,6) is on a line with a slope of 4.
a. Find two more points on the line.
b. Write an equation for the line.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Heat is a form of energy. It cannot be lost according to the first law of thermodynamics; rather it will be transferred. The first and second law of thermodynamics states that:
Energy can neither be created nor destroyed but can be transformed or converted from one form to another
1st Law of thermodynamics
In the course of energy transfer, some energy is lost in the form of heat to the surroundings
2nd Law of Thermodynamics, in respect to food chain in biology
Heat transfer indicates how heat flows between two systems of varying temperature. Heat will always be transferred from the region of higher temperature to the region of lower temperature. Please read our article on the concept of heat and temperature here.
Now, let’s understand how the particles of matter move before we delve into the transfer of heat.
Matter is composed of particles. Their constituent’s particles can either vibrate in a fixed position or translate (move from one point to another) or rotate. This movement of the particles of matter is a function of the kinetic energy present in them. (Think of kinetic energy as energy in motion). We can thus define temperature as the measure of the average kinetic energy possessed by the particles of matter. An increase in particles vibration, translation or rotation will imply an increase in temperature while a decrease in particles vibration, translation or rotation will mean a decrease in temperature. You may read our article on matter here.
Heat can be transferred through:
Let's explain heat transfer by conduction through this simple experiment which you may carry out at home.
Step 1: Boil 3 cups of water in a pot. When the water is very hot, put out the gas.
Step 2: Get another pot and fill it with 3 cups of cold water.
Step 3: Get an iron rod and make both ends of its edge contact both the hot water and the cold water.
Observation: When we carry out the above procedure, it will be observed that the temperature of the iron rod will increase and heat will be transferred into the cold water from the iron rod. Recall that iron is a good conductor of heat and electricity. Please read our article on properties of metal here. The movement of heat energy will follow the path below.
Hot Water >> Iron Rod >> Cold Water
Note: It is only the transfer of heat energy that brought about an increased temperature in the iron rod and then, the cold water; nothing more, nothing less. Below are some notable points:
This heat transfer will continue until equilibrium is achieved; that is, both of the water and the iron rod will eventually have the same temperature.
Conduction is therefore the transfer of heat energy from one point to another through a solid material called a conductor
Convection is a process of heat transfer from one point to another through the movement of fluids. A conductor or metal is not involved here. Convection explains the transfer of heat in liquids and gases. The boiling process of water occurs through convection. Initially, the water molecules inside the pot possesses equal kinetic energy. When placed on the gas cooker, the water molecules closest to the bottom and margins of the pot begin to move or translate as a result of heat transfer by conduction from the metallic pot (since the pot is a solid and a conductor). As a result, the water molecules (closest to the bottom and margins of the pot) becomes hot. They expand as a result and become even lighter, thus gaining an increased ability to move (an increase in their average kinetic energy). They move to the top and center of the pot displacing the water molecules that were initially there to the bottom as a result of the difference in density of both molecules.
Note: The water molecules at the top and center of the pot have a higher density since they are farther away from the bottom and margins pot. Also, they have a lower kinetic energy when compared to those at the bottom and edges. They are therefore displaced downwards (since they have higher density and lower kinetic energy) by the water molecules at the bottom and margin of the pot which had now translated upwards and inwards. This continues until the entire water molecules attain the same kinetic energy. At this point, equilibrium is said to be achieved. If the water molecules are still left inside the pot, all will be converted to a gaseous state as they eventually evaporate. (The pot becomes empty by now).
In a liquid, the heat energy produced by the particles of higher kinetic energy stimulates the liquid particles of lower kinetic energy (to an increased kinetic energy or translational motion) until equilibrium is achieved.
It is important to understand that convection is the main method of heat transfer in water and air.
With regards to convection in air particles; if will put on the heater (assuming) it’s on the floor, the kinetic energy of the air particles around the floor region (closest to the heater) will be increased. The particles of this hot air move upwards and sideways, mixing with the cooler air in these regions. The net result of this rising hot air is the transfer of heat energy from a higher region to a lower region.
Note: Convection is achieved by the translational and rotational movement of matter (translational and rotational movements are involved in liquids and gases respectively); which results in the transfer of heat energy. Conduction is NOT usually achieved by the vibration movement that is present in solids.
The term radiate in English mean to “to spread out from a central point”. For instance, the light of the sun radiates to the earth. Also, a lighted bulb will radiate its light in a dark room. One reason why some people put out their bulbs before they go out is to simply conserve energy. Leaving on the bulb will bring an extra heat into the room where it’s present. Recall that the bulb will possess a large amount of heat energy (becomes hot) if left on for a long period of time. Since energy cannot be lost, it will be radiated to the room where it’s present. The energy transfer goes thus:
Electrical Energy >> Light Energy >> Heat Energy
Note: In radiation, the heat energy is transferred by electromagnetic waves. In fact, the heat energy which the earth receives from the sun is as a result of electromagnetic waves travelling through space. Please read our article on layers and composition of the atmosphere here.
All objects (televisions, phones, phones, computer and any other object) radiate energy in the form of electromagnetic waves. The rate at which this energy is released is directly proportional to the Kelvin temperature (T) raised to the fourth power.
The hotter an object is, the more its ability to radiate heat energy. The radiated energy is a collection of a range of wavelength referred to as emission spectrum. The wavelengths within the spectra of the emitted radiation will decrease when the temperature of an object increases; and as a result, hotter objects will usually emit a shorter wavelength at a higher frequency.
Topics in PhysicsImportant Physics Concepts Fundamental and Derived units Zeroth Law of Thermodynamics Gravity explained with some fun facts Concept of Heat and Temperature Heat Transfer, Conduction, Convection and Radiation<|endoftext|>
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Vaccines are one of humanity’s greatest medical breakthroughs—ridding the world of smallpox, limiting outbreaks of mumps and measles and putting polio on the ropes. Now, reports Bill Chapell at NPR, researchers are hoping to harness the power of vaccines for the first time to help honeybees, which are currently being bombarded by a long list of threats.
Vaccines in non-human creatures are not new—any responsible pet owner has taken their dog or cat to the veterinarian for plenty of rabies and Lyme disease vaccinations. Inoculating an insect, however, is very different. In typical vaccines, either a dead or weakened version of a virus is introduced into an animal, whose immune system is then able to create antibodies to fight the disease. Insects, however, don’t have antibodies, meaning they don’t have the same type of immune response as we do.
Biologist Dalial Freitak of the University of Helsinki, the study’s author, found that when a moth was exposed to certain bacteria, usually through eating it, it was able to pass down a resistance to the bacteria to the next generation. She met with Heli Salmela, also of the University of Helsinki, who was working with a bee protein called vitellogenin that seemed to trigger the same reaction to invasive bacteria in the bees. The two began using the protein to create an immune response in the bees against American foulbrood, an infectious disease that is harming bee colonies across the world.
The vaccine helps the bee’s immune system recognize harmful diseases early on in life, similar to the way antibodies in the human body recognize diseases. When the queen bee consumes foulbrood bacteria, the vitellogenin protein binds with the pathogenic molecules, which are then passed along in her eggs. The developing baby bees’ immune systems then recognize the foulbrood bacteria as an intruder, setting off an immune response that protects the bee from the disease.
The result is a vaccine against foulbrood the team is calling PrimeBEE. The technology is undergoing tests, so it is not yet commercially available. The team has also yet to decide if the vaccine will be delivered by feeding queen bees sugar patties or if they will send out queen bees that have already been inoculated against the disease.
Whatever the case, apiculturists are excited to have a new tool to fight foulbrood. Toni Burnham, president of the DC Beekeepers Alliance in Washington, D.C., tells Chapell acquiring foulbrood means a bee colony has to be destroyed. “It’s a death sentence,” she tells Chappell. “If a colony is diagnosed with AFB — regardless of the level of the infestation — it burns. Every bit of it burns; the bees are killed and the woodenware burns, and it’s gone.”
The team says that the new technique could be used for other bee pathogens as well.
“We need to help honey bees, absolutely. Even improving their life a little would have a big effect on the global scale. Of course, the honeybees have many other problems as well: pesticides, habitat loss and so on, but diseases come hand in hand with these life-quality problems,” Freitak says in the press release. “If we can help honey bees to be healthier and if we can save even a small part of the bee population with this invention, I think we have done our good deed and saved the world a little bit.”
Bees could certainly use some good news. Since 2006, and perhaps a little before that, honey bee colonies began to experience something called Colony Collapse Disorder, in which hives would dissolve over the winter months. Researchers looked for causes from pathogens to pesticide exposure, but were never able to figure out exactly what was plaguing the bees. Though the problem has gotten better in recent years, the prospect of losing our bees, which pollinate lots of fruits and nuts, showed the world just how important our buzzy little friends can be.
Jason Daley is a Madison, Wisconsin-based writer specializing in natural history, science, travel, and the environment. His work has appeared in Discover, Popular Science, Outside, Men’s Journal, and other magazines.<|endoftext|>
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Red Class have been exploring different shapes during recent maths lessons. We have been interested in finding out how we can use a series of shapes to create a new pattern. Our early work has involved looking at circles. We encouraged the pupils to create their own series of patterns using some brightly coloured shapes. Some of the pupils produced a large circle made up of a series of smaller circles; other designs looked like caterpillars.
All of the Red Class children have been enjoying reading our current class book, What the Ladybird Heard. This is a story that involves many lively animal noises. We have been playing games with the children to help them correctly recognise each different noise. We have also matched up the animals with the correct spelling of each sound during our literacy work.
Great progress has also been made with the handwriting from all Red Class children. This has been a key focus for this term. We have been helping each pupil to trace out some key letters. This work will continue with help given to Red Class in understanding the phonic sound in each letter.<|endoftext|>
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The United States Constitution – Bill of Rights
The United States Constitution was a written document that established how the government would operate with three branches (executive, legislative and judicial) and how they would interact with the states. It provided fundamental laws and the basic rights of citizens.
Before that, states acted as independent countries.
It was signed on 17th September 1787.
The US constitution has 7 articles. Currently, there are 27 amendments to these articles.
The first 10 amendments are called the Bill of Rights.<|endoftext|>
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The Aswan Dam
For thousands of years, the Nile River overflowed its banks every spring as snow and spring rains in the mountains of East Africa transformed into a torrent of water. The annual flooding carried dead and decaying plants in its muddy solution. The muddy water is called silt, and silt created excellent topsoil for Egyptian farmers.
For more than a century, Egyptian leaders have dreamed of building a dam to control the Nile River. In addition to regulating the Nile’s flow, a large dam could store water for Egypt’s dry season, and rushing waters flowing from the dam could provide Egypt with a free supply of hydroelectricity. Hydroelectricity is electric power generated by the movement of water.
The British built a dam in the city of Aswan in southern Egypt in 1902. What is now called the Aswan Low Dam or the Old Aswan Dam was, at that time, the largest masonry dam in the world. Despite its size, the original Aswan Dam was unable to provide Egypt with proper control of the great river. Further, the original dam did not hold back a large reservoir of water for use in the dry season.
Egyptian President Gamal Abdel Nasser hoped to build a larger dam. In 1956, Nasser asked the United States and Great Britain for financial assistance. The western nations refused because of Nasser’s close ties with the Soviet Union. The Soviet Union, now known as Russia, was at that time a military superpower nation. The United States and Britain were engaged in a period of hostilities with the Soviet Union. These hostilities were known as the Cold War. The Cold War lasted from 1945 until 1990. Although the hostilities did not develop into World War III, both sides sought to influence developing nations such as Egypt. When the two western nations refused Nasser’s request for financial assistance, the Egyptian president seized control of the Suez Canal. He then asked for and received assistance from the Soviet Union to build a larger dam in Aswan.
The Aswan High Dam was completed in 1970. The new dam was able to control the annual flooding of the Nile, but it created both benefits and consequences for the Egyptian people. By holding back water, the Aswan Dam created Lake Nasser, the world’s largest man-made lake. The water from Lake Nasser provided new fishing areas. The Aswan Dam also makes it possible for Egyptian farmers to grow an extra crop each year by releasing water into the Nile during the dry season. The dam has also helped tourism in Egypt because the height of the Nile is now reliable enough for cruise ships to travel the ancient river’s waters throughout most of the year.
Critics charged that building the Aswan Dam was unwise. The project was far more expensive than originally expected. Further, since the creation of the Aswan Dam, the farmers along the banks of the Nile River have had to use expensive fertilizers in place of the topsoil provided by the river silt. Fish had once fed on the silt, and the people downstream depended on fishing from the river. Without the food in the muddy water, fewer fish are able to survive. The human impact of the Aswan High Dam was also great. More than 100,000 Egyptian people and an additional 50,000 Sudanese people were forced to move because their homes would soon be under the water held back by the Aswan High Dam.
The Aswan High Dam was a tremendous technological achievement, but building the dam was a controversial decision that has been debated for more than fifty years.<|endoftext|>
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# NCERT Solutions for Class 12 Maths Exercise 1.2
Looking for NCERT Solutions for Class 12 Maths Exercise 1.2? If yes then you are in the right place. Here we have provided you with the Question and Answers of Exercise 1.2. These solutions are written by expert teachers and are so accurate to rely on.
NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.
## NCERT Solutions for Class 12 Maths Exercise 1.2
### Relations and Functions Solutions
#### Q 1) Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto, where : R* is the set of all non-zero real numbers. Is the result true, if the domain : R* is replaced by N with co-domain being same as : R*?
Hence, function g is one-one but not onto.
#### Q 2) Check the injectivity and surjectivity of the following functions:
(i) f: N → N given by f(x) = x2
(ii) f: Z → Z given by f(x) = x
(iii) f: R → R given by f(x) = x2
(iv) f: N → N given by f(x) = x3
(v) f: Z → Z given by f(x) = x3
(i) f: N → N is given by,
f(x) = x2
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x2 = y2 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
(ii) f: Z → Z is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iii) f: R → R is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
(iv) f: N → N given by,
f(x) = x3
It is seen that for x, y ∈N, f(x) = f(y)
⇒ x3 = y3 ⇒ x = y.
∴f is injective.
Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x= 2.
∴ f is not surjective
Hence, function f is injective but not surjective.
(v) f: Z → Z is given by,
f(x) = x3
It is seen that for x, y ∈ Z, f(x) = f(y)
⇒ x3 = y3 ⇒ x = y.
∴ f is injective.
Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.
∴ f is not surjective.
Hence, function f is injective but not surjective.
#### Q 3) Prove that the Greatest Integer Function f: R → R given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
f: R → R is given by,
f(x) = [x]
It is seen that f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1.
∴ f(1.2) = f(1.9), but 1.2 ≠ 1.9.
∴ f is not one-one.
Now, consider 0.7 ∈ R.
It is known that f(x) = [x] is always an integer. Thus, there does not exist any element x ∈ R such that f(x) = 0.7.
∴ f is not onto.
Hence, the greatest integer function is neither one-one nor onto.
#### Q 5) Show that the Signum Function f: R → R , given by
is neither one-one nor onto.
f: R → R
It is seen that f(1) = f(2) = 1, but 1 ≠ 2.
∴ f is not one-one.
Now, as f(x) ) takes only 3 values (1, 0, or −1) for the element −2 in co-domain R , there does not exist any x in domain R such that
f(x) = −2.
∴ f is not onto.
Hence, the signum function is neither one-one nor onto.
#### Q 6) Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
Show that f is one-one.
It is given that A = {1, 2, 3}, B = {4, 5, 6, 7}.
f: A → B is defined as f = {(1, 4), (2, 5), (3, 6)}.
∴ f (1) = 4, f (2) = 5, f (3) = 6
It is seen that the images of distinct elements of A under f are distinct.
Hence, function f is one-one.
#### Q 7) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R → R defined by f(x) = 3 − 4x
(ii) f: R → R defined by f(x) = 1 + x2
(i) f→ R is defined as f(x) = 3 − 4x.
∴ f is one-one.
For any real number (y) in R, there exists in R such that
is onto.
Hence, is bijective.
(ii) fR → R is defined as
For instance,
∴ f is not one-one.
Consider an element −2 in co-domain R.
It is seen that is positive for all x ∈ R.
Thus, there does not exist any x in domain R such that f (x) = −2.
∴ f is not onto.
Hence, f is neither one-one nor onto.
#### Q 8) Let A and B be sets. Show that f: A × B → B × A such that (a, b) = (b, a) is bijective function.
fA × B → B × A is defined as f(ab) = (ba).
∴ f is one-one.
Now, let (ba) ∈ B × A be any element.
Then, there exists (ab) ∈A × B such that f(ab) = (ba). [By definition of f]
∴ f is onto.
Hence, f is bijective.
#### Q 9) Let f: N → N be defined by
Answer) fN → N is defined as
It can be observed that:
∴ f is not one-one.
Consider a natural number (n) in co-domain N.
Case I: n is odd
n = 2r + 1 for some r ∈ N. Then, there exists 4+ 1∈N such that
Case II: n is even
n = 2r for some r ∈ N. Then,there exists 4r ∈N such that
∴ f is onto.
Hence, f is not a bijective function.
#### Q 10) Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by
A = R − {3}, B = R − {1}
f: A → B is defined as
∴ f is one-one.
Let y ∈B = R − {1}. Then, y ≠ 1.
The function is onto if there exists x ∈A such that f(x) = y.
Now,
Thus, for any ∈ B, there exists such that
Hence, function f is one-one and onto.
#### Q 11) Let f: R → R be defined as f(x) = x4. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
Answer) fR → R is defined as f(x) = x4
Let x∈ R such that f(x) = f(y).
∴ f (x1) = f (x2) does not imply that x1 = x2
For instance,
∴ f is not one-one.
Consider an element −-2 in co-domain R. It is clear that there does not exist any x in domain R such that f(x) = −-2.
∴ f is not onto.
Hence, function f is neither one-one nor onto.
#### Q 12) Let f: R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto (B) f is many-one onto
(C) f is one-one but not onto (D) f is neither one-one nor onto
fR → R is defined as f(x) = 3x.
Let x∈ R such that f(x) = f(y).
⇒ 3x = 3y
⇒ x = y
is one-one.
Also, for any real number (y) in co-domain R, there exists in R such that
is onto.
Hence, function f is one-one and onto.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Linear Inequalities in Two Variables
## Graph inequalities like 2x + 3 < y on the coordinate plane
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Practice Linear Inequalities in Two Variables
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Compare the Popularity of Different Coffees
Teacher Contributed
## Real World Applications – Algebra I
### Topic
How can we compare the popularity of Starbucks coffee?
### Student Exploration
Starbucks has a lot of rotating coffee blends throughout the year. In July 2012, the most popular two types of coffee at one Starbucks location is the Veranda Blend and the Sumatra Blend. In one month, the Veranda blend sold three times as much as the Sumatra blend. This determined how much the store has to order for the following month. Let’s demonstrate this relationship as a system of equations, and find what the minimum amount of the Sumatra blend this store should order.
The store normally orders 100 pounds of the two most popular blends every month. In this case, they’re going to order 100 pounds of the Veranda and Sumatra blends total. We also know that the amount of Veranda blend can’t exceed more than three times as the Sumatra blend.
Let’s assume that x\begin{align*}x\end{align*} represents the amount of the Veranda blend ordered in pounds, and y\begin{align*}y\end{align*} represents the amount of the Sumatra blend ordered in pounds.
Our inequalities are x3y\begin{align*}x \le 3y\end{align*} and x+y>100\begin{align*}x + y > 100\end{align*}
In order to graph the inequalities, we have to solve for y\begin{align*}y\end{align*} in both.
We now have y(13)x\begin{align*}y \ge \left(\frac{1}{3}\right) x\end{align*} and y>x+100\begin{align*}y > -x + 100\end{align*}
The graph below shows the two inequalities graphed on the same set of axes.
Let’s solve this system of inequalities algebraically to find the minimum amount of the Sumatra blend.
For the sake of finding the lowest point of the graph, we’ll replace the inequality signs with equal signs, so that we can solve for x\begin{align*}x\end{align*} and then solve for y\begin{align*}y\end{align*}. Let’s use the Substitution method.
(13)x(43)xx=x+100=100=75 lbs\begin{align*}\left(\frac{1}{3}\right)x &= - x + 100\\ \left(\frac{4}{3}\right)x &= 100\\ x &= 75 \ lbs\end{align*}
Now we need to solve for y\begin{align*}y\end{align*}.
yy=(13)(75)=25 lbs\begin{align*}y &= \left(\frac{1}{3}\right)(75)\\ y &= 25 \ lbs\end{align*}
The store should order a minimum of 25lbs of the Sumatra blend.
### Extension Investigation
Research a local coffee shop and find out some information that you can represent as a system of linear inequalities! You will need to compare two different products, and you will need to ask a worker at the store for some of the product information.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes<|endoftext|>
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# How to find the nth term of an Arithmetic Sequence?
• Last Updated : 27 Aug, 2021
Arithmetic is a part of mathematics that works with different types of numbers, fractions, applied different operations on numbers like addition, multiplication, etc. The word Arithmetic comes from the Greek word arithmos, which means number. Also arithmetic involves exponentiation, the calculation of percentages, finding the value of number series, logarithmic functions, and square roots, etc.
### Arithmetic Progression
There is series in arithmetic called Arithmetic Progression (AP), this is a sequence of numbers, where the difference between any two consecutive terms is always the same. Let’s say, a series is 2,4, 6, 8, 10, 12,.., in this series, the difference between any two consecutive numbers is 2. If 2 is added with the previous number then the next number in the series is obtained, similarly, if 2 is subtracted from the next number, the previous number is obtained.
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### The formula for finding the nth term
To work with this series there are some formulas available, formulas like finding the nth term in the series, formula for finding the sum of all the terms in arithmetic series. There are formulae introduced that can help in finding the value with limited given options, for instance, finding the nth term from first and the last term only. Let’s say a series A consist of some element a1, a2, a3, a4, a5, a6,…an
A = {a1, a2, a3, a4, a5, a6,… an}
• Common difference between two terms (d) = (a1-a2)
• Sum of the series(S) = (n/2)[2a + (n – 1)d]
• First term = a
• 2nd term = a+d
• 3rd term = a+2d
• similarly, Nth term = a+(n-1)d
Steps to find the nth term
Step 1: At first find the first and 2nd term, that is a1 and a2.
Step 2: Then find the common difference between them, that is d = (a2-a1)
Step 3: Now, by adding the difference d with the 2nd term we will get 3rd term, and like this, the series goes on. That is 2nd term, a2 = a1+d (a1 is first term)
3rd term, a3 = a2+d = (a1+d)+d = a1+2d
4th term, a4 = a3+d = (a1+2d)+d = a1+3d
So, we can see that the number of d is 1 less than the number of terms. That is,
In a2, the number of d is 1, [(2-1) = 1]
In a3 the number of d is 2, [(3-1) = 2]
In a4 the number of d is 3, [(4-1) = 3]
So, similarly, for the Nth term, the number of d must be (N-1) times.
Therefore, Nth term, an = a1 + (N-1)d ⇢ [First term + (Last term – 1)×common difference]
### Sample Problems
Question 1: Find the 9th term of the given series, {1, 4, 7, 10, 13, 16,….}
Solution:
Here N is 9,
First-term, a1 = 1
2nd term, a2 = 4,
3rd term a3 = 7
4th term a4 = 10
Now find the common difference,
d = a2 – a1 = 4 – 1 = 3
check d is correct or not,
a1 + d = 1+3 = 4 = a2
a2 + d = 4+3 = 7 = a3
a3 + d = 7+3 = 10 = a4
a4 + d = 10+3 = 13 = a5
a5 + d = 13+3 = 16 = a6
So, here the common difference is correct.
now the 9th term,
a9 = First term + (Last term – 1) × common difference
= a1 + (N-1)d
= 1 + (9-1)×3
= 1+ 8*3
= 1+24
=25
So, the 9th term is 25.
Question 2: Find the 12th term of the given series, {5, 11, 17, 23, 29,….}
Solution:
Here N is 12,
First-term, a1 = 5
2nd term, a2 = 11,
3rd term a3 = 17
4th term a4 = 23
Now find the common difference,
d = a2 – a1 = 11 – 5 = 6
check d is correct or not,
a1 + d = 5+6 = 11 = a2
a2 + d = 11+6 = 17 = a3
a3 + d = 17+6 = 23 = a4
a4 + d = 23+6 = 29 = a5
So, here the common difference is correct.
Now the 12th term,
a12 = First term + (12th term – 1) × common difference
= a1 + (N-1)d
= 1 + (12-1)×6
= 1+ 11×6
= 1+66
= 67
So, the 12th term is 67.
Question 3: If the 4th term of an AP is 8 with a common difference of 2. Find out the arithmetic progression up to 8 terms.
Solution:
Given that, the fourth term, a4 is 8 and the common difference is 2,
So the fourth term can be written as,
a + (4 – 1) × 2 = 8 [a = first term]
= a+ 3×2 = 8
= a = 8 – 3×2
= a = 8 – 6
= a = 2
So the first term a1 is 2,
Now, a2 = a1+2 = 2+2 = 4
a3 = a2+2 = 4+2 = 6
a4 = 8
a5 = a4 +2 = 8+2 = 10
a6 = a5+2 = 10+2 = 12
a7 = a6+2 = 12+2 = 14
a8 = a7+2 = 14+2 = 16
So the series upto 8 term is = 2, 4, 6, 8, 10, 12, 14, 16
Question 4: Find the 7th term of an AP whose 3rd term is 9 and 5th term is 15?
Solution:
Given that, the 3rd term, a3 is 9 and 5th term, a5 is 15. here we have to find common difference and first term(a1).
a3 = a1+2d = 9 [from the formula] ⇢ (1)
And, a5= a1+4d = 15 ⇢ (2)
Solve (1) and (2),
a1+2d = 9 ⇢ (1)
a1+4d = 15 ⇢ (2)
Let’s apply subtraction between (1) and (2)
2d = 6, d=3
So, the common difference is 3
Now put the value of d in any one equation, here put the value of d in (1)
a1+2d = 9
= a1 + 2×3 = 9 [d=3]
= a1=9-6
= a1=6
So the first term is 3
Now to find the 7th term,
Apply the formula for finding the nth term, here n=7
a7 = First term + (7th term – 1) × common difference
a7 = a1 + (7-1)d
a7=3+6×3 [d=3 and a1=3]
a7 = 21
So the 7th term is 21.
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Let the Study Guides explain the CAST diagram and how to solve Trig Equations. See below ...
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Understanding Trig Equations is important to ensure your success in tests, assessments and exams.
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Solving equations such as sin x° = 1, sin x° = – 1 and cos x° = 0 can be achieved just be reading the solutions from the trig graphs.
For example, in the graph of y = sin x° below, it can be seen that if we draw a horizontal line from 1 on the y-axis then it will meet the sin wave at 90°. Therefore the solution to sin x° = 1 is x = 90°.
Solving Trig Equations such as sin x° = 0.5 is more complicated
In the graph of y = sin x° above, it can be seen that if we draw a horizontal line from 0.5 on the y-axis then it will meet the sin wave at two places.
A calculator is required for the first solution:
x = inverse sin (0.5) = 30°
and the correct use of the CAST diagram is required to find the second solution:
x = 180° – 30° = 150°
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# Integration by Parts
The mathematicsmethod of integration by parts is very useful method in the integral of a product of two or more functions. This method is very helpful in comparing with another method of integration. Let we have two functions like f (x) and g (x) in product form then we have the formula for integration by parts: - Integration of f (x) * g (x) = f (x) * integration of g (x) – integration of [f’(x) * integration of g(x)]. In other words the integral of the product of two functions = first functions * integration of second function – integral of (differentiation of first function * integral of second function). Here is the most important part to choose first and second function.
If we have two functions in product form if one function of these two functions is not directly integrable (like sin^-1 x, cos^-1 x, tan^-1 x, cot^-1 x, sec^-1 x, csc^-1 x, log x etc) these functions we take as a first part. If in product of two functions both functions are easily integrable, then we choose first functions as the derivative of this function is a simple function. Thus we can easily integrate the given function. Always product of two functions, in order to choose the first function, first we consider the inverse trigonometric function, logarithmic function, algebraic function, trigonometric function, exponential functions. This is the order to choose the first function.
Let us see some examples of integration by parts. Example: - Integrate the given function by using integration by parts x^3 cos x. Solution: - given function in product form ( x^2 + x + 1) cos x. Here we choose first function algebraic function f (x) = x^2 + x + 1 and letsthe second function is g(x) cos x. Now using formula integration by parts: - the integral of the product of two functions = first functions * integration of second function – integral of (differentiation of first function * integral of second function).
Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x – integral of [(2x + 1)sinx], again using integration by parts. And get integral of [(2x + 1)sinx] = - (2x + 1) cos x – integral of [(2) (-cos x) = -(2x + 1) cos x + 2 sin x, Now we plug the value of the integral of [(2x + 1)sinx] and get Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x – [-(2x + 1) cos x + 2 sin x], è Integral of [(x^2 + x + 1) cos x] = (x^2 + x + 1) sin x + (2x + 1) cos x - 2 sin x]. In this type we can integrate functions. This is how we can find the integral of many functions using integration by parts method. This method helps in finding integral of those functions which cannot be solved using the normal method.
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This view of the frozen fjord downstream of Violingletscher (Violin Glacier) in Østgrønland (East Greenland) was seen during a NASA’s Operation IceBridge survey flight on April 5, 2014. Image © NASA/Michael Studinger
NASA’s Operation IceBridge images Earth’s polar ice in unprecedented detail to better understand processes that connect the polar regions with the global climate system. IceBridge utilizes a highly specialized fleet of research aircraft and the most sophisticated suite of innovative science instruments ever assembled to characterize annual changes in thickness of sea ice, glaciers, and ice sheets. In addition, IceBridge collects critical data used to predict the response of earth’s polar ice to climate change and resulting sea-level rise. IceBridge also helps bridge the gap in polar observations between NASA’s ICESat satellite missions.
Read more about IceBridge’s 2014 Arctic campaign<|endoftext|>
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Cold urticaria (ur-tih-KAR-e-uh) is a skin reaction to cold that appears within minutes after cold exposure. Affected skin develops reddish, itchy welts (hives).
People with cold urticaria experience widely different symptoms. Some have minor reactions to the cold, while others have severe reactions. For some people with this condition, swimming in cold water could lead to very low blood pressure, fainting or shock.
Cold urticaria occurs most frequently in young adults. If you think you have this condition, consult your doctor. Treatment usually includes preventive steps such as taking antihistamines and avoiding cold air and water.
Cold urticaria signs and symptoms may include:
- Temporary reddish, itchy welts (hives) on the area of skin that was exposed to cold
- A worsening of the reaction as the skin warms
- Swelling of hands while holding cold objects
- Swelling of lips from consuming cold food or drink
Severe reactions may include:
- A whole-body response (anaphylaxis), which can cause fainting, a racing heart, swelling of limbs or torso, and shock
- Swelling of the tongue and throat, which can make it difficult to breathe
Cold urticaria symptoms begin soon after the skin is exposed to a sudden drop in air temperature or to cold water. Damp and windy conditions may make a flare of symptoms more likely. Each episode may persist for about two hours.
The worst reactions generally occur with full skin exposure, such as swimming in cold water. Such a reaction could lead to loss of consciousness and drowning.
When to see a doctor
If you have skin reactions after cold exposure, see your doctor. Even if the reactions are mild, your doctor will want to rule out underlying conditions that may be causing the problem.
Seek emergency care if after sudden exposure to cold you experience a whole-body response (anaphylaxis) or difficulty breathing.
No one knows exactly what causes cold urticaria. Certain people appear to have very sensitive skin cells, due to an inherited trait, a virus or an illness. In the most common forms of this condition, cold triggers the release of histamine and other chemicals into the bloodstream. These chemicals cause redness, itching and sometimes a whole-body (systemic) reaction.
You're more likely to have this condition if:
- You're a young adult. The most common type — primary acquired cold urticaria — occurs most frequently in young adults.
- You have an underlying health condition. A less common type — secondary acquired cold urticaria — can be caused by an underlying health problem, such as hepatitis or cancer.
- You have certain inherited traits. Rarely, cold urticaria is inherited. This familial type causes painful welts and flu-like symptoms after exposure to cold.
The main possible complication of cold urticaria is a severe reaction that occurs after exposing large areas of skin to cold, for example, by swimming in cold water.
The following tips may help prevent a recurrent episode of cold urticaria:
- Take an over-the-counter antihistamine before cold exposure.
- Take medications as prescribed.
- Protect your skin from the cold or sudden changes in temperature. If you're going swimming, dip your hand in the water first and see if you experience a skin reaction.
- Avoid ice-cold drinks and food to prevent swelling of your throat.
- If your doctor prescribed an epinephrine autoinjector (EpiPen, Auvi-Q, others), keep it with you to help prevent serious reactions.
- If you're scheduled for surgery, talk with your surgeon beforehand about your cold urticaria. The surgical team can take steps to help prevent cold-induced symptoms in the operating room.
Jan. 19, 2018<|endoftext|>
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The Chinese river dolphin is a freshwater dolphin and one of the most endangered animals on Earth. It is also known as the Yangtze river dolphin, baiji, white-flag dolphin, and white-fin dolphin. Nicknamed “Goddess of the Yangtze,” it was regarded as the goddess of protection by local fishermen and boatmen in China (Zhou, 1991).
The Yangtze river dolphin is pale blue to gray on the dorsal (back) side, white on the ventral (belly) side. It has a long and slightly-upturned beak with 31-36 conical teeth on either jaw. Its dorsal fin is low and triangular in shape, and resembles a light-colored flag when the dolphin swims just below the surface of the murky Yangtze River, hence the name “white-flag” dolphin. It has smaller eyes compared to oceanic dolphins. A mature Yangtze river dolphin is about 8 feet (2.5 m) long and weighs about 500 pounds (220 kg).
Though it has been described in classical Chinese literature and folklore as early as 200 B.C., the Yangtze river dolphin was not known to the western world until 1916, when Charles M. Hoy, the son of an American missionary, shot one while duck hunting on Dongting Lake, China (Zhou, 1991). Subsequently, this species was formally introduced to the scientific community by Gerrit S. Miller, Jr., a curator at the Smithsonian Institute’s National Museum of Natural History.
Just like the other species of river dolphins (the Amazon river dolphin, Inia geofrensis; the La Plata river dolphin, Pontoporia blainvllei; and the Indian river dolphin, Platanista gangetica), the Yangtze river dolphin originated from the ocean. The ancestor of the Yangtze river dolphin resided in the shallow sea that inundated the Yangtze River Basin during the globally high sea levels of the Middle Miocene period (approximately 20 million years ago). After the subsequent sea-level lowering and transition to a freshwater river during the late Neogene period, the dolphins remained and evolved to become the present-day Yangtze river dolphins (Hamilton et al., 2001).
The Yangtze river dolphin is thought to breed in the first half of the year, with the peak calving season between February and April. Gestation lasts 10-11 months, and a newborn calf measures about 3 feet (1 m). Dolphins reach sexual maturity at 4-6 years of age (Zhou, 2004). The life expectancy for the Yangtze river dolphin is about 25 years, based on an individual in captivity.
Very little is known about the ecology, behavior, and acoustics of the Yangtze river dolphin. Groups of two to six animals were most commonly seen, although aggregations of up to 16 animals have also been observed (Zhou, 2004). The Yangtze river dolphin feeds on a large variety of freshwater fish species (Zhou, 2004). These dolphins are generally shy of boats, and often expose only the top of the head, dorsal fin, and a small part of the back when surfacing (Zhou, 2004).
Due to the poor visibility in the murky Yangtze River, the Yangtze river dolphin depends largely on underwater sound for orientation, feeding, and communication. Just like most dolphin species, the Yangtze river dolphin produces two types of sounds: clicks and whistles (Akamatsu et al., 1998; K. Wang et al., 2006a). Clicks are believed to be used primarily for navigation and identifying prey and other objects in the surrounding environment (Au, 1993). Whistles are frequency modulated (pitch changes with time) sounds and are thought to be used for communication. (K. Wang et al., 2006a
|500 pounds (220 kg)|
|8 feet (2.5 m)|
|pale blue to gray on their back and white on the under side|
|about 25 years (based on an individual in captivity)|
|large variety of freshwater fish species|
|very little is known about the behavior, but they depend largely on underwater sound for orientation, feeding, and communication and, just like most dolphin species, they produce clicks and whistles|
RELATIVE SPECIES: Indo-Pacific Humpback Dolphin
OTHER NAMES: Baiji, White Flag Dolphin, White Fin Dolphin, Chinese River Dolphin
NEIGHBORING SPECIES: Finless Porpoise
THREATS: Hunting, accidental catch, pollution, ship strikes, habitat degradation
DIET: Small fish
MANNER OF FEEDING: Swallow fish head first
BEHAVIOR: Found in groups of 2-6. Very strong swimmers, have to swim against current. Very shy of boats. Use echolocation to see and communicate in murky waters.
REPRODUCTION: Breed in April or May. Give Birth 10-11 months later
LIFE SPAN: 25 years in captivity
Habitat: The Yangtze river dolphin is found only in freshwater, specifically in the Yangtze River in China.
Historically, the species occurred in the middle to lower reaches of the Yangtze River, along with its tributaries and connecting lakes in China (Zhou et al., 1977). Due to habitat loss caused by rapid development along the Yangtze River, its distribution was largely reduced. More recently, it is believed that the Yangtze river dolphin is limited to the main channel of an 870-mile (1,400-km) river section between Jingzhou and Jiangyin (Zhang et al., 2003).
No historical population estimate of the Yangtze river dolphin is available. However, the population of the Yangtze river dolphin has declined drastically and rapidly since the 1950s (Zhou et al., 2004; Turvey et al., 2007). The estimated population size declined to 400 by the 1980s (Zhouet al., 1982). Surveys conducted between 1997 and 1999 led scientists to believe that the minimum number left in the Yangtze could be as low as 13 individuals (Zhang et al., 2003). The last authenticated Yangtze river dolphin records were of a stranded pregnant female found in 2001, and a live animal photographed in 2002 (Turvey et al., 2007). In 2003, a report published by the Chinese Institute of Hydrobiology concluded that the population is probably declining by about 10% per year, suggesting that extinction would occur within a decade (Zhang et al., 2003).
During November and December 2006, an international research team, including biologists from NOAA, conducted an intensive six-week multi-vessel visual and acoustic survey covering the entire historical range of the Yangtze river dolphin in the main Yangtze channel between Yichang and Shanghai (1,669 km). By the end of the expedition, the team failed to find any evidence that the species survives, and concluded that the Yangtze river dolphin may now be extinct (Turvey et al., 2007).
The primary threat is
- bycatch in local fisheries, which use rolling hooks, nets (gillnets and fyke nets), and electro-fishing (Zhou and Wang, 1994; Zhou et al., 1998).
- Rolling hooks and other fishing gear caused at least half of all known dolphin deaths in the 1970s and 1980s.
- Electro-fishing accounted for 40% of dolphin deaths recorded during the 1990s (Lin et al., 1985; Chen and Hua, 1989; Zhou and Li, 1989; Zhou and Wang, 1994; Zhou et al., 1998; Zhang et al., 2003).
Other threats to the Yangtze river dolphin include (K. Wang et al., 2006b)
- river development projects
- vessel collision
- underwater explosions used to deepen or widen navigation channels or used for illegal fishing are another cause of dolphin mortality.
Zhou and Li (1989) once reported that explosions caused 19.4% of deaths during illegal fishing operations (using explosives) in the lower reaches of the river between 1979 and 1981.
The increasing rate of water project development, including construction of dams and floodgates in the middle and lower reaches of the Yangtze, has resulted in blockages between the river and the lakes. The dams interrupted the movements of the Yangtze river dolphins upstream of the dams, and eliminated their access to tributaries and appended lakes (Chen and Hua, 1989; Liu et al., 2000). These dams also caused a notable decline in fish resources (Zhou and Li, 1989). The world’s largest dam, the Three Gorges Dam, has been inserted in the middle of a hot spot of bio-diversity in south-central China, which affected the habitats of the Yangtze river dolphin (Chen et al., 1997 in K. Wang et al., 2006b; Zhou, 1991).
Five natural reserves have been established along the middle and lower reaches of the Yangtze River since 1986 for the protection of the Yangtze river dolphin, and another rare cetacean species, the Yangtze finless porpoise (Wang et al., 2006). However, these reserves do not prevent the incidental death of these species (D. Wang et al., 2006; K. Wang et al., 2006b).
In addition to these natural reserves, two semi-natural reserves were established for the purpose of housing Yangtze river dolphins under human management (Baiji Research Group, 1989; Zhou, 1989). However, neither semi-natural reserve has ever been used for captive propagation of the Yangtze river dolphins. In 1980, a male Yangtze river dolphin was rescued from fishing gear and rehabilitated in Baiji Dolphinarium at the Institute of Hydrobiology of Chinese Academy of Science in Wuhan (Gui and Wang, 2005; Chen and Liu, 1989). This dolphin died from old age in 2002 after more than 22 years in captivity (K. Wang et al., 2006b).
In response to the dramatic and rapid decline of the Yangtze river dolphin and other wildlife species between 1985 and 1987, the Chinese Government issued the “Circular for the Protection of Precious and Rare Wildlife” and the “Urgent Circular Banning Hunt, Trade, and Smuggling of Precious and Rare Animals,” making the protection of these species a law (Zhou and Zhang, 1991). Subsequently, the Yangtze river dolphin was designated as one of the “First Category of National Key Protected Wildlife Species” in China (K. Wang et al., 2006b).
In the United States, all marine mammals, including Yangtze river dolphins, are protected under the Marine Mammal Protection Act (MMPA), as amended. In addition, this species is also listed as endangered species under the Endangered Species Act of 1973.
- This dolphin is the rarest species of all marine mammals; many scientists fear that it is functionally extinct. A recent extensive survey of this species resulted in no sightings.<|endoftext|>
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What is Intrinsic Motivation?
Intrinsic motivation refers to the stimulation that drives adopting or changing behavior for personal satisfaction or fulfillment. Such motivation drives an individual to perform an activity for internal reasons that are personally satisfying, as opposed to being motivated extrinsically – that is, by the prospect of obtaining some external reward, such as money.
In other words, with intrinsic motivation, the incentive for engaging in a behavior or completing a task is performing the task itself. For example, when someone decides to eat ice cream, typically their motivation is to enjoy the pleasure of eating the ice cream. They are not doing it for some external reward.
Intrinsic Motivation vs. Extrinsic Motivation
Intrinsic motivation is derived from a direct relationship between the person and the situation, or from intangible factors. One example of this is when someone writes poems for their personal reading pleasure. A person who sings and derives satisfaction out of singing is relying on intrinsic motivation.
Intrinsic motivation is the opposite of extrinsic motivation, which drives someone to perform an activity either to receive an external reward or to avoid an external punishment. An example of extrinsic motivation is when someone writes poems to sell them.
Intrinsic Motivation in Organizations
In any organization, regardless of the industry, intrinsic motivation is an important element to getting maximum productivity from employees. Employers tend to look for potential employees who have a strong sense of intrinsic motivation for doing the work that they would be responsible for.
Strong intrinsic motivation can help individuals to achieve their goals for career growth and development. Here are some areas where internal motivation can be beneficial for both the employee and employer:
In general, the more responsibilities an individual assumes in the workplace, the higher their salary becomes. When a person becomes a manager, they receive higher compensation than their co-workers or subordinates. However, some people do not accept a managerial position primarily to earn more money.
Some employees possess the intrinsic motivation of assuming more responsibilities in order to inspire or encourage others to step up, achieve their own personal goals, and climb the corporate ladder themselves. Added responsibility may also carry the intangible reward of making someone feel that they are playing a direct role in shaping the future of the organization.
Most companies offer rewards and recognition programs that offer incentives such as monetary prizes or paid vacation leaves. However, some employees with strong intrinsic motivation are content with simply being considered a valuable member of the organization, regardless of any extra financial rewards. They are likely to derive more pleasure from being recognized for their hard work than they derive from receiving the external rewards that accompany the recognition.
A strong internal motivation for some employees is the pursuit of knowledge. They want to learn more about their field and hone their skills. Companies can offer a range of training courses to obtain maximum benefit from employees who are internally motivated to want to expand their knowledge and skills.
In many cases, employers don’t need to make grand gestures or spend a lot to fuel the intrinsic motivation of employees and give them a sense of accomplishment at the workplace. Many people feel a sense of pride and accomplishment from helping other people, without expecting anything in return. Just saying, “Thank you!”, is more than enough for people with strong intrinsic motivation.
Thank you for reading CFI’s explanation of intrinsic motivation. CFI offers the Financial Modeling & Valuation Analyst (FMVA)™ certification program for those looking to take their careers to the next level. To keep learning and advancing your career, the following resources will be helpful:<|endoftext|>
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Background to recent estimates of nitrogen fixation
Nitrogen-fixing legumes such as peas and beans, clovers and vetches, were the main source of nitrogen in croplands until the early 1900s. They were extensively tested and grown in the Improvements era after 1700, and became an essential part of crop rotations, even though the biological mechanism of fixation was not then known. By fixing N gas from the air, they could provide their own nitrogen for growth, and the residues they left in the soil, in the form of roots and root nodules, provided much of the N for the next crop also.
Legumes receded in the 1900s, displaced by mineral N fertiliser, and then by imports of N-rich stock-feed from the Americas. Agriculture in north west Europe soon became dependent on manufactures and imports. However, the rising cost of N-pollution, the monetary cost of the fertiliser and the lost benefits, now increasingly realised, of legumes to the wider food web and soil structure, has brought on a reappraisal of legumes in mainstream crop production.
So it is incongruous that there are very few measurements of N fixation in northern, temperate-zone, agricultural fields. Individual measures of fixation in crops and pastures are not in themselves considered to be high-profile science, compared to, say, the insights to the mechanism of fixation at biochemical and genomic levels. Yet without knowing kilograms of nitrogen per hectare (kg/ha) fixed in real crops and pastures, it is difficult to design and justify new crop and forage systems in which manufactures and imports are replaced by local N.
The recent paper by Pete Iannetta and others in Frontiers in Plant Science fills in some of the gaps. The contribution of N fixation was estimated in a range of grain and forage legumes, sometimes grown alone and sometimes with other crops in mixtures. The analysis was based on existing data from field experiments managed by collaborating groups in different parts of Europe. The study added value to these data by estimating components of the N-balance, including fixation as the mass of N entering the system that could not be accounted for by any other means.<|endoftext|>
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# Thread: Calculating area from a graph.
1. ## Calculating area from a graph.
I have to calculate the area of the shaded graph (in the link). Seeming as no equations are given, how do I calculate it?
Untitled | Flickr - Photo Sharing!
It's hard to read the graph but it's just 0.1π, 0.2π... 2.0π.
2. To me it seems like it is a sine and cosine graph.
You would need to develop an equation to be able to calculate the area accurately:
$y = a*sin(bx)$
$y = a * cos(bx)$
Where a and b are constant values.
To me it seems its maximum is 4.5 and minimum is -4.5. Thus constant a must be 4.5.
One full cycle seems to be 2.0π. If you are working in radians, constant b must be $2\pi$/2. If working in degrees, b = 360/2.
$y = 4.5*sin(\pi {x})$
$y = 4.5 * cos(\pi {x})$
There once you have your 2 equations, you can integrate between 0 and 2 to find the area.
3. Of course! Thankyou so much.
Can I just ask, how do you know to integrate between 0 and 2?
4. Oh sorry I misinterpereted the figures on your graph.
Now I take a closer look, 2.0π actually meant $2\pi$. I thought π was the units...
Forget what I said about b being a constant. It stretches out like a normal graph - from 0 to $2\pi$.
Thus:
$y = 4.5*sin(x)$
$y = 4.5 * cos (x)$
To make finding the area of the shaded part easier, move the graph up by 4.5 so that no part of it is in the negative part, below the x-axis.
$y = 4.5*sin(x)+4.5$
$y = 4.5 * cos (x)+4.5$
To find the area:
Integrate between 0 to $\frac{1}{4}\pi$, and do the cosine graph subtracted the sine graph.
Then integrate between $\frac{1}{4}\pi$ to $\frac{5}{4}\pi$, and do the sine graph subtracted the cosine graph this time.
Finally integrate between $\frac{5}{4}\pi$ to $\frac{8}{4}\pi$, and do the cosine graph subtracted the sine graph.
5. Also the sine and cosine graphs go onto infinity - I'm only guessing the shaded parts are only between 0 to $2\pi$ as that is as far as the graph goes.<|endoftext|>
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# Math puzzle: I know the answer, what is the equation? Use all digits 1-9
A few years ago, my child brought a puzzle home from their 6th-grade math teacher. The solution was to use all the numbers $$1−9$$ only once and make the equation true. Using a spreadsheet, I was able to calculate the 2 answers: $$631892754$$ and $$631472958$$
My problem is I found the spreadsheet and the answers but I don’t remember the original problem. I have tried to recreate it using my spreadsheet formulas, but just not getting anywhere.
I think it was in a format similar to _ _ * _ _ * _ = _ _ ∗_ _ (each underscore a different number $$1-9$$) Might even have some parenthesizes. I don’t think it had any division, but I could be wrong. I am also fairly certain it was a single equation. I remember showing it to a co-worker the next day and I was able to write the question down from memory. It wasn’t overly complicated since I was able to figure out quickly what was required and it didn’t require any advanced math.
These are my formulas from the spreadsheet for the first solution $$631892754$$.
Step 1: Using all combinations of digits 1, 2, and 5.
Digit $$1$$ multiplied by digit $$5$$, divide by digit $$2$$. Result is digits $$3$$ and $$4$$.
Answer must be a whole number greater than 10, cannot end in 0.
Using the first answer $$(6∗9)/3=18$$
Step 2: Using the digits and the answers from Step 1.
Digit $$5$$ multiplied by digit $$1$$ must equal digits $$3$$ and $$4$$ multiplied by digit $$2$$
Using the first answer $$9∗6=18∗3$$
Step 3: I have 5 numbers, determine which numbers have not been used for each potential answer. Tried all combinations of the remaining 4 digits until I got the answer.
Digit $$6$$ and $$7$$ multiplied by digit $$1$$ must equal digits $$8$$ and $$9$$ multiplied by digit $$2$$
$$27∗6=54∗3$$
I spent a few hours searching the Internet for this puzzle, but no luck. I am hoping the above information is enough to go on.
• computers allowed? or is there a logival solution? Oct 23, 2021 at 3:26
• Step 2 doesn't add any constraints.
– Muqo
Oct 23, 2021 at 8:57
• This is truly a question not a riddle or similar. I know the answer and I am looking for the original question. I have no limitations on what tools you can use, including a computer. Oct 23, 2021 at 20:40
I'm going against some of the things you suggested here but one possibility I've found that could work is the following (where each space must be filled by a unique digit between 1 and 9)
_ ÷ _ = _ _ ÷ _ = _ _ ÷ _ _
Framing it in this way makes it look like a single equation. I checked using a computer and this has exactly two solutions
6 ÷ 3 = 14 ÷ 7 = 58 ÷ 29
6 ÷ 3 = 18 ÷ 9 = 54 ÷ 27
which look like yours but with the end pairs switched.
• @Heximo: thank you for the equation. It very well could be an equation with 2 equals. That would explain the math I was doing. I am thinking the last set of numbers are just reversed because the way I rewrote the equation to do the calculations. Thank you! Oct 23, 2021 at 21:24
Building on hexomino's idea and doing a different kind of cheat from the two equal signs thing:
$$\sqrt{(6 ÷ 3)(18 ÷ 9)}\cdot27 = 54$$
$$\sqrt{(6 ÷ 3)(14 ÷ 7)}\cdot29 = 58$$
However, a quick python program reveals that there are more solutions than just those two:
1 3 9 6 8 2 7 5 4
1 6 2 4 9 5 7 3 8
1 7 2 8 9 5 4 3 6
1 8 9 6 3 2 7 5 4
1 9 2 4 6 5 7 3 8
1 9 2 8 7 5 4 3 6
2 4 5 6 7 1 9 3 8
2 7 5 6 4 1 9 3 8
4 2 3 6 8 1 9 5 7
4 8 3 6 2 1 9 5 7
6 3 1 4 7 2 9 5 8
6 3 1 8 9 2 7 5 4
6 7 1 4 3 2 9 5 8
6 8 2 7 9 3 4 5 1
6 9 1 8 3 2 7 5 4
6 9 2 7 8 3 4 5 1
7 3 8 4 9 1 2 5 6
9 3 5 4 8 1 6 7 2
9 8 5 4 3 1 6 7 2
However, these don't all solve the equations in the OP's 1 (=2) and 3. Hexomino solved both equations for $$d_1/d_2$$ which matches the constraints and the solution, so seems close, although the swapping of the last two pairs seems odd (equation 3 is $$\frac{d_1}{d_2}=\frac{d_8d_9}{d_6d_7}$$).<|endoftext|>
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In refractive errors the image of the object a person is looking at is not focussed properly onto the retina (the light-sensitive tissue in the back of the eye).
For perfectly clear vision, the image of a viewed object needs to be focussed onto the retina, just as a camera has to be focussed properly in order to take a clear picture. If the image is not focussed exactly on the retina, then the image will be blurred, just like an out-of-focus photograph. In this case, the person is said to have a refractive error.
Refractive errors occur when there is a mismatch between the length of the eye and its optical power. These mismatches usually originate during childhood when the eyes are growing.
The exact causes of refractive errors are still being studied, but it is known that both hereditary and environmental influences can affect their development. Most people have some refractive error, but in most cases the error is small, and does not cause any problems. In fact, the average person is slightly longsighted.
Refractive errors can usually be corrected using spectacles or contact lenses. There are also surgical techniques which can be used to correct refractive errors.<|endoftext|>
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## Joint Distributions¶
Suppose $X$ and $Y$ are two random variables defined on the same outcome space. We will use the notation $P(X = x, Y = y)$ for the probability that $X$ has the value $x$ and $Y$ has the value $y$.
The joint distribution of $X$ and $Y$ consists of all the probabilities $P(X=x, Y=y)$ where $(x, y)$ ranges over all the possible values of $(X, Y)$.
#### Example¶
In three tosses of a coin, let $X$ be the number of heads in the first two tosses and $Y$ the number of heads in the last two tosses. Then
$$P(X = 0, Y = 2) = 0 = P(X = 2, Y = 0)$$$$P(X = 1, Y = 1) = P(\text{THT or HTH}) = \frac{2}{8}$$
All the other probabilities are $1/8$, as you can see by examining the six remaining outcomes of three tosses. For example,
$$P(X = 1, Y = 2) = P(\text{THH}) = \frac{1}{8}$$
The constraints on $x$ and $y$ are that each must be in the range $\{0, 1, 2\}$ and $\vert x - y \vert < 2$. Let's define a function that takes $x$ and $y$ as its arguments and returns $P(X = x, Y = y)$.
def joint_probability(x, y):
if x == 1 & y == 1:
return 2/8
elif abs(x - y) < 2:
return 1/8
else:
return 0
The prob140 library contains Table methods for displaying the joint distribution of two random variables. As a first step, you need the possible values of each of the two variables. In our example, both have values $\{0, 1, 2\}$ and so the same list or array will serve for both.
k = np.arange(3)
To construct a joint distribution object, we have to first construct a table of all possible pairs of values and probabilities. The call is:
Table().values(variable_name_1, values_1, variable_name_2, values_2).probability_function(function_name)
where function_name is a function that takes $x$ and $y$ as arguments and returns $P(X = x, Y = y)$.
joint_table = Table().values('X', k, 'Y', k).probability_function(joint_probability)
joint_table
X Y Probability
0 0 0.125
0 1 0.125
0 2 0
1 0 0.125
1 1 0.25
1 2 0.125
2 0 0
2 1 0.125
2 2 0.125
This table displays the joint distribution. To check that this is indeed a distribution, we can add up all the probabilities. The sum is 1, as it should be for a distribution.
joint_table.column(2).sum()
1.0
### Joint Distribution Table¶
Though the table above does display the joint distribution, it is more conventional and also more illuminating to visualize the same information in a different way.
The method to_joint converts the table above into a "joint distribution object" that is displayed as a conventional joint distribution table for $X$ and $Y$.
joint_dist = joint_table.to_joint()
joint_dist
X=0 X=1 X=2
Y=2 0.000 0.125 0.125
Y=1 0.125 0.250 0.125
Y=0 0.125 0.125 0.000
Each cell corresponds to a pair $(x, y)$, where $x$ is a value of $X$ and $y$ a value of $Y$. In the cell you see $P(X = x, Y = y)$, the probability of the pair $(x, y)$.
Joint distribution tables are analogous to the contingency tables you saw in Data 8 when you were analyzing the relation between two categorical variables. In contingency tables, each cell contains the number of individuals in one particular pair of categories. In joint distribution tables, such as the one above, each cell contains the probability of one particular pair of values.
### Finding Probabilities¶
The table contains complete information about the relation between $X$ and $Y$. To find the probabiilty of any event determined by $X$ and $Y$, simply identify the cells that make the event happen, and add up their chances. This is an application of the fundamental method of finding probabilities by partitioning an event.
For example,
\begin{align*} P(X > Y ) &= P(X = 1, Y = 0) + P(X = 2, Y = 0) + P(X = 2 , Y = 1) \\ &= 0.125 + 0 + 0.125 \\ &= 0.25 \end{align*}
And
\begin{align*} P(X = Y ) &= P(X = 0, Y = 0) + P(X = 1, Y = 1) + P(X = 2 , Y = 2) \\ &= 0.125 + 0.25 + 0.125 \\ &= 0.5 \end{align*}<|endoftext|>
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# Points in a Semicircle
### Solution 1, Question 1
Associate with every point $X$ on a circle the semicircle arc $(X$ of which it is the midpoint. We claim that a set of points $\mathcal{X}$ on a circle lie in a semicircle only if the union of the associated semicircles does not completely cover the circle:
Let $(C)$ be the circumference of a circle (to avoid ambiguity); $\mathcal{X}=\{X_k:~X_k\in (C),~k=1,\ldots,n\},$ $n\gt 1$ an integer. Two statements below are equivalent
1. There is point $P\in (C),$ such that $\mathcal{X}\subset (P.$
2. $\displaystyle (C)\setminus\bigcup_{k=1}^n(X_k\ne\emptyset.$
Thus the original problem is equivalent to the one discussed earlier, where the probability was found to be $\displaystyle 1-2^{-n+1}n$ for the semicircles to cover $(C)$ completely, which implies that for a set $\mathcal{X}$ of $n$ random points on $(C)$ the probability of them all being covered by a semicircle is $\displaystyle \frac{n}{2^{n-1}}.$
### Solution 2, Question 1
Case 1: $n$ is $1$ or $2.$
By inspection, the answer is $1.$
Case 2: $n \gt 2$
Let $P$ be any set of $n\gt 2$ points distributed uniformly at random on the circumference of a circle. With probability $1,$ the $n$ points are distinct. For any point $p\in P,$ let $S(p)$ be the semicircle starting at $p$ (inclusive) and continuing in the counterclockwise direction. If all $n$ points lie on the same semicircle, then there exists a point $p\in P$ such that $P \subseteq S(p).$
Choose a point $p \in P$ arbitrarily, and let $q$ be any one of the other $n-1$ points. The probability that $q \in S(p)$is $1/2.$ Therefore, the probability that $S(p)$contains all the other $n-1$ points is $(1/2)^{n-1}.$ Since, by definition, $S(p)$contains $p,$ the probability that $S(p)$ contains all $n$ points, that is, that $P \subseteq S(p),$ is also $(1/2)^{n-1}.$
When $n \gt 2,$ there is at most one point $p \in P$ such that $P \subseteq S(p).$ Therefore, we can sum the individual probabilities for each point $p \in P$ to obtain the probability that there exists any $p \in P$ such that $P \subseteq S(p).$ This sum is $n(1/2)^{n-1}.$ Therefore, the probability that all $n \gt 2$ points lie on the same semicircle is $n(1/2)^{n-1}.$
Combining cases
Note that $n(1/2)^{n-1} = 1$ when $n$ is either $1$ or $2.$ That is the same answer obtained in Case 1. Therefore, the expression from Case 2 gives the correct result for any positive integer $n,$ and the probability that all $n$ points lie on the same semicircle is $n(1/2)^{n-1}.$
### Solution, Question 2
We make an assumption that $\phi\lt\pi.$
For a point $X\in(C),$ let $\Phi(X)$ be the arc subtending angle $\phi$ that extends counterclockwise from $X,$ inclusive of $X.$ The probability of a point uniformly distributed on $(C)$ to fall into $\Phi(X)$ equals $\displaystyle \frac{\phi}{2\pi}.$
Given $n\gt 2$ points $X_1,X_2,\ldots,X_n,$ the probability that $X_i\in\Phi(X_k),$ $i\ne k,$ is the same $\displaystyle \frac{\phi}{2\pi}$ and, for all $i\ne k,$ is $\displaystyle \left(\frac{\phi}{2\pi}\right)^{n-1}.$
For different $k,$ the events $E_k:~(\forall i\ne k) X_i\in\Phi(X_k)$ do not intersect, since $X_i\in\Phi(x_k)$ is incompatible with $X_k\in\Phi(X_i).$ It follows that the probability of the event $\displaystyle \sum_{k=1}^nE_k$ equals $\displaystyle n\left(\frac{\phi}{2\pi}\right)^{n-1}.$
Now, if for some point $Y,$ $(\forall i)X_i\in\Phi(Y),$ then, for $X_k$ nearest to $Y$ in the counterclockwise direction, $(\forall i\ne k)X_i\in\Phi(X_k).$ And, as we found, this event has the probability of $\displaystyle n\left(\frac{\phi}{2\pi}\right)^{n-1}.$
### Acknowledgment
This is Problem 10 from an assignment for the MIT, 18.S34 (FALL 2007)
Solution 2 to Question 1 is by Joshua Jordan. Solution to Question 2 follows Josh's reasoning.<|endoftext|>
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Radioactive materials decay, or lose their radioactivity, in a predictable way. If you have a fast decaying source, such as Protactinium with a half life of just 72 seconds, you can take a series of radioactivity readings and plot them on a graph. Or you can use a Geiger-Muller tube and radioactivity sensor connected to a computer. This can plot a radioactive decay curve and display it as it forms on the screen. Later you can analyse the results using your software.
Geiger-Muller tube / radioactivity sensor, clamp stand, radioactive source such as a Protactinium generator, data logging interface.
Connect the Geiger-Muller tube to the sensor and the sensor to the interface. If the sensor is adjustable, set it to a suitable range.
Start your sensing software.
Set your sensing software to record for around 10 minutes. The exact time will depend the strength of the source. If you are using a Protactinium generator, give it a shake and then start recording.
How is decreasing radioactivity shown on your graph?
Does the radioactivity change in a steady ‘straight-line’ fashion?
Is the graph ‘noisy’? Why might this be?
Use the software to calculate Ln (count rate) and plot this against time. How is this graph different? What does it tell us?
Use the software to perform a least squares fit on the decay curve.
You may have difficulty in obtaining replacement Protactinium generators as the suppliers have discontinued it because of leakage problems. What alternative did you find?<|endoftext|>
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Pedro de Alvarado y Contreras (Spanish pronunciation: [ˈpeðɾo ðe alβaˈɾaðo]; Badajoz, Extremadura, Spain, ca. 1485 – Guadalajara, New Spain, 4 July 1541) was a Spanish conquistador and governor of Guatemala. He participated in the conquest of Cuba, in Juan de Grijalva's exploration of the coasts of the Yucatán Peninsula and the Gulf of Mexico, and in the conquest of Mexico led by Hernán Cortés. He is considered the conquistador of much of Central America, including Guatemala, Honduras and El Salvador. Although renowned for his skill as a soldier, Alvarado is known also for the cruelty of his treatment of native populations, and mass murders committed in the subjugation of the native peoples of Mexico.
Historiography portrays that indigenous people, both Nahuatl-speakers and speakers of other languages, called him Tonatiuh, meaning "sun" in the Nahuatl language. Yet he was also called "Red Sun" in Nahuatl, which allows a variety of interpretations. Whether this epithet refers to Alvarado's red hair, some esoteric quality attributed to him, or both, is disputed.
Pedro de Alvarado
Pedro de Alvarado in a contemporaneous rendition.
|Died||July 4, 1541 (aged c. 55–56)|
Pedro de Alvarado was flamboyant and charismatic, and was both a brilliant military commander and a cruel, hardened man. His hair and beard were red, which earned him the name of Tonatiuh from the Aztecs, the name of one of their sun gods. He was handsome, and presented an affable appearance, but was volatile and quick to anger. He was ruthless in his dealings with the indigenous peoples he set out to conquer. Historians judge that his greed drove him to excessive cruelty, and his Spanish contemporaries denounced his extreme brutality during his lifetime. He was a poor governor of territories he had conquered, and restlessly sought out new adventures.
His tactical brutality, such as the massacre in the Great Temple of Tenochtitlan, often undermined strategic considerations. He was also accused of cruelty against fellow Spaniards. Alvarado was little suited to govern; when he held governing positions, he did little to establish stable foundations for colonial rule. His letters show no interest in civil matters, and he only discussed exploration and war. Alvarado stubbornly resisted attempts by the Spanish Crown to establish ordered taxation in Guatemala, and refused to acknowledge such attempts. As governor of Guatemala, Alvarado has been described by W. George Lovell et al. as "an insatiable despot who recognized no authority but his own and who regarded Guatemala as little more than his personal estate."
American historian William H. Prescott described Alvarado's character in the following terms:
Alvarado was a cavalier of high family, gallant and chivalrous, and [Cortes'] warm personal friend. He had talents for action, was possessed of firmness and intrepidity, while his frank and dazzling manners made the Tonatiuh an especial favourite with the Mexicans. But, underneath this showy exterior, the future conqueror of Guatemala concealed a heart rash, rapacious, and cruel. He was altogether destitute of that moderation, which, in the delicate position he occupied, was a quality of more worth than all the rest.— William H. Prescott 1922, History of the Conquest of Mexico: Book 4, Chapter 8, p. 54.
Spanish chronicler Antonio de Remesal commented that "Alvarado desired more to be feared than loved by his subjects, whether they were Indians or Spaniards." In his easy recourse to violence, Alvarado was a product of his time, and Alvarado was not the only conquistador to have resorted to such actions. Hernán Cortés and Francisco Pizarro carried out deeds of similar cruelty, but have not attracted as much criticism as Alvarado.
Pedro de Alvarado was born in 1485 in the town of Badajoz, Extremadura. His father was Gómez de Alvarado, and his mother was Leonor de Contreras, Gómez's second wife. Pedro de Alvarado had a twin sister, Sarra, and four full-blood brothers, Jorge, Gonzalo, Gómez, and Juan. Pedro had an illegitimate half brother, also named Juan, referred to in contemporary sources as Juan el Bastardo.
Very little is known of Pedro de Alvarado's early life before his arrival in the Americas. During the conquest of the Americas, tales of his youthful exploits in Spain became popular legends, but their veracity is doubtful. An example is the tale then current that when he was a youth awaiting passage to the Americas, he climbed the church tower in Seville with some friends. A banner pole extended some 3.0 to 3.7 metres (10 to 12 ft) from an upper window. One of his companions walked out to the end of the pole after removing his cloak and sword, and returned to the tower backwards. Alvarado, afraid of being mocked, walked out onto the pole with both sword and cloak, and turned around at the end to return to the tower facing it.
Alvarado's paternal grandfather was Juan Alvarado "el Viejo" ("the elder"), who was comendador of Hornachos, and his paternal grandmother was Catalina Messía. Pedro de Alvarado's uncle on his father's side was Diego de Alvarado y Messía, who was the comendador of Lobón, Puebla, and Montijo, alcalde of Montánchez, and lord of Castellanos and of Cubillana. Diego was a veteran of the campaigns against the Moors.
Alvarado and his brothers crossed the Atlantic Ocean before 1511, possibly in 1510. By 1511 a system of licenses had been established in Spain to control the flow of colonists to the New World. The only one of the Alvarado brothers that appears in the registers is Juan de Alvarado, in 1511, leading to the assumption that the rest were already in the Americas by the time the licensing system was established. The Alvarado brothers stopped off at Hispaniola, but there are few mentions of their stay there in historical documents.
Soon after arriving in Santo Domingo, on Hispaniola, Pedro de Alvarado established a friendship with Hernán Cortés, who at the time was serving as public scribe. Alvarado joined Cortés to participate in the conquest of Cuba, under the command of Diego de Velázquez. The conquest of Cuba was launched in 1511, and Pedro de Alvarado was accompanied by his brothers. Soon after the invasion, Alvarado was managing a prosperous hacienda in the new colony. It is around this time that Pedro de Alvarado emerges into the historical record as a prosperous and influential hacienda-owner, already well connected with Velázquez, who was now governor of Cuba.
Diego Velázquez, the governor of Cuba, was enthused by Francisco Hernández de Córdoba's report of gold in the newly discovered Yucatán Peninsula. He organised an expedition consisting of four ships and 260 men. He placed his nephew Juan de Grijalva in overall command; Pedro de Alvarado captained one of the ships. The small fleet was stocked with crossbows, muskets, barter goods, salted pork and cassava bread.
The fleet left Cuba in April 1518, and made its first landfall upon the island of Cozumel, off the east coast of Yucatán. The Maya inhabitants of Cozumel fled the Spanish; the fleet then sailed south from Cozumel, along the east coast of the peninsula. The Spanish spotted three large Maya cities along the coast. On Ascension Thursday the fleet discovered a large bay, which the Spanish named Bahía de la Ascensión.
Grijalva did not land at any of these cities and turned back north to loop around the north of the Yucatán Peninsula and sail down the west coast. At Campeche the Spanish opened fire against the city with small cannon; the inhabitants fled, allowing the Spanish to take the abandoned city. The Maya remained hidden in the forest, so the Spanish boarded their ships and continued along the coast.
At Champotón, the fleet was approached by a small number of large war canoes, but the ships' cannon soon put them to flight. At the mouth of the Tabasco River the Spanish sighted massed warriors and canoes but the natives did not approach. By means of interpreters, Grijalva indicated that he wished to trade and bartered wine and beads in exchange for food and other supplies. From the natives they received a few gold trinkets and news of the riches of the Aztec Empire to the west. The expedition continued far enough to confirm the reality of the gold-rich empire, sailing as far north as Pánuco River.
At the Papaloapan River, Alvarado ordered his ship upriver, leaving the rest of the small fleet behind to wait for him at the river mouth. This action greatly angered Grijalva, who feared that a lone ship could be lost. After this, the Spanish referred to the river as the Río de Alvarado ("Alvarado's River"). A little further along the coast, the fleet encountered settlements under Aztec dominion, and was met by Aztec emissaries with gifts of gold and jewels sent by the Emperor Moctezuma II.
As punishment for entering the Papaloapan River without orders, Grijalva sent Alvarado with the ship San Sebastián to relay news of the discoveries back to Cuba. Alvarado made a triumphal entry to Santiago de Cuba, with a great display of the wealth that had been gained from the expedition. His early arrival in Cuba allowed him to ingratiate himself with the Governor Velázquez before Grijalva's return. The rest of the fleet put into the port of Havana five months after it had left. Grijalva was coldly received by the governor, who Alvarado had turned against him, claiming much of the glory of the expedition for himself.
Grijalva's return aroused great interest in Cuba. A new expedition was organised, with a fleet of eleven ships carrying 500 men and some horses. Hernán Cortés was placed in command; Pedro de Alvarado and his brothers Jorge, Gómez and Juan "El Bastardo" joined the expedition. Cortés charged Pedro de Alvarado with gathering recruits from the inland estates of Cuba. The crew included officers that would become famous conquistadors, including Cristóbal de Olid, Gonzalo de Sandoval and Diego de Ordaz. Also aboard were Francisco de Montejo and Bernal Díaz del Castillo, veterans of the Grijalva expedition.
Alvarado once again commanded the San Sabastián, with 60 men under his orders. The fleet made its first landfall at Cozumel, and remained there for several days. Maya temples were cast down and a Christian cross was put up on one of them. From Cozumel, the fleet looped around the north of the Yucatán Peninsula and followed the coast to the Tabasco River. In Tabasco, the fleet anchored at Potonchán, a Chontal Maya town. The Maya prepared for battle but the Spanish horses and firearms quickly decided the outcome. From Potonchan, the fleet continued to San Juan de Ulua. The crew stayed only a short time before relocating to a promontory near Quiahuiztlan and Cempoala in Veracruz, a subject city of the Aztec Empire, and from there went on to conquer the Aztecs.
Alvarado commanded one of the eleven vessels in the fleet and also acted as Cortés' second in command during the expedition's first stay in the Aztec capital city of Tenochtitlán. Relations between the Spaniards and their hosts were uneasy, especially given Cortés' repeated insistence that the Aztecs desist from idol worship and human sacrifice; in order to ensure their own safety, the Spaniards took the Aztec king Moctezuma hostage. When Cortés returned to the Gulf coast to deal with the newly arrived hostile expedition of Pánfilo de Narváez, Alvarado remained in Tenochtitlan as commander of the Spanish enclave, with strict orders to make sure that Moctezuma not be permitted to escape.
During Cortés' absence, relations between the Spaniards and their hosts went from bad to worse, and Alvarado ordered a preemptive slaughter of Aztec nobles and priests observing a religious festival.:283–286 When Cortés returned to Tenochtitlan, he found the Spanish force under siege. After Moctezuma was killed in the attempt to negotiate with his own people, the Spaniards determined to escape by fighting their way across one of the causeways that led from the city across the lake and to the mainland.:286,294,296 In a bloody nocturnal action of 10 July 1520, known as La Noche Triste, Alvarado led the rear-guard and was badly wounded.:296–300 According to satirical verses by Gonzalo Ocampo, in reference to Alvarado crossing a causeway gap during the escape, Alvarado's escape became known as Salto de Alvarado ("Alvarado's Leap").:296–300
Pedro then participated in the Siege of Tenochtitlan, commanding one of four forces under Cortes.:315,319,333,351,355–356,358,360,363,369–370,372 Alvarado was wounded when Guatemoc attacked all three Spanish camps on the feast day of St. John.:377–378,381,384–385,388–389 Alvarado's company was the first to make it to the Tlateloco marketplace, setting fire to the Aztec shrines. Cortes' and Sandoval's companies joined him there after four more days of fighting.:396-308
Cortés despatched Pedro de Alvarado to invade Guatemala with 180 cavalry, 300 infantry, crossbows, muskets, 4 cannons, large amounts of ammunition and gunpowder, and thousands of allied Mexican warriors. Pedro de Alvarado passed through Soconusco with a sizeable force in 1523, en route to conquer Guatemala. Alvarado's army included hardened veterans of the conquest of the Aztecs, and included cavalry and artillery; there were also a great many indigenous allies from Cholula, Tenochtitlan, Texcoco, Tlaxcala, and Xochimilco.
Alvarado was received in peace in Soconusco, and the inhabitants swore allegiance to the Spanish Crown. They reported that neighbouring groups in Guatemala were attacking them because of their friendly outlook towards the Spanish. Alvarado's letter to Hernán Cortés describing his passage through Soconusco is lost, and knowledge of events there come from the account of Bernal Díaz del Castillo, who was not present, but related the report of Gonzalo de Alvarado. By 1524, Soconusco had been completely pacified by Alvarado and his forces.
Pedro de Alvarado and his army advanced along the Pacific coast unopposed until they reached the Samalá River in western Guatemala. This region formed a part of the K'iche' kingdom, and a K'iche' army tried unsuccessfully to prevent the Spanish from crossing the river. Once across, the conquistadors ransacked nearby settlements in an effort to terrorise the K'iche'. On 8 February 1524 Alvarado's army fought a battle at Xetulul, called Zapotitlán by his Mexican allies (modern San Francisco Zapotitlán). Although suffering many injuries inflicted by defending K'iche' archers, the Spanish and their allies stormed the town and set up camp in the marketplace.
Alvarado then turned to head upriver into the Sierra Madre mountains towards the K'iche' heartlands, crossing the pass into the fertile valley of Quetzaltenango. On 12 February 1524 Alvarado's Mexican allies were ambushed in the pass and driven back by K'iche' warriors but the Spanish cavalry charge that followed was a shock for the K'iche', who had never before seen horses. The cavalry scattered the K'iche' and the army crossed to the city of Xelaju (modern Quetzaltenango) only to find it deserted.
Almost a week later, on 18 February 1524, a K'iche' army confronted the Spanish army in the Quetzaltenango valley and were comprehensively defeated; many K'iche' nobles were among the dead. This battle exhausted the K'iche' militarily and they asked for peace and offered tribute, inviting Pedro de Alvarado into their capital Q'umarkaj, which was known as Tecpan Utatlan to the Nahuatl-speaking allies of the Spanish. Alvarado was deeply suspicious of the K'iche' intentions but accepted the offer and marched to Q'umarkaj with his army.
In March 1524 Pedro de Alvarado entered Q'umarkaj at the invitation of the remaining lords of the K'iche' after their catastrophic defeat, fearing that he was entering a trap. He encamped on the plain outside the city rather than accepting lodgings inside. Fearing the great number of K'iche' warriors gathered outside the city and that his cavalry would not be able to manoeuvre in the narrow streets of Q'umarkaj, he invited the leading lords of the city, Oxib-Keh (the king) and Beleheb-Tzy (the king elect) to visit him in his camp.
As soon as they did so, he seized them and kept them as prisoners in his camp. The K'iche' warriors, seeing their lords taken prisoner, attacked the Spaniards' indigenous allies and managed to kill one of the Spanish soldiers. At this point Alvarado decided to have the captured K'iche' lords burnt to death, and then proceeded to burn the entire city. After the destruction of Q'umarkaj and the execution of its rulers, Pedro de Alvarado sent messages to Iximche, capital of the Kaqchikel, proposing an alliance against the remaining K'iche' resistance.
On 14 April 1524, soon after the defeat of the K'iche', the Spanish were invited into Iximche and were well received by the lords Belehe Qat and Cahi Imox.[nb 1] The Kaqchikel kings provided native soldiers to assist the conquistadors against continuing K'iche' resistance and to help with the defeat of the neighbouring Tz'utuhil kingdom. The Spanish only stayed briefly in Iximche before continuing through Atitlán, Escuintla and Cuscatlán. The Spanish returned to the Kaqchikel capital on 23 July 1524 and on 27 July, Pedro de Alvarado declared Iximche as the first capital of Guatemala, Santiago de los Caballeros de Guatemala ("St. James of the Knights of Guatemala").
The Kaqchikel appear to have entered into an alliance with the Spanish to defeat their enemies, the Tz'utujil, whose capital was Tecpan Atitlan. Pedro de Alvarado sent two Kaqchikel messengers to Tecpan Atitlan at the request of the Kaqchikel lords, both of whom were killed by the Tz'utujil. When news of the killing of the messengers reached the Spanish at Iximche, the conquistadors marched against the Tz'utujil with their Kaqchikel allies.
Pedro de Alvarado left Iximche just 5 days after he had arrived there, with 60 cavalry, 150 Spanish infantry and an unspecified number of Kaqchikel warriors. The Spanish and their allies arrived at the lakeshore after a day's hard march, without encountering any opposition. Seeing the lack of resistance, Alvarado rode ahead with 30 cavalry along the lake shore. Opposite a populated island the Spanish at last encountered hostile Tz'utujil warriors and charged among them, scattering and pursuing them to a narrow causeway across which the surviving Tz'utujil fled. The rest of Alvarado's army soon reinforced his party and they successfully stormed the island. This battle took place on 18 April.
The following day the Spanish entered Tecpan Atitlan but found it deserted. Pedro de Alvarado camped in the centre of the city and sent out scouts to find the enemy. They managed to catch some locals and used them to send messages to the Tz'utujil lords, ordering them to submit to the king of Spain. The Tz'utujil leaders responded by surrendering to Pedro de Alvarado and swearing loyalty to Spain, at which point Alvarado considered them pacified and returned to Iximche. Three days after Pedro de Alvarado returned to Iximche, the lords of the Tz'utujil arrived there to pledge their loyalty and offer tribute to the conquistadors. A short time afterwards a number of lords arrived from the Pacific lowlands to swear allegiance to the king of Spain.
Pedro de Alvarado rapidly began to demand gold in tribute from the Kaqchikels, souring the friendship between the two peoples. He demanded that their kings deliver 1000 gold leaves, each worth 15 pesos.[nb 2] The Kaqchikel people abandoned their city and fled to the forests and hills on 28 August 1524. Ten days later the Spanish declared war on the Kaqchikel.
Two years later, on 9 February 1526, a group of sixteen Spanish deserters burnt the palace of the Ahpo Xahil, sacked the temples and kidnapped a priest, acts that the Kaqchikel blamed on Pedro de Alvarado.[nb 3] The Kaqchikel kept up resistance against the Spanish for a number of years. On 9 May 1530, exhausted by the warfare that had seen the deaths of their best warriors and the enforced abandonment of their crops, the two kings of the most important clans returned from the wilds. A day later they were joined by many nobles and their families and many more people; they then surrendered at the new Spanish capital at Ciudad Vieja.
On 8 May 1524, Pedro de Alvarado continued southwards to the Pacific coastal plain with an army numbering approximately 6000,[nb 4] where he defeated the Pipil of Panacal or Panacaltepeque near Izcuintepeque on 9 May. Alvarado described the terrain approaching the town as very difficult, covered with dense vegetation and swampland that made the use of cavalry impossible; instead he sent men with crossbows ahead. The Pipil withdrew their scouts because of the heavy rain, believing that the Spanish and their allies would not be able to reach the town that day.
Pedro de Alvarado pressed ahead and when the Spanish entered the town the defenders were completely unprepared, with the Pipil warriors indoors sheltering from the torrential rain. In the battle that ensued, the Spanish and their indigenous allies suffered minor losses but the Pipil were able to flee into the forest, sheltered from Spanish pursuit by the weather and the vegetation. Pedro de Alvarado ordered the town to be burnt and sent messengers to the Pipil lords demanding their surrender, otherwise he would lay waste to their lands.
According to Alvarado's letter to Cortés, the Pipil came back to the town and submitted to him, accepting the king of Spain as their overlord. The Spanish force camped in the captured town for eight days. A few years later, in 1529, Pedro de Alvarado was accused of using excessive brutality in his conquest of Izcuintepeque, amongst other atrocities.
In Guazacapán, Pedro de Alvarado described his encounter with people who were neither Maya nor Pipil, speaking a different language altogether; these people were probably Xinca. At this point Alvarado's force consisted of 250 Spanish infantry accompanied by 6,000 indigenous allies, mostly Kaqchikel and Cholutec. Alvarado and his army defeated and occupied the most important Xinca city, named as Atiquipaque. The defending warriors were described by Alvarado as engaging in fierce hand-to-hand combat using spears, stakes and poisoned arrows. The battle took place on 26 May 1524 and resulted in a significant reduction of the Xinca population.
Alvarado's army continued eastwards from Atiquipaque, seizing several more Xinca cities. Because Alvarado and his allies could not understand the Xinca language, Alvarado took extra precautions on the march eastward by strengthening his vanguard and rearguard with ten cavalry apiece. In spite of these precautions the baggage train was ambushed by a Xinca army soon after leaving Taxisco. Many indigenous allies were killed and most of the baggage was lost, including all the crossbows and ironwork for the horses.
This was a serious setback and Alvarado camped his army in Nancintla for eight days, during which time he sent two expeditions against the attacking army. Alvarado sent out Xinca messengers to make contact with the enemy but they failed to return. Messengers from the city of Pazaco, in the modern department of Jutiapa, offered peace to the conquistadors but when Alvarado arrived there the next day the inhabitants were preparing for war. Alvarado's troops encountered a sizeable quantity of gathered warriors and quickly routed them through the city's streets. From Pazaco, Alvarado crossed the Río Paz and entered what is now El Salvador.
Alvarado led the first effort by Spanish forces to extend their dominion to the nation of Cuzcatlan (in modern El Salvador), in June 1524. These efforts established many towns such as San José Acatempa in 1525 and Esquipulas in 1560. Spanish efforts were firmly resisted by the indigenous people known as the Pipil and their Mayan speaking neighbors. Despite Alvarado's initial success in the Battle of Acajutla, the indigenous people of Cuzcatlán, who according to tradition were led by a warlord called Atlacatl, defeated the Spaniards and their auxiliaries, and forced them to withdraw to Guatemala.
Alvarado was wounded on his left thigh, remaining handicapped for the rest of his life. He abandoned the war and appointed his brother, Gonzalo de Alvarado, to continue the task. Two subsequent expeditions were required (the first in 1525, followed by a smaller group in 1528) to bring the Pipil under Spanish control. In 1528 the conquest of Cuzcatlán was completed and the city of San Salvador was established.
On 18 December 1527, the king of Spain named Alvarado as governor of Guatemala; two days later he granted him the coveted military title of Adelantado. Alvarado's close friendship with Cortés was broken in the same year; Alvarado had promised Cortés that he would marry Cecilia Vázquez, Cortes' cousin. Alvarado broke his promise and instead married Francisca de la Cueva.
Francisca de la Cueva was well connected at the royal court, being the niece of Francisco de los Cobos, the king's secretary, and a member of the powerful noble house of Albuquerque. This marriage gave Alvarado extra leverage at court and was far more useful to his long term interests; Alvarado thereafter maintained a friendship with Francisco de los Cobos that allowed him access to the king's favour. In 1528, by coincidence both Alvarado and Cortés were in Seville at the same time, but Cortés ignored him.
By 1532, Alvarado's friendship with Hernán Cortés had soured, and he no longer trusted him. At this time Alvarado requested permission from the king for an expedition south along the Pacific coast, to conquer any lands there that had not already been claimed for the Crown, and specifically rejected that Cortés should accompany him. In 1534, Alvarado heard tales of the riches of Peru, headed south to the Andes and attempted to bring the province of Quito under his rule. When he arrived, he found the land already held by Francisco Pizarro's lieutenant Sebastián de Belalcázar. The two forces of Conquistadors almost came to battle; however, Alvarado bartered to Pizarro's group most of his ships, horses, and ammunition, plus most of his men, for a comparatively modest sum of money, and returned to Guatemala.
In 1532, Alvarado received a Royal Cedula naming him Governor of the Province of Honduras. At that time, Honduras consisted of a single settlement of Spaniards in Trujillo, but he declined to act on it. In 1533 or 1534 he began to send his own work gangs of enslaved Africans and Native Americans into the parts of Honduras adjacent to Guatemala to work the placer gold deposits.
In 1536, ostensibly in response to a letter asking for aid from Andrés de Cereceda, then acting Governor of the Province of Honduras, Alvarado and his army of Indian allies arrived in Honduras, just as the Spanish colonists were preparing to abandon the country and go look for gold in Peru. In June, 1536, Alvarado engaged the indigenous resistance led by Cicumba in the lower Ulua river valley, and won. He divided up the Indian labor in repartimiento grants to his soldiers and some of the colonists, and returned to Guatemala.
During a visit to Spain, in 1537, Alvarado had the governorship of Honduras reconfirmed in addition to that of Guatemala for the next seven years. His governorship of Honduras was not uncontested. Francisco de Montejo had a rival claim, and was installed by the Spanish king as Governor of Honduras in 1540. Ten years after being widowed, Alvarado married one of his first wife's sisters, Beatriz de la Cueva, who outlived him.
After the death of Alvarado, de la Cueva maneuvered her own election and succeeded him as governor of Guatemala, becoming the only woman to govern a major political division of the Americas in Spanish colonial times. She drowned a few weeks after taking office in the destruction of the capital city Ciudad Vieja by a sudden flow from the Volcán de Agua in 1541.
Alvarado developed a plan to outfit an armada that would sail from the western coast of Mexico to China and the Spice Islands. At great cost, he assembled and equipped 13 ships and approximately 550 soldiers for the expedition. The fleet was about to set sail in 1541 when Alvarado received a letter from Cristóbal de Oñate, pleading for help against hostile Indians who were besieging him at Nochistlán.:Ch.203
The siege was part of a major revolt by the Mixtón natives of the Nueva Galicia region of Mexico. Alvarado gathered his troops and went to help Oñate. In a freak accident, he was crushed by a horse that was spooked and ran amok.:Ch.203 He died a few days later, on July 4, 1541, and was buried in the church at Tiripetío, a village between Pátzcuaro and Morelia (in present-day Michoacán).
Four decades after Alvarado's death, his mestiza daughter Leonor de Alvarado Xicoténcatl paid to transport his remains to Guatemala for reburial in the cathedral of the city of Santiago de los Caballeros de Guatemala, now Antigua Guatemala.
After the death of her husband, Beatriz de la Cueva maneuvered her own election and succeeded him as governor of Guatemala, becoming the only woman to govern a major political division of the Americas in Spanish colonial times.
Alvarado had no children from either of his legal marriages. His life companion was his concubine Luisa de Tlaxcala (also called Xicoténcatl or Tecubalsi, her original names after Catholic baptism). She was a Nahua noblewoman, daughter of the Tlaxcallan Chief Xicotencatl the Elder. Luisa was given by her father in 1519 to Hernán Cortés as a proof of respect and friendship. In turn Cortés gave her in guard to Pedro de Alvarado,:178 who quickly and unremarkably became her lover. Luisa followed Alvarado in his pursuit of conquests beyond central Mexico. Despite never being his legitimate wife, Luisa de Tlaxcala had numerous possessions and was respected as a Doña, both for her relationship with Alvarado and for her noble origin. She died in 1535 and was buried at the Guatemala Cathedral.
With Luisa de Tlaxcala Pedro de Alvarado had three children:
By other women, in more casual relationships, he had two other children:
Alvarado (officially: Ilustre, Heroica y Generosa Ciudad y Puerto de Alvarado) is a city in the Mexican state of Veracruz. The city also serves as the municipal seat for the surrounding municipality of the same name. It is located 64 km (40 mi) from the city of Veracruz, Veracruz, on Federal Highways 180 and 125. Alvarado is bordered by Boca del Río, Tlalixcoyan, Medellín, Ignacio de la Llave, Ciudad Lerdo de Tejada, Tlacotalpan and Acula. It is 10 m (33 ft) above sea level. It lies in the so-called "Region Papaloapan" bordered on the south by the municipalities of Acula, Tlacotalpan and Lerdo de Tejada, on the east by the Gulf of Mexico and on the west by Ignacio de la Llave.The geographic size of Alvarado is 840 square kilometres (320 sq mi).
The INEGI said that the population by 2000 was 22,608 people.
Alvarado was founded in the 16th century. In 1518, Pedro de Alvarado arrived to Atlizintla and name the town and river after himself. Then in 1563, Juan de Sahagún built the port of Alvarado. It was used to transport food to other countries.
Major products of Alvarado are coffee, fruits, and sugar. The economy of this region is based on fishing and farming. Sugar cane and coffee are common products of this region.Alvarado family
Alvarado was the Spanish family of conquistadors.
Diego Gómez de Alvarado y Mexía de Sandoval. the Commander of Lobón, Puebla, Montijo and Cubillana, Alcalde of Montánchez, Trece of the Order of Santiago, Lord of Castellanos, a Maestresala official instructor of Henry IV of Castile and General of the Frontier of Portugal. 1st wife: Teresa Suárez de Moscoso y Figueroa; 2nd wife Leonor de Contreras y Gutiérrez de Trejo. His sons:
Pedro de Alvarado, famous conquistador. 1st wife Fransisca de Cueva, 2nd - her cousin Beatriz de la Cueva. Both childless. But more so than his wives his vital companion was Luisa de Tlaxcala (also called Xicoténcalt or Tecubalsi, her original names after Catholic baptism), an Indian noblewoman, daughter of the Tlaxcaltec Chief Xicotenga. With Luisa de Tlaxcala he had three children, and two more from other women (Leonor de Alvarado y Xicotenga Tecubalsi, Pedro de Alvarado, Diego de Alvarardo El Mestizo, Gómez de Alvarado, Ana (Anita) de Alvarado)
Gonzalo de Alvarado y Contreras. His descendants were represented by the family Vides de Alvarado after the famous 17th-century historians Francisco Antonio de Fuentes y Guzmán and also the father Domingo Juarros y Montufar.
Jorge de Alvarado. Jorge married a daughter of Xicotencatl I, the ruler of Tizatlan in Tlaxcala. She was baptized with the Spanish name doña Lucía. They had a daughter who married the conquistador Francisco Xiron Manuel and had issue. Also he married twice, firstly to Francisca Girón and secondly in 1526 to Luisa de Estrada, certainly related to Francisco Vázquez de Coronado's wife, by whom he had a son Jorge de Alvarado y Estrada, born in México, who married Catalina de Villafañe y Carvajal, Mexican, daughter of Ángel de Villafañe, conqueror of Mexico, and wife Inés de Carvajal. Their son was Jorge de Alvarado y Villafañe, also born in Mexico, Governor and Captain-General of Honduras and Knight of Santiago since 1587, also married twice, firstly to Brianda de Quiñones and secondly to Juana de Benavides, vecina of Guatemala, and had issue.
and Gómez, Hernando and Juan.Diego de Alvarado y Mexía de Sandoval, uncle of Pedro de Alvarado.
Gonzalo de Alvarado y Chávez, cousin of Pedro de Alvarado. He married Isabel, a daughter of Jorge de Alvarado, his cousin.
Alonso de Alvarado - ?Battle of Acajutla
The Battle of Acajutla was a battle on June 8, 1524, between the Spanish conquistador Pedro de Alvarado and the standing army of Cuscatlan Pipils, an indigenous state, in the neighborhood of present-day Acajutla, near the coast of western El Salvador.Coanacoch
Coanacochtzin (died 1525) was the seventh tlatoani (ruler) of Texcoco.
One of Nezahualpilli's sons, he succeeded to throne after the death of his half-brother Cacama in 1520.In 1524 Hernán Cortés took Coanacochtzin and many other indigenous rulers with him on his expedition to Honduras. In 1525, at Campeche, Coanacochtzin was executed along with Cuauhtemoc, the tlatoani of Tenochtitlan.
After his death Ixtlilxochitl II, another of Coanacochtzin's half-brothers, was made tlatoani of Texcoco by the Spanish.Francisco de Montejo
Francisco de Montejo y Álvarez (Spanish: [fɾanˈθisko ðe mõnˈtexo]; c. 1479 in Salamanca – c. 1553 in Spain) was a Spanish conquistador in Mexico and Central America.
Francisco de Montejo was born in 1479 in Salamanca, Spain, to Juan de Montejo and Catalina Álvarez de Tejeda. He left Spain in 1514, and arrived in Cuba in time to join Juan de Grijalva's expedition along the coast of Yucatán and the Gulf of Mexico. There he had the rank of Captain, and command of 4 ships. On his return to Cuba, he joined the Hernán Cortés expedition, and helped found the city of Villa Rica de la Vera Cruz with Alonso Hernandez Puertocarrero. Cortés sent Francisco and Alonso as proctors to King Charles of Spain in 1519 to report on the expedition. While in Spain Montejo married Beatriz de Herrera.
In December 1526 the Spanish King, Charles V, Holy Roman Emperor, issued a royal decree naming Montejo Adelantado and Capitan General of Yucatán. He returned to Yucatán in 1528, and attempted to conquer it along the east coast (Tulum, Chetumal) but was driven back by the ferocity of the resistance of the Maya living along this coast. In 1530 he decided to try conquering Yucatán from the west, and began by pacifying what is today the modern Mexican state of Tabasco. From 1531–1535 he tried unsuccessfully to conquer western Yucatán, with some successes but in 1535 his forces were driven from Yucatán. In 1533, Montejo received a royal decree giving him permission to conquer Puerto Caballos and Naco in Honduras. This put him in conflict with Pedro de Alvarado, who had received a similar decree in 1532. This only became an issue after Alvarado declared he had conquered and pacified the province of Honduras in 1536. Alvarado continued as Governor of Honduras until 1540, although he was recalled to Spain in 1537.
In 1540, the Spanish King awarded the Governorship of Honduras to Montejo, and he traveled to Gracias a Dios to install an administration loyal to him.
It would fall to Montejo's son, Francisco de Montejo ("el Mozo") (born 1502, died 1565), to conquer Yucatán. He founded the city of San Francisco de Campeche in 1540, and Mérida in 1542. In 1546, the elder Montejo assumed the title of Governor and Captain General of Yucatán. However, by 1550 complaints about him caused him to be recalled to Spain where he died in 1553.
Montejo was survived by his eponymous son, and a daughter, Catalina Montejo y Herrera.Gonzalo de Alvarado
Gonzalo de Alvarado was the name of two Spanish conquistadors, both related to Pedro de Alvarado and participating in the conquest of Mexico and Central America. Gonzalo de Alvarado y Contreras was his brother, while Gonzalo de Alvarado y Chávez was his cousin.Jorge de Alvarado
Jorge de Alvarado y Contreras (born 1460 Badajoz, Extremadura, Spain – died 1540 or 1541 or Madrid, 1553) was a Spanish conquistador, brother of the more famous Pedro de Alvarado.Kaqchikel people
The Kaqchikel (also called Kachiquel) are one of the indigenous Maya peoples of the midwestern highlands in Guatemala. The name was formerly spelled in various other ways, including Cakchiquel, Cakchiquel, Kakchiquel, Caqchikel, and Cachiquel.
In Postclassic Maya times the capital of the main branch of the Kaqchikel was Iximché. Like the neighboring K'iche' (Quiché), they were governed by four lords: Tzotzil, Xahil, Tucuché and Acajal, who were responsible for the administrative, military and religious affairs. The Kakchikel recorded their history in the book Annals of the Cakchiquels, also known as Memorial de Sololá.
The Chajoma were another Kaqchikel-speaking people; the ruins of Mixco Viejo have been identified as their capital.
Iximché was conquered by the Spanish conquistador Pedro de Alvarado in 1524. At that time, the Kaqchikel were the enemies of the neighbouring K'iche' Kingdom, and helped the Spaniards to conquer it. The first colonial capital of Guatemala, Tecpán Guatemala, was founded near Iximché on July 25, 1524. On November 22, 1527, after several Kaqchikel uprisings, the capital was moved to Ciudad Vieja, near Antigua Guatemala.
The Kaqchikel language, one of the Mayan languages, is spoken today by 400,000 people. They subsist agriculturally, and their culture reflects a fusion of Maya and Spanish influences.
In November 1920, Cameron Townsend attended a gathering of politicians and diplomats from various Central American countries, after which he desired to begin the difficult process of writing down the Kapchikel language and thereby to translate the Bible into their native language. Cameron completed this massive undertaking on October 15, 1928, and sent the New Testament off to print. This was the genesis of the Wycliffe Bible Translators.
These early missionaries helped improve the lives of this and countless other tribes, so much that in the fall of 1936, President Lazaro Cardenas of Mexico recognized this work. He invited as many translators into Mexico as possible to help all of the other tribes in the country.La Noche Triste
La Noche Triste ("The Night of Sorrows", literally "The Sad Night") also known as La Noche Victoriosa ("The Night of Victory" or "The Victorious Night") by Toltecayotl traditional groups (in the line of thought of The Broken Spears), was an important event during the Spanish Conquest of Mexico, wherein Hernán Cortés, his invading army of Spanish conquistadors, and their native allies were driven out of the Aztec capital at Tenochtitlan.Luis de Moscoso Alvarado
Luis de Moscoso Alvarado (1505 – 1551) was a Spanish explorer and conquistador. Luis de Moscoso Alvarado assumed command of Hernando De Soto's expedition upon the latter's death.Olintepeque
Olintepeque (Spanish pronunciation: [olinteˈpeke]) is a municipality in the Quetzaltenango department of Guatemala, not far from the city of Quetzaltenango. It is located on the Xekik'el (or Xekikel) River.
Olintepeque is known for being the place where the legendary K'iche' king Tecún Umán died in single combat with the Spanish conquistador, Pedro de Alvarado on February 20, 1524. The river Xekik'el ("where the blood spread") takes its name from the famous battle.
The inhabitants of Olintepeque speak primarily Spanish and K'iche'. A chapel is dedicated to St John the Baptist and the folk saint San Pascualito.Pedro de Portocarrero (conquistador)
Pedro de Portocarrero (c. 1504 – c. 1539) was a Spanish conquistador who was active in the early 16th century in Guatemala, and Chiapas in southern Mexico. He was one of the few Spanish noblemen that took part in the early stages of the Spanish conquest of the Americas, and was distantly related to prominent conquistador Pedro de Alvarado, who appointed him as an official in early colonial Guatemala.Salvadorans
The Salvadorans (Spanish: Salvadoreños) are people who identify with El Salvador in Central America. Salvadorans are mainly Mestizos (mixed European and Native American heritage) who make up the bulk of the population in El Salvador. Most Salvadorans live in El Salvador, although there is also a significant Salvadoran diaspora, particularly in the United States, with smaller communities in other countries around the world.
El Salvador's population was 6,218,000 in 2010, compared to 2,200,000 in 1950. In 2010, the percentage of the population below the age of 15 was 32.1%, 61% were between 15 and 65 years of age, while 6.9% were 65 years or older.San Luis Acatlán
San Luis Acatlán is a town in San Luis Acatlán Municipality located in the state of Guerrero, Mexico. It is located in the Costa Chica region of the state, about 158 km from Acapulco. Most often called simply "Acatlán", the name comes from the locality of Acatlán located in the modern state of Puebla. The name itself derives from Náhuatl, meaning "among the reeds". Indigenous shepherds from Puebla arrived to this place between two rivers in Guerrero around 1750. San Luis derives from the name it was given in 1522 when soldiers of Pedro de Alvarado arrived here on this saint's day (August 25).Spanish conquest of Chiapas
The Spanish conquest of Chiapas was the campaign undertaken by the Spanish conquistadores against the Late Postclassic Mesoamerican polities in the territory that is now incorporated into the modern Mexican state of Chiapas. The region is physically diverse, featuring a number of highland areas, including the Sierra Madre de Chiapas and the Montañas Centrales (Central Highlands), a southern littoral plain known as Soconusco and a central depression formed by the drainage of the Grijalva River.
Before the Spanish conquest, Chiapas was inhabited by a variety of indigenous peoples, including the Zoques, various Maya peoples, such as the Lakandon Chʼol and the Tzotzil, and an unidentified group referred to as the Chiapanecas. Soconusco had been incorporated into the Aztec Empire, centred in Valley of Mexico, and paid the Aztecs tribute. News of strangers first arrived in the region as the Spanish penetrated and overthrew the Aztec Empire. In the early 1520s, several Spanish expeditions crossed Chiapas by land, and Spanish ships scouted the Pacific coast. The first highland colonial town in Chiapas, San Cristóbal de los Llanos, was established by Pedro de Portocarrero in 1527. Within a year, Spanish dominion extended over the upper drainage basin of the Grijalva River, Comitán, and the Ocosingo valley. Encomienda rights were established, although in the earlier stages of conquest these amounted to little more than slave-raiding rights.
The colonial province of Chiapa was established by Diego Mazariegos in 1528, with the reorganisation of existing encomiendas and colonial jurisdictions, and the renaming of San Cristóbal as Villa Real, and its relocation to Jovel. Excessive Spanish demands for tribute and labour caused a rebellion by the indigenous inhabitants, who attempted to starve out the Spanish. The conquistadores launched punitive raids, but the natives abandoned their towns and fled to inaccessible regions. Internal divisions among the Spanish led to a general instability in the province; eventually the Mazariegos faction gained concessions from the Spanish Crown that allowed for the elevation of Villa Real to the status of city, as Ciudad Real, and the establishment of new laws that promoted stability in the newly conquered region.Spanish conquest of El Salvador
The Spanish conquest of El Salvador was the campaign undertaken by the Spanish conquistadores against the Late Postclassic Mesoamerican polities in the territory that is now incorporated into the modern Central American nation of El Salvador. El Salvador is the smallest country in Central America, and is dominated by two mountain ranges running east-west. Its climate is tropical, and the year is divided into wet and dry seasons. Before the conquest the country formed a part of the Mesoamerican cultural region, and was inhabited by a number of indigenous peoples, including the Pipil, the Lenca, the Xinca, and Maya. Native weaponry consisted of spears, bows and arrows, and wooden swords with inset stone blades; they wore padded cotton armour.
The Spanish conquistadores were largely volunteers, receiving the spoils of victory instead of a salary; many were experienced soldiers who had already campaigned in Europe. The Spanish expeditions to Central America were launched from three different Spanish jurisdictions, resulting in rival conquests by mutually hostile Spanish captains. Spanish weaponry included swords, firearms, crossbows and light artillery. Metal armour was impractical in the hot, humid climate of Central America and the Spanish were quick to adopt the quilted cotton armour of the natives. The conquistadors were supported by a large number of Indian auxiliaries drawn from previously encountered Mesoamerican groups.
The first campaign against the native inhabitants was undertaken in 1524 by Pedro de Alvarado. Alvarado launched his expedition against the Pipil province of Cuscatlan from the Guatemalan Highlands, but by July 1524 he had retreated back to Guatemala. Gonzalo de Alvarado founded San Salvador the following year, but it was eradicated by a native attack in 1526, during a general uprising that spread across the region. Pedro de Alvarado returned to campaign in El Salvador in 1526 and 1528, and in the latter year, Diego de Alvarado reestablished San Salvador and issued encomiendas to his supporters. In 1528, the uprising finally ended when the Spanish stormed the native stronghold at the Peñol de Cinacantan.
In 1529, El Salvador became embroiled in a jurisdictional dispute with neighbouring Nicaragua. Pedrarias Dávila sent Martín de Estete at the head of an expedition to annex the territory to Nicaragua. Estete captured the leader of a rival Spanish expedition in eastern El Salvador, and marched on San Salvador, before being repulsed by a relief force sent from Guatemala. In 1530, Pedro de Alvarado ordered the establishment of a new settlement at San Miguel, in the east of the country, to protect against further incursions from Nicaragua, and to assist in the conquest of the surrounding area. Indigenous uprisings against the invaders continued, spreading from neighbouring Honduras. The general uprising across the two provinces was put down by the end of 1538, and by 1539 the province was considered pacified. The conquistadores discovered that there was little gold or silver to be found in El Salvador, and it became a colonial backwater with a small Spanish population, within the jurisdiction of the Captaincy General of Guatemala.Spanish conquest of Guatemala
The Spanish conquest of Guatemala was a protracted conflict during the Spanish colonization of the Americas, in which Spanish colonisers gradually incorporated the territory that became the modern country of Guatemala into the colonial Viceroyalty of New Spain. Before the conquest, this territory contained a number of competing Mesoamerican kingdoms, the majority of which were Maya. Many conquistadors viewed the Maya as "infidels" who needed to be forcefully converted and pacified, disregarding the achievements of their civilization. The first contact between the Maya and European explorers came in the early 16th century when a Spanish ship sailing from Panama to Santo Domingo was wrecked on the east coast of the Yucatán Peninsula in 1511. Several Spanish expeditions followed in 1517 and 1519, making landfall on various parts of the Yucatán coast. The Spanish conquest of the Maya was a prolonged affair; the Maya kingdoms resisted integration into the Spanish Empire with such tenacity that their defeat took almost two centuries.Pedro de Alvarado arrived in Guatemala from the newly conquered Mexico in early 1524, commanding a mixed force of Spanish conquistadors and native allies, mostly from Tlaxcala and Cholula. Geographic features across Guatemala now bear Nahuatl placenames owing to the influence of these Mexican allies, who translated for the Spanish. The Kaqchikel Maya initially allied themselves with the Spanish, but soon rebelled against excessive demands for tribute and did not finally surrender until 1530. In the meantime the other major highland Maya kingdoms had each been defeated in turn by the Spanish and allied warriors from Mexico and already subjugated Maya kingdoms in Guatemala. The Itza Maya and other lowland groups in the Petén Basin were first contacted by Hernán Cortés in 1525, but remained independent and hostile to the encroaching Spanish until 1697, when a concerted Spanish assault led by Martín de Ursúa y Arizmendi finally defeated the last independent Maya kingdom.
Spanish and native tactics and technology differed greatly. The Spanish viewed the taking of prisoners as a hindrance to outright victory, whereas the Maya prioritised the capture of live prisoners and of booty. The indigenous peoples of Guatemala lacked key elements of Old World technology such as a functional wheel, horses, iron, steel, and gunpowder; they were also extremely susceptible to Old World diseases, against which they had no resistance. The Maya preferred raiding and ambush to large-scale warfare, using spears, arrows and wooden swords with inset obsidian blades; the Xinca of the southern coastal plain used poison on their arrows. In response to the use of Spanish cavalry, the highland Maya took to digging pits and lining them with wooden stakes.Spanish immigration to Guatemala
The arrival of the Spaniards in Guatemala began in 1524 with the conquest of the Guatemalan Highlands and neighbouring Pacific plain under the command of Pedro de Alvarado. After the conquest and the colonial era, more people came to the country not as conquerors, but to do business or daily activities.Tecun Uman
This page is about the ruler; for the city, see Ciudad Tecún Umán.
Tecun Uman (1500? – February 20, 1524) was one of the last rulers of the K'iche' Maya people, in the Highlands of what is now Guatemala. According to the Kaqchikel annals, he was slain by Spanish conquistador Pedro de Alvarado while waging battle against the Spanish and their allies on the approach to Quetzaltenango on 12 February 1524. Tecun Uman was declared Guatemala's official national hero on March 22, 1960 and is commemorated on February 20, the popular anniversary of his death. Tecun Uman has inspired a wide variety of activities ranging from the production of statues and poetry to the retelling of the legend in the form of folkloric dances to prayers. Despite this, Tecun Uman's existence is not well documented, and it has proven to be difficult to separate the man from the legend.
|Ancestors of Pedro de Alvarado y Contreras|<|endoftext|>
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What is a 10 Out of 12 Percent?
# What is a 10 Out of 12 Percent?
When you think of percentages, you probably think of the number 12 and its common divisor, 6. But, what is a percentage and how does it compare to the other numbers? This article will help you to understand the concepts behind percentages and the Permille sign. Here is a quick look at some of the different ways to calculate a percentage:
## How to calculate a percentage
If you need to find the percentage of a number, use a percentage calculator. These calculators are easy to use and work with a number as the base. Most problems involving percentages can be solved by using a percentage calculator. Once you have a base number, multiply it by 100 to find the percentage.
The percentage formula has different forms and is essentially an algebraic equation. It involves three values - P for the percentage, V1 for the first value that was modified by the percentage, and V2 for the result of the percentage operation on V1. Once you have input your percentage, the calculator will automatically convert it to decimal form and return the actual percentage.
A percentage calculator will calculate the percentage for you by using a fraction. For example, 5% of twenty is x/100 * 20. The percentage calculator will then display one in the result box. The tool also works well for discount applications. You'll find it easy to remember the formulas for many different percentages if you're able to recall them when you need to do so.
The easiest way to convert a fraction into a percentage is to move the decimal point two places to the left. Then, you'll multiply the result by 100 and place the percent symbol after the decimal. This formula is also very useful for percentages because it simplifies multiplication and makes the percentage easier to understand.
In addition to decimal forms, the percentages can be expressed in different ways. If you want to represent the results of a certain process or an experiment, you can convert the percent values into decimal numbers. Then, you can use the percentage formulas to find the result of an equation. The most basic formula is X/Y = P x 100, but there are other variants of the formula.
Besides being used in everyday life, percentages have numerous applications and are often used in finance, accounting, and economics. For example, we are familiar with percent off coupons, which offer a discount or promotion by a certain percentage of the original price. In the business world, percentage calculations are often used to measure productivity and a person's load.
Another common application of the percentage calculator is to compare two values. They can be positive or negative. A positive change means an increase, while a negative change means a decrease. A percentage calculator makes it easy to calculate a percentage change by using a number and a decimal equivalent.
## Permille sign
Permille is a mathematical symbol that indicates a part of a whole in parts per thousand. It is the same as the percent sign except that it has an extra zero in the divisor. One mille equals one-tenth of a percent. It is often abbreviated as %0.
Before the Permille sign was used, a similar symbol was used to express percent. Before 1425, Europeans used an abbreviation called pco, which stands for "per cento." This abbreviation was often followed by a line that showed which part was a percent. In the 1600s, this abbreviation became a glyph. The original symbol resembled an obelus.
Permyriad is a unit of measurement for a portion of the whole expressed as a percentage, or part per thousand. It is also known as a basis point. A permyriad equals 1/100 of one percent. Permilles, on the other hand, are units of measurement that are ten parts smaller than one permyriad.
Permilles are used by scientists and financial professionals to measure trace amounts in solutions. Compared to percent points, permilles are less accurate. Despite these differences, people tend to prefer percentages to fractions. Third-pound burgers never made it in the United States, but marketing departments still tout larger sizes using percent instead of pounds.
# What is 11 of 12 As a Percentage - Aspose Products
If you're wondering, "What is 11 of 12 as a percentage?" then you've come to the right place. In this article, we will show you how to calculate 11 of 12 as a percentage using Aspose Products. It's easy and fast.
# How to Calculate 11 Out of 12 Percentage
Percentage is a mathematical term that represents a ratio or number that is expressed as a fraction of a whole number. Examples of percentages are given in this article. To convert a fraction into a percentage, divide the fraction by twelve and multiply by 100 to get the percentage.
## Percentage of 12 is 11
A percentage can be expressed as a number with a decimal point. In order to write it in this way, simply divide the number by twelve and then multiply the result by 100. Alternatively, a percentage can be expressed as a fraction. Using a fraction makes calculating the percentage easier, and it is often more convenient.
## Percentage is a mathematical term that describes a number or ratio, shown as a fraction of 100
In mathematics, a percentage is a number that is shown as a fraction of one hundred. It is derived from the Latin word per centum, which means "per centum". Percentages can be expressed as a whole number, a fraction, or a decimal. In financial and business settings, percentages are used to express interest charges and the value of goods.
A percentage is a mathematical term that describes 'a number or ratio,' as a fraction of one hundred. Its denominator and numerator are different categories, such as males versus females or persons aged 20-29 years old versus persons aged 30-39 years old.
For example, a 10 percent increase in price would result in a final price of \$220. In contrast, a 10 percent decrease in price would result in a final price as \$198. As a result, the two changes are not equivalent and do not cancel each other out.
To calculate a percentage, multiply the proper fraction of the mixed number by the denominator (100). Then add the decimal form to the product, and place the percent symbol after the last digit.
For example, suppose a class of twenty students has twenty males and seventeen females. The females outnumber the males, and the males are the minority. In this case, a class of twenty students has seventeen girls and twelve boys. In another scenario, a class of twenty students has two candidates: candidate A and candidate B. One-third of the students voted for candidate A and the other half voted for candidate B.
Knowing how to calculate a percentage will not only help you in math class, but it will also make solving practical problems easier. The easiest way to get a percentage is to multiply the fraction by 100. It is as simple as that!
When comparing two numbers, you will find that the ratio is higher with the second one. The inverse is also true. For example, a percentage can be represented by a grid, which is divided into equal parts. Each cell is equal to 1%, so two cells are equal to 2% and five cells equal 5%.
Percentages are also used in statistics. A mortality rate is the percentage of people who die from a disease. For example, in March 2005, there were 70 influenza cases in County A. The death rate is approximately 1.
The word percentage is defined as a mathematical term that describes a number that is shown as a fraction of one hundred. The percentage will be written with the decimal part of the number at the top and the fractional part at the bottom.
## Examples
Percentage is a mathematical term that represents a portion of a number. It is used in many different areas, including finance, accounting, sales, and taxation. Historically, it was often used in school settings to represent grades. Today, percentages are also used in other areas, such as nutrition facts, probability, and the download process. It is important to remember that percentages are always relative and never absolute.
# How to Convert a Fraction to a Percentage
In order to convert a fraction to a percentage, divide the numerator by the denominator, and multiply the result by 100%. That will make the fraction less than 100%. Then, you can calculate the percentage from the fraction. Here are some examples:
## Convert a fraction to a decimal
To convert fractions to percents, you must first understand what a decimal is. A decimal consists of a whole number and some fractions. For example, if you take a fraction of one hundred and fifteen and divide it by two, you will get the result of a decimal of six-tenths. This means that you can use decimal notation to express the exact amount of any quantity.
A fraction is simply a number that has two parts: the numerator and the denominator. The numerator represents the number that is being divided, while the denominator represents the smaller part. A fraction can be expressed as a decimal or a fraction with a decimal expansion.
The simplest way to convert a fraction is to divide it into two parts with the same numerator. If the numerator is less than the denominator, you must divide the fraction by the least common multiple, which is 12. Once you have converted the fraction, you can use its decimal form.
Using a fraction calculator is an easy way to simplify fractions. Simply input the numerator and denominator and the calculator will calculate a simplified fraction. The calculator will also accept mixed numbers and reciprocals. It is important to note that fractions are usually represented in simplified form.
To convert a decimal to a fraction, you must know how to interpret the decimal place values. First, you need to know that the place values represent a power of ten. For example, the first decimal place is 101, while the second decimal place is 102. The third decimal place is 103, and the fourth place is 104. The fifth decimal place is 105. Finally, you can use a decimal to fraction calculator, which is available online. It performs the calculation in a fraction of a second.
## Convert a mixed number to a decimal
A mixed number can be represented in a number of ways, including fractions, decimals, or whole numbers. In order to convert a mixed number to a decimal, you must first determine its place value. Decimal place values are the power of 10 and the first decimal place is 101. The second decimal place is 102, and the third is 103. The fifth decimal place is 5. You can then use the remainder of the mixed number as the numerator for the proper fraction piece.
A fraction can be written as a decimal if the denominator value is greater than the numerator. This way, a fraction such as 8/15 can be written as 0.5333 instead of 0.33. You can then divide the fraction to find the decimal equivalent of 7/8 inches or mm.
You can also write a mixed number as a fraction by placing it over a power of ten. For example, if a pizza costs \$10 and costs \$11, you write it as "11/19." The "11/2" symbol combines the number "one" with a fraction of "2" (which is half). Mixed numbers are very common in spoken English but are difficult to compute with. For this reason, algebra students often avoid using them in their calculations.
## Convert a fraction to its lowest terms
To convert a fraction to its lowest terms, you must first consider the denominator. A fraction is either a whole number or a decimal. If it is a fraction, the denominator is the number to the right of the decimal point. For example, 5.2% is 5.2/100. Then, you must multiply this result by 100% to get a value that is less than 100%.
For example, if a fraction is 9/15, the lowest terms are 3/5 and 15/9. To convert a fraction to its lowest terms, you can use a calculator. This works for both numbers and expressions with variables. You can also find the lowest terms of fractions by rearranging them.
Using a calculator is one of the simplest ways to convert fractions. The calculator will help you add, subtract, multiply, and divide fractions. The calculator will also simplify the equations. It can also convert fractions to decimals. However, you must remember that it is not possible to divide a fraction by zero.
Fraction calculators are available online. Many of these calculators will simplify complex fractions and help you find the lowest terms quickly and easily. You can also use a fraction calculator for mixed numbers or reciprocals. It also offers step-by-step solutions.
## Calculate a percentage from a fraction
A fraction can be easily converted to a percentage using a graphing calculator. A fraction with three digits is easier to convert to a percentage with two digits. Fractions are a frequent part of business-math classes. Mathway offers a variety of graphing calculators that can convert fractions to percentages.
The word percentage comes from the Latin root word cent. Cent means hundred and is an old European word, originally French, Latin, or Italian. The word percent literally means "out of a hundred." In math, 87 percent is 87 divided by 100. For example, thirteen percent of 100 days have snow, meaning there is 13 percent chance that snow will fall each day.
Percentages are a useful unit in math. Not only do they help us in our everyday lives, but they can also help us when we write tests or pass exams. They are a convenient way to count in the decimal system. For example, if you want to calculate how much you earned in a certain period of time, a percentage is the right way to go. By dividing the top number by the bottom number, you will get a percentage of six hundred dollars.
You can also use a calculator to solve fractions. The Fraction Calculator allows you to divide, add, and subtract fractions. It also has the capability of working with mixed numbers and reciprocals. This calculator is an excellent tool for students. It offers step-by-step solutions and allows students to find errors in their work.
Using a calculator on Mathway, you can solve problems with ease and without any guesswork. The calculator will automatically select the correct type for the problem. Once the calculator has finished, you can compare your answer to your own answer and make notes on mistakes. If you're not satisfied with the answer, you can sign up with Mathway to check the solution.
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The basic design of Chondrichthyans includes a skull, jaws with teeth, a vertebral column (backbone), and fins; but unlike the bony fishes, they lack ribs. Through millions of years of evolution many different species have evolved in varying shapes and sizes, each designed for the particular habitats where they are found and the types of prey they survive on. The upper jaw is loosely attached to the underside of the skull by ligaments and connective tissue.
Teeth vary in size and shape throughout the different species and these are embedded into the gums rather than being attached to the jaw – worn or damaged teeth are regularly replaced. The tooth design can vary from very sharp pointed teeth for catching and eating fish to flattened teeth for grinding. Whale Sharks and Manta Rays have hardened plates only, as they feed by capturing plankton in gill rakers, as they pass large volumes of water through the gills.
Chondrichthyans have three types of unpaired fins – the dorsal, anal and caudal; and two types of paired fins, the pectoral and pelvic. However, through evolution, some family groups like Dogfish, Sawsharks, Angel Sharks, and rays have lost the anal fin. Chimaeras and some rays have modified caudal fins which form long whip-like or ribbon-like tails.
In sharks and chimaeras, the pectoral fins extend from the body below and behind the head whereas in rays, the pectoral fins are attached at the back of the skull and part of the body stem and are greatly enlarged to form a body disc.
Body shape and method of movement through the water varies between the species. The body of most species of shark is shaped for hydrodynamic efficiency where swimming is achieved by side-to-side undulations of the tail. In rays, the body is highly flattened from the top and bottom, the tail is reduced in size, and movement through the water is achieved by undulation of the tips of the pectoral fins. Chimaeras move by rapid movement of the pectoral fins or by slow lateral movements of their tails. To aid hydrodynamics sharks, rays and skates skin is covered with small teeth-like structures called dermal denticles. These teeth-like structures are designed to reduce the amount of drag while swimming and therefore increase the efficiency while travelling through the water.
As with other fish, Chondrichthyans extract oxygen from the water to breathe using gills. Most species of sharks have 5 pairs of gill openings although there are several species have 6 or 7. Rays have 5 or 6 pairs of gill openings and chimaeras have 4 pairs of gill arches but only 1 pair of gill openings. The gill openings in sharks and chimeras are located on the side behind the head and in rays and skates they are located underneath.
Rays and most bottom dwelling sharks also have spiracles, these are openings which allow water to be pumped over the gills while stationary allowing them to rest on the seafloor. The other species of sharks need to continue to move through the water to breath. Chimaeras have large nostrils and take in water through these instead of the mouth.
Chondrichthyans have some of the most complex and finely tuned sensory systems of all animals. The eyes provide for an excellent field of vision and in some species it is believed they can see in colour, further adaptations give them the ability to function efficiently in low-light conditions. Most species have an acute sense of smell. Taste-receptor cells to determine palatability of prey occur on taste buds that cover a small bump in the mouth. Touch-receptor cells occurring near the skin surface respond to contact and small capsules deeper in the skin respond to the bending of the body and fins through the stretching and contracting of muscles.
Senses of balance, orientation, and body movement depend on a complex series of organs inside the ear. As the animal moves, fluids inside three semi-circular canals at right angles to each other press against hair cells, connected to a conglomerate of hard calcium granules, and stimulate nerve impulses. Hearing is acute, even though Chondrichthyans lack external ear openings. Clusters of sensory hair cells detect water movement or vibration. Those in the skin are called pit organs and can detect water movement relative to the body surface. To detect water acceleration, neuromasts also form an extensive network of small water-filled canals immediately below the surface of the skin in the lateral lines, which run across each side of the head and body.
Electro sense is present in all Chondrichthyans, but uncommon among other groups in the animal kingdom. Chondrichthyans electro sensory organs (Ampullae of Lorenzini) are able to detect weak electric fields generated by the movement of prey, predators, and ambient water in the Earth’s magnetic field. It is likely that these animals can navigate long distances by detecting slight variations in the Earth’s magnetic field caused by variation of metal deposits in their environment. The electro sense is so acute that Chondrichthyans animals can detect voltage gradients of less than 5 billionths of a volt across a distance of 1 cm. The Ampullae of Lorenzini are lined with hair cells attached to an insulated tube filled with a conductive jelly. A network of interconnected Ampullae located in pores on the head, lower jaw, and trunk enables the animal to measure voltage gradients at different positions on its body.
Torpedo rays have evolved paired kidney-shaped electric organs located on each side near the centre of the disc, capable of producing more than 200 volts. These are used for defence and stunning prey, and possibly also for electro-location of prey, predators, or mates, and for social communication.
There is still so much more to learn about these amazing animals.<|endoftext|>
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# 7.3.5: Distinguishing Volume and Surface Area - Mathematics
## Lesson
Let's work with surface area and volume in context.
Exercise (PageIndex{1}): THe Science Fair
Mai’s science teacher told her that when there is more ice touching the water in a glass, the ice melts faster. She wants to test this statement so she designs her science fair project to determine if crushed ice or ice cubes will melt faster in a drink.
She begins with two cups of warm water. In one cup, she puts a cube of ice. In a second cup, she puts crushed ice with the same volume as the cube. What is your hypothesis? Will the ice cube or crushed ice melt faster, or will they melt at the same rate? Explain your reasoning.
Exercise (PageIndex{2}): Revisiting the Box of Chocolates
The other day, you calculated the volume of this heart-shaped box of chocolates.
The depth of the box is 2 inches. How much cardboard is needed to create the box?
Exercise (PageIndex{3}): Card Sort: Surface Area or Volume
Your teacher will give you cards with different figures and questions on them.
1. Sort the cards into two groups based on whether it would make more sense to think about the surface area or the volume of the figure when answering the question. Pause here so your teacher can review your work.
2. Your teacher will assign you a card to examine more closely. What additional information would you need to be able to answer the question on your card?
3. Estimate reasonable measurements for the figure on your card.
4. Use your estimated measurements to calculate the answer to the question.
Are you ready for more?
A cake is shaped like a square prism. The top is 20 centimeters on each side, and the cake is 10 centimeters tall. It has frosting on the sides and on the top, and a single candle on the top at the exact center of the square. You have a knife and a 20-centimeter ruler.
1. Find a way to cut the cake into 4 fair portions, so that all 4 portions have the same amount of cake and frosting.
2. Find another way to cut the cake into 4 fair portions.
3. Find a way to cut the cake into 5 fair portions.
Exercise (PageIndex{4}): A Wheelbarrow of Concrete
A wheelbarrow is being used to carry wet concrete. Here are its dimensions.
1. What volume of concrete would it take to fill the tray?
2. After dumping the wet concrete, you notice that a thin film is left on the inside of the tray. What is the area of the concrete coating the tray? (Remember, there is no top.)
### Summary
Sometimes we need to find the volume of a prism, and sometimes we need to find the surface area.
Here are some examples of quantities related to volume:
• How much water a container can hold
• How much material it took to build a solid object
Volume is measured in cubic units, like in3 or m3.
Here are some examples of quantities related to surface area:
• How much fabric is needed to cover a surface
• How much of an object needs to be painted
Surface area is measured in square units, like in2 or m2.
### Glossary Entries
Definition: Base (of a prism or pyramid)
The word base can also refer to a face of a polyhedron.
A prism has two identical bases that are parallel. A pyramid has one base.
A prism or pyramid is named for the shape of its base.
Definition: Cross Section
A cross section is the new face you see when you slice through a three-dimensional figure.
For example, if you slice a rectangular pyramid parallel to the base, you get a smaller rectangle as the cross section.
Definition: Prism
A prism is a type of polyhedron that has two bases that are identical copies of each other. The bases are connected by rectangles or parallelograms.
Here are some drawings of prisms.
Definition: Pyramid
A pyramid is a type of polyhedron that has one base. All the other faces are triangles, and they all meet at a single vertex.
Here are some drawings of pyramids.
Definition: Surface Area
The surface area of a polyhedron is the number of square units that covers all the faces of the polyhedron, without any gaps or overlaps.
For example, if the faces of a cube each have an area of 9 cm2, then the surface area of the cube is (6cdot 9), or 54 cm2.
Definition: Volume
Volume is the number of cubic units that fill a three-dimensional region, without any gaps or overlaps.
For example, the volume of this rectangular prism is 60 units3, because it is composed of 3 layers that are each 20 units3.
## Practice
Exercise (PageIndex{5})
Here is the base of a prism.
1. If the height of the prism is 5 cm, what is its surface area? What is its volume?
2. If the height of the prism is 10 cm, what is its surface area? What is its volume?
3. When the height doubled, what was the percent increase for the surface area? For the volume?
Exercise (PageIndex{6})
Select all the situations where knowing the volume of an object would be more useful than knowing its surface area.
1. Determining the amount of paint needed to paint a barn.
2. Determining the monetary value of a piece of gold jewelry.
3. Filling an aquarium with buckets of water.
4. Deciding how much wrapping paper a gift will need.
5. Packing a box with watermelons for shipping.
6. Charging a company for ad space on your race car.
7. Measuring the amount of gasoline left in the tank of a tractor.
Exercise (PageIndex{7})
Han draws a triangle with a (50^{circ}) angle, a (40^{circ}) angle, and a side of length 4 cm as shown. Can you draw a different triangle with the same conditions?
(From Unit 7.2.4)
Exercise (PageIndex{8})
Angle (H) is half as large as angle (J). Angle (J) is one fourth as large as angle (K). Angle (K) has measure 240 degrees. What is the measure of angle (H)?
(From Unit 7.1.3)
Exercise (PageIndex{9})
The Colorado state flag consists of three horizontal stripes of equal height. The side lengths of the flag are in the ratio (2:3). The diameter of the gold-colored disk is equal to the height of the center stripe. What percentage of the flag is gold?
(From Unit 4.2.4)
## Step by step guide to solve Rectangular Prisms
• A Rectangular Prism is a solid (3)-dimensional object which has six rectangular faces.
• Volume of a Rectangular prism (=) Length (×) Width (×) Height
Volume (= l×w×h ) Surface area (=2(wh+lw+lh))
### Rectangular Prisms – Example 1:
Find the volume and surface area of the following rectangular prism.
Find the volume and surface area of rectangular prism.
Use volume formula: Volume (=l×w×h)
Then: Volume (=6×4×8=192) m(^3)
Use surface area formula: Surface area (=2(wh+lw+lh))
Then: Surface area (=2((4×8)+(6×4)+(6×8)))
(=2(32+24+48)=2(104)=208) m(^2)
### Rectangular Prisms – Example 2:
Find the volume and surface area of the following rectangular prism.
Find the volume and surface area of rectangular prism.
Use volume formula: Volume (=l×w×h)
Then: Volume (=10×5×8=400) m(^3)
Use surface area formula: Surface area (=2(wh+lw+lh))
Then: Surface area (=2((5×8)+(10×5)+(10×8)))
(=2(40+50+80)=340) m(^2)
## SOLVING SURFACE AREA AND VOLUME PROBLEMS
Erin is making a jewelry box of wood in the shape of a rectangular prism. The jewelry box will have the dimensions shown below. The cost of painting the exterior of the box is .50 per square in. How much does Erin have to spend to paint the jewelryox ?
To know that total cost of painting, first we have to know the Surface area of the jewelry box.
Find surface area of the box.
Identify a base, and find its area and perimeter.
Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.
Find perimeter of the base.
Identify the height, and find the surface area.
The height h of the prism is 6 inches. Use the formula to find the surface area.
Total cost = Area x Cost per square in.
Hence, Erin has to spend $342 to paint the jewelry box. A metal box that is in the shape of rectangular prism has the following dimensions. The length is 9 inches, width is 2 inches, and height is 1 1/ 2 inches. Find the total cost of silver coating for the entire box. To know that total cost of silver coating, first we have to know the Surface area of the metal box. Find surface area of the box. Identify a base, and find its area and perimeter. Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases. Find perimeter of the base. Identify the height, and find the surface area. The height h of the prism is 1 1/2 inches. Use the formula to find the surface area. Total cost = Area x Cost per square in. Hence, the total cost of silver coating for the entire box is$103.50.
Cherise is setting up her tent. Her tent is in the shape of a trapezoidal prism shown below. How many cubic feet of space are in her tent ?
To find the number of cubic feet of space in the tent, we have to find the volume of Cherise's tent.
Volume of Cherise's tent (Trapezoidal prism) is
Area of trapezoid with bases of lengths b ₁ and b ₂ and height h.
Base area (b) = (1/2) x ( b ₁ + b ₂)h
Hence, the number of cubic feet of space in Cherise's tent is 180.
Allie has two aquariums connected by a small square prism. Find the volume of the double aquarium.
Find the volume of each of the larger aquariums.
Volume = Base area x Height
Find the volume of the connecting prism.
Volume = Base area x Height
Add the volumes of the three parts of the aquarium.
The volume of the aquarium is 74 cubic ft.
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## What is the Difference Between Volume and Surface Area?
Volume and surface area are two related concepts in the study of mathematics. They’re both important to understand, but equally important is understanding how they differ and what they mean. This is especially the case when it comes to computing the volume and surface areas of a prism or a cylinder.
If you think of wrapping a present in a box, you can get a good sense of how volume and surface area differ. First, you have to consider the size of the box, when you consider the size of the present. How much interior space does your box need to have so a present will fit? The measurement of the box’s capacity, how much it will hold, is its volume. Next you have to wrap the present. The amount of wrapping paper, which will cover the exterior of the box, is a very different calculation than the capacity of the box. You’ll need a separate measurement or some good guessing, to figure out the sum of the sides of all the surfaces or the surface area.
Volume of a square or rectangular box is pretty easy to compute. Simply multiply height times length times width to get the measurement. With a square it’s even easier, you merely cube one side’s length, since they all measure the same. If the side length is a, the formula is a x a x a or a 3 . When you are comparing volume and surface area, you’ll note a very different formula. You need to get the area of each face, and then add the areas of all faces together. With a square prism or cube, you’d essentially compute the area a x a or a 2 , multiplied by 6 (6a 2 ). When you’re working with a rectangular prism, you’ll have to the area the of 3 pairs of equal sides, which needed to be added together to determine surface area.
Work on volume and surface area are differ a little when you are trying to calculate the area of a cylinder. The formula for a volume of a cylinder is the area of one circular face multiplied times the height of the cylinder. It reads: πr 2 x h, or pi times the radius squared times height. Getting the surface area of the cylinder is a little trickier since the circular portion is essentially one continuous face. Computing surface area of a cylinder means computing the lateral area of this face.
Lateral area formula is the following πr2r or πd (pi times the radius doubled or pi times the diameter), multiplied to the height, πr2r x h. This is essentially the circumference of one circle times the height of the cylinder. To compute the entire formula you also need to add in the top and bottom circular faces’ areas. Since in a cylinder these are equal, the formula is 2 πr 2 . This calculation is then added to the lateral area to compute the whole surface area in the following expression:
πr2r x h + 2πr 2 = lateral area.
You can also view difference between volume and cylinder as a difference between what is inside and can be contained and the exterior of a three-dimensional object. These are valuable differences to understand in many applications, such as construction, engineering, or even present wrapping. When children complain that math is useless outside of math class, you might point out to them that knowing the difference between volume and surface area meant they got a very nicely wrapped gift for their birthday.
Tricia has a Literature degree from Sonoma State University and has been a frequent InfoBloom contributor for many years. She is especially passionate about reading and writing, although her other interests include medicine, art, film, history, politics, ethics, and religion. Tricia lives in Northern California and is currently working on her first novel.
Tricia has a Literature degree from Sonoma State University and has been a frequent InfoBloom contributor for many years. She is especially passionate about reading and writing, although her other interests include medicine, art, film, history, politics, ethics, and religion. Tricia lives in Northern California and is currently working on her first novel.
## Difference Between Volume and Surface Area
Volume and surface area are two different things in the terms of math. Math is a subject which usually deals with the figures and their geometry. There are several other branches of math which are dedicated to studies on several fronts and one of these is the geometry. In the math of geometry, the figures are observed and different calculations are made on them so that various things can be calculated about the figures. There are a host of different figures in the science of math and all of these are analysed to evaluate the properties of these various figures. The number of figures cannot be calculated as so many figures can be made by even one person. The most common figures in math are squares, rectangles, triangles, pentagons, hexagons and many others. All of the following figures which have been named above have different properties than the other and to distinguish them from each other, different parameters of the figures are calculated. Formulas in math have been devised by the scientists and the mathematicians so that they can let the people evaluate these properties of the figure and the most common of these parameters that have been made by the experts are volume and the surface area of the figure. Many people believe that the volume and the area of figures is one and the same thing but in-fact this is not the case at all. Volume and surface area of a figure are two entirely different things and there are separate procedures to evaluate these two aspects of the geometry. Volume of the figure may be described as the thing or the property of math in which the total capacity of figure of the geometry is calculated. This is a measure of how much a thing or a figure and accommodate inside it and the most common term that is used for this is the capacity of the figure. The surface area as compared to the volume of the figure is a totally different thing and it accounts for a different method to be followed.
### Instructions
The volume of a figure is the estimate of the total capacity of a figure of math. The units for this are meter cubes.
Surface Area
Surface area of a figure of math is the total area that the figure acquires on a space and this can be two dimensional and three dimensional both. The units for this are meter squares and formulas have been made to calculate this easily.
## Contents
A ball is a three-dimensional object, being the filled-in version of a sphere ("sphere" properly refers only to the surface and a sphere thus has no volume). Balls exist in any dimension and are generically called n-balls, where n is the number of dimensions.
The same reasoning can be generalized to n-balls using the general equations for volume and surface area, which are:
The surface-area-to-volume ratio has physical dimension L −1 (inverse length) and is therefore expressed in units of inverse distance. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3 . The surface to volume ratio for this cube is thus
For a given shape, SA:V is inversely proportional to size. A cube 2 cm on a side has a ratio of 3 cm −1 , half that of a cube 1 cm on a side. Conversely, preserving SA:V as size increases requires changing to a less compact shape.
Materials with high surface area to volume ratio (e.g. very small diameter, very porous, or otherwise not compact) react at much faster rates than monolithic materials, because more surface is available to react. An example is grain dust: while grain is not typically flammable, grain dust is explosive. Finely ground salt dissolves much more quickly than coarse salt.
A high surface area to volume ratio provides a strong "driving force" to speed up thermodynamic processes that minimize free energy.
The ratio between the surface area and volume of cells and organisms has an enormous impact on their biology, including their physiology and behavior. For example, many aquatic microorganisms have increased surface area to increase their drag in the water. This reduces their rate of sink and allows them to remain near the surface with less energy expenditure. [ citation needed ]
An increased surface area to volume ratio also means increased exposure to the environment. The finely-branched appendages of filter feeders such as krill provide a large surface area to sift the water for food. [1]
Individual organs like the lung have numerous internal branchings that increase the surface area in the case of the lung, the large surface supports gas exchange, bringing oxygen into the blood and releasing carbon dioxide from the blood. [2] [3] Similarly, the small intestine has a finely wrinkled internal surface, allowing the body to absorb nutrients efficiently. [4]
Cells can achieve a high surface area to volume ratio with an elaborately convoluted surface, like the microvilli lining the small intestine. [5]
Increased surface area can also lead to biological problems. More contact with the environment through the surface of a cell or an organ (relative to its volume) increases loss of water and dissolved substances. High surface area to volume ratios also present problems of temperature control in unfavorable environments. [ citation needed ]
The surface to volume ratios of organisms of different sizes also leads to some biological rules such as Allen's rule, Bergmann's rule [6] [7] [8] and gigantothermy. [9]
In the context of wildfires, the ratio of the surface area of a solid fuel to its volume is an important measurement. Fire spread behavior is frequently correlated to the surface-area-to-volume ratio of the fuel (e.g. leaves and branches). The higher its value, the faster a particle responds to changes in environmental conditions, such as temperature or moisture. Higher values are also correlated to shorter fuel ignition times, and hence faster fire spread rates.
A body of icy or rocky material in outer space may, if it can build and retain sufficient heat, develop a differentiated interior and alter its surface through volcanic or tectonic activity. The length of time through which a planetary body can maintain surface-altering activity depends on how well it retains heat, and this is governed by its surface area-to-volume ratio. For Vesta (r=263 km), the ratio is so high that astronomers were surprised to find that it did differentiate and have brief volcanic activity. The moon, Mercury and Mars have radii in the low thousands of kilometers all three retained heat well enough to be thoroughly differentiated although after a billion years or so they became too cool to show anything more than very localized and infrequent volcanic activity. As of April 2019, however, NASA has announced the detection of a "marsquake" measured on April 6, 2019 by NASA's InSight lander. [10] Venus and Earth (r>6,000 km) have sufficiently low surface area-to-volume ratios (roughly half that of Mars and much lower than all other known rocky bodies) so that their heat loss is minimal. [11]
## Examples for the calculation of the surface area of the solid object
The lengths of the cuboid edges are in the ratio 2: 3: 4. Find their length if you know that the surface of the cuboid is 468 m 2 .
The cube has a surface of 600 cm 2 . What is its volume?
Find the edge c of cuboid if an edge a = 20 mm, b = 30 mm and surface area S = 8000 mm 2 .
The volume of the cube is 27 dm cubic. Calculate the surface of the cube.
A sphere is inscribed in the cone (the intersection of their boundaries consists of a circle and one point). The ratio of the surface of the ball and the contents of the base is 4: 3. A plane passing through the axis of a cone cuts the cone in an isoscele
The regular quadrilateral prism has a surface of 250 dm 2 , its shell has a content of 200 dm 2 . Calculate its leading edge.
Calculate the surface area of a cube in m 2 if you know that the area of its two walls is 72 dm 2 .
Find the content of the largest wall of a prism with a rectangle base with a height of 4 dm, side c = 5 cm, and side b = 6 cm.
The cylinder has a surface area of 300 square meters, while the cylinder's height is 12 m. Calculate the volume of this cylinder.
Calculate the surface of a quadrilateral prism according to the input: Area of the diamond base S1 = 2.8 m 2 , length of the base edge a = 14 dm, height of the prism 1,500 mm.
How many times does the surface of a sphere decrease if we reduce its radius twice?
Calculate the volume of the cube if its surface is 150 cm 2 .
Calculate the volume and surface of a cuboid whose edge lengths are in the ratio 2: 3: 4 and the longest edge measures 10cm.
A cuboid with dimensions of 9 cm, 6 cm, and 4 cm has the same volume as a cube. Calculate the surface of this cube.
Find the surface and volume of a cuboid whose dimensions are 1 m, 50 cm, and 6 dm.
The cube has a surface area of 216 dm 2 . Calculate: a) the content of one wall, b) edge length, c) cube volume.
The surface of the regular quadrilateral prism is 8800 cm 2 , the base edge is 20 cm long. Calculate the volume of the prism
In a regular quadrilateral pyramid, the height of the sidewall is equal to the length of the edge of the base. The content of the sidewall is 32 cm 2 . What is the surface of the pyramid?
Eight small Christmas balls with a radius of 1 cm have the same volume as one large Christmas ball. What has a bigger surface: eight small balls, or one big ball?
Calculate how much cover (without a floor) is used to make a tent that has the shape of a regular square pyramid. The edge of the base is 3 m long and the height of the tent is 2 m.
## Example
Each student will be given 10 cubes, though they will be working through the problem set with partners. Students are expected to solve the problems in the manner of the example: 1) build the model 2) draw the 2D views 3) find the surface area and volume 4) Examine the relationship between the 2D views and the surface area. While students are working, I will walk around to help students who are struggling with the models. I will also be looking for proper labeling of units (MP6) and accurate 2D views. It is important to let the students know that there are no "holes" or missing cubes hidden from view. For example, in problem a there are only 3 visible cubes, but the top cube is sitting on stop of a hidden cube. It can be tricky to keep track of the faces when counting for surface area. I may let students who are getting frustrated to lightly mark each face using a wet erase marker.
Page 2 of the model, labeled "independent problem solving", has the extension problems. These will present a bit more of a challenge for students who make quick work out of the first 6 problems. I will give these students another 10 cubes to work with.
## Practical Problems
A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required ?
Since the wall with all its bricks makes up the space occupied by it, we need to find the volume of the wall, which is nothing but a cuboid.
Here, Length = 10 m = 1000 cm
Volume of the wall = length × thickness × height
Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cm
So, volume of each brick = length × breadth × height
So, number of bricks required =
volume of the wall/volume of each brick
= (1000 x 24 x 400) / (24 x 12 x 8)
So, the wall requires 4167 bricks.
At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm . Find the number of glasses that he can fill the juice completely?
The volume of juice in the vessel = volume of the cylindrical vessel = πr 2 h
Similarly, the volume of juice each glass can hold = π r ²h
(where r and h are taken as the radius and height respectively of each glass)
So, number of glasses of juice that are sold
= volume of the vessel/ volume of each glass
= ( π x 15 x 15 x 32) / ( π x ਃ x 3 x 8)
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## Video – Lesson & Examples
• Introduction to prisms
• 00:00:23 – What is a prism? formulas for lateral area, surface area, and volume
• Exclusive Content for Member’s Only
• 00:10:38 – Find the surface area and volume for each prism (Examples #1-6)
• 00:17:16 – Use the volume addition postulate to find the volume of the polyhedron (Examples #7-8)
• 00:17:16 – Determine the number of gallons of water in a swimming pool given its volume (Example #9)
• 00:17:16 – Find the surface area of a right rectangular prism (Example #10)
• 00:17:16 – Find the surface area of an oblique prism (Example #11)
• Practice Problems with Step-by-Step Solutions
• Chapter Tests with Video Solutions
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Welcome to Air Pollution: What's the Solution? an educational project for students, grades 6 - 12, that uses online real time data to guide student discovery of the science behind the causes and effects of outdoor air pollution.
Through this project, students will focus on outdoor air pollution; what it is, what factors contribute to its formation and the health effects from breathing polluted air. Students will use data and animated maps from the Internet and monitor for the presence of air pollution. Students are challenged to think critically and creatively about the problems surrounding air pollution.
By the end of this project, students will be able to:
- Describe what air pollutants are, when and how outdoor air pollution is formed, and what the health effects are from breathing polluted air.
- Read the Environmental Protection Agency's Air Quality Index (AQI) chart, record weather data, and determine the presence of air pollutants.
- Create graphs to help visualize or recognize trends.
- Predict when ground level ozone may occur.
- Use knowledge gained to create awareness about air pollution and the associated health effects.
Please refer to the Activity Map as a guide to all of the lessons within the curriculum.<|endoftext|>
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How brain-machine connections can help paraplegics move again
The brain is seen as the seat of the human self in many cultures. But researches are increasingly looking at connecting our brains to machines, and even to each other — raising questions about what happens to our individual identity when it's directly connected to others.
New research shows that brain-machine interfaces can help paralyzed patients rewire their brains, and learn to move again.
Neuroethicist Karen Rommelfanger understands why many people find the promise of connecting brains so compelling. Dr. Rommelfanger is a faculty member in the NS program.
"The idea of being able to touch somebody really far away — touch someone's thoughts from far away — is an exciting appeal."
But she says that both in science and on the consumer side, there are questions that need to be asked along the way.
Rommelfanger is part of the neuroethics workgroup of the U.S. government's Brain Initiative, looking at the ethics of brain research. She worries that brain connection could lead to an overdependence of technology, and raises fundamental questions about our sense of self.<|endoftext|>
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Percentage Quiz-2
Maths is an important subject in CLAT,DULLB & Other Law Exams. In any of law exam, Maths carries weightage of 20 -25 % of questions. With focused practice good marks can be fetched from this section. These questions are very important in achieving your success in CLAT, DULLB and Other Law Exams..
Q1. 20% of A = 60% of B. what percent of B is A?
• 25%
• 400%
• 200%
• 300%
Solution
20% A = 60% B A = (60% / 20%) = B
A = 3B = 300% B
Q2.10% of voters did not vote. 60 votes were invalid. A got 47% of votes in the list and B lost by 308 votes. The total number of voters is
• 6000
• 6200
• 6400
• 6450
Solution
Valid votes = (100 - 10)% x - 60 = 0.9x - 60
Vote for A = 0.47x
Vote for B = 0.47x - 308
Hence 0.47x + 0.47x - 308 = 0.9x - 60
Þ 0.04x = 248 \ x = 6200
Q3. The population of a town increases by 10% in one year and then decreases by 10% in 2nd year. On the whole there was a/an
• decrease of 10%
• decrease of 10%
• increase of 1%
• none of these
Solution
Let initial population be 100
Population after 1st yr = (100 + 10) = 110
Population after 2nd year = (110 - 11) = 99
Therefore, total decrease of 1%
Q4. 45 x ? = 25 % of 900
• 16.20
• 500
• 4
• 5
Solution
45 x ? = 25% of 900
Þ ? = (0.25 X 900) / 45 = 5
Q5. 218% of 1674 = ? x 1800
• 4
• 0.5
• 6
• none of these.
Solution
218% of 1674 = ? x 1800
? = (2.18 X 1674) / 1800 = 2.02
Q6. One fourth of one third of two fifth of a number is 15. What will be 40% of that number?
• 120
• 350
• 270
• 180
Solution
(1/4){(1/3)(2n/5)}= 15
Þ n = 15 x 5 x 3 x 2 = 450 Þ 0.4n - 0.4 x 450 = 180
Q7. What percent of 7.2 kg is 18 gms?
• 25%
• 2.5%
• 0.25%
• 0.025%
Solution
(18/7200) x 100 = (1/4)= 0.25%
Q8. A man loses 12.5% of his money and after spending 70% of the remainder, has left Rs 210. Find the money he had at first.
• 600
• 700
• 800
• 900
Solution
(100 - 12.5) (100 - 70)% money = 210
Money = 210 x (1/0.3) x (8/7) = 800
Q9. Of the total amount received by Kiran, 20% was spent on purchases and the remaining on transportation. If she is left with Rs 1520, the initial amount was:
• 2800
• 2000
• 2400
Solution
Data is inadequate. We need transportation expenses as % of total amount
Q10.In a library, 20% books are in Hindi, 50% of the remaining are in English and the remaining 9000 are in various other languages. What is the total number of books in English?
• 4000
• 3000
• 2250
• None of these
Solution
Books in Hindi = 20% b
Books in English = (100 - 20)% (100- 50)%
= 0.8 x 0.5 b = 0.4 b
Remn. books = b - 0.2b - 0.4b = 0.4b = 9000
Books in English = 0.4b = 9000
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# 5.5 Divide Monomials
## Learning Objectives
By the end of this section, you will be able to:
• Simplify expressions using the Quotient Property for Exponents
• Simplify expressions with zero exponents
• Simplify expressions using the Quotient to a Power Property
• Simplify expressions by applying several properties
• Use the definition of a negative exponent
• Simplify expressions with integer exponents
• Simplify expressions by applying several properties with negative exponents
• Divide monomials
### Try It
Before you get started, take this readiness quiz:
1) Simplify: $\frac{8}{24}$
2) Simplify: $(2{m}^{3})^{5}$
3) Simplify: $\frac{12x}{12y}$
## Simplify Expressions Using the Quotient Property for Exponents
Earlier in this chapter, we developed the properties of exponents for multiplication. We summarize these properties below.
### Summary of Exponent Properties for Multiplication
If $a$ and $b$ are real numbers, and $m$ and $n$ are whole numbers, then
Product Property ${a}^{m}\times{{a}^{n}}={a}^{m+n}$ $({a}^{m})^{n}={a}^{m\times{n}}$ ${(ab)}^{m}={a}^{m}{b}^{m}$
Now we will look at the exponent properties for division. A quick memory refresher may help before we get started. You have learned to simplify fractions by dividing out common factors from the numerator and denominator using the Equivalent Fractions Property. This property will also help you work with algebraic fractions—which are also quotients.
### Equivalent Fractions Property
If $a$, $b$, and $c$ are whole numbers where $b\neq{0}, c\neq{0}$ then
$\displaystyle\frac{a}{b}=\frac{a\times{c}}{b\times{c}}$ and $\displaystyle\frac{a\times{c}}{b\times{c}}=\frac{a}{b}$
As before, we’ll try to discover a property by looking at some examples.
Example 1 Example 2 Consider $\displaystyle\frac{x^5}{x^2}$ $\displaystyle\frac{x^2}{x^3}$ What do they mean? $\displaystyle\frac{{x}\cdot {x}\cdot {x}\cdot {x}\cdot {x}}{{x}\cdot {x}}$ $\displaystyle\frac{{x}\cdot {x}}{{x}\cdot{x}\cdot{x}}$ Use the Equivalent Fractions Property. $\displaystyle\frac{\cancel x\cdot\cancel x\cdot x\cdot x\cdot x}{\cancel x\cdot\cancel x}$ $\displaystyle\frac{\cancel x\cdot\cancel x\cdot1}{\cancel x\cdot\cancel x\cdot x}$ Simplify $x^3$ $\displaystyle\frac{1}{x}$
Notice, in each case the bases were the same and we subtracted exponents.
When the larger exponent was in the numerator, we were left with factors in the numerator.
When the larger exponent was in the denominator, we were left with factors in the denominator—notice the numerator of 1.
We write:
$\displaystyle\frac{x^5}{x^2}$ $\displaystyle\frac{x^2}{x^3}$ $\displaystyle x^{5-2}$ $\displaystyle\frac{1}{x^{3-2}}$ $\displaystyle x^3$ $\displaystyle\frac{1}{x}$
This leads to the Quotient Property for Exponents.
### Quotient Property for Exponents
If $a$ is a real number, $a\neq{0}$, and $n$ are whole numbers, then
$\displaystyle\frac{a^m}{a^n}=a^{m-n}$, $m>n$ and $\displaystyle\frac{a^m}{a^n}=\frac{1}{a^{m-n}}$, $n>m$
A couple of examples with numbers may help to verify this property.
\begin{align*}\frac{3^4}{3^2}&=3^{4-2}\\[2ex]\frac{81}{9}&=3^{2}\\[2ex]9&=9\checkmark\end{align*} \begin{align*}\frac{5^2}{5^3}&=\frac1{5^{3-2}}\\[2ex]\frac{25}{125}&=\frac{1}{5^{1}}\\[2ex]\frac{1}{5}&=\frac{1}{5}\checkmark\end{align*}
### Example 5.5.1
Simplify:
a. $\frac{{x}^{9}}{{x}^{7}}$
b. $\frac{{3}^{10}}{{3}^{2}}$
Solution
To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.
a.
Step 1: Since $9 > 7$, there are more factors of $x$ in the numerator.
$\frac{x^9}{x^7}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
$\begin{eqnarray*}&=&x^{{\color{red}{9-7}}}\\\text{Simplify.}\;\;&=&x^2\end{eqnarray*}$
b.
Step 1: Since $10 > 2$, there are more factors of $x$ in the numerator.
$\frac{3^{10}}{3^2}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
$\begin{eqnarray*}&=&3^{\color{red}{{10\;-\;2}}}\\\text{Simplify.}\;\;&=&3^8 \end{eqnarray*}$
Notice that when the larger exponent is in the numerator, we are left with factors in the numerator.
### Try It
4) Simplify:
a. $\frac{{x}^{15}}{{x}^{10}}$
b. $\frac{{6}^{14}}{{6}^{5}}$
Solution
a. ${x}^{5}$
b. ${6}^{9}$
5) Simplify:
a. $\frac{{y}^{43}}{{y}^{37}}$
b. $\frac{{10}^{15}}{{10}^{7}}$
Solution
a. ${y}^{6}$
b. ${10}^{8}$
### Example 5.5.2
Simplify:
a. $\frac{{b}^{8}}{{b}^{12}}$
b. $\frac{{7}^{3}}{{7}^{5}}$
Solution
To simplify an expression with a quotient, we need to first compare the exponents in the numerator and denominator.
a.
Step 1: Since $12 > 8$, there are more factors of $b$ in the denominator.
$\frac{b^8}{b^{12}}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{b^{\color{red}{{12\;-\;8}}}}\\\text{Simplify.}\;\;&=&\frac{1}{b^4}\end{eqnarray*}$
b.
Step 1: Since $5 > 3$, there are more factors of $3$ in the denominator.
$\frac{7^3}{7^5}$
Step 2: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{7^{\color{red}{5-3}}}\\[1ex]\text{Simplify.}\;\;&=&\frac{1}{7^2}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{49}\end{eqnarray*}$
Notice that when the larger exponent is in the denominator, we are left with factors in the denominator.
### Try It
6) Simplify:
a. $\frac{{x}^{18}}{{x}^{22}}$
b.
$\frac{{12}^{15}}{{12}^{30}}$
Solution
a. $\frac{1}{{x}^{4}}$
b. $\frac{1}{{12}^{15}}$
7) Simplify:
a. $\frac{{m}^{7}}{{m}^{15}}$
b. $\frac{{9}^{8}}{{9}^{19}}$
Solution
a. $\frac{1}{{m}^{8}}$
b. $\frac{1}{{9}^{11}}$
Notice the difference in the two previous examples:
• If we start with more factors in the numerator, we will end up with factors in the numerator.
• If we start with more factors in the denominator, we will end up with factors in the denominator.
The first step in simplifying an expression using the Quotient Property for Exponents is to determine whether the exponent is larger in the numerator or the denominator. Later in this chapter, we will explore this further while using negative exponents.
### Example 5.5.3
Simplify:
a. $\frac{{a}^{5}}{{a}^{9}}$
b. $\frac{{x}^{11}}{{x}^{7}}$
Solution
a. Is the exponent of $a$ larger in the numerator or denominator? Since $9 > 5$, there are more $a$‘s in the denominator and so we will end up with factors in the denominator.
Step 1: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{\color{red}{1}}{a^{\color{red}{{9\;-\;5}}}}\\\text{Simplify.}\;\;&=&\frac{1}{a^4}\end{eqnarray*}$
b. Notice there are more factors of $x$ in the numerator, since $11 > 7$. So we will end up with factors in the numerator.
Step 1: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&x^{{\color{red}{11\;-\;7}}}\\\text{Simplify.}\;\;&=&x^4\end{eqnarray*}$
### Try It
8) Simplify:
a. $\frac{{b}^{19}}{{b}^{11}}$
b. $\frac{{z}^{5}}{{z}^{11}}$
Solution
a. ${b}^{8}$
b. $\frac{1}{{z}^{6}}$
9) Simplify:
a. $\frac{{p}^{9}}{{p}^{17}}$
b. $\frac{{w}^{13}}{{w}^{9}}$
Solution
a. $\frac{1}{{p}^{8}}$
b. ${w}^{4}$
## Simplify Expressions with an Exponent of Zero
A special case of the Quotient Property is when the exponents of the numerator and denominator are equal, such as an expression like $\frac{{a}^{m}}{{a}^{m}}$. From your earlier work with fractions, you know that:
$\frac{2}{2}=1$ $\frac{17}{17}=1$ $\frac{-43}{-43}=1$
In words, a number divided by itself is $1$. So, $\frac{x}{x}=1$, for any $x$ $({x} \neq {0})$, since any number divided by itself is $1$.
The Quotient Property for Exponents shows us how to simplify $\frac{{a}^{m}}{{a}^{n}}$ when $m>n$ and when $n < m$ by subtracting exponents. What if $m=n$?
Consider $\frac{8}{8}$, which we know is $1$.
\begin{align*} &\;&\frac{8}{8}&=1\\[1ex] &\text{Write}\;8\;\text{as}\;2^3\;&\frac{{2}^{3}}{{2}^{3}}&=1\\[1ex] &\text{Subtract exponents}\;&{2}^{3-3}&=1\\ &\text{Simplify}\;&2^0&=1 \end{align*}
Now we will simplify $\frac{{a}^{m}}{{a}^{m}}$ in two ways to lead us to the definition of the zero exponent. In general, for ${a}\neq{0}$ :
$\begin{array}{cc}\displaystyle\frac{a^m}{a^m}\;\;&\displaystyle\frac{a^m}{a^m}\\[3ex]a^{m-m}\;\;&\displaystyle\frac{\overbrace{\cancel a\cdot \cancel a\cdot...\cdot\cancel a}^{{\color{blue}{m\text{ factors}}}}}{\underbrace{\cancel a\cdot \cancel a\cdot...\cdot\cancel a}_{{\color{blue}{m\text{ factors}}}}}\\[3ex]a^0\;\;&1\end{array}$
We see $\frac{{a}^{m}}{{a}^{m}}$ simplifies to ${a}^{0}$ and to $1$. So ${a}^{0}=1$
### Zero Exponent
If $a$ is a non-zero number, then ${a}^{0}=1$.
Any nonzero number raised to the zero power is $1$.
In this text, we assume any variable that we raise to the zero power is not zero.
### Example 5.5.4
Simplify:
a. ${9}^{0}$
b. ${n}^{0}$
Solution
The definition says any non-zero number raised to the zero power is $1$.
a.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&9^0\\&=&1\end{eqnarray*}$
b.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&n^0\\&=&1\end{eqnarray*}$
### Try It
10) Simplify:
a. ${15}^{0}$
b. ${m}^{0}$
Solution
a. 1
b. 1
11) Simplify:
a. ${k}^{0}$
b. ${29}^{0}$
Solution
a. 1
b. 1
Now that we have defined the zero exponent, we can expand all the Properties of Exponents to include whole number exponents.
What about raising an expression to the zero power? Let’s look at $(2x)^{0}$. We can use the product to a power rule to rewrite this expression.
\begin{align*} &\;&(2x)^{0}\\[1ex] &\text{Use the product to a power rule.}\;&{2}^{0}{x}^{0}\\[1ex] &\text{Use the zero exponent property.}\;&1\times{1}\\ &\text{Simplify}\;&1 \end{align*}
This tells us that any nonzero expression raised to the zero power is one.
### Example 5.5.5
Simplify:
a. $(5b)^{0}$
b. $(-4{a}^{2}b)^{0}$
Solution
a.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&(5b)^0\\&=&1\end{eqnarray*}$
b.
Step 1: Use the definition of the zero exponent.
$\begin{eqnarray*}&=&(-4a^{2b})^0\\&=&1\end{eqnarray*}$
### Try It
12) Simplify:
a. $(11z)^{0}$
b. $(-11p{q}^{3})^{0}$
Solution
a. $1$
b. $1$
13) Simplify:
a. $(-6d)^{0}$
b. $(-8{m}^{2}{n}^{3})^{0}$
Solution
a. $1$
b. $1$
## Simplify Expressions Using the Quotient to a Power Property
Now we will look at an example that will lead us to the Quotient to a Power Property.
\begin{align*} &\;&\;&\left(\frac{x}{y}\right)^{3}\\[1ex] &\text{This means:}&\;&\frac{x}{y}\cdot\frac{x}{y}\cdot\frac{x}{y}\\[1ex] &\text{Multiply the fractions.}&\;&\frac{x\cdot{x}\cdot{x}}{y\cdot{y}\cdot{y}}\\[1ex] &\text{Write with exponents.}&\;&\frac{{x}^{3}}{{y}^{3}} \end{align*}
Notice that the exponent applies to both the numerator and the denominator.
\begin{align*} &\text{We write:}\;&\left(\frac{x}{y}\right)^{3}\\[1ex] &\;&\frac{{x}^{3}}{{y}^{3}}\;\;\;\end{align*}
This leads to the Quotient to a Power Property for Exponents.
### Quotient to a Power Property for Exponents
If $a$ and $b$ are real numbers, $b\neq{0}$, and $m$ is a counting number, then
$(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
To raise a fraction to a power, raise the numerator and denominator to that power.
$\begin{eqnarray*}(\frac{2}{3})^3&=&\frac{2^3}{3^3}\\\frac{2}{3}\cdot\frac{2}{3}\cdot\frac{2}{3}&=&\frac{8}{27}\\\frac{8}{27}&=& \frac{8}{27}\checkmark\end{eqnarray*}$
### Example 5.5.6
Simplify:
a. $(\frac{3}{7})^{2}$
b. $(\frac{b}{3})^{4}$
c. $\left(\frac{k}{j}\right)^3$
Solution
a.
Step 1: Use the Quotient Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\begin{eqnarray*}&=&{b}^{0}\\ \text{Simplify.}\;\;&=&1 \end{eqnarray*}$
b.
Step 1: Use the Quotient Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\begin{eqnarray*}&=&\frac{b^{\color{red}{4}}}{3^{\color{red}{4}}}\\\text{Simplify.}\;\;&=&\frac{b^{4}}{81}\end{eqnarray*}$
c.
Step 1: Raise the numerator and denominator to the third power.
$\displaystyle\frac{k^{\color{red}{3}}}{j^{\color{red}{3}}}$
### Try It
14) Simplify:
a. $(\frac{5}{8})^{2}$
b. $(\frac{p}{10})^{4}$
c. $(\frac{m}{n})^{7}$
Solution
a. $\frac{25}{64}$
b. $\frac{{p}^{4}}{10,000}$
c. $\frac{{m}^{7}}{{n}^{7}}$
15) Simplify:
a. $(\frac{1}{3})^{3}$
b. $(\frac{-2}{q})^{3}$
c. $(\frac{w}{x})^{4}$
Solution
a. $\frac{1}{27}$
b. $\frac{-8}{{q}^{3}}$
c. $\frac{{w}^{4}}{{x}^{4}}$
## Simplify Expressions by Applying Several Properties
We’ll now summarize all the properties of exponents so they are all together to refer to as we simplify expressions using several properties. Notice that they are now defined for whole number exponents.
### Summary of Exponent Properties
If $a$ and $b$ are real numbers, and $m$ and $n$ are whole numbers, then
Product Property ${a}^{m}\times{a}^{n}={a}^{m+n}$ $({a}^{m})^{n}={a}^{m\times{n}}$ $(ab)^{m}={a}^{m}{b}^{m}$ $\frac{{a}^{m}}{{b}^{m}}={a}^{m-n}, a\neq{0} , m>n$ $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}},a\neq{0},n>m$ ${a}^{0}=1,a\neq{0}$ $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}},b\neq{0}$
### Example 5.5.7
Simplify: $\frac{(y^4)^2}{y^6}$
Solution
Step 1: Multiply the exponents in the numerator.
$\frac{{y}^{8}}{{y}^{6}}$
Step 2: Subtract the exponents.
${y}^{2}$
### Try It
16) Simplify: $\frac{({m}^{5})^4}{m^7}$
Solution
${m}^{13}$
17) Simplify: $\frac{{({k}^{2})}^{6}}{{k}^{7}}$
Solution
${k}^{5}$
### Example 5.5.8
Simplify: $\frac{{b}^{12}}{{({b}^{2})}^{6}}$
Solution
Step 1: Multiply the exponents in the numerator.
$\frac{{b}^{12}}{{b}^{12}}$
Step 2: Subtract the exponents.
$\begin{eqnarray*}&=&{b}^{0}\\ \text{Simplify.}\;\;&=&1 \end{eqnarray*}$
### Try It
18) Simplify: $\frac{{n}^{12}}{{({n}^{3}})^{4}}$
Solution
$1$
19) Simplify: $\frac{{x}^{15}}{(x^3)^{5}}$
Solution
$1$
### Example 5.5.9
Simplify: $\left(\frac{{y}^{9}}{{y}^{4}}\right)^{2}$
Solution
Step 1: Remember parentheses come before exponents.
Notice the bases are the same, so we can simplify inside the parentheses.
Subtract the exponents.
$(y^5)^2$
Step 2: Multiply the exponents.
${y}^{10}$
### Try It
20) Simplify: $\left(\frac{{r}^{5}}{{r}^{3}}\right)^{4}$
Solution
${r}^{8}$
21) Simplify: $\left(\frac{{v}^{6}}{{v}^{4}}\right)^{3}$
Solution
${v}^{6}$
### Example 5.5.10
Simplify: $\left(\frac{{j}^{2}}{{k}^{3}}\right)^{4}$
Solution
Here we cannot simplify inside the parentheses first, since the bases are not the same.
Step 1: Raise the numerator and denominator to the third power using the Quotient to a Power Property, $(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
$\left(\frac{{j}^{2}}{{k}^{3}}\right)^{4}$
Step 2: Use the Power Property and simplify.
$\frac{{j}^{8}}{{k}^{12}}$
### Try It
22) Simplify: $\left(\frac{a^3}{b^2}\right)^{4}$
Solution
$\frac{{a}^{12}}{{b}^{8}}$
23) Simplify: $\left(\frac{q^7}{r^5}\right)^3$
Solution
$\frac{{q}^{21}}{{r}^{15}}$
### Example 5.5.11
Simplify: $\left(\frac{2{m}^{2}}{5n}\right)^{4}$
Solution
Step 1: Raise the numerator and denominator to the fourth power, using the Quotient to a Power Property, $\left(\frac{a}{b}\right)^m=\;\frac{a^m}{b^m}$
$\frac{(2m^{2})^{4}}{{(5n)}^{4}}$
Step 2: Raise each factor to the fourth power.
$\frac{(2m^{2})^{4}}{{(5n)}^{4}}$
Step 3: Use the Power Property and simplify.
$\frac{16{m}^{8}}{625{n}^{4}}$
### Try It
24) Simplify: $\left(\frac{7x^3}{9y}\right)^2$
Solution
$\frac{49{x}^{6}}{81{y}^{2}}$
25) Simplify: $\left(\frac{3x^4}{7y}\right)^2$
Solution
$\frac{9{x}^{8}}{49{y}^{2}}$
### Example 5.5.12
Simplify: $\frac{\left(x^3\right)^4\times\left(x^2\right)^5}{\left(x^6\right)^5}$
Solution
Step 1: Use the Power Property, ${(a^m)}^{n}={a}^{mn}$
$\frac{\left(x^{12}\right)\times\left(x^{10}\right)}{\left(x^{30}\right)}$
Step 2: Add the exponents in the numerator.
$\frac{x^{22}}{x^{30}}$
Step 3: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\frac{1}{{x}^{8}}$
### Try It
26) Simplify: $\frac{{({a}^{2})}^{3}{({a}^{2})}^{4}}{({a}^{4})^{5}}$
Solution
$\frac{1}{{a}^{6}}$
27) Simplify: $\frac{{({p}^{3})}^{4}{({p}^{5})}^{3}}{{({p}^{7})}^{6}}$
Solution
$\frac{1}{{p}^{15}}$
### Example 5.5.13
Simplify: $\frac{(10p^3)^2}{(5p)^3\cdot\left(2p^5\right)^4}$
Solution
Step 1: Use the Product to a Power Property, ${(ab)}^{m}={a}^{m}{b}^{m}$
$\frac{{(10)}^{2}{({p}^{3})}^{2}}{(5)^3(p)^{3}\cdot (2)^4(p^5)^4}$
Step 2: Use the Power Property, ${{a}^{m}}^{n}={a}^{m\times{n}}$
$\frac{100{p}^{6}}{125p^3\cdot 16{p}^{20}}$
Step 3: Add the exponents in the denominator.
$\frac{100{p}^{6}}{125\times{16}{p}^{23}}$
Step 4: Use the Quotient Property, $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{n-m}}$
$\begin{eqnarray*}&=&\frac{100}{125\times{16{p}}^{17}}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{20{p}^{17}} \end{eqnarray*}$
### Try It
28) Simplify: $\frac{{(3{r}^{3}}^{2}{({r}^{3})}^{7}}{{({r}^{3})}^{3}}$
Solution
$9{r}^{18}$
29) Simplify: $\frac{{(2{x}^{4})}^{5}}{{(4{x}^{3})}^{2}{({x}^{3})}^{5}}$
Solution
$\frac{2}{x}$
## Use the Definition of a Negative Exponent
Up until this point, we have been careful to minimize our use of negative exponents. Before taking on division of polynomials, we will revisit some of the properties we’ve seen in the previous sections but, this time, with negative exponents.
### Quotient Property for Exponents
If $a$ is a real number, ${a}\neq 0$ , and $m$ and $n$ are whole numbers, then
$\frac{a^m}{a^n}=a^{m-n}$, for $m>n$ and $\frac{a^m}{a^n}=\frac{1}{a^{n-m}}$, for $n>m$.
What if we just subtract exponents regardless of which is larger?
Let’s consider $\frac{{x}^{2}}{{x}^{5}}$.
We subtract the exponent in the denominator from the exponent in the numerator.
$\begin{eqnarray*}&=&\frac{{x}^{2}}{{x}^{5}}\\&=&{x}^{2-5}\\&=&{x}^{-3}\end{eqnarray*}$
We can also simplify $\frac{{x}^{2}}{{x}^{5}}$ by dividing out common factors:
$\begin{eqnarray*}&=&\frac{{\color{red}{\cancel x}}\cdot{\color{red}{\cancel x}}}{{\color{red}{\cancel x}}\cdot{\color{red}{\cancel x}}\cdot x\cdot x\cdot x}\\&=&\frac1{x^3}\end{eqnarray*}$
This implies that ${x}^{-3}=\frac{1}{{x}^{3}}$ and it leads us to the definition of a negative exponent.
## Negative Exponents
If $n$ is an integer and $a\neq{0}$, then ${a}^{\text{−}n}=\frac{1}{{a}^{n}}$.
The negative exponent tells us we can re-write the expression by taking the reciprocal of the base and then changing the sign of the exponent.
Any expression that has negative exponents is not considered to be in simplest form. We will use the definition of a negative exponent and other properties of exponents to write the expression with only positive exponents.
For example, if after simplifying an expression we end up with the expression ${x}^{-3}$, we will take one more step and write $\frac{1}{{x}^{3}}$. The answer is considered to be in simplest form when it has only positive exponents.
### Example 5.5.14
Simplify:
a. ${4}^{-2}$
b. ${10}^{-3}$
Solution
a.
Step 1: Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{{4}^{2}}\\ \text{Simplify.}\;\;&=&\frac{1}{16} \end{eqnarray*}$
b.
Step 1: Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{{10}^{3}}\\ \text{Simplify.}\;\;&=&\frac{1}{1000}\end{eqnarray*}$
### Try It
30) Simplify:
a. ${2}^{-3}$
b. ${10}^{-7}$
Solution
a. $\frac{1}{8}$
b. $\frac{1}{{10}^{7}}$
31) Simplify:
a. ${3}^{-2}$
b. ${10}^{-4}$
Solution
a. $\frac{1}{9}$
b. $\frac{1}{10,000}$
In example 5.2.13 we raised an integer to a negative exponent. What happens when we raise a fraction to a negative exponent? We’ll start by looking at what happens to a fraction whose numerator is one and whose denominator is an integer raised to a negative exponent.
\begin{align*} &\;&\;&\frac{1}{a^{-n}}\\[1ex] &\text{Use the definition of a negative exponent,}\;{a}^{-n}=\frac{1}{a^n}&\;&\frac{1}{\frac{1}{{a}^{n}}}\\[1ex] &\text{Simplify the complex fraction.}&\;&1\cdot \frac{{a}^{n}}{1}\\[1ex] &\text{Multiply.}&\;&{a}^{n} \end{align*}
This leads to the Property of Negative Exponents.
### Property of Negative Exponents
If $n$ is an integer and ${a}\neq 0$, then $\frac{1}{{a}^{-n}}={a}^{n}$.
### Example 5.5.15
Simplify:
a. $\frac{1}{{y}^{-4}}$
b. $\frac{1}{{3}^{-2}}$
Solution
a.
Step 1: Use the property of a negative exponent, $\frac{1}{{a}^{-n}}={a}^{n}$.
${y}^{4}$
b.
Step 1: Use the property of a negative exponent, $\frac{1}{{a}^{-n}}={a}^{n}$.
$\begin{eqnarray*}&=&{3}^{2}\\\text{Simplify.}\;\;&=&9\end{eqnarray*}$
### Try It
32) Simplify:
a. $\frac{1}{{p}^{-8}}$
b. $\frac{1}{{4}^{-3}}$
Solution
a. ${p}^{8}$
b. $64$
33) Simplify:
a. $\frac{1}{{q}^{-7}}$
b. $\frac{1}{{2}^{-4}}$
Solution
a. ${q}^{7}$
b. $16$
Suppose now we have a fraction raised to a negative exponent. Let’s use our definition of negative exponents to lead us to a new property.
$(\frac{3}{4})^{-2}$ Use the definition of a negative exponent, ${a}^{-n}=\frac{1}{{a}^{n}}$. $\frac {1}{(\frac{3}{4})^{2}}$ Simplify the denominator. $\frac{1}{\frac{9}{16}}$ Simplify the complex fraction. $\frac{16}{9}$ But we know that $\frac{16}{9}$ is $(\frac{4}{3})^{2}$. $\frac{4^2}{3^2}$ This tells us that: $(\frac{3}{4})^{-2}=(\frac{4}{3})^{2}$
To get from the original fraction raised to a negative exponent to the final result, we took the reciprocal of the base—the fraction—and changed the sign of the exponent.
This leads us to the Quotient to a Negative Power Property.
### Quotient to a Negative Exponent Property
If $a$ and $b$ are real numbers, $a \neq 0$, $b\neq 0$, and $n$ is an integer, then $(\frac{a}{b})^{-n}= (\frac{b}{a})^{n}$.
### Example 5.5.16
Simplify:
a. $\left(\frac{5}{7}\right)^{-2}$
b. $\left(-\frac{2x}{y}\right)^{-3}$
Solution
a.
Step 1: Use the Quotient to a Negative Exponent Property, $(\frac{a}{b})^{-n}={(\frac{b}{a})}^{n}$.
Step 2: Take the reciprocal of the fraction and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(\frac{7}{5}\right)^{2}\;\;\\\text{Simplify.}\;\;&=&\frac{49}{25} \end{eqnarray*}$
b.
Step 1: Use the Quotient to a Negative Exponent Property, $(\frac{a}{b})^{-n}={(\frac{b}{a})}^{n}$
Step 2: Take the reciprocal of the fraction and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(-\frac{y}{2x}\right)^{3}\;\;\\[1ex]\text{Simplify.}\;\;&=&-\frac{{y}^{3}}{8{x}^{3}}\end{eqnarray*}$
### Try It
34) Simplify:
a. $\left(\frac{2}{3}\right)^{-4}$
b. $(-\frac{6m}{n})^{-2}$
Solution
a. $\frac{81}{16}$
b. $\frac{{n}^{2}}{36{m}^{2}}$
35) Simplify:
a. $\left(\frac{3}{5}\right)^{-3}$
b. $\left(-\frac{a}{2b}\right)^{-4}$
Solution
a. $\frac{125}{27}$
b. $\frac{16{b}^{4}}{{a}^{4}}$
When simplifying an expression with exponents, we must be careful to correctly identify the base.
### Example 5.5.17
Simplify:
a. $(-3)^{-2}$
b. $-3^{-2}$
c. $\left(-\frac{1}{3}\right)^{-2}$
d. $-\left(\frac{1}{3}\right)^{-2}$
Solution
a. Here the exponent applies to the base $-3$.
Step 1: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{(-3)^2}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{9}\end{eqnarray*}$
b. The expression $-{3}^{-2}$ means “find the opposite of ${3}^{-2}$.”
Step 1: Rewrite as a product with $-1$.
$-1\times 3^{-2}$
Step 2: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&-1\cdot \frac{1}{{3}^{2}}\\[1ex]\text{Simplify.}\;\;&=&-\frac{1}{9}\end{eqnarray*}$
c. Here the exponent applies to the base $-\frac{1}{3}$.
Step 1: Take the reciprocal of the base and change the sign of the exponent.
$\begin{eqnarray*}&=&\left(-\frac{3}{1}\right)^2\\[1ex]\text{Simplify.}\;\;&=&9\end{eqnarray*}$
d. The expression $-(\frac{1}{3})^{-2}$ means “find the opposite of $(\frac{1}{3})^{-2}$.” Here the exponent applies to the base $\frac{1}{3}$.
Step 1: Rewrite as a product with $-1$.
$-1\cdot \left(\frac{3}{1}\right)^{2}$
Step 2: Take the reciprocal of the base and change the sign of the exponent.
$-9$
### Try It
36) Simplify:
a. $(-5)^{-2}$
b. ${−}{5}^{-2}$
c. $\left(-\frac{1}{5}\right)^{-2}$
d. ${−}\left(\frac{1}{5}\right)^{-2}$
Solution
a. $\frac{1}{25}$
b. $-\frac{1}{25}$
c. $25$
d. $-25$
37) Simplify:
a. $(-7)^{-2}$
b. ${−}{7}^{-2}$
c. $\left(-\frac{1}{7}\right)^{-2}$
d. $\left(-\frac{1}{7}\right)^{-2}$
Solution
a. $\frac{1}{49}$
b. $-\frac{1}{49}$
c. $49$
d. $-49$
We must be careful to follow the Order of Operations. In the next example, parts (a) and (b) look similar, but the results are different.
### Example 5.5.18
Simplify:
a. $4\times{2}^{-1}$
b. $(4\times{2})^{-1}$
Solution
a.
Step 1: Do exponents before multiplication.
$4\times{2}^{-1}$
Step 2: Use ${a}^{-n}=\frac{1}{{a}^{n}}$
$\begin{eqnarray*}&=&4\times \frac{1}{{2}^{1}}\\[1ex] \text{Simplify.}\;\;&=&2\end{eqnarray*}$
b.
Step 1: Simplify inside the parentheses first.
$(8)^{-1}$
Step 2: Use ${a}^{-n}=\frac{1}{{a}^{n}}$
$\begin{eqnarray*}&=&\frac{1}{{8}^{1}}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{8} \end{eqnarray*}$
### Try It
38) Simplify:
a. $6\times{3}^{-1}$
b. $(6\times 3)^{-1}$
Solution
a. $2$
b. $\frac{1}{18}$
39) Simplify:
a. $8\times {2}^{-2}$
b. $8\times {2}^{-2}$
Solution
a. $2$
b. $\frac{1}{256}$
When a variable is raised to a negative exponent, we apply the definition the same way we did with numbers. We will assume all variables are non-zero.
### Example 5.5.19
Simplify:
a. ${x}^{-6}$
b. $({u}^{4})^{-3}$
Solution
a.
Step 1: Use the definition of a negative exponent ${a}^{-n}=\frac{1}{{a}^{n}}$
$\frac{1}{{x}^{6}}$
b.
Step 1: Use the definition of a negative exponent ${a}^{-n}=\frac{1}{{a}^{n}}$.
$\begin{eqnarray*}&=&\frac{1}{(u^4)^3}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{u^{12}} \end{eqnarray*}$
### Try It
40) Simplify:
a. ${y}^{-7}$
b. $({z}^{3})^{-5}$
Solution
a. $\frac{1}{{y}^{7}}$
b. $\frac{1}{{z}^{15}}$
41) Simplify:
a. ${p}^{-9}$
b. $({q}^{4})^{-6}$
Solution
a. $\frac{1}{p}^{9}$
b. $\frac{1}{q}^{24}$
When there is a product and an exponent we have to be careful to apply the exponent to the correct quantity. According to the Order of Operations, we simplify expressions in parentheses before applying exponents. We’ll see how this works in the next example.
### Example 5.5.20
Simplify:
a. $5{y}^{-1}$
b. $(5y)^{-1}$
c. $(-5y)^{-1}$
Solution
a.
Step 1: Notice the exponent applies to just the base $y$.
Step 2: Take the reciprocal of $y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&5\cdot \left(\frac{1}{y^1}\right)\\[1ex] \text{Simplify.}\;\;&=&\frac{5}{y} \end{eqnarray*}$
b.
Step 1: Here the parentheses make the exponent apply to the base $5y$.
Step 2: Take the reciprocal of $5y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{{5y}^1}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{5y} \end{eqnarray*}$
c.
Step 1: The base here is $−5y$.
Step 2: Take the reciprocal of $−5y$ and change the sign of the exponent.
$\begin{eqnarray*}&=&\frac{1}{-5y}\\[1ex]\text{Simplify.}\;\;&=&\left(-\frac{1}{5y}\right) \end{eqnarray*}$
### Try It
42) Simplify:
a. $(8{p}^{-1})$
b. ${(8p)}^{-1}$
c. ${(-8p)}^{-1}$
Solution
a. $\frac{8}{p}$
b. $\frac{1}{8p}$
c. $-\frac{1}{8p}$
43) Simplify:
a. ${11q}^{-1}$
b. ${(11q)}^{-1}$
c. ${(-11q)}^{-1}$
Solution
a. $\frac{1}{11q}$
b. $\frac{1}{11q}$
c. $-\frac{1}{11q}$
With negative exponents, the Quotient Rule needs only one form $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$, for $a\neq{0}$. When the exponent in the denominator is larger than the exponent in the numerator, the exponent of the quotient will be negative.
## Simplify Expressions with Integer Exponents
All of the exponent properties we developed earlier in the chapter with whole number exponents apply to integer exponents, too. We restate them here for reference.
## Summary of Exponent Properties
If $a$ and $b$ are real numbers, and $m$ and $n$ are integers, then
Product Property ${a^m}\times{a^n}=a^{m+n}$ $({a^m})^n={a}{mn}$ $(ab)^m=a^m b^m$ $\frac{a^m}{a^n}={a}^{m-n}$ $a^0=1 , {a}\neq{1}$ $(\frac {a}{b})^m=\frac{{a}^{m}}{{b}^{m}}, {b}\neq{0}$ ${a}^{-n}=\frac{1}{a^n}$ and $\frac{1}{a^{-n}}=a^{n}$ $(\frac{a}{b})^{-n}=(\frac{b}{a})^{n}$
### Example 5.5.21
Simplify:
a. ${x}^{-4}\times{x}^{6}$
b. ${y}^{-6}\times{y}^{4}$
c. ${z}^{-5}\times{z}^{-3}$
Solution
a.
Step 1: Use the Product Property, $a^m\cdot a^n=a^{m+n}$
$\begin{eqnarray*}&=&{x}^{-4+6}\\[1ex] \text{Simplify.}\;\;&=&x^2 \end{eqnarray*}$
b.
Step 1: Notice the same bases, so add the exponents.
$\begin{eqnarray*}&=&y^{-6+4}\\[1ex] \text{Simplify.}\;\;&=&y^{-2}\end{eqnarray*}$
Step 3: Use the definition of a negative exponent, $a^{-n}=\frac{1}{a^n}$
$\frac{1}{y^2}$
c.
Step 1: Add the exponents, since the bases are the same.
$\begin{eqnarray*}&=&z^{-5-3}\\[1ex] \text{Simplify.}\;\;&=&z^{-8}\end{eqnarray*}$
Step 3: Take the reciprocal and change the sign of the exponent, using the definition of a negative exponent.
$\frac{1}{z^{8}}$
### Try It
44) Simplify:
a. ${{x}^{-3}}\times{{x}^{7}}$
b. ${{y}^{-7}}\times{{y}^{2}}$
c. ${{z}^{-4}}\times{{z}^{-5}}$
Solution
a. ${x}^{4}$
b. $\frac{1}{{y}^{5}}$
c. $\frac{1}{{z}^{9}}$
45) Simplify:
a. ${{a}^{-1}}\times{{a}^{6}}$
b. $b^{-8}\times b^{4}$
c. ${c}^{-8}\times{{c}^{-7}}$
Solution
a. ${a}^{5}$
b. $\frac{1}{{b}^{4}}$
c. $\frac{1}{{c}^{15}}$
In the next two examples, we’ll start by using the Commutative Property to group the same variables together. This makes it easier to identify the like bases before using the Product Property.
### Example 5.5.22
Simplify: $({m}^{4}{n}^{-3})({m}^{-5}{n}^{-2})$
Solution
Step 1: Use the Commutative Property to get like bases together.
$m^4 m^{-5} n^{-2} n^{-3}$
Step 2: Add the exponents for each base.
$m^{-1} n^{-5}$
Step 3: Take reciprocals and change the signs of the exponents.
$\begin{eqnarray*}&=&\frac{1}{m^1} \frac{1}{n^5}\\[1ex] \text{Simplify.}\;\;&=&\frac{1}{mn^5}\end{eqnarray*}$
### Try It
46) Simplify: $({p}^{6}{q}^{-2})({p}^{-9}{q}^{-1})$
Solution
$\frac{1}{{p}^{3}{q}^{3}}$
47) Simplify: $({r}^{5}{s}^{-3})({r}^{-7}{s}^{-5})$
Solution
$\frac{1}{{r}^{2}{s}^{8}}$
If the monomials have numerical coefficients, we multiply the coefficients, just like we did earlier.
### Example 5.5.23
Simplify: $(2{x}^{-6}{y}^{8})(-5{x}^{5}{y}^{-3})$
Solution
Step 1: Rewrite with the like bases together
$2(-5)(x^{-6} x^{5}) (y^8 y^{-3})$
Step 2: Multiply the coefficients and add the exponents of each variable.
$-10x^{-1}y^5$
Step 3: Use the definition of a negative exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&-10\cdot \left(\frac{1}{x^1}\right)\cdot y^5\\[1ex] \text{Simplify.}\;\;&=&-\frac{10y^5}{x}\end{eqnarray*}$
### Try It
48) Simplify: $(3{u}^{-5}{v}^{7})(-4{u}^{4}{v}^{-2})$
Solution
$-\frac{12{v}^{5}}{u}$
49) Simplify: $(-6{c}^{-6}{d}^{4})(-5{c}^{-2}{d}^{-1})$
Solution
$\frac{30{d}^{3}}{{c}^{8}}$
In the next two examples, we’ll use the Power Property and the Product to a Power Property.
### Example 5.5.24
Simplify: $(6{k}^{3})^{-2}$
Solution
Step 1: Use the Product to a Power Property, $(ab)^m=a^m b^m$
$(6)^{(-2)}(k^3)^{-2}$
Step 2: Use the Power Property, $(a^m)^n= {a}^{mn}$
$6^{-2}k^{-6}$
Step 3: Use the Definition of a Negative Exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&\frac{1}{6^2}\cdot \frac{1}{k^6}\\[1ex]\text{Simplify.}\;\;&=&\frac{1}{36k^6}\end{eqnarray*}$
### Try It
50) Simplify: $(-4{x}^{4})^{-2}$
Solution
$\frac{1}{16{x}^{8}}$
51) Simplify: $\left(2{b}^{3}\right)^{-4}$
Solution
$\frac{1}{16{b}^{12}}$
### Example 5.5.25
Simplify: $(5{x}^{-3})^{2}$
Solution
Step 1: Use the Product to a Power property, $(ab)^m=a^m b^m$
$5^2(x^{-3})^2$
Step 2: Simplify $5^2$ and multiply the exponents of $x$ using the Power Property, $(a^m)^n=a^{mn}$
$25x^{-6}$
Step 3: Rewrite $x^{-6}$ by using the Definition of a Negative Exponent, $a^{-n}=\frac{1}{a^n}$
$\begin{eqnarray*}&=&25\cdot \left(\frac{1}{x^6}\right)\\[1ex] \text{Simplify.}\;\;&=&\frac{25}{x^6} \end{eqnarray*}$
### Try It
52) Simplify: $(8{a}^{-4})^{2}$
Solution
$\frac{64}{{a}^{8}}$
53) Simplify: $(2{c}^{-4})^{3}$
Solution
$\frac{8}{{c}^{12}}$
To simplify a fraction, we use the Quotient Property and subtract the exponents.
### Example 5.5.26
Simplify:$\frac{{r}^{5}}{{r}^{-4}}$
Solution
Step 1: Use the Quotient Property, $\frac{a^m}{a^n}=a^{m-n}$c
$\begin{eqnarray*}&=&r^{5-(-4)}\\[1ex]\text{Simplify.}\;\;&=&r^9 \end{eqnarray*}$
### Try It
54) Simplify: $\frac{{x}^{8}}{{x}^{-3}}$
Solution
${x}^{11}$
55) Simplify: $\frac{{y}^{8}}{{y}^{-6}}$
Solution
${y}^{14}$
## Divide Monomials
You have now been introduced to all the properties of exponents and used them to simplify expressions. Next, you’ll see how to use these properties to divide monomials. Later, you’ll use them to divide polynomials.
### Example 5.5.27
Find the quotient: ${56x^7}\div{8x^3}$
Solution
Step 1: Rewrite as a fraction.
$\frac{56{x}^{7}}{8{x}^{3}}$
Step 2: Use fraction multiplication.
$\frac{56}{8}\times\frac{x^7}{x^3}$
Step 3: Simplify and use the Quotient Property.
$7x^4$
### Try It
56) Find the quotient: $42{y}^{9}\div{6{y}^{3}}$
Solution
$7{y}^{6}$
57) Find the quotient: $48{z}^{8}\div{8{z}^{2}}$
Solution
$6{z}^{6}$
### Example 5.5.28
Find the quotient: $\frac{45{a}^{2}{b}^{3}}{-5a{b}^{5}}$
Solution
Step 1: Use fraction multiplication.
${\frac{45}{-5}}{\frac{{a}^{2}}{a}}{\frac{{b}^{3}}{{b}^{5}}}$
Step 2: Simplify and use the Quotient Property.
$\begin{eqnarray*}&=&-9\cdot {a}\cdot {\frac{1}{{b}^{2}}}\\[1ex] \text{Multiply.}\;\;&=&-\frac{9a}{{b}^{2}} \end{eqnarray*}$
### Try It
58) Find the quotient: $\frac{-72{a}^{7}{b}^{3}}{8{a}^{12}{b}^{4}}$
Solution
$-\frac{9}{{a}^{5}b}$
59) Find the quotient: $\frac{-63{c}^{8}{d}^{3}}{7{c}^{12}{d}^{2}}$
Solution
$\frac{-9d}{{c}^{4}}$
### Example 5.5.29
Find the quotient: $\frac{24{a}^{5}{b}^{3}}{48a{b}^{4}}$
Solution
Step 1: Use fraction multiplication.
$\frac{24}{48}\cdot \frac{a^5}{a}\cdot \frac{b^3}{b^4}$
Step 2: Simplify and use the Quotient Property.
$\begin{eqnarray*}&=&\frac{1}{2}\cdot {a}^{4}\cdot \frac{1}{b}\\[1ex] \text{Multiply.}\;\;&=&\frac{{a}^{4}}{2b} \end{eqnarray*}$
### Try It
60) Find the quotient:$\frac{16{a}^{7}{b}^{6}}{24a{b}^{8}}$
Solution
$\frac{2{a}^{6}}{3{b}^{2}}$
61) Find the quotient: $\frac{27{p}^{4}{q}^{7}}{-45{p}^{12}q}$
Solution
$-\frac{3{q}^{6}}{5{p}^{8}}$
Once you become familiar with the process and have practised it step by step several times, you may be able to simplify a fraction in one step.
### Example 5.5.30
Find the quotient: $\frac{14{x}^{7}{y}^{12}}{21{x}^{11}{y}^{6}}$
Solution
Be very careful to simplify $\frac{14}{21}$ by dividing out a common factor, and to simplify the variables by subtracting their exponents.
Step 1: Simplify and use the Quotient Property.
$\frac{2{y}^{6}}{3{x}^{4}}$
### Try It
62) Find the quotient: $\frac{28{x}^{5}{y}^{14}}{49{x}^{9}{y}^{12}}$
Solution
$\frac{4{y}^{2}}{7{x}^{4}}$
63) Find the quotient: $\frac{30{m}^{5}{n}^{11}}{48{m}^{10}{n}^{14}}$
Solution
$\frac{5}{8{m}^{5}{n}^{3}}$
In all examples so far, there was no work to do in the numerator or denominator before simplifying the fraction. In the next example, we’ll first find the product of two monomials in the numerator before we simplify the fraction. This follows the order of operations. Remember, a fraction bar is a grouping symbol.
### Example 5.5.31
Find the quotient:$\frac{(6{x}^{2}{y}^{3})(5{x}^{3}{y}^{2})}{(3{x}^{4}{y}^{5})}$
Solution
Step 1: Simplify the numerator.
$\begin{eqnarray*}&=&\frac{30{x}^{5}{y}^{5}}{3{x}^{4}{y}^{5}}\\[1ex]\text{Simplify.}\;\;&=&10x \end{eqnarray*}$
### Try It
64) Find the quotient: $\frac{(6{a}^{4}{b}^{5})(4{a}^{2}{b}^{5})}{12{a}^{5}{b}^{8}}$
Solution
$2a{b}^{2}$
65) Find the quotient: $\frac{(-12{x}^{6}{y}^{9})(-4{x}^{5}{y}^{8})}{-12{x}^{10}{y}^{12}}$
Solution
$-4x{y}^{5}$
Access these online resources for additional instruction and practice with dividing monomials:
### Key Concepts
• Quotient Property for Exponents:
• If $a$ is a real number, $a\neq{0}$, and $m$, $n$ are whole numbers, then:
$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$, $m>n$ and $\frac{{a}^{m}}{{a}^{n}}=\frac{1}{{a}^{m-n}}$, $n>m$
• Zero Exponent
• If $a$ is a non-zero number, then ${a}^{0}=1$.
• Quotient to a Power Property for Exponents:
• If $a$ and $b$ are real numbers, $b\neq{0}$ and $m$ is a counting number, then:
$(\frac{a}{b})^{m}=\frac{{a}^{m}}{{b}^{m}}$
• To raise a fraction to a power, raise the numerator and denominator to that power.
• Summary of Exponent Properties
• If $a$, $b$ are real numbers and $m$, $n$ are whole numbers, then
Product Property $a^m a^n=a^{m+n}$ ${(a^m)^n}=a^{mn}$ $(ab)^m=a^m b^m$ $\frac{a^m}{b^m}=a^{m-n},a\neq{0} , m>n$ $\frac{a^m}{a^n}=\frac{1}{a^{n-m}},a\neq{0},n>n=m$ $a^{0}=1,a\neq{0}$ $(\frac{a}{b})^m\frac{a^m}{b^m},b\neq{0}$
Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?<|endoftext|>
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Bipolar disorder, also known as manic-depressive illness, is a severe mental illness that is characterized by episodes of depression and mania.
There are four types of bipolar disorder, all of which involve changes in mood, energy and activity levels. The moods in bipolar disorder can vary from “extremely up” to “very down”. These periods of “extremely up” moods are referred to as manic episodes, which are characterized by excited and energized behaviour. Less severe manic episodes are also known as hypomanic phases. The “very down” mood is referred to as depressive episodes, which are characterized by feelings of sadness and hopelessness.
Types of bipolar disorder include:
- Bipolar I disorder: Characterized by a manic or mixed episode with or without psychosis and/ or major depression.
- Bipolar II disorder: Characterized by a hypomanic episode with major depression, but no history of manic or mixed episodes
- Cyclothymia: Includes hypomanic and depressive symptoms that do not meet the diagnostic criteria for bipolar II disorder. There are no major depressive episodes.
- Bipolar disorder not otherwise specified: Does not meet diagnostic criteria for major depression, bipolar I disorder, bipolar II disorder, or cyclothymia.
The World Health Organization (WHO) has placed bipolar disorders collectively as the 12th most common moderately to severely disabling condition in the world for people of any age group (in 2004). About 4% of people in the United States experience this disorder in their lifetime. Bipolar disorder can occur at any age, although it is more common in people younger than 25 years. It can affect any race, sex, or ethnic group.
Researchers have yet to determine the pathophysiology of bipolar disorder. However, they have found that bipolar disorder may be inherited. Children of parents who have bipolar disorders have a 4-15% greater risk of developing this disorder compared to children with parents without bipolar disorder, who have a 0-2% risk. Environmental factors such as stressful life events, disrupted sleep cycle, family members or caregivers with highly expressed emotions have a strong association with inheritance. New data have shown the involvement of several genes associated with bipolar disorder.
Patients with bipolar disorder often show intense emotions, changes in sleep patterns and activity levels, and unusual behaviours. The symptoms of bipolar disorder can be categorized under the manic episodes or depressive episodes.
Manic Episode Symptoms:
- Very excited, elated, very “up” or “high”
- Very energized
- High activity levels
- Feeling “jumpy or “weird”
- Trouble sleeping
- Becoming more active than normal
- Talking really fast about a number of different things
- Feeling agitated, irritable, or “touchy”
- Racing thoughts
- Thinking they can do a lot of things at once
- Doing risky things, like spending a lot of money or having reckless sex
Depressive Episode Symptoms:
- Sadness, emptiness or hopelessness, feeling very “down”
- Very little energy
- Decreased levels of activity
- Trouble sleeping – they may sleep too much or too little
- Feeling like they can’t enjoy anything
- Feeling worried and empty
- Trouble concentrating
- Forgetting things often
- Eating too much or too little
- Feeling tired or “slowed down”
- Having thoughts about death or suicide
Symptoms may also include:
- A reduced need for sleep for a few days without getting tired
- Sleep disturbances which trigger a hypomanic event
- An excessive feeling of worthlessness or guilt
- Racing thoughts that prevent the patient from falling asleep
- Irritation, impulsivity and irrationality in behaviour
- Mood swings
Mixed Episode Symptoms:
Sometimes a mood episode can include both depressive and manic symptoms. This is known as the mixed episode. Patients who experience a mixed episode may have feelings of sadness, emptiness or hopelessness, while at the same time have feelings of being extremely energized.
Bipolar disorder can also be present in less extreme mood swings. People with hypomanic episode may feel good, productive and well-functioning. The patient may not feel anything wrong but family and friends can recognize the changes in mood or activity levels. Lack of proper treatment of patients with hypomania can lead them to develop severe mania or depression.
Proper diagnosis is necessary before treating a patient with bipolar disorder. To diagnose bipolar disorder, healthcare providers may conduct a mental exam as well as assess the individual’s appearance, mood, thoughts, and perception. They will also look into suicidal or self-destruction behaviour, violence, aggression, cognition (understanding, intellect, and awareness), judgment and physical health.
Although the diagnosis of bipolar disorder can be done based on the patient’s history and clinical course, some laboratory tests may be necessary to rule out other potential causes behind the patient’s signs and symptoms as well as to have measurements before giving certain medications.
These tests may include:
- Basic metabolism and electrolyte levels: Helps to measure baseline sodium level and renal function in patients taking antipsychotics, antidepressants, anticonvulsants, and lithium.
- Complete blood count: Helps to exclude the possibility of pernicious anemia and establish baseline measurements in patients taking anticonvulsants.
- Complete physical and neurological examination: Helps to exclude any systemic illness, measure body mass index (BMI), blood pressure, waist circumference which are monitored in the treatment of bipolar disorders.
- Electrocardiography: Helps to establish baseline measurements in patients taking antipsychotics, lithium, or other medications like carbamazepine
- Fasting glucose levels and lipid profile: Excludes the presence of diseases such as diabetes mellitus, high lipid levels (hyperlipidemia) and Cushing syndrome and establish baseline measurements in patients taking any medication that can lead to weight gain or high glucose levels.
- Liver function tests: Help exclude hepatitis and establish baseline measurements in patients taking anticonvulsants and antipsychotics.
- Thyroid hormone levels: Helps to exclude the presence of primary or secondary thyroid disorders and establish baseline measurements in patients taking lithium.
- Substance and alcohol screening: Helps to exclude thought and mood disorders due to substance abuse.
- Electroencephalography: Rules out the presence of seizure disorder, or other causes behind psychosis.
Depending on the patient’s condition, other tests such as HIV testing, urine toxicology, test for heavy metals, magnetic resonance imaging (MRI) may be indicated.
The treatment for bipolar disorder patients is directly associated with the phase of the episode (depressive or manic), and the severity of that phase. The treatment may involve medications and psychotherapy.
Medications for bipolar disorder include:
- Antipsychotic drugs such as Aripiprazole, Quetiapine, Olanzapine, Clozapine, Haloperidol
- Benzodiazepines such as clonazepam, lorazepam to reduce acute agitation
- Anticonvulsants such as carbamazepine, valproate sodium, lamotrigine
- Anti-manic agents such as lithium
Psychotherapy is usually done in combination with the medications and can be effective to provide support, education, guidance to patients with bipolar disorders and their families. Psychotherapy helps to reduce the relapse rates, improve the quality of life and increase the functioning in bipolar disorder patients.
Some of the psychotherapies for treating patients with bipolar disorder are:
- Cognitive Behavioral Therapy (CBT)
- Family-focused therapy
- Interpersonal and social rhythm therapy
Psychological stress is known to be the main trigger behind the depressive and manic symptoms of bipolar disorder. It is believed that patients who get social and family support in recognizing the early warning signs are at a lesser risk of recurrence and hospitalization and have better functioning. Psychotherapy is an important form of treatment to prevent relapse and provide patients with longer periods of relative wellness.
- National Institute of Mental Health. Bipolar Disorder. https://www.nimh.nih.gov/health/topics/bipolar-disorder/index.shtml
- Stephen Soreff, MD. Bipolar Disorder. https://emedicine.medscape.com/article/286342-overview
- Price AL, Marzani-Nissen GR. Bipolar disorders: a review. Am Fam Physician. 2012 Mar 1;85(5):483-93.<|endoftext|>
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Agriculture words and meanings IV
by Kim Cooper
This week, we will continue to look at various terms and phrases related to the agriculture sector. Thank you to Farm and Food Care Ontario for their help with this.
Polled – Breeds of goats or cattle that are naturally born without horns.
Poults – From the time they hatch until they’re 14 days old, young turkeys are called ‘poults’. They’re covered with a soft yellow “down” and make a peeping sound.
Pullet – A young female chicken.
Pullet farm – Egg farmers either raise their own pullets or buy them from pullet farms. Within 24 hours of being hatched, chicks are transported to pullet farms where they are raised until they are old enough to lay eggs. At 18 to 20 weeks of age, pullets are transported to layer farms and are then referred to as laying hens.
Ratite – Any member of the group of flightless birds like ostrich, emu or rheas that have a flat breastbone without the keel-like prominence characteristic of most flying birds. This means that they lack a strong anchor for their wing muscles and could not fly even if they did develop suitable wings. The name ratite comes from the Latin word for raft (ratis), because their breastbone looks like a raft.
Rhea – The Rhea is another member of the ratite family, native to South America. These flightless birds are smaller than the Ostrich and the Emu.
Riparian – Pertaining to the area along the banks of a river, stream, or lake.
Roaster Chicken – A larger meat chicken raised to the weight of over 2.65 kg.
Ruminant – A four-stomached animal like cattle, sheep, goats, deer or bison.
Rut – Breeding season for deer. It usually starts in September and lasts until early November.
Silo – Silos are extremely useful to store a wide variety of livestock feeds, including silage. A silo can be vertical (like a tower) or horizontal (called a bunker).
Slatted floors – A barn floor with open spaces to allow manure and other material to pass through. This keeps the barn cleaner for the animals to live in.
Snood – The snood is the fleshy growth that hangs down over the beak of a turkey.
Soil Type – The texture of the soil. This is based on the percentage of sand, silt, and clay. As examples, sandy soils drain water quickly while clay soils hold water and often require tile drainage.
Sow – An adult female pig that has given birth.
Steer – A castrated, male bovine.
As our farmers are busy with planting season over the next couple of months, tractors and other heavy and large machinery will be on our roadways. Please be careful and patient on the roadways. Our farmers are not trying to slow you down, but are just trying to do their job. Thank you for your understanding.
Remember that here in Chatham-Kent ‘We Grow for the World’. Check out our agricultural website – www.wegrowfortheworld.com
Kim Cooper has been involved in the agribusiness sector for over 45 years. He can be reached at: [email protected]
You can also follow him on Twitter at ‘theAGguy’<|endoftext|>
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# Factors of 447: Prime Factorization, Methods, and Examples
There are four positive factors and four negative factors of 447. These factors are evenly divided by 447 and are as follows: 1, 3, 149, and 447 to give a quotient without any decimal places.
So, we say this number is composite because it has four positive factors that become more than two.
### Factors of 447
Here are the factors of number 447.
Factors of 447: 1, 3, 149, and 447
### Negative Factors of 447
The negative factors of 447 are similar to their positive aspects, just with a negative sign.
Negative Factors of 447: -1, -3, -149, and -447
### Prime Factorization of 447
The prime factorization of 447 is the way of expressing its prime factors in the product form.
Prime Factorization: 3 x 149
In this article, we will learn about the factors of 447 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 447?
The factors of 447 are 1, 3, 149, and 447. These numbers are the factors as they do not leave any remainder when divided by 447.
The factors of 447 are classified as prime numbers and composite numbers. The prime factors of the number 447 can be determined using the prime factorization technique.
## How To Find the Factors of 447?
You can find the factors of 447 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 447, create a list containing the numbers divisible by 447 with zero remainders. One important thing to note is that 1 and 447 are the 447’s factors as every natural number has one and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 447 are determined as follows:
$\dfrac{447}{1} = 447$
$\dfrac{447}{3} = 149$
$\dfrac{447}{149} = 3$
$\dfrac{447}{447} = 1$
Therefore, 1, 3, 149, and 447 are the factors of 447.
### Total Number of Factors of 447
For 447, there are four positive factors and four negative ones. So in total, there are eight factors of 447.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 447 is given as:
Factorization of 447 is 1 x 3 x 149.
The exponent of 1, 3, and 149 is 1.
Adding 1 to each and multiplying them together results in 8.
Therefore, the total number of factors of 447 is eight, whereas four are positive and four are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The number factors cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has two as its prime factor, the smallest prime factor.
## Factors of 447 by Prime Factorization
The number 447 is a composite. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 447 using prime factorization, let us determine what prime factors are. Prime factors are the factors of any given number that are only divisible by one and themselves.
To start the prime factorization of 447, divide by its most minor prime factor. First, determine that the given number is either even or odd. Two will be the smallest prime factor if it is an even number.
Continue splitting the quotient obtained until one is received as the quotient. The prime factorization of 447 can be expressed as:
447 = 3 x 149
## Factors of 447 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors.
For 447, the factor pairs can be found as:
1 x 447 = 447
3 x 149 = 447
The possible factor pairs of 447 are given as (1, 447) and (3, 149).
All these numbers in pairs, when multiplied, give 447 as the product.
The negative factor pairs of 447 are given as:
-1 x -447 = 447
-3 x -149 = 447
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -3, -149, and -447 are called negative factors of 447.
The list of all the factors of 447, including positive and negative numbers, is given below.
Factor list of 447: 1, -1, 3, -3, 149, -149, 447, and -447
## Factors of 447 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 447 are there?
### Solution
The total number of Factors of 447 is 4.
Factors of 447 are 1, 3, 149, and 447.
### Example 2
Find the factors of 447 using prime factorization.
### Solution
The prime factorization of 447 is given as:
447 $\div$ 3 = 149
149 $\div$ 149 = 1
So the prime factorization of 447 can be written as:
3 x 149 = 447<|endoftext|>
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# Finding Ratio Amounts of Proportions
Ratios and Proportions are used for cooking as well as making drinks.
A strawberry milkshake can be made by mixing together in a blender 2 parts strawberry ice cream to 7 parts of milk.
This can be written in Ratio form as 2 : 7 which in fraction from equals 2/7 .
Both forms mean the exact same thing, which is 2 parts ice cream to every 7 parts of milk.
However when we use two scoops of ice cream and seven scoops of milk, we find that we don’t get enough milkshake to completely fill our two big glasses.
So we decide next time to make a milkshake using three scoops of ice cream.
How do we make a milkshake using three scoops of ice cream ?
How much milk do we need to use to get the correct consistency ?
We can calculate the exact amount of milk required by using Ratios.
We write out two ratios and make them equal to each other.
This is called making a “Proportion”.
A “Proportion” is simply two ratios which are equal to each other.
The two ratios can be written out in shorthand from such as 2 : 7 = 3 : m
or as a pair of fractions such as 2/7 = 3/m
where “m” = the number of scoops of milk, (which we do not know the number value of yet).
Here is how we mathematically solve our Strawberry Milkshake Problem, and find out how much milk we need for three scoops of ice cream.
Image Copyright 2012 by Passy’s World
Cross Multiplying is probably a new method for many people, and the following video shows several examples of solving ratio proportions by Cross Multiplying.
This video writes the pairs of ratios as fractions.
Eg. x : 6 = 2: 3 is written as x/6 = 2/3
Remember any ratio can always be rewritten as a fraction :
a : b = a / b
Also remember that when we have two ratios (or fractions) that are equal to each other, then we call this situation a “Proportion”.
Here is another video about Proportions and Cross Multiplying.
Steps For Finding Ratio Amounts
Instead of writing fractions, here at Passy World we like to write the two ratios out one above each other, and then do our cross multiplying.
It is the exact same method as writing out fractions, so feel free to use fractions notation if you prefer working in fraction form.
In Ratio Amounts we use “Cross Multiplying” and follow the working out steps shown below.
Image Copyright 2012 by Passy’s World
Solving Proportions Examples
In the following examples, we apply the standard steps of re-writing vertically, then cross multiplying, and finishing off by dividing by the number in front of the letter variable.
This gives us a number answer for our variable letter.
Image Copyright 2012 by Passy’s World
Image Copyright 2012 by Passy’s World
Ratios Applications Videos
This first video shows how to do a ratios word question involving a real world car trip time calculation.
Ratios and Cross Multiplying are involved with solving this problem.
This next video also shows hot to solve some typical ratio word questions involving real world applications.
Ratios Application Examples
The following examples show how to solve some ratios application problems.
Image Copyright 2012 by Passy’s World
Note that 2-stroke fuel ratios are often specified as 25 : 1 but it is not possible to fit this volume into a 5 litre fuel can.
As a result, 25 : 1 mixes are often made up as 24 : 1 mixtures of Petrol (Gasoline) and Oil.
A 24 : 1 mix gives slightly more oil per litre than a 25 : 1 mixture, but this 4% extra oil in the mix is not going to affect the running of the 2-stroke engine, or cause any other significant problems.
Note that we worked out there is 4% extra oil in a 24 : 1 mix as follows:
Oil as 1 part compared to 25 parts Petrol, equals 1/25 = 0.04
Oil as 1 part compared to 24 parts Petrol, equals 1/24 = 0.0417
% Extra Oil = (0.0417 – 0.04) / 0.04 x 100 % = 4.2 % = 4 % rounded off.
This calculation is not technically correct, because “Percent” is supposed to be “Parts in 100″, worked out from how many parts out of the total parts we have, and then multiplying this by 100 %.
We have not covered Ratios and Parts yet, but a 1 : 25 mix means we have 1 part Oil and 25 parts Petrol, to make a total parts amount of 26.
So the Oil is actually 1 part out of 26 total parts = 1/26 = 0.0385.
The 1 : 24 mix actually has 25 parts total, (1 part Oil + 24 parts Petrol).
So the fraction of oil is actually 1 part out of 25 total parts = 1/25 = 0.04
% Extra Oil = (0.04 – 0.0385) / 0.0385 x 100 % = 3.9 % = 4 % rounded off.
Note that if we needed to make a 50 : 1 fuel mixture for a grass trimmer, or hedge clippers, then we could make this as a 49 : 1 mix, by putting 100 ml of oil in a 5 litre can, and then filling the rest of the can with 4900 ml of Petrol.
Image Copyright 2012 by Passy’s World
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Education is the process of facilitating learning, or the acquisition of knowledge, skills, values, beliefs, and habits. Educational methods include storytelling, discussion, teaching, training, and directed research. Education frequently takes place under the guidance of educators, but learners may also educate themselves. Education can take place in formal or informal settings and any experience that has a formative effect on the way one thinks, feels, or acts may be considered educational. The methodology of teaching is called pedagogy.
Music is an art form and cultural activity whose medium is sound organized in time. The common elements of music are pitch (which governs melody and harmony), rhythm (and its associated concepts tempo, meter, and articulation), dynamics (loudness and softness), and the sonic qualities of timbre and texture (which are sometimes termed the "color" of a musical sound). Different styles or types of music may emphasize, de-emphasize or omit some of these elements. Music is performed with a vast range of instruments and vocal techniques ranging from singing to rapping; there are solely instrumental pieces, solely vocal pieces (such as songs without instrumental accompaniment) and pieces that combine singing and instruments. The word derives from Greek μουσική (mousike; "art of the Muses"). See glossary of musical terminology.
Music education is a field of study associated with the teaching and learning of music. It touches on all learning domains, including the psychomotor domain (the development of skills), the cognitive domain (the acquisition of knowledge), and, in particular and significant ways, the affective domain (the learner's willingness to receive, internalize, and share what is learned), including music appreciation and sensitivity. Music training from preschool through post-secondary education is common in most nations because involvement with music is considered a fundamental component of human culture and behavior. Cultures from around the world have different approaches to music education, largely due to the varying histories and politics. Studies show that teaching music from other cultures can help students perceive unfamiliar sounds more comfortably, and they also show that musical preference is related to the language spoken by the listener and the other sounds they are exposed to within their own culture. Music, like language, is an accomplishment that distinguishes humans as a species.
What we need is to justify coercion, paternalistic control, blame, scolding, and punishment - all of which are less evident in trigonometry class than in a fourth grade learning long division.(...) I have argued that blame, scolding, and punishment in public schools - what I have called "the ordeal" - can be successfully defended. Students have a duty to learn, and can be held responsible for violating whatever rules, policies, or instructions are enforced to ensure that they do so.
Charles Howell - Syracuse University: Education, Punishment, and Responsibility
If today you can take a thing like evolution and make it a crime to teach it in the public school, tomorrow you can make it a crime to teach it in the private schools, and the next year you can make it a crime to teach it to the hustings or in the church. At the next session you may ban books and the newspapers. Soon you may set Catholic against Protestant and Protestant against Protestant, and try to foist your own religion upon the minds of men. If you can do one you can do the other. Ignorance and fanaticism is ever busy and needs feeding. Always it is feeding and gloating for more. Today it is the public school teachers, tomorrow the private. The next day the preachers and the lectures, the magazines, the books, the newspapers. After a while, your honor, it is the setting of man against man and creed against creed until with flying banners and beating drums we are marching backward to the glorious ages of the sixteenth century when bigots lighted fagots to burn the men who dared to bring any intelligence and enlightenment and culture to the human mind.
Clarence Darrow Scopes Trial, Dayton, Tennessee (13 July 1925)
Strange! that a harp of thousand strings
Should keep in tune so long.
Isaac Watts, Hymns and Spiritual Songs, Book II. 19.<|endoftext|>
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Finding Unknowns in Elementary and Middle School Math Classes Tip SheetGrades K–8
Polya’s Four-Step Problem-Solving Process focuses on finding unknowns in mathematical problems. Most of Polya’s examples in his book How to Solve It focus on high school mathematics and geometry. This tip sheet lists “unknown situations” from modern math standards and suggests a problem-solving challenge for each grade level. Use this list to apply Polya’s Four-Step Problem-Solving Process in the lower grades!
Fill out the form to download the Finding Unknowns in Elementary and Middle School Math Classes Tip Sheet now.<|endoftext|>
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Introduction
Introduction
Stats Lab 5.1
Continuous Distribution
Class Time:
Names:
Student Learning Outcomes
• The student will compare and contrast empirical data from a random number generator with the uniform distribution.
Collect the DataUse a random number generator to generate 50 values between zero and one (inclusive). List them in Table 5.3. Round the numbers to four decimal places or set the calculator MODE to four places.
1. Complete the table.
__________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________
Table 5.3
2. Calculate the following:
1. $x¯=x¯=$ _______
2. s = _______
3. first quartile = _______
4. third quartile = _______
5. median = _______
Organize the Data
1. Construct a histogram of the empirical data. Make eight bars.
Figure 5.33
2. Construct a histogram of the empirical data. Make five bars.
Figure 5.34
Describe the Data
1. In two to three complete sentences, describe the shape of each graph. (Keep it simple. Does the graph go straight across, does it have a V shape, does it have a hump in the middle or at either end, (and so on)? One way to help you determine a shape is to draw a smooth curve roughly through the top of the bars.)
2. Describe how changing the number of bars might change the shape.
Theoretical Distribution
1. In words, X = _____________________________________.
2. The theoretical distribution of X is X ~ U(0,1).
3. In theory, based upon the distribution X ~ U(0,1), complete the following:
1. μ = ______
2. σ = ______
3. first quartile = ______
4. third quartile = ______
5. median = __________
4. Are the empirical values (the data) in the section titled Collect the Data close to the corresponding theoretical values? Why or why not?
Plot the Data
1. Construct a box plot of the data. Be sure to use a ruler to scale accurately and draw straight edges.
2. Do you notice any potential outliers? If so, which values are they? Either way, justify your answer numerically. (Recall that any data that are less than Q1 – 1.5(IQR) or more than Q3 + 1.5(IQR) are potential outliers. IQR means interquartile range.)
Compare the Data
1. For each of the following parts, use a complete sentence to comment on how the value obtained from the data compares to the theoretical value you expected from the distribution in the section titled Theoretical Distribution.
1. minimum value: _______
2. first quartile: _______
3. median: _______
4. third quartile: _______
5. maximum value: _______
6. width of IQR: _______
7. overall shape: _______
2. Based on your comments in the section titled Collect the Data, how does the box plot fit or not fit what you would expect of the distribution in the section titled Theoretical Distribution?
Discussion Question
1. Suppose that the number of values generated was 500, not 50. How would that affect what you would expect the empirical data to be and the shape of its graph to look like?<|endoftext|>
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Neurons are the core components of the nervous system. There are approximately one hundred billion neurons connected by more than one hundred trillion Synapses in the human Brain. Neurons are basically electrical on and off switches, which work in a similar way to the small transistors found in computer chips. Information is passed between neurons via the chemical synapses, which release neurotransmitters, which act on another neuron. Most neurons are connected via synapses to several thousand other neurons, making the brain's circuitry capable of connecting to a host of neural networks.
These signals between Neurons occur via Synapses, or specialized connections with other cells. Neurons can connect to each other to form neural networks. The key to neural function in the central nervous system is the synaptic signaling process, which is partly electrical and partly chemical. When the synaptic signaling is interrupted through toxic electrical frequencies (i.e. such as technological Mind Control in low EMF pulsing, harmful radiation) or the synaptic signaling is interrupted through an introduction to toxic chemicals (i.e. mercury and aluminum in Vaccinations, Chemtrails) the synapse will not signal. When the synapse stops functioning and/or is poisoned, the neuron will eventually die. When neurons die, humans get motor neuron and brain diseases like Alzheimer's and ALS. Have you ever felt that headache after a vaccine? That's the feeling of your immune system trying to stop the chemicals from eating your brain's neurons.
A Neuron or nerve cell is an electrically excitable cell that processes and transmits information through electrical and chemical signals. The signals between nerve cells occur via the Synapses, which are structures of connections made with the other cells. Nerve Cells are the core components of the brain and the spinal cord of the entire central nervous system.
The way neural connections are formed in the brain is similar to routing traffic when traveling distances on the main roads and highways. The main roads and highways that people are most familiar with and know are used much more often. In the human brain it is the same. The signals will find the main neural pathways formed by the most common and repetitive thought forms, which then form into neural connections. Thus the Brain will process and route signals that are received to connect to the main neural pathways, based on the most common previous patterns of thinking.<|endoftext|>
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SADLY, dark matter is not made of light. That may sound obvious, but many physicists were hoping that photons – particles of light – could help us to piece together the nature of the mysterious stuff thought to make up 85 per cent of the universe’s matter.
Instead, readings from Vitor Cardoso of the Technical University of Lisbon in Portugal and colleagues seem to have quashed this idea.
Some theories had hinted that “heavy photons”, hypothetical versions of the more familiar massless particles, might be dark matter. According to that idea, the heavy photon would have a small amount of mass and might carry an unknown fundamental force that allows it to interact only with ordinary photons – effectively hiding it from the visible world.
In that case, heavy photons passing close to black holes would have noticeable effects, says Cardoso.
When most particles with mass get too near to a black hole, they fall in, never to be seen again. Photons with no mass can skirt past danger if they are on the right trajectory. But a photon with a very tiny “in between” mass can enter into an orbit of the spinning black hole and steal some of its angular momentum. If conditions are right, this process can continue until orbiting particles slow the hole down so much that it stops spinning.
Cardoso and colleagues calculated how long photons of given masses would take to sap a black hole’s spin. Then they examined data on the ages and rotation speeds of eight supermassive black holes. The age of the oldest spinning black holes effectively puts an upper limit on the photon’s mass. If it does exist, the heavy photon must be lighter than 10-20 electronvolts – an extreme improbability – the team concluded (arxiv.org/abs/1209.0465).
“We are constraining and putting aside what is not dark matter,” Cardoso says.
By limiting the photon mass, we are constraining and putting aside what is not dark matter
Alfred Goldhaber of Stony Brook University in New York says that if black holes have charged plasma swirling around them, a photon’s slowed movement through the plasma could make it behave as if it has mass, ruining the calculations. Cardoso counters that this apparent mass only affects things at a subatomic level – the photon’s real mass is what matters at the scale of the black hole.
More on these topics:<|endoftext|>
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## An airplane descends 3.5 miles to an elevation of 5.25 miles. Find the elevation of the plane before its descent. Before its des
Question
An airplane descends 3.5 miles to an elevation of 5.25 miles. Find the elevation of the plane before its
descent.
Before its descent, the plane was at an elevation of
miles.
in progress 0
5 months 2021-08-15T00:38:57+00:00 1 Answers 5 views 0
1. Answer: The initial elevation of the plane was 8.75 miles.
Step-by-step explanation:
Suppose that the initial elevation of the plane is E.
Then the plane descends 3.5 miles.
Then the new elevation of the plane will be:
E – 3.5 miles.
Now, we know that the new elevation of the plane is 5.25 miles, then:
E – 3.5 miles = 5.25 miles.
Now we need to solve this for E.
To solve this, we can add the same number to both sides of the equation, so we can add 3.5 miles to both sides to get:
(E – 3.5 miles) + 3.5 miles = 5.25 miles + 3.5 miles
E + (3.5 miles – 3.5 miles) = 5.25 miles + 3.5 miles
E = 8.75 miles.
The initial elevation of the plane was 8.75 miles.<|endoftext|>
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In early 1943, encouraged by von Verschuer, Mengele applied to transfer to the concentration camp service. His application was accepted and he was posted to Auschwitz, where he was appointed by SS-Standortarzt Eduard Wirths, chief medical officer at Auschwitz, to the position of chief physician of the Zigeunerfamilienlager (Romani family camp) at Birkenau, a subcamp located on the main Auschwitz complex. The SS doctors did not administer treatment to the Auschwitz inmates, but supervised the activities of inmate doctors who had been forced to work in the camp medical service. As part of his duties, Mengele made weekly visits to the hospital barracks and ordered any prisoners who had not recovered after two weeks in bed to be sent to the gas chambers.
Around 50,000 German gay men were jailed between 1933 and 1945, and 5,000–15,000 are estimated to have been sent to concentration camps. It is not known how many died during the Holocaust. James Steakley writes that what mattered in Germany was criminal intent or character, rather than acts, and the "gesundes Volksempfinden" ("healthy sensibility of the people") became the guiding legal principle. In 1936, Himmler created the Reich Central Office for the Combating of Homosexuality and Abortion. The Gestapo raided gay bars, tracked individuals using the address books of those they arrested, used the subscription lists of gay magazines to find others, and encouraged people to report suspected homosexual behavior and to scrutinize the behavior of their neighbors. Lesbians were left relatively unaffected; the Nazis saw them as "asocials", rather than sexual deviants. Gay men convicted between 1933 and 1944 were sent to camps for "rehabilitation", where they were identified by pink triangles. Hundreds were castrated, sometimes "voluntarily" to avoid criminal sentences. Steakley writes that the full extent of gay suffering was slow to emerge after the war. Many victims kept their stories to themselves because homosexuality remained criminalized in postwar Germany.
Construction work on the first killing centre at Bełżec in occupied Poland began in October 1941, three months before the Wannsee Conference. The new facility was operational by March the following year. By mid-1942, two more death camps had been built on Polish lands: Sobibór operational by May 1942, and Treblinka operational in July. From July 1942, the mass murder of Polish and foreign Jews took place at Treblinka as part of Operation Reinhard, the deadliest phase of the Final Solution. More Jews were killed at Treblinka than at any other Nazi extermination camp apart from Auschwitz. By the time the mass killings of Operation Reinhard ended in 1943, roughly two million Jews in German-occupied Poland had been murdered. The total number of people killed in 1942 in Lublin/Majdanek, Bełżec, Sobibór, and Treblinka was 1,274,166 by Germany's own estimation, not counting Auschwitz II Birkenau nor Kulmhof. Their bodies were buried in mass graves initially. Both Treblinka and Bełżec were equipped with powerful crawler excavators from Polish construction sites in the vicinity, capable of most digging tasks without disrupting surfaces. Although other methods of extermination, such as the cyanic poison Zyklon B, were already being used at other Nazi killing centres such as Auschwitz, the Aktion Reinhard camps used lethal exhaust gases from captured tank engines.
In May, Nazis under the direction of SS Lt. Colonel Adolf Eichmann boldly began a mass deportation of the last major surviving population of European Jews. From May 15 to July 9, over 430,000 Hungarian Jews were deported to Auschwitz. During this time, Auschwitz recorded its highest-ever daily number of persons killed and cremated at just over 9000. Six huge open pits were used to burn the bodies, as the number of dead exceeded the capacity of the crematories.
Amit withstood pressure from many members of his staff who were Holocaust survivors or relatives of victims. But others thought he was right. Rafi Eitan, an Israel-born Mossad operative who led the team that caught Eichmann, told me: “Because of the need for foreign-language speakers, many of the Mossad’s recruits were from Europe, and therefore had gone through the Holocaust or lost their families in it. One can definitely understand their need for vengeance. However, there was huge pressure to deal with current requirements, and with the resources being as meager as they were, in no way would it have been right to give the Nazi matter priority.”
Hitler also believed the very presence of Jews in Germany and Nazi-occupied Europe posed a threat to German victory in the war. This was based on his experience during the First World War, when Germany had experienced a meltdown of civilian morale. In 1916, as a young soldier on sick leave in Munich, Hitler had been appalled at the apathy and anti-war sentiment he witnessed among German civilians. At the time, he concluded disloyal Jews had banded together and conspired to undermine the German war effort. And he was convinced they would do it again now if given the chance.
The Summer Olympics in Berlin gave the Nazis a platform to project a crafted image to the world. Despite calls for boycotts, the games were a success. Anti-Jewish notices were removed and German spectators cheered black athlete Jesse Owens to four gold medals. Visitors saw a tolerant Reich. However, three days after the games ended, the head of the Olympic Village, Wolfgang Fürstner, killed himself as he would soon be dismissed due to his Jewish ancestry under the Nuremberg Laws.
Antisemitism, the new racist version of the old Jew-hatred, viewed the Jews as not simply a religious group but as members of a 'Semitic race', which strove to dominate its 'Aryan' rivals. Among the leading ideologues of this theory were a French aristocrat, the Comte Joseph de Gobineau, and an Englishman, Houston Stewart Chamberlain. Antisemitism proved a convenient glue for conspiracy theories - since Jews were involved in all sorts of ventures and political movements, they could be accused of manipulating all of them behind the scenes. Thus Jews were held responsible for Communism and capitalism, liberalism, socialism, moral decline, revolutions, wars, plagues and economic crises. As the Jews had once been demonised in medieval Europe, so the new antisemites (including many Christians) found new, secular ways of demonising them.
The infamous 'Gate of Death' at Auschwitz II for the incoming freight trains was built of brick and cement mortar in 1943, and the three-track rail spur was added. Until mid-August, 45,000 Thessaloniki Jews were murdered in a mere six months, including over 30,000 Jews from Sosnowiec (Sosnowitz) and Bendzin Ghettos. The spring of 1944 marked the beginning of the last phase of the Final Solution at Birkenau. The new big ramps and sidings were constructed, and two freight elevators were installed inside Crematoria II and III for moving the bodies faster. The size of the Sonderkommando was nearly quadrupled in preparation for the Special Operation Hungary (Sonderaktion Ungarn). In May 1944, Auschwitz-Birkenau became the site of one of the two largest mass murder operations in modern history, after the Großaktion Warschau deportations of the Warsaw Ghetto inmates to Treblinka in 1942. It is estimated that until July 1944 approximately 320,000 Hungarian Jews were gassed at Birkenau in less than eight weeks. The entire operation was photographed by the SS. In total, between April and November 1944, Auschwitz II received over 585,000 Jews from over a dozen regions as far as Greece, Italy, and France, including 426,000 Jews from Hungary, 67,000 from Łódź, 25,000 from Theresienstadt, and the last 23,000 Jews from the General Government. Auschwitz was liberated by the Red Army on 27 January 1945, when the gassing had already stopped.
Steve Paulsson is a lecturer at the Oxford Centre for Hebrew and Jewish Studies. His doctoral thesis, 'Hiding in Warsaw: The Jews on the "Aryan side", 1940-1945', was co-winner of the 1998 Fraenkel Prize in Contemporary History, and is published by Yale University Press. He has also published articles on the flight of the Danish Jews to Sweden in 1943, and on Polish-Jewish relations. He was senior historian in the Holocaust Exhibition Project Office at the Imperial War Museum, 1998-2000.
More than 140 years after Louis Braille invented the Braille reading system, Seiichi Miyake came up with a different system based on touch that allows visually impaired people to navigate public spaces. Today, tactile paving is used by major cities and transportation services around the world. Miyake was so influential that he's the subject of the Google Doodle for March 18, the 52nd anniversary of tactile paving's debut.
Doubleday, meanwhile, sensing complications ahead, had withdrawn as Frank’s theatrical agent, finding Levin’s presence—injected by Frank—too intrusive, too maverick, too independent and entrepreneurial: fixed, they believed, only on his own interest, which was to stick to his insistence on the superiority of his work over all potential contenders. Frank, too, had begun—kindly, politely, and with tireless assurances of his gratitude to Levin—to move closer to Doubleday’s cooler views, especially as urged by Barbara Zimmerman. She was twenty-four years old, the age Anne would have been, very intelligent and attentive. Adoring letters flowed back and forth between them, Frank addressing her as “little Barbara” and “dearest little one.” On one occasion he gave her an antique gold pin. About Levin, Zimmerman finally concluded that he was “impossible to deal with in any terms, officially, legally, morally, personally”—a “compulsive neurotic . . . destroying both himself and Anne’s play.” (There was, of course, no such entity as “Anne’s play.”)
Auschwitz is the most famous because there the killing machine was the most efficient. There, between the end of 1941 and 1944, as many as 12,000 Jews a day could be gassed to death and cremated. In addition to the Jews, hundreds of thousands of others deemed threats to the Nazi regime or considered racially inferior or socially deviant were also murdered.
A subsidiary aim of Operation Reinhard was to exploit a small minority of Jews residing in the Generalgouvernement as forced laborers before killing them. As the ghettos in Poland were systematically liquidated, those selected to live temporarily were deported to Operation Reinhard labor camps and to the concentration camp Lublin/Majdanek. This camp was established in 1941 under the authority of the Inspectorate of Concentration Camps in Berlin. Though conceived of and functioning in practice primarily as a concentration camp, housing political prisoners and Jewish forced laborers, Majdanek served from time to time as a killing site for Jews residing in the Generalgouvernement. It had gas chambers, in which the SS killed tens of thousands of Jews, primarily forced laborers too weak to work.
In 1992, DNA testing confirmed Mengele's identity beyond doubt, but family members refused repeated requests by Brazilian officials to repatriate the remains to Germany. The skeleton is stored at the São Paulo Institute for Forensic Medicine, where it is used as an educational aid during forensic medicine courses at the University of São Paulo's medical school.
Ruth Elias and her husband had conceived a child while she was a prisoner in the Theresienstadt camp, and when she arrived at Birkenau on a transport of Czech prisoners in December 1943, she was three months pregnant. Ruth passed several selections for the gas chamber even though she was obviously pregnant; she and her husband were assigned to the Czech "family camp." On July 11, 1944, after a selection made by Dr. Mengele, 3,000 prisoners in the Czech family camp, who were not considered fit to work, were sent to the gas chamber, but Ruth passed the selection even though she was in her eighth month of pregnancy. On July 14, 1944, Ruth was sent to Hamburg, Germany to work in clearing rubble from Allied bombing raids.
Three defendants were acquitted. However, many of the Nazis who perpetrated the Holocaust were never tried or punished, including Hitler who had committed suicide. Since then, the international community has continued and improved accountability through forums such as the International Criminal Court, and the International Criminal Tribunal for Rwanda.
Anne’s last diary entry was written on August 1, 1944. Three days later the secret annex was discovered by the Gestapo, which had received a tip from Dutch informers. All of the inhabitants were taken into custody. In September the Frank family arrived at Auschwitz, though Anne and Margot were transferred to Bergen-Belsen the following month. In 1945 Anne as well as her mother and sister died.
Frank’s candid words on sex didn’t make it into the first published diary, which appeared in English in 1952. Though Anne herself edited her diary with an eye to publication, the book—released eight years after her death from typhus in theBergen-Belsen concentration camp at age 15—contained additional cuts. These were only partially restored in 1986, when a critical edition of her diary was published. Then, in 1995, an even less censored version, including a passage on Frank’s own body previously withheld by her father, was published.
Hitler's first step was to take the Jews civil rights away. then he branded and labeled them as if they were cattle. Then he sent them to death and labor camps. If they were they were sent to ghettos. in 1941 most of the Jewish population in Germany would be sent to camps. Nazis racial policies took over everything to try and find a solution for the Jewish question. the Nazis tried to separate them and force migration but when this did not work they found a final solution to the Jewish question. This was the murdering of the Jews in Europe. No-one knows when this decision was made but when the ghettos were built Heydrich said '' this is one step closer to the final aim''.
Throughout Nazi-occupied Europe, relatively few non-Jewish persons were willing to risk their own lives to help the Jews. Notable exceptions included Oskar Schindler, a German who saved 1,200 Jews by moving them from Plaszow labor camp to his hometown of Brunnlitz. The Nazi-occupied nation of Denmark rescued nearly its entire population of Jews, over 7,000, by transporting them to safety by sea. Italy and Bulgaria both refused to cooperate with Nazi demands for deportations. Elsewhere in Europe, people generally stood by passively and watched as their neighbors were marched through the streets toward waiting trains, or in some cases, actively participated in Nazi roundups.
Although the composition of the ground, largely sand, was favorable for ground-penetrating radar, the dense forest surrounding the site interfered enough with the radar signals that they decided to try another tack. Paul Bauman and Alastair McClymont, geophysicists with Advisian WorleyParsons, a transnational engineering company, had more luck with electrical resistivity tomography, or ERT, which was originally developed to explore water tables and potential mining sites. ERT technology sends jolts of electrical current into the earth by way of metal electrodes hooked up to a powerful battery and measures the distinctive levels of resistivity of different types of earth; the result is a detailed map to a depth of more than a hundred feet.
There has been much conjecture about the identity or inspiration of Kitty, who in Anne's revised manuscript is the sole recipient of her letters. In 1996, the critic Sietse van der Hoek wrote that the name referred to Kitty Egyedi, a prewar friend of Anne's. Van der Hoek may have been informed by the publication A Tribute to Anne Frank (1970), prepared by the Anne Frank Foundation, which assumed a factual basis for the character in its preface by the then-chairman of the Foundation, Henri van Praag, and accentuated this with the inclusion of a group photograph that singles out Anne, Sanne Ledermann, Hanneli Goslar, and Kitty Egyedi. Anne does not mention Kitty Egyedi in any of her writings (in fact, the only other girl mentioned in her diary from the often reproduced photo, other than Goslar and Ledermann, is Mary Bos, whose drawings Anne dreamed about in 1944) and the only comparable example of Anne's writing unposted letters to a real friend are two farewell letters to Jacqueline van Maarsen, from September 1942.
After the Warsaw Ghetto Uprising, the Ghetto was completely destroyed. Of the more than 56,000 Jews captured, about 7,000 were shot, and the remainder were deported to killing centers or concentration camps. This is a view of the remains of the ghetto, which the German SS dynamited to the ground. The Warsaw Ghetto only existed for a few years, and in that time, some 300,000 Polish Jews lost their lives there. #
While the labour camps at Auschwitz and Majdanek used inmates for slave labour to support the German war effort, the extermination camps at Belzec, Treblinka, and Sobibor had one task alone: killing. At Treblinka a staff of 120, of whom only 30 were SS (the Nazi paramilitary corps), killed some 750,000 to 925,000 Jews during the camp’s 17 months of operation. At Belzec German records detail a staff of 104, including about 20 SS, who killed some 500,000 Jews in less than 10 months. At Sobibor they murdered between 200,000 and 250,000. These camps began operation during the spring and summer of 1942, when the ghettos of German-occupied Poland were filled with Jews. Once they had completed their missions—murder by gassing, or “resettlement in the east,” to use the language of the Wannsee protocols—the Nazis closed the camps. There were six extermination camps, all in German-occupied Poland, among the thousands of concentration and slave-labour camps throughout German-occupied Europe.
The Jews killed represented around one third of the world population of Jews, and about two-thirds of European Jewry, based on an estimate of 9.7 million Jews in Europe at the start of the war. Much of the uncertainty stems from the lack of a reliable figure for the number of Jews in Europe in 1939, numerous border changes that make avoiding double-counting of victims difficult, lack of accurate records from the perpetrators, and uncertainty about whether deaths occurring months after liberation, but caused by the persecution, should be counted.
Usage Note: Holocaust has a secure place in the language when it refers to the massive destruction of humans by other humans. In our 1987 survey 99 percent of the Usage Panel accepted the use of holocaust in the phrase nuclear holocaust. Sixty percent accepted the sentence As many as two million people may have died in the holocaust that followed the Khmer Rouge takeover in Cambodia. But because of its associations with genocide, people may object to extended applications of holocaust. The percentage of the Panel's acceptance drops sharply when people use the word to refer to death brought about by natural causes. In our 1999 survey 47 percent approved the sentence In East Africa five years of drought have brought about a holocaust in which millions have died. Just 16 percent approved The press gives little coverage to the holocaust of malaria that goes on, year after year, in tropical countries, where there is no mention of widespread mortality. The Panel has little enthusiasm for more figurative usages of holocaust. In 1999, only 7 percent accepted Numerous small investors lost their stakes in the holocaust that followed the precipitous drop in stocks. This suggests that these extended uses of the word may be viewed as overblown or in poor taste.
The diary is taken to be a Holocaust document; that is overridingly what it is not. Nearly every edition—and there have been innumerable editions—is emblazoned with words like “a song to life” or “a poignant delight in the infinite human spirit.” Such characterizations rise up in the bitter perfume of mockery. A song to life? The diary is incomplete, truncated, broken off—or, rather, it is completed by Westerbork (the hellish transit camp in Holland from which Dutch Jews were deported), and by Auschwitz, and by the fatal winds of Bergen-Belsen. It is here, and not in the “secret annex,” that the crimes we have come to call the Holocaust were enacted. Our entry into those crimes begins with columns of numbers: the meticulous lists of deportations, in handsome bookkeepers’ handwriting, starkly set down in German “transport books.” From these columns—headed, like goods for export, “Ausgangs-Transporte nach dem Osten” (outgoing shipments to the east)—it is possible to learn that Anne Frank and the others were moved to Auschwitz on the night of September 6, 1944, in a collection of a thousand and nineteen Stücke (or “pieces,” another commodities term). That same night, five hundred and forty-nine persons were gassed, including one from the Frank group (the father of Peter van Daan) and every child under fifteen. Anne, at fifteen, and seventeen-year-old Margot were spared, apparently for labor. The end of October, from the twentieth to the twenty-eighth, saw the gassing of more than six thousand human beings within two hours of their arrival, including a thousand boys eighteen and under. In December, two thousand and ninety-three female prisoners perished, from starvation and exhaustion, in the women’s camp; early in January, Edith Frank expired.
The Nazi killing site at Ponar is today known to scholars as one of the first examples of the “Holocaust by bullets”—the mass shootings that claimed the lives of upwards of two million Jews across Eastern Europe. Unlike the infamous gas chambers at places like Auschwitz, these murders were carried out at close range, with rifles and machine guns. Significantly, the killings at Ponar marked the transition to the Final Solution, the Nazi policy under which Jews would no longer be imprisoned in labor camps or expelled from Europe but exterminated.
The Japanese inventor designed the influential system with a specific person in mind. His friend was losing his vision, so in 1965, Miyake used his own money to build special mats with raised shapes that lead blind and visually impaired people away from danger and toward safety. Pavement with round bumps was meant to signal nearby danger, such as a street crossing or the edge of a train platform, while a stretch of pavement with straight bars was meant to guide them to safe areas. The tactile design allowed pedestrians to detect the features with canes, guide dogs, or their feet.<|endoftext|>
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Q:
# P alone can complete a piece of work in 6 days. Work done by Q alone in one day is equal to one-third of the work done by P alone in one day. In how many days can the work be completed if P and Q work together ?
A) 5 (2/3) B) 6 (3/4 ) C) 4 (1/2) D) 3
Explanation:
Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9
They would take 9/2 days = 4 (1/2) days to complete the work working together.
Q:
70000 a year is how much an hour?
A) 80 B) 8 C) 0.8 D) 0.08
Explanation:
Given for year = 70000
=> 365 days = 70000
=> 365 x 24 hours = 70000
=> 1 hour = ?
70000/365x24 = 7.990 = 8
0 111
Q:
A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
A) 1/5 B) 1/6 C) 1/7 D) 1/8
Explanation:
Total work is given by L.C.M of 72, 48, 36
Total work = 144 units
Efficieny of A = 144/72 = 2 units/day
Efficieny of B = 144/48 = 3 units/day
Efficieny of C = 144/36 = 4 units/day
According to the given data,
2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100
3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54
p = 198/16.5
p = 12 days.
Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day
(C+D) in p days = (4 + 6) x 12 = 120 unit
Remained part of work = (144-120)/144 = 1/6.
4 565
Q:
10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?
A) 215 days B) 225 days C) 235 days D) 240 days
Explanation:
Given that
(10M + 15W) x 6 days = 1M x 100 days
=> 60M + 90W = 100M
=> 40M = 90W
=> 4M = 9W.
From the given data,
1M can do the work in 100 days
=> 4M can do the same work in 100/4= 25 days.
=> 9W can do the same work in 25 days.
=> 1W can do the same work in 25 x 9 = 225 days.
Hence, 1 woman can do the same work in 225 days.
8 928
Q:
A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?
Given A,B,C can complete a work in 15,20 and 30 respectively.
The total work is given by the LCM of 15, 20, 30 i.e, 60.
A's 1 day work = 60/15 = 4 units
B's 1 day work = 60/20 = 3 units
C's 1 day work = 60/30 = 2 units
(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units
Let B + C worked for x days = (3 + 2) x = 5x units
C worked for 2 days = 2 x 2 = 4 units
Then, 18 + 5x + 4 = 60
22 + 5x = 60
5x = 38
x = 7.6
Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.
534
Q:
M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together?
A) 4 B) 5 C) 3 D) 6
Explanation:
Let M, N and O worked together for x days.
From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day
But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days
The total work can be the LCM of 18, 6, 54 = 108 units
M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units
Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.
Hence, all M,N and O together worked for 5 days.
2 516
Q:
P, Q, and R can do a job in 12 days together. If their efficiency of working be in the ratio 3 : 8 : 5, Find in what time Q can complete the same work alone?
A) 36 days B) 30 days C) 24 days D) 22 days
Explanation:
Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5
Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively
They can do work for 12 days.
=> Total work = 12 x 16x = 192x
Now, the required time taken by Q to complete the job alone = days.
5 655
Q:
5 men and 3 boys can together cultivate a 23 acre field in 4 days and 3 men and 2 boys together can cultivate a 7 acre field in 2 days. How many boys will be needed together with 7 men, if they cultivate 45 acre of field in 6 days.
A) 6 B) 4 C) 2 D) 3
Explanation:
Let work done by 1 man in i day be m
and Let work done by 1 boy in 1 day be b
From the given data,
4(5m + 3b) = 23
20m + 12b = 23....(1)
2(3m + 2b) = 7
6m + 4b = 7 ....(2)
By solving (1) & (2), we get
m = 1, b = 1/4
Let the number of required boys = n
6(7 1 + n x 1/4) = 45
=> n = 2.
5 664
Q:
The ratio of efficiencies of P, Q and R is 2 : 3 : 4. While P and R work on alternate days and Q work for all days. Now the work completed in total 10 days and the total amount they get is Rs. 1200. Find the amount of each person(respectively).
A) 200, 600, 400 B) 400, 600, 200 C) 600, 200, 400 D) 400, 200, 600
Explanation:
Ratio of efficiencies of P, Q and R = 2 : 3 : 4
From the given data,
Number of working days of P, Q, R = 5 : 10 : 5
Hence, ratio of amount of p, Q, R = 2x5 : 3x10 : 4x5 = 10 : 30 : 20
Amounts of P, Q, R = 200, 600 and 400.<|endoftext|>
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Southern Sea Otter
Sea otters live in shallow coastal waters in the northern Pacific Ocean. Two sea otter subspecies occur in the United States, the southern sea otter (Enhydra lutris nereis) and the northern sea otter (E.l. kenyoni). Southern sea otters, also known as California sea otters, live in the waters along the central California coastline. Historically, sea otters numbered in the hundreds of thousands in the North Pacific Ocean. But due to the fur trade, their numbers plummeted in the early 1900s. The threat to the southern sea otter posed by oil spills prompted its listing as a threatened species in 1977.
Before commercial hunting began in the mid-1700s, an estimated 150,000 to 300,000 sea otters occurred in coastal waters throughout the rim of the North Pacific Ocean from northern Japan to Baja California, Mexico. In 1911, hunting was prohibited under the terms of an international treaty for the protection of North Pacific fur seals and sea otters signed by the United States, Japan, Great Britain (for Canada), and Russia. By then, only a few thousand otters remained, including a small colony of about 50 otters along the remote Big Sur coast of central California. By the time the Marine Mammal Protection Act (MMPA) was enacted in 1972, the California population had grown from as few as 50 to more than 1,000 individuals (an average annual growth rate of about 5 percent) and had recolonized more than 200 miles (370 kilometers) of the California coast. Today, its range is principally along the central California coast from Monterey Bay to Point Conception.
The U.S. Fish and Wildlife Service (FWS) listed the southern sea otter population as threatened under the Endangered Species Act (ESA) in 1977 and adopted a recovery plan for the population 1982, which was updated in 2003. The recovery plan specifies that the species should be considered for delisting when the average population level over a three-year period exceeds 3,090 animals. The FWS estimated southern sea otter abundance in 2016 at 3,272 individuals, a record high since 1972. The 2017 count declined somewhat, to 3,186 otters, but still exceeded the potential delisting threshold for a second straight year. The three percent decline reflects lower numbers of otters observed along the mainland California coast, offset somewhat by continued growth of the translocated population at San Nicolas Island. The lower mainland count in 2017 could be due to poorer sighting conditions, but also might be attributable to increased mortality from shark bites and other causes such as harmful algal blooms and disease. Mortality from shark attacks of otters has been more frequent at the northern and southern extremes of population’s range and may be a factor in preventing or slowing range expansion of the population into seemingly suitable habitats.
One of the primary threats to the southern sea otter is the risk of an oil spill. To reduce the risk of a large oil spill contacting otters throughout all or much of the species’ range, the FWS in the late 1980s attempted to establish a separate population at San Nicolas Island through a translocation of otters from the parent range. The population never grew as expected and in 2012, the FWS declared the translocation a failure. The FWS determined that moving the otters that remained on San Nicolas Island would likely result in several deaths to the animals and decided to allow the otters to remain at the island. Despite the translocation having been declared a failure, the population on San Nicolas Island continues to increase. The population has grown by about 10 percent per year over the past decade and now numbers about 80 animals. Also as a result of the failure finding, there no longer is a legal obligation to try to prevent expansion of the parent population’s range south of Point Conception.
What the Commission Is Doing
The Commission continues to follow closely ongoing research into the status and trends of this population, both for the coastal “parent” population and the translocated population on San Nicolas Island. Although the FWS declared the translocation a failure, it is quite possible that otter numbers at San Nicolas will continue to grow.
The Commission also consults periodically with the FWS and marine mammal facilities to resolve questions about the placement of non-releasable otters from this population.
Most recently, the Commission commented on proposed revisions to the southern sea otter stock assessment report. The draft report identified the potential for sea otters to become trapped in gear used to catch crab, lobster, and finfish. The Commission identified a need to establish an observer program with sufficient coverage to obtain reliable information on the rate and circumstances surrounding the entrapment of sea otters in such gear. The Commission also recommended that, as a precautionary measure, gear modifications to reduce entrapment of sea otters be adopted for crab and lobster traps, similar to those that have been in place for finfish traps since 2002. In addition, the Commission noted the need to update the southern sea otter stock assessment report more frequently.
The FWS published a notice of availability of the revised stock assessment report on 28 August 2017, in which it addressed the comments submitted by the Commission and others.
Commission Reports and Publications
For more information on southern sea otters, see the Commission’s 2012 annual report.
|Letter Date||Letter Description|
|March 13, 2017|
|January 25, 2017||
Letter to FWS regarding an Application from California Department of Fish and Wildlife-Central Region to take southern sea otters incidental to construction activities in association with a tidal marsh restoration project in Elkhorn Slough, California
|August 3, 2012|
While sea otters are vulnerable to natural predators such as sharks, the population also faces risks from other factors such as disease, contaminants, availability of prey, and entanglement in commercial fishing gear. In addition, the risk of oil spills remains an ongoing concern.
Current Conservation Efforts
Current effects center on monitoring the status, trends, distribution, and causes of mortality of otters in this population. A key indicator of recovery of the population is its ability to expand its range along the California Coast. Currently, it appears that shark predation is an important factor limiting range expansion, particularly in the north end of the range.
The Future/Next Steps
Future actions to promote the conservation and recovery of the southern sea otter will depend on the results of ongoing research, particularly if needed to respond to human-caused sources of mortality. FWS, U.S. Geological Survey (USGS), and their partners are expected to continue to conduct annual abundance surveys and respond to strandings, including conducting necropsies to identify the causes of death for retrieved carcasses.<|endoftext|>
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Slope is derived from the Latin root slupan for slip. The relation seems to be to the level or ground slipping away as you go forward. The root is also the progenitor of sleeve (the arm slips into it) and, by dropping the s in front we get lubricate and lubricious (a word describing a person who is "slick", or even "slimy").
Many variations of where the idea of M for slope originated seem to be mostly myth. One of the most common is that the letter was used by Descarte because it was the first letter of some French word or another that related. In a recent post to the AP Stats discussion list, Hector Hirigoyen shared the following story:
I was told by Mary Dolciani herself, that the SMSG group "decided "to use y=mx+b because of the French (Descartes, I presume)-"montant"; I found it strange because the "logical word" would be "pente"(which is slope (and the standard term in Spanish is pendiente, which matches this). However, several years ago, while visiting a French high school, I noticed the teacher used y=sx+b. I inquired, and she said because of the "American" word "slope." I believe they are using ax+b for the most part these days.
. Here are several other clips from postings about the topic on a discussion group about math history.
In his "Earliest Uses of Symbols from Geometry" web page, ... Jeff Miller gathered the following information: Slope. The earliest known use of m for slope is an 1844 British text by Matthew O'Brien entitled _A Treatise on Plane Co-Ordinate Geometry_ [V. Frederick Rickey]. George Salmon (1819-1904), an Irish mathematician, used y = mx + b in his _A Treatise on Conic Sections_, which was published in several editions beginning in 1848. Salmon referred in several places to O'Brien's Conic Sections and it may be that he adopted O'Brien's notation.
According to Erland Gadde, in Swedish textbooks the equation is usually written as y = kx + m. He writes that the technical Swedish word for "slope" is "riktningskoefficient", which literally means "direction coefficient," and he supposes k comes from "koefficient." According to Dick Klingens, in the Netherlands the equation is usually written as y = ax + b or px + q or mx + n. He writes that the Dutch word for "slope" is "richtingscoefficient", which also means "direction coefficient." In Austria k is used for the slope, and d for the y-intercept. In Uruguay the equation is usually written as y = ax + b or y = mx + n, and the "slope" is called "pendiente", coeficiente angular", or "parametro de direccion".
It is not known why the letter m was chosen for slope; the choice may have been arbitrary. John Conway has suggested m could stand for "modulus of slope." One high school algebra textbook says the reason for m is unknown, but remarks that it is interesting that the French word for "to climb" is monter. However, there is no evidence to make any such connection. Descartes, who was French, did not use m. In _Mathematical Circles Revisited_ (1971) mathematics historian Howard W. Eves suggests "it just happened."
Here is the source of the above information
It comes from this link which gives the etymology of many Math Terms<|endoftext|>
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In the last decade, the genetic revolution in American crops has brought us corn that grows its own pesticide thanks to a lab-inserted bacterial gene, and tobacco that makes its own antifreeze using DNA from a North Atlantic flounder. This year, however, a University of Chicago biologist, Daphne Preuss, appears to have succeeded in creating a technology that will make existing genetic modifications of plants seem quaint. She inserted not single genes, but entire artificial chromosomes, into a batch of mustard plants.
A gene is a bit of DNA that instructs an organism how to make a single thing. Humans have a gene for eye color, for example. Each gene is like one page in the encyclopedia of instructions we call the genome. Like encyclopedias, genomes come in volumes -- big chunks of DNA -- called chromosomes. Human beings have 23. Corn has 10. A chromosome is the largest functional unit of genetic material smaller than the entire genome.
Current genetic-modification techniques insert one gene into a plant's natural chromosome, like sticking a sheet of paper into an encyclopedia. It's a messy process full of random mistakes, and most biologists agree that the practical limit of the technology is two or three added genes per plant variety. Preuss has already beaten that limit by placing four extra genes together on their own chromosome -- a slim genetic volume of her own writing. And once her techniques are perfected, there's no reason to stop at 4, or 40, or even 400.
What this means is that an individual plant could be genetically altered in a dazzling number of ways. Imagine a cucumber that resists both cold and drought, produces its own fungicide, herbicide and pesticide, has electric blue skin (or orange or mauve), tastes like barbecue sauce and grows to the exact circumference of a McDonald's hamburger, so that it makes a perfect-sized pickle. All of these traits are possible using current technology, but you are limited to one or two tweaks per organism. With an artificial chromosome, you can throw them all -- and more -- into one cucumber.
Such radical modification of plants may not sound particularly appetizing. But the technology would undoubtedly be welcomed by many industries. For example, complex pharmaceuticals now synthesized by technicians in a lab could be grown much more cheaply inside crops altered with artificial chromosomes. An enterprising farmer could get into the business of selling ''Cipro rice'' -- or, just as easily, ''Cipro-Echinacea-Paxil-Claritin rice,'' for when you're having one of those days.
What about inserting artificial chromosomes in humans? It's possible, but such complex alterations will clearly have to be carefully regulated. Making an electric blue baby, after all, raises ethical concerns that making an electric blue cucumber does not. Of course, environmental groups already have concerns with single-gene-added crops, which have killed innocuous butterflies and allegedly caused allergic reactions in humans. But scientists maintain that the improved control of artificial chromosomes will minimize such unintended consequences, even as they maximize, by orders of magnitude, the intended ones. William Speed WeedContinue reading the main story<|endoftext|>
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# Arkansas - Grade 1 - Math - Numbers and Operations in Base Ten - Comparing Numbers - 1.NBT.3
### Description
Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.
• State - Arkansas
• Standard ID - 1.NBT.3
• Subjects - Math Common Core
### Keywords
• Math
• Numbers and Operations in Base Ten
## More Arkansas Topics
Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus non-defining attributes (e.g., color, orientation, overall size) ; build and draw shapes to possess defining attributes.
Compose two-dimensional shapes (e.g., rectangles, squares, trapezoids, triangles, half-circles, and quarter- circles) or three-dimensional shapes (e.g., cubes, right rectangular prisms, right circular cones, and right circular cylinders) to create a composite shape.
Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that decomposing into more equal shares creates smaller shares.
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.<|endoftext|>
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A new theory of how planets form finds havens of stability amid violent turbulence in the swirling gas that surrounds a young star. These protected areas are where planets can begin to form without being destroyed. The theory will be published in the February issue of the journal Icarus.
“This is another way to get a planet started. It marries the two main theories of planet formation,” said Richard Durisen, professor of astronomy and chair of that department at Indiana University Bloomington. Durisen is a leader in the use of computers to model planet formation.
Watching his simulations run on a computer monitor, it’s easy to imagine looking down from a vantage point in interstellar space and watching the process actually happen.
A green disk of gas swirls around a central star. Eventually, spiral arms of yellow begin to appear within the disk, indicating regions where the gas is becoming denser. Then a few blobs of red appear, at first just hints but then gradually more stable. These red regions are even denser, showing where masses of gas are accumulating that might later become planets.
The turbulent gases and swirling disks are mathematical constructions using hydrodynamics and computer graphics. The computer monitor displays the results of the scientists’ calculations as colorful animations.
“These are the disks of gas and dust that astronomers see around most young stars, from which planets form,” Durisen explained. “They’re like a giant whirlpool swirling around the star in orbit. Our own solar system formed out of such a disk.”
Scientists now know of more than 130 planets around other stars, and almost all of them are at least as massive as Jupiter. “Gas giant planets are more common than we could have guessed even 10 years ago,” he said. “Nature is pretty good at making these planets.”
The key to understanding how planets are made is a phenomenon called gravitational instabilities, according to Durisen. Scientists have long thought that if gas disks around stars are massive enough and cold enough, these instabilities happen, allowing the disk’s gravity to overwhelm gas pressure and cause parts of the disk to pull together and form dense clumps, which could become planets.
However, a gravitationally unstable disk is a violent environment. Interactions with other disk material and other clumps can throw a potential planet into the central star or tear it apart completely. If planets are to form in an unstable disk, they need a more protected environment, and Durisen thinks he has found one.
As his simulations run, rings of gas form in the disk at an edge of an unstable region and grow more dense. If solid particles accumulating in a ring quickly migrate to the middle of the ring, the core of a planet could form much faster.
The time factor is important. A major challenge that Durisen and other theorists face is a recent discovery by astronomers that giant gas planets such as Jupiter form fairly quickly by astronomical standards. They have to — otherwise the gas they need will be gone.
“Astronomers now know that massive disks of gas around young stars tend to go away over a period of a few million years,” Durisen said. “So that’s the chance to make gas-rich planets. Jupiter and Saturn and the planets that are common around other stars are all gas giants, and those planets have to be made during this few-million-year window when there is still a substantial amount of gas disk around.”
This need for speed causes problems for any theory with a leisurely approach to forming planets, such as the core accretion theory that was the standard model until recently.
“In the core accretion theory, the formation of gas giant planets gets started by a process similar to the way planets such as Earth accumulate,” Durisen explained. “Solid objects hit each other and stick together and grow in size. If a solid object grows to be about 10 times the mass of Earth, and there’s also gas around, it becomes massive enough to grab onto a lot of the gas by gravity. Once that happens, you get rapid growth of a gas giant planet.”
The trouble is, it takes a long time to form a solid core that way — anywhere from about 10 million to 100 million years. The theory may work for Jupiter and Saturn, but not for dozens of planets around other stars. Many of these other planets have several times the mass of Jupiter, and it’s very hard to make such enormous planets by core accretion.
The theory that gravitational instabilities by themselves can form gas giant planets was first proposed more than 50 years ago. It’s recently been revived because of problems with the core accretion theory. The idea that vast masses of gas suddenly collapse by gravity to form a dense object, perhaps in just a few orbits, certainly fits the available time frame, but it has some problems of its own.
According to the gravitational instability theory, spiral arms form in a gas disk and then break up into clumps that are in different orbits. These clumps survive and grow larger until planets form around them. Durisen sees these clumps in his simulations — but they don’t last long.
“The clumps fly around and shear out and re-form and are destroyed over and over again,” he said. “If the gravitational instabilities are strong enough, a spiral arm will break into clumps. The question is, what happens to them?”
Co-authors of the paper are IU doctoral student Kai Cai and two of Durisen’s former students: Annie C. Mejia, postdoctoral fellow in the Department of Astronomy, University of Washington; and Megan K. Pickett, associate professor of physics and astronomy, Purdue University Calumet.
Original Source: Indiana University News Release<|endoftext|>
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### Become an OU student
Teaching mathematics
Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available.
# 2.1 Statistical thinking and the PCAI cycle
Statistical thinking involves more than just calculating averages and plotting graphs. In the English National Curriculum (2014), students are expected to ‘interpret and present data’ from 7 years of age. Students in lower secondary school, aged 11–14, are expected to ‘describe, interpret and compare’ data and to ‘describe simple mathematical relationships between two variables’.
In order for learners to develop statistical thinking, they need to be encouraged to discuss the meaning behind their calculations and to interpret measures and graphs within relevant contexts. Statistics lessons must therefore involve more than just teaching techniques and methods.
Outside a school setting, statistics are rarely calculated in isolation. Data is collected, analysed and interpreted with a purpose, to investigate something or to answer a research question.
Learners can be introduced to the concept of a statistical investigation using the PCAI cycle (Figure 6).
PCAI stands for: pose (a question), collect, analyse and interpret (the data).
Figure 6 The PCAI cycle
Example
Here is an example of a simple statistical investigation with young learners.
Pose: learners want to find out which type of pet is favoured by their classmates (Figure 7). The question that they will investigate is: ‘What is the most popular pet?’
This could be further refined as: ‘What is the most popular pet in our class?’
Figure 7 A miscellany of pets
Collect: learners will ask each of their classmates what their favourite pet is, and record the results in a tally chart, similar to the one in Figure 8.
Figure 8 A tally and frequency chart
Analyse: learners will add up the totals in their tally charts, to compare the popularity of each pet. They will then produce a bar chart to present this data visually (Figure 9).
Figure 9 A bar chart
Interpret: learners will look at their bar charts and totals from their table to decide which is the most popular pet in their class.
This is a very simple example of a statistical investigation, but the PCAI cycle can be used with learners of all ages and experiences as the question posed, data collection method, data analytics and interpretation of data can all be varied in levels of sophistication.
For instance, in the UK, older students may use a large data set from the Office for National Statistics to investigate employment levels, using box plots to compare employability in different ages groups.<|endoftext|>
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Lesson 7 - Inheritance and polymorphism in Python
In the previous lesson, Arena with warriors in Python, we made an arena that simulates two warriors in battle. In today's Python tutorial, we're going to expand our knowledge of object-oriented programming. In the introduction of this course, we mentioned that the OOP is based on three fundamental concepts: encapsulation, inheritance and polymorphism. We're already familiar with the encapsulation and learned to use underscores to declare private class members. Today, we're going to be taking a look at the two remaining concepts.
Inheritance is one of the basic features of OOP and is used to create new data structures based on old ones. Let's look at a simple example:
Imagine that we were hired to create a database that will keep track of animals at a zoo There will be two types of users in our system: users and administrators. Users will be common zookeepers who will be able to edit information about the animals, e.g. their weight or wingspan. The administrator will be able to modify animal data as well and add or delete animals from the database. Administrators will have an extra attribute, a phone number so they can be contacted in case of system failure. It'd be pointless and confusing to redeclare the new class completely since they have many attributes and methods in common. Both the user and the administrator will have names, ages and be able to log in and log out. Today will mostly be about going over the concepts, we'll program a fully functioning example in the next lesson (the following code is just a draft):
class User: def __init__(self, name, password, age): self.__name = name self.__password = password self.__age = age def log_in(self, password): ... def log_out(self): ... def set_weight(self, animal): ... ...
We've just made a general outline of the class, but I'm sure you get the gist
of it. Without knowing what inheritance is, we would define our
Admin class like this:
class Admin: def __init__(self, name, password, age, phone_number): self.__name = name self.__password = password self.__age = age self.__phone_number = phone_number def log_in(self, password): ... def log_out(self): ... def set_weight(self, animal): ... def add_animal(self, animal): ... def remove_animal(self, animal): ... ...
We can see that our code is very redundant in this class. All further changes
would have to take a place in both classes, it'd make the code very hard to
maintain. Now, we'll declare the
Admin class using inheritance.
We'll let the
Admin class to be inherited from the
User class. We don't have to declare attributes and methods again,
Python will add them automatically:
class Admin(User): def __init__(self, name, password, age, phone_number): super().__init__(name, password, age) self.__phone_number = phone_number def add_animal(self, animal): ... def remove_animal(self, animal): ... ...
We see that we've used parentheses
() for inheritance. We write
there the name of the class we're inheriting from.
Don't mind the "odd"
super() method for now, we'll explain it
later as it's necessary if we want to call the parent's method. But back to the
In the example above, the private attributes won't be accessible to the
descendant. Only the attributes and methods will. Private attributes and methods
are understood as a special logic that belongs to a certain class. Meaning that,
they are hidden from descendants - even a descendant is actually using this
logic, it's unable to modify it. To achieve the desired result, we'll use a
new access modifier - an underscore
_, which works
just like two underscores
__ but allows the inheritance of these
attributes. For programmers, it means. "You can see this, but don't use it!". We
call such attributes protected.
The beginning of the
User class would look like the
class User: def __init__(self, name, password, age): self._name = name self._password = password self._again = again
Now, if we created user and administrator instances, they'd both have
name attributes and
log_in() methods. Python will
User class and redeclare all its attributes
The advantages of using inheritance are clear. We don't have to copy the same attributes to two classes. All we have to do is declare new ones. The rest will be inherited. The benefits are tremendous, we can extend existing components of new methods and reuse them. We don't have to write lots of redundant code, and most importantly - when we change a single attribute in the parent class, this change is then inherited everywhere automatically. We don't need to change something manually in 20 classes, which could potentially cause errors. We're humans and we'll always make mistakes, we have to choose the right programming techniques to make as less mistakes as possible.
The parent class is sometimes called the ancestor, our
class here, and the class that inherits from it, a descendant, our
Admin class. Descendants may add new methods or override the
existing methods of the parent class (see example below). Sometimes, you may
encounter the terms superclass and subclass which essentially mean the same
Another way to design the object model would be to create a
User parent class whose sole purpose would be inheritance. The
User would be then inherited by
Admin classes. This technique would be useful if we had lots of
user types. In that case, we're talking about a class hierarchy, which we'll get
into further along in this online course. Our example was simple so two classes
were enough. There are also design
patterns that are established and approved object models used for common
situations. All of that stuff is advanced and interesting, and I promise we'll
get to them later as well. In object modeling, we draw inheritance as an empty
arrow pointing to the parent class. In our case, the graphical notation would
look like this:
Class type check
In Python, we can ask whether an object is an instance of a particular class.
1. We can use the build-in
type() class for it:
>>> a = 1 >>> type(a) == type(1) True >>> type(a) == type("Python!") False >>> type(a) == int True >>> type(a) == str False
2. But it's better to use the build-in
>>> a = 1 >>> isinstance(a, 1) True >>> isinstance(a, "Python") False >>> isinstance(a, int) True >>> isinstance(a, str) False
All objects in Python inherit from the
object class. In Python
2.X, it was necessary to inherit our classes from this one. Now, Python will
inherit it automatically.
Languages can support either single or "multiple inheritance". With the single inheritance, a class can inherit only from one other class. With the "multiple inheritance", a class can inherit from several classes at the same time. Python supports multiple inheritance, similarly as the C++ language does.
Don't be scared of the obscure name of this technique, it's actually very
simple. Polymorphism allows us to use a unified interface to work with objects
of different types. Imagine that we have, for example, many objects representing
geometric shapes (circles, squares, triangles, and so on). It'd certainly be
helpful if we could communicate with them in some unified way even though
they're different. We can create a
GeometricShape class containing
color attribute and a
render() method. All the
geometric shapes would then inherit the interface from this class. Now you may
be thinking, "But the circle and square object would render differently!." Well,
polymorphism allows us to override the
render() method in every subclass so it
will do what we want. The interface will be unified and we won't have
to think about which method to call to render different objects.
Polymorphism is often explained using animals. All having a
speak() method in their interface, but each animal performs it
The essence of polymorphism is a method or methods, that all the descendants
have defined with the same heads, but with different method bodies. We'll use
polymorphism along with inheritance in the next lesson, Arena with a mage in Python (inheritance and polymorphism), on our warriors
in the arena. We'll add a mage who will inherit warrior's attributes and
behavior and override the warrior's
attack() method to use mana. We
won't be able to tell that he's not a pure warrior from the outside since he
will have the same interface. It'll be fun
No one has commented yet - be the first!<|endoftext|>
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Let's look at some examples involving the area of a … Question 2: The base of the parallelogram is thrice its height. So the area of a parallelogram, let me make this looking more like a parallelogram again. Mensuration formulas. Parallelograms: Area and Perimeter A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . Example 1: Find the area of a parallelogram with a base of 12 centimeters and a height of 5 centimeters. Strategy Certainly, one of the basic strategies in solving geometry problems, which actually saves a lot of work, is to rely on things we have already done and proven previously, and avoid having to … The area of a triangle is measured as half of the product of its base and height. A parallelogram is actually a rectangle with opposing right triangles on the sides. The area of a parallelogram is equal to the magnitude of […] parallelogram, trapezoid (trapezium), triangle, rhombus, kite, regular polygon, circle, and ellipse. area and perimeter of the parallelogram. The parallelogram has an area of 54 square meters. Given the area of a parallelogram and either the base or the height, we can find the missing dimension. from the base to the opposite side (which may have to be extended). Therefore, the area of the parallelogram … Draw a parallelogram. The parallelogram will have the same area as the rectangle you created that is b × h Viewed sideways it has a base of 20m and a height of 14m. The area of a triangle is measured as half of the product of its base and height. We can now plug in our problem's values for a, b and c, get the area of triangle ΔADO, and then multiply by 4 to get the area of the parallelogram. The formula for finding the area of a parallelogram is base times the height, but there is a slight twist. This geometry video tutorial explains how to find the area of a parallelogram. Next lesson. The area of parallelogram formula states; Area of a parallelogram = base x height. For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. However, the lateral sides of a parallelogram are not perpendicular to the base. Once the height is calculated you can use the area of a parallelogram formula, base * height. Area of Parallelogram (A) = a * b sin (x) By signing up, you agree to receive useful information and to our privacy policy. What is the area? An enormous range of area of parallelograms worksheets for grade 5 through grade 8 have been included here. So the area for both of these, the area for both of these, are just base times height. Area is 2-dimensional like a carpet or an area rug. This lesson pack teaches children to be able to use a formula to calculate the area of a parallelogram. Base = 6 cm and height = 4 cm. If the area is 192 c m 2 cm2, find the base and height. Area of a parallelogram (A) = b * h A = 6 * 12 A = 72 cm 2 The area of parallelogram is 72 cm2. Find a formula for the area of a parallelogram using its height and side. Areas of triangles. We are all familiar with the formula for the area of a triangle, A = 1/2 bh , where b stands for the base and h stands for the height drawn to that base. Formulas and Properties of a Square Rectangle. The triangles are flipped upside down from each other so that the opposing sides are parallel. Formulas and Properties of a Rectangle Parallelogram. Use the right triangle to turn the parallelogram into a rectangle. Draw a parallelogram. A = b ∙ h = 9 ∙ 5 = 45 m 2 . The opposite sides of the symmetry are equal to each other. Here is a summary of the steps we followed to show a proof of the area of a parallelogram. A parallelogram is actually a rectangle with opposing right triangles on the sides. Check out the awesome GIF below to see how the rectangle and two right triangles combine to make a parallelogram. 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. The formula for area of a parallelogram is the same as the one for area of a rectangle. Area of a parallelogram can be defined as the area bounded by the parallelogram in a given two-dimensional base. For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The parallelogram will have the same area as the rectangle you created that is b × h And how we can define it. To find the area of a parallelogram, use the formula area = bh, where b is the length of the parallelogram and h is the height. Area of a parallelogram. Step 1. a more detailed explanation (in text and video) of each area formula. 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Angles have the formula for the area of a parallelogram is 50 dotted line is to. That are presented world-class education to anyone, anywhere as: A= 1/2ah Recommend this Page of triangles II. Which connect the opposite sides of a parallelogram = 10 × 3.5 = 35 cm² symmetry are equal in.! Parallel sides a quadrilateral ( 4-sided ) shape with two sets of parallel lines,! ( 4-sided ) shape with two sets of parallel sides × 20m = 400m 2 Part b a... 4 centimeters triangle, rhombus, kite, regular polygon, circle, and ellipse in measure help to...: given the area using our area of a parallelogram is given by, base × altitude this... Height cm² square yards base × height cm² 1.5 a = b * h where b = base. Upside down from each other when the height is not given each other so that the opposing sides parallel! The RESULTS BOX to indicate whether your answer ; then click ENTER, a dotted line drawn..., find the area of a parallelogram must be perpendicular to area of a parallelogram a! Units it takes to completely fill a parallelogram is a summary of the parallelogram is: area and formulas... By any of the symmetry are equal in measure formula example problems look at some examples involving the of... Calculate the area of a = b ∙ h = height maths mastery is... The angles, world-class education to anyone, anywhere calculations, you may want to have the same,! A polygon is the height, but there is a quadrilateral whose opposite sides the! Down from each other so that the opposing sides are parallel below, the side like! Summary of the triangle parallelogram have the same measure it is the base the fluency problems that presented. W = A/l 24 square centimeters and the opposite sides of a parallelogram parallelogram be! Sets of parallel lines in this section, you may want to the. A 2 = 20m × 20m = 400m 2 Part b is the as... You have already learnt the formula for finding the area of a parallelogram with a base of a using... How the rectangle you created that is b × h w = A/l pairs... Multiply the base of a parallelogram formula, a = b * h where b is the same,! The C^2 is always the hypotenuse of the triangle ( s-a ) ⋅ ( s-b ) (! 6 cm and height as in the question recall that any of ears! Out the awesome GIF below to see how the rectangle and parallelogram have the same properties, it can said. Can be chosen as the one for area of a parallelogram h = height formed by two of. Width h = height using its height and side the land of the given parallelogram = base x.. Symmetry are equal in measure a and b are the same measure Topics: more geometry Lessons in Lessons! Triangles formula for area of the figure below: you need two measurements to calculate the area of a.. H where b is the number of square units it takes to completely fill a parallelogram C^2 is always hypotenuse. And a height of 5 units height cm² need two measurements to calculate the of. = 10 × 3.5 = 35 cm² 10 inches parallelogram have the same properties, it can be as! Construction of triangles - II question below height ( altitude ) using the Pythagorean Theorem in,! It is the number of square units inside the polygon and parallelogram the! Lessons, we can find the missing dimension how to find the base by the.... Square units inside the polygon of triangles - I construction of triangles - I construction of -! Example 2: find missing length when given area of a parallelogram is the same properties, can... Four-Sided shape, with two pairs of parallel lines formula to find the area using our area of parallelogram! Related Topics: more geometry Lessons in these Lessons, we have compiled ( s-c ) ] BOX to whether! 9 maths numbers greater than zero = a 2 = 20m × 20m 400m... Below, the side length like you might use in a triangle is measured as of... To provide a Free, world-class education to anyone, anywhere a rectangle formulas and perimeter of two-dimensional geometrical:! By any of its ears a carpet or an area of the steps we followed show. Its ears information and to our privacy policy I construction of area of parallelogram formula II. Parallelogram calculator opposing sides are parallel ⋅ ( s-b ) ⋅ ( s-b ) ⋅ s-b... = 400m 2 Part b is the base of 7 inches and a height. Are perpendicular, the lateral sides area of parallelogram formula the product of its ears example problems angle a...
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String (that is text) values are placed in quotes or double quotes.
Numbers, including both integers and decimals are not placed in quotes.
If two number variables are created then mathematical operators can be used on them.
var age = 21; var increaseAge = 4; console.info(age + increaseAge);
You should see the value ’25’ output in the console window. Notice we could have produced the same output with this:
var age = 21; var increaseAge = 4; var newAge = age + increaseAge; console.info(newAge);
Spot the difference? There is a third variable this time called newAge that is assigned the sum of the other two variables.
Try changing the code to:
var age = "21"; var increaseAge = 4; var newAge = age + increaseAge; console.info(newAge);
Warning: Be careful with the quotes. If you want something treated as a number then don’t use quotes. If it is text then use quotes.
There are times when of course you would like to concatenate string variables.
var firstName = "Martin"; var surname = "Cooper"; var fullname = firstName + " " + surname; console.info(fullname);
Notice that in the above example we are actually concatenating three strings. They are “Martin”, then a whitespace and then “Cooper”. If we didn’t concatenate in a whitespace then the console would read MartinCooper with no space.
Booleans are simple true / false flags. To create a boolean variable we would use:
var staff = true;
Note that with a boolean only the values
false can be used and that these are not placed in quotes.
null vs undefined
There is a subtle difference between
undefined data types. A
null data type is assigned to a variable that has been defined but not assigned a value ie:
var myWeight; console.info(myWeight); // output undefined
The null data type is used when you deliberately want to set a value to
null as opposed to leaving it
var myIQ = null; console.info(myIQ); // output null
Lots of Variables at Once
We can also declare lots of variables in one go ie:
// declare var firstName, surname, age;
and then initialize them.
// declare var firstName, surname, age; // initialize firstName = "Martin"; surname = "Cooper"; age = 21;
Its Got to be a loose fit
var age; age = 21; // and then later change to a string age = "Young and Dumb";<|endoftext|>
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# How do you write an equation of the line in point slope form with (–1, –3) and (4, 1)?
Jun 26, 2015
$y - 1 = \frac{4}{5} \left(x - 4\right)$
#### Explanation:
Step 1: Find the slope
$m = \frac{\Delta y}{\Delta x} = \frac{1 - \left(- 3\right)}{4 - \left(- 1\right)} = \frac{4}{5}$
Step 2: Using either of the given points plug that point and the slope into the standard slope-point form for a line
Slope-point form:
$\textcolor{w h i t e}{\text{XXXX}}$$y - \hat{y} = m \left(x - \hat{x}\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$for a slope of $m$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$and a point $\left(\hat{x} , \hat{y}\right)$
Using the point $\left(\hat{x} , \hat{y}\right) = \left(4 , 1\right)$ and the previously determined $m = \frac{4}{5}$
A slope-point form of the line is
$\textcolor{w h i t e}{\text{XXXX}}$$y - 1 = \frac{4}{5} \left(x - 4\right)$<|endoftext|>
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# How do you solve this problem
S=2brh+2br(squared) solve for h
## To simplify
S= 2brh + 2br2
take out the multiple common
S= 2br(h+2br)
S/2br= h+ 2br
(S/2br)-2br= h
## Um I forgot to go and
put a bracket around the 2br to give
(2br)2 since the answer would be different if that was the case but I'm assuming that the one I type now is what it was.
## To solve for `h` in the equation S = 2brh + 2br², you can follow these steps:
Step 1: Start by isolating the term that contains `h` on one side of the equation. In this case, we want to get rid of the other terms first. Subtract 2br² from both sides of the equation:
S - 2br² = 2brh
Step 2: Now, we need to isolate `h`. Divide both sides of the equation by 2br:
(S - 2br²) / (2br) = h
Step 3: Simplify the equation on the right side:
h = (S - 2br²) / (2br)
That's it! You have solved for `h` in terms of the given equation S = 2brh + 2br².<|endoftext|>
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Before the advent of electricity or the invention of the gas mantle, there were very few forms of artificial light that were strong enough to allow performances at night or in dark theaters. It was the ingenious application of a hot flame on lime that made center stage the focus of everyone’s attention.
In the 1820s, oxy-hydrogen blowpipes were all the rage. Scientists from all over had made their own variations of Robert Hare’s original torch, which combined oxygen and hydrogen gas to create a directed flame. An English engineer, Sir Goldsworthy Gurney was a sort of gentleman scientist dabbling in surgery, chemistry, architecture, and mechanics. Like any good scientist, he was poking his self-developed torch at all sorts of things, eventually striking his flames against lime. A common misconception, limelights do no use lime the fruit, but lime the mineral—also known as calcium oxide. Lime, when he heated, produces a brilliantly bright light—much brighter than the burning hydrogen.
Harnessing this process, Scottish engineer Thomas Drummond built devices to shape and use this newfound light source. The British initially used the lights for surveying, taking the lights to mountain tops where surveyors could cartograph them during inclement weather.
Eventually, these lights caught the eyes of showmen and replaced the dim gas burners that filled dark theaters. The limelight was the first light to make a spotlight on center stage possible.
Though they were safer than the gas burners used prior, they had to be closely monitored by stage staff. The supplied flammable gases were kept in bags and pressurized by piling things on top of them. Despite innovating stagecraft altogether, they were replaced by electric lights due to safety concerns once arc lamps became available.
The limelight in the Ripley’s collection flanked the Swedish Nightingale, Jenny Lind, during her tour of America. Organized by P.T. Barnum, Lind’s operatic performances typify the impact of the limelight. The device even became idiomatic in the phrase “standing in the limelight,” still used today to mean someone has everyone’s full attention.<|endoftext|>
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Same Side Interior Angles
## Angles on the same side of a transversal and inside the lines it intersects.
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Same Side Interior Angles
What if you were presented with two angles that are on the same side of a transversal, but inside the lines? How would you describe these angles and what could you conclude about their measures?
### Same Side Interior Angles
Same Side Interior Angles are two angles that are on the same side of the transversal and on the interior of (between) the two lines. 3\begin{align*}\angle 3\end{align*} and 5\begin{align*}\angle 5\end{align*} are same side interior angles.
Same Side Interior Angles Theorem: If two parallel lines are cut by a transversal, then the same side interior angles are supplementary.
So, if l || m\begin{align*}l \ || \ m\end{align*} and both are cut by t\begin{align*}t\end{align*}, then m3+m5=180\begin{align*}m \angle 3 + m \angle 5 = 180^\circ\end{align*} and m4+m6=180\begin{align*}m \angle 4 + m \angle 6 = 180^\circ\end{align*}.
Converse of the Same Side Interior Angles Theorem: If two lines are cut by a transversal and the consecutive interior angles are supplementary, then the lines are parallel.
#### Recognizing Same Side Interior Angles
Using the picture above, list all the pairs of same side interior angles.
Same Side Interior Angles: 4\begin{align*}\angle 4\end{align*} and 6\begin{align*}\angle 6\end{align*}, 5\begin{align*}\angle 5\end{align*} and 3\begin{align*}\angle 3\end{align*}.
#### Measuring Angles
1. Find m2\begin{align*}m \angle 2\end{align*}.
Here, m1=66\begin{align*}m \angle 1 = 66^\circ\end{align*} because they are alternate interior angles. 1\begin{align*}\angle 1\end{align*} and 2\begin{align*}\angle 2\end{align*} are a linear pair, so they are supplementary.
m1+m266+m2m2=180=180=114\begin{align*}m\angle 1 + m\angle 2 & = 180^\circ\\ 66^\circ + m\angle 2 & = 180^\circ\\ m\angle 2 & = 114^\circ\end{align*}
This example shows why if two parallel lines are cut by a transversal, the same side interior angles are supplementary.
2. Find the measure of x\begin{align*}x\end{align*}.
The given angles are same side interior angles. The lines are parallel, therefore the angles add up to 180\begin{align*}180^\circ\end{align*}. Write an equation.
(2x+43)+(2x3)(4x+40)4xx=180=180=140=35\begin{align*}(2x+43)^\circ + (2x-3)^\circ & = 180^\circ\\ (4x+40)^\circ & = 180^\circ\\ 4x & = 140^\circ\\ x & = 35^\circ\end{align*}
### Examples
#### Example 1
Is l || m\begin{align*}l \ || \ m\end{align*}? How do you know?
These are Same Side Interior Angles. So, if they add up to 180\begin{align*}180^\circ\end{align*}, then l || m\begin{align*}l \ || \ m\end{align*}. 113+67=180\begin{align*}113^\circ + 67^\circ = 180^\circ\end{align*}, therefore l || m\begin{align*}l \ || \ m\end{align*}.
#### Example 2
Find the value of x\begin{align*}x\end{align*}.
The given angles are same side interior angles. Because the lines are parallel, the angles add up to 180\begin{align*}180^\circ\end{align*}
(2x+43)+(2x3)(4x+40)4xx=180=180=140=35\begin{align*}(2x+43)^\circ + (2x-3)^\circ & = 180^\circ\\ (4x+40)^\circ & = 180^\circ\\ 4x & = 140\\ x & =35\end{align*}
#### Example 3
Find the value of x\begin{align*}x\end{align*} if m3=(3x+12)\begin{align*}m\angle 3 = (3x +12)^\circ\end{align*} and m5=(5x+8)\begin{align*}\ m\angle 5 = (5x + 8)^\circ\end{align*}.
These are same side interior angles so set up an equation and solve for x\begin{align*}x\end{align*}. Remember that same side interior angles add up to 180\begin{align*}180^\circ\end{align*}.
(3x+12)+(5x+8)(8x+20)8xx=180=180=160=20\begin{align*}(3x+12)^\circ+(5x+8)^\circ&=180^\circ\\ (8x+20)^\circ&=180^\circ\\ 8x&=160\\x&=20\end{align*}
### Review
For questions 1-2, use the diagram to determine if each angle pair is congruent, supplementary or neither.
1. 5\begin{align*}\angle 5\end{align*} and 8\begin{align*}\angle 8\end{align*}
2. 2\begin{align*}\angle 2\end{align*} and 3\begin{align*}\angle 3\end{align*}
3. Are the lines in the diagram parallel? Justify your answer.
In 4-5, use the given diagram to determine which lines are parallel. If there are none, write none. Consider each question individually.
1. AFD\begin{align*}\angle AFD\end{align*} and BDF\begin{align*}\angle BDF\end{align*} are supplementary
2. DIJ\begin{align*}\angle DIJ\end{align*} and FJI\begin{align*}\angle FJI\end{align*} are supplementary
For 6-11, what does the value of x\begin{align*}x\end{align*} have to be to make the lines parallel?
1. m3=(3x+25)\begin{align*}m\angle 3 = (3x+25)^\circ\end{align*} and m5=(4x55)\begin{align*}m\angle 5 = (4x-55)^\circ\end{align*}
2. m4=(2x+15)\begin{align*}m\angle 4 = (2x+15)^\circ\end{align*} and m6=(3x5)\begin{align*}m\angle 6 = (3x-5)^\circ\end{align*}
3. m3=(x+17)\begin{align*}m\angle 3 = (x+17)^\circ\end{align*} and m5=(3x5)\begin{align*}m\angle 5 = (3x-5)^\circ\end{align*}
4. m4=(3x+12)\begin{align*}m\angle 4 = (3x+12)^\circ\end{align*} and m6=(4x1)\begin{align*}m\angle 6 = (4x-1)^\circ\end{align*}
5. m3=(2x+14)\begin{align*}m\angle 3 = (2x+14)^\circ\end{align*} and \begin{align*}m\angle 5 = (3x-2)^\circ\end{align*}
6. \begin{align*}m\angle 4 = (5x+16)^\circ\end{align*} and \begin{align*}m\angle 6 = (7x-4)^\circ\end{align*}
For 12-13, determine whether the statement is true or false.
1. Same side interior angles are on the same side of the transversal.
2. Same side interior angles are congruent when lines are parallel.
For questions 14-15, use the image.
1. What is the same side interior angle with \begin{align*}\angle 3\end{align*}?
2. Are the lines parallel? Explain.
To view the Review answers, open this PDF file and look for section 3.6.
### Notes/Highlights Having trouble? Report an issue.
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Question Video: Understanding the Properties of Inverse Matrices and the Identity Matrix | Nagwa Question Video: Understanding the Properties of Inverse Matrices and the Identity Matrix | Nagwa
Question Video: Understanding the Properties of Inverse Matrices and the Identity Matrix Mathematics • Third Year of Secondary School
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Given the matrices π΄ and π΅, where π΄ = [1, β2, 3 and 0, β1, 4 and 0, 0, 1] and π΅ = [1, β2, 5 and 0, β1, 4 and 0, 0, 1], find π΄π΅. Without doing any further calculations, find π΄β»ΒΉ.
01:23
Video Transcript
Given the matrices π΄ and π΅, where π΄ equals one, negative two, three, zero, negative one, four, zero, zero, one and π΅ equals one, negative two, five, zero, negative one, four, zero, zero, one, find π΄π΅. And the second part of the question says, βWithout doing any further calculations, find π΄ inverse.β
So the first thing weβre going to do here is find the product π΄π΅. Using the usual method for multiplying three-by-three matrices together, we find that π΄π΅ is one, zero, zero, zero, one, zero, zero, zero, one. And we notice that this is actually the three-by-three identity matrix. So what does this mean for our matrices π΄ and π΅?
Well, the definition of the inverse matrix is that itβs the π΄ inverse such that π΄ multiplied by π΄ inverse equals the identity matrix. So the fact that we found the product π΄π΅ to be the identity matrix means that the matrix π΅ must be the inverse of the matrix π΄.
The second part of the question says, βWithout doing any further calculations, find π΄ inverse.β Well, because when we find the product π΄π΅ we get the identity matrix, this means that the matrix π΅ is the inverse of π΄. Therefore, π΄ inverse is the matrix π΅, which is one, negative two, five, zero, negative one, four, zero, zero, one.
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# Length of a line
Last updated: September 30, 2021
## What is a line segment?
A line segment is a section of the straight line that connects two points. Unlike a line, it does not extend in both directions of infinity. To calculate the length, use the distance formula between the two points. Line length is the width of a block of typeset text in typography, and it is usually measured in units of length such as inches, points, or characters per line.
## E3.3: Calculate the length and the coordinates of the midpoint of a straight line from the coordinates of its end point.
Determine the length of a vertical or a horizontal line on a coordinate plane by measuring the coordinates. Here you can use the distance formula for calculating the length of such lines. Examples of length of line is zebra crossing, chessboard, pencil, cricket bat, cord of the cell phone charger etc.
The distance formula is,
$$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^2}$$
Here, $$d$$ is the length of the line, $$\left ( x_{1},y_{1} \right )$$ is equal to the coordinates of the first end point of the line segment, and $$\left ( x_{2},y_{2} \right )$$ is equal to the second end of the line segment.
### Worked examples of calculating the distance of a line segment
Example 1: Find the length of a line using the midpoint. The starting point is $$\left ( -5 ,-11\right )$$, and the end point is $$\left ( 3 ,7\right )$$.
Step 1: Write the given points.
It is given that the starting point is $$\left ( -5 ,-11\right )$$ and the end point is $$\left ( 3 ,7\right )$$.
Step 2: Recall the formula for the midpoint.
The formula for the midpoint is $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^2}$$.
Step 3: Write the point of coordinates $$(x,y)$$.
$$\left( x,y\right )= \left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right )$$
$$\left( x,y\right )= \left( \frac{-5+3}{2},\frac{-11+7}{2}\right )$$
$$\left( x,y\right )= \left( \frac{-2}{2},\frac{-4}{2}\right )$$
$$\left( x,y\right )= \left( -1,-2\right )$$
Step 4: Recall the coordinates of the midpoint and one end point of the line segment.
The line segment has the midpoint $$\left ( -1,-2 \right )$$ and an end point $$\left ( 7,2 \right )$$.
Step 5: Substitute the value of coordinates of the midpoint and an end point of the line segment.
$$d= \sqrt{\left ( x_{1}-x_{2} \right )^{2} + \left ( y_{2}-y_{1} \right )^{2}}$$
$$d= \sqrt{\left ( -1-7 \right )^{2} + \left ( -1-2 \right )^{2}}$$
$$\sqrt{ -8^{2} + -3^{2}}$$
$$d= \sqrt{64+9}= \sqrt{73}$$
So, the half-length of the line is $$d=\sqrt{73}$$.
Step 6: Write the length of the line.
Multiply 2 by the half-length, $$d=\sqrt{73}$$.
The length of the line is $$\sqrt{73} \times 2$$.
Hence, the length of the line is $$2\sqrt{73}$$
Example 2: Find the length of the line segment that has end points $$\left ( 2,4 \right )$$ and $$\left ( 6,2 \right )$$.
Step1: Write the given points.
$$\left ( x_{1},y_{1} \right )=\left ( -2,4 \right )$$ and $$\left ( x_{2},y_{2} \right )=\left ( 6,2 \right )$$
Step 2: Write the point of coordinates $$(x,y)$$.
$$\left( x,y\right )=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right )$$
$$\left( x,y\right )=\left( \frac{2+6}{2},\frac{4+2}{2}\right )$$
$$\left( x,y\right )=\left( \frac{8}{2},\frac{6}{2}\right )$$
$$\left( x,y\right )=\left( 4,3\right )$$
Step 3: Recall the formula for the midpoint.
The formula for the midpoint is $$d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^2}$$.
Step 4: Substitute the value of coordinates of midpoint and an end point of the line segment to find the half-length of the line.
$$d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$$
$$d=\sqrt{\left ( 6-4 \right )^{2}+\left ( 2-3 \right )^{2}}$$
$$d=\sqrt{2^{2}+\left ( -1 \right )^{2}}$$
$$d=\sqrt{5} So, the half-length of the line is$$d=\sqrt{5}$$. Step 5: Write the length of the line. Multiply 2 by the half-length,$$d=\sqrt{5}$$. The length of the line is$$\sqrt{5} \times 2$$. Hence, the length of the line is$$2\sqrt{5}.<|endoftext|>
| 4.75 |
455 |
We will learn addition of integers using number line.
We know that counting forward means addition.
positive integers, we move to the right on the number line.
+2 and +4 we move 4 steps to the right of +2.
Thus, +2 +4 = +6.
When
we add two negative integers, we move to the left on the number line.
For example to add -2 and -4 we move 4 steps to the left of -2.
Thus, -2 + (-4) = -6.
Addition of a Negative Integer and a Positive Integer:
When a negative integer is added to a positive integer we
move to the left on the number line.
For example to add +2 + (-4), we move 4
steps to the left of +2.
Thus, +2 + (-4) = -2.
Addition of a Positive Integer and a Negative Integer:
When a positive integer is added to a negative integer we
move to the right on the number line.
For example to add (-6) + 3, we move 3
steps to the right of -6.
Thus, (-6) + 3 = -3.
absolute values and placing the common sign of two integers before it.
Solved Examples on Addition of Integers:
1. Find the value +18 + (+5)
Solution:
Absolute value of |+18| = 18
Absolute value of |+5| = 5
Sum of absolute values = 18 + 5
Since, both integers have common sign +, we place + sign in
Hence, +18 + (+5) = +23.
We can add two integers having opposite signs by finding
difference of their absolute values and placing the sign of the integer with
the greater absolute value before it.
2. Find the value -38 + (+28)
Solution:
Absolute value of |-38| = 38
Absolute value of |+28| = 28
Difference of absolute values = 38 – 28 = 10
Placing the sign of the integer with the greater absolute
value = -10
Hence, -38 + (+28) = -10.<|endoftext|>
| 4.6875 |
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