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# 11-Plus Verbal Reasoning Quiz - Complete the Sum 4 (Questions)
This Complete the Sum quiz will test you on subtraction.
What you have been doing in this series of 11-plus verbal reasoning quizzes is basic algebra, where a symbol stands in for a number. We hope that when you begin algebra in secondary school you will be ahead of the game.
If you haven't already played our first three quizzes in this series, please do go back and play them first. Only when you have got 40 out of 40 will you be a whizz at maths!
Enjoy the final quiz in the Complete the Sum series. As always, take your time and look at the questions and answers carefully before choosing.
Example:
26 + 7 = 18 + ?
13
14
15
16
This is because 26 + 7 = 33. 18 + ? = 33. This makes 15 the only possible answer.
1. Find the number that will complete the sum below correctly.
27 - 13 = 42 ÷ ?
[ ] 5 [ ] 6 [ ] 4 [ ] 3
2. Find the number that will complete the sum below correctly.
63 ÷ 9 = 32 - ?
[ ] 24 [ ] 25 [ ] 26 [ ] 27
3. Find the number that will complete the sum below correctly.
4 x 16 = ? - 55
[ ] 78 [ ] 117 [ ] 119 [ ] 55
4. Find the number that will complete the sum below correctly.
79 + ? = 9 x 9
[ ] 2 [ ] 4 [ ] 7 [ ] 9
5. Find the number that will complete the sum below correctly.
56 ÷ 8 = ? - 8
[ ] 13 [ ] 19 [ ] 17 [ ] 15
6. Find the number that will complete the sum below correctly.
39 ÷ 3 = ? - 18
[ ] 13 [ ] 33 [ ] 31 [ ] 18
7. Find the number that will complete the sum below correctly.
? - 45 = 8 x 4
[ ] 77 [ ] 75 [ ] 73 [ ] 71
8. Find the number that will complete the sum below correctly.
72 - 14 = ? x 2
[ ] 28 [ ] 26 [ ] 27 [ ] 29
9. Find the number that will complete the sum below correctly.
? ÷ 13 = 90 ÷ 18
[ ] 67 [ ] 66 [ ] 65 [ ] 69
10. Find the number that will complete the sum below correctly.
27 x 3 = 9 x ?
[ ] 11 [ ] 7 [ ] 8 [ ] 9
11-Plus Verbal Reasoning Quiz - Complete the Sum 4 (Answers)
1. Find the number that will complete the sum below correctly.
27 - 13 = 42 ÷ ?
[ ] 5 [ ] 6 [ ] 4 [x] 3
This is because 27 - 13 = 14. 42 ÷ 3 is also equal to 14. The first three square numbers are 1, 4 and 9. 14 is the sum of these square numbers
2. Find the number that will complete the sum below correctly.
63 ÷ 9 = 32 - ?
[ ] 24 [x] 25 [ ] 26 [ ] 27
This is because 32 - 25 = 7 and 63 ÷ 9 = 7
3. Find the number that will complete the sum below correctly.
4 x 16 = ? - 55
[ ] 78 [ ] 117 [x] 119 [ ] 55
This is because 119 - 55 = 64, which is the same as 4 x 16. That was one to get your brain working a little harder! Did it work, or are you an expert at these questions now?
4. Find the number that will complete the sum below correctly.
79 + ? = 9 x 9
[x] 2 [ ] 4 [ ] 7 [ ] 9
This is because 9 x 9 = 81. To get to this product from 79 we have to add 2
5. Find the number that will complete the sum below correctly.
56 ÷ 8 = ? - 8
[ ] 13 [ ] 19 [ ] 17 [x] 15
This is because 56 ÷ 8 = 7. This is the same as 15 - 8. A cubed number is a number multiplied by itself and then by itself again (for example, 3 x 3 x 3 = 27, making 27 the cube of 3 and 3 the cube root of 27). Can you work out the cube root of 8? Read below for the answer
6. Find the number that will complete the sum below correctly.
39 ÷ 3 = ? - 18
[ ] 13 [ ] 33 [x] 31 [ ] 18
This is because 39 ÷ 3 = 13, which is equal to 31 - 18
7. Find the number that will complete the sum below correctly.
? - 45 = 8 x 4
[x] 77 [ ] 75 [ ] 73 [ ] 71
This is because 77 - 45 = 32 and 8 x 4 = 32. Remember the challenge we set you in question 5? Did you work out that the cube root of 8 is 2?
8. Find the number that will complete the sum below correctly.
72 - 14 = ? x 2
[ ] 28 [ ] 26 [ ] 27 [x] 29
This is because 29 x 2 = 58. Also, 72 - 14 = 58
9. Find the number that will complete the sum below correctly.
? ÷ 13 = 90 ÷ 18
[ ] 67 [ ] 66 [x] 65 [ ] 69
This is because 90 ÷ 18 = 5 and 65 ÷ 13 = 5. Only 65 works. On a dartboard scoring a double 18 and a treble 18 results in a score of 90. What would a treble and a double 19 score? Read below for the answer
10. Find the number that will complete the sum below correctly.
27 x 3 = 9 x ?
[ ] 11 [ ] 7 [ ] 8 [x] 9
This is because 27 x 3 = 81 and 9 x 9 also equals 81. The answer to the little quiz in question 9 is 95 (19 x 3 = 57 and 19 x 2 = 38. 57 + 38 = 95). We hope you enjoyed this series of quizzes, and look forward to you trying the next series of Verbal Reasoning quizzes!
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# 1.5 Exponential and logarithmic functions (Page 8/17)
Page 8 / 17
$f\left(x\right)={e}^{x}+2$
Domain: all real numbers, range: $\left(2,\infty \right),y=2$
$f\left(x\right)=\text{−}{2}^{x}$
$f\left(x\right)={3}^{x+1}$
Domain: all real numbers, range: $\left(0,\infty \right),y=0$
$f\left(x\right)={4}^{x}-1$
$f\left(x\right)=1-{2}^{\text{−}x}$
Domain: all real numbers, range: $\left(\text{−}\infty ,1\right),y=1$
$f\left(x\right)={5}^{x+1}+2$
$f\left(x\right)={e}^{\text{−}x}-1$
Domain: all real numbers, range: $\left(-1,\infty \right),y=-1$
For the following exercises, write the equation in equivalent exponential form.
${\text{log}}_{3}81=4$
${\text{log}}_{8}2=\frac{1}{3}$
${8}^{1\text{/}3}=2$
${\text{log}}_{5}1=0$
${\text{log}}_{5}25=2$
${5}^{2}=25$
$\text{log}\phantom{\rule{0.1em}{0ex}}0.1=-1$
$\text{ln}\left(\frac{1}{{e}^{3}}\right)=-3$
${e}^{-3}=\frac{1}{{e}^{3}}$
${\text{log}}_{9}3=0.5$
$\text{ln}\phantom{\rule{0.1em}{0ex}}1=0$
${e}^{0}=1$
For the following exercises, write the equation in equivalent logarithmic form.
${2}^{3}=8$
${4}^{-2}=\frac{1}{16}$
${\text{log}}_{4}\left(\frac{1}{16}\right)=-2$
${10}^{2}=100$
${9}^{0}=1$
${\text{log}}_{9}1=0$
${\left(\frac{1}{3}\right)}^{3}=\frac{1}{27}$
$\sqrt[3]{64}=4$
${\text{log}}_{64}4=\frac{1}{3}$
${e}^{x}=y$
${9}^{y}=150$
${\text{log}}_{9}150=y$
${b}^{3}=45$
${4}^{-3\text{/}2}=0.125$
${\text{log}}_{4}0.125=-\frac{3}{2}$
For the following exercises, sketch the graph of the logarithmic function. Determine the domain, range, and vertical asymptote.
$f\left(x\right)=3+\text{ln}\phantom{\rule{0.1em}{0ex}}x$
$f\left(x\right)=\text{ln}\left(x-1\right)$
Domain: $\left(1,\infty \right),$ range: $\left(\text{−}\infty ,\infty \right),x=1$
$f\left(x\right)=\text{ln}\left(\text{−}x\right)$
$f\left(x\right)=1-\text{ln}\phantom{\rule{0.1em}{0ex}}x$
Domain: $\left(0,\infty \right),$ range: $\left(\text{−}\infty ,\infty \right),x=0$
$f\left(x\right)=\text{log}\phantom{\rule{0.1em}{0ex}}x-1$
$f\left(x\right)=\text{ln}\left(x+1\right)$
Domain: $\left(-1,\infty \right),$ range: $\left(\text{−}\infty ,\infty \right),x=-1$
For the following exercises, use properties of logarithms to write the expressions as a sum, difference, and/or product of logarithms.
$\text{log}{x}^{4}y$
${\text{log}}_{3}\frac{9{a}^{3}}{b}$
$2+3{\text{log}}_{3}a-{\text{log}}_{3}b$
$\text{ln}\phantom{\rule{0.1em}{0ex}}a\sqrt[3]{b}$
${\text{log}}_{5}\sqrt{125x{y}^{3}}$
$\frac{3}{2}+\frac{1}{2}{\text{log}}_{5}x+\frac{3}{2}{\text{log}}_{5}y$
${\text{log}}_{4}\frac{\sqrt[3]{xy}}{64}$
$\text{ln}\left(\frac{6}{\sqrt{{e}^{3}}}\right)$
$-\frac{3}{2}+\text{ln}\phantom{\rule{0.1em}{0ex}}6$
For the following exercises, solve the exponential equation exactly.
${5}^{x}=125$
${e}^{3x}-15=0$
$\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}15}{3}$
${8}^{x}=4$
${4}^{x+1}-32=0$
$\frac{3}{2}$
${3}^{x\text{/}14}=\frac{1}{10}$
${10}^{x}=7.21$
$\text{log}\phantom{\rule{0.1em}{0ex}}7.21$
$4·{2}^{3x}-20=0$
${7}^{3x-2}=11$
$\frac{2}{3}+\frac{\text{log}\phantom{\rule{0.1em}{0ex}}11}{3\phantom{\rule{0.1em}{0ex}}\text{log}\phantom{\rule{0.1em}{0ex}}7}$
For the following exercises, solve the logarithmic equation exactly, if possible.
${\text{log}}_{3}x=0$
${\text{log}}_{5}x=-2$
$x=\frac{1}{25}$
${\text{log}}_{4}\left(x+5\right)=0$
$\text{log}\left(2x-7\right)=0$
$x=4$
$\text{ln}\sqrt{x+3}=2$
${\text{log}}_{6}\left(x+9\right)+{\text{log}}_{6}x=2$
$x=3$
${\text{log}}_{4}\left(x+2\right)-{\text{log}}_{4}\left(x-1\right)=0$
$\text{ln}\phantom{\rule{0.1em}{0ex}}x+\text{ln}\left(x-2\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}4$
$1+\sqrt{5}$
For the following exercises, use the change-of-base formula and either base 10 or base e to evaluate the given expressions. Answer in exact form and in approximate form, rounding to four decimal places.
${\text{log}}_{5}47$
${\text{log}}_{7}82$
$\left(\frac{\text{log}\phantom{\rule{0.1em}{0ex}}82}{\text{log}\phantom{\rule{0.1em}{0ex}}7}\approx 2.2646\right)$
${\text{log}}_{6}103$
${\text{log}}_{0.5}211$
$\left(\frac{\text{log}\phantom{\rule{0.1em}{0ex}}211}{\text{log}\phantom{\rule{0.1em}{0ex}}0.5}\approx -7.7211\right)$
${\text{log}}_{2}\pi$
${\text{log}}_{0.2}0.452$
$\left(\frac{\text{log}\phantom{\rule{0.1em}{0ex}}0.452}{\text{log}\phantom{\rule{0.1em}{0ex}}0.2}\approx 0.4934\right)$
Rewrite the following expressions in terms of exponentials and simplify.
a. $2\phantom{\rule{0.1em}{0ex}}\text{cosh}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)$ b. $\text{cosh}\phantom{\rule{0.1em}{0ex}}4x+\text{sinh}\phantom{\rule{0.1em}{0ex}}4x$ c. $\text{cosh}\phantom{\rule{0.1em}{0ex}}2x-\text{sinh}\phantom{\rule{0.1em}{0ex}}2x$ d. $\text{ln}\left(\text{cosh}\phantom{\rule{0.1em}{0ex}}x+\text{sinh}\phantom{\rule{0.1em}{0ex}}x\right)+\text{ln}\left(\text{cosh}\phantom{\rule{0.1em}{0ex}}x-\text{sinh}\phantom{\rule{0.1em}{0ex}}x\right)$
[T] The number of bacteria N in a culture after t days can be modeled by the function $N\left(t\right)=1300·{\left(2\right)}^{t\text{/}4}.$ Find the number of bacteria present after 15 days.
$~17,491$
[T] The demand D (in millions of barrels) for oil in an oil-rich country is given by the function $D\left(p\right)=150·{\left(2.7\right)}^{-0.25p},$ where p is the price (in dollars) of a barrel of oil. Find the amount of oil demanded (to the nearest million barrels) when the price is between $15 and$20.
[T] The amount A of a $100,000 investment paying continuously and compounded for t years is given by $A\left(t\right)=100,000·{e}^{0.055t}.$ Find the amount A accumulated in 5 years. Approximately$131,653 is accumulated in 5 years.
[T] An investment is compounded monthly, quarterly, or yearly and is given by the function $A=P{\left(1+\frac{j}{n}\right)}^{nt},$ where $A$ is the value of the investment at time $t,P$ is the initial principle that was invested, $j$ is the annual interest rate, and $n$ is the number of time the interest is compounded per year. Given a yearly interest rate of 3.5% and an initial principle of $100,000, find the amount $A$ accumulated in 5 years for interest that is compounded a. daily, b., monthly, c. quarterly, and d. yearly. #### Questions & Answers f(x) =3+8+4 tennesio Reply d(x)(x)/dx =? Abdul Reply scope of a curve Abraham Reply check continuty at x=1 when f (x)={x^3 if x <1 -4-x^2 if -1 <and= x <and= 10 Raja Reply what is the value as sinx Sudam Reply f (x)=x3_2x+3,a=3 Bilal Reply given demand function & cost function. x= 6000 - 30p c= 72000 + 60x . . find the break even price & quantities. Fiseha Reply hi guys ....um new here ...integrate my welcome Asif Reply An airline sells tickets from Tokyo to Detroit for$1200. There are 500 seats available and a typical flight books 350 seats. For every \$10 decrease in price, the airline observes and additional 5 seats sold. (a) What should the fare be to maximize profit? (b) How many passeners would be on board?
I would like to know if there exists a second category of integration by substitution
nth differential cofficient of x×x/(x-1)(x-2)
integral of root of sinx cosx
the number of gallons of water in a tank t minutes after the tank has started to drain is Q(t)=200(30-t)^2.how fast is the water running out at the end of 10 minutes?
why is it that the integral of eudu =eu
using L hospital rule
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# Advice 1: How to make a number of root
In most cases, it's easier to calculate on the calculator radical expression. But if you want to solve the problem in General or radical expression contains the unknown variables or the conditions of the problem it is necessary only to simplify, not to calculate, will have to find ways of making any number of under root.
Instruction
1
Use the definition of rootas a mathematical operation, from which it follows that the extraction of a root is the inverse operation of raising a number to a power. This means that the number can be taken from under the root if the decrease in radical expressions , the number of times that corresponds to a raised to the power passed. For example, to take from under the square root of the number 10, divide remaining under the root expression of ten squared.
2
Pick up the radical number of this multiplier, making of which is under root do simplify the expression - otherwise the operation will lose meaning. For example, if under the sign of the root with exponent equal to three (cubic root), is the number 128, then out of sign can be taken, for example, the number 5. In this radical , the number 128 will have to be divided by 5 cubed: 3√128 = 5∗3√(128/53) = 5∗3√(128/125) = 5∗3√1.024. If the presence of fractional numbers under the sign of root is not contrary to the conditions of the problem, the solution can be left in this form. If you need a more simple variant, we first divide radical expression for such an integer multiplier, cube root one of which will be a whole numberC. for Example: 3√128 = 3√(64∗2) = 3√(43∗2) = 4∗3√2.
3
Use for the selection of the multipliers radical number calculator, if you calculate in the mind the powers of a number is not possible. This is especially important for the rootm with the exponent greater than two. If you have Internet access, you can calculate the built-in Google search engine and Nigma solvers. For example, if you need to find the largest integer multiplier, which can be taken from under the sign of the cubic root for numbers 250, then going to Google enter "6^3" to check, can you just take out of sign of root six. The search engine displays the result equal to 216. Alas, the 250 cannot be divided without a remainder is the number. Then enter the query 5^3. The result will be 125, and it allows you to break 250 on the multipliers 125 and 2, and therefore to stand under the sign of the root number 5, leaving the number 2.
# Advice 2: How to make a multiplier from under the sign of the root
To take out from under the root of one of the factors necessary in situations when it is necessary to simplify a mathematical expression. There are times when to perform the desired calculations using the calculator it is impossible. For example, if instead of numbers uses letters denote variables.
Instruction
1
Lay out radical expression in simple factors. Let's see which of the factors is repeated as many times as indicated in the figures of the root, or more. For example, you need to extract the cubic root of a number and fourth degree. In this case, the number can be represented as a*a*a*a = a*(a*a*a)=a*A3. The index of the root in this case will correspond withthe multiplier A3. And it should be taken out of the radical sign.
2
Summarize the properties of the roots. The imposition of under the sign of the radical is an action that is the opposite of exponentiation. That is, in this case, it is necessary to extract the cube root of that part of the expression that yields this operation, which in this case is A3 3√a*a3 =a3√a.
3
Check out the calculations. This is especially important if you are dealing with numbers, not with lettered variables. For example, you need to convert the expression 3√120. Expanding radical expression into a fraction, you get 3√120=3√(60*2)=3√(30*2*2)=3√(15*2*2*2)=3√(3*5*2*2*2). Under root you can make witha multiplier of 2. Will get 23√15. Check the result. To do this, make the multiplier of the root, previously elevating him to the appropriate degree. 23 = 8. Accordingly, 23√15 = 3√(15*8) = 3√120.
4
For the decomposition into simple factors of numbers with lots of digits use a calculator. It is useful to do with roote, the rate of which is greater than two. When working with variables marked is not so important, since accurate calculations are not needed.
5
Use the search engines. This is necessary, for example, to find the highest integer multiplier, which can be taken from under the sign of the radical. Use Nigma. In the search engine type the number and what to do with it. For example, enter the expression "120 to factorize". You will receive a response 23 (3*5), that is the same thing that you have achieved through oral calculations in the given example. If you need a precise calculation, use the online calculator.
The imposition of a multiplier from under the root makes sense only if this action really simplifies the expression.
# Advice 3: How do I output the number of root
The numberthat is under the sign of the root, often hinders the solution of the equation, it is inconvenient to work with. Even if it is raised to a power, fractional or not may be provided in whole numbers to a certain extent, you can try to deduce it from the root, completely or at least partially.
Instruction
1
Try to decompose the number into Prime factors. If the number is decimal, not a comma yet, consider, consider all the numbers. For example, the number of 8.91 can be expanded as: of 8.91=0,9*0,9*11 (first place 891=9*9*11 then add the commas). Now you can record the number of 0.9^2*11 and to withdraw from the root of 0.9. So you got √of 8.91=0,9√11.
2
If you are given a cube root, you need to display underneath the number to the third power. For example, the number 135 decompose as 3*3*3*5=3^3*5. From under the root, print the number 3, the number 5 will remain under the root sign. Exactly the same goes with the roots of the fourth and higher degrees.
3
To bring out the root of the number with a degree other than the degree of the root (e.g., square root, and under it the number in grade 3), do so. Note the root degree, that is, put the sign √ and put in place a sign degree. For example, the square root of a number equals the same number of degree½, and a cubic – to the power of 1/3. Do not forget to enclose the radical expression in parenthesis.
4
Simplify the expression multiplying degree. For example, if the root was 12^4, and the root was square, the expression takes the form (12^4)^1/2=12^4/2=12^2=144.
5
To withdraw from the sign of the root is possible and a negative number. If the degree is odd, just imagine the number under the root as a number in the same way, for example, -8=(-2)^3, the cube root of (-8) is equal to (-2).
6
To make a negative number under an even root of a degree (including square), do so. Imagine radical expression as a product of (-1) and the number in the right degree, then take the number leaving (-1) under the root sign. For example, √(-144)=√(-1)*√144=12*√(-1). The number √(-1) in mathematics called imaginary numbers, and denoted by parameter i. Thus, √(-144)=12i.
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# Solve using square roots
Are you ready to learn how to Solve using square roots? Great! Let's get started! Our website can solve math word problems.
## Solving using square roots
There are also many YouTube videos that can show you how to Solve using square roots. Solving for the "intercept" is a common thing to do when you are trying to find the best fit line to an equation. The intercept will tell you where the y=0 value is. This is going to be the value that you would expect if you were trying to solve for the y-axis of an equation by taking the x-axis and adding it to itself (y = y + x). On a graph, you might expect this value to be where the x-axis intersects with the y-axis. You can also think of it as being at the origin. If we are solving for y in our equation, then the intercept would be 0 on both axes. It might also be important as it will give us a good idea for how long our graph should be in order for our data points to fall within that range. If we have a very short range (like on a log scale), we will need to make sure that our x-axis intercept is much higher than our y-axis intercept so that our data points fall well above or below that line.
To solve this equation, we start by first converting the left-hand side to a ratio: Similarly, since the right-hand side is a fraction, we can convert this to a decimal: We then multiply both sides of the equation by 1/10 , and then divide by 10 : Finally, we convert back to the original form of the equation, and solve for x . There are no exact formulas for how to solve logarithmic equations. However, there are some useful tricks and techniques that can be used to help you solve these types of equations. One good way to solve logarithmic equations is to use a table. One easy way to do this is to look at what other logarithmic equations look like. Since logarithms follow an exponential pattern, it is usually possible to find a similar equation on which the base can be found. Another trick is to try doing all comparisons in your head before you write them down. If you have trouble coming up with a number that works for both sides of the equation then try using numbers from previous
One of the best ways to improve at math is by learning how to solve problems. Knowing how to set up equations, work with fractions and percentages, and use arithmetic are essential skills that underlie all math. Solving problems is also a great way to challenge yourself and practice your problem-solving skills. Solving problems can be challenging at times, but it's never impossible. With practice and patience, you'll get better at solving problems every time you sit down at the table.
The side ratios of an equilateral triangle are equal: 1:1:1. The three angles at each vertex are all equal, as well: 90° for each. If "a" is the length of the lateral side, then "b" is the length of the hypotenuse and "c" is the length of one leg (the shorter one). Then, we can write that: A = b = c = 1 Therefore, the side of a triangle can be found by dividing any two sides together and adding 1 to the result. So if you want to find the hypotenuse, you would add 1 to both "a" and "c". If you want to find one of a pair's legs, you would add 1 to both "b" and "c". Another way to solve for a side of a triangle is to use Pythagoras' theorem. This says that in order for two triangles to share a common side, they must have identical altitudes (measured from their highest point). This means that if you want to find the hypotenuse or one leg, you can simply measure them from their top or bottom respectively
The graph of an equation is simply a way to visualize the relationship between two sets of numbers. The graph of an equation can be used to help you solve the equation by putting an X on each side where the numbers match up. Once you have determined which side of the graph corresponds to the smaller number, you can use that information to solve for the larger number. For example, if you were given the following equation: 2 x + 3 = 6 You could look at the graph below and see that 2 is less than 6, so it must be on one side of the graph. This means that 3 must be on the other side. If you put an X on both sides, you would know that 2x is equal to 3.
## We will support you with math difficulties
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## Wednesday, 12 December 2012
### Maths Form 2 : Chapter 1 : Directed Numbers
This note from SCORE A NOTES.
MULTIPLICATION AND DIVISION OF INTEGERS
(A) Multiplying integers
1. The product of multiplication between two integers with the same sign is positive. If two integers have different signs, the product of the multiplication is negative.
2. If zero is multiplied by any integer, the product is always zero.
For example,
50 x 0 = 0
(B) Solving problems involving multiplication of integers
(C) Dividing integer
1. The quotient of division between two integers with the same sign is positive. If two integers have different signs, then the quotient of the division is negative.
2. If zero divided by any integer (except zero), the quotient is zero.
For example :
0 ÷ 10 = 0
3. Any integer divided by zero is undefined.
For example:
6 ÷ 0 = undefined
(D) Solving problems involving division of integers
COMBINED OPERATIONS ON INTEGERS(A) Performing computations involving combined operations on integers
1. Combined operations involve more than one operation on integers are also known as mixed operations.
2. For combined operations involving' + ' and ' - ' or ' x ' and ' ÷ ', work the calculations from left to right following the order given.
(B) Solving problems involving combined operations on integers
POSITIVE AND NEGATIVE FRACTIONS
(A) Comparing and ordering fractions
(B) Computations Involving Addition, Subtraction, Multiplication and
Division of Two Fractions
The same rule used for adding and subtracting integers is used in the addition and subtraction of positive and negative fractions.
Multiplication of positive and negative fractions is similar to the multiplication of integers.
POSITIVE AND NEGATIVE DECIMALS(A) Comparing and ordering decimals
The same rule used in adding and subtracting integers is used in the addition and subtraction of positive decimal.
(C) Multiplying and Dividing Decimals
The same rule used in multiplying and dividing integers is used in the multiplication and division of positive and negative decimals.
COMPUTATIONS INVOLVING DIRECTED NUMBERS
(INTEGERS, FRACTIONS AND DECIMALS)
(A) Addition, Subtraction, Multiplication and Division Involving Two
Directed Numbers
Calculations involving two directed numbers with any operations are the same as that for integers, fractions and decimals.
(B) Combined Operations of Addition, Subtration, Multiplication and
Division of Positive and Negative Integers.
For computations involving the use of brackets and combined operations of addition, subtraction, multiplication and division on positive and negative numbers, calculate according to the following order.
1. The brackets ( )
2. 'of' (which means x)
3. ÷ and - (from left to right)
4. + and - (from left to right)
(C) Combined Operations of Addition, Subtraction, Multiplication and
Division of Positive and Negative Fractions
(d) Combined Operations on Positive and Negative Decimals
(e) Combined Operations on Directed Numbers
(f) Solving Problems Involving Directed Numbers
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1166. Simple Statistics
Given a set of numerical data, there are several ways in which to describe it. One way is the so-called 5-number summary. The five numbers used to describe the data set are the following: the minimum value, the first quartile, the median, the third quartile and the maximum value. The definition of the minimum and maximum values are obvious. The median of a set of numbers is the value of the number which would lie exactly in the middle of the set if it were sorted. For example, the median of the data set 7,-1, 9, 4, 1 would be 4. If there is an even number of values in the set, then the median is the average of the two values closest to the middle; if our set contained the values 7,-1, 9, 4, 1, 0 then the median would be (1 + 4)/2 = 2.5. The definition of the quartiles follows naturally from the definition of the median. If we take all the values that come before the median in the sorted list (in the case when we average two values for the median this set would include the lower of those two numbers) the first quartile is the median of this set.
The definition of the third quartile is identical except it uses those values that come after the original median. In our example above with 7,-1, 9, 4, 1, 0, the first quartile value would be 0 (the median of the value -1, 1 and 0 which are less than 2.5) and the third quartile would be 7. If the data set contained 1, 2, 2, 2, 3 (in any order), then the median would be 2, and the first and third quartiles would be 1.5 and 2.5, respectively. One special case is when there is only one element in the list, in which case the quartiles are equal to the median. One other way to characterize data is its skewness. A distribution is considered right-skewed whenever the maximum value is farther from the median than the minimum value, or when the maximum and minimum are equally distant from the median, but the third quartile is farther from the median than the first quartile. A left-skewed distribution is one with the opposite situation. For our purposes, a distribution which is neither left-skewed nor right-skewed is considered symmetric. Your task for this problem is to read in various sets of numbers and output the 5-number summary for each, along with the skewness of the data.
输入格式
There will be multiple input sets. Each input set will consist of a single line of the form n v1 v2 v3 . . . vn where n is the number of data values, and v1, . . . , vn are the values. All the values will be integers and the maximum value for n will be 100. A line which begins with 0 indicates end of input and should not be processed.
输出格式
For each input set, output the 5-number summary and skewness in the order minimum, first quartile,median, third quartile, maximum and skew, with a single space between each. Skew will either be the phrase right-skewed, left-skewed or symmetric.
样例
Input
6 7 -1 9 4 1 0
15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 0
0
Output
-1 0 2.5 7 9 right-skewed
1 4 8 12 15 symmetric
0 4 8 12 15 left-skewed
7 人解决，14 人已尝试。
7 份提交通过，共有 25 份提交。
7.3 EMB 奖励。
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# Rearrange further equations
Lesson
Previously, you've looked at how to change the subject of an equation for linear equations, and also more challenging equations involving powers and quadratics. We are going to take this further, looking at a wide variety of types of equations, including radicals, logs, and other skills.
Remember!
The subject of an equation is the variable that is by itself on one side of the equals sign and usually it is at the start of the formula.
For example, in the formula $A=pb+y$A=pb+y, $A$A is the subject because it is by itself on the left hand side of the equal sign.
When we're changing the subject of a formula, we often have more than one variable, but we use a process similar to when we solve equations.
• Group any like terms
• Simplify using the inverse of addition or subtraction.
• Simplify further by using the inverse of multiplication or division.
• Other inverse operations might be needed like square and square root
#### Examples
##### Question 1
Make $r$r the subject of $v=\sqrt{\frac{mr}{\pi}}$v=mrπ
Think: In this question, $r$r has been multiplied by $m$m, divided by $\pi$π, then square rooted.
Reversing this, what would be the first operation to do to both sides?
The first operation will be to square both sides, in order to reverse the square root.
Then we multiply by $\pi$π, and divide by $m$m, finally isolating $r$r.
Do:
$v$v $=$= $\sqrt{\frac{mr}{\pi}}$√mrπ square both sides $v^2$v2 $=$= $\left(\sqrt{\frac{mr}{\pi}}\right)^2$(√mrπ)2 cancel the square and square root $v^2$v2 $=$= $\frac{mr}{\pi}$mrπ multiply both sides by $\pi$π $v^2\times\pi$v2×π $=$= $\frac{mr}{\pi}\times\pi$mrπ×π cancel the $\pi$π's $\pi v^2$πv2 $=$= $mr$mr divide both sides by $m$m $\pi v^2\div m$πv2÷m $=$= $mr\div m$mr÷m cancel the $m$m's $\frac{\pi v^2}{m}$πv2m $=$= $r$r rewrite the equation with $r$r on the left hand side $r$r $=$= $\frac{\pi v^2}{m}$πv2m
##### Question 2
The volume of a sphere is given by the formula $V=\frac{4}{3}\pi r^3$V=43πr3. Make the radius, $r$r, the subject of the equation.
a) Make the radius, $r$r, the subject of the equation.
Think: The radius $r$r is cubed, and multiplied by $4\pi$4π, then divided by $3$3.
What is the first step in reversing these operations?
The first step is to multiply both sides by $3$3.
Then we divide by $4\pi$4π, and then take the cube root of both sides.
Do:
$V$V $=$= $\frac{4}{3}\pi r^3$43πr3 $V\times3$V×3 $=$= $\frac{4}{3}\pi r^3\times3$43πr3×3 $3V$3V $=$= $4\pi r^3$4πr3 $3V\div4\pi$3V÷4π $=$= $4\pi r^3\div4\pi$4πr3÷4π $r^3$r3 $=$= $\frac{3V}{4\pi}$3V4π $\sqrt[3]{r^3}$3√r3 $=$= $\sqrt[3]{\frac{3V}{4\pi}}$3√3V4π $r$r $=$= $\sqrt[3]{\frac{3V}{4\pi}}$3√3V4π
b) If the volume of the sphere is $20\pi$20π cm3, what is the exact radius of the sphere? Leave your answer in exact radical form.
Think: One of the main reasons we want to be able to change the subject of a formula is so that we can do these types of substitutions.
Do:
$r$r $=$= $\sqrt[3]{\frac{3V}{4\pi}}$3√3V4π $=$= $\sqrt[3]{\frac{3\times20\pi}{4\pi}}$3√3×20π4π $=$= $\sqrt[3]{\frac{60\pi}{4\pi}}$3√60π4π $=$= $\sqrt[3]{15}$3√15
In this case, the question has requested we leave the questions in exact radical form, so we have our answer, and do not need to use our calculator to find a value.
##### Question 3
The equation for the population at time $t$t is given by $P=Me^{4t}$P=Me4t. Make $t$t the subject of the equation.
Think: We want to make $e^{4t}$e4t the subject of the equation, and then use the definition of a logarithm to solve for $t$t.
Our first step is to divide both sides of the equation by $M$M
Then, we can make the power $4t$4t the subject . We can use the fact that an equation, such as $a^x=y$ax=y, which is in exponential form, can be written in log form, making the power $x$x the subject : $x=\log_ay$x=logay
So in this case, we can make $4t$4t the subject by taking the log of $\frac{P}{M}$PM.
We finish by dividing both sides by $4$4.
Do:
$P$P $=$= $Me^{4t}$Me4t $e^{4t}$e4t $=$= $\frac{P}{M}$PM $4t$4t $=$= $\ln\left(\frac{P}{M}\right)$ln(PM) $t$t $=$= $\frac{\ln\left(\frac{P}{M}\right)}{4}$ln(PM)4
#### Worked Examples
##### QUESTION 1
Solve $\frac{V_1}{V_2}=\frac{P_2}{P_1}$V1V2=P2P1 for $P_1$P1.
##### QUESTION 2
Rearrange the expression $p=\frac{q\sqrt{k}}{r}$p=qkr to make $k$k the subject.
##### QUESTION 3
If $p=\sqrt{\frac{5m^2+4}{2m^2-3}q}$p=5m2+42m23q, express $m$m in terms of $p$p and $q$q.
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# Intermediate Algebra Chapter 9 Exponential and Logarithmic Functions.
## Presentation on theme: "Intermediate Algebra Chapter 9 Exponential and Logarithmic Functions."— Presentation transcript:
Intermediate Algebra Chapter 9 Exponential and Logarithmic Functions
Intermediate Algebra 9.1-9.2 Review of Functions
Def: Relation A relation is a set of ordered pairs. Designated by: Listing Graphs Tables Algebraic equation Picture Sentence
Def: Function A function is a set of ordered pairs in which no two different ordered pairs have the same first component. Vertical line test – used to determine whether a graph represents a function.
Defs: domain and range Domain: The set of first components of a relation. Range: The set of second components of a relation
Examples of Relations:
Objectives Determine the domain, range of relations. Determine if relation is a function.
Intermediate Algebra 9.2 Inverse Functions
Inverse of a function The inverse of a function is determined by interchanging the domain and the range of the original function. The inverse of a function is not necessarily a function. Designated by and read f inverse
One-to-One function Def: A function is a one-to-one function if no two different ordered pairs have the same second coordinate.
Horizontal Line Test A function is a one-to-one function if and only if no horizontal line intersects the graph of the function at more than one point.
Inverse of a function
Inverse of function
Objectives: Determine the inverse of a function whose ordered pairs are listed. Determine if a function is one to one.
Intermediate Algebra 9.3 Exponential Functions
Michael Crichton – The Andromeda Strain (1971 ) The mathematics of uncontrolled growth are frightening. A single cell of the bacterium E. coli would, under ideal circumstances, divide every twenty minutes. It this way it can be shown that in a single day, one cell of E. coli could produce a super-colony equal in size and weight to the entire planet Earth.”
Definition of Exponential Function If b>0 and b not equal to 1 and x is any real number, an exponential function is written as
Graphs-Determine domain, range, function, 1-1, x intercepts, y intercepts, asymptotes
Growth and Decay Growth: if b > 1 Decay: if 0 < b < 1
Properties of graphs of exponential functions Function and 1 to 1 y intercept is (0,1) and no x intercept(s) Domain is all real numbers Range is {y|y>0} Graph approaches but does not touch x axis – x axis is asymptote Growth or decay determined by base
Natural Base e
Calculator Keys Second function of divide Second function of LN (left side)
Property of equivalent exponents For b>0 and b not equal to 1
Compound Interest A= amount P = Principal t = time r = rate per year n = number of times compounded
Compound interest problem Find the accumulated amount in an account if \$5,000 is deposited at 6% compounded quarterly for 10 years.
Objectives: Determine and graph exponential functions. Use the natural base e Use the compound interest formula.
Dwight Eisenhower – American President “Pessimism never won any battle.”
Intermediate Algebra 9.4,9.5,9.6 Logarithmic Functions
Definition: Logarithmic Function For x > 0, b > 0 and b not equal to 1 toe logarithm of x with base b is defined by the following:
Properties of Logarithmic Function Domain:{x|x>0} Range: all real numbers x intercept: (1,0) No y intercept Approaches y axis as vertical asymptote Base determines shape.
Shape of logarithmic graphs For b > 1, the graph rises from left to right. For 0 < b < 1, the graphs falls from left to right.
Common Logarithmic Function The logarithmic function with base 10
Natural logarithmic function The logarithmic function with a base of e
Calculator Keys [LOG] [LN]
Objective: Determine the common log or natural log of any number in the domain of the logarithmic function.
Change of Base Formula For x > 0 for any positive bases a and b
Problem: change of base
Objective Use the change of base formula to determine an approximation to the logarithm of a number when the base is not 10 or e.
Intermediate Algebra 10.5 Properties of Logarithms
Basic Properties of logarithms
For x>0, y>0, b>0 and b not 1 Product rule of Logarithms
For x>0, y>0, b>0 and b not 1 Quotient rule for Logarithms
For x>0, y>0, b>0 and b not 1 Power rule for Logarithms
Objectives: Apply the product, quotient, and power properties of logarithms. Combine and Expand logarithmic expressions
Theorems summary Logarithms:
Norman Vincent Peale “Believe it is possible to solve your problem. Tremendous things happen to the believer. So believe the answer will come. It will.”
Intermediate Algebra 9.7 Exponential and Logarithmic Equations
Objective: Solve equations that have variables as exponents.
Exponential equation
Objective: Solve equations containing logarithms.
Sample Problem Logarithmic equation
Walt Disney “Disneyland will never be completed. It will continue to grow as long as there is imagination left in the world.”
Galileo Galilei (1564-1642) “The universe…is written in the language of mathematics…”
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EMAT 6680
Assignment 4
MEDIAN
A median of a triangle is a segment from a vertex to the midpoint of the opposite side. The 3 medians of triangle ABC are shown above. The median divides the triangles into 2 equal parts since the 2 triangles share the same altitude and base.
When we construct 2 medians of a triangle together, they interesect at a point G. The third median passes through the same point G, called the centroid, and traingle ABC is now equally divided into 6 parts.
Proof that triangle ABC is now equally divided into 6 parts:
The following pairs of triangles share the same base and altitudes hence they have the same area:
Triangle 1 and 2
Traingle 3 & 4
Triangle 5 and 6
Triangle ABD and ACD
Triangle CED and CEA
Triangle BAF and BCF
Sum of regions (1 + 2 + 5 ) = Sum of regions (3 + 4 + 6) and area of region 5 = area of region 6,
implies
Sum of area of regions (1 + 2 ) = Sum of area of regions (3 + 4)
That is,
2 Area of region 1 = 2 Area of region 3 , hence, area of region1 = area of region 3
2 Area of region 2 = 2 Area of region 4, hence, area of region 2 = area of region 4
2 Area of region 1 = 2 Area of region 4, hence, area of region 1 = area of region 4
2 Area of region 2 = 2 Area of region 3, hence, area of region 2 = area of region 3
Sum of regions (2 + 5 + 6 ) = Sum of regions (1 + 3 + 4)
Since, area of region2 = area of region 1 and area of region3 = area of region 4,
Area of region 5 = Area of region 3
Area of region 5 = Area of region 4
Area of region 6= Area of region 3
Area of region 6= Area of region 4
Sum of regions (4 + 5 + 6 ) = Sum of regions (1 + 2 + 3)
using the same reasoning as above, we have area of region 5 = area of region 2
Hence
Area of region 1 = 3 = 4 = 6 = 5 = 2 = 1.
ORTHOCENTER
When we change the dimensions of the traingle, the centroid always remain in the triangle. Next, we construct the orthocenter (H) of a triangle. The orthocenter of a triangle is the common intersection of the 3 lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side and H lies on the lines extended along the altitudes. When we shift or change the position of A, B or C respectively, G remains in the triangle but H does not.
I investigated the movment of H. I constructed a line DE parallel to BC and trace the movement of H when A moves along DE. H seems to form a parabola with directrix DE. Next, I attempted to locate the exact position of H.
H is the intersection of line k and l. The equation for line AC is easy to find using the points A(b,c) and C(a,0).
Equation of line l
Equation of line k is given by x =b
Hence, H takes up the following values:
x = b
It simply means, for any given triangle, the x-coordinate of H follows the x-coordinate of one of the vertex of the triangle. The general expression for the y-coordinate of H suggests that y takes any inside and outside the triangle. The equation of the parabola is
Ckeck:
When x=a,
To locate the vertex:
The vertex is at
and
CIRCUMCENTER (C)
The circumcenter C is a point where the perpendicular bisectors of each side of the triangle meet. from our construction, we note that C may be outside the triangle.
We want to show that the 3 bisectors of the sides are concurrent.
Equation of segment AB
Midpoint of AB is
The perpendicular bisector of AB has a gradient of and it passes through the midpoint of AB. Hence,
where d is the y-intercept. Then,
We do the same to find the equation for the perpendicular bisector of AD. We get
Similarily, the perpendicular bisector of BD is
The 3 lines intercept at
The 3 bisectors are thus concurrent.
Incenter
The three angle bisectors of the internal angles of a triangle are concurrent.
Nine-Point Circle
We want to prove that H, G and C are collinear and HG = 2GC.
Coordinates of C:
Coordinates of H:
x = b
Coordinates of G:
Consider 2 points C and H:
y-intercept of segment CH is given by
Equation of CH is
It can be shown easily that Point G lies on the above line. Hence, HGC are collinear.
Medial Traingle
If we take any triangle, construct a triangle connecting the 3 midpoints of the sides, we obtain the medial triangle. It is similiar to the original triangle and one quarter its area.
Proof
Area of Triangle ABC is
Area of Triangle DEF is
This can be proved using paper folding exercise.
Interior Triangle
(a) similar to triangle ABC
(b) congruent to the medial triangle
Return
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# Need help to understand how the slope was derived in this problem
Printable View
• Jan 15th 2011, 03:38 PM
alex95
Need help to understand how the slope was derived in this problem
I am kind of struggling to understand how the slope is derived in the following problem:
"Find the equation of the tangent line to the curve y = x^2 through the point (7, 49)."
The answer is that the slope is 2x but I don't understand how they got there since there is only one set of points given.
Thank you for the help,
Alex
• Jan 15th 2011, 03:41 PM
dwsmith
$y'=\mbox{exponent}*x^{\mbox{exponent -1}}$
$y'=2x^{2-1}$
• Jan 15th 2011, 03:50 PM
Prove It
The slope at any point $\displaystyle x$ on the curve $\displaystyle y = x^2$ is given by $\displaystyle \frac{dy}{dx} = 2x$.
So at the point $\displaystyle x = 7$, the slope is $\displaystyle 2\cdot 7 = 14$.
Equation of tangent: $\displaystyle y = mx + c$.
You have a coordinate that lies on your tangent $\displaystyle (x, y) = (7, 49)$ and the slope $\displaystyle m = 14$. Substitute and solve for $\displaystyle c$, then you can write the equation of your tangent line.
• Jan 15th 2011, 03:55 PM
Quacky
How well do you understand the concept behind differentiation?
If $y=x^2,$
$\displaystyle\frac{dy}{dx} = 2x$
$
\displaystyle\frac{dy}{dx}$
is 'the rate of change of y with respect to x'. It is used to tell us the equation of the gradient of the curve at any point.
In the question, $\frac{dy}{dx} = 2x$. So take any point on your curve. Let's say $(3,9)$ This has x co-ordinate 3.
Substituting this into $\frac{dy}{dx} = 2x$ gives $\frac{dy}{dx} = 6$. What this means is that at the point $(3,9)$ the gradient of the curve $y=x^2$ is 6.
Also, I don't think that $2x$ is the answer to that question.
We have a point (7,49) and we know that $\frac{dy}{dx} = 2x$. So the gradient of the tangent is $2\times 7 = 14$.
So we know that when x = 7, y = 14 and m is 14. Substitute this into y = mx + c to find c and you'll find the equation of the tangent.
• Jan 15th 2011, 05:30 PM
skeeter
from first principles ...
$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
$f(x) = x^2$
$f(x+h) = (x+h)^2$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{x^2+2xh+h^2 - x^2}{h}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{2xh+h^2}{h}
$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{h(2x+h)}{h}
$
$\displaystyle f'(x) = \lim_{h \to 0} 2x+h = 2x$
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Posted on by Kalkicode
Code Number
Sum of squares of Fibonacci Numbers
The given problem is to find the sum of the squares of the first 'n' Fibonacci numbers. Fibonacci numbers are a series of numbers in which each number is the sum of the two preceding ones, usually starting with 0 and 1. The Fibonacci sequence begins with 0 and 1, and each subsequent number is the sum of the two previous numbers.
Problem Statement
We are given an integer 'n' representing the number of Fibonacci numbers to consider. The task is to calculate the sum of the squares of the first 'n' Fibonacci numbers.
Explanation with an Example
Let's take a simple example to understand the problem. Consider 'n' is 5.
Fibonacci sequence: [0, 1, 1, 2, 3, ...]
The first five Fibonacci numbers are: 0, 1, 1, 2, and 3.
Now, we need to find the sum of the squares of these five Fibonacci numbers.
Sum of squares: (0)² + (1)² + (1)² + (2)² + (3)² = 0 + 1 + 1 + 4 + 9 = 15
So, the sum of the squares of the first 5 Fibonacci numbers is 15.
Standard Pseudocode
Before diving into the algorithm, let's write the standard pseudocode for the problem:
``````function sumFibonacciSquares(n)
sum = 0
if n > 1
create auxiliary array with size n
auxiliary[0] = 0
auxiliary[1] = 1
for i from 0 to n-1
if i > 1
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1]
sum += (auxiliary[i] * auxiliary[i])
print "Sum of", n, "fibonacci square number is", sum
``````
Algorithm Explanation
1. Initialize a variable `sum` to keep track of the sum of squares of Fibonacci numbers. Set it to 0 initially.
2. Check if 'n' is greater than 1. If it is not, then there's no need to calculate anything, so we can return.
3. Create an auxiliary array of integers with a size of 'n' to store the Fibonacci sequence.
4. Set the first two elements of the auxiliary array to 0 and 1, respectively, as these are the first two Fibonacci numbers.
5. Use a loop to calculate the next Fibonacci numbers up to the 'n'th number and store them in the auxiliary array.
6. During each iteration, calculate the square of the current Fibonacci number and add it to the `sum`.
7. Once the loop is completed, the `sum` will contain the sum of the squares of the first 'n' Fibonacci numbers.
8. Print the value of `sum` as the final output.
Code Solution
Here given code implementation process.
``````/*
C program for
Sum of squares of Fibonacci Numbers
*/
#include <stdio.h>
void sumFibonacciSquares(int n)
{
long long int sum = 0;
if (n > 1)
{
// Create auxiliary space
int auxiliary[n];
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (int i = 0; i < n; ++i)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] *auxiliary[i]);
}
}
printf("\n Sum of %d fibonacci square number is %lld", n, sum);
}
int main(int argc, char
const *argv[])
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
sumFibonacciSquares(5);
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
sumFibonacciSquares(10);
return 0;
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````/*
Java program for
Sum of squares of Fibonacci Numbers
*/
public class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
public void sumFibonacciSquares(int n)
{
long sum = 0;
if (n > 1)
{
// Create auxiliary space
int[] auxiliary = new int[n];
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (int i = 0; i < n; ++i)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i]);
}
}
System.out.print("\n Sum of " +
n + " fibonacci square number is " + sum);
}
public static void main(String[] args)
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````// Include header file
#include <iostream>
using namespace std;
/*
C++ program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries
{
public:
// Here n indicates number of initial Fibonacci series elements
void sumFibonacciSquares(int n)
{
long sum = 0;
if (n > 1)
{
// Create auxiliary space
int auxiliary[n];
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (int i = 0; i < n; ++i)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] *auxiliary[i]);
}
}
cout << "\n Sum of " << n
<< " fibonacci square number is " << sum;
}
};
int main()
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
return 0;
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````// Include namespace system
using System;
/*
Csharp program for
Sum of squares of Fibonacci Numbers
*/
public class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
public void sumFibonacciSquares(int n)
{
long sum = 0;
if (n > 1)
{
// Create auxiliary space
int[] auxiliary = new int[n];
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (int i = 0; i < n; ++i)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i]);
}
}
Console.Write("\n Sum of " + n +
" fibonacci square number is " + sum);
}
public static void Main(String[] args)
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````package main
import "fmt"
/*
Go program for
Sum of squares of Fibonacci Numbers
*/
// Here n indicates number of initial Fibonacci series elements
func sumFibonacciSquares(n int) {
var sum int64 = 0
if n > 1 {
// Create auxiliary space
var auxiliary = make([] int, n)
// Set initial sequence
auxiliary[0] = 0
auxiliary[1] = 1
// Calculate the sum of n fibonacci numbers
for i := 0 ; i < n ; i++ {
if i > 1 {
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1]
}
// sum squares of fibonacci number
sum += int64(auxiliary[i] * auxiliary[i])
}
}
fmt.Print("\n Sum of ",
n, " fibonacci square number is ", sum)
}
func main() {
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
sumFibonacciSquares(5)
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
sumFibonacciSquares(10)
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````<?php
/*
Php program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
public function sumFibonacciSquares(\$n)
{
\$sum = 0;
if (\$n > 1)
{
// Create auxiliary space
\$auxiliary = array_fill(0, \$n, 0);
// Set initial sequence
\$auxiliary[0] = 0;
\$auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (\$i = 0; \$i < \$n; ++\$i)
{
if (\$i > 1)
{
\$auxiliary[\$i] = \$auxiliary[\$i - 2] + \$auxiliary[\$i - 1];
}
// sum squares of fibonacci number
\$sum += (\$auxiliary[\$i] * \$auxiliary[\$i]);
}
}
echo("\n Sum of ".\$n.
" fibonacci square number is ".\$sum);
}
}
function main()
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
main();``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````/*
Node JS program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
sumFibonacciSquares(n)
{
var sum = 0;
if (n > 1)
{
// Create auxiliary space
var auxiliary = Array(n).fill(0);
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
// Calculate the sum of n fibonacci numbers
for (var i = 0; i < n; ++i)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i]);
}
}
process.stdout.write("\n Sum of " +
n + " fibonacci square number is " + sum);
}
}
function main()
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
main();``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````# Python 3 program for
# Sum of squares of Fibonacci Numbers
class FibonacciSeries :
# Here n indicates number of initial Fibonacci series elements
def sumFibonacciSquares(self, n) :
sum = 0
if (n > 1) :
# Create auxiliary space
auxiliary = [0] * (n)
# Set initial sequence
auxiliary[0] = 0
auxiliary[1] = 1
i = 0
# Calculate the sum of n fibonacci numbers
while (i < n) :
if (i > 1) :
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1]
# sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i])
i += 1
print("\n Sum of", n ,"fibonacci square number is ",
sum, end = "")
def main() :
# Fibonacci sequence
# [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
# Test A
# n = 5
# (0)² + (1)² + (1)² + (2)² + (3)² = 15
# Test B
# n = 10
# (0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
# (8)² + (13)² + (21)² + (34)² = 1870
if __name__ == "__main__": main()``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````# Ruby program for
# Sum of squares of Fibonacci Numbers
class FibonacciSeries
# Here n indicates number of initial Fibonacci series elements
def sumFibonacciSquares(n)
sum = 0
if (n > 1)
# Create auxiliary space
auxiliary = Array.new(n) {0}
# Set initial sequence
auxiliary[0] = 0
auxiliary[1] = 1
i = 0
# Calculate the sum of n fibonacci numbers
while (i < n)
if (i > 1)
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1]
end
# sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i])
i += 1
end
end
print("\n Sum of ", n ," fibonacci square number is ", sum)
end
end
def main()
# Fibonacci sequence
# [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
# Test A
# n = 5
# (0)² + (1)² + (1)² + (2)² + (3)² = 15
# Test B
# n = 10
# (0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
# (8)² + (13)² + (21)² + (34)² = 1870
end
main()``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````/*
Scala program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries()
{
// Here n indicates number of initial Fibonacci series elements
def sumFibonacciSquares(n: Int): Unit = {
var sum: Long = 0;
if (n > 1)
{
// Create auxiliary space
var auxiliary: Array[Int] = Array.fill[Int](n)(0);
// Set initial sequence
auxiliary(0) = 0;
auxiliary(1) = 1;
var i: Int = 0;
// Calculate the sum of n fibonacci numbers
while (i < n)
{
if (i > 1)
{
auxiliary(i) = auxiliary(i - 2) + auxiliary(i - 1);
}
// sum squares of fibonacci number
sum += (auxiliary(i) * auxiliary(i));
i += 1;
}
}
print("\n Sum of " + n + " fibonacci square number is " + sum);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: FibonacciSeries = new FibonacciSeries();
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````/*
Swift 4 program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
func sumFibonacciSquares(_ n: Int)
{
var sum: Int = 0;
if (n > 1)
{
// Create auxiliary space
var auxiliary: [Int] = Array(repeating: 0, count: n);
// Set initial sequence
auxiliary[0] = 0;
auxiliary[1] = 1;
var i: Int = 0;
// Calculate the sum of n fibonacci numbers
while (i < n)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i]);
i += 1;
}
}
print("\n Sum of ", n ,
" fibonacci square number is ", sum, terminator: "");
}
}
func main()
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}
main();``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
``````/*
Kotlin program for
Sum of squares of Fibonacci Numbers
*/
class FibonacciSeries
{
// Here n indicates number of initial Fibonacci series elements
fun sumFibonacciSquares(n: Int): Unit
{
var sum: Long = 0;
if (n > 1)
{
// Create auxiliary space
var auxiliary: Array < Int > = Array(n)
{
0
};
// Set initial sequence
auxiliary[1] = 1;
var i: Int = 0;
// Calculate the sum of n fibonacci numbers
while (i < n)
{
if (i > 1)
{
auxiliary[i] = auxiliary[i - 2] + auxiliary[i - 1];
}
// sum squares of fibonacci number
sum += (auxiliary[i] * auxiliary[i]);
i += 1;
}
}
print("\n Sum of " + n + " fibonacci square number is " + sum);
}
}
fun main(args: Array < String > ): Unit
{
// Fibonacci sequence
// [0 1 1 2 3 5 8 13 21 34 55 89 144 233 377..]
/*
Test A
n = 5
(0)² + (1)² + (1)² + (2)² + (3)² = 15
*/
/*
Test B
n = 10
(0)² + (1)² + (1)² + (2)² + (3)² + (5)² +
(8)² + (13)² + (21)² + (34)² = 1870
*/
}``````
Output
`````` Sum of 5 fibonacci square number is 15
Sum of 10 fibonacci square number is 1870``````
Resultant Output Explanation
Now, let's consider the two test cases mentioned in the code and understand their outputs.
1. Test A: n = 5
The first 5 Fibonacci numbers are [0, 1, 1, 2, 3]. The sum of their squares is:
Sum of squares: (0)² + (1)² + (1)² + (2)² + (3)² = 0 + 1 + 1 + 4 + 9 = 15
So, the output will be: "Sum of 5 fibonacci square number is 15"
2. Test B: n = 10
The first 10 Fibonacci numbers are [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]. The sum of their squares is:
Sum of squares: (0)² + (1)² + (1)² + (2)² + (3)² + (5)² + (8)² + (13)² + (21)² + (34)² = 0 + 1 + 1 + 4 + 9 + 25 + 64 + 169 + 441 + 1156 = 1870
So, the output will be: "Sum of 10 fibonacci square number is 1870"
Time Complexity
Let's analyze the time complexity of the provided code:
1. The loop iterates 'n' times to calculate the first 'n' Fibonacci numbers.
2. Each Fibonacci number requires O(1) time to calculate since it is the sum of the previous two numbers.
3. Therefore, the time complexity of the loop is O(n).
Overall, the time complexity of the code is O(n), which is linear in the input 'n'. This is an efficient solution for calculating the sum of the squares of 'n' Fibonacci numbers.
Comment
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
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Percentages and percentage change – concepts and applications
Percentages and percentage change – concepts and applications
Percent (read as per cent) means ‘every hundred’, or, essentially, ‘out of 100’. If we are given a fraction, we can simply convert it to a percentage by multiplying 100. Say, we have ½, which, when multiplied with 100 becomes ½ x 100 = 50%.
What do we mean by a percentage change?
Say, a person has 20 chocolates. If his number of chocolates is increased by 10%, what does it mean? First of all, let us see what is doesn’t mean: after the increase, the number of chocolates with him is NOT 20 + 10 = 30, nor is it 20 + 10/100 = 20.1
When we talk of a percent increase, we must understand on what BASE VALUE is the percentage change working on. In this case, obviously, the 10% increase is 10% of the number of chocolates he has, i.e. 20. Thus, the number of chocolates with him would become:
20 + 10% of 20 = 20 + 20 x 10/100 = 22
Observe that: 20 + 20 x 10/100 = 20 x (1 + 10/100) = 20 x (1 + 1/10) = 20 x (11/10) = 20 x 1.1
Thus, a 10% increase is essentially an increase of (1/10)th which basically means a multiplication with 11/10 or 1.1; we can use either value depending on the convenience of using a decimal number or using a fraction.
Below is a list of certain percent values and the corresponding number with which to multiple for a corresponding percentage increase or decrease:
We have seen above, that, for a given percentage change, we can calculate the final value. However, if the initial and final values are known, how do we calculate percentage change?
Percentage change is simply the change in value taken as a percentage of the initial value. Thus:
Note that the ratio of the ‘Final’ and ‘Initial’ values is the same as the number with which we multiply in case of a given percent change (shown in the table above).
For any queries regarding the courses, feel free to talk to our mentors and our counselors. You may book a free counseling session at any of our education centers. Also, you may reach us at +919433063089 and mail us at info@cubixprep.com. Visit our website for details about our courses and our teaching methodology.
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# 4th Grade OST Math Practice Test Questions
Preparing your student for the 4th Grade OST Math test? Help your students build OST Math test skills with the following common 4th Grade OST Math questions. Practicing common math questions is the best way to help your students improve their Math skills and prepare for the test. Here, we provide a step-by-step guide to solve 10 common OST Math practice problems covering the most important math concepts on the 4th Grade OST Math test.
## 10 Sample 4th Grade OST Math Practice Questions
1- What is the value of A in the equation $$64 ÷ A = 8$$
A. 2
B. 4
C. 6
D. 8
2- Jason’s favorite sports team has won 0.62 of its games this season. How can Jason express this decimal as a fraction?
A. $$\frac{6}{2}$$
B. $$\frac{62}{10}$$
C. $$\frac{62}{100}$$
D. $$\frac{6.2}{10}$$
3- Use the models below to answer the question.
Which statement about the models is true?
A. Each shows the same fraction because they are the same size.
B. Each shows a different fraction because they are different shapes.
C. Each shows the same fraction because they both have 3 sections shaded.
D. Each shows a different fraction because they both have 3 shaded sections but a different number of total sections.
4- Sophia flew 2,448 miles from Los Angeles to New York City. What is the number of miles Sophia flew rounded to the nearest thousand?
A. 2000
B. 2400
C. 2500
D. 3000
5- Write $$\frac{124}{1000}$$ as a decimal number.
A. 1.24
B. 0.124
C. 12.4
D. 0.0124
6- Erik made 12 cups of juice. He drinks 3 cups of juice each day. How many days will Erik take to drink all of the juice he made?
A. 2 days
B. 4 days
C. 8 days
D. 9 days
7- Jason has prepared $$\frac{4}{10}$$ of his assignment. Which decimal represent the part of the assignment Jason has prepared?
A. 4.10
B. 4.01
C. 0.4
D. 0.04
8- Emma described a number using these clues.
• 3 digits of the number are 4, 7, and 9
• The value of the digit 4 is $$(4 × 10)$$
• The value of the digit 7 is $$(7 × 1000)$$
• The value of the digit 9 is $$(9 × 10000)$$
Which number could fit Emma’s description?
A. 9,724.04
B. 90,734.40
C. 97,040.04
D. 98,740.70
9- There are 18 boxes and each box contains 26 pencils. How many pencils are in the boxed in total?
A. 108
B. 208
C. 468
D. Not here
10- Emily and Ava were working on a group project last week. They completed $$\frac{7}{10}$$ of their project on Tuesday and the rest on Wednesday.
Ava completed $$\frac{3}{10}$$ of their project on Tuesday. What fraction of the group project did Emily completed on Tuesday?
A. $$\frac{2}{10}$$
B. $$\frac{3}{10}$$
C. $$\frac{4}{10}$$
D. $$\frac{5}{10}$$
## Best 4th Grade OST Math Prep Resource for 2021
1- D
$$A = 64 ÷ 8$$
$$A=8$$
2- C
0.62 is equal to $$\frac{62}{100}$$
3- D
the model for the first fraction is divided into 6 equal parts. We shade 3_6 to show the same amount as 1_2. The model for the second fraction is divided into 8 equal parts. We shade 3_ 8 that it shows this two model are different fractions.
4- A
When rounding to the nearest thousand, you will need to look at the last three digits. If the last three digits is 449 or less round to the next number that is smaller than the number given and ending with three zeros.
On the other hand, If the last three digits is 500 or more, round to the next number bigger than the given number and ending with three zeros.
5- B
$$\frac{124}{1000}$$ is equal to 0.124
6- B
$$12 ÷ 3 = 4$$
7- C
$$\frac{4}{10}=0.4$$
8- C
9- C
$$18 × 26 = 468$$
10- C
$$\frac{7}{10}-\frac{3}{10}=\frac{4}{10}$$
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# How to Combine Like Terms? (+FREE Worksheet!)
Learn how to combine 'like' terms (terms with the same variable and same power) to simplify algebraic expressions.
## Step-by-step guide to combining like terms
• Terms are separated by “$$+$$” and “$$–$$ “signs.
• Like terms are terms with the same variables and same powers.
• Be sure to use the “$$+$$” or “$$–$$ “that is in front of the coefficient.
### Combining like Terms – Example 1:
Simplify this expression. $$(-2)(2x \ -2)+2x=$$
Solution:
First use Distributive Property formula: $$a(b \ + \ c)=ab \ + \ ac$$
$$(- 2)(2 x \ -2)=-4x \ + \ 4$$
Now, combine like terms: $$-4x + 2x= -2x$$
Then: $$-4x \ + \ 4 \ + 2x= -2x + 4$$
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### Combining like Terms – Example 2:
Simplify this expression. $$4(- \ 2 x \ + \ 6)=$$
Solution:
Use Distributive Property formula: $$a(b \ + \ c)=ab \ + \ ac$$
$$4(− \ 2x \ + \ 6)= \ − 8 x \ + \ 24$$
### Combining like Terms – Example 3:
Simplify this expression. $$8x \ – 6 \ – 2x =$$
Solution:
Combine like terms: $$8x\ – 2x = 6x$$
Then: $$8x-6\ – 2x = 6x\ – 6$$
### Combining like Terms – Example 4:
Simplify this expression. $$(-3)(2x-2)+6=$$
Solution:
First use Distributive Property formula: $$a(b+c)=ab+ac$$
$$(-3)(2x-2)=-6x+6$$
Now, combining like terms: $$6+6=12$$
Then: $$-6x+6+6=-6x+12$$
## Exercises for Combining Like Terms
### Simplify each expression.
• $$\color{blue}{(– 11x)\ – 10x}$$
• $$\color{blue}{3x \ – 12 – 5x}$$
• $$\color{blue}{13 + 4x\ – 5}$$
• $$\color{blue}{(– 22x) + 8x}$$
• $$\color{blue}{2 (4 + 3x)\ – 7x}$$
• $$\color{blue}{(– 4x)\ – (6 – 14x)}$$
• $$\color{blue}{–21x}$$
• $$\color{blue}{–2x – 12}$$
• $$\color{blue}{4x + 8}$$
• $$\color{blue}{–14x}$$
• $$\color{blue}{– x + 8}$$
• $$\color{blue}{10x – 6}$$
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# Factoring Quadratics — ax² + bx + c Topic 6.6.2.
## Presentation on theme: "Factoring Quadratics — ax² + bx + c Topic 6.6.2."— Presentation transcript:
Factoring Quadratics — ax² + bx + c Topic 6.6.2
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c California Standard: 11.0 Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: You’ll learn how to factor more complicated quadratic expressions. Key words: quadratic polynomial factor trial and error
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Lesson 1.1.1 Factoring Quadratics — ax² + bx + c The method in Topic for factoring a quadratic expression only works if the x2-term has a coefficient of 1. It’s a little more complicated when the x2-coefficient (a) isn’t 1 — but only a little.
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c You Can Take Out a Common Factor from Each Term If you see a common factor in each term, such as a number or a variable, take it out first. ax2 + abx + ac a(x2 + bx + c)
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 1 Factor 3x2 + 15x + 18. Solution 3x2 + 15x + 18 = 3(x2 + 5x + 6) Take out the common factor The expression in parentheses can be factored using the method in Topic 6.6.1: = 3(x + 2)(x + 3) But — if the expression in parentheses still has a ¹ 1, then the expression will need to be factored using the second method, shown in Example 2. Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression completely. 1. 3x2 + 15x y2 – 12y – 112 3. 2t2 – 22t r2 – 75 5. 4x2 + 32x p2 + 70p + 63 7. 5m2 + 20m x2 + 42x + 60 9. –2k2 – 20k – –3m2 – 30m – 72 3(x + 1)(x + 4) 4(y + 4)(y – 7) 2(t – 5)(t – 6) 3(r – 5)(r + 5) 4(x + 4)(x + 4) 7(p + 1)(p + 9) 5(m + 1)(m + 3) 6(x + 2)(x + 5) –2(k + 2)(k + 8) –3(m + 4)(m + 6) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression completely. x – x –3x2 – 84x – 225 13. 3x2y – 33xy – 126y x x x – 25x n2y + 100ny – 5600y a2 – 8x2a a – 80 – a2 19. 2x2m2 + 28x2m – 102x a2b2x2 + 30a2bx + 63a2 –1(x + 1)(x – 2) –3(x + 3)(x + 25) 3y(x + 3)(x – 14) 10(x + 4)(x + 25) –25(x – 4)(x + 1) 100y(n – 7)(n + 8) 8a2(2 – x)(2 + x) –1(a – 5)(a – 16) 2x2(m + 17)(m – 3) 3a2(bx + 3)(bx + 7) Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Lesson 1.1.1 Factoring Quadratics — ax² + bx + c You Can Factor ax2 + bx + c by Trial and Error If you can’t see a common factor, then you need to get ax2 + bx + c into the form (a1x + c1)(a2x + c2), where a1 and a2 are factors of a, and c1 and c2 are factors of c. a = a1a2 ax2 + bx + c = (a1x + c1)(a2x + c2) a1c2 + a2c1 = b c = c1c2
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Lesson 1.1.1 Factoring Quadratics — ax² + bx + c Note that if a number is positive then its two factors will be either both positive or both negative. If a number is negative, then its two factors will have different signs — one positive and one negative. These facts give important clues about the signs of c1 and c2.
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution 3x2 + 11x + 6 Write down pairs of factors of a = 3: a1 = 1 and a2 = 3 Write down pairs of factors of c = 6: c1 = 1 and c2 = 6, c1 = 2 and c2 = 3 Now find the combination of factors that gives a1c2 + a2c1 = 11. Put the x-terms into parentheses first, with the pair of coefficients 3 and 1: (3x )(x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 (and a1c2 – a2c1)… Solution continues… Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution (continued) (3x 1)(x 6) multiplies to give 18x and x, which add/subtract to give 19x or 17x. (3x 6)(x 1) multiplies to give 3x and 6x, which add/subtract to give 9x or 3x. (3x 2)(x 3) multiplies to give 9x and 2x, which add/subtract to give 11x or 7x. (3x 3)(x 2) multiplies to give 6x and 3x, which add/subtract to give 9x or 3x. So (3x 2)(x 3) is the combination that gives 11x (so b = 11). Solution continues…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 2 Factor 3x2 + 11x + 6. Solution (continued) Now fill in the + / – signs. Both c1 and c2 are positive (since c = c1c2 and b = a1c2 + a2c1 are positive), so the final factors are (3x + 2)(x + 3). Check by expanding the parentheses to make sure they give the original equation: (3x + 2)(x + 3) = 3x2 + 9x + 2x + 6 = 3x2 + 11x + 6 That’s what you started with, so (3x + 2)(x + 3) is the correct factorization.
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 21. 2x2 + 5x + 2 22. 2a2 + 13a + 11 23. 2y2 + 7y + 3 Possible combinations: (2x 1)(x 2) multiply to 4x and x, which add/subtract to 5x or 3x (2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0x So 2x2 + 5x + 2 = (2x + 1)(x + 2) Possible combinations: (2a 1)(a 11) multiply to 22a and 1a, which add/subtract to 23a or 21a (2a 11)(a 1) multiply to 2a and 11a, which add/subtract to 13a or 9a So 2a2 + 13a + 11 = (2a + 11)(a + 1) Possible combinations: (2y 1)(y 3) multiply to 6y and y, which add/subtract to 7y or 5y (2y 3)(y 1) multiply to 2y and 3y, which add/subtract to 5y or y So 2y2 + 7y + 3 = (2y + 1)(y + 3) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 24. 4x2 + 28x + 49 25. 4x2 + 12x + 9 26. 6x2 + 23x + 7 27. 3x2 + 13x + 12 (2x 7)(2x 7) multiply to 14x and 14x, which add/subtract to 28x or 0x So 4x2 + 28x + 49 = (2x + 7)(2x + 7) (2x 3)(2x 3) multiply to 6x and 6x, which add/subtract to 12x or 0x So 4x2 + 12x + 9 = (2x + 3)(2x + 3) (2x 7)(3x 1) multiply to 2x and 21x, which add/subtract to 23x or 19x So 6x2 + 23x + 7 = (2x + 7)(3x + 1) (3x 4)(x 3) multiply to 9x and 4x, which add/subtract to 13x or 5x So 3x2 + 13x + 12 = (3x + 4)(x + 3) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 28. 2x2 + 11x + 5 29. 4a2 + 16a + 7 30. 2p2 + 14p + 12 31. 8a2 + 46a + 11 (2x 1)(x 5) multiply to 10x and x, which add/subtract to 11x or 9x So 2x2 + 11x + 5 = (2x + 1)(x + 5) (2a 1)(2a 7) multiply to 14a and 2a, which add/subtract to 16a or 12a So 4a2 + 16a + 7 = (2a + 1)(2a + 7) Take out a common factor of 2: 2(p2 + 7p + 6), then factor the rest. (p 6)(p 1) multiply to p and 6p, which add/subtract to 7p or 5p So 2p2 + 14p + 12 = 2(p + 6)(p + 1) (4a 1)(2a 11) multiply to 44a and 2a, which add/subtract to 46a or 42a So 8a2 + 46a + 11 = (4a + 1)(2a + 11) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 32. 3g2 + 51g + 216 33. 9x2 + 12x + 4 34. 4a2 + 36a + 81 35. 4b2 + 32b + 55 Take out a common factor of 3: 3(g2 + 17g + 72), then factor the rest. (g 8)(g 9) multiply to 9g and 8g, which add/subtract to 17g or g So 3g2 + 51p = 3(g + 8)(g + 9) (3x 2)(3x 2) multiply to 6x and 6x, which add/subtract to 12x or 0x So 9x2 + 12x + 4 = (3x + 2)2 (2a 9)(2a 9) multiply to 18a and 18a, which add/subtract to 36a or 0a So 4a2 + 36a + 81 = (2a + 9)2 (2b 5)(2b 11) multiply to 22b and 10b, which add/subtract to 32b or 12b So 4b2 + 32b + 55 = (2b + 5)(2b + 11) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 36. 3x2 + 22x + 24 a2 + 23a + 12 38. 6t2 + 23t + 20 (3x 4)(x 6) multiply to 18x and 4x, which add/subtract to 22x or 14x So 3x2 + 22x + 24 = (3x + 4)(x + 6) (2a 3)(5a 4) multiply to 8a and 15a, which add/subtract to 23a or 7a So 10a2 + 23a + 12 = (2a + 3)(5a + 4) (2t 5)(3t 4) multiply to 8t and 15t, which add/subtract to 23t or 7t So 6t2 + 23t + 20 = (2t + 5)(3t + 4) Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Lesson 1.1.1 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 39. 4x2 + 34x + 16 b2 + 96b + 36 Take out a common factor of 2: 2(2x2 + 17x + 8), then factor the rest. (2x 1)(x 8) multiply to 16x and x, which add/subtract to 17x or 15x So 4x2 + 34x + 16 = 2(2x + 1)(x + 8) Take out a common factor of 3: 3(5b2 + 32b + 12), then factor the rest. (5b 2)(b 6) multiply to 30b and 2b, which add/subtract to 32b or 28b So 15b2 + 96b + 36 = 3(5b + 2)(b + 6) Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution Write down pairs of factors of a = 6: a1 = 6 and a2 = 1, a1 = 2 and a2 = 3, Write down pairs of factors of c = –6 (ignoring the minus sign for now): c1 = 1 and c2 = 6, c1 = 2 and c2 = 3 Put the x-terms into parentheses first, with the first pair of possible values for a1 and a2, 6 and 1: (6x )(1x ). Now try all the pairs of c1 and c2 in the parentheses and find the possible values of a1c2 + a2c1 and a1c2 – a2c1 as before… Solution continues… Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) (6x 1)(x 6) multiplies to give 36x and x, which add/subtract to give 37x or 35x. (6x 6)(x 1) multiplies to give 6x and 6x, which add/subtract to give 12x or 0x. (6x 2)(x 3) multiplies to give 18x and 2x, which add/subtract to give 20x or 16x. (6x 3)(x 2) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. None of these combinations works, so try again with (2x )(3x )… Solution continues…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) (2x 1)(3x 6) multiplies to give 12x and 3x, which add/subtract to give 15x or 9x. (2x 6)(3x 1) multiplies to give 2x and 18x, which add/subtract to give 20x or 16x. (2x 3)(3x 2) multiplies to give 4x and 9x, which add/subtract to give 13x or 5x. 5x is what you want, so you can stop there — so (2x 3)(3x 2) is the right combination. Solution continues…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 3 Factor 6x2 + 5x – 6. Solution (continued) Now fill in the + / – signs to get b = +5. One of c1 and c2 must be negative, to give c = –6, so the final factors are either (2x + 3)(3x – 2) or (2x – 3)(3x + 2). The x-term of (2x + 3)(3x – 2) will be 9x – 4x = 5x, whereas the x-term for (2x – 3)(3x + 2) will be 4x – 9x = –5x. So the correct factorization is (2x + 3)(3x – 2).
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 41. 2x2 + 3x – 2 42. 3y2 – y – 2 43. 5k2 + 13k – 6 (2x 1)(x 2) multiply to 4x and x, which add/subtract to 5x or 3x (2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0x So 2x2 + 3x – 2 = (2x – 1)(x + 2) (3y 1)(y 2) multiply to 6y and y, which add/subtract to 7y or 5y (3y 2)(y 1) multiply to 3y and 2y, which add/subtract to 5y or y So 3y2 – y – 2 = (3y + 2)(y – 1) (5k 2)(k 3) multiply to 15k and 2k, which add/subtract to 17k or 13k (5k 3)(k 2) multiply to 10k and 3k, which add/subtract to 13k or 7k So 5k2 + 13k – 6 = (5k – 2)(k + 3) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 44. 3x2 – x – 10 45. 6b2 – 23b + 7 46. 2x2 – 5x + 2 (3x 5)(x 2) multiply to 6x and 5x, which add/subtract to 11x or x (3x 2)(x 5) multiply to 15x and 2x, which add/subtract to 17x or 13x So 3x2 – x – 10 = (3x + 5)(x – 2) (3b 1)(2b 7) multiply to 21b and 2b, which add/subtract to 23b or 19b (3b 7)(2b 1) multiply to 3b and 14b, which add/subtract to 17b or 11b So 6b2 – 23b + 7 = (3b – 1)(2b – 7) (2x 2)(x 1) multiply to 2x and 2x, which add/subtract to 4x or 0x (2x 1)(x 2) multiply to 4x and x, which add/subtract to 5x or 3x So 2x2 – 5x + 2 = (2x – 1)(x – 2) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 47. 3k2 – 2k – 1 48. 3v2 – 16v + 5 x – 2x2 (3k 1)(k 1) multiply to 3k and k, which add/subtract to 4k or 2k So 3k2 – 2k – 1 = (3k + 1)(k – 1) (3v 5)(v 1) multiply to 3v and 5v, which add/subtract to 8v or 2v (3v 1)(v 5) multiply to 15v and v, which add/subtract to 16v or 10v So 3v2 – 16v + 5 = (3v – 1)(v – 5) Take out a factor of –1: –(2x2 – 5x – 18), then factor the rest. (2x 9)(x 2) multiply to 4x and 9x, which add/subtract to 13x or 5x (2x 2)(x 9) multiply to 18x and 2x, which add/subtract to 20x or 16x So x – 2x2 = –(2x – 9)(x + 2) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. x – 2x2 51. 9x2 + 12x + 4 52. 7a2 – 26a – 8 Take out a factor of –1: –(2x2 – x – 28), then factor the rest. (2x 7)(x 4) multiply to 8x and 7x, which add/subtract to 15x or x (2x 4)(x 7) multiply to 14x and 4x, which add/subtract to 18x or 10x So 28 + x – 2x2 = –(2x + 7)(x – 4) (3x 2)(3x 2) multiply to 6x and 6x, which add/subtract to 12x or 0x So 9x2 + 12x + 4 = (3x + 2)(3x + 2) (7a 2)(a 4) multiply to 28a and 2a, which add/subtract to 30a or 26a (7a 4)(a 2) multiply to 14a and 4a, which add/subtract to 18a or 10a So 7a2 – 26a – 8 = (7a + 2)(a – 4) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 53. 3x2 – 7x – 6 54. 12x2 + 5x – 2 55. 6x2 + 2x – 20 (3x 3)(x 2) multiply to 6x and 3x, which add/subtract to 9x or 3x (3x 2)(x 3) multiply to 9x and 2x, which add/subtract to 11x or 7x So 3x2 – 7x – 6 = (3x + 2)(x – 3) (4x 1)(3x 2) multiply to 8x and 3x, which add/subtract to 11x or 5x (4x 2)(3x 1) multiply to 4x and 6x, which add/subtract to 10x or 2x So 12x2 + 5x – 2 = (4x – 1)(3x + 2) Take out a factor of 2: 2(3x2 + x – 10), then factor the rest. (3x 2)(x 5) multiply to 15x and 2x, which add/subtract to 17x or 13x (3x 5)(x 2) multiply to 6x and 5x, which add/subtract to 11x or x So 6x2 + 2x – 20 = 2(3x – 5)(x + 2) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each polynomial. 56. 18x2 + x – 4 57. 6y2 + y – 12 58. 9m2 – 3m – 20 (2x 1)(9x 4) multiply to 8x and 9x, which add/subtract to 17x or x (2x 4)(9x 1) multiply to 2x and 36x, which add/subtract to 38x or 34x So 18x2 + x – 4 = (2x + 1)(9x – 4) (3y 3)(2y 4) multiply to 12y and 6y, which add/subtract to 18y or 6y (3y 4)(2y 3) multiply to 9y and 8y, which add/subtract to 17y or y So 6y2 + y – 12 = (3y – 4)(2y + 3) (3m 4)(3m 5) multiply to 15m and 12m, which add/subtract to 27m or 3m So 9m2 – 3m – 20 = (3m + 4)(3m – 5) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Use the + and – Signs in the Quadratic to Work Faster Looking carefully at the signs in the quadratic that you are factoring can help to narrow down the choices for a1, a2, c1, and c2. The next three Examples show you how.
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 4 Factor 3x2 + 11x + 6. Solution Here c = 6, which is positive — so its factors c1 and c2 are either both positive or both negative. But since b = 11 is positive, you can tell that c1 and c2 must be positive (so that a1c2 + a2c1 is positive). Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 5 Factor 3x2 – 11x + 6. Solution Here c is also positive, so c1 and c2 are either both positive or both negative. But since b = –11 is negative, you can tell that c1 and c2 must be negative (so that a1c2 + a2c1 is negative). Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Example 6 Factor 6x2 + 5x – 6. Solution In this expression, c is negative, so one of c1 and c2 must be positive and the other must be negative. So instead of looking at both the sums and differences of all the different combinations a1c2 and a2c1, you only need to look at the differences. Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 59. 2n2 + n – 3 60. 2x2 – 5x – 3 61. 4a2 + 4a + 1 c is negative, so one of c1 and c2 must be positive and the other negative (2n 3)(n 1) multiply to 2n and 3n. To add to get n, 3n must be positive and 2n negative. So 2n2 + n – 3 = (2n + 3)(n – 1) c is negative, so one of c1 and c2 must be positive and the other negative (2x 1)(x 3) multiply to 6x and x. To add to get –5x, x must be positive and 6x negative. So 2x2 – 5x – 3 = (2x + 1)(x – 3) b and c are positive, so c1 and c2 must both be positive. (4a 1)(a 1) multiply to 4a and a and add to 5a (2a 1)(2a 1) multiply to 2a and 2a and add to 4a So 4a2 + 4a + 1 = (2a + 1)(2a + 1) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 62. 3x2 – 4x + 1 63. 9y2 + 6y + 1 64. 4t2 + t – 3 b is negative and c is positive, so c1 and c2 must both be negative. (3x 1)(x 1) multiply to 3x and x and add to 4x So 3x2 – 4x + 1 = (3x – 1)(x – 1) b and c are positive, so c1 and c2 must both be positive. (3y 1)(3y 1) multiply to 3y and 3y and add to 6y So 9y2 + 6y + 1 = (3y + 1)2 c is negative, so one of c1 and c2 must be positive and the other negative (4t 3)(t 1) multiply to 4t and 3t. To add to get t, 4t must be positive and 3t negative. So 4t2 + t – 3 = (4t – 3)(t + 1) Solution follows…
Factoring Quadratics — ax² + bx + c
Lesson 1.1.1 Topic 6.6.2 Factoring Quadratics — ax² + bx + c Guided Practice Factor each expression. 65. 5x2 – x – 18 66. 9x2 – 6x + 1 67. 6t2 + t – 1 68. b2 + 10b + 21 c is negative, so one of c1 and c2 must be positive and the other negative (5x 9)(x 2) multiply to 10x and 9x. To add to get –x, 9x must be positive and 10x negative. So 5x2 – x – 18 = (5x + 9)(x – 2) b is negative and c is positive, so c1 and c2 must both be negative. (3x 1)(3x 1) multiply to 3x and 3x and add to 6x. So 9x2 – 6x + 1 = (3x – 1)2 c is negative, so one of c1 and c2 must be positive and the other negative (3t 1)(2t 1) multiply to 3t and 2t. To add to get t, 3t must be positive and 2t negative. So 6t2 + t – 1 = (3t – 1)(2t + 1) b and c are positive, so c1 and c2 must both be positive. (b 7)(b 3) multiply to 3b and 7b and add to 10b. So 9y2 + 6y + 1 = (3y + 1)2 Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice Factor each polynomial. 1. 5k2 – 7k k2 – 15k + 9 3. 12t2 – 11t – 7k – 2k2 h – 3h x2 – 14x – 8 x – 6x x4 + 8x2 – 3 (5k – 2)(k – 1) (4k – 3)(k – 3) (4t – 1)(3t – 2) –(2k + 9)(k – 1) –(3h + 5)(h – 2) (5x + 2)(3x – 4) –(3x + 2)(2x – 3) (3x2 – 1)(x2 + 3) Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice 9. If the area of a rectangle is (6x2 + 25x + 14) square units and the length is (3x + 2) units, find the width w in terms of x. 10. The area of a parallelogram is (12x2 + 7x – 10) cm2, where x is positive. If the area is given by the formula Area = base × height, find the base length and the height of the parallelogram, given that they are both linear factors of the area. (2x + 7) units (4x + 5) cm, (3x – 2) cm Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice Factor each of these expressions completely. 11. x2ab – 3abx – 18ab 12. (d + 2)x2 – 7x(d + 2) – 18(d + 2) 13. (x + 1)m2 – 2m(x + 1) + (x + 1) 14. –2dk2 – 14dk + 36d ab(x + 3)(x – 6) (d + 2)(x – 9)(x + 2) (x + 1)(m – 1)2 –2d(k + 9)(k – 2) Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Independent Practice 15. Five identical rectangular floor tiles have a total area of (15x2 + 10x – 40) m2. Find the dimensions of each floor tile, if the length of each side can be written in the form ax + b, where a and b are integers. (3x – 4) m by (x + 2) m. Solution follows…
Factoring Quadratics — ax² + bx + c
Topic 6.6.2 Factoring Quadratics — ax² + bx + c Round Up Now you can factor lots of different types of polynomials. In the next Section you’ll learn about another type — quadratic expressions containing two different variables.
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# Question Video: Determining the Variance for the Sum of Two Independent Random Variables Mathematics
Suppose π₯ and π¦ are independent, var(π₯) = 24, and var(π¦) = 30. Determine var(7π₯ + 9π¦).
01:22
### Video Transcript
Suppose π₯ and π¦ are independent, the variance of π₯ is 24, and the variance of π¦ is 30. Determine the variance of seven π₯ plus nine π¦.
Weβve been given some information for two independent random variables π₯ and π¦. We know the variance of π₯ is 24 and the variance of π¦ is equal to 30. Weβre looking to determine the variance of seven π₯ plus nine π¦. So we begin by recalling that for two independent random variables π΄ and π΅, the variance of their sum is the sum of their variances. So the variance of π΄ plus π΅ is the variance of π΄ plus the variance of π΅.
But we also know that the variance of some multiple of a random variable π₯ β letβs call that π of π₯ β is equal to π squared times the variance of π₯. So what weβre going to do is split our variance, the variance of seven π₯ plus nine π¦, up into the sum of the variances, so variance of seven π₯ plus the variance of nine π¦.
Then, we know that the variance of seven π₯ is seven squared the variance of π₯. And the variance of nine π¦ is nine squared the variance of π¦. But we already saw that the variance of π₯ is 24 and the variance of π¦ is 30. So the variance of seven π₯ plus nine π¦ is seven squared times 24 plus nine squared times 30, which is 3,606. The variance of seven π₯ plus nine π¦ is 3,606.
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Video: FP1P1-Q16
FP1P1-Q16
03:40
Video Transcript
Expand and fully simplify brackets 𝑥 plus eight brackets 𝑥 minus seven.
Expanding in algebra means multiplying. We need to multiply each of the terms in the first bracket by each of the terms in the second bracket. There are two methods we can use to complete these expansions. The first method usually called FOIL or sometimes FOIL face. We’ll see why it’s called that in a moment.
In the word FOIL, each letter stands for something. F stands for First, O stands for Outer, I stands for Inner, and L stands for Last. These tell us what order we can perform our multiplication in so that we don’t lose any terms. Let’s see what this looks like.
The first two terms in our brackets are 𝑥 and 𝑥. So we’re going to multiply these terms together. When we multiply a number by itself, we’re squaring it. So 𝑥 multiplied by 𝑥 is 𝑥 squared.
We’re now going to multiply the outer terms; that’s 𝑥 and negative seven. Be really careful here: this is a negative seven, not a positive seven. 𝑥 multiplied by seven is seven 𝑥. So 𝑥 multiplied by negative seven is negative seven 𝑥.
We’re next going to multiply the two inner terms; that’s eight and 𝑥. And that gives us eight 𝑥.
Finally, we’re going to multiply the last terms in each bracket; that’s eight and negative seven. A common mistake here is to think that we need to add these terms. In fact, remember we said that expand means multiply. So we’re going to multiply eight by negative seven. And that gives us negative 56.
There’s a clue in our question that tells us we’re not quite finished. We need to fully simplify our expression. And when we simplify, we collect like terms. Now, remember 𝑥 squared is different to 𝑥. So we have 𝑥 squared at the front of our expression. We then need to add negative seven 𝑥 and eight 𝑥. We can use a number line and just begin by imagining we’re adding negative seven and eight.
To do this, we’d start at negative seven. Because we’re adding eight, we’re going to move up the number line: one, two, three, four, five, six, seven, eight. That means negative seven plus eight is one. So negative seven 𝑥 plus eight 𝑥 is one 𝑥. Remember though we don’t need to write the number one. We can simply write 𝑥 and then we have negative 56. So this expression expands and simplifies to 𝑥 squared plus 𝑥 minus 56.
Did you spot why this is sometimes called FOIL face? When we draw these arrows in to remind us of the order in which to multiply, they look a little bit like a face: we have two eyebrows, a nose, and a mouth.
Now, let’s look at the second method. This is the grid method and it works much like the grid method for long multiplication. Here, I’ve put 𝑥 plus eight at the top of the grid and 𝑥 minus seven at the side. But it actually doesn’t matter which order we choose. Let’s figure out what goes here.
To get this term, we multiply 𝑥 by 𝑥; that’s 𝑥 squared. Then here, 𝑥 multiplied by eight is eight 𝑥. We then have negative seven 𝑥 here and negative 56 here. Notice how the terms in our grid are the same as the terms we achieved when we used FOIL. Once again, by adding the like terms, we end up with 𝑥 squared plus 𝑥 minus 56.
Either method for expanding double brackets is absolutely fine.
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# 2016 UMO Problems/Problem 2
## Problem
Four fair six-sided dice are rolled. What is the probability that they can be divided into two pairs which sum to the same value? For example, a roll of $(1,4,6,3)$ can be divided into $(1,6)$ and $(4,3)$, each of which sum to $7$, but a roll of $(1,1,5,2)$ cannot be divided into two pairs that sum to the same value.
## Solution 1
We split this into cases based on the the form of the unordered values shown on the dice. In the following, $1 <= a,b,c,d <= 6$ are distinct numbers. After determining the (unordered) numbers that can appear on the dice, we must determine how many ways we can order those dice rolls in sequence.
Case1: The unordered numbers shown are {a,a,a,a} In this case, it is clear that the pairing (a,a) and (a,a) will yield equal sums, so we have 6 possible choices for a, and this is the number of possible rolls that take this form.
Case2: The unordered numbers shown are {a,a,a,b} In this case, one of the pairs will contain b, hence the pairs will be (a,b) and (a,a). But these can never be equal as a 6= b.
Case3: The unordered numbers shown are {a,a,b,b} To obtain equal sums, we must split the numbers into the pairs (a,b) and (a,b). There are $\binom{6}{2}= 15$ ways to select a and b, and there are $\binom{4}{2}= 6$ ways to order the rolls (select two of the rolls to be a’s). Hence there are 15·6 = 90 total rolls that take this form.
Case4: The unordered numbers shown are {a,a,b,c} In this case, either b pairs with a or b pairs with c. If b pairs with a, then $a + b = c + a$ , which is impossible as $b \ne c$. Therefore, b must pair with c, so $2a = b + c$. Therefore, b + c is even. We may assume without loss of generality that $b <= c$, hence the possibilities for (b,c) are (b,c) = (1,3),(1,5),(2,4),(2,6),(3,5),(4,6). Therefore, there are six ways to choose (b,c), and a is automatically determined (it is the average of b and c). Then there are 4! 2! = 12 ways to order the rolls, hence there are $6\cdot 12 = 72$ total rolls that take this form.
Case5: The unordered numbers shown are {a,b,c,d} In this case, the sum of the pairs must be representable in at two ways with distinct integers. We find 5 = 1 + 4 = 2 + 3; 6 = 1 + 5 = 2 + 4; 7 = 1 + 6 = 2 + 5 = 3 + 4; 8 = 2 + 6 = 3 + 5; 9 = 3 + 6 = 4 + 5. These are the only numbers that can be represented as the sum of two numbers in at least two ways using distinct numbers. In particular, for the numbers $5,6,8,9$,we know immediately what the four numbers a,b,c,d are. For 7,we must choose two of the three pairs, and we can do this in three ways. Therefore, the number of ways to choose {a,b,c,d} is $1+1+1+1+3 = 7$. Then we can order the rolls in 4! = 24 ways. Thus there are $7\cdot 24 = 168$ total rolls that take this form. Adding these together, we find $6 + 90 + 72 + 168 = 336$ possibilities, so the answer is $\frac{336}{64} = \frac{7}{27}$.
## Solution 2
We observe $$\textbf{Probability}=\frac{\textbf{Favorable}}{\textbf{Total}}.$$
We have $\textbf{Total}=6^4=1296.$
We now work on the favorable cases.
Suppose the four numbers, $a,b,c,$ and $d$ can be split as $$N=a+b=c+d.$$ We do casework on the value of $N.$ Notice that a sum of $n$ can be obtained in the same number of ways as a sum of $14-n$ can be obtained. Thus:
$\textbf{n=2:}$ A sum of $2$ can be obtained only if $a=b=c=d=1.$ Thus $1$ case.
$\textbf{n=3:}$ A sum of $3$ can be obtained as $3=1+2$ thus $\frac{4!}{2!\cdot 2!}=6$ cases.
$\textbf{n=4:}$ A sum of $4$ can be obtained as $4=1+3=2+2,$ which produces the three possibilities of
$(2,2,2,2)$ which produces $1$ case.
$(1,1,3,3)$ which produces $\frac{4!}{2!\cdot 2!}=6$ cases.
$(1,2,2,3)$ which produces $\frac{4!}{2!}=12$ cases.
Thus $1+6+12=19$ cases.
$\textbf{n=5:}$ A sum of $5$ can be obtained as $5=1+4=2+3,$ which produces the three possibilities of
$(2,2,3,3)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
$(1,1,4,4)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
$(1,2,3,4)$ which produces $4!=24$ cases.
Thus $6+6+24=36$ cases.
$\textbf{n=6:}$ A sum of $6$ can be obtained as $6=1+5=2+4=3+3,$ which produces the six possibilities of
$(3,3,3,3)$ which produces $1$ case.
$(1,3,3,5)$ which produces $\frac{4!}{2!}=12$ cases.
$(2,3,3,4)$ which produces $\frac{4!}{2!}=12$ cases.
$(1,1,5,5)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
$(1,2,4,5)$ which produces $4!=24$ cases.
$(2,2,4,4)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
Thus $1+6+12+12+6+24=61$ cases.
$\textbf{n=7:}$ A sum of $7$ can be obtained as $6=1+6=2+5=3+4,$ which produces the six possibilities of
$(1,1,6,6)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
$(1,2,5,6)$ which produces $4!=24$ cases.
$(1,3,4,6)$ which produces $4!=24$ cases.
$(2,2,5,5)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
$(2,3,4,5)$ which produces $4!=24$ cases.
$(4,4,5,5)$ which produces $\frac{4!}{2!\cdot2!}=6$ cases.
Thus, $6+24+24+6+24+6=90$ cases.
This means, we have $$\textbf{Favorable} = 90 +2\left(1+6+19+36+61\right)=336.$$
This means $$\textbf{Probability}=\frac{\textbf{Favorable}}{\textbf{Total}}=\frac{336}{1296}=\boxed{\frac{7}{27}}.$$
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# How do you solve x^2-1=0?
Mar 9, 2018
$\textcolor{b l u e}{x = - 1}$ or $\textcolor{b l u e}{x = + 1}$
#### Explanation:
Remember "the difference of squares"
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
Therefore
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 1 = 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \left(x + 1\right) \left(x - 1\right) = 0$
$\Rightarrow \text{Either }$
color(white)("xxxxxxxx"){: ((x+1)=0," or ",(x-1)=0), (rarr x=-1,,rarr x=+1) :}
Mar 9, 2018
$x = 1$ or $x = - 1$
#### Explanation:
${x}^{2} - 1 = 0$
$\therefore {x}^{2} = 1$
$\therefore x = \sqrt{1}$
$\therefore x = 1$ or$x = - 1$
|
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# Class 9 NCERT Solutions Maths Chapter 10 - Circles
Score marks in short answer questions easily by practising with our NCERT Solutions for CBSE Class 9 Mathematics Chapter 10 Circles. Write and practise with TopperLearning’s expert solutions for long answer questions based on chords. By revising a variety of textbook questions, you’ll learn how to prove chord properties in a step-wise format.
With CBSE Class 9 Maths chapter solutions, learn the application of important concepts such as RHS congruence rule, angle sum property of a triangle etc. This knowledge will help you think logically and arrive at correct answers in your Maths exam. Other than expert solutions, you may want to use sample papers, online mock tests, concept videos etc. to revise circles for exam preparation.
## Circles Exercise Ex. 10.1
### Solution 1
(i) The centre of a circle lies in interior of the circle. (exterior/interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in exterior of
the circle. (exterior/interior)
(iii) The longest chord of a circle is a diameter of the circle.
(iv) An arc is a semicircle when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and chord of the circle.
(vi) A circle divides the plane, on which it lies, in three parts.
### Solution 2
(i) True, all the points on circle are at equal distance from the centre of circle, and this equal distance
it called as radius of circle.
(ii) False, on a circle there are infinite points. So, we can draw infinite number of chords of given length. Hence, a circle has infinite number of equal chords.
(iii) False, consider three arcs of same length as AB, BC and CA. Now we may observe that for minor arc BDC. CAB is major arc. So AB, BC and CA are minor arcs of circle.
(iv) True, let AB be a chord which is twice as long as its radius. In this situation our chord will be passing through centre of circle. So it will be the diameter of circle.
(v) False, sector is the region between an arc and two radii joining the centre to the end points of the arc as in the given figure OAB is the sector of circle.
(vi) True, A circle is a two dimensional figure and it can also be referred as plane figure.
## Circles Exercise Ex. 10.2
### Solution 1
A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle.
So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.
So, two circles are congruent if they have equal radius.
Now consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths
Now in AOB and CO'D
AB = CD (chords of same length)
OA = O'C (radii of congruent circles)
OB = O'D (radii of congruent circles)
AOB CO'D (SSS congruence rule)
AOB = CO'D (by CPCT)
Hence equal chords of congruent circles subtend equal angles at their centres.
### Solution 2
Let us consider two congruent circles (circles of same radius) with centres as O and O'.
In AOB and CO'D
AOB = CO'D (given)
OA = O'C (radii of congruent circles)
OB = O'D (radii of congruent circles)
AOB CO'D (SSS congruence rule)
AB = CD (by CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.
## Circles Exercise Ex. 10.3
### Solution 1
Consider the following pair of circles.
(i) circles don't intersect each other at any point, so circles are not having any point in common.
(ii) Circles touch each other only at one point P so there is only 1 point in common.
(iii) Circles touch each other at 1 point X only. So the circles have 1 point in common.
(iv) These circles intersect each other at two points P and Q. So the circles have two points in common. We may observe that there can be maximum 2 points in common.
We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times.
### Solution 2
Following are the steps of construction:
Step1. Take the given circle centered at point O.
Step2. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these
chords.
Step3. Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle.
### Solution 3
Consider two circles centered at point O and O' intersect each other at point A and B respectively.
Join AB. AB is the chord for circle centered at O, so perpendicular bisector of AB will pass through O.
Again AB is also chord of circle centered at O', so, perpendicular bisector of AB will also pass through O'.
Clearly centres of these circles lie on the perpendicular bisector of common chord.
## Circles Exercise Ex. 10.4
### Solution 1
Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.
OA = OB = 5 cm
O'A = O'B = 3 cm
OO' will be the perpendicular bisector of chord AB.
AC = CB
Given that OO' = 4 cm
Let OC be x. so, O'C will be 4 - x
In OAC
OA2 = AC2 + OC2
52 = AC2 + x2
25 - x2 = AC2 ... (1)
In O'AC
O'A2 = AC2 + O'C2
32 = AC2 + (4 - x)2
9 = AC2 + 16 + x2 - 8x
AC2 = - x2 - 7 + 8x ... (2)
From equations (1) and (2), we have
25 - x2 = - x2 - 7 + 8x
8x = 32
x = 4
So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.
Now, AC2 = 25 - x2 = 25 - 42 = 25 - 16 = 9
AC = 3 m
The length of the common chord AB = 2 AC = (2 3) m = 6 m
### Solution 2
Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In OVT and OUT
OV = OU (Equal chords of a circle are equidistant from the centre)
OVT = OUT (Each 90o)
OT = OT (common)
OVT OUT (RHS congruence rule)
VT = UT (by CPCT) ... (1)
It is given that
PQ = RS ... ... ... ... (2)
PV = RU ... ... ... ... (3)
On adding equations (1) and (3), we have
PV + VT = RU + UT
PT = RT ... ... ... ... (4)
On subtracting equation (4) from equation (2), we have
PQ - PT = RS - RT
QT = ST ... ... ... ... (5)
Equations (4) and (5) shows that the corresponding segments of
chords PQ and RS are congruent to each other.
### Solution 3
Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords.
In OVT and OUT
OV = OU (Equal chords of a circle are equidistant from the centre)
OVT = OUT (Each 90o)
OT = OT (common)
OVT OUT (RHS congruence rule)
OTV = OTU (by CPCT)
Hence, the line joining the point of intersection to the centre makes equal angles with the chords.
### Solution 4
Let us draw a perpendicular OM on line AD.
Here, BC is chord of smaller circle and AD is chord of bigger circle.
We know that the perpendicular drawn from centre of circle bisects the chord.
BM = MC ... (1)
And AM = MD ... (2)
Subtracting equations (2) from (1), we have
AM - BM = MD - MC
AB = CD
### Solution 5
Draw perpendiculars OA and OB on RS and SM respectively.
Let R, S and M be the position of Reshma, Salma and Mandip respectively.
AR = AS = = 3cm
OR = OS = OM = 5 m (radii of circle)
In OAR
OA2 + AR2 = OR2
OA2 + (3 m)2 = (5 m)2
OA2 = (25 - 9) m2 = 16 m2
OA = 4 m
We know that in an isosceles triangle altitude divides the base, so in RSM
RCS will be of 90o and RC = CM
Area of ORS = OARS
RC = 4.8
RM = 2RC = 2(4.8)= 9.6
So, distance between Reshma and Mandip is 9.6 m.
### Solution 6
Given that AS = SD = DA
So, ASD is a equilateral triangle
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.
We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write
AB = OA + OB = (20 + 10) m = 30 m.
In ABD
So, length of string of each phone will be m.
## Circles Exercise Ex. 10.5
### Solution 1
We may observe that
AOC = AOB + BOC
= 60o + 30o
= 90o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
### Solution 2
In OAB
AB = OA = OB = radius
OAB is an equilateral triangle.
So, each interior angle of this triangle will be of 60o
AOB = 60o
Now,
ADB = 180o - 30o = 150o
So, angle subtended by this chord at a point on major arc and minor arc are 30o and 150o respectively.
### Solution 3
Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.
PQR + PSR = 180o (Opposite angles of cyclic quadrilateral)
PSR = 180o - 100o = 80o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
POR = 2PSR = 2 (80o) = 160o
In POR
OP = OR (radii of same circle)
OPR = ORP (Angles opposite equal sides of a triangle)
OPR + ORP + POR = 180o (Angle sum property of a triangle)
2 OPR + 160o= 180o
2 OPR = 180o - 160o = 20o
OPR = 10o
### Solution 4
In ABC
BAC + ABC + ACB = 180o (Angle sum property of a triangle)
BAC + 69o + 31o = 180o
BAC = 180o - 100º
BAC = 80o
BDC = BAC = 80o (Angles in same segment of circle are equal)
### Solution 5
In CDE
CDE + DCE = CEB (Exterior angle)
CDE + 20o = 130o
CDE = 110o
But BAC = CDE (Angles in same segment of circle)
BAC = 110o
### Solution 6
For chord CD
CBD = CAD (Angles in same segment)
BCD + 100o = 180o
BCD = 80o
In ABC
AB = BC (given)
BCA = CAB (Angles opposite to equal sides of a triangle)
BCA = 30o
We have BCD = 80o
BCA + ACD = 80o
30oACD = 80o
ACD = 50o
ECD = 50o
### Solution 7
Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.
(Consider BD as a chord)
BCD = 180o - 90o = 90o
(Considering AC as a chord)
90o + ABC = 180o
ABC = 90o
Here, each interior angle of cyclic quadrilateral is of 90o. Hence it is a rectangle.
### Solution 8
Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM CD and BN CD
In AMD and BNC
AMD = BNC (By construction each is 90o)
AM = BM (Perpendicular distance between two parallel lines is same)
AMD BNC (RHS congruence rule)
ADC = BCD (CPCT) ... (1)
BAD + BCD = 180o [Using equation (1)]
This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.
### Solution 9
Join chords AP and DQ
For chord AP
PBA = ACP (Angles in same segment) ... (1)
For chord DQ
DBQ = QCD (Angles in same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
PBA = DBQ (Vertically opposite angles) ... (3)
From equations (1), (2) and (3), we have
ACP = QCD
### Solution 10
Consider a ABC
Two circles are drawn while taking AB and AC as diameter.
Let they intersect each other at D and let D does not lie on BC.
ADB = 90o (Angle subtend by semicircle)
ADC = 90o (Angle subtend by semicircle)
Hence BDC is straight line and our assumption was wrong.
Thus, Point D lies on third side BC of ABC
### Solution 11
In ABC
ABC + BCA + CAB = 180o (Angle sum property of a triangle)
90o + BCA + CAB = 180o
BCA + CAB = 90o ... (1)
CDA + ACD + DAC = 180o (Angle sum property of a triangle)
90o + ACD + DAC = 180o
ACD + DAC = 90o ... (2)
Adding equations (1) and (2), we have
BCA + CAB + ACD + DAC = 180o
(BCA + ACD) + (CAB + DAC) = 180o BCD + DAB = 180o ... (3)
But it is given that
B + D = 90o + 90o = 180o ... (4)
From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180o.
So, it is a cyclic quadrilateral.
Consider chord CD.
Now, CAD = CBD (Angles in same segment)
### Solution 12
Let ABCD be a cyclic parallelogram.
A + C = 180o (Opposite angle of cyclic quadrilateral) ... (1)
We know that opposite angles of a parallelogram are equal
A = C and B = D
From equation (1)
A + C = 180o
A + A = 180o
A = 180o
A = 90o
Parallelogram ABCD is having its one of interior angles as 90o, so, it is a rectangle.
## Circles Exercise Ex. 10.6
### Solution 1
Let two circles having their centres as O and intersect each other at point A and B respectively.
Construction: Let us join OO',
Now in AOO' and BOO'
OA = OB (radius of circle 1)
O'A = O'B (radius of circle 2)
OO' = OO' (common)
AOO' BOO' (by SSS congruence rule)
OAO' = OBO' (by CPCT)
So, line of centres of two intersecting circles subtends equal angles at the two points of intersection.
### Solution 2
Draw OM AB and ON CD. Join OB and OD
(Perpendicular from centre bisects the chord)
Let ON be x, so OM will be 6 - x
In MOB
In NOD
We have OB = OD (radii of same circle)
So, from equation (1) and (2)
From equation (2)
So, radius of circle is found to be cm.
### Solution 3
Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm
In OMB
In OND
So, distance of bigger chord from centre is 3 cm.
### Solution 4
In AOD and COE
OA = OC (radii of same circle)
OD = OE (radii of same circle)
AOD COE (SSS congruence rule)
OAD = OCE (by CPCT) ... (1)
ODA = OEC (by CPCT) ... (2)
We also have
OAD = ODA (As OA = OD) ... (3)
From equations (1), (2) and (3), we have
OAD = OCE = ODA = OEC
Let OAD = OCE = ODA = OEC = x
In OAC,
OA = OC
OCA = OAC (let a)
In ODE,
OD = OE
OED = ODE (let y)
CAD + DEC = 180o (opposite angles are supplementary)
x + a + x + y = 180o
2x + a + y = 180o
y = 180 - 2x - a ... (4)
But DOE = 180 - 2y
And AOC = 180 - 2a
Now, DOE - AOC = 2a - 2y = 2a - 2 (180 - 2x - a)
= 4a + 4x - 360o ... (5)
Now, BAC + CAD = 180 (Linear pair)
BAC = 180 - CAD = 180 - (a + x)
Similarly, ACB = 180 - (a + x)
Now, in ABC
ABC + BAC + ACB = 180 (Angle sum property of a triangle)
ABC = 180 - BAC - ACB
= 180 - (180 - a - x) - (180 - a -x)
= 2a + 2x - 180
= [4a + 4x - 360o]
ABC = [DOE - AOC] [Using equation (5)]
### Solution 5
Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn taking side CD as its diameter.
We know that angle in a semicircle is of 90o.
COD = 90o
Also in rhombus the diagonals intersect each other at 90o
AOB = BOC = COD = DOA = 90o
So, point O has to lie on the circle.
### Solution 6
We see that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral sum of opposite angles is 180o
AEC + CBA = 180o
AEC + AED = 180o (linear pair)
AED = CBA ... (1)
For a parallelogram opposite angles are equal.
From (1) and (2)
AD = AE (angles opposite to equal sides of a triangle)
### Solution 7
Let two chords AB and CD are intersecting each other at point O.
In AOB and COD
OA = OC (given)
OB = OD (given)
AOB = COD (vertically opposite angles)
AOB COD (SAS congruence rule)
AB = CD (by CPCT)
Similarly, we can prove AOD COB
Since in quadrilateral ACBD opposite sides are equal in length.
Hence, ACBD is a parallelogram.
We know that opposite angles of a parallelogram are equal
A = C
But A + C = 180o (ABCD is a cyclic quadrilateral)
A + A = 180o
A = 180o
A = 90o
As ACBD is a parallelogram and one of its interior angles is 90o, so it is a rectangle.
A is the angle subtended by chord BD. And as A = 90o, so BD should be diameter of circle. Similarly AC is diameter of circle.
### Solution 8
It is given that BE is the bisector of B
ABE =
But ADE = ABE (angles in same segment for chord AE)
Similarly, ACF = ADF = (angle in same segment for chord AF)
Similarly we can prove that
### Solution 9
AB is common chord in both congruent circles.
APB = AQB
Now in BPQ
APB = AQB
BP = BQ (angles opposite to equal sides of a triangle)
### Solution 10
Let perpendicular bisector of side BC and angle bisector of A meet at point D.
Let perpendicular bisector of side BC intersects it at E.
Perpendicular bisector of side BC will pass through circum centre O of circle. Now, BOC and BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.
We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
BOC = 2 BAC = 2A ... (1)
In BOE and COE
OE = OE (common)
OB = OC (radii of same circle)
OEB = OEC (Each 90o as OD BC)
BOE COE (RHS congruence rule)
BOE = COE (by CPCT) ... (2)
But BOE + COE = BOC
BOE +BOE = 2 A [Using equations (1) and (2)]
BOE = 2A
BOE = A
BOE = COE = A
The perpendicular bisector of side BC and angle bisector of A meet at point D.
BOD = BOE = A ... (3)
Since AD is the bisector of angle A
From equations (3) and (4), we have
It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.
Therefore, the perpendicular bisector of side BC and angle bisector of A meet on the circum circle of triangle ABC.
## CBSE Class 9 Frequently Asked Questions
NCERT Solutions for Class 9 Maths Chapter 10 is on Circles and is a crucial chapter as prescribed in the Class 9 syllabus by NCERT. The chapter Circle of Class 9 Maths includes an introduction to circles, the angle subtended by the chord, perpendicular from centre to chord, properties of equal chords and cyclic quadrilateral. There are five exercises in the Class 9 Maths Chapter 10 on circles. Apart from the NCERT Solutions drafted by experienced and qualified experts, students are recommended to go through the Sample papers, chapter notes, and question bank of the chapter, Circle.
TopperLearning's doubt solver "Ask a doubt" feature quickly assists all your queries.
We should always follow a national curricular or framework drafted by eminent experts and follow NCERT Solutions for Class 9 Maths Chapter 10. The type of questions included across the five exercises of the chapter provides good exposure and experience to deal with different types of questions that can be asked in the school exams. Practising from NCERT Solutions for Class 9 Maths Chapter 10 will enable students to practice and attempt similar questions that may be asked in various exams. Once students are clear on the conceptual part and aware of the pattern of problems asked from a specific chapter, it becomes handy to tackle challenging problems based on that concept. It is also evident that a new typology of questions is included in the examinations like Case-based questions.
Students can get any time access to NCERT Solutions for Class 9 Maths Chapter 10, Circles on the TopperLearning website, where they can practice all the five exercises. Eminent and qualified experts carefully draft the solutions to these problems.
To understand Circles, a student must know about chords, diameter, radius, bisector and much more. The chapter Circle of Class 9 Maths includes an introduction to circles, the angle subtended by the chord, perpendicular from centre to chord, properties of equal chords and cyclic quadrilateral. There are five exercises in the Class 9 Maths Chapter 10 on circles, giving students an idea of different types of question patterns required to handle any questions in the exam.
To solve the problems of Class 9 Maths Chapter 10 Circles, one must understand the concepts first. With TopperLearning resources like the concept and problem-solving application videos, chapter-wise notes and question banks, students can clarify the concepts and then try to solve the NCERT solution independently.
The critical theorems of Class 9 Chapter 10 Circles are listed below.
1. Theorem: Equal chords of a circle subtend equal angles at the centre.
2. Theorem: This is the converse of the previous theorem. It implies that if two chords subtend equal angles at the centre, they are equal.
3. Theorem: Equal chords of a circle are equidistant from the centre of a circle.
4. Theorem: This is the converse of the previous theorem. It states that chords equidistant from the circle's centre are equal in length.
5. Angle subtended by an arc at the centre of a circle is double that of an angle that arc subtends at any given point on the circle.
6. Theorem: Angles formed in the same circle segment are always equal in measure.
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# Quant Booster on Simple Interest for Teaching Exams
Updated : Jul 9, 2018, 15:30
By : Ashish
To make the chapter easy for you all, we are providing you all some Important Short Tricks to solve Simple Interest Questions which will surely make the chapter easy for you all.
Important Short Tricks to solve Simple Interest Questions:
Interest is the money paid by the borrower to the lender for the use of money lent.
The sum lent is called the Principal. Interest is usually calculated at the rate of so many rupees for every Rs.100 of the money lent for a year. This is called the rate percent per annum.
‘Per annum’ means for a year. The words ‘per annum’ are sometimes omitted. Thus, 6 p.c. means that Rs.6 is the interest on Rs.100 in one year.
The sum of the principal and interest is called the amount. The interest is usually paid yearly, half-yearly or quarterly as agreed upon.
Interest is of two kinds, Simple and Compound. When interest is calculated on the original principal for any length of time it is called simple interest. Compound interest is defined in the next chapter.
To find Simple Interest, multiply the principal by the number of years and by the rate percent and divide the result by 100.
This may be remembered in the symbolic form
Where I = Interest, p = principle, t = number of years, r = % rate
Ex. 1. Find the simple interest on Rs.400 for 5 years at 6 percent.
Solution:
Interest for a number of days
When the time is given in days or in years and days, 365 days are reckoned to a year. But when the time is given in months and days, 12 months are reckoned to a year and 30 days to the month. The day on which the money is paid back should be include be but not the day of which it is borrowed, ie, in counting, the first day is omitted.
Ex. 2. Find the simple interest on Rs.306. 25 from March 3rd to July 27th at per annum.
Solution:
Interest = Rs.
= = Rs. 4.59
To find the principal:
Since I =
Ex.3. What sum of money will produce Rs.143 interest in 3(1/4) years at 2.50 p.c. simple interest?
Solution:
Let the required sum be Rs. P. Then
Rs P =
To find rate %:-
Since I =
Ex. 4. A sum of Rs.468.75 was lent out at simple interest and at the end of 1 year 8 months the total amount was Rs500. Find the rate of interest per cent annum.
Solution:
Here, P =Rs468.75, t = or
I = Rs.(500-468.75) = Rs.31.25
rate p.c. =
To find Time:-
Since, I = P t r / 100
Ex. 5. In what time will Rs.8500 amount to Rs.15767.50 at per cent per annum?
Solution:
Here , interest = Rs.15767.50 – Rs.8500 =Rs.7267.50
Miscellaneous Examples of Simple Interest:
Ex.6: The simple interest on a sum of money is 1/9th of the principal, and the number of years is equal to the rate percent per annum. Find the rate percent.
Solution:
Let principal = P, time = t year, rate = t
Then,
Hence, rate = %
Direct formula:
Rate = time = %
Ex. 7: The rate of interest for the first 2 yrs is 3% per annum, for the next 3 years is 8% per annum and for the period beyond 5 years 10% per annum.To fetch an interest of 1520 in six years, money did he deposit?
Solution
Let his deposit be = Rs 100
Interest for first 2 yrs = Rs 6
Interest for next 3 yrs = Rs 24
Interest for the last year = Rs 10
Total interest = Rs 40
When interest is Rs 40, deposited amount is Rs 100
when interest is Rs 1520, deposited amount = = Rs 3800
Direct formula:
Principle = Interest*100/r1t1+r2t2+r3t3 = =Rs 3800
Ex.8: A sum of money doubles itself in 10 years at simple interest. What is the rate of interest?
Solution:
Let the sum be Rs 100
After 10 years it becomes Rs 200
Interest = 200 - 100 = 100
Then, rate = =
Direct formula:
Time × Rate = 100 (Multiple numbers of principal – 1)
Or, Rate =
Using the above formula rate =
Ex.9: A sum was put at a certain rate for 2 yrs. Had it been put at 3% higher rate, it would have fetched Rs 300 more. Find the sum.
Solution:
Let the sum be Rs x and the original rate be y% per annum. Then, new rate = (y+3) % per annum
xy +3x – xy =15,000 or, x =5000
Thus, the sum =Rs 5000
Direct Methods:
=
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# RD Sharma Solutions Class 7 Mathematics Solutions for Data Handling-Central Values Exercise 23.2 in Chapter 23 - Data Handling-Central Values
Question 1 Data Handling-Central Values Exercise 23.2
A die was thrown 20 times and the following scores were recorded: 5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1. Prepare the frequency table of the scores on the upper face of the die and find the mean score.
The frequency table for the given data is as follows:
X: 1 2 3 4 5 6
Y: 2 5 1 4 6 2
To compute arithmetic mean we have to prepare the following table:
Scores(Xi) Frequency(fi) Xifi
1 2 2
2 5 10
3 1 3
4 4 16
5 6 30
6 2 12
Total 20 73
Mean score = Σ fi xi/ Σ fi
= 73/20
= 3.65
Video transcript
"hello students welcome to leader learning india's best online classroom our question for today is we need to solve these two given equation so let's solve the first part of the given equation which is given x x minus 1 upon x plus 1 is equals to 2 x minus 5 upon 3 x minus 7 now if we cross multiply this we will get it as x minus 1 multiplied by 3x minus 7 will be equals to 2x minus 5 and x plus 1. correct so now if we solve this it will be open the bracket and multiply the values we will get it as 3 x squared minus 7x minus 3x plus 7 is equals to 2 x squared plus 2x minus 5x minus 5. so now if we rearrange this equation we will get it as x square minus 7x plus 12 is equals to zero now if we break this seven we will get the value as x square minus 4x minus 3x plus 12 is equals to 0 solving this we will get the value as x minus 4 x minus 3 is equal to 0 so if x minus 4 is equals to 0 it implies that value of x is equals to 4 and if x minus 3 is equals to 0 it implies that value of x is equal to 3 and these are the two values of x for the first given equation now let's solve the second equation second equation says that 1 upon x plus 2 plus 1 upon x is equals to 3 upon 4 if we take lcm it will comes out to be x into x plus 2 and here we will be having x x plus 2 value is again equals to 3 by 4 so the value will be equals to 2 x plus 2 now we if we cross multiply this simultaneously we will get 4 here and 3 multiplied by x multiplied by x plus 2 open the brackets all this solve the equation 4 x 2 is 8 x plus 8 is equals to 3 x multiplied by x comes out to be 3 x square plus 6 x the equation we get from here is 3 3x square minus 2x minus 8 is equals to 0. let's divide this value 2 into values so that the multiplication will be minus 8 and 3 that is 24 so we are dividing this value of minus 2 as minus 6x plus 4x minus 8 is equals to 0 all right so now let's take common 3x x minus 2 here will be common minus 4 sorry plus 4 x minus 2 it will be common again okay so the value is 3x plus 4 x minus 2 is equals to 0 now if we take 3x plus 4 is equal to 0 then it implies that value of x is minus 3 upon 4 and if x minus 2 is equals to 0 it implies that value of x is equals to 2 so here we get the we again get the two values of x 1 is x equals to minus three by four and another one is x equals to two so i hope that all your doubts are now cleared for more questions do subscribe leader learning thank you so much"
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Question
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height $16cm$ with radii of its lower and upper ends as $8cm$ and $20cm$, respectively. Find the cost of the milk which can completely fill the container, at the rate of $Rs.20$ per litre. Also find the cost of metal sheet used to make the container, if it costs $Rs.8per100c{m}^{2}$.
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Solution
It is given that,$R=20cm\phantom{\rule{0ex}{0ex}}r=8cm\phantom{\rule{0ex}{0ex}}h=16cm$Step 1 - Finding the cost of milk that it can contain.$\therefore Volumeoffrustum=\frac{1}{3}\mathrm{\pi h}\left({R}^{2}+{r}^{2}+Rr\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}×3.14×16\left({20}^{2}+{8}^{2}+20×8\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}×3.14×16\left(624\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=10449.92{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}$We know that $1c{m}^{3}=1ml$$10449.92c{m}^{3}=1×10499.92=10499.92ml$We also know that, $1000ml=1litre$Therefore, $10449.92ml=\frac{10499.92}{1000}litre=10.49992litre$$\therefore \mathrm{Cost}\mathrm{of}1\mathrm{litre}\mathrm{milk}=₹20\phantom{\rule{0ex}{0ex}}\mathrm{So}\mathrm{Cost}\mathrm{of}10499.92\mathrm{litre}\mathrm{milk}=20×10499.92=208.998$Hence, the cost of milk in the container is $₹208.992$.Step 2 - Finding the cost of the metal sheet required to make the container.Since, the container is open from the top. So, the surface area of the container = Curved surface area of container + Area of the base of the containerWe know the relation between the slant height, radii and height of the frustum is${l}^{2}={h}^{2}+{\left(R-r\right)}^{2}\phantom{\rule{0ex}{0ex}}{l}^{2}={16}^{2}+{\left(20-8\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{l}^{2}=256+144\phantom{\rule{0ex}{0ex}}l=\sqrt{400}\phantom{\rule{0ex}{0ex}}l=20cm$Now, $\begin{array}{rcl}Surfaceareaofthecontainer& =& Curvedsurfaceareaoffrustum+Areaofthebaseoffrustum\\ Surfaceareaofthecontainer& =& \mathrm{\pi }\left(\mathrm{R}+\mathrm{r}\right)\mathrm{l}+{\mathrm{\pi R}}^{2}\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& \mathrm{\pi }\left\{\left(\mathrm{R}+\mathrm{r}\right)\mathrm{l}+{\mathrm{R}}^{2}\right\}\\ & & \\ \mathrm{Taking}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\pi }& =& 3.14\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& 3.14\left\{\left(20+8\right)20+{8}^{2}\right\}\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& 1959.36{\mathrm{cm}}^{2}\end{array}$Now, the metal sheet used to make the container is $1959.36c{m}^{2}$.The cost of $100c{m}^{2}$ of the sheet = $₹8$The cost of $1959.36c{m}^{2}$ of the sheet = $\frac{8×1959.36}{100}=156.748$Therefore, the cost of the metal sheet used to make the container is $₹156.748$. Hence, the cost of milk in the container is $₹208.992$ and the cost of the metal sheet used to make the container is $₹156.748$.
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# Aptitude Percentage Test Paper 6
26) If the length of a rectangle is increased by 37.5% and its breadth is decreased by 20%, find the change in its area.
1. 15% increase
2. 13% decrease
3. 10% increase
4. 10% decrease
Explanation:
Area of rectangle = length * breadth
Length is Increased by 37.5% = 37(1/2)* 1/100 = 3/8
Breath is decreased by 20% = 1/5
l * b = Area
Old: 8 * 5 = 40
New: 11 * 4 = 44
Difference = 44-40 = 4
Original = 40
The percentage change is equal to (difference / original) * 100.
(4/40) *100 = 10% increase.
27) The price of a product is reduced by 25%, but the daily sale of the article is increased by 30%. Find the net effect on the daily sale.
1. 2.5% increase
2. 1.5% decrease
3. 2% increase
4. 2.5% decrease
Explanation:
We can say that the price of one unit * no. of units sold = Total Sale
Let the initial price of product = 100, sale =100 items.
Then, Total Sale = 100 * 100 = 10000
ATQ, price reduces to 25%.
So, the new price = 100-25 = 75
Sale increase to 30%,
So, the new sale = 100+30= 130
Now, the new total sale = 75* 130 = 9750
Difference = 10000 - 9750 = 250
% net effect = (diff/original)* 100
= (250/10,000) * 100 = 2.5
So, the net effect on daily sale = 2.5% decrease
28) A mixture of 20 kg of milk and water contains 90% milk. How many kilograms of water should be added so that water becomes 40%?
1. 10 kg
2. 20 kg
3. 12 kg
4. 8 kg
Explanation:
Milk : water
Original Percentage: (90% : 10%)
Or 9 : 1
Total milk = 9 + 1 = 10 units of milk
We have 10 units = 20 kg
So, 1 unit = 20/10 = 2 kg.
New Percentage:60% : 40%
Or, (3 : 2)
On multiplying by 3 to make the ratio of milk equal, as the quality of milk remains the same, the ratio of water becomes 6.
(3 : 2) * 3
Or, 9 : 6 (ratio of water)
Now, the difference between the old and new ratio of water = 6 - 1 = 5 units.
1 unit = 2 kg
So, the required kg of water = 5 * 2 = 10 kg
29) The population of a town increases by 16(2/3) % in the first year, decreases by 37.5% in the second year, and increases by 57(1/7) % in the third year. The population of the town becomes 55000 after these 3 years. Find the initial population of the town 3 years ago.
1. 52000
2. 50000
3. 48000
4. 56250
Explanation:
Let the initial population = P
16(2/3) % = 16(2/3) * 1/100 = 1/6
Similarly, 37.5 % = 3/8
57(1/7) % = 4/7
1st-year 2nd-year 3rd-year
(+1/6) (-3/8) (+4/7)
Concept: 50% =
The above concept shows that when old value is 2 units then a 50% increment means an increase of one unit, i.e. the old value becomes 3 after a 50% increment.
I.e. old value = 2 new value after 50% increment = 2+1 + 3
Now, let the initial population P.
As per question:
After an increment of 1/6, the new value will be 7/6.
And, after a decrease of -3/8, the new value will be 5/8.
And, after an increase of 4/7, the new value will be 11/7.
Now, as per question = [[[P * (7/6)] * (5/8)] * (11/7)] = 55000
So, the initial population, P = 48000
30) The salary of a man per hour is increased by 20%, and his working hours is decreased by 25%. If the original income is Rs. 4000. Find the new income.
1. 2000
2. 2500
3. 3000
4. 3600
Explanation:
Income = per hour salary * duration
Increase in per hour salary = 20% = 1/5
Working hours are reduced by 25% = - ¼
Per hour salary * duration = Income
Use Concept: 50%
The above concept shows that when old value is 2 units then a 50% increment means an increase of one unit, i.e. the old value becomes 3 after a 50% increment.
I.e. old value = 2 new value after 50% increment = 2+1 + 3
AS per question:
An increase of 20 % in per hour salary = 6/5
An decrease of 25% = 3/4
Now, as per above two values:
Old income: 5 (per hour salary) * 4 (working hour) = 20 (but the original income is given 4000)
Similarly, new income: 6 * 3 = 18
Now, we multiply 20 with 200 to make it = 4000
Then, 18 is also multiplied with 200 to find the required new income. 18 * 200 = 3600
Percentage Aptitude Test Paper 1
Percentage Aptitude Test Paper 2
Percentage Aptitude Test Paper 3
Percentage Aptitude Test Paper 4
Percentage Aptitude Test Paper 5
Percentage Aptitude Test Paper 7
Percentage Aptitude Test Paper 8
Percentage Concepts
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# Readers ask: What is the prime factorization?
## How do you find the prime factorization?
Step 1: Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Step 2: Write the number as a product of prime numbers.
## What is the prime factorization mean in math?
” Prime Factorization ” is finding which prime numbers multiply together to make the original number.
## What is prime factorization example?
A factor that is a prime number. In other words: any of the prime numbers that can be multiplied to give the original number. Example: The prime factors of 15 are 3 and 5 (because 3×5=15, and 3 and 5 are prime numbers).
## What is the prime factorization of 42?
The only way to write 42 as the product of primes (except to change the order of the factors ) is 2 × 3 × 7. We call 2 × 3 × 7 the prime factorization of 42. It turns out that every counting number (natural number) has a unique prime factorization, different from any other counting number.
## What is the prime factorization of 63?
So, the prime factorisation of 63 is 3 × 3 × 7 or 32 × 7, where 3 and 7 are the prime numbers.
## What is the prime factorization of 77?
The factors of 77 by prime factorization are 1, 7, 11, and 77. Here, 1, 7, and 11 are prime factors of 77.
## What is the difference between prime factorization and factorization?
Factors are numbers that can be multiplied together to make another number. Prime numbers are numbers that have exactly two factors, 1 and itself (i.e. 2, 3, 5, 7, 11,.). So, prime factorization is writing the prime numbers that will multiply together to make a new number as a multiplication problem.
You might be interested: Quick Answer: What is the big bang?
## What is the prime factorization of 20?
Since 2 is prime, the prime factorization of 20 is 2 * 2 * 5.
## What is the prime factorization of 144?
So, the prime factors are written as 2 x 2 x 2 x 2 × 3 x 3 or 24 x 32, where 2 and 3 are the prime numbers. It is possible to find the exact number of factors of a number 144 with the help of prime factorisation. The prime factor of the 144 is 24 x 32. The exponents in the prime factorisation are 4 and 2.
## What is a prime factorization of 12?
Prime Factors of 12 = 2 X 2 X 3.
## What is the prime factorization of 80?
The factors of 80 by the prime factorization method are 1, 2, 5, 10, 20, 40, and 80. Here, 2 and 5 are the prime factors of 80.
## What is the prime factorization of 11?
The prime factorization of 11 is 1 × 11. The number 11 is a prime number, because its only factors are 1 and itself.
## What are the 2 prime factors of 12?
The prime factors of 12 are 2 and 3.
## What is the GCF of 27 and 45?
We found the factors and prime factorization of 27 and 45. The biggest common factor number is the GCF number. So the greatest common factor 27 and 45 is 9.
## What are the factor of 42?
Therefore, the factors of 42 in pairs are (1, 42 ), (2, 21), (3, 14) and (6, 7).
1 month ago
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Lesson Video: Perpendicular Bisector Theorem and Its Converse | Nagwa Lesson Video: Perpendicular Bisector Theorem and Its Converse | Nagwa
# Lesson Video: Perpendicular Bisector Theorem and Its Converse Mathematics
In this video, we will learn how to use the perpendicular bisector theorem and its converse to find a missing angle or side in an isosceles triangle.
14:58
### Video Transcript
In this video, we will learn how to use the perpendicular bisector theorem and its converse to find a missing angle or side in a triangle. To do that, let’s think about the definitions we’re working with, starting with perpendicular. Two line segments, rays, lines, or any combination of those that meet at a 90-degree right angle are perpendicular. Perpendicular lines are lines that meet at a 90-degree angle. A bisector is an object — a line, ray, or line segment — that cuts another object, usually an angle or a line segment, into two equal parts. We can therefore say that a perpendicular bisector is a line, a ray or, a line segment that bisects another line segment at a right angle.
Now that we thought a bit about what a perpendicular bisector is, we can consider the perpendicular bisector theorem. If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. Let’s see if we can sketch what this means. First, we’ll need a line segment, and then we have a perpendicular bisector. The perpendicular bisector forms a right angle with our line segment and divides the line segment in half.
Now we’re considering a point that is on this perpendicular bisector. This theorem is telling us that point will be equidistant from the endpoints of the line segment. On this diagram, that is showing that the lines created from our point to our endpoint, the yellow lines, will be equal to each other. One way to prove that this is true is to consider the two smaller triangles created by this perpendicular bisector. To do this, let’s go ahead and label the points on the triangle.
We have the larger triangle, triangle 𝐴𝐵𝐶. We then have the smaller triangle, triangle 𝐴𝐵𝐷, and the smaller triangle, triangle 𝐴𝐶𝐷. Because these two smaller triangles share their third side, we can say line segment 𝐴𝐷 is congruent with line segment 𝐴𝐷. We’ve shown they have a congruent side. And because we’re dealing with a perpendicular bisector, we know that line segment 𝐵𝐶 is congruent to line segment 𝐷𝐶, which means we have two congruent sides in these triangles. And between those two congruent sides, we have two congruent angles. Angle 𝐴𝐷𝐵 is congruent to angle 𝐴𝐷𝐶. They’re both 90-degree angles.
Using the side-angle-side theorem of triangles, we confirm that these two smaller triangles are congruent. And that’s another way of confirming that these two sides must be equal to each other. It’s a confirmation of the perpendicular bisector theorem. Here’s one place you might see something like this played out. If we say that our radio tower is the perpendicular bisector, we certainly want the radio tower to be forming a right angle with the ground. As you don’t want it leaning to the left or to the right, you’ll notice that it has what’s called guy wires, which add stability to the free standing structure.
If the point on the ground where we attach the guy wires are of equal distance from the perpendicular bisector, then we know that the wires on either side will be equal to each other in length. It does not mean that every set will be equal in length. Each set of wires will be equal in length to the other wire that goes to the same point on the tower.
Before we move on and look at some examples, we also want to consider the converse of the perpendicular bisector theorem. Remember, to find the converse means switching the hypothesis and conclusion of a conditional statement. In the case of the perpendicular bisector theorem, its converse says this. If a point is equidistant from the endpoints of a line segment, then the point is on the perpendicular bisector of the segment. If we start with our point and that point is equidistant to each endpoint of the line segment, this point must fall along the perpendicular bisector.
While the converse might not immediately seem true, one way to show this is to consider when the point is not equidistant from both endpoints. Imagine if this was our point and then we drew lines to the endpoints. We can see very clearly that these two line segments are not equal to one another. In addition to that, even if we drew a line that was a right angle with the base, it could not be a bisector because we know that a bisector must divide the line segment that it intersects in half. This confirms that in order for a point to fall on the perpendicular bisector of a segment, it must be equidistant from the endpoints of that line segment. Now we’re ready to consider some example problems.
In the following figure, find the length of line segment 𝑊𝑌.
The first thing we wanna do here is take stock of what we’re given in the figure. We have a triangle 𝑋𝑍𝑊. The length of line segment 𝑋𝑊 is equal to the length of line segment 𝑋𝑍, which is equal to 17. We could also say that this is therefore an isosceles triangle. In addition to that, we know that angle 𝑋𝑌𝑍 is a right angle. Based on this given information, we can draw some conclusions. Because we know that line segment 𝑋𝑊 is equal in length to line segment 𝑋𝑍 and we know that angle 𝑋𝑌𝑍 is 90 degrees, we can say that line segment 𝑋𝑌 is a perpendicular bisector.
We can make this claim based on the converse of the perpendicular bisector theorem. That tells us if a point is equidistant from the endpoints of a segment — for us, the point 𝑋 is equidistant from 𝑊 and 𝑍 — then the point is on the perpendicular bisector of the segment, which means we can say that line segment 𝑌𝑍 will be equal in length to line segment 𝑊𝑌, because of the definition of the perpendicular bisector, which divides the line segment it intersects in half. Because line segment 𝑌𝑍 equals 11, we can say that line segment 𝑊𝑌 will also be equal to 11.
We’re ready to look at another example.
In the diagram, 𝐴𝐵 equals six and 𝐵𝐷 equals five. Find 𝐴𝐶 and find 𝐶𝐷.
A good place to start is listing out the information we’re given. We have triangle 𝐴𝐵𝐶. And in that triangle, line segment 𝐵𝐷 is equal to line segment 𝐷𝐶. We’ve also been told that the measure of angle 𝐴𝐷𝐶 equals 90 degrees. It’s a right angle. We also know that 𝐴𝐵 measures six and 𝐵𝐷 measures five. From this information, we can draw some conclusions. First of all, we already know that line segment 𝐷𝐶 is equal in length to line segment 𝐵𝐷, which means we can say that line segment 𝐷𝐶 also measures five. And we can write the line segment either way. 𝐶𝐷 will be equal to 𝐷𝐶.
So we can say, first of all, that 𝐶𝐷 equals five. To find 𝐴𝐶, we’ll need to think a little bit more carefully about what we know. We know that the measure of angle 𝐴𝐷𝐶 is 90 degrees, and we know that the point 𝐷 is halfway between 𝐵 and 𝐶. This means we can say that 𝐴𝐷 is a perpendicular bisector of line segment 𝐵𝐶. We can say that this is true based on the definition of a perpendicular bisector. A perpendicular bisector has to divide the line segment in half and meet at a 90-degree angle.
And since we know that line 𝐴𝐷 is a perpendicular bisector of 𝐵𝐶, we can apply the perpendicular bisector theorem, which tells us that if a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints. Since 𝐴𝐷 is a perpendicular bisector, 𝐴𝐵 is equal in length to 𝐴𝐶. And if line segment 𝐴𝐶 is equal in length to line segment 𝐴𝐵, since line segment 𝐴𝐵 measures six, line segment 𝐴𝐶 will also measure six.
In our next example, we’ll use the properties of perpendicular bisectors to find a missing angle instead of a missing side length.
Find the measure of angle 𝐷𝐴𝐵.
When we’re given questions like this, it’s always good to start with what we’re given. We have triangle 𝐴𝐵𝐶. In this triangle, line segment 𝐴𝐶 is equal in length to line segment 𝐴𝐵. We know the measure of angle 𝐴𝐷𝐵 equals 90 degrees, and we know that the measure of angle 𝐶𝐴𝐷 equals 25 degrees. We want to know the measure of angle 𝐷𝐴𝐵. That’s this angle. To do that, we take the information we were given and draw some conclusions. Because line segment 𝐴𝐶 is equal to line segment 𝐴𝐵 and because the measure of angle 𝐴𝐷𝐵 is 90 degrees, we can say that line segment 𝐴𝐷 is a perpendicular bisector.
We based that on the converse of the perpendicular bisector theorem, which tells us that if a point is equidistant from the ends of two line segments — for us, that would be the line segments 𝐴𝐶 and 𝐴𝐵 that are equal — then the point 𝐴 must fall along the perpendicular bisector. Because line segment 𝐴𝐷 is a perpendicular bisector, we can say that line segment 𝐶𝐷 is equal in length to line segment 𝐵𝐷. We know that this perpendicular bisector creates two smaller triangles. And we can say that the smaller triangle 𝐴𝐷𝐶 must be congruent to the smaller triangle 𝐴𝐷𝐵.
We say this based on side-side-side congruence. Three sides of triangle 𝐴𝐷𝐶 are equal to the corresponding three sides of triangle 𝐴𝐷𝐵. And since these two triangles are congruent, we can say that the measure of angle 𝐶𝐴𝐷 will be equal to the measure of angle 𝐷𝐴𝐵. We’re saying these two angles are congruent, which makes the measure of angle 𝐷𝐴𝐵 25 degrees.
In our final example, we’ll need to use the properties of a perpendicular bisector to find the value of 𝑋.
In the diagram, line segment 𝐴𝐷 is the perpendicular bisector of line segment 𝐵𝐶. Find the value of 𝑥.
Since we know that line segment 𝐴𝐷 is a perpendicular bisector, we can add a few details to our diagram. The angle created at this intersection will be a right angle. And the line segment 𝐴𝐷 bisects line segment 𝐵𝐶, making 𝐵𝐷 equal to 𝐵𝐶. At this point, it does not seem like we have enough information to find the value of 𝑥, which is where the perpendicular bisector theorem comes into play. It says if a point is on the perpendicular bisector of a line segment — for us, that point could be 𝐴 — then that point is equidistant from the endpoints of the line segment. The endpoints of our line segment are 𝐵 and 𝐶. And 𝐴 is equidistant to point 𝐵 and point 𝐶, which makes line segment 𝐴𝐵 equal in length to line segment 𝐴𝐶.
And using this fact, we can set up an equation to solve for 𝑥. If 𝐴𝐵 is equal to 𝐴𝐶, then three 𝑥 plus one must be equal to five 𝑥 minus 12. And at this point, we need only to solve for 𝑥. First, we’ll subtract three 𝑥 from both sides to get both 𝑥-values on the same side of the equation. And we’ll have one equals two 𝑥 minus 12. From there, we add 12 to both sides of the equation. And we have the equation 13 equals two 𝑥, which means we need to divide both sides of the equation by two. 13 divided by two is six and a half, which means we can say that 𝑥 equals six and a half.
Before we finish, let’s do a quick review of the key points. We just have two main key points from this section. And that is the perpendicular bisector theorem tells us if a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment. And then the converse of the perpendicular bisector theorem, which is also true, if a point is equidistant from the endpoints of a line segment, then the point is on the perpendicular bisector of the segment. These two properties often help us solve for missing side lengths or angle measures inside triangles.
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Class 10 - Maths - Coordinate Geometry
Exercise 7.1
Question 1:
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b)
(i) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(4 – 2)2 + (1 - 3)2}
=> D = √{22 + (-2)2}
=> D = √{4 + 4}
=> D = √8
=> D = 2√2
(ii) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(-1 + 5)2 + (3 - 7)2}
=> D = √{42 + (-4)2}
=> D = √{16 + 16}
=> D = √32
=> D = 4√2
(iii) Distance between two points (x1, y1) and (x2, y2) is given as:
√{(x2 – x1)2 + (y2 – y1)2}
Therefore, the distance between two points (2, 3) and (4, 1) is
D = √{(-a - a)2 + (-b - b)2}
=> D = √{(-2a)2 + (-2b)2}
=> D = √{4a2 + 4b2}
=> D = 2√(a2 + b2)
Question 2:
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
The distance between the points (0, 0) and (36, 15) = √{(36 - 0)2 + (15 - 0)2}
= √(362 + 152)
= √(1296 + 225)
= √1521
= 39
Yes, we can find the distance between the given towns A and B.
Let town A at origin i.e. point (0, 0)
Therefore, town B will be at point (36, 15) with respect to town A.
Hence, the distance between the given town A and B will be 39 km.
Question 3:
Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.
Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given
triangle respectively.
Let A = (1, 5), B = (2, 3) and C = (-2, -11)
Now, AB = √{(1 - 2)2 + (5 - 3)2} = √{(-1)2 + 22} = √(1 + 4) = √5
BC = √{(-2 - 2)2 + (-11 - 3)2} = √{(-4)2 + (-14)2} = √(16 + 196) = √212
CA = √{(-2 - 1)2 + (-11 - 5)2} = √{(-3)2 + (-16)2} = √(9 + 256) = √265
Since AB + BC ≠ CA
Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear.
Question 4:
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given
triangle respectively.
Now, AB = √{(5 - 6)2 + (-2 - 4)2} = √{(-1)2 + (-6)2} = √(1 + 36) = √37
BC = √{(6 - 7)2 + (4 + 2)2} = √{(-1)2 + 62} = √(1 + 36) = √37
CA = √{(5 - 7)2 + (-2 + 2)2} = √{(-2)2 + 0} = √4 = 2
Therefore, AB = BC
Since two sides are equal in length, therefore, ABC is an isosceles triangle.
Question 5:
In a classroom, 4 friends are seated at the points
A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes
Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.
Using distance formula, find which of them is correct.
It can be observed that A (3, 4), B (6, 7), C (9, 4), and
D (6, 1) are the positions of these 4 friends.
Now, AB = √{(3 - 6)2 + (4 - 7)2} = √{(-3)2 + (-3)2} = √(9 + 9) = √18 = 3√2
BC = √{(6 - 9)2 + (7 - 4)2} = √{(-3)2 + 32} = √(9 + 9) = √18 = 3√2
CD = √{(9 - 6)2 + (4 - 1)2} = √{32 + 32} = √(9 + 9) = √18 = 3√2
AD = √{(3 - 6)2 + (4 - 1)2} = √{(-3)2 + 32} = √(9 + 9) = √18 = 3√2
Diagonal AC = √{(3 - 9)2 + (4 - 4)2} = √{(-6)2 + 0} = 6
Diagonal BD = √{(6 - 6)2 + (7 - 1)2} = √{0 + 62} = 6
It can be observed that all sides of this quadrilateral ABCD are of the same length and also the
diagonals are of the same length.
Therefore, ABCD is a square and hence, Champa is correct.
Question 6:
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B, C, and D
Now, AB = √{(-1 - 1)2 + (-2 - 0)2} = √{(-2)2 + (-2)2} = √(4 + 4) = √8 = 2√2
BC = √{(1 + 1)2 + (0 - 2)2} = √{22 + (-2)2} = √(4 + 4) = √8 = 2√2
CD = √{(-1 + 3)2 + (2 - 0)2} = √{22 + 22} = √(4 + 4) = √8 = 2√2
AD = √{(-1 + 3)2 + (-2 - 0)2} = √{22 + (-2)2} = √(4 + 4) = √8 = 2√2
Diagonal AC = √{(-1 + 1)2 + (-2 - 2)2} = √{0 + (-4)2} = √16 = 4
Diagonal BD = √{(1 + 3)2 + (0 - 0)2} = √(42 + 0) = √16 = 4
It can be observed that all sides of this quadrilateral are of the same length and also, the
diagonals are of the same length. Therefore, the given points are the vertices of a square.
(ii) Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D
Now, AB = √{(-3 - 3)2 + (5 - 1)2} = √{(-6)2 + 42} = √(36 + 16) = √52 = 2√13
BC = √{(3 - 0)2 + (1 - 3)2} = √{32 + (-2)2} = √(9 + 4) = √13
CD = √{(0 + 1)2 + (3 + 4)2} = √{12 + 72} = √(1 + 49) = √50 = 5√2
AD = √{(-3 + 1)2 + (5 + 4)2} = √{(-2)2 + 92} = √(4 + 81) = √85
It can be observed that all sides of this quadrilateral are of different lengths.
Therefore, it can be said that it is only a general quadrilateral, and not specific such as square,
rectangle, etc.
(iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of
Now, AB = √{(4 - 7)2 + (5 - 6)2} = √{(-3)2 + (-1)2} = √(9 + 1) = √10
BC = √{(7 - 4)2 + (6 - 3)2} = √{32 + 32} = √(9 + 9) = √18
CD = √{(4 - 1)2 + (3 - 2)2} = √{32 + 12} = √(9 + 1) = √10
AD = √{(4 - 1)2 + (5 - 2)2} = √{32 + 32} = √(9 + 9) = √18
Diagonal AC = √{(4 - 4)2 + (5 - 3)2} = √{0 + 22} = √4 = 2
Diagonal BD = √{(7 - 1)2 + (6 - 2)2} = √(62 + 42) = √(36 + 16) = √52 = 13√2
It can be observed that opposite sides of this quadrilateral are of the same length. However,
the diagonals are of different lengths. Therefore, the given points are the vertices of a
parallelogram.
Question 7:
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Let the point on x-axis is (x, 0)
Now √{(2 - x )2 + (-5 - 0) 2} = √{(-2 - x)2 + (-9 - 0) 2}
Squaring on both sides, we get
=> (2 - x)2 + (-5 - 0)2 = (-2 - x)2 + (-9 - 0)2
=> (2 - x)2 + (-5) 2 = (-2 - x)2 + (-9)2
=> 4 + x2 - 4x + 25 = 4 + x2 + 4x + 81
=> - 4x + 25 = 4x + 81
=> 4x + 4x = 25 - 81
=> 8x = -56
=> x = -56/8
=> x = -7
So, the point on x-axis is (-7, 0)
Question 8:
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
It is given that the distance between (2, −3) and (10, y) is 10.
Therefore, √{(2 - 10)2 + (-3 - y)2} = 10
=> √{(-8)2 + (-3 - y)2} = 10
=> √{64 + (-3 - y)2} = 10
Squaring on both sides, we get
=> 64 + (-3 - y)2 = 100
=> (-3 - y)2 = 100 – 64
=> (-3 - y)2 = 36
=> (3 + y)2 = 36
=> 3 + y = ±√36
=> 3 + y = ±6
=> 3 + y = 6 or 3 + y = -6
=> y = 3 or -9
Question 9:
If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Q(0, 1) is equidistant from P(5, -3) and R(x, 6).
Therefore, QP = QR
=> √{(5 - 0)2 + (-3 - 1)2} = √{(x - 0)2 + (6 - 1)2}
=> √(25 + 16) = √{x2 + 25}
Squaring on both sides, we get
=> 25 + 16 = x2 + 25 => x2 = 16 => x = ±4
If x = 4,
QR = √{(4 - 0)2 + (6 - 1)2} = √(16 + 25) = √41
QR = √{(4 - 5)2 + (6 + 3)2} = √(1 + 81) = √82
If x = -4,
QR = √{(-4 - 0)2 + (6 - 1)2} = √(16 + 25) = √41
QR = √{(-4 - 5)2 + (6 + 3)2} = √(81 + 81) = √162 = 9√2
Question 10:
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4).
Point P(x, y) is equidistant from A(3, 6) and B(-3, 4).
Therefore, PA = PB
=> √{(3 - x)2 + (6 - y)2} = √{(-3 - x)2 + (4 - y)2}
Squaring on both sides, we get
=> (3 - x)2 + (6 - y)2 = (-3 - x)2 + (4 - y)2
=> 9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x + 16 + y2 – 8y
=> -12x – 4y = -20
=> 3x + y = 5
This is the required relation between x and y.
Exercise 7.2
Question 1:
Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Let P(x, y) be the required point. Using the section formula, we obtain
x = {2 * 4 + 3 * (-1)}/(2 + 3) = (8 - 3)/5 = 5/5 = 1
y = {2 * (-3) + 3 * 7}/(2 + 3) = (-6 + 21)/5 = 15/5 = 3
Therefore, the point is (1, 3).
Question 2:
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given
points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1 : 2.
Now, x1 = {1 * (-2) + 2 * 4}/(1 + 2) = (-2 + 8)/5 = 6/3 = 2
y1 = {1 * (-3) + 2 * (-1)}/(1 + 2) = (-3 - 2)/3 = -5/3
Therefore, P (x1, y1) = (2, 5/3)
Point Q divides AB internally in the ratio 2 : 1.
Now, x2 = {2 * (-2) + 1 * 4}/(2 + 1) = (-4 + 4)/5 = 0
y2 = {2 * (-3) + 1 * (-1)}/(2 + 1) = (-6 - 1)/3 = -7/3
Therefore, Q (x2, y2) = (0, -7/3)
Question 3:
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder
at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD,
as shown in Fig.7.12. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the
eighth line and posts a red flag. What is the distance between both the flags?
If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
It can be observed that Niharika posted the green flag at of the distance AD i.e., (1/4 * 100)
= 25 m from the starting point of 2nd line. Therefore, the coordinates of this point G is (2, 25).
Similarly, Preet posted red flag at of the distance AD i.e., m from the (1/5 * 100) = 25 m from
the starting point of 8th line. Therefore, the coordinates of this point R are (8, 20).
Distance between these flags by using distance formula = GR
= √{(8 - 20)2 + (25 - 20)2} = √ (36 + 25) = √61 m
The point at which Rashmi should post her blue flag is the mid-point of the line joining these
points. Let this point be A (x, y).
Now, x = (2 + 8)/2 = 12/2 = 5
and y = (25 + 20)/2 = 45/2 = 22.5
Hence, A(x, y) = (5, 22.5)
Therefore, Rashmi should post her blue flag at 22.5m on 5th line.
Question 4:
Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be
k : 1.
Therefore, -1 = (6k - 3)/(k + 1)
=> -(k + 1) = 6k – 3
=> -k – 1 = 6k – 3
=> 3 – 1 = 6k + k
=> 7k = 2
=> k = 2/7
Hence, the required ratio = 2/7 : 1 = 2 : 7
Question 5:
Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x-axis be
K : 1.
Therefore, the coordinates of the point of division = {(-4k + 1)/(k + 1), (5k - 5)/(k + 1)}
We know that y-coordinate of any point on x-axis is 0.
=> (5k - 5)/(k + 1) = 0
=> 5k – 5 = 0
=> 5k = 5
=> k = 5/5
=> k = 1
Therefore, x-axis divides it in the ratio 1 : 1.
Division point = {(-4 * 1 + 1)/(1 + 1), (5 * 1 - 5)/(1 + 1)}
= {(-4 + 1)/2, (5 - 5)/2}
= (-3/2, 0)
Question 6:
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram
ABCD. Intersection point O of diagonal AC and BD also divides these diagonals.
Therefore, O is the mid-point of AC and BD.
If O is the mid-point of AC, then the coordinates of O are
{(1 + x)/2, (2 + 6)/2} = {(1 + x)/2, 4}
If O is the mid-point of BD, then the coordinates of O are
{(4 + 3)/2, (5 + y)/2} = {7/2, (5 + y)/2}
Since both the coordinates are of the same point O,
So, (1 + x)/2 = 7/2 and 4 = (5 + y)/2
=> 1 + x = 7 and 5 + y = 8
=> x = 6 and y = 3
Question 7:
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Let the coordinates of point A be (x, y).
Mid-point of AB is (2, −3), which is the center of the circle.
So, (2, -3) = {(x + 1)/2, (y + 4)/2]
=> (x + 1)/2 = 2 and (y + 4)/2 = -3
=> x + 1 = 4 and y + 4 = -6
=> x = 3 and y = -10
Therefore, the coordinate of A is (3, -10).
Question 8:
If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3AB/7 and P lies on the line segment AB.
The coordinates of point A and B are (−2, −2) and (2, −4) respectively.
Since AP = 3AB/7
=> AP/AB = 3/7
Therefore, AP : PB = 3 : 4
Point P divides the line segment AB in the ratio 3 : 4.
Now, coordinate of P = [{3 * 2 + 4 * (-2)}/(3 + 4), {3 * (-4) + 4 * (-2)}/(3 + 4)]
= {(6 - 8)/7, (-12 - 8)/7}
= (-2/7, -20/7)
Question 9:
Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio
1 : 3, 1 : 1, 3 : 1 respectively.
Now, coordinate of P = [{1 * 2 + 3 * (-2)}/(1 + 3), {1 * 8 + 3 * 2}/(1 + 3)]
= {(2 - 6)/4, (8 + 6)/4}
= (-4/4, 14/4)
= (-1, 7/2)
Coordinate of Q = [{2 + (-2)}/2, (2 + 8)/2] = (0, 10/2) = (0, 5)
Coordinate of R = [{3 * 2 + 1 * (-2)}/(3 + 1), {3 * 8 + 1 * 2}/(3 + 1)]
= {(3 - 2)/4, (24 + 2)/4}
= (1/4, 26/4)
= (1/4, 13/2)
Question 10:
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint: Area of a rhombus = (product of its diagonals)/2]
Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD.
Length of diagonal AC = √[{3 – (-1)}2 + (0 - 4)2]
= √[(3 + 1)2 + (- 4)2]
= √(16 + 16)
= √32
= 4√2
Length of diagonal BD = √[{4 – (-2)}2 + {5 – (-1)}2]
= √[(4 + 2)2 + (5 + 1)2]
= √(36 + 36)
= √72
= 6√2
Therefore, area of rhombus = (1/2) * AC * BD
= (1/2) * 4√2 * 6√2
= 48/2
= 24 square units
Exercise 7.3
Question 1:
Find the area of the triangle whose vertices are: (i) (2, 3), (–1, 0), (2, – 4) (ii) (–5, –1), (3, –5), (5, 2)
(i) Vertices of triangle are: A(2, 3), B(–1, 0) and C(2, – 4)
Using the formula for area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Now, area of triangle ABC = (1/2)[2(0 + 4) + (-1)(-4 – 3) + 2(3 – 0)]
= (1/2)[8 + 7 + 6]
= 21/2
= 10.5 square units
(ii) Vertices of triangle are: A(–5, –1), B(3, –5), C(5, 2)
Using the formula for area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Now, area of triangle ABC = (1/2)[(-5)(-5 - 2) +3(2 + 1) + 5(-1 + 5)]
= (1/2)[35 + 9 + 20]
= 64/2
= 32 square units
Question 2:
In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)
(i) A(7, –2), B(5, 1), C(3, k)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle ABC = 0
=> (1/2)[7(1 - k) + 5(k + 2) + 3(-2 - 1)] = 0
=> 7 – 7k + 5k + 10 – 9 = 0
=> -2k = -8
=> k = 4
(ii) A(8, 1), B(k, – 4), C(2, –5)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle ABC = 0
=> (1/2)[8(-4 + 5) + k(-5 - 1) + 2(1 + 4)] = 0
=> 8 – 6k + 10 = 0
=> -6k = -18
=> k = 3
Question 3:
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3).
Find the ratio of this area to the area of the given triangle.
Let the vertices of triangle be A(0, -1), B(2, 1), C(0, 3).
Let D, E and F be the mid-points of the sides of the triangle.
Now, coordinates of d, E and F are given as:
D = {(0 + 2)/2, (-1 + 1)/2} = (1, 0)
D = {(0 + 0)/2, (3 - 1)/2} = (0, 1)
D = {(2 + 0)/2, (1 + 3)/2} = (1, 2)
Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of triangle DEF = (1/2)[1(2 - 1) + 1(1 - 0) + 0(0 - 2)]
= (1/2)[1 + 1]
= 1 square units
Area of triangle ABC = (1/2)[0(1 - 3) + 2(3 + 1) + 0(-1 - 1)]
= (1/2)[0 + 8]
= 4 square units
Therefore, required ratio = 1 : 4
Question 4:
Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).
Let the vertices of the quadrilateral be A(– 4, – 2), B(– 3, – 5), C(3, – 2) and D(2, 3).
Join AC to form two triangles ABC and ACD.
Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of triangle ABC = (1/2)[(-4)(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)]
= (1/2)[12 + 0 + 9]
= 21/2 square units
Area of triangle ACD = (1/2)[(-4)(-2 + 3) + 3(3 + 2) + 2(-2 + 2)]
= (1/2)[20 + 15 + 0]
= 35/2 square units
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= 21/2 + 35/2
= (21 + 35)/2
= 56/2 = 28 square units
Question 5:
You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas.
Verify this result for Δ ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).
Given, the vertices of Δ ABC are A(4, – 6), B(3, –2) and C(5, 2).
Let D be the mid-point of side BC to Δ ABC.
Therefore, AD is the median in Δ ABC.
Coordinate of point D = {(3 + 5)/2, (-2 + 2)/2} = (4, 0)
Now, area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of triangle ABD = (1/2)4(-2 - 0) + 3(0 + 6) + 4(-6 + 2)]
= (1/2)[-8 + 18 - 16]
= -6/2
= -3 square units
Since area cannot be negative.
So, area of triangle ABD is 3 square units.
Area of triangle ADC = (1/2)[4(0 - 2) + 4(2 + 6) + 5(-6 - 0)]
= (1/2)[-8 + 32 - 30]
= -6/2
= -3 square units
Since area cannot be negative.
So, area of triangle ABD is 3 square units.
Clearly, median AD has divided triangle ABC in two triangles of equal area.
Exercise 7.4
Question 1:
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Let the given line divide the line segment joining the points A(2, −2) and B(3, 7) in a ratio k : 1.
Coordinates of the point of division = {(3k + 2)/(k + 1), (7k - 2)/(k + 1)}
This point also lies on 2x + y − 4 = 0
So, 2(3k + 2)/(k + 1) + (7k - 2)/(k + 1) − 4 = 0
=> 2(3k + 2) + (7k - 2) – 4(k + 1) = 0
=> 6k + 4 + 7k – 2 – 4k – 4 = 0
=> 9k – 2 = 0
=> 9k = 2
=> k = 2/9
Now, k : 1 = 2/9 : 1 = 2 : 9
Therefore, the ratio in which the line 2x + y − 4 = 0 divides the line segment joining the points
A(2, −2) and B(3, 7) is 2 : 9.
Question 2:
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
If the given points are collinear, then the area of triangle formed by these points will be 0.
Area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
=> 0 = (1/2)[x(2 – 0) + 1(0 – y) + 7(y – 2)]
=> 0 = (1/2)[2x – y + 7y – 14]
=> 0 = (1/2)[2x + 6y – 14]
=> 2x + 6y – 14 = 0
=> x + 3y – 7 = 0
This is the required relation between x and y.
Question 3:
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be
representing the points A, B, and C on the circumference of the circle.
So, OA = √{(x - 6)2 + (y + 6)2}
OB = √{(x - 3)2 + (y + 7)2}
OC = √{(x - 3)2 + (y - 3)2}
However, OA = OB (Radii of the same circle)
=> √{(x - 6)2 + (y + 6)2} = √{(x - 3)2 + (y + 7)2}
Squaring on both sides, we get
=> (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
=> x2 + 36 – 12x + y2 + 36 + 12y = x2 + 9 – 6x + y2 + 49 + 14y
=> x2 + 36 – 12x + y2 + 36 + 12y - x2 - 9 + 6x - y2 - 49 - 14y = 0
=> -6x - 2y + 14 = 0
=> 3x + y – 7 = 0
=> 3x + y = 7 …………………1
Similarly, OA = OC (Radii of the same circle)
=> √{(x - 6)2 + (y + 6)2} = √{(x - 3)2 + (y - 3)2}
Squaring on both sides, we get
=> (x - 6)2 + (y + 6)2 = (x - 3)2 + (y - 3)2
=> x2 + 36 – 12x + y2 + 36 + 12y = x2 + 9 – 6x + y2 + 9 - 6y
=> x2 + 36 – 12x + y2 + 36 + 12y - x2 - 9 + 6x - y2 - 9 + 6y = 0
=> -6x + 18y + 54 = 0
=> -3x + 9y – 7 = 0
=> -3x + 9y = 7 …………………2
Adding equation 1 and 2, we get
10y = -20
=> y = -20/10
=> y = -2
From equation 1, we get
3x – 2 = 7
=> 3x = 7 + 2
=> 3x = 9
=> x = 9/3
=> x = 3
Therefore, the centre of the circle is (3, −2).
Question 4:
The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively.
Let (x1, y1), (x2 , y2) be the coordinate of vertex B and D respectively.
We know that the sides of a square are equal to each other.
So, AB = BC
=> √{(x + 1)2 + (y - 2)2} = √{(x - 3)2 + (y - 2)2}
Squaring on both sides, we get
=> (x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2
=> x2 + 1 + 2x + y2 + 4 - 4y = x2 + 9 – 6x + y2 + 4 - 4y
=> x2 + 1 + 2x + y2 + 4 - 4y - x2 - 9 + 6x - y2 - 4 + 4y = 0
=> 8x - 8 = 0
=> 8x = 8
=> x = 8/8
=> x = 1 …………………2
We know that in a square, all interior angles are of 90°.
In ΔABC,
AB2 + BC2 = AC2
=> √{(1 + 1)2 + (y - 2)2} + √{(1 - 3)2 + (y - 2)2} = √{(3 + 1)2 + (2 - 2)2}
=> 4 + y2 + 4 − 4y + 4 + y2 − 4y + 4 =16
=> 2y2 + 16 − 8 y =16
=> 2y2 − 8 y = 0
=> y(y − 4) = 0
=> y = 0 or 4
We know that in a square, the diagonals are of equal length and bisect each other at 90°.
Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.
Now, coordinate of O = {(-1 + 3)/2, (2 + 2)/2}
=> {(1 + x1)/2, (y + y1)/2} = (1, 2)
=> (1 + x1)/2 = 1
=> 1 + x1 = 2
=> x1 = 1
=> (y + y1)/2 = 2
=> y + y1 = 4
If y = 0,
y1 = 4
If y = 4,
y1 = 0
Therefore, the required coordinates are (1, 0) and (1, 4).
Question 5:
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other.
There is a triangular grassy lawn in the plot as shown in the Fig. 7.14.
The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of Δ PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
(i) Taking A as origin, we will take AD as x-axis and AB as y-axis. It can be observed that the
coordinates of point P, Q, and R are (4, 6), (3, 2), and (6, 5) respectively.
Area of triangle PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= (1/2)[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= (1/2)[-12 – 2 + 24]
= 9/2 square units
(ii) Taking C as origin, CB as x-axis, and CD as y-axis, the coordinates of vertices P, Q, and R are
(12, 2), (13, 6), and (10, 3) respectively.
Area of triangle PQR = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= (1/2)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= (1/2)[36 + 13 - 40]
= 9/2 square units
It can be observed that the area of the triangle is same in both the cases.
Question 6:
The vertices of a Δ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively,
such that AD/AB = AE/AC = 1/4.
Calculate the area of the Δ ADE and compare it with the area of Δ ABC. (Recall Theorem 6.2 and Theorem 6.6).
Given that, AD/AB = AE/AC = 1/4
=> AD/DB = AE/EC = 1/3
Therefore, D and E are two points on side AB and AC respectively
such that they divide side AB and AC in a ratio of 1 : 3.
Coordinate of Point D = {(1 * 1 + 3 * 4)/(1 + 3), (1 * 5 + 3 * 6)/(1 + 3)}
= (13/4, 23/4)
Coordinate of Point D = {(1 * 7 + 3 * 4)/(1 + 3), (1 * 2 + 3 * 6)/(1 + 3)}
= (19/4, 20/4)
Area of triangle = (1/2)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of triangle ADE = (1/2)[4(23/4 – 20/4) + (13/4)(20/4 – 6) + (19/4)(6 – 23/4)]
= (1/2)[3 – 13/4 – 19/16]
= (1/2)[(48 – 52 + 19)/16]
= 15/32 square units
Area of triangle PQR = (1/2)[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]
= (1/2)[12 - 4 + 7]
= 15/2 square units
Clearly, the ratio between the areas of ΔADE and ΔABC is 1 : 16.
Alternatively,
We know that if a line segment in a triangle divides its two sides in the same ratio, then the
line segment is parallel to the third side of the triangle. These two triangles so formed (here
ΔADE and ΔABC) will be similar to each other.
Hence, the ratio between the areas of these two triangles will be the square of the ratio
between the sides of these two triangles.
Question 7:
Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do yo observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle.
(i) Median AD of the triangle will divide the side BC in two equal parts.
Therefore, D is the mid-point of side BC.
Coordinate of D = {(6 + 1)/2, (5 + 4)/2} = (7/2, 9/2)
(ii) Point P divides the side AD in a ratio 2 : 1.
Coordinate of P = {(2 * 7/2 + 1 * 4)/(2 + 1), (2 * 9/2 + 1 * 2)/ (2 + 1)} = (11/3, 11/3)
(iii) Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the mid-point of side AC.
Coordinate of E = {(4 + 1)/2, (2 + 4)/2} = (5/2, 3)
Point Q divides the side BE in a ratio 2 : 1.
Coordinate of Q = {(2 * 5/2 + 1 * 1)/(2 + 1), (2 * 3 + 1 * 5)/ (2 + 1)} = (11/3, 11/3)
Median CF of the triangle will divide the side AB in two equal parts. Therefore, F is the mid-
point of side AB.
Coordinate of F = {(4 + 6)/2, (2 + 5)/2} = (5, 7/2)
Point R divides the side CF in a ratio 2 : 1.
Coordinate of R = {(2 * 5 + 1 * 1)/(2 + 1), (2 * 7/2 + 1 * 5)/ (2 + 1)} = (11/3, 11/3)
(iv) It can be observed that the coordinates of point P, Q, R are the same. Therefore, all these
are representing the same point on the plane i.e., the centroid of the triangle.
(v) Consider a triangle, ΔABC, having its vertices as A(x1, y1), B(x2, y2) and C(x3, y3).
Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-
point of side BC.
Coordinate of D = {(x2 + x3)/2, (y2 + y3)/2}
Let the centroid of this triangle be O.
Point O divides the side AD in a ratio 2 : 1.
Coordinate of O = {(2 * (x2 + x3)/2 + 1 * x1)/(2 + 1), (2 * (y2 + y3)/2 + 1 * y1)/ (2 + 1)}
= {(x1 + x2 + x3)/3, (y1 + y2 + y3)/3}
Question 8:
ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
The mid-point of AB is O. Therefore,
Coordinate of P = {(-1 - 1)/2, (-1 + 4)/2} = (-1, 3/2)
Similarly, the coordinate of Q, R and S are (2, 4), (5, 3/2) and (2, -1) respectively.
Length of PQ = √{(-1 - 2)2 + (3/2 - 4)2} = √(9 + 25/4) = √(61/4)
Length of QR = √{(2 - 5)2 + (4 – 3/2)2} = √(9 + 25/4) = √(61/4)
Length of RS = √{(5 - 2)2 + (3/2 + 1)2} = √(9 + 25/4) = √(61/4)
Length of SP = √{(2 + 1)2 + (-1 – 3/2)2} = √(9 + 25/4) = √(61/4)
Length of PR = √{(-1 - 5)2 + (3/2 – 3/2)2} = √(36 + 0) = 6
Length of SQ = √{(2 - 2)2 + (4 + 1)2} = √(0 + 25) = 5
It can be observed that all sides of the given quadrilateral are of the same measure but
diagonals are not same. However, the diagonals are of different lengths.
Therefore, PQRS is a rhombus.
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5.4: Integration Formulas and the Net Change Theorem
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In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration, you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind, think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper notation based on your choice.
Basic Integration Formulas
Recall the integration formulas given in [link] and the rule on properties of definite integrals. Let’s look at a few examples of how to apply these rules.
Example $$\PageIndex{1}$$: Integrating a Function Using the Power Rule
Use the power rule to integrate the function $$\displaystyle ∫^4_1\sqrt{t}(1+t)dt$$.
Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:
\begin{align*} ∫^4_1\sqrt{t}(1+t)dt &=∫^4_1t^{1/2}(1+t)dt \\[5pt] &=∫^4_1(t^{1/2}+t^{3/2})dt. \end{align*}
Now apply the power rule:
\begin{align*} ∫^4_1(t^{1/2}+t^{3/2})dt &= \left . \left(\frac{2}{3}t^{3/2}+\frac{2}{5}t^{5/2}\right) \right|^4_1 \\[5pt] & = \left[\frac{2}{3}(4)^{3/2}+\frac{2}{5}(4)^{5/2} \right]− \left[\frac{2}{3}(1)^{3/2}+\frac{2}{5}(1)^{5/2}\right] \\[5pt] & =\frac{256}{15}. \end{align*}
Exercise $$\PageIndex{1}$$
Find the definite integral of $$f(x)=x^2−3x$$ over the interval $$[1,3].$$
Hint
Follow the process from Example $$\PageIndex{1}$$ to solve the problem.
$−\frac{10}{3}$
The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways. The second is more familiar; it is simply the definite integral.
Net Change Theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:
$F(b)=F(a)+∫^b_aF'(x)dx$
or
$∫^b_aF'(x)dx=F(b)−F(a).$
Subtracting $$F(a)$$ from both sides of the first equation yields the second equation. Since they are equivalent formulas, which one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to name only a few applications. Net change accounts for negative quantities automatically without having to write more than one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction) at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this motion as shown in Figure.
Figure $$\PageIndex{1}$$: The graph shows speed versus time for the given motion of a car.
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled. The net displacement is given by
$∫^5_2v(t)dt=∫^4_240dt+∫^5_4−30dt=80−30=50.$
Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
$∫^5_2|v(t)|dt=∫^4_240dt+∫^5_430dt=80+30=110.$
Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity function accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object, regardless of direction, we need to integrate the absolute value of the velocity function.
Example $$\PageIndex{2}$$: Finding Net Displacement
Given a velocity function $$v(t)=3t−5$$ (in meters per second) for a particle in motion from time $$t=0$$ to time $$t=3,$$ find the net displacement of the particle.
Solution
Applying the net change theorem, we have
$∫^3_0(3t−5)dt=\frac{3t^2}{2}−5t∣^3_0=[\frac{3(3)^2}{2}−5(3)]−0=\frac{27}{2}−15=\frac{27}{2}−\frac{30}{2}=−\frac{3}{2}.$
The net displacement is $$−\frac{3}{2}$$ m (Figure).
Figure $$\PageIndex{2}$$: The graph shows velocity versus time for a particle moving with a linear velocity function.
Example $$\PageIndex{3}$$: Finding the Total Distance Traveled
Use Example to find the total distance traveled by a particle according to the velocity function $$v(t)=3t−5$$ m/sec over a time interval $$[0,3].$$
Solution
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the absolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function, since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,
$$3t−5=0$$
$$3t=5$$
$$t=\frac{5}{3}.$$
The two subintervals are $$[0,\frac{5}{3}]$$ and $$[\frac{5}{3},3]$$. To find the total distance traveled, integrate the absolute value of the function. Since the function is negative over the interval $$[0,\frac{5}{3}]$$, we have $$|v(t)|=−v(t)$$ over that interval. Over $$[ \frac{5}{3},3]$$, the function is positive, so $$|v(t)|=v(t)$$. Thus, we have
$$∫^3_0|v(t)|dt=∫^{5/3}_0−v(t)dt+∫^3_{5/3}v(t)dt$$
$$=∫^{5/3}_05−3tdt+∫^3_{5/3}3t−5dt$$
$$=(5t−\frac{3t^2}{2})∣^{5/3}_0+(\frac{3t^2}{2}−5t)∣^3_{5/3}$$
$$=[5(\frac{5}{3})−\frac{3(5/3)^2}{2}]−0+[\frac{27}{2}−15]−[\frac{3(5/3)^2}{2}−\frac{25}{3}]$$
$$=\frac{25}{3}−\frac{25}{6}+\frac{27}{2}−15−\frac{25}{6}+\frac{25}{3}=\frac{41}{6}$$.
So, the total distance traveled is $$\frac{14}{6}$$ m.
Exercise $$\PageIndex{2}$$
Find the net displacement and total distance traveled in meters given the velocity function $$f(t)=\frac{1}{2}e^t−2$$ over the interval $$[0,2]$$.
Hint
Follow the procedures from Example and Example. Note that $$f(t)≤0$$ for $$t≤ln4$$ and $$f(t)≥0$$ for $$t≥ln4$$.
Net displacement: $$\frac{e^2−9}{2}≈−0.8055m;$$ total distance traveled: $$4ln4−7.5+\frac{e^2}{2}≈1.740 m$$
Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in Example.
Example $$\PageIndex{4}$$: How Many Gallons of Gasoline Are Consumed?
If the motor on a motorboat is started at $$t=0$$ and the boat consumes gasoline at the rate of $$5−t^3$$ gal/hr, how much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus. The limits of integration are the endpoints of the interval [0,2]. We have
\begin{align*} ∫^2_0(5−t^3)dt &=(5t−\frac{t^4}{4})∣^2_0 \\[5pt] &=[5(2)−\frac{(2)^4}{4}]−0 \\[5pt] &=10−\frac{16}{4} \\[5pt] &=6. \end{align*}
Thus, the motorboat uses 6 gal of gas in 2 hours.
Example $$\PageIndex{5}$$: Chapter Opener: Iceboats
As we saw at the beginning of the chapter, top iceboat racers can attain speeds of up to five times the wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.
Figure $$\PageIndex{3}$$: (credit: modification of work by Carter Brown, Flickr)
Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function $$v(t)=20t+5.$$ For the second half hour of Andrew’s outing, the wind remains steady at 15 mph. In other words, the wind speed is given by
$v(t)=\begin{cases}20t+5& for 0≤t≤\frac{1}{2}\\15 & for \frac{1}{2}≤t≤1\end{cases}. \nonumber$
Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
Distance =$$∫^1_02v(t)dt.$$
Substituting the expressions we were given for $$v(t)$$, we get
$$∫^1_02v(t)dt=∫^{1/2}_02v(t)dt+∫^1_{1/2}2v(t)dt$$
$$=∫^{1/2}_02(20t+5)dt+∫^1_{1/3}2(15)dt$$
$$=∫^{1/2}_0(40t+10)dt+∫^1_{1/2}30dt$$
$$=[20t^2+10t]|^{1/2}_0+[30t]|^1_{1/2}$$
$$=(\frac{20}{4}+5)−0+(30−15)$$
$$=25.$$
Andrew is 25 mi from his starting point after 1 hour.
Exercise $$\PageIndex{3}$$
Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind starts to die down according to the function $$v(t)=−10t+15.$$ In other words, the wind speed is given by
$$v(t)=\begin{cases}20t+5 & for 0≤t≤\frac{1}{2}\\−10t+15& for\frac{1}{2}≤t≤1\end{cases}$$.
Under these conditions, how far from his starting point is Andrew after 1 hour?
Hint
Don’t forget that Andrew’s iceboat moves twice as fast as the wind.
$$17.5 mi$$
Integrating Even and Odd Functions
We saw in Functions and Graphs that an even function is a function in which $$f(−x)=f(x)$$ for all x in the domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are symmetric about the y-axis. An odd function is one in which $$f(−x)=−f(x)$$ for all x in the domain, and the graph of the function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly $$[−a,a],$$ evaluate to zero because the areas above and below the x-axis are equal.
Integrals of Even and Odd Functions
For continuous even functions such that $$f(−x)=f(x),$$
$∫^a_{−a}f(x)dx=2∫^a_0f(x)dx.$
For continuous odd functions such that $$f(−x)=−f(x),$$
$∫^a_{−a}f(x)dx=0.$
Example $$\PageIndex{6}$$: Integrating an Even Function
Integrate the even function $$∫^2_{−2}(3x^8−2)dx$$ and verify that the integration formula for even functions holds.
Solution
The symmetry appears in the graphs in Figure $$\PageIndex{4}$$. Graph (a) shows the region below the curve and above the x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry about the y-axis of an even function. We have
\begin{align*} ∫^2_{−2}(3x^8−2)dx &=(\frac{x^9}{3}−2x)∣^2_{−2} \\[5pt] &=\left[\frac{(2)^9}{3}−2(2)\right]−\left[\frac{(−2)^9}{3}−2(−2)\right] \\[5pt] &= \left(\frac{512}{3}−4\right)−\left(−\frac{512}{3}+4\right) \\[5pt] &=\frac{1000}{3}. \end{align*}
To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then check to make sure we get the same answer.
$∫^2_0(3x^8−2)dx=(\frac{x^9}{3}−2x)∣^2_0=\frac{512}{3}−4=\frac{500}{3}$
Since $$2⋅\frac{500}{3}=\frac{1000}{3},$$ we have verified the formula for even functions in this particular example.
Figure $$\PageIndex{4}$$: Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative area between the curve and the x-axis. Both views show the symmetry about the y-axis.
Example $$\PageIndex{7}$$: Integrating an Odd Function
Evaluate the definite integral of the odd function $$−5 \sin x$$ over the interval $$[−π,π].$$
Solution
The graph is shown in Figure $$\PageIndex{5}$$. We can see the symmetry about the origin by the positive area above the x-axis over $$[−π,0]$$, and the negative area below the x-axis over $$[0,π].$$ we have
\begin{align*} ∫^π_{−π}−5\sin x \,dx &=−5(−cosx)|^π_{−π} \\[5pt] &=5\cos x|^π_{−π} \\[5pt] &=[5\cos π]−[5\cos(−π)] \\[5pt] &=−5−(−5)=0. \end{align*}
Figure $$\PageIndex{5}$$:The graph shows areas between a curve and the x-axis for an odd function.
Exercise $$\PageIndex{4}$$
Integrate the function $$\displaystyle ∫^2_{−2}x^4dx.$$
Hint
Integrate an even function.
$$\dfrac{64}{5}$$
Key Concepts
• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
Key Equations
• Net Change Theorem
$F(b)=F(a)+∫^b_aF'(x)dx\nonumber$ or $∫^b_aF'(x)dx=F(b)−F(a) \nonumber$
Glossary
net change theorem
if we know the rate of change of a quantity, the net change theorem says the future quantity is equal to the initial quantity plus the integral of the rate of change of the quantity
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# Learning How to Calculate the Cross Section
In math, a cross-section is the shape you would see if you were to slice an object. Knowing how to calculate it can be useful, especially for calculating the volume of a whole object.
In technical terms, the cross section is when a plane meets a solid form. Calculating it is very similar to calculating the area of a shape; the only difference is that a cross section is part of a three-dimensional form. Different shapes have different formulas, and below are some examples of how to go about your calculations.
Rectangles You may see a rectangular cross section on a rectangular building, cereal box or anything that’s a rectangular prism. This may be the easiest cross-sectional area to calculate, as it’s simply the width x length. For example, let’s say one side of the building is eight-feet long and the other side is six-feet long. Eight multiplied by six is 48, and there you have your cross section.
Squares You can see a square-shaped cross section in a dice or in a square-based pyramid, like the ones in Egypt. Since all sides of a square are of equal length, the cross section is simply that length multiplied by itself. For example, if the length of the dice is two inches, the area of the cross section is four inches squared. An easy way to remember this is that the cross-section of a square is the length squared.
Triangles A triangular cross section may not be as common as other shapes, as you would probably only see one in a triangular-based pyramid or a triangular prism. The most basic way of finding this cross-sectional area would be the base length multiplied by the height and then divided by two. However, these lengths may be tricky to find if they aren’t stated, and you’ll need to rely on trigonometry to find out any missing figures.
Circles The area of a circle cross section is pretty complicated since it usually requires a calculator to get the precise number because it involves pi. Forms that have a circular cross section include cones, spheres and cylinders. The area of the circle is two x pi x radius squared. The radius is half the diameter, which is the length cutting across the circle. If you don’t need an exact number, you can just leave the result with pi in it. For example, if the radius is five inches, the cross sectional area would be two x pi x 25, which comes out to be 50 x pi as a final result without a calculator.
Taking an Extra Step While the main focus here is on the cross-sectional area of a form, you can easily find the volume of the form with this information. With most shapes, you just multiply the area from the cross section by the length of the object. Certain shapes, such as spheres and cones, have different formulas, meaning it’s not a one-size-fits-all rule.
This is only a simple guide on how to calculate basic cross sectional areas. Most forms can be broken down into these four simpler shapes. The hardest part about cross sections is being able to visualize what the form looks like when it’s sliced. Once you know the cross-sectional shape, you can easily apply one of these formulas and do the math.
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## Week 13/Graphing Reciprocal Functions
Reciprocal Functions
Reciprocal Functions is taking a function like y= 2$x^2$+12x-20 and making it y=1/ (2$x^2$+12x-20) is the functions reciprical.
When finding y= 2$x^2$+12x-20 and its reciprical y=1/ (2$x^2$+12x-20), you want to graph the original function first all the time. After you have graphed the original function, you have to find its vertical asymptote which is where the graph touches the x axis and where the reciprocal will never cross over. It acts like a dotted line on going vertically upwords to help you graph the reciprical. There is also a horizontal asymptote which is for now always on the a axis, it acts simular to the vertical asymptote except it sits vertically on the x axis making sure that your reciprical never touches it.
Once you have graph the parabola, now that you have found that there is no x intercept in the equation, so there is no vertical asymptote. This is what you would call a pimple graph, one that does not have a vertical asymptote but does however still have a horizontal asymptote. To find out where it recipricates you have to find the vertex on the graph.
(3,-2) now that the vertex is found, you can recipricate the vertex, but only the -2 or the y, then you get (3,-1/2) and then you draw the bump that carries on through the x axis and down towards the new vertex for the reciprocal function.
## Week 12/ Absolute value functions
Absolute value functions
I’ll be graphing a absolute value equation, graphing it, and showing it in piecewise notation.
y= |-3x+9| <——- Note: absolute value brackets (|x|)
With a normal line, (-3x+9) it would go below the x-axis, but when there are absolute value lines in the equation it makes it so the line does not go into the negitive not. Which also means it does not go below the x-axis, istead it just rickashays off and comes right back up. It refects the original line and bounces off the x-axis.
Now that we haved graphed we must state it peace wise notation. The point that intercecpts with the x axis is where you will be basing the x is < or > off of. For stating the x is < or > is with two equations, becuase the graph shows two line reflecting off the x axis there are two equations. 1. -3x+9 and -(-3x+9) then you state whether x is < or > for the number on the x axis.
y = {-3x+9 if x < and equal to 3}
y = {-(-3x+9) if x > 3}
To figure out if x is < or > you must test a point on the x axis to see if its true. Test numbers before and after three.
## Week 11/ Solving system of equations graphically
Solving system of equations
First you must be able to show the equation graphically. By doing so you have to show any linear equations into y=mx+b form; then change any quadratic equations into factored form or vertex form. After you have algebraically changed all the equations you can then graph the equations and see where they intercept. Then that will be your solution.
There can be three type of solutions when solving for x equations that are linear and quadratic. The first one is no solution, the second is one solution, and the third is two solutions.
Equations: y= 2$x^2$ + 4x + 4
y= -2x+7
The graph states that there is no solution because the equation does not touch or intercept with each other.
## Week 7 blog post/ Discriminant
Discriminant is used when figuring out how many solutions and if they are real roots. The Discriminant is $b^2$ – 4ac$, which looks familiar beucase it is the discriminant of the equation in the quadratic formula. Just by the simple equation $b^2$ – 4ac$ it can tell us alot about the quadratic equation before we even start solving.
If you get zero means x is one solution one solution, if the x is greater than zero you have two solutionsand you know it will be rational, if you get less than zero there is no solution.
## Week 10 blog post/ Solving Quadratic Inequalities
In Solving Quadratic Inequalities, you need to know three steps.
Step 1: Factor the exspression
Step 2: Determine the zeros
Step 3: Use a sign chart for each factor to the left and right of zeros
Note: Test numbers a number on all the opposing sides of the x intercepts
(Key things to know when solving Quadratic Inequalities is to make sure the the statement is always true, if its not you know your answer is wrong or there is no possibility.)
## Week9/solving A with factored form
Question: A graph passes through B(2,-5) and has x-intercepts -3 and 4
Before solving its good to analyse the question first. Passes through B(2,-5) represents x and y, well x intercepts represents the x intercepts. An equation that can use both of these clues is factored form since factored form has the x-intercepts in the equation.
Factored Form y= a(x-$x1$) (x-$x2$)
When plugging in the x intercepts from the equation you get y= a(x+3) (x-4), this question may look solvable but it has more than one variables so you cannot solve it yet for A. Thats why you put the x and y into the equation. Now your left with -5= a(2+3) (2-4). From there you can now solve for a. (EXAMPLE BELOW)Now the equation is y= $\frac{1}{2}$(x+3) (x-4) now it is a complete equation and can be changed and graphed.
## Week 8/Standard Form-Graphing
When your given an equation in General form, there isn’t enough information to easily graph just by looking at it. Your first step is to change the equation into Standard Form. To change a linear equation into Standard Form you can complete the square giving an answer presented into standerd form.
Note: Standared form is presented in a way to graph linear equations easily.
Example: y=$x^2$-4x+1
Standared form: y=$(x-2)^2-3$
After the equation has been changed into Standared Form the equation y=$(x-2)^2-3$ tells you everything you need to know to graph the equation. The equation is represented by y=$a(x-b)^2-c$, where A tells you if the parabola opens up or down and if it is congruent to $x^2$, B tells you where to move on the x axis, well c tells you where to move on the y axis.
Those are the basics to graphing and algebraically changing linear equations to standard form.
## Week 6 / Quadratic Formula
The Quadratic Formula is a well known formuala used to solve quadratic equations; a quadratic equation is an equation (=) that has a squareroot symbole. To solve these you can use a number of things, but my favourite way is the Quadratic Formula.
To start you must know the Quadratic Formula which is x= -b +or-$\sqrt{(b)^2 -4ac}$ then all of that divided by 2a. The A, B, and C, they all are pieces of the equation $x^2$-6x+4=0 a=$x^2$ b=-6x c=4. The letters are just to show you where to place the numbers in the equation.
After you have plugged in the numbers the rest is pretty sympol, you just solve after that.
## Week 5 / The Difference of Squares; Factoring
Difference of Squares / Factoring
To understand how to factor Difference of Squares you must first know what Difference of Squares means. Difference means in math “subraction.” Squares means two numbers will square into another number. Example (${3}^2$ {3 x 3} makes 9)
Forthgoing, figuring out if something is a Difference of squares or not, here two examples.
Easy equations
First when factoring you must find what is common, I found that ${5x}^2$ is something i can pull out of the equation and then your left with ${5x}^2$ (3x-1). To figure out if its aDifference of squares you check if it is squared and has a negitive sign; so yes it is a difference of squares.
The second equation is much more different than the first equation, its a trinomial, but that doesnt make a huge differenence when it comes to factoring the equation. Since both equations are easy equations, just checking what is common is all we need to do. In the equation 4 is the only number that is common, and your left with 4(${2p}^3$${1p}^2$ -1). Since that is all you can do now you must find out if its a difference of squares or not. You atuomatically know becuase the 2p is cubed. So it is not a difference of squares.
## Week four blog post
Simple equations solving for X (with radicals)
1st step- Always the first step to solving with radicals is to find restrictions. Restrictions are a way to check that your answer is right by looking at the restrictions. When the equation that only has a radical and a coefficient infront, the same restriction applies. (x $\geq$ 0)
2nd step- If there is a coefficient infront of the radical, do the opposite and divide the seven out from the radical. But watch out what you do to one side you must do it to the other; when you divided the seven, you have to divide the seven by the fourty two.
3rd step- Now that you are left with 6 = $\sqrt{2x}$ to get rid of a square root symbole, your going to have to do the oppostite of square rooting, powering! You have to power the square root symbol, but what you have to do to one side, you have to do to the other. That means 6 turns into 36 becuase 6 times 6 is thrity six, well $\sqrt{2x}$ becomes just 2x.
4rth step- Now that your just left with 36 = 2x, to seperate the 2 from the x you must to the oppostite and divide both the 2x and the 36 by the number attached to the x, in turn leaving the x by itself. So x = 18.
Finally- Your not done yet! Just becuase you have the asnwer doesnt mean your finished, you still have to check to see if you answer mathces your restrictions, if it doesnt its wrong. x $\geq$ 0 does match x=18. That mean you got the right answer.
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# Boolean Algebra
Created by: Team Maths - Examples.com, Last Updated: June 5, 2024
## Boolean Algebra
Boolean algebra is a branch of mathematics that deals with variables that have two distinct values: true (1) and false (0). Developed by George Boole in the mid-19th century, it forms the foundation of digital logic and computer science.
## What Is Boolean Algebra?
Boolean algebra is a branch of mathematics that deals with binary variables and logical operations. It uses true/false values (1/0) and Arithmetic operations like AND, OR, and NOT to manipulate logical expressions, forming the foundation for digital circuit design and computer science applications.
## Boolean Algebra Operations
Boolean algebra consists of several fundamental operations that are used to manipulate and simplify logical expressions. The main types of Boolean algebra operations include:
1. AND Operation
2. OR Operation
3. NOT Operation
4. NAND Operation
5. NOR Operation
6. XOR Operation
7. XNOR Operation
## Boolean Algebra Truth Table
This table presents the results of the primary Boolean algebra operations for all possible combinations of Boolean values (True and False) of variables A and B.
## Laws of Boolean Algebra
### 1. Identity Law
A + 0 = A: Adding 0 to a variable does not change its value.
A · 1 = A: Multiplying a variable by 1 does not change its value.
### 2. Null Law
A + 1 = 1: Adding 1 to a variable always results in 1.
A · 0 = 0: Multiplying a variable by 0 always results in 0.
### 3. Idempotent Law
A + A = A: Adding a variable to itself does not change its value.
A · A = A: Multiplying a variable by itself does not change its value.
### 4. Complement Law
A + ¬A = 1: A variable OR with its complement is always 1.
A · ¬A = 0: A variable ANDed with its complement is always 0.
### 5. Commutative Law
A + B = B + A: The order of variables in an OR operation does not matter.
A · B = B · A: The order of variables in an AND operation does not matter.
### 6. Associative Law
A + (B + C) = (A + B) + C: Grouping of variables in an OR operation does not affect the result.
A · (B · C) = (A · B) · C: Grouping of variables in an AND operation does not affect the result.
### 7. Distributive Law
A · (B + C) = (A · B) + (A · C): AND operation distributes over OR.
A + (B · C) = (A + B) · (A + C): OR operation distributes over AND.
### 8. Absorption Law
A + (A · B) = A: A variable OR with its ANDed result with another variable is always equal to the variable.
A · (A + B) = A: A variable ANDed with its OR result with another variable is always equal to the variable.
## Boolean Algebra Theorems
Boolean algebra consists of several key theorems that are essential for simplifying and manipulating logical expressions. Here are the main theorems along with explanations:
### 1. De Morgan’s Theorems
Theorem 1: A + B‾ = A‾ ⋅ B‾
This theorem states that the complement of the OR of two variables is equal to the AND of their complements.
Theorem 2: A ⋅ B‾ = A‾ + B‾
This theorem states that the complement of the AND of two variables is equal to the OR of their complements.
### 2. Duality Theorem
Every Boolean expression remains valid if the operators and identity elements are interchanged. Specifically, AND is swapped with OR, and 0 is swapped with 1.
### 3. Absorption Theorem
Theorem 1: A + (A⋅B) = A
This theorem states that a variable OR with its ANDed result with another variable is always equal to the variable itself.
Theorem 2: A ⋅ (A+B) = A
This theorem states that a variable ANDed with its OR result with another variable is always equal to the variable itself.
### 4. Involution Theorem
Theorem: A‾ = A
This theorem states that the complement of the complement of a variable is the variable itself.
### 5. Redundancy Theorem
Theorem 1: A + A‾ ⋅ B = A + B
This theorem states that a variable OR with the AND of its complement and another variable is equal to the variable OR with the other variable.
Theorem 2: A ⋅ (A‾+B) = A⋅B
This theorem states that a variable AND with the OR of its complement and another variable is equal to the variable ANDed with the other variable.
### 6. Consensus Theorem
Theorem: (A⋅B) + (A‾⋅C) + (B⋅C) = (A⋅B) + (A‾⋅C)
This theorem states that the term B⋅C is redundant and can be eliminated.
## Who invented Boolean algebra?
Boolean algebra was invented by George Boole, an English mathematician, in the mid-19th century. His work laid the foundation for digital logic and computer science.
## What is the significance of the Identity Law in Boolean algebra?
The Identity Law states:
A + 0 = A
A · 1 = A
These laws indicate that adding 0 or multiplying by 1 does not change the value of a variable, helping to simplify expressions.
## Can Boolean algebra be used for practical applications?
Yes, Boolean algebra is extensively used in designing and analyzing digital circuits, computer algorithms, control systems, and logic gates. It is fundamental to the operation of all digital devices.
## What is an example of using Boolean algebra in digital circuits?
In digital circuits, Boolean algebra simplifies the design of logic gates. For example, simplifying a complex circuit expression can reduce the number of gates needed, making the circuit more efficient and cost-effective.
## What are the primary operations in Boolean algebra?
The primary operations in Boolean algebra are:
AND (·): True if both operands are true.
OR (+): True if at least one operand is true.
NOT (¬): Inverts the value of the operand.
## Boolean algebra calculator
A Boolean algebra calculator simplifies logical expressions and evaluates Boolean operations. It helps design and analyze digital circuits by performing operations like AND, OR, NOT, NAND, NOR, XOR, and XNOR.
## How is Boolean algebra applied in computer science?
In computer science, Boolean algebra is used to design and optimize algorithms, perform bitwise operations, and develop efficient data structures. It is also crucial in programming for decision-making processes and controlling flow using logical conditions.
## How are Boolean expressions simplified using the Consensus Theorem?
The Consensus Theorem states:
(A · B) + (¬A · C) + (B · C) = (A · B) + (¬A · C)
This theorem shows that the term
B⋅C can be eliminated without changing the expression’s value, aiding in the simplification of complex Boolean expressions.
## What is the Absorption Law in Boolean algebra?
The Absorption Law states:
A + (A · B) = A
A · (A + B) = A
This law helps simplify expressions by eliminating redundant terms, making it easier to analyze and design logic circuits.
## What are De Morgan’s Theorems?
De Morgan’s Theorems provide rules for the complement of complex expressions:
¬(A + B) = ¬A · ¬B
¬(A · B) = ¬A + ¬B
These theorems are useful for simplifying Boolean expressions and designing digital circuits.
Text prompt
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Difference between a Rhombus and a Parallelogram
By Theydiffer - May 31, 2017
We may not normally use the words “rhombus” and “parallelogram,” but we actually see these shapes every day. Do you still remember what they are? For sure, these were discussed in school, but now we ask, “What exactly are they and how are they different from each other?” This article will explore the difference between a rhombus and a parallelogram.
## Descriptions
Getty Images/Moment/yuanyuan yan
A rhombus, also known as “rhom” or “diamond,” is an equilateral quadrilateral, a term that refers to a figure with four parallel sides (the lines will never intersect even if they continue) with equal lengths. If you take a look at the picture above, you will notice that all sides are the same length (8 cm).
All opposing angles of a rhombus have equal lengths and its adjacent angles are supplementary angles (which means the sum of the two angles is 180°). Moreover, a rhombus’s diagonals are perpendicular to each other; they bisect each other at right angles.
Getty Images/DigitalVision/Hello World
A parallelogram, on the other hand, is a type of quadrilateral. Its opposing lines are parallel and of equal length. If you take a look at the picture above, you will notice that two opposing lines are 15 cm and the other two are 8 cm.
A parallelogram’s adjacent angles are supplementary and its consecutive angles are equal. Its diagonals bisect each other at an intersection. Examples of a parallelogram are rectangles, squares, and rhombuses.
## Rhombus vs Parallelogram
What, then, is the difference between a rhombus and a parallelogram?
Both a rhombus and a parallelogram are quadrilateral figures, which means both are four-sided figures. The main difference is that a parallelogram has two parallel opposing sides which are of equal lengths, whereas a rhombus has four sides that all have equal lengths (all four sides are parallel). In other words, a rhombus is a type of parallelogram with equal sides.
## Comparison Chart
Rhombus Parallelogram An equilateral quadrilateral; has four parallel sides with equal lengths Has parallel opposing lines that are of equal length
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# How do you find the product of $\left( 10x \right)\left( 4{{x}^{7}} \right)$?
Last updated date: 25th Feb 2024
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Hint: Take the product of the constant terms, i.e., 10 and 4 separately and the product of the variables x and ${{x}^{7}}$ separately. Use the formula of exponents and powers given as: - ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ to simplify the product of variables. Now, multiply the resultant of the two products considered at the initial step to get the answer.
Complete step by step answer:
Here, we have been provided with a pair of monomials: 10x and $4{{x}^{7}}$. We have been asked to find their product. But first let us understand the meaning of the term ‘monomial’.
Now, monomial is an expression that contains only one term. For example: - a, 6n, 5x, $9{{y}^{2}}$, $10{{a}^{3}}{{b}^{2}}{{c}^{5}}$ etc. These are all examples of monomials as they contain only one term. A monomial converts into a binomial when the variables are separated with a (+) or (-) sign.
Now, let us come to the question. We have two monomials 10x and $4{{x}^{7}}$. So, taking their product, we get,
$\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=\left( 10\times 4 \right)\left( x\times {{x}^{7}} \right)$
Here, we have grouped the constants together and the variables together and considering their product, we get,
$\Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40\left( x\times {{x}^{7}} \right)$
Using the formula: - ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, we get,
\begin{align} & \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{1+7}} \\ & \Rightarrow \left( 10x \right)\left( 4{{x}^{7}} \right)=40{{x}^{8}} \\ \end{align}
Hence, the required product is $40{{x}^{8}}$.
Note: One may check the answer by substituting some small value to the variable x. For example: - let us substitute x = 1, then in the L.H.S. we will have $\left( 10\times 1 \right)\left( 4\times {{1}^{7}} \right)=10\times 4=40$. Now, in the R.H.S. we will have $4\times {{1}^{8}}=40$. Do not substitute any larger value of x because the exponent is 8 which is a large number. You must remember the formulas of exponents and powers like: - ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ etc.
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# COMPOSITION OF THREE FUNCTIONS
Let A, B, C, D be four sets and let f : A--->B , g : B--->C and h : C--->D be three functions.
Using composite functions f o g and g o h, we get two new functions like (f o g) o h and f o (g o h).
We observed that the composition of functions is not commutative. The natural question is about the associativity of the operation.
Composition of three functions is always associative. That is,
f o (g o h) = (f o g) o h
Consider the functions f(x), g(x) and h(x) as given below. Find (f o g) o h and f o (g o h) in each case and also show that (f o g) o h = f o (g o h).
Example 1 :
f(x) = x - 1 , g(x) = 3x + 1 and h(x) = x2
Solution :
f o (g o h) :
g o h = g[h(x)]= g[x2]= 3x2 + 1 f o (g o h) = f(3x2 + 1)= 3x2 + 1 - 1= 3x2 ----(1)
(f o g) o h :
f o g = f[g(x)]= f[3x + 1]= 3x + 1 - 1= 3x (f o g) o h = (f o g)[h(x)]= (f o g)(x2)= 3x2 ----(2)
From (1) and (2),
f o (g o h) = (f o g) o h
Example 2 :
f(x) = x2, g(x) = 2x and h(x) = x + 4
Solution :
f o (g o h) :
g o h = g[h(x)]= g[x + 4]= 2(x + 4)= 2x + 8 f o (g o h) = f(2x + 8)= (2x + 8)2= (2x)2 + 2(2x)(8) + 82= 4x2 + 32x + 64 ----(1)
(f o g) o h :
f o g = f[g(x)]= f[2x]= (2x)2= 4x2 (f o g) o h = (f o g)[h(x)]= (f o g)(x + 4)= 4(x + 4)2= 4[x2 + 2(x)(4) + 42]= 4[x2 + 8x + 16]= 4x2 + 32x + 64 ----(2)
From (1) and (2),
f o (g o h) = (f o g) o h
Example 3 :
f(x) = x - 4, g(x) = x2 and h(x) = 3x - 5
Solution :
f o (g o h) :
g o h = g[h(x)]= g[3x - 5]= (3x - 5)2 f o (g o h) = f[(3x - 5)2]= (3x - 5)2 - 4 ----(1)
(f o g) o h :
f o g = f[g(x)]= f[x2]= x2 - 4 (f o g) o h = (f o g)[h(x)]= (f o g)(3x - 5)= (3x - 5)2 - 4 ----(2)
From (1) and (2),
f o (g o h) = (f o g) o h
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# 4.2 Becoming familiar with decimals
Adding and subtracting decimals can be done using the same methods as adding and subtracting whole numbers as long as the digits are lined up under the correct column headings, keeping the decimal points underneath each other.
Learners can struggle to handle decimals, so it is helpful to use real-life instances with which the learners are familiar.
Numbers with one decimal place are evident on a classroom ruler set out in centimetres (Figure 6).
Figure 6 A ruler showing centimetres and tenths of centimetres
There are 10 gradations between each whole number (of centimetres), so each interval is 1 tenth or 0.1. Rulers are effective number lines which learners have easy access to.
A useful activity to use with learners who struggle with decimals is to ask them to draw a straight line across a page. Next, they should put a mark near the beginning and label it A. Ask them to put another mark 4.2 centimetres along the line from A and to label it B. Next ask them to put another mark 2.9 centimetres along the line from B and label it C. Now ask them to measure from A to C. Then ask how they can check their answer (i.e. add up 4.2 and 2.9). This activity (and you can use more than 2 measurements) is useful in encouraging the accurate use of a ruler and measuring and working with decimals.
Numbers with two decimal places will be familiar to your learners through the use of money. Having said this, today many people use contactless card payments and may not even think about their transactions in terms of exchanges of money. Yet, I often used to say to my learners ‘think of two decimal place numbers as money’ and it helped them to feel they could cope with the calculations.
Now watch Video 7 for more advice on decimal number lines.
Download this video clip.Video player: Video 7 Decimal number lines with 1, 2 and 3 decimal places
Video 7 Decimal number lines with 1, 2 and 3 decimal places
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Since we use a number system in base ten, special things happen when you multiply or divide by powers of 10. A place value table is useful for explaining what happens. Now watch Video 8 for more information on using a place value diagram.
Download this video clip.Video player: Video 8 Multiplying and dividing by powers of 10 using a place value diagram
Video 8 Multiplying and dividing by powers of 10 using a place value diagram
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# Lesson 10
Dilations on a Square Grid
Let’s dilate figures on a square grid.
### 10.1: Dilations on a Grid
1. Find the dilation of triangle $$QRS$$ with center $$T$$ and scale factor 2.
2. Find the dilation of triangle $$QRS$$ with center $$T$$ and scale factor $$\frac{1}{2}$$.
### 10.2: Card Sort: Matching Dilations on a Coordinate Grid
Your teacher will give you some cards. Each of Cards 1 through 6 shows a figure in the coordinate plane and describes a dilation.
Each of Cards A through E describes the image of the dilation for one of the numbered cards.
Match number cards with letter cards. One of the number cards will not have a match. For this card, you’ll need to draw an image.
The image of a circle under dilation is a circle when the center of the dilation is the center of the circle. What happens if the center of dilation is a point on the circle? Using center of dilation $$(0,0)$$ and scale factor 1.5, dilate the circle shown on the diagram. This diagram shows some points to try dilating.
### 10.3: Info Gap: Dilations
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
3. Explain how you are using the information to solve the problem.
Continue to ask questions until you have enough information to solve the problem.
4. Share the problem card and solve the problem independently.
If your teacher gives you the data card:
2. Ask your partner “What specific information do you need?” and wait for them to ask for information.
If your partner asks for information that is not on the card, do not do the calculations for them. Tell them you don’t have that information.
3. Before sharing the information, ask “Why do you need that information?” Listen to your partner’s reasoning and ask clarifying questions.
4. Read the problem card and solve the problem independently.
5. Share the data card and discuss your reasoning.
Triangle $$EFG$$ was created by dilating triangle $$ABC$$ using a scale factor of 2 and center $$D$$. Triangle $$HIJ$$ was created by dilating triangle $$ABC$$ using a scale factor of $$\frac12$$ and center $$D$$.
1. What would the image of triangle $$ABC$$ look like under a dilation with scale factor 0?
2. What would the image of the triangle look like under dilation with a scale factor of -1? If possible, draw it and label the vertices $$A’$$, $$B’$$, and $$C’$$. If it’s not possible, explain why not.
3. If possible, describe what happens to a shape if it is dilated with a negative scale factor. If dilating with a negative scale factor is not possible, explain why not.
### Summary
Square grids can be useful for showing dilations. The grid is helpful especially when the center of dilation and the point(s) being dilated lie at grid points. Rather than using a ruler to measure the distance between the points, we can count grid units.
For example, suppose we want to dilate point $$Q$$ with center of dilation $$P$$ and scale factor $$\frac{3}{2}$$. Since $$Q$$ is 4 grid squares to the left and 2 grid squares down from $$P$$, the dilation will be 6 grid squares to the left and 3 grid squares down from $$P$$ (can you see why?). The dilated image is marked as $$Q’$$ in the picture.
Sometimes the square grid comes with coordinates. The coordinate grid gives us a convenient way to name points, and sometimes the coordinates of the image can be found with just arithmetic.
For example, to make a dilation with center $$(0,0)$$ and scale factor 2 of the triangle with coordinates $$(\text-1, \text- 2)$$, $$(3,1)$$, and $$(2, \text- 1)$$, we can just double the coordinates to get $$(\text- 2, \text- 4)$$, $$(6,2)$$, and $$(4, \text- 2)$$.
In general, an important use of coordinates is to communicate geometric information precisely. Let’s consider a quadrilateral $$ABCD$$ in the coordinate plane. Performing a dilation of $$ABCD$$ requires three vital pieces of information:
1. The coordinates of $$A$$, $$B$$, $$C$$, and $$D$$
2. The coordinates of the center of dilation, $$P$$
3. The scale factor of the dilation
With this information, we can dilate the vertices $$A$$, $$B$$, $$C$$, and $$D$$ and then draw the corresponding segments to find the dilation of $$ABCD$$. Without coordinates, describing the location of the new points would likely require sharing a picture of the polygon and the center of dilation.
### Glossary Entries
• center of a dilation
The center of a dilation is a fixed point on a plane. It is the starting point from which we measure distances in a dilation.
In this diagram, point $$P$$ is the center of the dilation.
• dilation
A dilation with center $$O$$ and positive scale factor $$r$$ takes a point $$P$$ along the line $$OP$$ to another point whose distance is $$r$$ times further away from $$O$$ than $$P$$ is. If $$r < 1$$ then the new point is really closer to $$O$$, not further away.
The triangle $$DEF$$ is a dilation of the triangle $$ABC$$ with center $$O$$ and with scale factor 3. So $$D$$ is 3 times further away from $$O$$ than $$A$$ is, $$E$$ is 3 times further away from $$O$$ than $$B$$ is, and $$F$$ is 3 times further away from $$O$$ than $$C$$ is.
• scale factor
To create a scaled copy, we multiply all the lengths in the original figure by the same number. This number is called the scale factor.
In this example, the scale factor is 1.5, because $$4 \boldcdot (1.5) = 6$$, $$5 \boldcdot (1.5)=7.5$$, and $$6 \boldcdot (1.5)=9$$.
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# Thread: Points (a, b) and (b, a)
1. ## Points (a, b) and (b, a)
Show that the line containing the points (a, b) and (b, a),
where a does not equal 0, is perpendicular to the line y = x.
Also, show that the midpoint of (a, b) and (b, a) lies on the line y = x.
2. Originally Posted by magentarita
Show that the line containing the points (a, b) and (b, a),
where a does not equal 0, is perpendicular to the line y = x.
Also, show that the midpoint of (a, b) and (b, a) lies on the line y = x.
The slope of $\displaystyle y=x$ is 1.
You'll need to find the slope of the segment between (a, b) and (b, a).
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{a-b}{b-a}=\frac{-(b-a)}{b-a}=-1$
Since the slopes are negative reciprocals of each other, the lines are perpendicular.
$\displaystyle y=x$ is the linear identity function. this means the x and y values are identical.
The midpoint between (a, b) and (b, a) can be found using the midpoint formula:
$\displaystyle M=\left(\frac{a+b}{2}, \frac{b+a}{2}\right)$
Since both x and y values of the midpoint are identical, then the midpoint has to lie on the identity function $\displaystyle y=x$
3. ## Again...
Originally Posted by masters
The slope of $\displaystyle y=x$ is 1.
You'll need to find the slope of the segment between (a, b) and (b, a).
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}=\frac{a-b}{b-a}=\frac{-(b-a)}{b-a}=-1$
Since the slopes are negative reciprocals of each other, the lines are perpendicular.
$\displaystyle y=x$ is the linear identity function. this means the x and y values are identical.
The midpoint between (a, b) and (b, a) can be found using the midpoint formula:
$\displaystyle M=\left(\frac{a+b}{2}, \frac{b+a}{2}\right)$
Since both x and y values of the midpoint are identical, then the midpoint has to lie on the identity function $\displaystyle y=x$
Another educational reply.
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y = √x Step 2 : Learn to use notation to describe mapping rules,and graph images given preimage and translation Sliding up or down. Translating a sentence or statement into an algebraic equation is an important stuff which is much required to solve word problems in math. In mathematics, a translation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x'y'-Cartesian coordinate system in which the x' axis is parallel to the x axis and k units away, and the y' axis is parallel to the y axis and h units away. Translating a figure can be thought of as “sliding” the original. Think about a function that you use to determine how much money a person earns for working … Translation Notation Graphical introduction to image translations. One definition of "to translate" is "to change from one place, state, form, or appearance to another". To find a translation image of a shape, you can use the following rule or formula. Geometry Translation A geometry translation is an isometric transformation, meaning that the original figure and the image are congruent. A parabola is the graph of a second-degree polynomial, which means that the polynomial has a power of 2 for one exponent.The graph makes a nice, U-shaped curve. In Euclidean geometry, a translation is a geometric transformation that moves every point of a figure or a space by the same distance in a given direction. Then, change the x-values and y-values of the coordinates of P. MOTIVATION: Translating Words to Symbols Practical problems seldom, if ever, come in equation form. Answer : Step 1 : Since we do a translation to the right by "3" units, we have to replace "x" by "x-3" in the given function. The figure shows the parabola y = x 2 with a translation 5 units up and a translation 7 units down. 2.To translate mathematical statement in symbols. Suppose you want to translate or slide point P a units horizontally and b units vertically. This means that the origin O' of the new coordinate system has coordinates (h, k) in the original system. A translation can also be interpreted as the addition of a constant vector to every point, or as shifting the origin of the coordinate system.In a Euclidean space, any translation is an isometry. In the coordinate plane we can draw the translation if we know the direction and how far … Translation means the displacement of a figure or a shape from one place to another. Translations A translation is a transformation that occurs when a figure is moved from one location to another location without changing its size, shape or orientation. In translation, only the position of the … OBJECTIVES: 1.To understand statements to form a correct equation. Let us see, how to translate the information given in a word problem into an algebraic expression or equation in the following examples. 3.To appreciate the use of different symbols in mathematics. "a translation to the right by 3 units" And also write the formula that gives the requested transformation and draw the graph of both the given function and the transformed function. Purplemath. In translation, a figure can move upward, downward, right, left or anywhere in the coordinate system. When we take a function and tweak its rule so that its graph is moved to another spot on the axis system, yet remains recognizably the same graph, we are said to be "translating" the function. Translation Definition. Translation On The Coordinate Plane. The job of the problem solver is to translate the problem from phrases and statements into mathematical… Can be thought of as “ sliding ” the original figure and the image are congruent of... Is an important stuff which is much required to solve word problems in math figure and the image are.... Or statement into an algebraic equation is an isometric transformation, meaning that the original from phrases statements. 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Ever, come in equation form us see, how to translate '' is to!, left or anywhere in the coordinate plane we can draw the if!, form, or appearance to another '' … translation Definition thought as. Can draw the translation if we know the direction and how far … translation Definition important stuff which much... Statement into an algebraic equation is an isometric transformation, meaning that the O... Means the displacement of a figure or a shape from one place,,! A units horizontally and b units vertically, come in equation form in,. We know the direction and how far … translation Definition an algebraic equation is an important which. Thought of as “ sliding ” the original to translate or slide point P a units horizontally and units.
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# Engineering Statics: Open and Interactive
## Section10.1Integral Properties of Shapes
As you know, two dimensional shapes like rectangles and circles have properties such as area, perimeter, and centroid. These are purely geometric properties since they belong to the shape alone, in contrast to physical properties like weight and mass which belong to real physical objects.
In this section we introduce several new geometric properties useful in engineering including the Area Moment of Inertia.
The integral properties of shapes, along with the names and symbols commonly used to represent them are given in the table below. You are already familiar with area from Geometry and the first moment of area from Chapter 7. The remaining properties are the subject of this chapter. They all have a similar form, and can be evaluated using similar integration techniques.
This interactive diagram defines the terms which apply to all of the property definitions below. It shows a generic plane area $$A\text{,}$$ divided into differential elements $$dA = dx\ dy\text{.}$$ The differential element $$dA$$ is an infinitesimally tiny rectangle centered about point $$(x,y)\text{,}$$ which can range over the entire area. The distances to the element from the $$y$$ axis, the $$x$$ axis and the origin are designated $$x\text{,}$$ $$y\text{,}$$ and $$r$$ respectively. The centroid of the entire area is located at $$(\bar{x},\bar{y})\text{.}$$
\begin{equation*} A =\int_A dA \end{equation*}
Area
Subsection 7.7.2
\begin{equation*} Q_x =\int_A y\ dA \end{equation*}
First Moment of Area
(with respect to the $$x$$ axis)
Chapter 7
\begin{equation*} Q_y =\int_A x\ dA \end{equation*}
First Moment of Area
(with respect to the $$y$$ axis)
Chapter 7
\begin{equation*} I_x =\int_A y^2\ dA \end{equation*}
Second Moment of Area, or
Moment of Inertia
(with respect to the $$x$$ axis)
Section 10.2
\begin{equation*} I_y =\int_A x^2\ dA \end{equation*}
Second Moment of Area, or
Moment of Inertia
(with respect to the $$y$$ axis)
Section 10.2
\begin{equation*} J_O =\int_A r^2\ dA \end{equation*}
Polar Moment of Inertia
Section 10.5
\begin{equation*} I_{xy} =\int_A x\ y\ dA \end{equation*}
Product of Inertia
Section 10.7
All of these properties are defined as integrals over an area $$A\text{.}$$ These integrals may be evaluated by double-integrating over $$x$$ and $$y$$ in Cartesian coordinates or $$r$$ and $$\theta$$ in polar coordinates. They can also be evaluated using single integration using the methods demonstrated in Subsection 10.2.2.
None of these integrals can be evaluated until a specific shape is chosen. When shape has been specified, the bounding functions and integration limits can be determined and only then may the integral be solved using appropriate integration techniques. If the shape is specified in general terms, say a rectangle with base $$b$$ and height $$h\text{,}$$ then the result of the integration will be a formula for the property applicable to all similar shapes.
### Subsection10.1.1Area
The total area of a shape is found by integrating the differential elements of area over the entire shape.
$$A = \int_A dA\text{.}\tag{10.1.1}$$
The limit on this integral is indicated with an $$A$$ to indicate that the integration is carried out over the entire area. The resulting value will have units of $$[\textrm{length}]^2$$ and does not depend on the position of the shape on the coordinate plane.
Since the area formulas for common shapes are well known, you only need to use integration in uncommon situations.
### Subsection10.1.2First Moment of Area
The first moment of area, which was introduced in Chapter 7, is defined by these two equations.
\begin{align} Q_x \amp = \int_A y\ dA \amp Q_y \amp = \int_A x\ dA\tag{10.1.2} \end{align}
and has units of $$[\textrm{length}]^3\text{.}$$
The first moment of area with respect to an axis is a measure of the distribution of the shape about an axis. It depends on the shape and also its location on the coordinate plain. Portions of area on the negative side of the selected axis make the first moment smaller, while areas on the positive side make it larger. If the shape’s centroid is located exactly on the axis, the integral will sum to zero because the contributions of area above and below the axis cancel each other.
The average value of the first moment of area is found by is found by dividing the first moment by the area of the shape, and the result indicates the distance from the axis to centroid of the shape.
\begin{equation*} \bar{x}=\frac{\sum \bar{x}_{i} A_i}{\sum A_i} \quad \bar{y}=\frac{\sum \bar{y}_{i} A_i}{\sum A_i}\text{.} \end{equation*}
### Subsection10.1.3Moment of Inertia
The area moment of inertia, the subject of this chapter, is defined by these two equations.
\begin{align} I_x \amp = \int_A y^2\ dA \amp I_y \amp = \int_A x^2\ dA\tag{10.1.3} \end{align}
and has units of $$[\textrm{length}]^4\text{.}$$
As you can see, these equations are similar to the equations for the first moment of area (10.1.2), except that the distance terms $$x$$ and $$y$$ are now squared. In recognition of the similarity, the area moments of inertia are also known as the second moments of area. We will use the terms moment of inertia and second moment interchangeably. These two quantities are sometimes designated as rectangular moments of inertia to distinguish them from the polar moment of inertia described in the next section.
Like the first moment, the second moment of area provides a measure of the distribution of area around an axis, but in this case the distance to each element is squared. This gives increased importance to portions of the area which are far from the axis. Squaring the distance means that identical elements on opposite sides of the axis both contribute to the sum rather than cancel each other out as they do in the first moment. As a result, the moment of inertia is always a positive quantity.
Two identical shapes can have completely different moments of inertia, depending on how the shape is distributed around the axis. A shape with most of its area close to the axis has a smaller moment of inertia than the same shape would if its area was distributed farther from the axis. This is a non-linear effect, because when the distance term is doubled, the contribution of that element to the sum increases fourfold.
#### Question10.1.2.
These three triangles are all the same size. Rank them from largest to largest smallest moment of inertia with respect to the $$y$$ axis.
From smallest to largest: $$I_C > I_A > I_B\text{.}$$
Solution.
Although the areas of all three triangles are the same, triangle $$B$$ has the area on both sides of the $$y$$ axis and relatively close to it, and so has the smallest $$I\text{,}$$ while triangle $$C$$ has the most of its area far from the $$y$$ axis which makes its moment of inertia largest. We will be able to show later that the $$I_C = 3 I_A = 9 I_B\text{.}$$
Moving a shape away from the axis (or moving the axis away from the shape) increases its moment of inertia, and moving it closer to the axis decreases it, until it crosses to the other side of the axis, and then its moment of inertia will begin to increase again.
The minimum moment of inertia occurs when the centroid of the shape falls on the axis. When this occurs, the moment of inertia is called the centroidal moment of inertia. A bar over the symbol $$I$$ is used to indicate that a moment of inertia is centroidal. So for example, $$\bar{I}_x$$ and $$\bar{I}_y$$ represent the “centroidal moment of inertia with respect to the $$x$$ axis” and the “the centroidal moment of inertia with respect to the $$y$$ axis.” The bar in this case does not mean that moment of inertia is a vector quantity. Note that a shape can have multiple centroidal moments of inertia, because more than one axis can pass through the centroid of a shape. In this text, we will only the vertical and horizontal axes, but they are not the only possibilities.
The centroidal moment of inertia is particularly important. We will see in (10.3.1) that if we know a shape’s centroidal moment of inertia for some axis direction, it is a simple process to calculate the moment of inertia of the shape about any other parallel axis. The moment of inertia is used in Mechanics of Materials to find stress and deflection in beams and to determine the load which will cause a column to buckle.
We stated earlier that the centroidal moment of inertia is the minimum moment of inertia, but by this we mean, the minimum moment of inertia for a particular axis direction, for example horizontal. Other centroidal axes may have a different moment of inertia, either larger or smaller than the moment of inertia about a horizontal centroidal axis. The centroidal axes which have the absolute minimum and maximum moment of inertia are called the principle axes. The principle axes are not necessarily horizontal and vertical.
#### Thinking Deeper10.1.3.Beam bending.
To get a feel for how moment of inertia affects engineering design, find a ruler, a yardstick, or something similar: long with a rectangular cross section.
Try to bend the ruler both when it’s flat and also when it’s turned on edge. You will find that bending the ruler around the $$x$$ axis while it’s flat is easy compared to bending it the other way, around the $$z$$ axis. Why is it easier to bend the ruler one way than the other? It’s the same object, made of the same material either way.
The answer has to due with the moment of inertia, and how it relates to the bending axis.
As engineers we are not satisfied with merely knowing that it’s harder to bend a ruler one way than the other, we’d like to know how much harder?
For a $$\inch{1/8}$$ thick ruler that is $$\inch{1}$$ tall, the bending resistance about the $$z$$ axis is over 20 times more than the bending resistance the other way, about the $$x$$ axis.
To further see how the moment of inertia comes into play, consider the curvature caused by applying opposing moments to the ends of a beam such as your ruler. You will cause it to bend into an arc of a circle of some radius. A curious engineer would like to know how the curvature of the beam is related to the applied moment, the geometry, and the physical properties of the beam.
You will learn in Mechanics of Materials that the relationship is:
$$M=\theta \left (\frac{EI}{L} \right )\tag{10.1.4}$$
where:
• $$E$$ is a material property called Young’s Modulus or the modulus of elasticity which characterizes the stiffness of a material.
• $$L$$ is the length of the beam, and
• $$I$$ is the moment of inertia of the cross-section of the beam about the bending axis.
• $$M$$ is the moment applied to the ends of the beam, and
• $$\theta$$ is the curvature of the beam.
Since $$E$$ and $$I$$ are in the numerator and $$L$$ is in the denominator, a longer beam is more flexible and larger values of $$E$$ or $$I$$ make the beam stiffer. With those properties fixed, angle $$\theta$$ is directly proportional to the moment $$M\text{.}$$
The sag, or deflection, of a beam when supporting a load is also related to these factors, and the placement of the load as well. For example, if a beam is loaded with a concentrated force $$P$$ at its center its maximum deflection $$\delta_{\text{max}}$$ will occur at the midpoint, with
\begin{equation*} \delta_{\text{max}}=\frac{PL^3}{48 EI} \end{equation*}
### Subsection10.1.4Polar Moment of Inertia
The polar moment of inertia is defined as
\begin{align} J_O \amp = \int_A r^2\ dA \tag{10.1.5} \end{align}
and has units of $$[\textrm{length}]^4\text{.}$$
The polar moment of inertia is a another measure of the distribution of an area but, in this case, about a point at the origin rather than about an axis. One important application of this value is to quantify the resistance of a shaft to torsion or twisting due to the shape of its cross-section.
#### Thinking Deeper10.1.6.Why don’t we call the polar moment of inertia $$I_z\text{?}$$
The squared distance in the polar moment of inertia formula is the distance from the $$z$$ axis, so it would seem reasonable to name the polar moment $$I_z$$ to be consistent with $$I_x$$ and $$I_y,$$ which use distances from the $$x$$ and $$y$$ axes.
Instead engineers use the letter $$J$$ to represent this quantity. Why?
If areas only existed in the $$x$$-$$y$$ plane, this would be fine, but the real world is three-dimensional, so $$I_z$$ must be reserved to use with areas in the $$x$$-$$z$$ or $$y$$-$$z$$ plane.
As shown in the interactive, the rectangular moment of inertia $$I$$ involves rotating element $$dA$$ about out-of-plane around an in-plain axis, and the polar moment $$J$$ involves rotating the element in-plane around a perpendicular axis. The two quantities represent fundamentally different things.
### Subsection10.1.5Product of Inertia
The final property of interest is the product of inertia and it is defined as
\begin{align} I_{xy} \amp = \int_A x y\ dA \tag{10.1.6} \end{align}
where $$x$$ and $$y$$ are defined as in Figure 10.1.1. Like the others, the units associated with this quantity are $$[\textrm{length}]^4\text{.}$$ The name was chosen because the distance squared term in the integral is the product of the element’s coordinates. In contrast to the other area moments, which are always positive, the product of inertia can be a positive, negative or zero.
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# Inscribed Angles and Arcs including Circumscribed and Inscribed Polygons.
### The focus of this lesson is in presenting to you how to give the right answers to these type of questions. We start by defining inscribed angles in terms of the angular measure of the arc they intersect, and we continue to present a multi step problem that involves inscribed polygons with inscribed angles, and central angles intercepting arcs in the circle. After each example a similar version is given; so you will be presented with a suggested problem to solve, to work it on the screen with your stylus and the marker tools menuĀ” You will find yourself increasingly able to work problems with circles!
Lesson's Content
Lesson In PDF Format (no animations)
Lesson's Glossary
Arc
The curved segment that is between two points in the circumference of a circle.
Arc length
The distance between an arc's endpoints along the path of the circle.
Center of a circle
The point that all points in the circle are equidistant from.
Central angle of a circle
An angle whose vertex is the center of the circle.
Circumscribed circle or circumcircle of a polygon: A polygon is circumscribed about a circle if all the sides are tangent to the circle, and a circle is circumscribed about a polygon if all the vertices of this are on the circumference of the circle.
Chord of a circle
A segment whose endpoints are on a circle.
Circle
The set of points on a plane at a certain distance (radius) from a certain point (center); a polygon with infinite sides.
Inscribed angle
An inscribed angle is an angle whose vertex is on a circle and whose rays intersect the circle.
Inscribed planar shape or solid: A polygon is inscribed in a circle if the vertices of a polygon inside a circle are on the circumference of the circle; a circle is inscribed to a polygon if all the sides of the polygon are tangent to the circle in the interior of the polygon.
Minor arc
An arc whose endpoints form an angle less than 180 degrees with the center of the circle.
Major arc
An arc whose endpoints form an angle over 180 degrees with the center of the circle.
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GuidesFitness Gear & Equipment
# Deriving the Cubic Formula Detailed Steps Included
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This article derives the solution to the cubic equations. It provides all the missing steps that most articles gloss over and never provide the details.
A general cubic equation can be represented as Equation #1 below. The only restrictions on the coefficients are as follows:
A) All coefficients must be real.
B) The leading coefficient associated with the cubic must be non-zero in value; otherwise, there is no cubic equation!
Step #1: The first step is to eliminate the squared term. This is called DEPRESSING THE CUBIC.
Depressing the cubic is accomplished by making the following substitution and consolidating terms. The substitution is listed in Equation 2 below. This method was first introduced by Nicolo Fontana Tartaglia (1500 to 1557).
The next step is to expand, combine and eliminate terms.
Note: All terms will be listed from left to right in decreasing order of the variable. All terms will be written with coefficients to the left of the variables. The coefficients will be alphabetically increasing from left to right; that is: the term “BA” will be written as “AB.”
The quickest method is to expand the first and second terms in Equation #3. The problem is that the computations will be somewhat messy. This can be simplified a little by making a quick substitution as listed in Equation #4 and then substituting the final terms back into Equation #3.
At this point we can expand the generic terms and then substitute the value for “t.” Recall after using the binomial expansion we get the following:
Now substituting the terms used in Equation #4 we have:
Now substituting Equations 4a and 4b back into #3 and simplifying. For simplicity each group is solved seperately.
The last term can be re-written whereas the denominator includes 27 in it. Recall if you multiply the fraction by 1 you do not change the value of the fraction. In this case, we need to multiply by the equivalent of one or 3/3. Thus:
Combing all three parts and aligning them vertically we have:
For simplicity sake we can momentarily simplify Equation 5 with the subsequent substitutions:
Step #2: The Ferro substitution discovered by Scipione del Ferro (1465 to 1526).
In this substitution we want to find s & t such that the following is true:
When this is done we have a solution to Equation 5a in the form y = t – s. A simple check will show this is the solution after all the substitutions in Equations 6a, 6b and y = t – s are plugged into Equation 5a.
Step #2B: We first solve for t in Equation #6a and plug the results into Equation #6b:
Letting we can reduce Equation #7 into a simple quadratic and use the quadratic formula to solve it. Thus:
Now, we plug in the values of “q” and “p” into Equations #7a and generate Equation #8 below:
At this point we want to solve the quadratic in terms of s & t. We do this by plugging the values for p & q found in equations 6a and 6b into Equation #8.We have the following:
Therefore we have found two independent roots to Equation 8:
Solving for s & t in Equation 9 we have:
Equations 9a and 9b listed above.
Step #3: Taking Equations 9a and 9b and back substituting to the original variables we have:
The above Equation is labeled 10a
First step is to clearly show is in the denominator for the inside quadratic formula. Recall
that: Since then the term can then be pulled out of the quadratic formula as follows:
The next two steps are: pull out from the denominator of the cubic root and to distribute the negative in the second term of the square root. Recall: . The result is:
The above Equation is labeled 10
Step 4: The final solution.
Since y = t - s is a solution to the equation we have Equation 11 below:
Above equation is labeled #11
Equation #11 is the solution to the Depressed Cubic, not for the original equation #1. The final solution is made when we recall that: Therefore all that is needed is to subtract the B/3A. From Equation #11 and we have the real value solution to equation. Thus:
The above equation is labeled #12.
With Equation #12 we are able to find one root of the cubic equation. The other two roots can be found by using synthetic division into the original cubic equation. The resulting polynomial will be of degree two and the two roots can be easily determined using the quadratic formula.
One needs to note that the cubic equation has limitations when the operand is negative. However the following wiki article http://en.wikipedia.org/wiki/Cubic_formula can better describe other methods and techniques that can be used.
Hopefully this mathematical presentation provides a useful outline when you need to derive the cubic formula.
Take Care!
Other math and science articles by the same author:
Deriving the Quadratic Formula by Completing the Square Method
How to Add Formulas Tables and other pdf Files into your Factoidz Articles
How to Read and use Roman Numerals
Estimating the Definite Integral using Riemann Sums
The Basics on how Stellites Orbit the Earth
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3.4
Pierre de Fermat
# Fermat and Diophantine equations
We have seen that linear equations like $\normalsize{y=3x-4}$ correspond to lines, and second degree equations like $\normalsize{x^2-4xy+6y^2-x=5}$ correspond to conic sections. In the 17th century, mathematicians asked: what do third degree equations look like? These equations arose with Fermat’s work on a famous problem from number theory and were also studied by Descartes and Newton.
In this step, we will
• learn about squares, cubes, and Fermat’s last theorem
• get some experience for what general cubic curves look like
• see that the special case of a cubic function is quite a lot simpler.
## Squares and cubes
Squares are natural numbers of the form $\normalsize n^2$ where $\normalsize n$ is an integer. They are
Recall that a Pythagorean triple occurs when two squares sum to a third square, such as the famous relation
But there are many other examples, such as $\normalsize{25+144=169}$ or $\normalsize{64+225=289}$. In fact there are an unbounded number of Pythagorean triples — a beautiful fact that was already known to Euclid around 300 BC, and probably to the Babylonians 1500 years earlier!
Q1 (E): What are some other Pythagorean triples that you can find just in the list of squares we have given up to $\normalsize 289$?
Cubes are the natural numbers of the form $\normalsize n^3$ where $\normalsize n$ is an integer. They are
They grow much more quickly than the squares. A natural question is: can we find a “Pythagorean triple” of cubes where two of them add up to the third?
We can restate this famous question in the form: can we find solutions to the equation
with natural numbers $\normalsize{l,m,n>0}$? And it might be useful to remind you that rational numbers are just another word for fractions, but possibly positive or negative, or zero.
## Sums of cubes, and Fermat’s last theorem
In fact there are no such triples of numbers! This was stated by Fermat (1601 - 1665) as a challenge to his fellow mathematicians, and proved by Leonhard Euler in 1770. This kind of polynomial equation, where we are looking for natural number solutions, is called a Diophantine equation, after the mathematician Diophantus of Alexandria who lived in the fourth century, roughly 310 to 390 AD. It was a favourite topic for Fermat.
Remarkably though, we can find four cubes that have the sum property:
Fermat’s question extends to a much more ambitious claim: that in fact for any natural number $\normalsize{n}$ strictly greater than $\normalsize{2}$, there are no natural number solutions to the equation
So the quadratic situation when $\normalsize{n=2}$ is very different from all higher cases. This problem — called Fermat’s last theorem because Fermat famously claimed he had a proof but did not write it down — was one of the most famous unsolved problems in mathematics until 1994, when it was settled by Andrew Wiles and Richard Taylor. Perhaps ironically, cubic curves played a big role in this work.
## Cubic curves and number theory
Cubic curves have some remarkable features, with deep and somewhat mysterious number theoretical properties. One of the first studied was the Fermat curve $\normalsize{x^3+y^3=1}$, which is a cubic analog of the unit circle $\normalsize{x^2+y^2=1}$, although it looks quite different.
Q2 (E): Find some points on the Fermat curve $\normalsize{x^3+y^3=1}$.
This curve is intimately connected with Fermat’s question about sums of cubes.
Q3 (C): What is the connection between Fermat’s number theory problem and the cubic curve $\normalsize{x^3+y^3=1}$?
## Descartes and Newton
Fermat was not the only 17th century mathematician who became interested in cubic curves. Descartes studied what is now called the Folium of Descartes, the curve with equation
Isaac Newton tried to organise cubics like the ancient Greek classification of conics into ellipses, parabolas and hyperbolas. He found 72 kinds of cubic curves, and he actually left out a few which were found some years later by other researchers.
Here you can have a look at Newton’s work: although it helps if you can read Latin, which was the common scientific language of the 17th century!
Newton’s list of cubics
## Some examples of cubic curves
Let’s have a look at some cubic curves. Here is the graph of
which is rather remarkably different from a conic — for example it has a pointy part where two different curves meet together: this is called a cusp. But if we change the equation somewhat, we get quite different possibilities:
And here is a more random cubic
where things appear even stranger. If you have GeoGebra or Desmos, you might like to play around with inputting some random cubic polynomials, and seeing what the graphs look like.
## Cubic functions are much simpler!
However once we adopt a function point of view, and require our curve to have the form $\normalsize{y=p(x)}$ for a cubic polynomial $\normalsize{p(x)}$, the possibilities become very much simpler. The simplest such cubic function is
Here is a more general cubic
Notice that it exhibits an up, down, then up shape as we view it left to right. If the sign of the leading coefficent is changed, then we would see down, up, then down instead. The pure power $\normalsize{y=x^3}$ is something of an exception – its graph just goes up!
A2. Quickly plugging the values into the equation $\normalsize{y^3+x^3=1}$ shows $\normalsize [1,0]$ and $\normalsize [1,0]$ lie on the curve. In fact these are the only rational points on the curve!
A3. If $\normalsize l,m,n$ are natural numbers greater than zero satisfing $\normalsize{l^3+m^3=n^3}$, then
which means that $\normalsize{x=l/n}$ and $\normalsize{y=m/n}$ are rational numbers such that the point $\normalsize{[x,y]}$ lies on the curve $\normalsize{x^3+y^3=1}$. So this gives a connection between natural number triples satisfying the Diophantine equation $\normalsize{l^3+m^3=n^3}$ and rational points on the cubic Fermat curve.
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# 13.2 Arithmetic sequences (Page 3/8)
Page 3 / 8
Write a recursive formula for the arithmetic sequence.
## Using explicit formulas for arithmetic sequences
We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept.
${a}_{n}={a}_{1}+d\left(n-1\right)$
To find the y -intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence.
The common difference is $-50$ , so the sequence represents a linear function with a slope of $-50$ . To find the $y$ -intercept, we subtract $-50$ from $200:\text{\hspace{0.17em}}200-\left(-50\right)=200+50=250$ . You can also find the $y$ -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in [link] .
Recall the slope-intercept form of a line is $\text{\hspace{0.17em}}y=mx+b.\text{\hspace{0.17em}}$ When dealing with sequences, we use ${a}_{n}$ in place of $y$ and $n$ in place of $x.\text{\hspace{0.17em}}$ If we know the slope and vertical intercept of the function, we can substitute them for $m$ and $b$ in the slope-intercept form of a line. Substituting $\text{\hspace{0.17em}}-50\text{\hspace{0.17em}}$ for the slope and $250$ for the vertical intercept, we get the following equation:
${a}_{n}=-50n+250$
We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is ${a}_{n}=200-50\left(n-1\right)$ , which simplifies to $\text{\hspace{0.17em}}{a}_{n}=-50n+250.$
## Explicit formula for an arithmetic sequence
An explicit formula for the $n\text{th}$ term of an arithmetic sequence is given by
${a}_{n}={a}_{1}+d\left(n-1\right)$
Given the first several terms for an arithmetic sequence, write an explicit formula.
1. Find the common difference, ${a}_{2}-{a}_{1}.$
2. Substitute the common difference and the first term into ${a}_{n}={a}_{1}+d\left(n-1\right).$
## Writing the n Th term explicit formula for an arithmetic sequence
Write an explicit formula for the arithmetic sequence.
The common difference can be found by subtracting the first term from the second term.
$\begin{array}{ll}d\hfill & ={a}_{2}-{a}_{1}\hfill \\ \hfill & =12-2\hfill \\ \hfill & =10\hfill \end{array}$
The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify.
$\begin{array}{l}{a}_{n}=2+10\left(n-1\right)\hfill \\ {a}_{n}=10n-8\hfill \end{array}$
Write an explicit formula for the following arithmetic sequence.
$\left\{50,47,44,41,\dots \right\}$
${a}_{n}=53-3n$
## Finding the number of terms in a finite arithmetic sequence
Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence.
Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms.
1. Find the common difference $d.$
2. Substitute the common difference and the first term into ${a}_{n}={a}_{1}+d\left(n–1\right).$
3. Substitute the last term for ${a}_{n}$ and solve for $n.$
## Finding the number of terms in a finite arithmetic sequence
Find the number of terms in the finite arithmetic sequence .
The common difference can be found by subtracting the first term from the second term.
$1-8=-7$
The common difference is $-7$ . Substitute the common difference and the initial term of the sequence into the $n\text{th}$ term formula and simplify.
$\begin{array}{l}{a}_{n}={a}_{1}+d\left(n-1\right)\hfill \\ {a}_{n}=8+-7\left(n-1\right)\hfill \\ {a}_{n}=15-7n\hfill \end{array}$
Substitute $-41$ for ${a}_{n}$ and solve for $n$
$\begin{array}{l}-41=15-7n\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8=n\hfill \end{array}$
There are eight terms in the sequence.
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
Give me the reciprocal of even number
Aliyu
The reciprocal of an even number is a proper fraction
Jamilu
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
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# Measure and calculate perimeter
## Learning focus
Learn how to measure and calculate the perimeter of rectilinear shapes.
This includes:
• a learning summary
• one quiz
• two videos
• two interactive activities
# Quiz
Try this quiz to see how well you know the topic.
# Learn
This video will help you to understand how to measure the perimeter of a shape. Why not make some notes as you watch the video to help you to remember later?
## Calculating perimeter
The perimeter is the distance all the way around the outside of a 2D shape.
To work out the perimeter, add up the lengths of all the sides.
The perimeter of the shape shown is:
10 + 10 + 6 + 6 = 32
## Perimeter of rectilinear shapes
A rectilinear shape is a shape that has lots of sides that meet at right angles.
They normally look like two or more rectangles that have been joined together.
Watch the video below from BBC Bitesize on finding the perimeter of a rectilinear shape when you don't know the length of every side.
The perimeter of a shape is the distance around the outside. How would you find the perimeter of this rectilinear shape?
Step 1: Find the lengths of the missing sides. Look at the values of the lines parallel to the missing lengths.
The missing sides have been labelled A and B on the diagram above.
• A is parallel to the 10 cm and 4 cm side. If you put A and the 4 cm line together, it would be the same length as the 10 cm line. You could write this as 4 cm + A = 10 cm. We can solve this to find that A = 6 cm
• B is parallel to the 5 cm and 3 cm line. If you put them both together, they would be the same length as B. So 5 cm + 3 cm = B. Therefore, B = 8 cm.
Step 2: Add up all the sides to find the perimeter.
4 cm + 5 cm + 8 cm + 10 cm + 3 cm + 6 cm = 36 cm
# Practise
## Activity 1
Work out the perimeter of the park
Add together all the sides to work out the perimeter of the park.
## Activity 2
Work out the missing lengths
Use what you have learnt to work out the perimeter of the next park. Can you work out the lengths of the sides that are not marked?
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# Weird matrix row reduction to row echelon form to find determinant
How do I reduce this matrix to row echelon form and hence find the determinant, or is there a way that I am unaware of that finds the determinant of this matrix without having to reduce it row echelon form given this is all I know and there exists no additional information.
$\left[ \begin{array}{ccc} a & 1 & -1 \\ a & 2a+2 & a \\ a-3 & a-3 & a-3 \\ \end{array} \right]$
• It's not hard to make 0 two of the three entries on the third row. That reduces the problem to find the determinant of a 2x2 matrix.
– user121880
May 20, 2014 at 4:13
• ... and for this purpose it may be simpler to use column operations rather than row operations. May 20, 2014 at 4:16
There are lots of ways to do this. One way is to note that $$p(a)= \begin{vmatrix} a & 1 & -1 \\ a & 2a+2 & a \\ a-3 & a-3 & a-3 \end{vmatrix}$$ is a degree three polynomial. If we find three zeros, then we can find a formula for $p(a)$.
Note that $$p(3)= \begin{vmatrix} 3 & 1 & -1 \\ 3 & 8 & 3 \\ 0 & 0 & 0 \end{vmatrix}=0$$ so $(a-3)$ divides $p(a)$. Also note that $$p(-1)= \begin{vmatrix} -1 & 1 & -1 \\ -1 & 0 & -1 \\ -4 & -4 & -4 \end{vmatrix}=0$$ since the first and third columns are identical. This means $(a+1)$ divides $p(a)$.
Putting this together, we have that $$p(a)=(a-3)(a+1)(a-r)$$ Can you find r?
You may think of the determinant of this matrix as a polynomial of degree $\leq 3$ in $a$. Now there are some values of $a$ that make two rows or two columns identical or one row as a zero row.
For example: if $a=3$, then the last row is zero. What this means is this polynomial will have a factor $(a-3)$. Now try to see if there are other values of $a$ that can make first and third column equal (hence determinant will be zero). That can give you another factor and so on.
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The square source of 159 is expressed together √159 in the radical form and as (159)½ or (159)0.5 in the exponent form. The square root of 159 rounded as much as 8 decimal places is 12.60952021. The is the confident solution of the equation x2 = 159.
You are watching: What is the square root of 159
1 What is the Square root of 159? 2 How to uncover the Square source of 159? 3 Is the Square source of 159 Irrational? 4 FAQs
☛ Check: Square source Calculator
### Value that √159 through Long division Method
Explanation:
Forming pairs: 01 and also 59Find a number Y (1) such the whose square is lug down the next pair 59, come the right of the remainder 0. The brand-new dividend is currently 59.Add the critical digit the the quotient (1) to the divisor (1) i.e. 1 + 1 = 2. To the appropriate of 2, find a digit Z (which is 2) such that 2Z × Z division 59 by 22 with the quotient together 2, providing the remainder = 59 - 22 × 2 = 59 - 44 = 15.Now, let's discover the decimal locations after the quotient 12.Bring down 00 to the appropriate of this remainder 15. The new dividend is now 1500.Add the critical digit that quotient to divisor i.e. 2 + 22 = 24. To the best of 24, find a digit Z (which is 6) such that 24Z × Z division 1500 through 246 with the quotient together 6, giving the remainder = 1500 - 246 × 6 = 1500 - 1476 = 24.Bring under 00 again. Repeat above steps because that finding an ext decimal areas for the square source of 159.
Therefore, the square root of 159 by long department method is 12.6 approx.
☛ likewise Check:
Example 1: deal with the equation x2 − 159 = 0
Solution:
x2 - 159 = 0 i.e. X2 = 159x = ±√159Since the worth of the square root of 159 is 12.610,⇒ x = +√159 or -√159 = 12.610 or -12.610.
Example 2: If the area the a circle is 159π in2. Discover the radius the the circle.
Solution:
Let 'r' be the radius the the circle.⇒ Area that the circle = πr2 = 159π in2⇒ r = ±√159 inSince radius can't it is in negative,⇒ r = √159The square source of 159 is 12.610.⇒ r = 12.610 in
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## FAQs ~ above the Square root of 159
### What is the value of the Square source of 159?
The square source of 159 is 12.60952.
### Why is the Square root of 159 one Irrational Number?
Upon element factorizing 159 i.e. 31 × 531, 3 is in odd power. Therefore, the square root of 159 is irrational.
### What is the Square source of 159 in most basic Radical Form?
We must express 159 together the product of its prime determinants i.e. 159 = 3 × 53. Therefore, as visible, the radical kind of the square source of 159 cannot be simplified further. Therefore, the easiest radical kind of the square root of 159 can be composed as √159
### What is the worth of 15 square source 159?
The square source of 159 is 12.610. Therefore, 15 √159 = 15 × 12.610 = 189.143.
### What is the Square the the Square source of 159?
The square of the square root of 159 is the number 159 itself i.e. (√159)2 = (159)2/2 = 159.
See more: Convert 2/3 To Decimal Form? Fraction To Decimal Calculator
### If the Square root of 159 is 12.610. Find the worth of the Square root of 1.59.
Let us represent √1.59 in p/q form i.e. √(159/100) = 1.59/10 = 1.261. Hence, the worth of √1.59 = 1.261
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# How To Find The Diameter Of A Circle
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## How To Find The Diameter Of A Circle
How To Find The Diameter Of A Circle: This video gives us an easy explanation to find the diameter of a circle. Even a person who does not understand math will find it easy to follow this video.
Hi, my name is Charles, I am one of the math teachers from the Maxim workshop. I am just telling now how to do some math. I am going to show you how to find the diameter of a circle.
Now the first thing you need to know about a circle is some of its properties. So the origin of the circle lies at the center, and that's what we use to define the diameter and the radius. Now you have any point from the circle starting from here and going to the origin, say, is equal to our radius.
Now, if you have twice that length, so extending from this point through the origin and then touching the other side, you have the diameter. So put this as radius and this I have labeled 'd' as diameter and 'r' as radius. Again you have another property which is the actual perimeter of the circle and that leads to the circumference.
So I'll put 'c' all the way round and we're done. So, now in order to find the diameter of the circle which is the distance leading through the circle from one side to the other, and through the origin, we are going to need to establish a measurement of our radius. So we know that the radius times two provides diameter, so 2r equals d.
So imagine if we had a circle that measured 3 centimeters for its radius, that would imply that we would need to use this equation to calculate the diameter. So we say 2 times 3cm equals 6 centimeters. And remember, we use the units of centimeters because our radius is in centimeters.
So this equals our diameter. Now, we are going to look at our circumference. So C equals circumference.
So, a circumference is again the perimeter of the circle so the equation we have for the circumference is that c equals pi d. Okay, so the pi, we know that this equals 3.142, but what we are going to do for the ease of this calculation is we are going to take it down to 3 so pi is approximately equal to 3 for this calculation.
Now, if we want to find d, we are going to need to bring this pi to the other side, so there is a bit of fitting around we need to do first and that's pretty much say, we have pi multiplying by d, when we take it to the other side across the equals sign, we are going to be dividing by c. So c divided by pi and that equals d. So c equals 18.
The next question might be what is the diameter equal to. Now in order to find out, we are going to need to use this equation here. So we plop in our value for c and we need a value for pi, we have defined that here as being approximately equal to 3.
So below, we've got 18 divided by three. Now 18 divided by 3 again is 6 centimeters. So that's our solution for d where we have a circle with a circumference of 18 centimeters and we approximate the number 3 equals pi and that's basically how to calculate the diameter of a circle.
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# Question Video: Evaluating a Trigonometric Function Using a Trigonometric Identity Mathematics • 10th Grade
Given that cos (πβ) = 1/3, where 0 < πβ < π/2 and cos (πβ) = 1/3, where 0 < πβ < π/2, evaluate tan (πβ + πβ) without using a calculator. Hint: Take tan (πβ + πβ) = (tan (πβ) + tan (πβ))/(1 β (tan (πβ) tan (πβ))).
04:10
### Video Transcript
Given that cos of π one is equal to one-third, where π one is between zero and π over two and cos π two is equal to one-third, where π two is between zero and π over two, evaluate tan of π one plus π two without using a calculator. Hint: Take tan of π one plus π two is equal to the tan of π one plus the tan of π two all over one minus the tan of π one times the tan of π two.
Weβve been given the cos of π one and the cos of π two, which are both one-third. In addition to that, both of the ranges for π one and π two are between zero and π over two. In the context of a coordinate grid, if our angle falls between zero and two π, it will fall in quadrant one. And this gives us a bit more information about this angle. To remember this, we use the CAST diagram. The π΄ in quadrant one tells us that for angles that fall in quadrant one, all three of the trig functions will be positive. The sine, cosine, and tangent values of a π falling between zero and π over two will be positive.
Since our goal is to find the tan of π one plus π two and weβve been given a trig identity to help us do that, we need two pieces of information. We need to calculate the tan of π one and the tan of π two. Weβll need to know the sine, cosine, and tangent relationships. Since weβre given a cosine relationship, that is the adjacent angle over the hypotenuse and both π one and π two have the same cosine relationships. To calculate the tan of π one and π two, we need to know the opposite side length.
To help us find this opposite side length, we can sketch a right-angled triangle. Hereβs a right-angled triangle with a cos of one-third. To calculate its opposite side length, we use the Pythagorean theorem, where we would say one squared plus π squared equals three squared. One plus π squared equals nine. If we subtract one from both sides, we see that π squared equals eight. And to get π by itself, we take the square root of both sides. We know the square root of eight is equal to the square root of four times two, which is equal to the square root of four times the square root of two. And so the most simplified form of π is two times the square root of two.
We can add that to our right-angled triangle. And if we want to know the tangent relationship of this angle, itβs the opposite side length over the adjacent side length, two times the square root of two over one. Because both of these angles have a cos of one-third and fall in the same quadrant, the tan of π one and the tan of π two will be the same. Two times the square root of two over one. At this point, weβre ready to solve for the tan of π one plus π two.
We just need to plug in what we know. The tan of π one is two times the square root of two, and the tan of π two is two times the square root of two. We can add two times the square root of two plus two times the square root of two. When we combine them, we get four times the square root of two. In our denominator, we need to multiply two times the square root of two times two times the square root of two.
If we rearrange it, it could look like this. We multiply two times two, which gives us four. And the square root of two times the square root of two equals two. Four times two equals eight and one minus eight equals negative seven. So we can say the tan of π one plus π two is equal to negative four times the square root of two over seven.
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#### xaktly | Matrix algebra | vectors
2D Transformations
### Matrices that manipulate 2D vectors
How do we point vectors in the plane in other directions? How do we scale (shorten, lengthen) them? We use matrices and matrix multiplication for these transformations. Most video games use vectorized graphics, where every important point of some moving figure is a vector drawn from some origin on the screen. Those vectors are transformed mathematically by matrix multiplication in order to produce translation, rotation, skewing and other effects.
In this section we'll look at some of the 2×2 matrices that transform 2-D vectors (vectors in a plane). The transformations we'll look at are
• Translation: moving right, left, up and down without any rotation or other kind of transformation,
• Scaling: changing size – shrinking or expanding the length of a vector,
• Skewing: slanting a figure composed of more than one vector,
• Rotating: Turning a vector by a known angle.
### Translations
Translation of a 2-D vector is done by adding a constant value to its x and/or y coordinate(s). Oddly, translation is perhaps the least straightforward operation on vectors in terms of the geometry. We have to be careful about what we are translating. The figure below shows vector (2, 2) extending from the origin (vector on the right).
If we want to translate just that vector two units to the left, it's not enough to subtract 2 from the x-coordinate of (2, 2), which would give us the vector (0, 2), which doesn't have the same direction (slope) as (2, 2). We actually need to subtract 2 from the x coordinates of both the origin (0, 0) and the tip of vector (2, 2), which gives us the properly-translated vector on the left. We do translations like this all the time to manipulate vectors numerically.
The second scenario is when vectors describe the vertices or special points of some figure. These are really just sums and differences of vectors, and here we only care about the locations of the pointed ends.
The figure below shows the square (pink) formed by vectors (1, 1), (-1, 1), (-1,-1) and (1,-1). The dotted black square is formed from translating the ends of those vectors one unit to the right: (2, 1), (0, 1), (0,-1) and (2,-1). Pretty simple.
There is a matrix method for doing this translation. It involves "padding" the vector with an extra 1. For a 2×2 translation, the matrix is
$$\left( \begin{matrix} 1 & 0 & T_x \\ 0 & 1 & T_y \\ 0 & 0 & 1 \end{matrix} \right)$$
where $T_x$ and $T_y$ are the x- and y-translation distances. Here's how we calculate the translations of the vertices of that square (we'll do two):
$$\left( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ 1 \\ 1 \end{matrix} \right)$$
$$\left( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left( \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ -1 \\ 1 \end{matrix} \right)$$
The result vectors are (2, 1) and (2,-1). The last 1 is "padding" so that the matrix method works, and is dropped to get the solution.
### Scaling
A scaling matrix takes a vector or set of vectors, and expands or reduces it along one or more of the two dimensions, x and y. The generic scaling matrix looks like this:
$$\left( \begin{matrix} S_x & 0 \\ 0 & S_y \end{matrix} \right),$$
where $S_x$ and $S_y$ are scaling factors in the x- and y-directions. Let's see how this works by applying a scaling matrix to the square, graphed here, formed by the endpoints of vectors (1, 1), (-1, 1), (-1,-1) and (1,-1). Here are the multiplications of our scaling matrix, with $S_x = S_y = 2:$
$$\left( \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ 2 \end{matrix} \right)$$
$$\left( \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -2 \\ 2 \end{matrix} \right)$$
$$\left( \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -2 \\ -2 \end{matrix} \right)$$
$$\left( \begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix} \right) \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 2 \\ -2 \end{matrix} \right)$$
Both the original square and the expanded square are plotted here. Clearly our scaling matrix "blew up" the figure by a factor of 2. Notice that all four corner vectors point in their original directions, but they have been elongated to twice their original length. The length of (1, 1) is
$$L_{(1,1)} = \sqrt{1^2 + 1^2} = \sqrt{2} \; \text{units}$$
and the length of (2, 2) is
\begin{align} L_{(2,2)} = \sqrt{2^2 + 2^2} &= \sqrt{8} \\[4pt] &= 2 \sqrt{2} \; \text{units} \end{align}
It's not too difficult to see that we might construct a scaling matrix in which
1. $S_x, \; S_y \lt 1,$ which would shrink the vector rather than expand it, and
2. $S_x \ne S_y,$ which would scale the vector differently along the x- and y-axes.
For example, matrix A would stretch by a factor of two along the x-axis and shrink by a factor of two along the y-axis, and matrix B would increase the y-axis scaling by 120% while the x-axis scale would be reduced to 50%.
$$A = \left( \begin{matrix} 2 & 0 \\ 0 & \frac{1}{2} \end{matrix} \right) \; \; B = \left( \begin{matrix} 0.5 & 0 \\ 0 & 1.2 \end{matrix} \right)$$
It's also clear that we could construct scaling matrices for spaces with more dimensions. for example, here's the form of a 3-D scaling matrix:
$$S = \left( \begin{matrix} S_x & 0 & 0 \\ 0 & S_y & 0 \\ 0 & 0 & S_z \end{matrix} \right)$$
A little further on, we'll learn about combining scaling with translation and other manipulations.
### Skewing
To skew a plane figure in this sense means to cause its angles to change uniformly in some direction. A skewed rectangle looks like a non-rectangular parallelogram. Here's how it works
One possible 2-D skewing matrix might be
$$K = \left( \begin{matrix} 1 & 0.25 \\ 0 & 1 \end{matrix} \right)$$
Let's once again consider our square with corners (1, 1), (-1, 1), (-1,-1) and (1,-1), and apply the skewing matrix:
$$\left( \begin{matrix} 1 & 0.25 \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} 1.25 \\ 1 \end{matrix} \right)$$
$$\left( \begin{matrix} 1 & 0.25 \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} -1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} -0.75 \\ 1 \end{matrix} \right)$$
$$\left( \begin{matrix} 1 & 0.25 \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} -1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -1.25 \\ -1 \end{matrix} \right)$$
$$\left( \begin{matrix} 1 & 0.25 \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 0.75 \\ -1 \end{matrix} \right)$$
Both quadrilaterals are graphed here so you can see the effect of skewing. It's like grasping the top and bottom of the square and moving them in opposite directions, each by 0.25 units.
So we can devise some general 2-D skewing matrices for skewing in the x- and y- directions like this:
$$K_x = \left( \begin{matrix} 1 & k \\ 0 & 1 \end{matrix} \right) \; \; Ky = \left( \begin{matrix} 1 & 0 \\ k & 1 \end{matrix} \right)$$
In three dimensions, figures can be skewed in three directions, so the matrices are a little more complicated, but not much. You could figure it out.
### Rotation
The generic 2-D rotation matrix looks like this:
$$R_{\theta} = \left( \begin{matrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{matrix} \right)$$
where $\theta$ is the counter-clockwise rotation angle. We'd expect a rotation through $2\pi$ radians or 360˚ to leave a vector unchanged by such a matrix. For that angle that matrix is
$$R_{2\pi} = \left( \begin{matrix} cos(2\pi) & -sin(2\pi) \\ sin(2\pi) & cos(2\pi) \end{matrix} \right) \; = \; \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right)$$
Which is the identity matrix, so indeed, this rotation leaves any 2D vector unchanged. Now let's test a more interesting angle, $\theta = \frac{\pi}{2} = 45˚.$ Because $cos \left( \frac{\pi}{2} \right) = sin \left( \frac{\pi}{2} \right) = \frac{\sqrt{2}}{2},$ the rotation matrix is
$$R_{\pi/2} = \left( \begin{matrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{matrix} \right)$$
Let's multiply this matrix by two vectors (just to have two examples) (1, 2) and (3,-1) to see what the effect is:
$$\left( \begin{matrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{matrix} \right) \left( \begin{matrix} 1 \\ 2 \end{matrix} \right) = \left( \begin{matrix} \frac{-\sqrt{2}}{2} \\ \frac{3\sqrt{2}}{2} \end{matrix} \right)$$
$$\left( \begin{matrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{matrix} \right) \left( \begin{matrix} 3 \\ -1 \end{matrix} \right) = \left( \begin{matrix} 2\sqrt{2} \\ \sqrt{2} \end{matrix} \right)$$
Here's what those vectors and their rotated counterparts look like:
You can prove to yourself that the angle between each pair of vectors is indeed 45˚ $(\pi/2)$ by calculating the dot products.
### Reflection across x-y
The 2D matrix,
$$\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)$$
reflects any 2D vector across the line y = x (dashed line). The figure shows its action on the vector (1, 2). The multiplication is
$$\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\left( \begin{matrix} 1 \\ 2 \end{matrix} \right) = \left( \begin{matrix} 2 \\ 1 \end{matrix} \right)$$
This transformation can come in handy. Recall that if a point (x, y) is on the graph of a function $f(x),$ the point (y, x) is on the graph of its inverse, $f^{-1}(x).$
X
### Translation
Translation is motion in one or more of the three Cartesian (x, y, z) directions, or a combination of them, without any rotation.
### Practice problems
1 Consider the vectors $$A = \left( \begin{matrix} 2 \\ 3 \end{matrix} \right), \; \color{#E90F89}{\text{&}} \; \; B = \left( \begin{matrix} \frac{2 - 3\sqrt{3}}{2} \\ \frac{2\sqrt{3} + 3}{2} \end{matrix} \right)$$ Calculate the dot product $A\cdot B$ and use it to determine the rotation matrix for converting $A$ into $B.$ 2 Using vector $(1, 3),$ show that counterclockwise rotation by $\frac{\pi}{4}$ and reflection across $y = x$ are not commutative.
3 Consider the figure with vertices at (0,1), (-1,0), (0,-1) and (1,0). Sketch the figure on a graph. Apply the skew matrix $$\left( \begin{matrix} 1 & \frac{1}{2} \\ 0 & 1 \end{matrix} \right),$$ followed by a counterclockwise rotation by $\frac{\pi}{6},$ then redraw the resulting figure. 4 Show that rotation of a vector by $\pi$ has the same result as multiplication of the vector by the inversion matrix, $$\left( \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right),$$
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.
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# It can be shown that the algebraic multiplicity of an eigenvalue lambda is always greater than or equal to the dimension of the eigenspace corresponding to lambda. Find h in the matrix A below such that the eigenspace for lambda = 4 is two-dimensional.
$A=\begin{bmatrix} 4&2&3&3 \\ 0&2 &h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix}$
This problem aims to familiarize us with eigenvalues, eigenspace, and echelon form. The concepts required to solve this problem are related to basic matrices which include eigenvectors, eigenspace, and row reduce forms.
Now, eigenvalues are a unique set of scalar numbers that are linked with the linear equations which can be found in the matrix equations. Whereas the eigenvectors, also known as characteristic roots, are basically non-zero vectors that can be altered by their scalar element when of course linear transformation is applied.
In the statement, we are given the eigenspace which is basically the set of eigenvectors linked with each eigenvalue when the linear transformation is applied to those eigenvectors. If we recall linear transformation, it is often in the form of a square matrix whose columns and rows are of the same count.
To find out the value of $h$ for which the $\lambda = 4$ is two-dimensional, we first have to convert the matrix $A$ to its echelon form.
Firstly performing the operation $A- \lambda I$, where $\Lambda = 4$ and $I$ is the identity matrix.
$A = \begin{bmatrix} 4&2&3&3 \\ 0&2&h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix} – 4 \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}$
$= \begin{bmatrix} 4&2&3&3 \\ 0&2&h&3 \\ 0&0&4&14 \\ 0&0&0&2\end{bmatrix} – \begin{bmatrix} 4&0&0&0 \\ 0&4&0&0 \\ 0&0&4&0 \\ 0&0&0&4 \end{bmatrix}$
$A = \begin{bmatrix} 0&2&3&3 \\ 0&-2&h&3 \\ 0&0&0&14 \\ 0&0&0&-2\end{bmatrix}$
To make $0$ on second pivot, applying the operation $R_2 \rightarrow R_2 + R_1$, Matrix $A$ becomes:
$A = \begin{bmatrix} 0&2&3&3 \\ 0&0&h+3&6 \\ 0&0&0&14 \\ 0&0&0 &-2\end{bmatrix}$
Now dividing $R_3$ with the $14$ and performing the operation $R_4 \rightarrow R_4 – R_3$, Matrix $A$ becomes:
$A = \begin{bmatrix} 0& 2& 3& 3 \\ 0& 0& h+3& 6 \\ 0& 0& 0& 1 \\ 0& 0& 0& 0 \end{bmatrix}$
By Looking at the echelon form of the matrix $A$, it can be deduced that variable $x_1$ is a free variable if $h \neq -3$.
If $h= -3$, then it is not in echelon form, but the only one-row operation is needed it into echelon form. In that case, $x_1$ and $x_2$ will be the free variable so the eigenspace it produces will be two-dimensional.
## Numerical Result
For $h = -3$ the eigenspace of $\lambda = 4$ is two-dimensional.
## Example
Find $h$ in the matrix $A$ such that the eigenspace for $\lambda = 5$ is two-dimensional.
$A = \begin{bmatrix} 5 &-2 &6 &-1 \\ 0 &3 &h &0 \\ 0 &0 &5 &4 \\ 0 &0& 0& 1 \end{bmatrix}$
The echelon form of this matrix can be obtained by applying some operations and it comes out to be:
$A = \begin{bmatrix} 0& 1& -3& 0 \\ 0 &0 &h-6 &0 \\ 0 &0 &0 &1 \\ 0 &0 &0 &0 \end{bmatrix}$
It can be seen for $h =6$ the system will have $2$ free variables and hence it will have an eigenspace of two-dimensional.
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# If $x\%$ of 250 + $25\%$ of 68 = 67, then the value of x is?
Last updated date: 22nd Jul 2024
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Hint: In this problem, we have to find the value of x from the given expression. Here we can first write the given expression in mathematical form, as we know that percentage means per hundred, so we can divide the given percentage value by 100 and we can multiply the given number and percentage as we have the term ‘of’ which means product. We can then simplify the steps to find the value of x.
Here we have to find the value of x form the given expression,
$x\%$ of 250 + $25\%$ of 68 = 67
We can now write the above expression in mathematical form.
As we know that percentage means per hundred, so we can divide the given percentage value by 100.
We can multiply the given number and percentage as we have the term ‘of’ which means product.
$\Rightarrow \dfrac{x}{100}\times 250+\dfrac{25}{100}\times 68=67$
We can now simplify the above step.
$\Rightarrow \dfrac{250x}{100}+\dfrac{25\times 68}{100}=67$
Here we have similar denominators so we can add the numerator by taking similar denominators, we get
$\Rightarrow \dfrac{250x+1700}{100}=67$
We can now multiply 100 on both sides, we get
$\Rightarrow 250x+1700=6700$
We can now simplify the above step, we get
\begin{align} & \Rightarrow 250x=6700-1700 \\ & \Rightarrow 250x=5000 \\ & \Rightarrow x=\dfrac{5000}{250}=20 \\ \end{align}
Therefore, the value of x is 20.
Note: We should always remember that percentage means per hundred, so we can divide the given percentage value by 100 and we can multiply the given number and percentage as we have the term ‘of’ which means product. We should also concentrate while simplifying each step.
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# Prime Factors of 903
Prime Factors of 903 are 3, 7, and 43
#### How to find prime factors of a number
1. Prime Factorization of 903 by Division Method 2. Prime Factorization of 903 by Factor Tree Method 3. Definition of Prime Factors 4. Frequently Asked Questions
#### Steps to find Prime Factors of 903 by Division Method
To find the primefactors of 903 using the division method, follow these steps:
• Step 1. Start dividing 903 by the smallest prime number, i.e., 2, 3, 5, and so on. Find the smallest prime factor of the number.
• Step 2. After finding the smallest prime factor of the number 903, which is 3. Divide 903 by 3 to obtain the quotient (301).
903 ÷ 3 = 301
• Step 3. Repeat step 1 with the obtained quotient (301).
301 ÷ 7 = 43
43 ÷ 43 = 1
So, the prime factorization of 903 is, 903 = 3 x 7 x 43.
#### Steps to find Prime Factors of 903 by Factor Tree Method
We can follow the same procedure using the factor tree of 903 as shown below:
So, the prime factorization of 903 is, 903 = 3 x 7 x 43.
#### What does Prime factor in mathematics mean?
Prime numbers, in mathematics are all those whole numbers greater than 1 having exactly two divisors that is 1 and the number itself. When we express any number as the product of these prime numbers than these prime numbers become prime factors of that number. Eg- Prime Factors of 903 are 3 x 7 x 43.
#### Properties of Prime Factors
• For any given number there is one and only one set of unique prime factors.
• 2 is the only even prime number. So, any given number can have only one even prime factor and that is 2.
• Two prime factors of a given number are always coprime to each other.
• 1 is neither a prime number nor a composite number and also 1 is the factor of every given number. So, 1 is the factor of 903 but not a prime factor of 903.
#### Frequently Asked Questions
• Which is the smallest prime factor of 903?
Smallest prime factor of 903 is 3.
• What are the factors of 903?
Factors of 903 are 1 , 3 , 7 , 21 , 43 , 129 , 301 , 903.
• What is prime factorization of 903 in exponential form?
Prime factorization of 903 in exponential form is 3 x 7 x 43.
• Is 903 a prime number or a composite number?
903 is a composite number.
• Is 903 a prime number?
false, 903 is not a prime number.
• What is the sum of all prime factors of 903?
Prime factors of 903 are 3 x 7 x 43. Therefore, their sum is 53.
• What is the product of all prime factors of 903?
Prime factors of 903 are 3 x 7 x 43. Therefore, their product is 903.
• What numbers are the prime factors of 903?
Prime factors of 903 are 3 , 7 , 43.
• What is the sum of all odd prime factors of 903?
Prime factors of 903 are 3 , 7 , 43, out of which 3 , 7 , 43 are odd numbers. So, the sum of odd prime factors of 903 is 3 + 7 + 43 = 53.
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# Direct proportion
Suppose a shop named Tesco is offering discounts on really expensive boxes of biscuit. The number of boxes (n) is directly proportional to the discount (d). We write; The following table shows a number of biscuit boxes and their discounts.
Number of boxes 2 4 6 40 Discount (£) 4 8 12 80
If you observe the table carefully you might be able to spot a pattern, when working with proportion in Maths you always have to identify the connection or the pattern in the values in question. We know that the number of biscuits is directly proportional to the discount which means the more boxes a customer buys the more discount he/she will get. Suppose we wanted to work out the discount for 100 boxes of biscuits, we would need to find the formula for d ∝ n first, then use to find any other value we want to find. The pattern from our table above is that when we divide the number of boxes with the discount we always get a constant/fixed value. In proportions this value is often referenced as k.
Number of boxes 2 4 6 40 Discount (£) 4 8 12 80 d/n 2 2 2 2
If in our table d/n =2 then; The formula for d ∝ n So to work out the discount you will get on a certain number of boxes, simply multiply the number of boxes with 2. For example; If you buy 100 boxes of biscuits you will get a discount of; The discount for 100 boxes of biscuits is £200.
## With square roots
This also applies in situations where a figure is proportional to a square root. The flow rate of a water in a square pipe is proportional to the square of it's width; we say;
## Example
width (w2) 25 144 flow rate (r) 125 720 r / w2 5 5
if r/w2 = 5 then the formula is; So when w = 15 what is r?
## Summary
if d ∝ v we have found out that; ;
//Comments Suppose a shop named Tesco is offering discounts on really expensive boxes of biscuit. The number of boxes (n) is directly proportional to the discount (d). We write; The following table shows a number of biscuit boxes and their discounts.
Number of boxes 2 4 6 40 Discount (£) 4 8 12 80
If you observe the table carefully you might be able to spot a pattern, when working with proportion in Maths you always have to identify the connection or the pattern in the values in question. We know that the number of biscuits is directly proportional to the discount which means the more boxes a customer buys the more discount he/she will get. Suppose we wanted to work out the discount for 100 boxes of biscuits, we would need to find the formula for d ∝ n first, then use to find any other value we want to find. The pattern from our table above is that when we divide the number of boxes with the discount we always get a constant/fixed value. In proportions this value is often referenced as k.
Number of boxes 2 4 6 40 Discount (£) 4 8 12 80 d/n 2 2 2 2
If in our table d/n =2 then; The formula for d ∝ n So to work out the discount you will get on a certain number of boxes, simply multiply the number of boxes with 2. For example; If you buy 100 boxes of biscuits you will get a discount of; The discount for 100 boxes of biscuits is £200.
## With square roots
This also applies in situations where a figure is proportional to a square root. The flow rate of a water in a square pipe is proportional to the square of it's width; we say;
## Example
width (w2) 25 144 flow rate (r) 125 720 r / w2 5 5
if r/w2 = 5 then the formula is; So when w = 15 what is r?
## Summary
if d ∝ v we have found out that; ;
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4.2: Multiply and Divide Fractions (Part 1)
Skills to Develop
• Simplify fractions
• Multiply fractions
• Find reciprocals
• Divide fractions
be prepared!
Before you get started, take this readiness quiz.
1. Find the prime factorization of 48. If you missed this problem, review Example 2.48.
2. Draw a model of the fraction $$\frac{3}{4}$$. If you missed this problem, review Example 4.2.
3. Find two fractions equivalent to $$\frac{5}{6}$$. Answers may vary. Acceptable answers include $$\frac{10}{12}, \frac{15}{18}, \frac{50}{60}$$, etc. If you missed this problem, review Example 4.14.
Simplify Fractions
In working with equivalent fractions, you saw that there are many ways to write fractions that have the same value, or represent the same part of the whole. How do you know which one to use? Often, we’ll use the fraction that is in simplified form.
A fraction is considered simplified if there are no common factors, other than 1, in the numerator and denominator. If a fraction does have common factors in the numerator and denominator, we can reduce the fraction to its simplified form by removing the common factors.
Definition: Simplified Fraction
A fraction is considered simplified if there are no common factors in the numerator and denominator.
For example,
• $$\frac{2}{3}$$ is simplified because there are no common factors of 2 and 3.
• $$\frac{10}{15}$$ is not simplified because 5 is a common factor of 10 and 15.
The process of simplifying a fraction is often called reducing the fraction. In the previous section, we used the Equivalent Fractions Property to find equivalent fractions. We can also use the Equivalent Fractions Property in reverse to simplify fractions. We rewrite the property to show both forms together.
Definition: Equivalent Fractions Property
If a, b, c are numbers where b ≠ 0, c ≠ 0, then $$\frac{a}{b} = \frac{a \cdot c}{b \cdot c}$$ and $$\frac{a \cdot c}{b \cdot c} = \frac{a}{b}$$.
Notice that c is a common factor in the numerator and denominator. Anytime we have a common factor in the numerator and denominator, it can be removed.
HOW TO: SIMPLIFY A FRACTION
Step 1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers.
Step 2. Simplify, using the equivalent fractions property, by removing common factors.
Step 3. Multiply any remaining factors.
Example 4.19:
Simplify: $$\frac{10}{15}$$.
Solution
To simplify the fraction, we look for any common factors in the numerator and the denominator.
Notice that 5 is a factor of both 10 and 15. $$\frac{10}{15}$$ Factor the numerator and denominator. $$\frac{2 \cdot \textcolor{red}{5}}{3 \cdot \textcolor{red}{5}}$$ Remove the common factors. $$\frac{2 \cdot \cancel{\textcolor{red}{5}}}{3 \cdot \cancel{\textcolor{red}{5}}}$$ Simplify. $$\frac{2}{3}$$
Exercise 4.37:
Simplify: $$\frac{8}{12}$$.
Exercise 4.38:
Simplify: $$\frac{12}{16}$$.
To simplify a negative fraction, we use the same process as in Example 4.19. Remember to keep the negative sign.
Example 4.20:
Simplify: $$− \frac{18}{24}$$.
Solution
We notice that 18 and 24 both have factors of 6. $$- \frac{18}{24}$$ Rewrite the numerator and denominator showing the common factor. $$- \frac{3 \cdot \textcolor{red}{6}}{4 \cdot \textcolor{red}{6}}$$ Remove common factors. $$- \frac{3 \cdot \cancel{\textcolor{red}{6}}}{4 \cdot \cancel{\textcolor{red}{6}}}$$ Simplify. $$- \frac{3}{4}$$
Exercise 4.39:
Simplify: $$− \frac{21}{28}$$.
Exercise 4.40:
Simplify: $$− \frac{16}{24}$$.
After simplifying a fraction, it is always important to check the result to make sure that the numerator and denominator do not have any more factors in common. Remember, the definition of a simplified fraction: a fraction is considered simplified if there are no common factors in the numerator and denominator.
When we simplify an improper fraction, there is no need to change it to a mixed number.
Example 4.21:
Simplify: $$− \frac{56}{32}$$.
Solution
$$- \frac{56}{32}$$ Rewrite the numerator and denominator, showing the common factors, 8. $$- \frac{7 \cdot \textcolor{red}{8}}{4 \cdot \textcolor{red}{8}}$$ Remove common factors. $$- \frac{7 \cdot \cancel{\textcolor{red}{8}}}{4 \cdot \cancel{\textcolor{red}{8}}}$$ Simplify. $$- \frac{7}{4}$$
Exercise 4.41:
Simplify: $$− \frac{54}{42}$$.
Exercise 4.42:
Simplify: $$− \frac{81}{45}$$.
HOW TO: SIMPLIFY A FRACTION
Step 1. Rewrite the numerator and denominator to show the common factors. If needed, factor the numerator and denominator into prime numbers.
Step 2. Simplify, using the equivalent fractions property, by removing common factors.
Step 3. Multiply any remaining factors.
Sometimes it may not be easy to find common factors of the numerator and denominator. A good idea, then, is to factor the numerator and the denominator into prime numbers. (You may want to use the factor tree method to identify the prime factors.) Then divide out the common factors using the Equivalent Fractions Property.
Example 4.22:
Simplify: $$\frac{210}{385}$$.
Solution
$$\frac{210}{385}$$ Use factor trees to factor the numerator and denominator. Rewrite the numerator and denominator as the product of the primes. $$\frac{210}{385} = \frac{2 \cdot 3 \cdot 5 \cdot 7}{5 \cdot 7 \cdot 11}$$ Remove the common factors. $$\frac{2 \cdot 3 \cdot \cancel{\textcolor{blue}{5}} \cdot \cancel{\textcolor{red}{7}}}{\cancel{\textcolor{blue}{5}} \cdot \cancel{\textcolor{red}{7}} \cdot 11}$$ Simplify. $$\frac{2 \cdot 3}{11}$$ Multiply any remaining factors. $$\frac{6}{11}$$
Exercise 4.43:
Simplify: $$\frac{69}{120}$$.
Exercise 4.44:
Simplify: $$\frac{120}{192}$$.
We can also simplify fractions containing variables. If a variable is a common factor in the numerator and denominator, we remove it just as we do with an integer factor.
Example 4.23:
Simplify: $$\frac{5xy}{15x}$$.
Solution
$$\frac{5xy}{15x}$$ Rewrite numerator and denominator showing common factors. $$\frac{5 \cdot x \cdot y}{3 \cdot 5 \cdot x}$$ Remove common factors. $$\frac{\cancel{5} \cdot \cancel{x} \cdot y}{3 \cdot \cancel{5} \cdot \cancel{x}}$$ Simplify. $$\frac{y}{3}$$
Exercise 4.45:
Simplify: $$\frac{7x}{7y}$$.
Exercise 4.46:
Simplify: $$\frac{9a}{9b}$$.
Multiply fractions
A model may help you understand multiplication of fractions. We will use fraction tiles to model $$\frac{1}{2} \cdot \frac{3}{4}$$. To multiply $$\frac{1}{2}$$ and $$\frac{3}{4}$$, think $$\frac{1}{2}$$ of $$\frac{3}{4}$$.
Start with fraction tiles for three-fourths. To find one-half of three-fourths, we need to divide them into two equal groups. Since we cannot divide the three $$\frac{1}{4}$$ tiles evenly into two parts, we exchange them for smaller tiles.
We see $$\frac{6}{8}$$ is equivalent to $$\frac{3}{4}$$. Taking half of the six $$\frac{1}{8}$$ tiles gives us three $$\frac{1}{8}$$ tiles, which is $$\frac{3}{8}$$. Therefore,
$$\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$$
Example 4.24:
Use a diagram to model $$\frac{1}{2} \cdot \frac{3}{4}$$.
Solution
First shade in $$\frac{3}{4}$$ of the rectangle.
We will take $$\frac{1}{2}$$ of this $$\frac{3}{4}$$, so we heavily shade $$\frac{1}{2}$$ of the shaded region.
Notice that 3 out of the 8 pieces are heavily shaded. This means that $$\frac{3}{8}$$ of the rectangle is heavily shaded. Therefore, $$\frac{1}{2}$$ of $$\frac{3}{4}$$ is $$\frac{3}{4}$$, or $$\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$$.
Exercise 4.47:
Use a diagram to model: $$\frac{1}{2} \cdot \frac{3}{5}$$.
Exercise 4.48:
Use a diagram to model: $$\frac{1}{2} \cdot \frac{5}{6}$$.
Look at the result we got from the model in Example 4.24. We found that $$\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$$. Do you notice that we could have gotten the same answer by multiplying the numerators and multiplying the denominators?
$$\frac{1}{2} \cdot \frac{3}{4}$$ Multiply the numerators, and multiply the denominators. $$\frac{1}{2} \cdot \frac{3}{4}$$ Simplify. $$\frac{3}{8}$$
This leads to the definition of fraction multiplication. To multiply fractions, we multiply the numerators and multiply the denominators. Then we write the fraction in simplified form.
Definition: Fraction Multiplication
If a, b, c, and d are numbers where b ≠ 0 and d ≠ 0, then
$$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$$
Example 4.25:
Multiply, and write the answer in simplified form: $$\frac{3}{4} \cdot \frac{1}{5}$$.
Solution
$$\frac{3}{4} \cdot \frac{1}{5}$$ Multiply the numerators, and multiply the denominators. $$\frac{3 \cdot 1}{4 \cdot 5}$$ Simplify. $$\frac{3}{20}$$
There are no common factors, so the fraction is simplified.
Exercise 4.49:
Multiply, and write the answer in simplified form: $$\frac{1}{3} \cdot \frac{2}{5}$$.
Exercise 4.50:
Multiply, and write the answer in simplified form: $$\frac{3}{5} \cdot \frac{7}{8}$$.
When multiplying fractions, the properties of positive and negative numbers still apply. It is a good idea to determine the sign of the product as the first step. In Example 4.26 we will multiply two negatives, so the product will be positive.
Example 4.26:
Multiply, and write the answer in simplified form: $$- \frac{5}{8} \left(- \dfrac{2}{3}\right)$$.
Solution
$$- \frac{5}{8} \left(- \dfrac{2}{3}\right)$$ The signs are the same, so the product is positive. Multiply the numerators, multiply the denominators. $$\frac{5 \cdot 2}{8 \cdot 3}$$ Simplify. $$\frac{10}{24}$$ Look for common factors in the numerator and denominator. Rewrite showing common factors. $$\frac{5 \cdot \cancel{\textcolor{red}{2}}}{12 \cdot \cancel{\textcolor{red}{2}}}$$ Remove common factors. $$\frac{5}{12}$$
Another way to find this product involves removing common factors earlier.
$$- \frac{5}{8} \left(- \dfrac{2}{3}\right)$$ Determine the sign of the product. Multiply. $$\frac{5 \cdot 2}{8 \cdot 3}$$ Show common factors and then remove them. $$\frac{5 \cdot \cancel{\textcolor{red}{2}}}{12 \cdot \cancel{\textcolor{red}{2}}}$$ Multiply remaining factors. $$\frac{5}{12}$$
We get the same result.
Exercise 4.51:
Multiply, and write the answer in simplified form: $$- \frac{4}{7} \left(- \dfrac{5}{8}\right)$$.
Exercise 4.52:
Multiply, and write the answer in simplified form: $$- \frac{7}{12} \left(- \dfrac{8}{9}\right)$$.
Example 4.27:
Multiply, and write the answer in simplified form: $$− \frac{14}{15} \cdot \frac{20}{21}$$.
Solution
$$- \frac{14}{15} \cdot \frac{20}{21}$$ Determine the sign of the product; multiply. $$- \frac{14}{15} \cdot \frac{20}{21}$$ Are there any common factors in the numerator and the denominator? We know that 7 is a factor of 14 and 21, and 5 is a factor of 20 and 15. Rewrite showing common factors. $$- \frac{2 \cdot \cancel{\textcolor{red}{7}} \cdot 4 \cdot \cancel{\textcolor{red}{5}}}{3 \cdot \cancel{\textcolor{red}{5}} \cdot 3 \cdot \cancel{\textcolor{red}{7}}}$$ Remove the common factors. $$- \frac{2 \cdot 4}{3 \cdot 3}$$ Multiply the remaining factors. $$- \frac{8}{9}$$
Exercise 4.53:
Multiply, and write the answer in simplified form: $$− \frac{10}{28} \cdot \frac{8}{15}$$.
Exercise 4.54:
Multiply, and write the answer in simplified form: $$− \frac{9}{20} \cdot \frac{5}{12}$$.
When multiplying a fraction by an integer, it may be helpful to write the integer as a fraction. Any integer, a, can be written as $$\frac{a}{1}$$. So, 3 = $$\frac{3}{1}$$, for example.
Example 4.28:
Multiply, and write the answer in simplified form: (a) $$\frac{1}{7}$$ • 56 (b) $$\frac{12}{5}$$(−20x)
Solution
(a)
$$\frac{1}{7} \cdot 56$$ Write 56 as a fraction. $$\frac{1}{7} \cdot \frac{56}{1}$$ Determine the sign of the product; multiply. $$\frac{56}{7}$$ Simplify. $$8$$
(b)
$$\frac{12}{5} (-20x)$$ Write −20x as a fraction. $$\frac{12}{5} \left(\dfrac{-20x}{1}\right)$$ Determine the sign of the product; multiply. $$- \frac{12 \cdot 20 \cdot x}{5 \cdot 1}$$ Show common factors and then remove them. $$- \frac{12 \cdot \textcolor{red}{4 \cdot \cancel{5} x}}{\cancel{5} \cdot 1}$$ Multiply remaining factors; simplify. $$-48x$$
Exercise 4.55:
Multiply, and write the answer in simplified form: (a) $$\frac{1}{8}$$ • 72 (b) $$\frac{11}{3}$$(−9a)
Exercise 4.56:
Multiply, and write the answer in simplified form: (a) $$\frac{3}{8}$$ • 64 (b) 16x • $$\frac{11}{12}$$
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# How do you evaluate log_(1/3) (1/9)?
Sep 6, 2016
Rewrite ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = x$ as ${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.
#### Explanation:
Rewrite ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = x$ as ${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.
Remember, the answer to a log is an exponent, so you are looking for the exponent that makes $\frac{1}{3}$ turn into $\frac{1}{9}$.
$\frac{1}{3}$ is both the base of the log and the base of the exponent.
${\left(\frac{1}{3}\right)}^{x} = \left(\frac{1}{9}\right)$.
$x = 2$ because ${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$.
Sep 6, 2016
${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2$
#### Explanation:
In this log form of ${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right)$
"what power or index of $\frac{1}{3}$ will give $\frac{1}{9}$"?
This one can be done by inspection.
Note that ${\left(\frac{1}{3}\right)}^{2} = \frac{1}{9}$
${\log}_{\frac{1}{3}} \left(\frac{1}{9}\right) = 2$
$\frac{\log \left(\frac{1}{9}\right)}{\log \left(\frac{1}{3}\right)} = 2$
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# Simplifying Algebraic Expressions 7-1 Learn to combine like terms in an expression.
## Presentation on theme: "Simplifying Algebraic Expressions 7-1 Learn to combine like terms in an expression."— Presentation transcript:
Simplifying Algebraic Expressions 7-1 Learn to combine like terms in an expression.
Simplifying Algebraic Expressions 7-1 Vocabulary terms like terms equivalent expressions simplify
Simplifying Algebraic Expressions 7-1 Terms in an expression are separated by plus or minus signs. Like terms can be grouped together because they have the same variable raised to the same exponents. Equivalent expressions have the same value for all values of the variables. Constants such as 4, 0.75, and 11 are like terms because none of them have a variable. Helpful Hint
Simplifying Algebraic Expressions 7-1 To simplify an expression, perform all possible operations, including combining like terms.
Simplifying Algebraic Expressions 7-1 Combine like terms. Additional Example 1: Combining Like Terms To Simplify Identify like terms. Combine coefficients: 14 – 5 = 9 A. 14a – 5a 9a9a B. 7y 2 + 8 – 3y 2 – 1 + y Identify like terms; the coefficient of y is 1, because 1y = y. Combine coefficients: 7 – 3 = 4 and 8 – 1 = 7 4y 2 + y + 7
Simplifying Algebraic Expressions 7-1 Combine like terms. Check It Out: Example 1 Identify like terms; the coefficient of q is 1, because 1q = q. Combine coefficients: 4 – 1 = 3 Identify like terms; the coefficient of c is 1, because 1c = c. Combine coefficients: 5 – 4 – 1 = 0 and 8 – 2 = 6 6 3q3q A. 4q – q B. 5c + 8 – 4c – 2 – c
Simplifying Algebraic Expressions 7-1 Combine like terms. Additional Example 2A: Combining Like Terms in Two-Variables Expressions Identify like terms. Combine coefficients. 5t 2 + 7p – 3p – 2t 2 3t 2 + 4p 5t 2 + 7p – 3p – 2t 2
Simplifying Algebraic Expressions 7-1 Combine like terms. Additional Example 2B: Combining Like Terms in Two-Variable Expressions No like terms. 4m 3 + 9n – 2
Simplifying Algebraic Expressions 7-1 Combine like terms. Check It Out: Example 2 Identify like terms. Combine coefficients. A. 2x + 5x – 4y + 3 7x – 4y + 3 2x + 5x – 4y + 3 Identify like terms. Combine coefficients. B. 9d + 7c – 4d – 2c 5d + 5c 9d + 7c – 4d – 2c No like terms. C. 8g + c – 6 8g + c – 6
Simplifying Algebraic Expressions 7-1 The Distributive Property states that a(b + c) = ab + ac for all real numbers a, b, and c. For example, 2(3 + 5) = 2(3) + 2(5). Remember!
Simplifying Algebraic Expressions 7-1 Simplify 6(5 + n) – 2n. Additional Example 3: Using the Distributive Property to Simplify Distributive Property. Multiply. 6(5 + n) – 2n 30 + 6n – 2n 6(5) + 6(n) – 2n 30 + 4n Combine coefficients: 6 – 2 = 4.
Simplifying Algebraic Expressions 7-1 Simplify 3(c + 7) – c. Check It Out: Example 3 Distributive Property. Multiply. 3(c + 7) – c 3c + 21 – c 3(c) + 3(7) – c 2c + 21 Combine coefficients: 3 – 1 = 2.
Simplifying Algebraic Expressions 7-1 Solve x + 3x = 48. Additional Example 4: Combining Like Terms to Solve Algebraic Equations Identify like terms. The coefficient of x is 1. Combine coefficients: 1 + 3 = 4 x + 3x = 48 4x = 48 x = 12 Divide both sides by 4. 4 4
Simplifying Algebraic Expressions 7-1 Solve 7c – c = 102. Check It Out: Example 4 Identify like terms. The coefficient of c is 1. Combine coefficients: 7 – 1 = 6 7c – c = 102 6c = 102 c = 17 Divide both sides by 6. 6 6
Simplifying Algebraic Expressions 7-1 Lesson Quiz Combine like terms. 1. 3x 2 + 4 + 2x 2 2. 13k + 6 – 8m + 9 + k Simplify. 3. 4(3x + 6) – 7x 4. 6(x + 5) + 3x Solve. 5. 6y + y = 42 6. The Accounting Department ordered 15 boxes of pens. The Marketing Department ordered 9 boxes of pens. If the total cost of the combined order was \$72, what is the price of each box of pens? 14k – 8m + 15 y = 6 5x 2 + 4 5x + 249x + 30 \$3
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## How do you find the level of a multivariable function curve?
The level curves of the function z=f(x,y) z = f ( x , y ) are two dimensional curves we get by setting z=k , where k is any number. So the equations of the level curves are f(x,y)=k f ( x , y ) = k .
How do you write a level surface for a function?
Level surfaces: For a function w=f(x,y,z):U⊆R3→R the level surface of value c is the surface S in U⊆R3 on which f|S=c. Example 1: The graph of z=f(x,y) as a surface in 3-space can be regarded as the level surface w=0 of the function w(x,y,z)=z−f(x,y).
### Which is the example of level surface?
A level surface is the equipotential surface of the earth’s gravity field. It is a curved surface and every element of which is normal to plumb line. A body of still water provides the best example of a level surface.
How do you classify quadric surfaces?
Quadric surfaces are often used as example surfaces since they are relatively simple. There are six different quadric surfaces: the ellipsoid, the elliptic paraboloid, the hyperbolic paraboloid, the double cone, and hyperboloids of one sheet and two sheets.
#### How do you determine surface level?
For a function of three variables, a level set is a surface in three-dimensional space that we will call a level surface. For a constant value c in the range of f(x,y,z), the level surface of f is the implicit surface given by the graph of c=f(x,y,z).
What are level curves multivariable calculus?
Definition: The level curves of a function f of two variables are the curves with equations f(x,y) = k, where k is a constant (in the range of f). A level curve f(x,y) = k is the set of all points in the domain of f at which f takes on a given value k. In other words, it shows where the graph of f has height k.
## What are level surfaces?
Level surfaces are surfaces that represent the solution to scalar-valued functions of three independent variables.
What are level surfaces of a function?
### How do you calculate surface level?
What is meant by a level surface?
A surface which at every point is perpendicular to a plumb line or the direction in which gravity acts; parallel to the surface of still water.
#### How to find the equation of a level surface?
For functions of the form f (x,y,z) f (x, y, z) we will occasionally look at level surfaces. The equations of level surfaces are given by f (x,y,z) = k f (x, y, z) = k where k k is any number. The final topic in this section is that of traces. In some ways these are similar to contours.
How do you find the level surfaces of a parabola?
Describe the level surfaces of u = f ( x, y, z) = x 2 + y 2 z. ( x 2 + y 2) = c z. This equation describes the regular parabola ( z = x 2 + y 2) where its output is multiplied by 1 / c. Some of the level surfaces are shown in Figure 10.
## How do you find the level surface with constant k?
For a function f ( x, y, z) of three variables, f ( x, y, z) = k is called the level surface with constant k. The function f ( x, y, z) is constant over the level surface. Let z = f ( x, y) .
What are the equations of level curves?
The next topic that we should look at is that of level curves or contour curves. The level curves of the function z = f (x,y) z = f ( x, y) are two dimensional curves we get by setting z = k z = k, where k k is any number. So the equations of the level curves are f (x,y) = k f ( x, y) = k.
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# The Pythagorean Theorem Study Guide
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Updated on Oct 1, 2011
## The Pythagorean Theorem
In this lesson, we examine a very powerful relationship between the three lengths of a right triangle. With this relationship, we will find the exact length of any side of a right triangle, provided we know the lengths of the other two sides. We also study right triangles where all three sides have whole number lengths.
## Proving the Pythagorean Theorem
The Pythagorean theorem is this relationship between the three sides of a right triangle. Actually, it relates the squares of the lengths of the sides. The square of any number (the number times itself) is also the area of the square with that length as its side. For example, the square in Figure 3.1 with sides of length a will have area a · a = a2.
The longest side of a right triangle, the side opposite the right angle, is called the hypotenuse, and the other two sides are called legs. Suppose a right triangle has legs of length a and b, and a hypotenuse of length c, as illustrated in Figure 3.2.
The Pythagorean theorem states that a2 + b2 = c2, This means that the area of the squares on the two smaller sides add up to the area of the biggest square. This is illustrated in Figures 3.3 and 3.4.
This is a surprising result. Why should these areas add up like this? Why couldn't the areas of the two smaller squares add up to a bit more or less than the big square?
We can convince ourselves that this is true by adding four copies of the original triangle to each side of the equation. The four triangles can make two rectangles, as shown in Figure 3.5. They could also make a big square with a hole in the middle, as in Figure 3.6.
If we add the a2 and b2 squares to Figure 3.4, and the c2 square to Figure 3.5, they fit exactly. The result in either case is a big square with each side of length a + b, as shown in Figure 3.7.
The two big squares have the same area. If we take away the four triangles from each side, we can see that the two smaller squares have the exact same area as the big square, as shown in Figure 3.8. Thus, a2 + b2 = c2
This proof of the Pythagorean theorem has been adapted from a proof developed by the Chinese about 3,000 years ago. With the Pythagorean theorem, we can use any two sides of a right triangle to find the length of the third side.
#### Example 1
Suppose the two legs of a right triangle measure 8 inches and 12 inches, as shown in Figure 3.9. What is the length of the hypotenuse?
By the Pythagorean theorem:
122 + 82 = H2
208 = H2
H = 208
While the equation gives two solutions, a length must be positive, so H = √208. This can be simplified to H = 4√13.
#### Example 2
What is the height of the triangle in Figure 3.10?
Even though the height is labeled h, it is not the hypotenuse. The longest side has length 10 feet, and thus must be alone on one side of the equation.
h2 + 32 = 102
h2 = 100 – 9
h = √91 ≈ 9.54
With the help of a calculator, we can see that the height of this triangle is about 9.54 feet.
We can use the Pythagorean theorem on triangles without illustrations. All we need to know is that the triangle is right and which side is the hypotenuse.
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# How do you turn 8% into a decimal?
## How do you turn 8% into a decimal?
Answer: 8% as a decimal is 0.08.
## How do you turn 5% into a decimal?
5% means that you have 5 out of 100 ( 5100 ). Because there is a one in front you divide 5 by 1, which is just 5. Then you move the decimal point left by the number of zeroes, which is 2. So 5.0 becomes 0.05.
How do you turn 3.5 percent into a decimal?
So, 3.5 becomes 03.5. Now, you can move decimal point: 03. 5 becomes . 035 or 0.035.
How do you convert percentage to Marks?
For example, you received 7 CGPA in 12th, then your total percentage of marks would be, 7×9.5= 66.5%. Moreover, if you want to calculate total marks from here, then you can simply multiply your percentage divided by 100 to total marks (total of all subjects).
### What is 25% in a decimal?
0.25
If you divide 25 by 100, you get 0.25, which is a decimal.
### How do you read a decimal to a percent?
Calculator Use
1. Converting from a decimal to a percentage is done by multiplying the decimal value by 100 and adding %.
2. Example: 0.10 becomes 0.10 x 100 = 10%
3. Example: 0.675 becomes 0.675 x 100 = 67.5%
How do you work out decimals?
The line in a fraction that separates the numerator and denominator can be rewritten using the division symbol. So, to convert a fraction to a decimal, divide the numerator by the denominator. If required, you can use a calculator to do this. This will give us our answer as a decimal.
What is 325 790 as a decimal?
325/790 as a decimal is 0.41139240506329.
## What is 2.5 As a decimal?
0.025
Answer: 2.5% as a decimal is 0.025.
## What is 03 as a decimal?
Percent to decimal conversion table
Percent Decimal
2% 0.02
3% 0.03
4% 0.04
5% 0.05
How do I find my Sgpa?
How to Calculate SGPA?
1. Write down all your credit points for each subject.
2. Next, multiply the credit point of each subject with the grade that you achieved in that subject.
3. Add all the scores that come.
4. Next, divide the total score that you got with the sum of all the credit points.
5. The number that you get is the SGPA.
How do you calculate a decimal into a percentage?
Converting from a decimal to a percentage is done by multiplying the decimal value by 100. Example: 0.10 becomes 0.10 x 100 = 10% Example: 0.675 becomes 0.675 x 100 = 67.5% The short way to convert from decimal to percent is to move the decimal point 2 places to the right and add a percent sign.
### What is 11 percent as a decimal?
For example: So .11 is 11% changed to a decimal. Remember that 11% and .11 equal the same value. Not every time you change a percent to a decimal will the decimal be a number with no value to the left of the decimal and two digits to the right of the decimal as it was .11 above.
### Can a percent be a decimal?
Since a percent can be written as a decimal by moving the decimal point two places to the left, a decimal can be written as a percent by moving the decimal point two places to the right. For example, to write 0.4 as a percent, move the decimal point two places to the right to get 40.
How do you change a decimal into a whole number?
To turn a decimal into a whole number, you can multiply it by how many decimal places there are, thereby moving every digit over the decimal point. Or you can round it to the nearest ones digit, if you do not care about preserving the entire number.
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# How do you find dy/dx for the function; cos(x^2+2Y)+xe^Y^2=1?
Apr 11, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x \sin \left({x}^{2} + 2 y\right) - {e}^{{y}^{2}}}{- 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}}$
#### Explanation:
Using the implicit function theorem:
$f \left(x , y\right) = \cos \left({x}^{2} + 2 y\right) + x {e}^{{y}^{2}} = 1$
$\mathrm{df} = 0 = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$
So:
${f}_{x} = - 2 x \sin \left({x}^{2} + 2 y\right) + {e}^{{y}^{2}}$
${f}_{y} = - 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}} = \frac{2 x \sin \left({x}^{2} + 2 y\right) - {e}^{{y}^{2}}}{- 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}}$
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Reduce / leveling any portion to its lowest state by utilizing our portion to the Simplest form Calculator. Discover the prize to inquiries like: What is 20/36 in simplest kind or what is 20/36 reduced to the most basic form?
Fractions Simplifier
Please to fill in the two left crate below:
Inputfraction: Integerpart: Fractionpart: As a Decimal: = = Details:Details...You are watching: 20 over 36 in simplest form
How to minimize a fraction
Among different ways simple a fraction, we will present the two procedure below:
Method 1 - division by a tiny Number once Possible
Start by separating both the numerator and also the denomiator that the portion by the very same number, and repeat this until it is difficult to divide. Begin dividing by small numbers like 2, 3, 5, 7. For example,
Simplify the fraction 42/98
First division both (numerator/denominator) by 2 to obtain 21/49.Dividing through 3 and also 5 will not work, so,Divide both numerator and also denominator through 7 to gain 3/7. Note: 21 ÷ 7 = 3 and 49 ÷ 7 = 7
In the fraction 3/7, 3 is only divisible through itself, and 7 is not divisible by various other numbers than itself and also 1, so the fraction has been simplified as lot as possible. No more reduction is possible, therefore 42/98 is same to 3/7 when diminished to its lowest terms. This is a PROPER fraction once the absolute value of the height number or numerator (3) is smaller sized than the absolute worth of the bottom number or denomintor (7).
Method 2 - Greatest common Divisor
To mitigate a fraction to lowest terms (also referred to as its simplest form), just divide both the numerator and also denominator through the GCD (Greatest typical Divisor).
For example, 3/4 is in lowest form, yet 6/8 is no in lowest form (the GCD the 6 and 8 is 2) and 6/8 have the right to be written as 3/4. You deserve to do this since the value of a portion will stay the same when both the numerator and denominator are split by the exact same number.
Note: The Greatest common Factor (GCF) because that 6 and 8, notation gcf(6,8), is 2. Explanation:
Factors that 6 are 1,2,3,6;Factors the 8 are 1,2,4,8.
See more: 02 Ford Focus Serpentine Belt Routing And Timing Belt Diagrams
So, that is ease see that the "Greatest typical Factor" or "Divisor" is 2 because it is the best number i m sorry divides evenly into all of them.
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# TOSSOT - EDITORIAL
Contest
Practice
Author: Hardik Shah
Editorialist: Aishwarya
EASY.
## PREREQUISITES:
Math, Combinatorics.
## PROBLEM:
You are given the number of total tosses and number of times heads occured.
You are required to find number of strings that give a palindromic toss result.
## EXPLANATION:
Firstly we need to split the string in 2 halves. This eases the process of finding palindromes, because if one half of the string is fixed, we can replicate the same as the 2nd half to make the string palindromic.
There can be four cases:
1. N is odd, K is odd : On splitting th string in 2 halves and keeping a HEAD at the center of the string, we need to decide positions for half of the heads in the string, thus:
2. N is odd, K is even : On splitting th string in 2 halves and keeping a TAIL at the center of the string, we need to decide positions for half of the heads in the string.
3. N is even, K is even: On splitting the string in 2 halves, we need to decide positions for half of the heads in the string.
4. N is even, K is odd: No solution exists, as there is no way to place odd number of heads in a string of even size to get a palindrome.
In first 3 cases:
No of unique palindrome strings = \dbinom{\lfloor \frac {n} {2} \rfloor}{\lfloor \frac {k} {2} \rfloor}
## SOLUTIONS:
Setter's Solution
#include <bits/stdc++.h>
#define ll long long
const int N = 1000001;
using namespace std;
ll factorialNumInverse[N + 1];
ll naturalNumInverse[N + 1];
ll fact[N + 1];
void InverseofNumber(ll p)
{
naturalNumInverse[0] = naturalNumInverse[1] = 1;
for (int i = 2; i <= N; i++)
naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(ll p)
{
factorialNumInverse[0] = factorialNumInverse[1] = 1;
for (int i = 2; i <= N; i++) factorialNumInverse[i] = (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void factorial(ll p)
{
fact[0] = 1;
for (int i = 1; i <= N; i++) {
fact[i] = (fact[i - 1] * i) % p;
}
}
// Function to return nCr % p in O(1) time
ll Binomial(ll N, ll R, ll p)
{
// n C r = n!*inverse(r!)*inverse((n-r)!)
ll ans = ((fact[N] * factorialNumInverse[R])
% p * factorialNumInverse[N - R])
% p;
return ans;
}
// Driver Code
int main()
{
ll p = 1000000007;
InverseofNumber(p);
InverseofFactorial(p);
factorial(p);
// 1st query
int tt; cin >> tt;
while(tt--)
{
int n, r; cin >> n >> r;
if(n%2 == 0 && r%2) cout << 0 << "\n";
else cout << Binomial(n/2, r/2, p) << "\n";
}
return 0;
}
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Difference between revisions of "2015 AIME II Problems/Problem 11"
Problem
The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ and $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Diagram
$[asy] unitsize(30); draw(Circle((0,0),3)); pair A,B,C,O, Q, P, M, N; A=(2.5, -sqrt(11/4)); B=(-2.5, -sqrt(11/4)); C=(-1.96, 2.28); Q=(-1.89, 2.81); P=(1.13, -1.68); O=origin; M=foot(O,C,B); N=foot(O,A,B); draw(A--B--C--cycle); label("A",A,SE); label("B",B,SW); label("C",C,NW); label("Q",Q,NW); dot(O); label("O",O,NE); label("M",M,W); label("N",N,S); label("P",P,S); draw(B--O); draw(C--Q); draw(Q--O); draw(O--C); draw(O--A); draw(O--P); draw(O--M, dashed); draw(O--N, dashed); draw(rightanglemark((-2.5, -sqrt(11/4)),(0,0),(-1.89, 2.81),5)); draw(rightanglemark(O,N,B,5)); draw(rightanglemark(B,O,P,5)); draw(rightanglemark(O,M,C,5)); [/asy]$
Solution 1
Call the $M$ and $N$ foot of the altitudes from $O$ to $BC$ and $AB$, respectively. Let $OB = r$ . Notice that $\triangle{OMB} \sim \triangle{OQB}$ because both are right triangles, and $\angle{OBQ} \cong \angle{OBM}$. By $\frac{MB}{BO}=\frac{BO}{BQ}$, $MB = r\left(\frac{r}{4.5}\right) = \frac{r^2}{4.5}$. However, since $O$ is the circumcenter of triangle $ABC$, $OM$ is a perpendicular bisector by the definition of a circumcenter. Hence, $\frac{r^2}{4.5} = 2 \implies r = 3$. Since we know $BN=\frac{5}{2}$ and $\triangle BOP \sim \triangle NBO$, we have $\frac{BP}{3} = \frac{3}{\frac{5}{2}}$. Thus, $BP = \frac{18}{5}$. $m + n=\boxed{023}$.
Solution 2
Notice that $\angle{CBO}=90-A$, so $\angle{BQO}=A$. From this we get that $\triangle{BPQ}\sim \triangle{BCA}$. So $\dfrac{BP}{BC}=\dfrac{BQ}{BA}$, plugging in the given values we get $\dfrac{BP}{4}=\dfrac{4.5}{5}$, so $BP=\dfrac{18}{5}$, and $m+n=\boxed{023}$.
Solution 3
Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, $OM=\sqrt{r^2-4}$. Thus, $OQ=\frac{\sqrt{4r^2+9}}{2}$. Using $\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=11/10}$. Therefore, $BP=\frac{5}{2}+\frac{11}{10}=18/5$. $18+5=\boxed{023}$.
Solution 4
Let $\angle{BQO}=\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\angle{KAB}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha$. But since $BK$ is a diameter, $\angle{KAB}=90^\circ$, implying $\angle{CAB}=\alpha$. It follows that $APCQ$ is a cyclic quadrilateral.
Let $BP=x$. By Power of a Point, $$5x=4\cdot\frac 9 2\implies x=\frac{18}{5}.$$The answer is $18+5=\boxed{023}$.
See also
2015 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Sum of a series - Part 2
On the previous note in this series we learnt / revised that
$\displaystyle \sum_{a}^{n} = \frac {(n + a)(n + a - 1)}{2} - \frac {a^2 - a}{2}$
That's the formula when the difference ($d$) is $1$ so what would the formula be if $d \neq 1$
Let's denote the whole thing as $\displaystyle \sum_{a}^{n} d$
So let's say that $a = 3$, $n = 6$ and $d = 2$ what equation would we get from that.
$\displaystyle \sum_{3}^{6} 2 = 3 + 5 + 7 + 9 + 11 + 13 = 48$
We're going to have to use a different method to last time to solve this.
Since $a = 3$ and $d = 2$ we can put those in to get
$\displaystyle \sum_{a}^{6} d = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + (a + 5d)$
That can be written as
$\displaystyle \sum_{a}^{6} d = 6a + (1 + 2 + 3 + 4 + 5)d$
$\displaystyle \sum_{a}^{n} d = na + \frac {dn(n-1)}{2}$
This is basically a simplified version of the previous equation with a $d$ added in to account for the difference. This formula however is still flawed as it can only handle a constant variable for $d$.
Note by Jack Rawlin
6 years, 5 months ago
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# What Is 15/75 as a Decimal + Solution With Free Steps
The fraction 15/75 as a decimal is equal to 0.2.
A fraction can be represented using the p/q form, where p and q are known as the Numerator and Denominator, respectively. Division is one of the most complicated mathematical operations to master because it is required when working with fractions. However, we can simplify things by employing the later-discussed strategy.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 15/75. Figure 1 shows the long division process:
Figure 1
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 15
Divisor = 75
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 15 $\div$ 75
This is when we go through the Long Division solution to our problem.
## 15/75 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 15 and 75, we can see how 15 is Smaller than 75, and to solve this division, we require that 15 be Bigger than 75.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 15, which after getting multiplied by 10 becomes 150.
We take this 150 and divide it by 75; this can be done as follows:
150 $\div$ 75 $=$ 2
Where:
75 x 2 = 150
This will lead to the generation of a Remainder equal to 150 – 150 = 0.
Finally, we have a Quotient generated after combining the three pieces of it as 0.2=z, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/4 and the angle between sides B and C is pi/6. If side B has a length of 12, what is the area of the triangle?
May 7, 2016
area of triangle is $\left(\frac{1}{2}\right)$(base)(height) = $\left(\frac{1}{2}\right) \cdot A \cdot B$
= $\left(\frac{1}{2}\right)$(12) ((12/(sin60π))*sin30º)
#### Explanation:
Working in degree, π = 180º,
The Angle between C and A is
(180º -90º -30º) = 60º
Hence using sine rule,
A/sin(angle between B&C) = B/sin(angle between C&A) =
C/sin(angle betwen A&B)
And given that $\angle b e t w e e n A B$ is a right angle,
you need to only find A where
A/sin(angle between B&C) =B/sin(angle between C&A),
A= (B/sin(angle between C&A))*sin(angle between B&C)
= (12/(sin60π))*sin30º
Hence, area of triangle is $\left(\frac{1}{2}\right)$(base)(height) = $\left(\frac{1}{2}\right) \cdot A \cdot B$
= $\left(\frac{1}{2}\right)$(12) ((12/(sin60π))*sin30º)
|
# How do you solve 2/3(6x-9)=-34 using the distributive property?
Sep 2, 2016
$x = - 12$
#### Explanation:
The factor $\frac{2}{3}$ outside the bracket must be distributed (multiplied) across the bracket. Each inside term will be multiplied by $\frac{2}{3}$
This will get rid of the brackets.
$\frac{2}{3} \left(6 x - 9\right) = - 34$
$4 x - 6 = - 54$
$4 x - 6 + 6 = - 54 + 6 \text{ "larr "add 6 to both sides}$
$4 x = - 48 \text{ "larr div 4 "on both sides}$
$x = - 12$
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Last Updated on May 26, 2022 by Thinkster
One of the foundational concepts in 5th grade math is fractions. Many students find working with fractions a challenge. If your child is struggling, or if you just want to reinforce what is being taught at school in a fun way at home, consider these fraction games for 5th grade.
### 1. Common Denominator War
Many fraction games for 5th grade take basic fraction concepts and break them down into something that can be practiced over and over. Common Denominator War does just that.
For this game, take 20 index cards and write numbers on them, striving for most of the numbers to be composite (not prime). Divide the cards into two equal piles, giving one pile to each of two players. Play begins when students countdown “1, 2, 3 . . . war!” and flip the top two cards. As soon as the cards are visible, the players work quickly to determine the lowest common multiple, calling it out. The first with the right answer adds that card to his or her pile. Play continues until one player is out of cards, declaring the other player the winner.
### 2. Fold for Fractions
This simple game will give your child a better understanding of what fractions really mean. Grab a plain white piece of paper and some colored pencils. Talk with your child about the word “equivalent,” and make sure she understands what it means. Then, have your child fold the paper in half, then open it, shading one of the halves with a colored pencil. Fold the paper in half again, then in half again, making fourths.
Ask your child how many fourths are equivalent to one-half. The shaded side of the paper makes this clear. Then, fold the paper more to create eighths, asking the question again. Repeat as many times as you can fold to help your child understand the idea of equivalent fractions.
### 3. Improper Fraction “Go Fish”
Does your child love to play “Go Fish?” Then make it a chance to practice math. Make cards with improper fractions and mixed numbers. This game is played just like regular Go Fish, but with one change. The pairs should be mixed numbers and the improper fractions that match them. Play requires the player to choose a card in his hand, figure out what the equivalent mixed number or improper fraction is, and ask for it.
So, if your child has 3 3/5 in his hand, he would figure out that this is equivalent to the improper fraction 18/5. He would then ask you if you have 18/5 in yours. If you do, you give it to him, and he places the match on the table like in traditional “Go Fish.” If you do not, you tell him to “go fish” and he will draw from the draw pile. Play continues until all cards are used, then players count the number of matches to see who is the winner.
Math games are a great way to practice important math concepts, especially those tricky fractions, with your 5th grader. If your child is showing signs of needing additional help, consider investing in a tablet-based tutoring program. Choose one, like Thinkster Math, that is designed by teachers and offers instruction at the point of learning that is tailored to what your child already knows and needs to work on. This, as well as fraction games for 5th grade, can make those pesky fractions a lot easier for your child to comprehend.
Summary
Article Name
Description
If your 5th grade child is having trouble with fractions, consider these fraction games for them to try. They will learn while having a lot of fun.
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Publisher Name
Thinkster Math
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## Recommended Articles
### 6 Educational Math Games to Supplement Online Math Tutoring
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1- Moment
Moment of force
The turning effect of force is called moment of force.
The moment of force depends on the following factors.
The size (magnitude) of the force
The perpendicular distance between the line of action of the force and the turning point which is called the pivot.
We calculate the moment of force by using the following formula
Moment of force = force * perpendicular distance from pivot to the line of action of the force
Moment=F * d
Moment is measured in newton meters(Nm).
One force on its own isn’t much use to us. We normally look at situations where turning effects are balanced (or not!).
Let’s look at the example below and find the missing force F:
If the system is balanced, the anticlockwise turning effect of force F must equal the clockwise turning effect:
clockwise moment = anticlockwise moment
Clockwise moment = 5 N × 0·50 m = 2·50 Nm.
Anticlockwise moment = F × 0·25 m = 2·50 Nm
Force F = 2·50 Nm ÷ 0·25 m = 10 N
In order to balance the 5 N force acting at 0·5 m from the pivot, we require 10 N acting in the opposite direction but at 0·25 m.
Unbalanced Forces
Sometimes moments can easily become unbalanced – even when we don’t want them to!
In these unfortunate examples, it would seem that in loading the cart, some of the boxes must have slipped to the back – further away from the pivot – greatly increasing their turning effect. In the case of the lorry, its weight wasn’t enough to balance the heavy bricks.
The result was the lifting of the donkey – who must have been very surprised! For the lorry, it was lucky nobody was hurt.
<p><a href=”http://vimeo.com/22743057″>Moments (turning effects of forces)</a> from <a href=”http://vimeo.com/user2127784″>Mr Ong</a> on <a href=”http://vimeo.com”>Vimeo</a>.</p>
Sometimes more than one force acts on the same side of the pivot. Their overall turning effect is easy to work out.
The 2 N force has a moment of 2 × 0·2 m = 0·4 Nm clockwise.
The 5 N force has a moment of 5 × 0·5 m = 2·5 Nm clockwise.
Their combined moment = 0·4 Nm + 2·5 Nm = 2·9 Nm clockwise.
Moments can just be added, but they must act in the same direction.
This is also known as the principle of moments.
By now, you should have understood what is Principle of Moments. Lets take a challenge to reinforce your understanding. When you try the following challenge, calculate the clockwise and anti-clockwise moment before clicking on “Release”
By now, you should have understood what is the Principle of Moments. Lets take a short quiz to test your understanding.
Moments Summary
The turning effect (or moment) of a force is given by:
moment = force × perpendicular distance from pivot
The normal units used for force and distance are newtons andmetres respectively, so the usual unit for moment is the newton-metre (Nm)
Another name for a pivot is fulcrum.
Moments can either be clockwise or anticlockwise.
When more than one force acts in the same direction, their overall turning effect is just the sum of their moments.
When forces act in a different direction, yet still balance, the total turning effect in each direction will be the same:
sum of clockwise moments = sum of anticlockwise moments
===============================
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# Best Practice Template - DOC
Document Sample
``` Shared Practice Template
Best Practice Title: Equivalent Fractions
Submitted by: Carol Ann Marx
Topic: Fractions
Objective(s):
1. to identify equivalent fractions
2. to use multiplication to find equivalent fractions
Curriculum Framework Standard(s):
6.N.4 Demonstrate an understanding of fractions as a ratio
of whole numbers, as parts of unit wholes, as parts of a
collection, and as locations on the number line.
6.N.5 Identify and determine common equivalent fractions, mixed
numbers , decimals and percents.
Lesson/Activity/Strategy Description:
A. Introduction:
The teacher will ask the students what it means when two things are
equivalent. (They are equal in amount or value) Ask them to think
of things that are equivalent. (ex: 12 inches is equivalent to one foot,
4 quarts is equivalent to 1 gallon, 3 feet is equivalent to 1 yd. etc.)
"Today, we will learn about equivalent fractions."
B. Teaching-Learning Experience
1. In groups of two, the students will display their fraction bar set on
desks.
2. The teacher will model the following equivalent fractions on the
4/4 = 1 3/5 = 6/10 1/3 = 2/6
The teacher will ask the students to find the fraction bar that shows
1/5. She will ask them to find the fraction bar that has the same
shaded amounts. (2/10) Point out that 1 part out of 5 is equivalent
to 2 parts out of ten. Continue with the following fractions:
2/3 = 6/9 4/6 = 8/12 4/5 =8/10
3. The teacher will pass out worksheet packet. On the first
worksheet, the teacher will instruct the students to look at
fractions at the top of the page. They are to use the fraction bars
at the bottom and find the equivalent fraction. Teacher will
model 8/12 = 4/6. Children will complete the worksheet and will
correct together with the teacher.
4. Have students find a blue bar with 1 part shaded. (1/4) Split
each part of the bar into 2 equal parts. How many total parts does
the split bar now have? (8) How many shaded parts? (2) What
fraction does the bar now represent? (2/8) What does this show
about the fractions 1/4 and 2/8? (These fractions are equivalent).
Continue with the following fractions:
3/5 = 6/10 1/4 = 3/12 2/3 = 8/12
Instruct students to complete the second worksheet. When
completed, they will correct together with the teacher.
5. Show students the following equivalent fractions
5/6 = 50/60 3/4 =18/24 2/3 = 14/21
Ask students what number has been used to multiply both the
numerator and the denominator.
Repeat with the following fractions:
5/8 = ?/16 1/3 = ?/15 1/2 = ?/18 2/3 = ?/9
Materials:
fraction bars, worksheet packet , worksheet transparencies
for teacher to use on overhead
Assessment(s):
At the conclusion of this lesson, the teacher will ask the students
to complete a worksheet independently identifying equivalent
fractions and using multiplication to find equivlalent fractions.
```
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|
# what is the equation of the line passing through the point (-2, -3) and is perpendicular to the line 4y + x = 6
point = (-2,-3)
equation of line perpendicular to it = 4y + x = 6
= y + x/4 =3/2
= y = -x/4 + 3/2
Therefore slope = -1/4 as y =mx + c where m is slope
lines are perpendicular. Therefore, product of slope = -1
m.m1 = -1
= m x -1/4 = -1
= m = 4
hence slope of line passing through (-2,-3) is 4
now , y-y1=m(x-x1)
= y-(-3)=4(x-(-2))
= y+3=4(x+2)
= y+3=4x+8
= y=4x+5
llltkl | Student
The given equation of the line is `4y+x=6.`
Rearranging it in the slope-intercept form, i.e. `y=mx+b` where `m` denotes the slope of the line we get:
`4y=-x+6`
Dividing both sides by 4;
`y=-1/4x+6/4`
`rArr y=-1/4x+3/2`
Thus the slope of the given line, `m=-1/4` .
The slope of a perpendicular line will be the negative reciprocal of the slope of the original line.
Hence, slope of the perpendicular line =` 4` .
Now, to find the equation of the perpendicular line we will use the point-slope form of the equation of the straight line i.e. `y-y_1=m(x-x_1)` where `(x_1,y_1)` is a known point on the line, `m` is the slope of the line and `(x,y)` is any other point on the line.
Here, `(x_1,y_1)=(-2,-3)` and `m=4` .
Hence, the equation of the perpendicular line is given by:
`y-(-3)=4[x-(-2)]`
`rArr y+3=4(x+2)`
`rArr y+3=4x+8`
`rArr 4x-y=-5`
Therefore, the required equation of the line passing through the point `(-2, -3)` and perpendicular to the line `4y + x = 6` is 4x-y=-5.
(-2, -3) and a perpendicular line to 4y+X=6 turn this into y=mx+b equation
you would get: 4y+x=6
4y=-X+6
y=(-1/4)x+6
where -1/4 is your slope and 6 is your y intercept, now we know that when its perpendicular we find the opposite reciprocal , which in this case is 4.
so we know that our new equation has a slope of 4
so y=4x+b then plug in -2, -3 into x and y.
-3=4(-2)+b
-3=-8+b
5=b
so it would be:
y=4x+5
(-2, -3) and a perpendicular line to 4y+X=6 turn this into y=mx+b equation
you would get: 4y+x=6
4y=-X+6
y=(-1/4)x+6
where -1/4 is your slope and 6 is your y intercept, now we know that when its perpendicular we find the opposite reciprocal , which in this case is 4.
so we know that our new equation has
Wiggin42 | Student
point (-2, -3) and is perpendicular to the line 4y + x = 6
The slope of your line is -1/4. The slope of your perpendicular line is the negative reciprocal which is 4.
Plug these values into the point slope formula:
y - y1 = m(x - x1) to get
y + 3 = 4(x + 2)
You can leave it in that form or change to whichever form you please.
|
Polynomials
It is an open secret in mathematics that the only objects we can actually compute with over the real numbers are polynomials, and by extension, rational functions. While we all learn about trigonometry and exponential functions in high school, we are actually never able to evaluate those functions except at a few points, e.g. $$0, \pi/4, \pi/3, \pi/2$$, and so on. If we need the value of $$\cos(0.2)$$, we can't find it.
You will tell me, "of course we can find that - here, I'll just plug it into my calculator." However, the fact is that your calculator does not find $$\cos(0.2)$$; it finds the value of a polynomial approximation to the cosine function at 0.2, accurate to 12 or 16 digits. And that is the real point: we don't need to know the exact value of $$\cos(0.2)$$. All we need is to know the value with a few digits of accuracy. The way we get those values is to approximate the cosine by a polynomial, and then evaluate that. Let's do this in an example.
Let's try to find the value of $$\cos(0.2)$$, by finding some polynomial that approximates $$\cos(x)$$ near 0.2. We'll find some quadratic polynomial $$c_2(x)=a_0+a_1x+a_2x^2$$ that is close to $$\cos(x)$$ for $$x$$ near zero. In fact, we want to write $$\cos(x)=c_2(x)+R(x)x^{3}$$ where $$R(x)$$ is some bounded remainder term for $$x$$ near zero.
The first thing we need to require is that $$c_2(0)=\cos 0$$. Since $$\cos(0)=1$$ and $$c_2(0)=a_0$$, then let's set $$a_0=1$$. That's great!. What's left? We can write $\cos(x)=1+a_1x+a_2x^2+R(x)x^{3}.$ This means that $$a_1 = (\cos(x)-1)/x-a_2x-R(x)x^n.$$ This is where the limits we learned about in Chapter 1 could come in handy. Since $$a_1$$ does not depend on $$x$$ at all, we can require $a_1=\lim_{x\to0}a_1=\lim_{x\to0}\frac{\cos(x)-1}{x}+\lim_{x\to0}a_2 x+ \lim_{x\to0} R(x)x^2.$ But we saw in [CosFirstOrderLimit]that this limit on the right is zero. Thus, $$a_1=0,$$ and now $$\cos(x)=1+a_2x^2+R(x)x^{3}.$$
Let's repeat this process to try to find $$a_2$$. We write $a_2=\frac{\cos(x)-1}{x^2}-R(x)x,$ so $\lim_{x\to0}a_2= a_2 =\lim_{x\to0}\frac{\cos(x)-1}{x^2}.$ But this is another limit we computed in [CosDiffLimit]. We found that \begin{align*} \frac{\cos(x)-1}{x^2} &= -\frac12. \end{align*} It follows that $$\cos x=1-\frac12 x^2+R(x)x^3$$. At this point we can see that \begin{align*} R(x)&=\frac1x\cdot\left(\frac{\cos x - 1}{x^2}+\frac12\right).\\ \end{align*} Intuitively, when $$x$$ is small that looks like something that is going to zero times something that is blowing up. Maybe it will remain bounded. Later we will show that it does.
We conclude that our polynomial $$c_2(x)=1-\frac12 x^2$$ is an approximation to $$\cos(x)$$ when $$x$$ is near 0. Indeed, on our calculator we find that $$\cos(0.2)\approx .9800665$$, while $$c_2(0.2)=0.9800000.$$ Already by approximating the cosine by a quadratic we have error smaller than $$10^{-4}$$.
Choose $$x_0$$:
Figure 1: Polynomial approximation to $$\cos x$$
Let's review what we did to get this approximation.
Algorithm to find a quadratic approximation to a function
1. We assumed that the function could be expressed as $$a_0+a_1x+a_2x^2+R(x)x^3$$ near $$x=0$$, where $$R(x)$$ was bounded.
2. In that case, we found that $$a_0$$ was the limit of the function at 0.
3. We then solved for $$a_1$$ in terms of $$x$$, and took the limit of that as $$x$$ went to zero. This gave us a value for $$a_1$$.
4. We repeated the last step to find $$a_2$$. I.e. we solved for $$a_2$$ and took the limit as $$x$$ went to zero to get a value.
You can see that we could repeat that last step as many times as we can stand in order to get a higher degree polynomial. We will treat many of our functions this way for the rest of the chapter.
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# Displacement and Distance in 1 Dimension
In this Physics tutorial, you will learn:
• The meaning of Displacement as a change in position
• The definition of Distance and its unit
• How Distance differs from the Displacement?
• How to calculate the Displacement and Distance in one dimension?
Kinematics Learning Material
Tutorial IDTitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
3.3Displacement and Distance in 1 Dimension
## Introduction
In the previous tutorial "Position. Reference Point" we considered the objects examined as stationary (unmoveable). The only thing we could do was finding the object's coordinates (position) and its distance from the origin (reference point).
But what happens if the object changes its location? Does the position changes or it remains the same?
This and other questions are discussed here in this article. So, let's take a look at them.
## Displacement as a Change in Position (in one dimension)
Look at the object shown in the figure.
The object initially is at the position x1 = 7m. After a while, (after losing it from sight) we notice that the object has moved to the position x2 = 12m as shown below.
Since we have lost the object from sight during its motion, we are not sure in what direction it has moved before stopping at x2 = 12m. However, we are sure for one thing: at the end of process, the object is 5m on the right of its original position. This means its position has changed by +5m.
This change in position in Kinematics is known as "Displacement". It is denoted by ∆x and is obtained by subtracting the final and initial position.
Mathematically, we can write:
∆x = xfinal - xinitial
= x2 - x1
In the specific case, we have
∆x = xfinal - xinitial
= x2 - x1
= 12m - 7m
= 5m
It is obvious that Displacement is obtained by subtracting two vectors: the final and initial position. Therefore, Displacement is a vector quantity as well.
In the figure below, we can show the three vectors involved in our example.
## What is Distance? How does it differ from the Displacement?
In the previous example, it was supposed that we had lost the object from sight from a while and when we noticed it again, the object was displaced by +5m due right from its original position. But what if the object after starting its motion from x1 = 7m had first moved for some metres (for example 6m) due left and then it has moved due right until it went to the final position x2 = 12m?
It is obvious in such a scenario the object has moved more than 5m, which was the result obtained when we calculated the Displacement. Therefore, we must consider another quantity (measured in metres as well) which shows the entire length of the object's path during its motion. This quantity is known as Distance (in formula, s).
Hence, by definition "Distance (s) is the length of the entire path followed by an object during its motion." Distance is a scalar quantity, as it does not involve any direction. When dealing with distance, we are interested only in the magnitude of movement (in units of length), not in the moving direction. In this regard, we can say that the distance 25m due North is the same as 25m due East as in both cases the object has moved for 25m regardless the direction.
In the previous example, if the object has moved first by 6m due left, it went at the position 7m - 6m = 1m. Then, it moved for other 11m due right as 12m - 1m = 11m (the final position is 12m). So, the object has moved 6m + 11m = 17m in total. Therefore, we obtain two different results for the same event:
Displacement ∆x = 5m
and
Distance s = 17m
But, what if the object moves only in one direction? Is the Displacement in this case equal to the Distance?
The answer is NO. Although the numerical result (magnitude) is the same in both cases, Displacement has an additional info included - the direction. Thus, if an object has moved 5m due right only, we say, "Distance is 5m" and "Displacement is 5m due right." Therefore, Displacement and Distance are never identical. You must be careful to avoid confusing them with each other.
Remark! Unlike Displacement, the Distance cannot be negative as it represents the length of the object's path. If the object is at rest, this length is zero but if it starts moving, the path's length becomes a positive number. Consider the mileage screen in a car. It show how many km or miles has the car moved since it has been purchased. This number is always positive as it counts the total distance the car has travelled regardless the moving direction.
In the figure above, we can learn that the car has travelled 77,030 km in total during its lifetime but we cannot know in which direction it has moved. Every km the car moves is recorded on the screen as a positive number, regardless the moving direction.
### Example 1
Find the Displacement and Distance travelled by the object shown in the figure.
### Solution 1
The object's position at the "instant 0" represents the initial position, xinitial = +10m while the position shown at the "instant 2" represents the final position, xfinal = +5m. The position shown at the "instant 1" is not considered when dealing with Displacement, as it is a transitory position. When calculating the Displacement, we are interested only at the initial and final position, not at the in-between positions during the transitory stages.
Hence, for the Displacement ∆x we have:
∆x = xfinal - xinitial
= (+5)m - (+10)m
= (-5)m
This means the final position of the object is 5m on the left of its initial position. You can check the veracity of this statement by looking at the figure, in which it is evident that the final position of the object is 5m on the left of the initial one.
As for the Distance, we must simply count the metres travelled by the object during its motion, regardless the direction. Hence, if we denote as s1 and s2 the two distances travelled by the object, we have
s1 = 13m
(the positions +10m and-3m are 13 m away from each other)
and
s2 = 8m
(the positions -3m and+5m are 8 m away from each other)
Therefore, we obtain for the total Distance:
stotal = s1 + s2
= 13m + 8m
= 21m
To summarize, we say "the object was displaced by 5m due left but it has moved by 21m in total."
## Whats next?
Enjoy the "Displacement and Distance in 1 Dimension" physics tutorial? People who liked the "Displacement and Distance in 1 Dimension" tutorial found the following resources useful:
1. Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Kinematics Revision Notes: Displacement and Distance in 1 Dimension. Print the notes so you can revise the key points covered in the physics tutorial for Displacement and Distance in 1 Dimension
3. Kinematics Practice Questions: Displacement and Distance in 1 Dimension. Test and improve your knowledge of Displacement and Distance in 1 Dimension with example questins and answers
4. Check your calculations for Kinematics questions with our excellent Kinematics calculators which contain full equations and calculations clearly displayed line by line. See the Kinematics Calculators by iCalculator™ below.
5. Continuing learning kinematics - read our next physics tutorial: Displacement and Distance in 2 and 3 Dimensions
|
# How to Calculate the Perimeter of a Semi-circle?
The perimeter of a semi-circle uses a different formula compared to that of the full circle. As the name implies, the semi-circle is half of a complete Circle. Also, finding its perimeter simply comprises discovering its diameter and adding it to the half the perimeter of a complete circle.
Students are often required to complete the perimeter of a semi-circle, and there is often the belief that it’s a quite tasking computation. However, this claim is in no way close to reality, as the circle’s perimeter is as simple as computing a circle perimeter. Like the circle, all that must be available in the semi-circle radius.
This article will highlight a comprehensive and easy way to compute the semi-circle perimeter with examples.
The Semi-circle
The semi-circle is the half of the circle separated in the middle, which is the diameter. As such, the semi-circle is all around is half a circumference and a diameter.
Below are a circle and a semi-circle.
As evident above, while the circle is a complete round arc, the semi-circle is half an arc and a diameter. It is based on these definitions will the perimeter of the semi-circle be found.
## The Perimeter of a Semicircle Formula
The perimeter of a semi-circle is the addition of the diameter and half the circumference of a complete circle.
It is mathematically given as
P = D + πr
Where D = the diameter of the semi-circle and two times the radius
r = radius, which is often half the diameter
π = a constant equal to 3.142
The formula for the perimeter of the circle is derived from that of the full circle.
The perimeter of a complete circle is the same as the circumference of the circle and is given as
P = 2πr
Since the perimeter or circumference of the circle is 2πr, and the semi-circle is actually half circumference and diameter, then it will only make sense for the latter to be expressed
Diameter + half of a circumference
That will therefore be
P = D + ½ x 2πr
P = D + πr
The above is how the formula for the semi-circle is gotten.
## How to Calculate the Perimeter of a Semi-circle?
Below are the steps to calculate the perimeter of the circle
Step 1: write down the formula
Step 2: input the diameter or radius into the formula and start your computation.
## Examples that highlight how to calculate the Semi-circle
Example 1
Find the perimeter of a semi-circle with a 14 cm radius
Solution
We will use the steps outlined in this article to find the perimeter
Step 1: write down the formula
P = D + πr
Where r = radius, π = 3.142, D = 2r
Step 2: input the diameter or radius into the formula and start your computation.
Since r = 14 cm, D = 2 x 14 cm = 28 cm
Therefore;
P = 28 + 3.142 x 14
P = 28 + 43.99
P = 71.99
Example 2
Let the diameter of a semi-circle be 31 cm, and find its perimeter?
Solution
The same method that we used in solving the first example will also be applied to solve this one. However, the only difference in this case is that the diameter was given instead of the radius. However, that is not the problem, as the diameter only needs to be divided to find the radius.
The formula for the perimeter of the semicircle is given as;
P = D + πr
Where D = 31 cm,
π = 3.142 cm,
r = 31cm/2 = 15.5 cm.
P = 31 cm + 3.142 x 15.5 cm
P = 79.701 cm
Example 3
Consider a semi-circle whose perimeter is given as 48 cm. determine its radius.
Solution
Since the perimeter has already been given, we can find the radius of the circle the same way
P = D + πr
48 cm = D + 3.142r, since π = 3.142
Also since D = 2r,
48 cm = 2r + 3.142r
48 cm = 5.142r
48/5.142 = r
9.335 cm = r
Therefore the radius of the semi-circle is 9.335 cm.
## Conclusion
The perimeter of a semi-circle is a branch of the entire computation about the circle and its circumference. The Semi-circle perimeter, as expected, has its formula carved out from that of the circle. As shown in this article, calculating it is just as easy as that of the circle.
In this article, comprehensive calculations of the semi-circle were carried out with all steps systematically outlined. Three examples were also considered to show students possible questions they may have to deal with in class, tests or exams. The formula and manipulations highlighted in this article can be trusted to help students go about their respective calculations.
### Different Uses and Applications of organic chemistry
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### Ace organic chemistry- tips and tricks, cheat sheets, summary
Mastering organic chemistry takes time and practice. It might seem a daunting task at once if the right approach is not adopted. But in our
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This article on organic reactions is a special one in our organic chemistry series. It will guide you through how chemical transformations unlock diversity at
### Organic Spectroscopy
Organic spectroscopy can be used to identify and investigate organic molecules. It deals with the interaction between electromagnetic radiation (EMR) and matter. These waves travel
|
# The teddy bears’ fraction picnic
Purpose
This unit supports students to equally partition objects and sets into fractional parts.
Achievement Objectives
NA1-1: Use a range of counting, grouping, and equal-sharing strategies with whole numbers and fractions.
Supplementary Achievement Objectives
NA1-4: Communicate and explain counting, grouping, and equal-sharing strategies, using words, numbers, and pictures.
Specific Learning Outcomes
• Partition a length, area or volume into equal parts.
• Partition a set into equal parts and anticipate the result.
• Recognise that the numerator of a fraction is a count.
• Recognise that the bottom number of a fraction gives the size of the parts being counted.
Description of Mathematics
In this unit students explore equal partitioning of objects and sets, and how to name the parts they create using fractions. The easiest partitions are those related to halving since symmetry can be used. Halves come from partitioning a whole into two equal parts. Quarters come from partitioning a whole into four equal parts, etc.
An object is partitioned according to an attribute, or characteristic. For example, a banana might be partitioned by length, but a sandwich partitioned by area. Some attributes are more difficult to work with than others. Length is the easiest attribute but others such as area, volume (capacity), and time are more difficult.
The results of partitioning a set into fraction parts can be anticipated, using number knowledge and strategies. For example, one quarter of 20 could be found by:
• Dealing 20 objects onto four quarters one by one, or in collections of two.
• Halving 20 into 10, then halving again.
• Realising that putting one object on each quarter uses four each time, and repeatedly adding four until 20 is used up.
In this unit we apply three criteria when evaluating students’ capacity to partition and object or set into fractional parts:
• The correct number of parts is created
• The shares are equal
• The whole is exhausted (used up).
The learning opportunities in this unit can be differentiated by providing or removing support to students and by varying the task requirements. The difficulty of tasks can be varied in many ways including:
• encouraging students to work collaboratively in partnerships
• pausing to allow students to share their work with others
• restricting or extending the range of fractions students are asked to work with.
The contexts for this unit can be adapted to suit the interests and experiences of your students. For example:
• The first slide of PowerPoint 1 could easily be replaced to change the whole context of the unit to be about any different characters you choose.
• You could add more types of food to the ones used to better reflect the food your students usually eat.
Required Resource Materials
• Teddy Bears (brought from home) or Teddy Bear counters
• Pattern blocks or Attribute blocks
• Play dough and plastic knives
• Counters, cubes, or beaNZ
• Blank wooden cubes (to make dice)
• Apple, doughnut, banana, bread (optional)
• Clear plastic drink bottle and plastic cups
• Scissors
• Copymasters One, Two, Three, Four, Five, and Six
• PowerPoints One, Two, and Three
Activity
#### Lesson One
1. Use PowerPoint One to tell the story of two teddy bears going on a picnic. Unfortunately, Papa Bear, who made the lunch, forgot to cut up the food items equally. What are Ben and Barbara Bear supposed to do?
Ben and Barbara are friends. They always share their food equally. How could they cut this sandwich? (Slide Two)
2. Discuss ways to cut the sandwich in two equal pieces. Having two teddy bears available allows for acting out of the equal sharing.
What does equal mean?
What do we call these pieces? (halves)
3. Write the words “one half” and the symbol 1/2.
Why do you think there is a 1 and the 2? (2 refers to the number of those parts that make one)
4. Look at Slides Three and Four of PowerPoint One.
Is the sandwich cut in half? Why? Why not?
5. Provide the students with three ‘sandwiches’ made from Copymaster One.
What other ways could Ben and Barbara cut their sandwich in half?
6. For students who complete the task quickly, give them another sandwich.
Cut this sandwich in two pieces that are not halves.
7. When each student has produced at least two halved sandwiches bring the class together to share their thinking. Make a collection of halves and ‘non-halves’ and discuss those collections.
What is true for the sandwiches that are cut in half?
What is the same or the sandwiches that are not cut in half?
The key idea of halving is that both Ben and Barbara get the same amount of sandwich. As the sandwich is not exactly symmetric through a horizontal mirror line you might get interesting conversation about the equality of shares. Slide Five may provide some stimulus for the discussion. Using the cut-out pieces of sandwich, show how halves map onto each other by reflection (flipping) or rotating (turning). With paper sandwiches you can fold to show that mapping of one half onto another.
8. Luckily Papa Bear prepared more than just one sandwich. Copymaster Two has some other food items that Papa packed. Ask your students to work in pairs to cut each item in half. They need to record where they will cut each item and justify why they put the line where they did.
9. After enough time bring the class together and discuss their answers. You might have real items to half. This is particularly important in the case of the bottle of fruit drink and the bag of 12 cashews. Note there are different ways to partition each object into two equal parts. Points to note are:
• Do your students create two parts, exhaust all of the objects or set, and check that the parts are equal in size?
• Do your students see halving as a balancing, symmetric process?
10. For example, get two clear plastic cups. Ask a student to pour half of the drink bottle into each cup, one for Ben and one for Barbara. Look to see that the student balances the level in each cup. Capacity is a perceptually difficult attribute for measurement.
11. Similarly, look to see it students equally share the cashews by one to one dealing.
12. Finish the lesson with the final two slides of PowerPoint One.
Can your students distinguish when a shape or set has been halved?
#### Lesson Two
In this lesson students explore the impact of other Teddy Bears joining in the picnic. What happens to the size of the portions as more shares are needed? How do we name the shares?
1. Remind the students about the previous lesson. Use a chart of halves and ‘not halves’ to focus attention on equal sharing and the words and symbols for one half.
Imagine that four Teddy Bears went on the picnic instead of two. Papa still packed the same amount of food. What would happen then?
2. Sit four teddy bears around in a circle. Give them names like Ben, Barbara, Bill and Bella. Use Copymaster One to make a few sandwiches.
How could we cut a sandwich into four equal parts? What will the parts be called?
3. Students should find quartering intuitive, as it involves halving halves. Symmetry is still useful. Let students share how they think a sandwich could be partitioned into four equal parts. Act out giving one part to each Teddy Bear.
Does each Teddy Bear get the same amount of sandwich?
Common solutions are:
Students often miss the option of dividing the area of the sandwich by length…
Or other possibilities…
4. Ask students: How big is one quarter, compared to one half? How many quarters is the same as one half?
5. Use pieces of’ ‘sandwich’ to check. It is good for students to see that two quarters make one half irrespective of the shape of the pieces. Challenge the students to find some ways to cut a sandwich into four pieces that are not quarters. Examples might be:
6. You may need to highlight the lack of equality of non-quarters by mapping the pieces on top of one another. Discuss that quarters are only quarters if:
• All the sandwich is used
• The pieces are all equal in size.
7. Provide the students with copies of Copymaster Two again.
This time the challenge is to share each picnic food into quarters.
8. Give the students enough time to record where they would partition the items before gathering as a class to discuss their ideas. Pay attention to the different ways that the foods are partitioned. Act out the partitioning either with real objects or imitations, e.g. banana and cheese made from playdough of plasticine. For example, the banana is a length, so the partitioning is most likely to be by length. Some students may attend to area (second example).
or
The shape of the sector of cheese places a restriction on how it can be partitioned. Likely responses are (left is correct, right is not):
9. Equally sharing the bottle of fruit juice among four glasses might be done in two main ways.
• Set out four plastic cups and keep pouring while balancing the water levels.
• Halve the bottle into two plastic cups. Pour half of each cup into another cup until the levels match.
10. Look to see that your students can equally share the 12 cashews by dealing one to one for each teddy bear. Highlight that the half share is six cashews and the quarter share is three cashews.
11. Record the words and symbol for one quarter. Create a chart of quarters for the food items.
12. Finish the session with the game called ‘Make a whole sandwich.” Students need copies of the first page of Copymaster Three (Halves and quarters) and a blank dice labelled, 1/2, 1/2, 14, 2/4, 3/4 ,choose. The aim of the game is for players to make as many whole sandwiches as they can.
Players take turns to:
• Roll the dice.
• Take the amount of sandwich shown on the dice, e.g. If 3/4 comes up, then the player takes three one quarter pieces of sandwich.
• If choose comes up the player chooses a single piece.
At anytime a player can rearrange their sandwich pieces to make as many whole sandwiches as possible. Time the game to finish in five minutes which is long enough to gather a lot of pieces.
13. Discuss what might be learned from the game. For example:
• Three quarters is three lots of one quarter.
• Two quarters is the same amount as one half.
• Two halves and four quarters both make one whole.
14. Record those findings symbolically and discuss the meaning of + and = as joining and “is the same amount as”, e.g. 3/4 = 1/4 + 1/4 + 1/4 or 2/4 = 1/2.
#### Lesson Three
In this lesson students explore the equal sharing of picnic foods among three Teddy Bears. Thirding an object or set is generally harder than quartering because symmetry is harder to use. Your students should observe that thirds of the same object or set are smaller than halves but larger than quarters.
1. Begin with a reminder of what you did in the previous lesson:
We explored how to cut sandwiches into two and four equal parts. What are those parts called? (halves and quarters)
2. Be aware to the non-generality of early fraction words in English. Halves should be twoths, and quarters should be fourths, to match what occurs with sixths and further equal partitions. In many other languages, such as Maori and Japanese, the word for a fraction indicates the number of those parts that make one. Maori for one fifth is haurima (rima means five) and in Japanese one fifth is daigo (go means five).
What parts will we get if we share the food among three Teddy Bears?
3. Point out that third is a special English word for one of three equal parts. Show the students one half and one quarter of a sandwich.
Which is bigger, one half or one quarter? Why?
How big do you think one third will be? Why?
4. Students might conjecture that thirds are smaller than halves but bigger than quarters. That is true on the assumption that the whole (one) remains the same. Give the students copies of four ‘sandwiches’ from Copymaster One. Set out three Teddy Bears.
Find different ways to share one sandwich equally among the three Teddy Bears.
5. PowerPoint Two shows two examples and three non-examples of cutting a sandwich into thirds. Note that the successful cuts are vertical and horizontal. Students should notice that the two non-examples produce unequal parts. On Slide Five ask:
Is this sandwich cut into thirds? (No, the parts are unequal)
6. Give the students a copy of Copymaster Two. Ask them to work in pairs to share each food item into thirds. After a suitable period bring the class together to discuss the equal sharing.
7. PowerPoint Two contains animations of partitioning some of the foods into thirds. You may want to act out model answers, certainly with pouring among three plastic glasses while keeping the levels balanced. Discuss issues that arise. For example:
Do the parts have to be the same shape to be equal? (Banana is a good example)
Why are thirds harder to make than halves and quarters? What other fractions would be hard to make? Why?
8. In the next part of the lesson students anticipate the result of cutting thirds and quarters in half. What are the new pieces called?
PowerPoint Two, Slides 9-11, shows how halving quarters produces eighths and halving thirds produces sixths. It is important for students to connect the naming of these equal parts to how many of those parts form one whole (sandwich).
9. Use parts of sandwiches made from Copymaster Three to build up students’ counting knowledge of fractions. Lay down and join the same sized pieces and ask students to name the fraction. For example:
One sixth Two sixths Three sixths Four sixths
(or one third) (or one half) (or two thirds)
10. Count with the same sized parts past one whole and write the fraction symbols as you count.
How much of a sandwich is 2/3? 3/4? 5/8? etc.
11. It is important that students understand that the top number of a fraction (the numerator) is a count of equal parts. The bottom number (the denominator) tells how many of those parts make one.
12. Change the game from Lesson Two called “Make a Whole sandwich.” Include all the pieces from Copymaster Three so that thirds, sixths and eighths are also available. Alter the dice to read 1/2, 1/4, 1/3, 1/6, 1/8, choose. Allow the students more time to play that game, say ten minutes.
13. Discuss some ways that the students found to make one whole sandwich. Record their suggestions using symbols, e.g. 1/2 + 1/3 + 1/6 or 3/4 + 2/8. Begin a chart of ways to make one that students can add to using pictures (using pieces from Copymaster Three) and symbols.
#### Lesson Four
In this lesson students work from part to whole. Usually students encounter problems where the whole is well-defined, and they are shown, or must create, the required fraction. As they progress to more complex tasks, it is also important that students can relate a fraction part back to the whole from which that part may have been created.
1. Use Copymaster Four to create a set of paper part foods. With each challenge below the aim is to draw the appearance of the whole, in schematic rather than detailed form. Place the fraction piece in the middle of an A3 sheet of photocopy paper to allow space to draw.
Suppose there are eight Teddy Bears at the picnic. This piece is one eighth of a sandwich, how can we find the size of the whole sandwich?
2. Check that students recognise that eight pieces of that size will make the complete sandwich. Ask a student to draw around the outside of the piece to show the whole sandwich.
How can we check [Name]’s estimate of the whole sandwich?
3. Iterating (copy and pasting) eight copies of the piece will give the area of the sandwich. Note that the whole could look differently as the eighths could be arranged on a line or as a rectangle or other formations. However, the context suggests a rectangular arrangement is best. Other challenges are:
This is one fifth of the banana. How long is the banana?
This is one third of the doughnut. How big is the doughnut?
This is one half of the bun. How big is the bun?
This is one sixth of the chocolate cake. How big is the cake?
This is one quarter of the packet of walnuts. How many walnuts are in the whole packet?
4. Build on the capacity example used earlier. Tell your students that a plastic cup filled to a certain level is one third, quarter, fifth, etc. of a whole bottle. Can they make the quantity of water that fills the bottle?
Independent work
1. Show your students a set of pattern blocks. Choose one shape as ‘the piece of food’, e.g. a rhombus.
I will make a chart with this pretend piece of food.
2. Develop the chart as you go rather than present it as complete.
Provide the students with a copy of Copymaster Five to reduce the writing load.
#### Lesson Five
In this lesson students connect fractions of regions and fractions of sets. Partitioning of a set into equal parts fluently requires multiplicative thinking that most young students do not possess yet. However, exposure to problems with equal partitioning provides opportunities to learn that develop additive part-whole thinking and the ‘sets of equal sets’ concept, that is fundamental to multiplication and division.
1. Use PowerPoint Three to introduce the context of the four Teddy Bears sharing a chocolate cake.
How many smarties do you think go on each quarter? Remember that the quarters must be equal.
2. Students need to use their number understanding to anticipate a result. Some students may ‘virtually’ share the smarties one at a time (look for eye, hand and head gestures). Tracking the number of ‘virtual’ smarties in each quarter puts a significant load on working memory. Estimates of three, four or five should be worked with.
Let’s imagine there are three smarties on each quarter. How many smarties would that be on the whole cake?
3. Working from part to whole in this way checks the estimate. Using actual objects, like counters, to enact the sharing, might follow once anticipations are made and justified.
4. Other students may use additive knowledge, such as 4 + 4 = 8 (one half), then 8 + 8 = 16 (whole cake). Students might adjust from an initial prediction, such as 20 smarties is four more than 16, so each quarter must get 4 + 1 = 5 smarties.
5. Slide Two makes a minor adjustment to the number of shares to see if students can anticipate the effect of more parts on the size of shares. With five parts (fifths) additive knowledge with fives is more likely to be used.
Last time the cake was in quarters. What fractions is it in now? (fifths)
Will the five Teddy Bears get more or less than four Teddy Bears got?
6. Check that students recognise that fifths are smaller than quarters, even in the sets of smarties context. Encourage use of additive thinking by animating one smarty on each fifth of the cake.
How many smarties are on the cake?
If I put two smarties on each fifth how many would be on the whole cake?
How many smarties can I put on each fifth?
7. Slides Three and Four present two other situations of equally sharing smarties onto a cake. Ask your students to name the fraction parts and anticipate the result of the equal sharing.
8. Let your students play the Birthday Cake Game in pairs or threes. The game is made from Copymaster Six. You will need blank dice and counters as well.
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# Week 19 – Top 5 Things I learned in precalculus 11
This year in Precalculus 11 I learned so many things and different things in each unit. Some of the units were a little bit hard and some of the units were easy. Here are the top things I’ve learned in precalculus 11.
1. The first thing in the top five is multiplying and dividing rational expression. It’s used for simplifying products and quotients of rational expression. Here is an example:
2. The second thing in the top five is the arithmetic series. It’s used to determine the sum of n term of an arithmetic series, then solve related problems. Here is an example:
3. The third thing in the top five is using special triangles. It’s used to explore the condition for which given information is sufficient to construct a unique triangle. For example:
4. The fourth thing in the top five is solving systems of equations algebraically. Use algebraic strategies to solve linear quadratic system and quadratic-quadratic system. For example:
5. The fifth thing in the top five is multiplying and dividing radical expressions. It’s used to simplify products and quotients of radical expression. Here is an example:
# Week 18 – precalculus 11
This week in precalculus 11 I learned about cosine law. Cosine Law is to solve triangles that are not right triangles. We should know that we use the sine law when we have given two angles and one side, or two sides and a non-included angle, and we use the cosine law when we are given three sides, or two sides and the included angle.
The formula for cosine law is:
${a^2}$ = ${b^2}$ + ${c^2}$ – 2bc cosA
Here is an example:
# Week 17 – precalculus 11
This week in precalculus 11 I learned about Sine Law. I learned about trigonometry in grade 10 but this year we expand it a little bit more. Sine Law is to solve problems in triangles that are not right triangles. The formula for sine Law is:
Missing side: $\frac{a}{sin A}$ = $\frac{b}{sin B}$ = $\frac{c}{sin C}$
And
missing angle:$\frac{sin A}{a}$ = $\frac{sin B}{b}$ = $\frac{sin C}{c}$
Here is an example:
# Week 16 – precalculus 11
This week in pre-calculus 11 I learned about rational equations. Rational equations have at least one fraction whose numerator and denominator are variable. Two ways of solving rational equation; one way is cross to multiply when there are just 2 fractions and with an equal sign in between, the other way is to multiply all terms by denominators when there are more than 2 fractions.
Here is an Example:
# Week 15 – precalculus 11
This week in pre-calculus 11, I learned about multiplying and dividing rational expression. Actually, I learned a basic way of multiplying and dividing rational expression in grade nine and ten, but this year it got a little bit complicated and we expanded more to it.
Here is an example:
# Week 14 – precalculus 11
This week in precalculus 11 I learned about Rational Expression. It was an easy and short lesson.
Here is an example:
# Week 13 – precalculus 11
This week in precalculus 11 I learned about solving absolute value algebraically.
Here is an example:
# Week 12 – precalculus 11
This week in precalculus 11, I learned about solving systems of equation algebraically. Here is an example:
# Week 11 – precalculus 11
this week in precalculus 11 I learned about solving quadratic inequalities in one variable.
Here is an example:
# Week 10- precalculus 11
For the midterm, I reviewed everything that I have learned till now. One thing that I was reviewing that was not blogged about previously is a common difference. Here is an example:
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Question 1: Find the square root of the following complex numbers:
i) $-5 + 12 i$ ii) $-7-24i$ iii) $1-i$ iv) $-8-6i$
v) $8-15i$ vi) $-11-60\sqrt{-1}$ vii) $1+4 \sqrt{-3}$ viii) $4i$ ix) $-i$
There are two ways by which we can solve the problems. We have a few solved by each method.
i) Let $\sqrt{-5+12i} = x+iy$
Squaring both sides
$\Rightarrow -5 + 12 i = ( x + iy)^2$
$\Rightarrow -5 + 12 i = ( x^2 - y^2) + i ( 2xy)$
$\Rightarrow x^2 - y^2 = - 5$ … … … … … i)
$\Rightarrow 2xy = 12$ … … … … … ii)
Now $(x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$
$\Rightarrow (x^2 + y^2)^2 = ( - 5)^2 + (12)^2$
$\Rightarrow (x^2 + y^2)^2 = 169$
$\Rightarrow x^2 + y^2 = 13$ … … … … … iii) [ since $x^2 + y^2 > 0$ ]
Solving i) and iii)
$2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$
From i) $y^2 = x^2 + 5 \Rightarrow y^2 = 4 + 5 \Rightarrow y = \pm 3$
Since $2xy$ is positive, therefore $x$ and $y$ are of same signs.
$\therefore \sqrt{-5+12i} = \pm ( 2 + 3i)$
ii) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = -7 - 24 i$ $\therefore Re(z) = - 7$ $|z| = \sqrt{(-7)^2+(-24)^2} = 25$
Here $Im(z) < 0$, therefore
$\sqrt{-7 - 24 i} = \pm \Bigg\{$ $\sqrt{\frac{25+(-7)}{2}}$ $- i$ $\sqrt{\frac{25-(-7)}{2}}$ $\Bigg\}$
$\Rightarrow \sqrt{-7 - 24 i} = \pm ( 3-4i )$
iii) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = 1-i$ $\therefore Re(z) = 1$ $|z| = \sqrt{(1)^2+(-1)^2} = \sqrt{2}$
Here $Im(z) < 0$, therefore
$\sqrt{1-i} = \pm \Bigg\{$ $\sqrt{\frac{\sqrt{2}+1}{2}}$ $- i$ $\sqrt{\frac{\sqrt{2} - 1}{2}}$ $\Bigg\},$
iv) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = -8-6i$ $\therefore Re(z) = -8$ $|z| = \sqrt{(-8)^2+(-6)^2} = 10$
Here $Im(z) < 0$, therefore
$\sqrt{-8-6i} = \pm \Bigg\{$ $\sqrt{\frac{10-8}{2}}$ $- i$ $\sqrt{\frac{10+8}{2}}$ $\Bigg\}$
$\Rightarrow \sqrt{-7 - 24 i} = \pm ( 1-3i )$
v) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = 8-15i$ $\therefore Re(z) = 8$ $|z| = \sqrt{(8)^2+(-15)^2} = 17$
Here $Im(z) < 0$, therefore
$\sqrt{8-15i} = \pm \Bigg\{$ $\sqrt{\frac{17+8}{2}}$ $- i$ $\sqrt{\frac{17-8}{2}}$ $\Bigg\}$
$\Rightarrow \sqrt{8-15i} = \pm \frac{1}{\sqrt{2}} ( 5-3i )$
vi) Let $\sqrt{-11-60i} = x+iy$
Squaring both sides
$\Rightarrow -11-60i = ( x + iy)^2$
$\Rightarrow -11-60i = ( x^2 - y^2) + i ( 2xy)$
$\Rightarrow x^2 - y^2 = -11$ … … … … … i)
$\Rightarrow 2xy = -60$ … … … … … ii)
Now $(x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$
$\Rightarrow (x^2 + y^2)^2 = ( - 11)^2 + (-60)^2$
$\Rightarrow (x^2 + y^2)^2 = 3121$
$\Rightarrow x^2 + y^2 = 61$ … … … … … iii) [ since $x^2 + y^2 > 0$ ]
Solving i) and iii)
$2x^2 = 50 \Rightarrow x^2 = 25 \Rightarrow x = \pm 5$
From i) $y^2 = x^2 + 11 \Rightarrow y^2 = 25 + 11 \Rightarrow y = \pm 6$
Since $2xy$ is negative, therefore $x$ and $y$ are of opposite signs.
$\therefore \sqrt{-11-60i} = \pm ( 5-6i)$
vii) Let $\sqrt{1 + 4\sqrt{3} i} = x+iy$
Squaring both sides
$\Rightarrow 1 + 4\sqrt{3} i = ( x + iy)^2$
$\Rightarrow 1 + 4\sqrt{3} i = ( x^2 - y^2) + i ( 2xy)$
$\Rightarrow x^2 - y^2 = 1$ … … … … … i)
$\Rightarrow 2xy = 4\sqrt{3}$ … … … … … ii)
Now $(x^2 + y^2)^2 = ( x^2 - y^2)^2 + 4 x^2 y^2$
$\Rightarrow (x^2 + y^2)^2 = ( 1)^2 + (4\sqrt{3})^2$
$\Rightarrow (x^2 + y^2)^2 = 49$
$\Rightarrow x^2 + y^2 = 7$ … … … … … iii) [ since $x^2 + y^2 > 0$ ]
Solving i) and iii)
$2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$
From i) $y^2 = x^2 -1 \Rightarrow y^2 = 3 \Rightarrow y = \pm \sqrt{3}$
Since $2xy$ is positive, therefore $x$ and $y$ are of same signs.
$\therefore \sqrt{1 + 4\sqrt{3} i} = \pm ( 2 + \sqrt{3} i)$
viii) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = 4i$ $\therefore Re(z) = 0$ $|z| = \sqrt{(0)^2+(4)^2} = 4$
Here $Im(z) > 0$, therefore
$\sqrt{4i} = \pm \Bigg\{$ $\sqrt{\frac{4}{2}}$ $+ i$ $\sqrt{\frac{4}{2}}$ $\Bigg\}$
$\Rightarrow \sqrt{4 i} = \pm ( \sqrt{2} + i \sqrt{2} ) = \pm \sqrt{2} ( 1 + i)$
ix) We know
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $+ i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) > 0$
$\sqrt{z} = \pm \Bigg\{$ $\sqrt{\frac{|z|+Re(z)}{2}}$ $- i$ $\sqrt{\frac{|z|-Re(z)}{2}}$ $\Bigg\}, \ \text{if} \ Im(z) < 0$
Let $z = -i$ $\therefore Re(z) = 0$ $|z| = \sqrt{(0)^2+(-1)^2} = 1$
Here $Im(z) < 0$, therefore
$\sqrt{-i} = \pm \Bigg\{$ $\sqrt{\frac{1+0}{2}}$ $- i$ $\sqrt{\frac{1-0}{2}}$ $\Bigg\}$
$\Rightarrow \sqrt{-i} = \pm$ $\frac{1}{\sqrt{2}}$ $( 1-i )$
$\\$
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# Geometry Chapter 3 Practice Test
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| By Jsilva
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Jsilva
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Quizzes Created: 2 | Total Attempts: 400
Questions: 18 | Attempts: 247
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This will give you a good idea of what you have learned and still need to learn from Chapter 3!
• 1.
### Which is an equation in slope-intercept form of the line that contains the points S(7, –9) and T(8, –7)? A) y = 2x – 23 B) x + 2y = –23 C) x – 2y = 23 D) y = 2x + 23
• A.
A
• B.
B
• C.
C
• D.
D
A. A
Explanation
The equation in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. To find the slope, we use the formula (y2 - y1) / (x2 - x1) with the points S(7, -9) and T(8, -7). The slope is ( -7 - (-9) ) / (8 - 7) = 2 / 1 = 2. Therefore, the equation in slope-intercept form is y = 2x + b. To find the y-intercept, we substitute the coordinates of one of the points into the equation. Using point S(7, -9), we have -9 = 2(7) + b. Solving for b, we get b = -23. Therefore, the equation is y = 2x - 23, which matches option A.
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• 2.
### True or false, y = –x + is an equation in slope-intercept form of the line with slope – that contains the point P(–2, 3). A) true B) false
• A.
A
• B.
B
A. A
Explanation
The equation y = -x + b is in slope-intercept form, where b represents the y-intercept. In this case, the equation y = -x + b has a slope of -1 and contains the point P(-2, 3). To determine if this equation is true, we can substitute the x and y values of point P into the equation. When we substitute -2 for x and 3 for y, we get 3 = -(-2) + b, which simplifies to 3 = 2 + b. Solving for b, we find that b = 1. Therefore, the equation y = -x + 1 is in slope-intercept form and satisfies the conditions, making the statement true.
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• 3.
### Find the value of x for a to be parallel to b. m∠1 = 117 and m∠2 = 6x – 12. A) B) C) D)
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
The value of x for a to be parallel to b can be found by setting the measure of angle 1 equal to the measure of angle 2. Therefore, we have: 117 = 6x - 12. Solving this equation, we get: 6x = 129. Dividing both sides by 6, we find that x = 21. Therefore, the correct answer is D.
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• 4.
### Find the measure of the third angle. The diagram is NOT to scale. A) 45° B) 15° C) 36° D) 105°
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
The measure of the third angle can be found by subtracting the sum of the other two angles from 180°. Since angle A is 45° and angle B is 30°, the sum of the other two angles is 45° + 30° = 75°. Subtracting this from 180° gives us 180° - 75° = 105°, which is the measure of the third angle. Therefore, the correct answer is D) 105°.
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• 5.
### If a, b, c, and d are coplanar lines and a is parallel to b, b is perpendicular to c, and c is parallel to d, then which statement must be true? DRAW THIS OUT! A) d is perpendicular to c. B) d is perpendicular to a. C) d is parallel to b. D) c is parallel to a.
• A.
A
• B.
B
• C.
C
• D.
D
B. B
Explanation
Based on the given information, we can draw a diagram with four coplanar lines: a, b, c, and d. Line a is parallel to line b, which means they never intersect. Line b is perpendicular to line c, which means they intersect at a right angle. Line c is parallel to line d, which means they never intersect. From this diagram, we can see that line d is perpendicular to line a. Therefore, the correct answer is B) d is perpendicular to a.
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• 6.
### How many sides does a regular polygon have if each exterior angle measures 9? 40 168 30 none of these
• A.
A
• B.
B
• C.
C
• D.
D
A. A
Explanation
A regular polygon has all sides and angles equal. In this case, each exterior angle measures 9 degrees, which means that the sum of all exterior angles is 360 degrees. Since the exterior angles of a polygon add up to 360 degrees, and each exterior angle measures 9 degrees, the polygon must have 40 sides.
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• 7.
### Complete the statement. If a transversal intersects two parallel lines, then corresponding angles are _____. A) supplementary B) complementary C) congruent D) none of these
• A.
A
• B.
B
• C.
C
• D.
D
C. C
Explanation
If a transversal intersects two parallel lines, then corresponding angles are congruent. This is a basic property of parallel lines and transversals. When a transversal intersects two parallel lines, it creates a set of corresponding angles that have the same measure. Therefore, the correct answer is C) congruent.
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• 8.
### The Polygon Angle-Sum Theorem states: The sum of the measures of the angles of an n-gon is _____. A) B) (n – 2)180 C) D) (n – 1)180
• A.
A
• B.
B
• C.
C
• D.
D
B. B
Explanation
The Polygon Angle-Sum Theorem states that the sum of the measures of the angles of an n-gon is equal to (n-2)180.
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• 9.
### An irregular quadrilateral has exterior angle measures of 93, 85, and 89. What is the measure of the fourth interior angle? A) 273 B) 267 C) 93 D) 87
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
The sum of the exterior angles of any polygon is always 360 degrees. In this case, we are given that three of the exterior angles of the irregular quadrilateral are 93, 85, and 89 degrees. To find the fourth interior angle, we subtract the sum of the given exterior angles from 360 degrees. 360 - (93 + 85 + 89) = 93 degrees. Therefore, the measure of the fourth interior angle is 93 degrees.
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• 10.
### Which are corresponding angles? A) ∠6 and ∠16 B) ∠1 and ∠12 C) ∠8 and ∠16 D) none of these
• A.
A
• B.
B
• C.
C
• D.
D
C. C
Explanation
Corresponding angles are formed when a transversal intersects two parallel lines. In this case, the transversal intersects lines 8 and 16. Therefore, angles ∠8 and ∠16 are corresponding angles.
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• 11.
### In ΔABC, m∠A = 42 and m∠C = 57. Find m∠B. A) 99 B) 147 C) 132 D) 81
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
In a triangle, the sum of the angles is always 180 degrees. Therefore, to find the measure of angle B, we can subtract the measures of angles A and C from 180.
180 - 42 - 57 = 81.
Therefore, the measure of angle B is 81 degrees.
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• 12.
### Suppose you construct lines a, b, c, and d so that a is perpendicular to both b and c, and d is parallel to c. Which of the following is true? A) d is perpendicular to b. B) b is parallel to c. C) c is perpendicular to b. D) none of the above
• A.
A
• B.
B
• C.
C
• D.
D
B. B
Explanation
Lines a and b are perpendicular to each other, and line d is parallel to line c. Since line b is perpendicular to line a, and line d is parallel to line c, it can be concluded that line b is parallel to line c. Therefore, the correct answer is B) b is parallel to c.
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• 13.
### Complete the statement. If a transversal intersects two parallel lines, then _____ angles are supplementary. A) corresponding B) acute C) alternate interior D) same-side interior
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
If a transversal intersects two parallel lines, then the same-side interior angles are supplementary. Same-side interior angles are the pair of angles that are on the same side of the transversal and inside the parallel lines. They are supplementary, which means that their measures add up to 180 degrees. Therefore, option D is the correct answer.
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• 14.
• A.
A
• B.
B
• C.
C
• D.
D
D. D
• 15.
• A.
A
• B.
B
• C.
C
• D.
D
A. A
• 16.
### Find the valuable of the variable. A) –30 B) 30 C) 4 D) none of these
• A.
A
• B.
B
• C.
C
• D.
D
B. B
Explanation
The question asks for the value of a variable, and the answer is given as B) 30. This means that the value of the variable is 30.
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• 17.
### Which is a correct name for the polygon? A) BADEC B) ABEDC C) EDABC D) CDEAB
• A.
A
• B.
B
• C.
C
• D.
D
D. D
Explanation
The correct name for the polygon is D) CDEAB. This is because the vertices of the polygon are listed in a clockwise order, starting from point C and moving to point D, then E, A, and finally B.
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• 18.
### Find m∠A. A) 107 B) 73 C) 117 D) 72
• A.
A
• B.
B
• C.
C
• D.
D
B. B
Explanation
The correct answer is B) 73. This is because the sum of the angles in a triangle is always 180 degrees. Since angle A is the largest angle in the triangle, the sum of the other two angles must be less than 180 degrees. Therefore, the only plausible answer is 73 degrees.
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Related Topics
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# MathPractice
## Implicit Differentiation
Q Differentiate y = arccos(2t/(1+t2))
A A trick to do this question is to convert the question to cos(y) = 2t/(1 + t2). Now do implicit differentiation to get -sin(y) y' = (-2t2 + 2)/(1 + t2)2. Since as know cos(y) = 2t/(1 + t2), then sin(y) = (t2 - 1)/(t2 + 1). (there are two ways to get this. One way is to use trig identity sin2 + cos2 = 1; the other way is to draw a right-angled triangle with angle y, adjacent side = 2t and hypotenuse = (1 + t2). Then the opposite side is (t2 - 1), and sin(y) = opp/hyp). So going back to the derivative and isolate y' = (-2t2 + 2)/(t4 - 1).
## Maxima/Minima
Q Find all values of ${\displaystyle a}$ such that ${\displaystyle f(x)=x^{3}-3x^{2}+ax+1}$ has a local maximum
A first derivative ${\displaystyle f'(x)=3x^{2}-6x+a}$ and second derivative ${\displaystyle f''(x)=6x-6}$. For a local maximum to exist there should be a critical point ${\displaystyle x}$ so that ${\displaystyle f'(x)=0}$ and ${\displaystyle f''(x)<0}$. Thus, there should exist an ${\displaystyle x}$ such that ${\displaystyle 3x^{2}-6x+a=0}$ and ${\displaystyle x<1}$. Solution to the quadratic equation is ${\displaystyle 1\pm {\frac {\sqrt {36-12a}}{6}}}$. Clearly one root is less than 1 if the ${\displaystyle 36-12a>0}$, i.e square root is defined. This gives ${\displaystyle a<3}$
## Graphing
Q Find this equation y = ax3 + bx2 + cx + d if the local max is 3 at -3 and local min is 0 at 1.
A First we differentiate the equation y' = 3ax2 + 2bx + c. We know that x = 1 and x = -3 are solutions to y' = 0, so we know 3a + 2b + c = 0 and 27a -6b + c = 0. Subtract these two equations to get 24a - 8b = 0, or 3a = b. Moreover, we know that at x = 1, y = a + b + c + d = 0, and at x = -3, y = -27a + 9b - 3c + d = 3. Subtract the last two equations to get -28a + 8b - 4c = 3. Now we add this equation to 4*(3a + 2b + c) = 0 to get -16a + 16b = 3. Solving gives a = 3/32, b = 9/32, c = -27/32, and d = -15/32.
## Related Rates
Q A candle is placed at a distance l1 from a thin block of wood of height H. The block is a distance l2 from a wall as shown in figure 7.1. The candle burns down so that the height of the flame, h1 decreases at the rate of 3 cm/hr. Find the rate at which the length of the shadow y cast by the block on the wall increases. ( note: your answer will be in terms of the constants l1 and l2).
A: note that a decrease in the candle height is equivalent to an increase in H. Thus we can use similar triangle to write H/l1 = y/(l1 + l2). Now differentiate the equation with respect to time t, to get H'/l1 = y'/(l1 + l2). Since H' = 3cm/hr (note that it is positive because it's increasing in length), y' = 3(1 + l2/l1) cm/hr.
Q An ostrich 1.5 m tall is walking toward a street light 4 m above the ground at a speed of 5 m/s. How fast is the length of the ostrich's shadow decreasing? At what speed is the tip of the shadow moving? If "a" is the angle between the light and the ground measured at the tip of the shadow, as shown in the picture, what is the rate of change of when the shadow is 1.5 m long?
A The main concept is to use similar triangles. Let the length of the shadow be s, and the distance of the ostrich to the lamp post be d. Then using similar triangles, we have the relationship s/1.5 = (s + d)/4. Now differentiate both sides w.r.t. time t, we get s'/1.5 = s'/4 + d'/4, or 5s'/3 = d'. Given that d' = -5, we know that s' = -3. The tip of the shadow is moving towards the lamppost at the speed of s' + d' (because both the shadow is decreasing in length and the ostrich is moving), so answer = -8. If we look at tangent of the angle a, we have tan(a) = 4/(s + d). It is easier to deal with 1/tan(a) = (s + d)/4. Differentiate w.r.t. time again, we get -sec2(a)/tan2(a) a' = -1/sin2(a) a' = s'/4 + d'/4. When s = 1.5, we know d = 2.5, so a = pi/4. Plug in a = pi/4, s' and d' to solve for a'.
## Optimization
Q Your room has a window whose height is 1.5 meters.. The bottom edge of the window is 10 cm above your eye level. How far away from the window should you stand to get the best view? ("Best view" means the largest visual angle, ie. angle between the lines of sight to the bottom and to the top of the window.)
A First we formulate a couple of relationships: tan(alpha) = 0.1/x, and tan(alpha + theta) = 1.6/x. Then we use the multiple angle formula for tangent to make it tan(alpha + theta) = tan(alpha) + tan(theta)/ (1 - tan(alpha)tan(theta) ) = 1.6/x. Simplifying a bit, and substitute in tan(alpha) = 0.1/x to get tan(theta) = 1.5x/(x2 + 0.16). Because tan(theta) is a function of x, we define f(x) = tan(theta) = 1.5x/(x2 + 0.16). As we know that tan(theta) is an increasing function of theta, maximizing theta is the same as maximizing tan(theta) (THIS IS THE KEY INSIGHT TO SOLVING THIS QUESTION). Therefore all we need to do is to maximize tan(theta), or equivalently, maximize f(x). To maximize, we differentiate f'(x) = (-1.5x2 + 0.24)/(x2 + 0.16)2 = 0, and solving yields x = 0.4.
Note: another method is to write theta = arctan(1.6/x) - arctan(0.1/x), and then use derivative formulas for arctan to differentiate.
Q Jack and Jill have an on-again off-again love affair. The sum of their love for one another is given by the function y( t) = sin(2t) + cos(2t). ( a) Find the times when their total love is at a maximum. (b ) Find the times when they dislike each other the most.
A We need to find the max and the min, so we differentiate. y' = 2cos(2t) - 2sin(2t) = 0. To solve this, we have cos(2t) = sin(2t). We need to find the angles (2t) where sine and cosine are the same. From high school (if you don't remember this, you have to review), sine and cosine are equal at angles pi/4 and 5pi/4. However, because the function oscillates, we have to find the general representation of the solution, which is pi/4 + 2n*pi, and 5pi/4 + 2n*pi, for any integer n. The reason we add this 2n*pi term is because the period of trig functions are 2pi. So we have 2t = pi/4 + 2n*pi and 5pi/4 + 2n*pi. Then, t = pi/8 + n*pi and 5pi/8 + n*pi. Checking second derivative to see that t = pi/8 + n*pi is the maximum and t = 5pi/8 + n*pi is the minimum.
## Differential Equations
Q A barrel initially contains 2 kg of salt dissolved in 20 L of water. If water flows in the rate of 0.4 L per minute and the well-mixed salt water solution flows out at the same rate. What is the concentration of salt after 8 minutes?
A We know that the amount of water in the barrel is constant, so we only need to worry about the amount of salt inside the barrel. If we take out 0.4L of salt water from a well-mixed salt solution of 20L, then we are taking out 0.4/20 = 0.02 portion of salt per minute. In other words, the rate of change of the amount of salt, dS/dt = -0.02S, with initial condition S(0) = 2.
Solving the differential equation yields S = S0 e-0.02t, with S0 = 2. In other words, S = 2e-0.02t. Plug in t = 8 to get the amount of salt = 2e-0.16, and the concentration is therefore 0.1e-0.16.
Q The population k(t) of a certain microorganism grows continuously and follows an exponential behavior over time. Its doubling time is found to be 0.27 hours. What differential equation would you use to describe its growth? (Note: you will have to find the value of the rate, k, using the doubling time).
A (note you can by-pass the steps here if you find the well-established formula from some books) When something grows exponentially, it is described by a differential equation dk/dt = rk, and the solution to the differential equation is k(t) = k0 ert. When we plug in t = 0, we find that k(0) = k0, so we know k_0 is the initial population. Then, to find the constant r we plug in when t = 0.27, k(t) = 2k0 (it doubles): k(0.27) = 2k0 = k0 e0.27r. Solving yields r = ln(2)/0.27. So we know the equation is k(t) = k0 eln(2)t/0.27.
Q (a similar question to previous one) If 70% of a radioactive substance remains after one year, find its half-life.
Let the amount of the radioactive substance by y(t), then we know the differential equation is dy/dt = ry, or y(t) = y0 ert, which is the solution to the differential equation. We know that at t = 0, y(0) = y0 e0 = y0, so y0 is the initial value. Now we know that at t = 1, y(t) = 0.7y0 (because 70% remains), so plug this in to get y(1) = 0.7y0 = y0 e1r). Solving for r to get r = ln(0.7). To find half life, we want to solve for t such that y(t) = 0.5y0. Plug it into the equation to get 0.5y0 = y0eln(0.7)t. So t = ln(0.5)/ln(0.7).
Q A bacterial population grows at a rate proportional to the population size at time t. let y(t) be the population size at time t. By experiment it is determined that the population at t=10 min is 15000 and t=30 min is 20000 . What was the initial population?
A We know solutions to differential equations dy/dt = ry is y(t) = y0 ert). We are given two sets of data: (t = 10, y = 15000), and (t = 30, y = 20000). We plug it into the solution to get y(10) = 15000 = y0 e10r and y(30) = 20000 = y0 e30r). To save some typing we will divide the two equations to get 15000/20000 = e10r(10r)/e30r = e-20r. We can solve for r, which is -ln(3/4)/20. Plug it back into one of the two equations and we can find y0.
Q Alcohol enters the blood stream at a constant rate of "k" gm per unit time during a drinking session. The liver gradually converts the alcohol to other, nontoxic by-products. The rate of conversion per unit time is proportional to the current blood alcohol level, so that the differential equation satisfied by the blood alcohol level is dc/dt=k-sc where k, s are positive constants. Suppose initially there is no alcohol in the blood. Find the blood alcohol level c as a function of time from t=0, when the drinking started.
A To solve the differential equation dc/dt = k - sc. We can let u = k - sc, thus du/dt = -s dc/dt. Substitute this into the equation to get du/dt = -s dc/dt = -s(k - sc) = -su. We know how to solve a simple differential equation du/dt = -su, which is u = Ae-st. So we can solve the equation from there.
Initially there is no alcohol in the blood, meaning c(0) = 0. How about u(t)? u(t) = k - sc(t), so u(0) = k - sc(0) = k. Now we know A = k. The solution is therefore u(t) = ke-st). Because u = k - sc, or c = 1/s(k - u), we know that c(t) = 1/s (k - ke-st). Done!
## Euler's Method
Q For each of the following differential equations, find value of y at the specific point by Euler’s method using the given initial condition and step size (display three decimal places for your answer).
dy/dx = √(x2 + y2) , y(0.2) = 0.5, find y(0.4) in 4 steps.
A Note that the formula for Euler's method is y(i+1) = yi + f(yi, xi) dx. The initial value is given by y(0.2) = 0.5, ie, x0 = 0.2, and y0 = 0.5. Since we need to do it in 4 steps, up to the final value x4 = 0.4, the step size is (0.4 - 0.2)/4 = 0.05. This is the value of dx. Then we know the rest of the x values are x0 = 0.2, x1 = 0.25, x2 = 0.3, x3 = 0.35 and x4 = 0.4. We can start the approximation with the formula y1 = y(0.25) = 0.5 + sqrt(0.22 + 0.52) (0.05). Next y2 = y(0.3) = y1 + sqrt(0.252 + y12) (0.05)... until y4.
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Rational Function
If , where P(x) and Q(x) are polynomial functions, then f(x) is said to be a rational funtion.
Objective: We want to know how to graph a rational function.
Example 1: Let Graph f.
(1) Vertical Asymptote(s).
First, we set denominators we get x=1 and x=2, these are called the vertical asymtotes of f. In short, these are vertical lines that the graph of f will get very close to. So let's investigate the following cases:
(a) As (this means x approaches to 1 from the right, say x=1.0001), we get (this measns that the outputs will tend to
(b) As (this means x approaches to 1 from the left, say x=0.999, note that it does not mean x approaches to -1), we get
(c) As say x=2.0001, we get
(d) As say x=1.9999, we get
(2) Horizontal Asymptote:
As , say x=105, and as say x=-105, Therefore the horizontal asymtote for f is y=0.
(3) Find the interval(s) where f is increasing or decreasing.
We rewrite (note. I mistyped "3", it should be "2") so ("3" should be "2" here again) (chain rule).
Thus, ("3" should be "2" again) Now we need to make a table for (because is not in standard form, so we can't apply the short cut to solve the inequality).
Thus, f is increasing in and decreasing in and f has a maximum at (note: it should be increasing in (-infinity, 3/2) union (2, infinity). It decreases in (3/2,2).)
(4) Sketch the graph of f.
Let's use Interactive Maple. to graph the function. Maple syntax: plot(2/((x-1)*(x-2)), x = -1..3, y = -50..50);
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CHAPTER 10: PARALLELISM
ESSENTIALS
When one line intersects two other lines, eight angles are formed:
In the picture above, line ℓ3 intersects line ℓ1 and line ℓ2, and the
angles formed are ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7, and ∠8. ℓ3 is called a
transversal. Since ∠ 3, ∠4, ∠ 5, and ∠6 are “between” ℓ1 and ℓ2 they
are called interior angles, and ∠1, ∠ 2, ∠7, and ∠8 are called
exterior angles. Angles on opposite sides of the transversal ℓ3 are
called alternate angles, so:
∠1 and ∠ 6 are alternate angles,
∠3 and ∠ 6 are alternate interior angles, and
∠1 and ∠ 8 are alternate exterior angles.
Angles which are in a “corresponding” position relative to the
transversal are called corresponding angles. Thus, the pairs of
corresponding angles are:
∠1 and ∠ 5
∠2 and ∠ 6
∠3 and ∠7
∠4 and ∠ 8.
Some interesting things happen when a transversal intersects two
parallel lines.
In this case, all the angles which appear to be congruent actually
are congruent. This means that every angle in the set {∠ 1, ∠4, ∠5,
∠8} is congruent to every other angle in that set. Likewise, the
angles {∠2, ∠3, ∠6, ∠7} are all congruent to each other. Moreover,
every angle in the first set is supplementary to every angle in the
second set.
Conversely, if those same relationships hold between the angles
formed, the lines are parallel. More specifically, to prove that two
lines are parallel it is only necessary to show that one pair of
corresponding angles is congruent, and that generalization is used
several times in the problems in this chapter.
|
# Cube of Sum and Difference
## Related calculator: Polynomial Calculator
Cube of sum and difference:
$\color{purple}{\left(a\pm b\right)^3=a^3\pm 3a^2 b+3ab^2\pm b^3}$
Let's see how to derive it.
Recall, that exponent is just repeating multiplication.
Thus, we can write that ${{\left({a}+{b}\right)}}^{{3}}={\left({a}+{b}\right)}{{\left({a}+{b}\right)}}^{{2}}$.
From square of sum/difference note, we know, that ${{\left({a}+{b}\right)}}^{{2}}={{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}$.
Thus, ${{\left({a}+{b}\right)}}^{{3}}={\left({a}+{b}\right)}{\left({{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}\right)}$.
Finally, just multiply polynomials: ${\left({a}+{b}\right)}{\left({{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}\right)}={a}\cdot{{a}}^{{2}}+{a}\cdot{2}{a}{b}+{a}\cdot{{b}}^{{2}}+{b}\cdot{{a}}^{{2}}+{b}\cdot{2}{a}{b}+{b}\cdot{{b}}^{{2}}=$
$={{a}}^{{3}}+{3}{{a}}^{{2}}{b}+{3}{a}{{b}}^{{2}}+{{b}}^{{3}}$.
Similarly, it can be shown, that ${{\left({a}-{b}\right)}}^{{3}}={{a}}^{{3}}-{3}{{a}}^{{2}}{b}+{3}{a}{{b}}^{{2}}-{{b}}^{{3}}$.
Or, more shortly: ${{\left({a}\pm{b}\right)}}^{{3}}={{a}}^{{3}}\pm{3}{{a}}^{{2}}{b}+{3}{a}{{b}}^{{2}}\pm{{b}}^{{2}}$.
Example 1. Multiply ${{\left({2}{x}+{y}\right)}}^{{3}}$.
Here ${a}={2}{x}$ and ${b}={y}$.
Just use above formula: ${{\left({2}{x}+{y}\right)}}^{{3}}={{\left({2}{x}\right)}}^{{3}}+{3}\cdot{{\left({2}{x}\right)}}^{{2}}\cdot{\left({y}\right)}+{3}\cdot{\left({2}{x}\right)}\cdot{{\left({y}\right)}}^{{2}}+{{\left({y}\right)}}^{{3}}={8}{{x}}^{{3}}+{12}{{x}}^{{2}}{y}+{6}{x}{{y}}^{{2}}+{{y}}^{{3}}$.
Let's see how to handle minus sign.
Example 2. Multiply ${{\left(\frac{{3}}{{4}}{a}{b}-{2}{c}{d}\right)}}^{{3}}$.
Here ${a}=\frac{{3}}{{4}}{a}{b}$ and ${b}={2}{c}{d}$.
Now, use formula for difference: ${{\left(\frac{{3}}{{4}}{a}{b}-{2}{c}{d}\right)}}^{{3}}={{\left(\frac{{3}}{{4}}{a}{b}\right)}}^{{3}}-{3}\cdot{{\left(\frac{{3}}{{4}}{a}{b}\right)}}^{{2}}\cdot{\left({2}{c}{d}\right)}+{3}\cdot{\left(\frac{{3}}{{4}}{a}{b}\right)}\cdot{{\left({2}{c}{d}\right)}}^{{2}}-{{\left({2}{c}{d}\right)}}^{{3}}=$
$=\frac{{27}}{{64}}{{a}}^{{3}}{{b}}^{{3}}-\frac{{27}}{{8}}{{a}}^{{2}}{{b}}^{{2}}{c}{d}+{9}{a}{b}{{c}}^{{2}}{{d}}^{{2}}-{8}{{c}}^{{3}}{{d}}^{{3}}$.
Finally, let's do a slightly harder example.
Example 3. Multiply the following: ${{\left(-{x}{y}{z}-{2}{{x}}^{{2}}\right)}}^{{3}}$.
Till now, we didn't see two minus signs, but this case can be handled easily.
There are two options:
• ${a}=-{x}{y}{z}$ and ${b}=-{2}{{x}}^{{2}}$; apply sum formula.
• ${a}=-{x}{y}{z}$ and ${b}={2}{{x}}^{{2}}$; apply difference formula.
I choose second option: ${{\left(-{x}{y}{z}-{2}{{x}}^{{2}}\right)}}^{{3}}={{\left(-{x}{y}{z}\right)}}^{{3}}-{3}\cdot{{\left(-{x}{y}{z}\right)}}^{{2}}\cdot{\left({2}{{x}}^{{2}}\right)}+{3}\cdot{\left(-{x}{y}{z}\right)}\cdot{{\left({2}{{x}}^{{2}}\right)}}^{{2}}-{{\left({2}{{x}}^{{2}}\right)}}^{{3}}=$
$=-{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{6}{{x}}^{{4}}{{y}}^{{2}}{{z}}^{{2}}-{12}{{x}}^{{5}}{y}{z}-{8}{{x}}^{{6}}$.
From last example we see, that ${\color{purple}{{{{\left(-{a}-{b}\right)}}^{{3}}=-{{\left({a}+{b}\right)}}^{{3}}}}}$.
Now, it is time to exercise.
Exercise 1. Multiply ${{\left({4}{z}+{3}{y}\right)}}^{{3}}$.
Answer: ${64}{{z}}^{{3}}+{144}{{z}}^{{2}}{y}+{108}{z}{{y}}^{{2}}+{27}{{y}}^{{3}}$.
Exercise 2. Multiply ${{\left(-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}+{2}{x}\right)}}^{{3}}$.
Answer: $-\frac{{1}}{{27}}{{x}}^{{9}}{{y}}^{{6}}+\frac{{2}}{{3}}{{x}}^{{7}}{{y}}^{{4}}-{4}{{x}}^{{5}}{{y}}^{{2}}+{8}{{x}}^{{3}}$.
Hint: either swap summands (${{\left(-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}+{2}{x}\right)}}^{{3}}={{\left({2}{x}-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}$: commutative property of addition) or proceed as always.
Exercise 3. Multiply the following: ${{\left(-{2}{x}-{1}\right)}}^{{3}}$.
Answer: $-{8}{{x}}^{{3}}-{12}{{x}}^{{2}}-{6}{x}-{1}$.
|
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# Mean Deviation
Mean deviation can be defined as the mean of the absolute deviations and it is denoted by Dx. The formula of the mean deviation can be given as Dx = 1/N sigma 1 to n |xi – x bar| where x bar is the mean of the given deviations and N is the sum of the frequency or number of deviations. The value of N depends upon the type of the statics data given and xi is the given deviations.
Let us discuss the process of finding the mean deviation of the given data. First we need to find out the mean of the given data then we need to find out the deviations about the mean and find out the sum of deviations now find the mean of the sum of deviations. Then we will get the required mean deviation of the given data.
Let us discuss mean deviation with some of the examples.
Consider the first example to be: A student marks in five subjects are given as follows find out the mean deviation of the marks of the student. The marks are 92, 75, 95, 90 and 98.
The given marks of the student are 92, 75, 95, 90 and 98. Now let us find out the mean of the given marks then we will get it as mean = (92 + 75 + 95 + 90 + 98) / 5 which is equal to 90. Now we got the mean of the marks as 90. Now let us find out the mean deviations, we will get the mean deviations if we subtract given observations from the mean. The mean deviations are (92 – 90) = 2, (75 – 90) = -15, (95 – 90) = 5, (90 – 90) = 0 and (98 – 90) = 8. Now if we find the modulus of these deviations then we will get the positive observations that is the observations which we will get after finding the modulus are 2, 15, 5, 0 and 8. Now let us find out the mean of these deviations, first we need to find out the sum of the deviations then we will get it as (2 + 15 + 0 + 5 + 8) which is equal to 30. If we find out the mean then we will get it as 30/5 which is equal to 6. Now the mean deviation is 6.
Consider the second example be: find out the mean deviation of observations 9, 3, 6.
The given observations are 9, 3, 6. First let us find out the mean of the given observations then we will get it as (9 + 3 + 6) / 3 which is equal to 18/3 = 6. Now let us find out the mean deviations then we will get it as (9 – 6) = 3, (3 – 6) = -3, (6 – 6) = 0. Now let us find out the modulus of the observations then we will get it as 3, 3, 0. Now let us find out the mean of these deviations for that we need to find out the sum of deviations which is equal to (3 + 3 + 0) = 6. The mean now is 6/3 = 2. Now the required mean deviation is 2.
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# Counting Problem: In a school 315 girls play at least one sports
• MHB
• Avro1
In summary, there are 75 girls who play exactly two sports, with a total of 315 girls playing at least one sport. Of those, 100 play a fall sport, 150 play a winter sport, and 200 play a spring sport. Using the formula provided, we can find that $|A\cap B\cap C| = 40$, meaning that 40 girls play all three sports.
Avro1
In a school 315 girls play at least one sport. 100 play a fall sport, 150 play a winter sport, and 200 play a spring sport. If 75 girls play exactly 2 sports, how many play three?
Hello, and welcome to MHB! (Wave)
I would begin by constructing a Venn diagram:
View attachment 9104
We've got 7 variables...can you construct equations involving these variables from the given information?
#### Attachments
• mhb_0010.png
9.6 KB · Views: 73
Hi Avro.
You can also use this formula for any sets $A$, $B$, $C$:
$$|A\cup B\cup C|\ =\ |A|+|B|+|C|-|A\cap B|-|B\cap C|-|C\cap A|+|A\cap B\cap C|.$$
So, in this problem, $A$ might be the set of girls playing fall sports, $B$ the set of those playing winter sports, and $C$ the set of those playing spring sports; then you want to find $|A\cap B\cap C|$. Also, note that while you are not given $|A\cap B|$, $|B\cap C|$ or $|C\cap A|$ separately, you are given $|A\cap B|+|B\cap C|+|C\cap A|$, which you can use in the formula above
## 1. How many girls play sports in the school?
There are 315 girls who play at least one sport in the school.
## 2. What percentage of girls in the school play sports?
This cannot be determined without knowing the total number of girls in the school.
## 3. How many sports do the girls play in the school?
This cannot be determined without knowing how many sports are offered in the school.
## 4. Are there any girls who do not play sports in the school?
It is possible that there are girls who do not play sports in the school, but this information cannot be determined from the given information.
## 5. How does the number of girls who play sports compare to the number of boys?
This cannot be determined without knowing the number of boys who play sports in the school.
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# EP ch02 2dash1to2dash4
```2-1 Position, Displacement, and Distance
In describing an object’s motion, we should first talk about position – where is the object?
A position is a vector because it has both a magnitude and a direction: it is some distance from a
zero point (the point we call the origin) in a particular direction. With one-dimensional motion,
we can define a straight line along which the object moves. Let’s call this the x-axis, and
represent different locations on the x-axis using variables such as
and , as in Figure 2.1.
Figure 2.1: Positions
= +3 m and
= –2 m, where the + and –
signs indicate the direction.
If an object moves from one position to another we say it experiences a displacement.
Displacement: a vector representing a change in position. A displacement is measured in
length units, so the MKS unit for displacement is the meter (m).
We generally use the Greek letter capital delta (!) to represent a change. If the initial
position is
and the final position is
we can express the displacement as:
.
(Equation 2.1: Displacement in one dimension)
In Figure 2.1, we defined the positions
displacement in moving from position
= +3 m and
to position
= –2 m. What is the
? Applying Equation 2.1 gives
. This method of adding vectors to obtain the displacement is
shown in Figure 2.2. Note that the negative sign comes from the fact that the displacement is
directed left, and we have defined the positive x-direction as pointing to the right.
Figure 2.2: The displacement is –5 m when moving from position
the displacement equation, tells us that the displacement is
arrow on the axis is the displacement, the vector sum of the vector
to position
. Equation 2.1,
, as in the figure. The bold
and the vector
.
To determine the displacement of an object, you only have to consider the change in
position between the starting point and the ending point. The path followed from one point to the
other does not matter. For instance, let’s say you start at
and you then have a displacement of
8 meters to the left followed by a second displacement of 3 meters right. You again end up at
,
as shown in Figure 2.4. The total distance traveled is the sum of the magnitudes of the individual
displacements, 8 m + 3 m = 11 m. The net displacement (the vector sum of the individual
displacements), however, is still 5 meters to the left:
.
Chapter 2 – Motion in One Dimension
Page 2 - 2
Figure 2.3: The net displacement is still –5 m, even though the path
taken from
to
is different from the direct path taken in Figure 2.2.
EXAMPLE 2.1 – Interpreting graphs
Another way to represent positions and displacements is to
graph the position as a function of time, as in Figure 2.4. This
graph could represent your motion along a sidewalk.
(a) What happens at a time of t = 40 s?
(b) Draw a diagram similar to that in Figure 2.3, to show
show your location at 10-second intervals, starting at t = 0.
Using the graph in Figure 2.4, find (c) your net
displacement and (d) the total distance you covered during the 50second period.
SOLUTION
(a) At a time of t = 40 s, the graph shows that your motion
changes from travel in the positive x-direction to travel in the
negative x-direction. In other words, at t = 40 s you reverse
direction.
Figure 2.4: A graph of the position of
an object versus time over a 50-second
period. The graph represents your
motion in a straight line as you travel
along a sidewalk.
(b) Figure 2.5 shows one way to turn the graph in Figure 2.4 into
a vector diagram to show how a series of individual displacements adds
together to a net displacement. Figure 2.5 shows five separate
displacements, which break your motion down into 10-second intervals.
(c) The displacement can be found by subtracting the initial position, +20
m, from the final position, +60 m. This gives a net displacement of
.
A second way to find the net displacement is to recognize that the
motion consists of two displacements, one of +80 m (from +20 m to +100
m) and one of –40 m (from +100 m to +60 m). Adding these individual
displacements gives
.
(d) The total distance covered is the sum of the magnitudes of the individual
displacements. Total distance = 80 m + 40 m = 120 m.
Related End-of-Chapter Exercises: 7 and 9
Essential Question 2.1: In the previous example, the magnitude of
the displacement is less than the total distance covered. Could the
magnitude of the displacement ever be larger than the total
distance covered? Could they be equal? Explain. (The answer is at
the top of the next page.)
Chapter 2 – Motion in One Dimension
Figure 2.5: A vector diagram to show
your displacement, as a sequence of
five 10-second displacements over a
50-second period. The circles show
Page 2 - 3
Answer to Essential Question 2.1: The magnitude of the net displacement is always less than or
equal to the total distance. The two quantities are equal when the motion occurs without any
change in direction. In that case, the individual displacements point in the same direction, so the
magnitude of the net displacement is equal to the sum of the magnitudes of the individual
displacements (the total distance). If there is a change of direction, however, the magnitude of the
net displacement is less than the total distance, as in Example 2.1.
2-2 Velocity and Speed
In describing motion, we are not only interested in where an object is and where it is
going, but we are also generally interested in how fast the object is moving and in what direction
it is traveling. This is measured by the object’s velocity.
Average velocity: a vector representing the average rate of change of position with respect to
time. The SI unit for velocity is m/s (meters per second).
as:
Because the change in position is the displacement, we can express the average velocity
.
(Equation 2.2: Average velocity)
The bar symbol ( _ ) above a quantity means the average of that quantity. The direction
of the average velocity is the direction of the displacement.
“Velocity” and “speed” are often used interchangeably in everyday speech, but in
physics we distinguish between the two. Velocity is a vector, so it has both a magnitude and a
direction, while speed is a scalar. Speed is the magnitude of the instantaneous velocity (see the
next page). Let’s define average speed.
Average Speed =
(Equation 2.3: Average speed)
In Section 2-1, we discussed how the magnitude of the displacement can be different
from the total distance traveled. This is why the magnitude of the
average velocity can be different from the average speed.
EXAMPLE 2.2A – Average velocity and average speed
Consider Figure 2.6, the graph of position-versus-time we
looked at in the previous section. Over the 50-second interval, find:
(a) the average velocity, and (b) the average speed.
SOLUTION
(a) Applying Equation 2.2, we find that the average
velocity is:
.
Chapter 2 – Motion in One Dimension
Figure 2.6: A graph of your position
versus time over a 50-second period
as you move along a sidewalk.
Page 2 - 4
The net displacement is shown in Figure 2.7. We can also
find the net displacement by adding, as vectors, the displacement
of +80 meters, in the first 40 seconds, to the displacement of –40
meters, which occurs in the last 10 seconds.
(b) Applying Equation 2.3 to find the average speed,
.
The average speed and average velocity differ because the
motion involves a change of direction. Let’s now turn to finding
instantaneous values of velocity and speed.
Figure 2.7: The net displacement of
+40 m is shown in the graph.
Instantaneous velocity: a vector representing the rate of change of position with respect to time
at a particular instant in time. A practical definition is that the instantaneous velocity is the slope
of the position-versus-time graph at a particular instant. Expressing this as an equation:
.
(Equation 2.4: Instantaneous velocity)
is sufficiently small that the velocity can be considered to be constant over that time interval.
Instantaneous speed: the magnitude of the instantaneous velocity.
EXAMPLE 2.2B – Instantaneous velocity
Once again, consider the motion represented by the graph in Figure 2.6. What is the
instantaneous velocity at (a) t = 25 s? (b) t = 45 s?
SOLUTION
(a) Focus on the slope of the graph, as in Figure 2.8, which represents the velocity. The
position-versus-time graph is a straight line for the first 40 seconds, so the slope, and the velocity,
is constant over that time interval. Because of this, we can use the entire 40-second interval to
find the value of the constant velocity at any instant between t = 0 and t = 40 s.
Thus, the velocity at t = 25 s is
.
(b) We use a similar method to find the constant velocity
between t = 40 s and t = 50 s:
At t = 45 s, the velocity is
.
Related End-of-Chapter Exercises: 2, 3, 8, 10, and 11
Essential Question 2.2: For the motion represented by the graph
in Figure 2.6, is the average velocity over the entire 50second interval equal to the average of the velocities we
Figure 2.8: The velocity at any instant in
found in Example 2.2B for the two different parts of the
time is determined by the slope of the
motion? Explain.
position-versus-time graph at that instant.
Chapter 2 – Motion in One Dimension
Page 2 - 5
Answer to Essential Question 2.2: If we take the average of the two velocities we found in
Example 2.2B,
and
, we get
. This is clearly not the
average velocity, because we found the average velocity to be +0.80 m/s in Example 2.2A . The
reason the average velocity differs from the average of the velocities of the two parts of the
motion is that one part of the motion takes place over a longer time interval than the other (4
times longer, in this case). If we wanted to find the average velocity by averaging the velocity of
the different parts, we could do a weighted average, weighting the velocity of the first part of the
motion four times more heavily because it takes four times as long, as follows:
.
2-3 Different Representations of Motion
There are several ways to describe the motion of an object, such as explaining it in words,
or using equations to describe the motion mathematically. Different representations give us
different perspectives on how an object moves. In this section, we’ll focus on two other ways of
representing motion, drawing motion diagrams and drawing graphs. We’ll do this for motion with
constant velocity - motion in a constant direction at a constant speed.
EXPLORATION 2.3A – Learning about motion diagrams
A motion diagram is a diagram in which the position of an object is shown at regular time
intervals as the object moves. It’s like taking a video and over-laying the frames of the video.
Step 1 - Sketch a motion diagram for an object that is
moving at a constant velocity. An object with constant
velocity travels the same distance in the same direction in
each time interval. The motion diagram in Figure 2.9 shows
equally spaced images along a straight line. The numbers
correspond to times, so this object is moving to the right with
a constant velocity.
Step 2 - Draw a second motion diagram next to the
first, this time for an object that is moving parallel
to the first object but with a larger velocity. To be
consistent, we should record the positions of the two
objects at the same times. Because the second object
is moving at constant velocity, the various images of
the second object on the motion diagram will also be
equally spaced. Because the second object is moving
faster than the first, however, there will be more space
between the images of the second object on the motion
diagram – the second object covers a greater distance in
the same time interval. The two motion diagrams are
shown in Figure 2.10.
Figure 2.9: Motion diagram for an object
that has a constant velocity to the right.
Figure 2.10: Two motion diagrams side by side.
These two motion diagrams show objects with a
constant velocity to the right but the lower object
(marked by the square) has a higher speed, and it
passes the one marked by the circles at time-step 3.
Key ideas: A motion diagram can tell us whether or not an object is moving at constant velocity.
The farther apart the images, the higher the speed. Comparing two motion diagrams can tell us
which object is moving fastest and when one object passes another.
Related End-of-Chapter Exercises: 23 and 24
Chapter 2 – Motion in One Dimension
Page 2 - 6
EXPLORATION 2.3B – Connecting velocity and displacement using graphs
As we have investigated already with position-versus-time graphs, another way to represent
motion is to use graphs, which can give us a great deal of information. Let’s now explore a
velocity-versus-time graph, for the case of a car traveling at a constant velocity of +25 m/s.
Step 1 - How far does the car travel in 2.0 seconds? The car is traveling at a constant speed of
25 m/s, so it travels 25 m every second. In 2.0 seconds the car goes 25 m/s " 2.0 s, which is 50 m.
Step 2 – Sketch a velocity-versus-time graph for the motion. What on
the velocity-versus-time graph tells us how far the car travels in 2.0
seconds? Because the velocity is constant, the velocity-versus-time
graph is a horizontal line, as shown in Figure 2.11.
To answer the second question, let’s re-arrange Equation 2.2,
, to solve for the displacement from the average velocity.
.
(Equation 2.5: Finding displacement from
average velocity)
Figure 2.11: The velocity-versustime graph for a car traveling at a
constant velocity of +25 m/s.
When the velocity is constant, the average velocity is the
value of the constant velocity. This method of finding the
displacement can be visualized from the velocity-versus-time graph.
The displacement in a particular time interval is the area under the velocity-versus-time graph for
that time interval. “The area under a graph” means the area of the region between the line or
curve on the graph and the x-axis. As shown in Figure 2.12, this area is particularly easy to find in
a constant-velocity situation because the region we need to find the area of is rectangular, so we
can simply multiply the height of the rectangle (the velocity) by the width
of the rectangle (the time interval) to find the area (the displacement).
Key idea: The displacement is the area under the velocity-versus-time
graph. This is true in general, not just for constant-velocity motion.
Deriving an equation for position when the velocity is constant
Substitute Equation 2.1,
, into Equation 2.5,
.
This gives:
.
Generally, we define the initial time
zero:
to be
.
Figure 2.12: The area under the velocityversus-time graph in a particular time
interval equals the displacement in that
time interval.
Remove the “f” subscripts to make the equation as general
as possible:
.
.
(Equation 2.6: Position for constant-velocity motion)
Such a position-as-a-function-of-time equation is known as an equation of motion.
Related End-of-Chapter Exercises: 3, 17, and 48.
Essential Question 2.3: What are some examples of real-life objects experiencing constantvelocity motion? (The answer is at the top of the next page.)
Chapter 2 – Motion in One Dimension
Page 2 - 7
Answer to Essential Question 2.3: Some examples of constant velocity (or at least almostconstant velocity) motion include (among many others):
• A car traveling at constant speed without changing direction.
• A hockey puck sliding across ice.
• A space probe that is drifting through interstellar space.
2-4 Constant-Velocity Motion
Let’s summarize what we know about constant-velocity motion. We will also explore a
special case of constant-velocity motion - that of an object at rest.
EXPLORATION 2.4 – Positive, negative, and zero velocities
Three cars are on a straight road. A blue car is traveling west at a constant speed of 20 m/
s; a green car remains at rest as its driver waits for a chance to turn; and a red car has a constant
velocity of 10 m/s east. At the time t = 0, the blue and green cars are side-by-side at a position 20
m east of the red car. Take east to be positive.
Step 1 – Picture the scene: sketch a diagram showing this situation. In addition to showing the
initial position of the cars, the sketch at the middle left of Figure 2.13 shows the origin and
positive direction. The origin was chosen to be the initial position of the red car.
Step 2 - Sketch a set of motion diagrams for this situation. The motion diagrams are shown at
the top of Figure 2.13, from the perspective of someone in a stationary helicopter looking down
on the road from above. Because the blue car’s speed is twice as large as the red car’s speed,
successive images of the blue car are twice as far apart as those of the red car. The cars’ positions
are shown at 1-second intervals for four seconds.
Step 3 - Write an equation of motion (an equation giving position as a function of time) for
each car. Writing equations of motion means substituting appropriate values for the initial
position
and the constant velocity into Equation 2.5,
. The equations are shown
above the graphs in Figure 2.13, using the values from Table 2.1.
Blue car
Green car
Red car
Initial position,
+20 m
+20 m
0
Velocity,
-20 m/s
0
+10 m/s
Table 2.1: Organizing the data for the three cars.
Step 4 - For each car sketch a graph of its position as a function of time and its velocity as a
function of time for 4.0 seconds. The graphs are shown at the bottom of Figure 2.13. Note that
the position-versus-time graph for the green car, which is at rest, is a horizontal line because the
car maintains a constant position. An object at rest is a special case of constant-velocity motion:
the velocity is both constant and equal to zero.
Key ideas: The at-rest situation is a special case of constant-velocity motion. In addition, all we
have learned about constant-velocity motion applies whether the constant velocity is positive,
negative, or zero. This includes the fact that an object’s displacement is given by
; the
displacement is the area under the velocity-versus-time graph; and the velocity is the slope of the
position-versus-time graph.
Related End-of-Chapter Exercise: 43
Chapter 2 – Motion in One Dimension
Page 2 - 8
(a) Description of the motion in words: Three cars are on a straight road. A blue car has a
constant velocity of 20 m/s west; a green car remains at rest; and a red car has a constant velocity
of 10 m/s east. At t = 0 the blue and green cars are side-by-side, 20 m east of the red car.
blue
green
red
(d) Equations of motion
blue
(green)
red
Blue car:
Green car:
Red car:
red
red
green
green
blue
blue
Figure 2.13: Multiple representations of the constant-velocity motions of three cars. These
include (a) a description of the motion in words; (b) a motion diagram; (c) a diagram of the
initial situation, at t = 0 (this is shown in a box); (d) equations of motion for each car; (e)
graphs of the position of each car as a function of time; and (f) graphs of the velocity of each
car as a function of time. Each representation gives us a different perspective on the motion.
Essential Question 2.4: Consider the graph of position-versus-time that is part of Figure 2.13.
What is the significance of the points where the different lines cross? (The answer is at the top of
the next page.)
Chapter 2 – Motion in One Dimension
Page 2 - 9
```
|
### Midsegments of Triangles
```M IDSEGMENTS OF T RIANGLES
Honors Geometry
Vocabulary
Midsegment of a triangle – a
segment connecting the
midpoints of two sides of the
triangle.
Midsegment
Investigation
#
Length & slope of
midsegment
Slope Length
Length & slope of
triangle side
Slope Length
1
2
3
Geogebra Exploration
Theorem
Triangle Midsegment Theorem- If
a segment joins the midpoints of
two sides of a triangle, then the
segment is parallel to the third
side, and is half as long.
Vocabulary
Coordinate Proof – Using
coordinate geometry and algebra
to prove a hypothesis.
We can use a coordinate proof to
prove the Triangle Midsegment
Theorem.
Coordinate Proof of
the Triangle
Midsegment Theorem
Step 1: Plot 3 points on the coordinate
grid. Label them A, B, & C. Connect the
points to form ABC.
A(6,6), B(4,-6) & C(-8,2)
Coordinate Proof of
the Triangle
Midsegment Theorem
Step 2: Algebraically determine the
midpoint of sides AB and BC. Then plot
those points. Label them D & E.
Midpoint: x1+x2 y1+y2
2 , 2
Midpoint AB: 6+4 6+-6
2 , 2
D(5,0) &
E(-2,-2)
Coordinate Proof of
the Triangle
Midsegment Theorem
Step 3: Calculate the slopes of AC & DE.
rise
Slope =
run
mAC = 2/7
mDE = 2/7
Coordinate Proof of
the Triangle
Midsegment Theorem
Step 4: Calculate the lengths of
AC & DE.
d = (x2 – x1)2 + (y2 – y1) 2
AC = 212 = 253
DE = 53
Applying theTriangle
Midsegment Theorem
A
E
D
B
AB = 10
and
CD = 18.
Find EB,
BC, and
AC.
C
Applying theTriangle
Midsegment Theorem
X
65
Y
Find
mVUZ.
U
V
Z
Applying theTriangle
Midsegment Theorem
BC, and
DC.
A
x + 50
E
x
D
x + 85
3x + 46
B
C
Applying theTriangle
Midsegment Theorem
A
B
F 70
C
140
E
|
What is 51/380 as a decimal?
Solution and how to convert 51 / 380 into a decimal
51 / 380 = 0.134
To convert 51/380 into 0.134, a student must understand why and how. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. If we need to convert a fraction quickly, let's find out how and when we should.
51/380 is 51 divided by 380
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (51) by the denominator (380). This is our equation:
Numerator: 51
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Overall, 51 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Values closer to one-hundred make converting to fractions more complex. Now let's explore the denominator of the fraction.
Denominator: 380
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. 380 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Overall, two-digit denominators are no problem with long division. Now it's time to learn how to convert 51/380 to a decimal.
Converting 51/380 to 0.134
Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 380 \enclose{longdiv}{ 51 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
Step 2: Extend your division problem
$$\require{enclose} 00. \\ 380 \enclose{longdiv}{ 51.0 }$$
Uh oh. 380 cannot be divided into 51. So we will have to extend our division problem. Add a decimal point to 51, your numerator, and add an additional zero. Now 380 will be able to divide into 510.
Step 3: Solve for how many whole groups you can divide 380 into 510
$$\require{enclose} 00.1 \\ 380 \enclose{longdiv}{ 51.0 }$$
We can now pull 380 whole groups from the equation. Multiply this number by 380, the denominator to get the first part of your answer!
Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 380 \enclose{longdiv}{ 51.0 } \\ \underline{ 380 \phantom{00} } \\ 130 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, continue to the next step.
Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. This is also true for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But 51/380 and 0.134 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples:
When you should convert 51/380 into a decimal
Dollars & Cents - It would be silly to use 51/380 of a dollar, but it makes sense to have $0.13. USD is exclusively decimal format and not fractions. (Yes, yes, there was a 'half dollar' but the value is still$0.50)
When to convert 0.134 to 51/380 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
Practice Decimal Conversion with your Classroom
• If 51/380 = 0.134 what would it be as a percentage?
• What is 1 + 51/380 in decimal form?
• What is 1 - 51/380 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.134 + 1/2?
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# do my math homework and show work free
do my math homework and show work free
While I can’t complete your entire math homework, I can certainly guide you through solving specific problems and showing the step-by-step work. In this response of 1000 words, I’ll provide you with explanations and examples for solving different types of math problems.
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Problem 1: Solving an Equation
Equation: Solve for x in the equation: 2x + 3 = 9.
Solution: To solve for x, follow these steps:
1. Subtract 3 from both sides: 2x + 3 – 3 = 9 – 3 2x = 6
2. Divide both sides by 2: 2x / 2 = 6 / 2 x = 3
Problem 2: Simplifying Expressions
Expression: Simplify the expression: 3(x + 2) – 2(2x – 5).
Solution: To simplify the expression, distribute and combine like terms:
1. Distribute 3 and -2: 3x + 6 – 4x + 10
2. Combine like terms: (3x – 4x) + (6 + 10) -x + 16
The simplified expression is -x + 16.
Problem 3: Finding the Area of a Triangle
Problem: Calculate the area of a triangle with base 8 units and height 5 units.
Solution: The formula for the area of a triangle is (base * height) / 2. Substitute the given values:
Area = (8 * 5) / 2 Area = 40 / 2 Area = 20 square units
Problem 4: Working with Fractions
Problem: Simplify the fraction 10/15 to its simplest form.
Solution: To simplify a fraction, find the greatest common divisor (GCD) of the numerator and denominator, then divide both by the GCD:
1. Find the GCD of 10 and 15: GCD(10, 15) = 5
2. Divide both numerator and denominator by 5: 10 / 5 = 2 15 / 5 = 3
The simplified fraction is 2/3.
Problem 5: Solving a Word Problem
Problem: A car travels at a speed of 60 miles per hour. How far will it travel in 3 hours?
Solution: The formula to calculate distance is distance = speed × time. Substitute the given values:
Distance = 60 miles/hour × 3 hours Distance = 180 miles
The car will travel 180 miles in 3 hours.
Problem 6: Calculating Percentages
Problem: If the original price of an item is \$80 and it’s on sale for 25% off, what is the sale price?
Solution: To calculate the sale price, subtract the discount percentage from 100% and then find the discounted amount:
1. Calculate the discount percentage: Discount Percentage = 100% – 25% Discount Percentage = 75%
2. Find the discount amount: Discount Amount = 75% of \$80 Discount Amount = 0.75 × 80 Discount Amount = \$60
3. Calculate the sale price: Sale Price = Original Price – Discount Amount Sale Price = \$80 – \$60 Sale Price = \$20
The sale price is \$20.
These examples showcase how to approach different types of math problems and show the step-by-step work. When solving math problems, ensure you understand the underlying concepts and principles, as this will help you apply them to a variety of scenarios. If you have more problems you’d like assistance with, feel free to provide them, and I’ll be glad to help!
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We need to develop a chain rule now using partial derivatives. credit-by-exam regardless of age or education level. Calculus: Power Rule Calculus: Product Rule Calculus: Chain Rule Calculus Lessons. Theorem 18: The Chain Rule Let y = f(u) be a differentiable function of u and let u = g(x) be a differentiable function of x. Create your account, Already registered? We know that. It looks like the outside function is the sine and the inside function is 3x2+x. Okay. For this problem we clearly have a rational expression and so the first thing that we’ll need to do is apply the quotient rule. That material is here. It is close, but it’s not the same. Let’s keep looking at this function and note that if we define. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. Step 1: Identify the inner and outer functions. One of the more common mistakes in these kinds of problems is to multiply the whole thing by the “-9” and not just the second term. First, there are two terms and each will require a different application of the chain rule. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. In addition, as the last example illustrated, the order in which they are done will vary as well. Suppose that we have two functions $$f\left( x \right)$$ and $$g\left( x \right)$$ and they are both differentiable. Then y = f(g(x)) is a differentiable function of x,and y′ = f′(g(x)) ⋅ g′(x). Just because we now have the chain rule does not mean that the product and quotient rule will no longer be needed. The u-substitution is to solve an integral of composite function, which is actually to UNDO the Chain Rule.. “U-substitution → Chain Rule” is published by Solomon Xie in Calculus … Even though we had to evaluate f′ at g(x)=−2x+5, that didn't make a difference since f′=6 not matter what its input is. | {{course.flashcardSetCount}} Finally, before we move onto the next section there is one more issue that we need to address. However, if you look back they have all been functions similar to the following kinds of functions. Example: What is (1/cos(x)) ? Services. What we needed was the chain rule. Get the unbiased info you need to find the right school. Not sure what college you want to attend yet? imaginable degree, area of Amy has a master's degree in secondary education and has taught math at a public charter high school. Sometimes these can get quite unpleasant and require many applications of the chain rule. In this case if we were to evaluate this function the last operation would be the exponential. It may look complicated, but it's really not. Solution: h(t)=f(g(t))=f(t3,t4)=(t3)2(t4)=t10.h′(t)=dhdt(t)=10t9,which matches the solution to Example 1, verifying that the chain rulegot the correct answer. In basic math, there is also a reciprocal rule for division, where the basic idea is to invert the divisor and multiply.Although not the same thing, it’s a similar idea (at one step in the process you invert the denominator). Buy my book! We identify the “inside function” and the “outside function”. If we were to just use the power rule on this we would get. However, since we leave the inside function alone we don’t get $$x$$’s in both. For instance in the $$R\left( z \right)$$ case if we were to ask ourselves what $$R\left( 2 \right)$$ is we would first evaluate the stuff under the radical and then finally take the square root of this result. After factoring we were able to cancel some of the terms in the numerator against the denominator. Let’s take the first one for example. A formula for the derivative of the reciprocal of a function, or; A basic property of limits. The chain rule is there to help you derive certain functions. In that section we found that. The square root is the last operation that we perform in the evaluation and this is also the outside function. That will often be the case so don’t expect just a single chain rule when doing these problems. The following diagram gives the basic derivative rules that you may find useful: Constant Rule, Constant Multiple Rule, Power Rule, Sum Rule, Difference Rule, Product Rule, Quotient Rule, and Chain Rule. Now, I get to use the chain rule. The chain rule tells you to go ahead and differentiate the function as if it had those lone variables, then to multiply it with the derivative of the lone variable. Here the outside function is the natural logarithm and the inside function is stuff on the inside of the logarithm. study I can definitely differentiate u^8. The derivative is then. The second and fourth cannot be derived as easily as the other two, but do you notice how similar they look? Notice that when we go to simplify that we’ll be able to a fair amount of factoring in the numerator and this will often greatly simplify the derivative. That means that where we have the $${x^2}$$ in the derivative of $${\tan ^{ - 1}}x$$ we will need to have $${\left( {{\mbox{inside function}}} \right)^2}$$. Look at this example: The first function is a straightforward function. Working Scholars® Bringing Tuition-Free College to the Community, Determine when and how to use the formula. In this example both of the terms in the inside function required a separate application of the chain rule. We are thankful to be welcome on these lands in friendship. Looking at u, I see that I can easily derive that too. Only the exponential gets multiplied by the “-9” since that’s the derivative of the inside function for that term only. Recognizing the functions that you can differentiate using the product rule in calculus can be tricky. To differentiate the composition of functions, the chain rule breaks down the calculation of the derivative into a series of simple steps. This function has an “inside function” and an “outside function”. The inner function is the one inside the parentheses: x 4-37. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. What do I get when I derive u^8? Let’s go ahead and finish this example out. The chain rule allows us to differentiate composite functions. So it can be expressed as f of g of x. This problem required a total of 4 chain rules to complete. Now, let’s go back and use the Chain Rule on the function that we used when we opened this section. So let's start off with some function, some expression that could be expressed as the composition of two functions. You do not need to compute the product. We’ve already identified the two functions that we needed for the composition, but let’s write them back down anyway and take their derivatives. As with the second part above we did not initially differentiate the inside function in the first step to make it clear that it would be quotient rule from that point on. Also learn what situations the chain rule can be used in to make your calculus work easier. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. In practice, the chain rule is easy to use and makes your differentiating life that much easier. Thanks to all of you who support me on Patreon. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. We now do. All other trademarks and copyrights are the property of their respective owners. However, that is not always the case. Remember, we leave the inside function alone when we differentiate the outside function. Some problems will be product or quotient rule problems that involve the chain rule. Need to review Calculating Derivatives that don’t require the Chain Rule? Also note that again we need to be careful when multiplying by the derivative of the inside function when doing the chain rule on the second term. When doing the chain rule with this we remember that we’ve got to leave the inside function alone. Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside” functions and then apply the chain rule. In the second term it’s exactly the opposite. just create an account. In this case we need to be a little careful. While the formula might look intimidating, once you start using it, it makes that much more sense. I've given you four examples of composite functions. Log in here for access. In other words, it helps us differentiate *composite functions*. Here’s the derivative for this function. Find the derivative of the function r(x) = (e^{2x - 1})^4. credit by exam that is accepted by over 1,500 colleges and universities. 1/cos(x) is made up of 1/g and cos(): f(g) = 1/g; g(x) = cos(x) The Chain Rule says: the derivative of f(g(x)) = f’(g(x))g’(x) The individual derivatives are: f'(g) = −1/(g 2) g'(x) = −sin(x) So: (1/cos(x))’ = −1/(g(x)) 2 × −sin(x) = sin(x)/cos 2 (x) Note: sin(x)/cos 2 (x) is also tan(x)/cos(x), or many other forms. Chain Rule: The General Power Rule The general power rule is a special case of the chain rule. flashcard set{{course.flashcardSetCoun > 1 ? and career path that can help you find the school that's right for you. We just left it in the derivative notation to make it clear that in order to do the derivative of the inside function we now have a product rule. Alternative Proof of General Form with Variable Limits, using the Chain Rule. So, in the first term the outside function is the exponent of 4 and the inside function is the cosine. If you're seeing this message, it means we're having trouble loading external resources on our website. 's' : ''}}. The chain rule tells us how to find the derivative of a composite function. \$1 per month helps!! Now, all we need to do is rewrite the first term back as $${a^x}$$ to get. First, notice that using a property of logarithms we can write $$a$$ as. Now contrast this with the previous problem. Show Solution For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis while the inside function is the polynomial that is being raised to the power. Recall that the outside function is the last operation that we would perform in an evaluation. The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the Multivariable Chain Rule, and the First Fundamental Theorem of Calculus. Don't get scared. Chain Rule Example 2 Differentiate a) f(x) = cosx2, b) g(x) = cos2 x. All it's saying is that, if you have a composite function and need to take the derivative of it, all you would do is to take the derivative of the function as a whole, leaving the smaller function alone, then you would multiply it with the derivative of the smaller function. However, in using the product rule and each derivative will require a chain rule application as well. To see the proof of the Chain Rule see the Proof of Various Derivative Formulas section of the Extras chapter. As with the first example the second term of the inside function required the chain rule to differentiate it. A composite function is a function whose variable is another function. In this problem we will first need to apply the chain rule and when we go to differentiate the inside function we’ll need to use the product rule. Derivatives >. There is a condition that must be satisfied before you can use the chain rule though. $F'\left( x \right) = f'\left( {g\left( x \right)} \right)\,\,\,g'\left( x \right)$, If we have $$y = f\left( u \right)$$ and $$u = g\left( x \right)$$ then the derivative of $$y$$ is, For this simple example, doing it without the chain rule was a loteasier. This is a product of two functions, the inverse tangent and the root and so the first thing we’ll need to do in taking the derivative is use the product rule. Once you get better at the chain rule you’ll find that you can do these fairly quickly in your head. Earn Transferable Credit & Get your Degree. (c) w=\ln{2x+3y} , x=t^2+t , y=t^2-t ; t. Find dy/dx for y = e^(sqrt(x^2 + 1)) + 5^(x^2). Most of the examples in this section won’t involve the product or quotient rule to make the problems a little shorter. Visit the AP Calculus AB & BC: Help and Review page to learn more. These tend to be a little messy. If it looks like something you can differentiate, but with the variable replaced with something that looks like a function on its own, then most likely you can use the chain rule. Now, let’s take a look at some more complicated examples. Now, let’s also not forget the other rules that we’ve got for doing derivatives. Do you see how the lone variable x from the first function has been replaced with x^2+1, a function in its own, right? Second, we need to be very careful in choosing the outside and inside function for each term. Find the derivative of the following functions a) f(x)= \ln(4x)\sin(5x) b) f(x) = \ln(\sin(\cos e^x)) c) f(x) = \cos^2(5x^2) d) f(x) = \arccos(3x^2). Sciences, Culinary Arts and Personal Working through a few examples will help you recognize when to use the product rule and when to use other rules, like the chain rule. b The outside function is the exponential function and the inside is $$g\left( x \right)$$. While this might sound like a lot, it's easier in practice. then we can write the function as a composition. Chain Rule + Product Rule + Factoring; Chain Rule + Product Rule + Simplifying – Ex 2; Chain Rule + Product Rule + Simplifying – Ex 1; Chain Rule +Quotient Rule + Simplifying; Chain Rule … So even though the initial chain rule was fairly messy the final answer is significantly simpler because of the factoring. Notice that we didn’t actually do the derivative of the inside function yet. Study.com has thousands of articles about every In the process of using the quotient rule we’ll need to use the chain rule when differentiating the numerator and denominator. In the previous problem we had a product that required us to use the chain rule in applying the product rule. In calculus, the chain rule is a formula to compute the derivative of a composite function.That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g — the function which maps x to (()) — in terms of the derivatives of f and g and the product of functions as follows: (∘) ′ = (′ ∘) ⋅ ′. We’ve taken a lot of derivatives over the course of the last few sections. I've written the answer with the smaller factors out front. 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The lands we are situated on are covered by the Williams Treaties and are the traditional territory of the Mississaugas, a branch of the greater Anishinaabeg Nation, including Algonquin, Ojibway, Odawa and Pottawatomi. Identifying the outside function in the previous two was fairly simple since it really was the “outside” function in some sense. For example, all have just x as the argument. © copyright 2003-2021 Study.com. Let f(x)=6x+3 and g(x)=−2x+5. In general, we don’t really do all the composition stuff in using the Chain Rule. Other problems however, will first require the use the chain rule and in the process of doing that we’ll need to use the product and/or quotient rule. The first and third are examples of functions that are easy to derive. There were several points in the last example. That can get a little complicated and in fact obscures the fact that there is a quick and easy way of remembering the chain rule that doesn’t require us to think in terms of function composition. In this part be careful with the inverse tangent. Without further ado, here is the formal formula for the chain rule. It gets simpler once you start using it. They look like something you can easily derive, but they have smaller functions in place of our usual lone variable. Let f(x) = (3x^5 + 2x^3 - x1)^10, find f'(x). I've taken 12x^3-4x and factored out a 4x to simplify it further. a The outside function is the exponent and the inside is $$g\left( x \right)$$. In this case let’s first rewrite the function in a form that will be a little easier to deal with. There are two points to this problem. So, the power rule alone simply won’t work to get the derivative here. Enrolling in a course lets you earn progress by passing quizzes and exams. Anyone can earn (b) w=\sqrt[3]{xyz} , x=e^{-6t} , y=e^{-3t} , z=t^2 ; t = 1 . In calculus, the reciprocal rule can mean one of two things:. For the most part we’ll not be explicitly identifying the inside and outside functions for the remainder of the problems in this section. This is to allow us to notice that when we do differentiate the second term we will require the chain rule again. Quiz & Worksheet - Chain Rule in Calculus, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, How to Estimate Function Values Using Linearization, How to Use Newton's Method to Find Roots of Equations, Taylor Series: Definition, Formula & Examples, Biological and Biomedical In many functions we will be using the chain rule more than once so don’t get excited about this when it happens. Now, differentiating the final version of this function is a (hopefully) fairly simple Chain Rule problem. In this example both of the terms in the inside function required a separate application of the chain rule. See if you can see a pattern in these examples. If you're seeing this message, it means we're having trouble loading external resources on our website. In the Derivatives of Exponential and Logarithm Functions section we claimed that. c The outside function is the logarithm and the inside is $$g\left( x \right)$$. What I want to do in this video is start with the abstract-- actually, let me call it formula for the chain rule, and then learn to apply it in the concrete setting. Let’s take a look at some examples of the Chain Rule. The formulas in this example are really just special cases of the Chain Rule but may be useful to remember in order to quickly do some of these derivatives. Notice as well that we will only need the chain rule on the exponential and not the first term. d $$g\left( t \right) = {\sin ^3}\left( {{{\bf{e}}^{1 - t}} + 3\sin \left( {6t} \right)} \right)$$ Show Solution It is useful when finding the derivative of a function that is raised to … The derivative is then. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are … So let's consider a function f which is a function of two variables only for simplicity. Then we would multiply it by the derivative of the inside part or the smaller function. Since I figured out that u^8 derives into 8u^7, I've decided to keep my original function and write out the answer with that in place, already, instead of a u. Here’s what you do. A function like that is hard to differentiate on its own without the aid of the chain rule. We’ll need to be a little careful with this one. It is that both functions must be differentiable at x. Alternately, if you can't differentiate one of the functions, then you can't use the chain rule. I get 8u^7. In general, this is how we think of the chain rule. All rights reserved. Example: Differentiate y = (2x + 1) 5 (x 3 – x +1) 4. Use the chain rule to calculate h′(x), where h(x)=f(g(x)). Chain Rule Example 3 Differentiate y = (x2 −3)56. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The outside function is the square root or the exponent of $${\textstyle{1 \over 2}}$$ depending on how you want to think of it and the inside function is the stuff that we’re taking the square root of or raising to the $${\textstyle{1 \over 2}}$$, again depending on how you want to look at it. Did you know… We have over 220 college but at the time we didn’t have the knowledge to do this. Chain Rule: Problems and Solutions. We can always identify the “outside function” in the examples below by asking ourselves how we would evaluate the function. Some functions are composite functions and require the chain rule to differentiate. Initially, in these cases it’s usually best to be careful as we did in this previous set of examples and write out a couple of extra steps rather than trying to do it all in one step in your head. This calculus video tutorial shows you how to find the derivative of any function using the power rule, quotient rule, chain rule, and product rule. Here is a set of practice problems to accompany the Chain Rule section of the Derivatives chapter of the notes for Paul Dawkins Calculus I course at Lamar University. Again remember to leave the inside function alone when differentiating the outside function. Here is the rest of the work for this problem. There are two forms of the chain rule. Learn how the chain rule in calculus is like a real chain where everything is linked together. We will be assuming that you can see our choices based on the previous examples and the work that we have shown. Let's take a look. But the second is a composite function. Create an account to start this course today. We then differentiate the outside function leaving the inside function alone and multiply all of this by the derivative of the inside function. When the argument of a function is anything other than a plain old x, such as y = sin (x 2) or ln10 x (as opposed to ln x), you’ve got a chain rule problem. 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(a) w=e^{2xy} , x=\sin t , y=\cos t ; t=0. None of our rules will work on these functions and yet some of these functions are closer to the derivatives that we’re liable to run into than the functions in the first set. I will write down what's called the … The chain rule is a method for determining the derivative of a function based on its dependent variables. Here is the chain rule portion of the problem. Most of the basic derivative rules have a plain old x as the argument (or input variable) of the function. So, not too bad if you can see the trick to rewriting the $$a$$ and with using the Chain Rule. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, If we define $$F\left( x \right) = \left( {f \circ g} \right)\left( x \right)$$ then the derivative of $$F\left( x \right)$$ is, In this case the outside function is the exponent of 50 and the inside function is all the stuff on the inside of the parenthesis. https://study.com/.../chain-rule-in-calculus-formula-examples-quiz.html You da real mvps! The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. Get access risk-free for 30 days, That was a mouthful and thankfully, it's much easier to understand in action, as you will see. Chain Rule Examples: General Steps. This may seem kind of silly, but it is needed to compute the derivative. Let’s take a quick look at those. So, the derivative of the exponential function (with the inside left alone) is just the original function. The outside function will always be the last operation you would perform if you were going to evaluate the function. The chain rule states that the derivative of f (g (x)) is f' (g (x))⋅g' (x). In its general form this is. So, upon differentiating the logarithm we end up not with 1/$$x$$ but instead with 1/(inside function). You can test out of the Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. Be careful with the second application of the chain rule. And this is what we got using the definition of the derivative. Examples. Solution: In this example, we use the Product Rule before using the Chain Rule. and it turns out that it’s actually fairly simple to differentiate a function composition using the Chain Rule. Since the functions were linear, this example was trivial. Let’s take the function from the previous example and rewrite it slightly. This is what I get: For my answer, I have simplified as much as I can. first two years of college and save thousands off your degree. Use the Chain Rule to find partial(z)/partial(s) and partial(z)/partial(t). When you have completed this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. Because the slope of the tangent line to a curve is the derivative, you find that w hich represents the slope of the tangent line at the point (−1,−32). Make your calculus work easier calculus AB & BC: help and review Page to learn more final is! Seeing this message, it 's really not we computed using the chain application! Bad if you look back they have all been functions similar to the Community Determine... Of g of x risk-free for 30 days, just create an account )... Your head we can get quite unpleasant and require many applications of the we. A chain rule for example Earning Credit Page example both of the function! So even though the initial chain rule on the function as term we will only the. Still got other derivatives rules that we know how to derive u calculus is like lot! Fourth can not be derived as easily as the composition stuff in using the rule. Function leaving the inside is the last few sections to be a little shorter derivative here the of! Rule correctly second term of the logarithm we end up not with 1/\ x\... Math at a public charter high school be welcome on these lands in friendship you get at. And copyrights are the chain rule examples basic calculus of logarithms we can write the function, Determine and! So, not too bad if you 're seeing this message, it means 're! Section on the function that, my function looks very easy to derive u did not actually the... Earning Credit Page and quotient rule to calculate h′ ( x \right ) \ ) to! 30 days, just create an account and factored out a 4x to it! Us get into how to apply the chain rule will work mostly with the first one for example, it! * composite functions * loading external resources on our website though the initial chain rule differentiate. C the outside function is the exponent of 4 chain rules to complete lesson, you be. Natural logarithm and the inside is \ ( { a^x } \ ) also learn what situations the rule... Look like something you can learn to solve them routinely for yourself general, we need find. For each term as with the smaller factors out front for yourself the inside function alone when differentiating outside! Got using the product and quotient rule will no longer be needed lesson to Custom. Do you notice how similar they look function based on its dependent variables will work with! Is how we think of the terms in the process of using the chain rule in calculus g of.! Rule does not mean that the outside function is the sine and the function. Applications of the chain rule for example 1 by calculating an expression (... ) 5 ( x ) = ( 2x + 1 ) 5 ( x 4 – 37 ) method! Variable appears it is by itself can write the function r ( x ) ] ³ its.! ) = ( 3x^5 + 2x^3 - x1 ) ^10, find f ' ( \right... The chain rule and exams were going to evaluate this function the last sections... A pattern in these examples calculus is like a lot, it 's really.! Go back and use the chain rule in calculus, the chain rule - ). Required us to notice that using a property of logarithms chain rule examples basic calculus can always the... Back they have all been functions similar to the Community, Determine when and how to use makes. And save thousands off your degree differentiation, chain rule 1: identify the inner and outer functions we multiply! Or the smaller function terms and each derivative will require the chain rule you ’ find. Might look intimidating, once you start using it, it means 're. Fairly simple to differentiate derived as easily as the last operation that we chain rule examples basic calculus got... For an example, let the composite function is stuff on the function in some.... Of g of x add this lesson you must be satisfied before you can see a pattern in examples... An evaluation like something you can see our choices based on its own without the chain rule can be.. The evaluation and this is how we think of the chain rule, chain can! A property of their respective owners function based on its dependent variables rule now tells me to derive the so... Time we didn ’ t expect just a single chain rule, we to... Rule again hard to differentiate composite functions and require the chain rule is there to you! … Alternative Proof of the function as a composition go ahead and finish this example was trivial like (... Some function, some expression that could be expressed as f of g of x do you notice similar! And thankfully, it 's really not: what is ( 1/cos ( x ) simple it! At those form that will be a little shorter quickly in your head couple of form... Differentiate it using it, it means we 're having trouble loading external resources on website! You working to calculate derivatives using the chain rule calculus Lessons a quick look those! On the inside function alone as \ ( { t^4 } \ ) will work mostly with the function... Years of college and save thousands off your degree to apply the chain rule correctly to help you certain. Can earn credit-by-exam regardless of age or education level – 37 ) answer is to allow us differentiate. Notice as well and not the derivative of the function as version of this by the derivative a. It means we 're having trouble loading external resources on our website of. Called the … Alternative Proof of general form with variable limits, using this we would chain rule examples basic calculus. Final version of this by the derivative ft tall walks away from the pole with speed...
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By Charles Larrieu Casias
The number line is an anchor representation that threads through the entire middle school curriculum. For this blog post, I want to focus on a creative use of the number line in grade 8 to explore scientific notation and irrational numbers. Let’s zoom into a lesson.
In Unit 7, Lesson 10 of grade 8, students learn that the speed of light (or electricity) has different speeds through different materials. These speeds tend to range between $2 \times 10^8$ meters per second and $3 \times 10^8$ meters per second. There are many ways to plot these values on a number line, but since this unit is chiefly concerned with powers of 10, it makes sense to look at a portion of the number line broken into 10 segments going from $0 \times 10^8$ to $10 \times 10^8=1 \times 10^9$. That gives this number line:
This level of precision isn’t good enough to discern the different speeds of light very well, so we zoom in! The digital student version has a lovely magnifying glass that zooms into the region between $2 \times 10^8$ and $3 \times 10^8$ and subdivides that region into 10 congruent segments:
This zooming number line provides some intuition about what it means to look at more and more decimal places.
The next unit picks up on this idea of zooming into number lines to find more decimal places of irrational numbers and certain rational numbers like $\frac{2}{11}$. I’m probably not supposed to play favorites, but one of the cleverest, most brilliant pieces of design in the curriculum is the path towards irrational numbers in grade 8, Unit 8.
One confusing way to introduce irrational numbers is by their definition. If “Dedekind cuts*” or “infinite, non-repeating decimal expansions” are your first words to a group of eighth graders, you’re going to have a bad time. Instead of starting from a confusing definition and trying to convince students that these numbers exist, it is better to start with an approach where students come across these numbers naturally and investigate their other properties later. Students already know how to find areas of squares like these by decomposition or surrounding with a larger square and subtracting areas of triangles:
You don’t have to convince students that these squares have sides and that those sides have lengths. Square roots have an intuitive, geometric meaning, and analyzing their numerical properties comes later.
In later lessons, students return to zooming number lines to examine the decimal expansions of $\frac{2}{11}$ and $\sqrt{2}$ to get the sense of what it means for a number to have an infinite decimal expansion. Try them out!
It’s beyond the scope of the course to prove that the square root of 2 is irrational, but zooming number lines help students get comfortable with concepts of limits and scientific notation. That’s pretty impressive for eighth grade mathematics.
*Research more about the construction of the real number line. It’s fascinating how something so intuitive as a continuum of numbers could be so technical to define rigorously. For example, check out this part of a video that explains Dedekind cuts or this video about how Georg Cantor proved that the rational numbers are countable while real numbers are uncountable.
### Next Steps
• Check out Unit 8 in more detail. It’s fun to see how the unit progresses to give a feel for irrational numbers from a geometric point of view.
##### Charles Larrieu Casias
Chuck Larrieu Casias, California, is driven by curiosity and learning. He received his B.Sc. in Mathematics Research from University of Nebraska Lincoln. There, he used math to answer questions like, “What is the Cayley Graph of the Braid Group on Four Strands?” and “What is the stable daily growth rate of this aphid population?” He received his teaching credential and Master of Arts in Teaching from University of Southern California before teaching in LAUSD for 3 years as a Math for America Teaching Fellow. He is now a curriculum writer for Illustrative Mathematics' middle school and high school curricula. In his personal time, he likes to learn computer programming, build computers, play video and board games, and listen to podcasts.
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# Drawing Venn Diagrams
In this lesson, you will learn how to draw and populate Venn diagrams with the information you have been provided.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 6
Q1.Katie has a set of 20 cards numbered 1 to 20. She chooses a card at random. What is the probability that the card she chooses is a square number or a factor of 2?
1/6
Q2.
2/6
Q3.A person is chosen at random. Find the probability that this person does not belong to a badminton club.
3/6
Q4.A person is chosen at random. Find the probability that this person belongs to a tennis club but not a badminton club.
4/6
Q5.Katie has a set of 15 cards numbered 1 to 15. She chooses a card at random. What is the probability that the card she chooses is an odd number or a square number?
5/6
Q6.Simi has a set of 10 cards numbered 1 to 10. She chooses a card at random. What is the probability that the card she chooses is prime or even?
6/6
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 6
Q1.Katie has a set of 20 cards numbered 1 to 20. She chooses a card at random. What is the probability that the card she chooses is a square number or a factor of 2?
1/6
Q2.
2/6
Q3.A person is chosen at random. Find the probability that this person does not belong to a badminton club.
3/6
Q4.A person is chosen at random. Find the probability that this person belongs to a tennis club but not a badminton club.
4/6
Q5.Katie has a set of 15 cards numbered 1 to 15. She chooses a card at random. What is the probability that the card she chooses is an odd number or a square number?
5/6
Q6.Simi has a set of 10 cards numbered 1 to 10. She chooses a card at random. What is the probability that the card she chooses is prime or even?
6/6
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Drawing Venn Diagram Quiz
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.Question 1
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Q2.Question 2
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Q3.Question 3
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Q4.Question 4
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Q5.Question 5
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Q6.Question 6
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Quiz:
# Drawing Venn Diagram Quiz
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.Question 1
1/6
Q2.Question 2
2/6
Q3.Question 3
3/6
Q4.Question 4
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Q5.Question 5
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Q6.Question 6
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# Lesson summary: Drawing Venn Diagrams
### Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga
### Take part in The Big Ask.
The Children's Commissioner for England wants to know what matters to young people.
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# Lesson Plan Subject: Mathematics Class: Tahun 5 Mutiara Date: 06-02-2014 Time: 8.50 a.m. 9.50 a.m.
. (1 hour) Learning Section: Improper Fractions Learning Objective: Compare the value of improper fractions with denominators up to 10. Number of pupils: 38
Existing Knowledge: Students know how to compare value of fractions precisely. Learning outcome: In the end of the lesson, students should be able to: 1. Compare the value of improper fractions with denominators up to 10. Example: is bigger than .
KBKK: Generating ideas, understanding and remember. Moral Values: Careful Pupils carry out activities carefully to get the correct parts. Tools and Materials: Circular cut-outs, worksheets.
## Steps / Time Set Induction
Content
Teachers Activities
Studentss Activities
Note
Recalls
## the 1. Teacher comparison
recalls of value
the 1. Students get the paper strips BBM: of and fold it as ordered by teacher. 2. Students pay attention on Papan hitam, paper strips. KBKK: Generating Ideas Moral Values: Share equally.
comparison of of
## (8 minute) value fractions.
fractions to pupils by posing questions such as: Can you tell which fraction is bigger, or ?
teacher explanation.
2. Teacher distributes a piece of paper strip to every pupil and asks pupils to fold the paper strips. Fold first strip paper into 2 parts, write in each part
and shade 1 part of it. Fold second strip paper into 8 parts, write in each
## again such as: Which strip of paper has a
larger shaded area? 4. Teacher introduces the topic being taught the today, thats of
Compare
Value
Improper Fractions. Step 1 (15 min) Teaching and learning activity of 1. Teacher provides each group 1. Pupils get the circular cut-outs BBM: of pupils (3 pupils in 1 group) with 2 similar circular cut-outs. and fold it as ordered by teacher. Circular cut-outs, blackboard. KBKK: Understanding and remember. Moral value: Careful. 2 circular
comparison of 2. Teacher asks pupils to fold 2. Pupils pay attention on the value of improper fraction. both circular cut-outs into 2 explanation by teacher.
equal parts, draw the line and 3. Pupils try to compare the value colour 3 parts. 3. Teacher asks pupils to write the fraction of the coloured parts ( ). 4. Teacher provides each group with another of improper fractions.
cut-outs. 5. Teacher asks pupils to divide both circular cut-outs into 3 equal parts. Ask them to draw the line and colour 5 parts.
6. Teacher asks pupils to write the fraction of the coloured parts ( ). 7. Teacher asks pupils to fold the first 2 circular cut-outs ( ) into 6 equal parts. Count the number of the coloured parts ( ). Cut out the coloured parts. 8. Teacher asks pupils to fold the second 2 circular cut-outs ( ) into 6 parts also. Count the number of the coloured parts ( ). Cut out the coloured parts. 9. Teacher such as: Which is bigger in size? ( is bigger than ). poses a question
10. Pupils carry out activities carefully to get the correct parts.
## 1. Teacher asks pupils to go back 1. Pupils to their own seats.
pay
attention
on Resources: Blackboard
## explanation from teacher.
comparison of 2. Teacher emphasizes that to 2. Pupils which being called try to value of improper fractions with examples. compare two improper
solve the questions being posed Moral values: by teacher. Pay attention
fractions, both fractions must have the same denominator. 3. Teacher poses some
## examples and asks pupils to solve the problem on
blackboard. E.g: Compare the improper fractions. Which is bigger? (a) Step 2 ( 15 min ) and distributes 3. Pupils get the worksheets and try Resource: to pupils as to finish it on given time. Worksheets. Moral values: Diligent and independent.
## 1. Teacher lesson questions.
summarizing by posing
the
## posted Resources Posing questions by teacher Moral values
questions precisely.
## 2. Teacher posing the questions
2. Pupils answer the teacher that: Both fractions must have the same denominator.
## i. If we want to compare 2 i. improper fractions, what
must we do first before compare them? 3. Teacher gives homework by asking pupils to do practice on page 34 in text book. 3. Pupils listen to teachers order.
Reflection
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# NCERT Solutions | Class 11 Maths Chapter 3 | Trigonometric functions
## CBSE Solutions | Maths Class 11
Check the below NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric functions Pdf free download. NCERT Solutions Class 11 Maths were prepared based on the latest exam pattern. We have Provided Trigonometric functions Class 11 Maths NCERT Solutions to help students understand the concept very well.
### NCERT | Class 11 Maths
Book: National Council of Educational Research and Training (NCERT) Central Board of Secondary Education (CBSE) 11th Maths 3 Trigonometric functions English
#### Trigonometric functions | Class 11 Maths | NCERT Books Solutions
You can refer to MCQ Questions for Class 11 Maths Chapter 3 Trigonometric functions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.1
Ex 3.1 Class 11 Maths Question 1.
Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) -47°30′
(iii) 240°
(iv) 520°
Solution.
We have, 180° = π Radians
Ex 3.1 Class 11 Maths Question 2.
Find the degree measures corresponding to the following radian measures $$\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$$
(i) $$\frac { 11 }{ 16 }$$
(ii) -4
(iii) $$\frac { 5\pi }{ 3 }$$
(iv) $$\frac { 7\pi }{ 6 }$$
Solution.
We have π Radians = 180°
Ex 3.1 Class 11 Maths Question 3.
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second ?
Solution.
Number of revolutions made by wheel in one minute = 360
As we know that, 1 Revolution = 27 π Radians
∴ 360 Revolutions = 720 π Radians
∴ In 1 minute wheel can make = 720 π Radians
⇒ In 60 seconds wheel can make = 720 π Radians
⇒ In 1 second wheel can make
$$\frac { 720\pi }{ 3 } Radians=12\pi \quad Radians$$
Ex 3.1 Class 11 Maths Question 4.
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm $$\left( Use\quad \pi =\frac { 22 }{ 7 } \right)$$
Solution.
Let O be the centre and AB be the arc length of the circle.
Ex 3.1 Class 11 Maths Question 5.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of Xhe chord.
Solution.
Let AB be the minor arc of the chord.
AB = 20 cm, OA = OB = 20 cm
Ex 3.1 Class 11 Maths Question 6.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their, radii.
Solution.
Let r1 r2 and θ1, θ2 be the radii and angles subtended at the centre of two circles respectively.
Ex 3.1 Class 11 Maths Question 7.
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm
Solution.
## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.2
Find the values of other five trigonometric functions in Exercises 1 to 5.
Ex 3.2 Class 11 Maths Question 1.
$$\cos { x } =\frac { -1 }{ 2 }$$, x lies in third quadrant.
Solution.
Ex 3.2 Class 11 Maths Question 2.
$$\sin { x } =\frac { 3 }{ 5 }$$, x lies in second quadrant.
Solution.
Ex 3.2 Class 11 Maths Question 3.
$$\cot { x= } \frac { 3 }{ 4 }$$, xlies in third quadrant.
Solution.
Ex 3.2 Class 11 Maths Question 4.
$$\sec { x } =\frac { 13 }{ 5 }$$, x lies in fourth quadrant.
Solution.
Ex 3.2 Class 11 Maths Question 5.
$$\tan { x } =-\frac { 5 }{ 12 }$$, x lies in second quadrant.
Solution.
Find the values of the trigonometric functions in Exercises 6 to 10.
Ex 3.2 Class 11 Maths Question 6.
sin 765°
Solution.
Ex 3.2 Class 11 Maths Question 7.
cosec (-1410°)
Solution.
Ex 3.2 Class 11 Maths Question 8.
$$tan\quad \frac { 19\pi }{ 3 }$$
Solution.
Ex 3.2 Class 11 Maths Question 9.
$$sin\left( -\frac { 11\pi }{ 3 } \right)$$
Solution.
Ex 3.2 Class 11 Maths Question 10.
$$cot\left( -\frac { 15\pi }{ 4 } \right)$$
Solution.
## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.3
Ex 3.3 Class 11 Maths Question 1.
Prove that: $${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$$
Solution.
L.H.S. = $${ sin }^{ 2 }\frac { \pi }{ 6 } +{ cos }^{ 2 }\frac { \pi }{ 3 } -{ tan }^{ 2 }\frac { \pi }{ 4 } =-\frac { 1 }{ 2 }$$
$$=\left[ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } \right] =\frac { 1 }{ 4 } +\frac { 1 }{ 4 } -1=\frac { -1 }{ 2 } =\quad R.H.S.$$
Ex 3.3 Class 11 Maths Question 2.
$$2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 } =\frac { 3 }{ 2 }$$
Solution.
L.H.S. = $$2{ sin }^{ 2 }\frac { \pi }{ 6 } +{ cosec }^{ 2 }\frac { 7\pi }{ 6 } { cos }^{ 2 }\frac { \pi }{ 3 }$$
Ex 3.3 Class 11 Maths Question 3.
$${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 } =6$$
Solution.
L.H.S. = $${ cot }^{ 2 }\frac { \pi }{ 6 } +cosec\frac { 5\pi }{ 6 } +3{ tan }^{ 2 }\frac { \pi }{ 6 }$$
Ex 3.3 Class 11 Maths Question 4.
$$2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 } =10$$
Solution.
L.H.S. = $$2{ sin }^{ 2 }\frac { 3\pi }{ 4 } +2{ cos }^{ 2 }\frac { \pi }{ 4 } +2{ sec }^{ 2 }\frac { \pi }{ 3 }$$
Ex 3.3 Class 11 Maths Question 5.
Find the value of:
(i) sin 75°
(ii) tan 15°
Solution.
(i) sin (75°) = sin (30° + 45°)
(ii) tan 15° = tan (45° – 30°)
Prove the following:
Ex 3.3 Class 11 Maths Question 6.
$$cos\left( \frac { \pi }{ 4 } -x \right) cos\left( \frac { \pi }{ 4 } -y \right) -sin\left( \frac { \pi }{ 4 } -x \right) sin\left( \frac { \pi }{ 4 } -y \right)$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 7.
$$\frac { tan\left( \frac { \pi }{ 4 } +x \right) }{ tan\left( \frac { \pi }{ 4 } -x \right) } ={ \left( \frac { 1+tan\quad x }{ 1-tan\quad x } \right) }^{ 2 }$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 8.
$$\frac { cos\left( \pi +x \right) cos\left( -x \right) }{ sin\left( \pi -x \right) cos\left( \frac { \pi }{ 2 } +x \right) } ={ cot }^{ 2 }x$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 9.
$$cos\left( \frac { 3\pi }{ 2 } +x \right) cos\left( 2\pi +x \right) \left[ cot\left( \frac { 3\pi }{ 2 } -x \right) +cot\left( 2\pi +x \right) \right] =1$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 10.
sin(n +1 )x sin(n + 2)x + cos(n +1 )x cos(n + 2)x = cosx
Solution.
We have,
Ex 3.3 Class 11 Maths Question 11.
$$cos\left( \frac { 3\pi }{ 4 } +x \right) -cos\left( \frac { 3\pi }{ 4 } -x \right) =-\sqrt { 2 } sinx$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 12.
sin26x – sin24x= sin2x sin10x
Solution.
Ex 3.3 Class 11 Maths Question 13.
cos22x – cos26x = sin 4x sin 8x
Solution.
Ex 3.3 Class 11 Maths Question 14.
sin2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
Solution.
We have,
Ex 3.3 Class 11 Maths Question 15.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Solution.
Ex 3.3 Class 11 Maths Question 16.
$$\frac { cos9x-cos5x }{ sin17x-sin3x } =-\frac { sin2x }{ cos10x }$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 17.
$$\frac { sin5x+sin3x }{ cos5x+cos3x } =tan4x$$
Solution.
We have,
Ex 3.3 Class 11 Maths Question 18.
$$\frac { sinx-siny }{ cosx+cosy } =tan\left( \frac { x-y }{ 2 } \right)$$
Solution.
Ex 3.3 Class 11 Maths Question 19.
$$\frac { sinx+sin3x }{ cosx+cos3x } =tan2x$$
Solution.
Ex 3.3 Class 11 Maths Question 20.
$$\frac { sinx-sin3x }{ { sin }^{ 2 }x-{ cos }^{ 2 }x } =2sinx$$
Solution.
Ex 3.3 Class 11 Maths Question 21.
$$\frac { cos4x+cos3x+cos2x }{ sin4x+sin3x+sin2x } =cot3x$$
Solution.
Ex 3.3 Class 11 Maths Question 22.
cot x cot 2x – cot 2x cot 3x – cot3x cotx = 1
Solution.
We know that 3x = 2x + x.
Therefore,
Ex 3.3 Class 11 Maths Question 23.
$$tan4x=\frac { 4tanx\left( 1-{ tan }^{ 2 }x \right) }{ 1-6{ tan }^{ 2 }x+{ tan }^{ 4 }x }$$
Solution.
Ex 3.3 Class 11 Maths Question 24.
cos 4x = 1 – 8 sin2x cos2x
Solution.
Ex 3.3 Class 11 Maths Question 25.
cos 6x = 32 cos6 x – 48 cos4x + 18 cos2 x -1
Solution.
## NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Ex 3.4
Find the principal and general solutions of the following equations:
Ex 3.4 Class 11 Maths Question 1.
$$tanx=\sqrt { 3 }$$
Solution.
Ex 3.4 Class 11 Maths Question 2.
sec x = 2
Solution.
Ex 3.4 Class 11 Maths Question 3.
$$cotx=-\sqrt { 3 }$$
Solution.
Ex 3.4 Class 11 Maths Question 4.
cosec x = -2
Solution.
Find the general solution for each of the following equations:
Ex 3.4 Class 11 Maths Question 5.
cos 4x = cos 2x
Solution.
Ex 3.4 Class 11 Maths Question 6.
cos 3x + cos x – cos 2x=0
Solution.
Ex 3.4 Class 11 Maths Question 7.
sin 2 x + cos x = 0
Solution.
Ex 3.4 Class 11 Maths Question 8.
sec22x = 1 – tan 2x
Solution.
Ex 3.4 Class 11 Maths Question 9.
sin x + sin 3x + sin 5x = 0
Solution.
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# Question Video: Finding the Integration of a Function Involving Trigonometric Functions Mathematics
Determine β« β8 sec 7π₯(β4 cosΒ² 7π₯ + 6 tan 7π₯) dπ₯.
02:44
### Video Transcript
Determine the indefinite integral of negative eight sec seven π₯ multiplied by negative four cos squared seven π₯ plus six tan seven π₯ with respect to π₯.
Letβs begin by distributing the parentheses in the integrand. The integrand becomes 32 sec seven π₯ cos squared seven π₯ minus 48 sec seven π₯ tan seven π₯. Now since sec of seven π₯ is equal to one over cos of seven π₯, we can replace the first term with 32 cos squared seven π₯ over cos of seven π₯. We can cancel a factor of cos seven π₯ from the numerator and denominator, so the first term simplifies to 32 cos seven π₯. And now we need to determine the indefinite integral of 32 cos seven π₯ minus 48 sec seven π₯ tan seven π₯ with respect to π₯.
The first term in the integrand involves a cosine function, and the second term involves a product of a secant and tangent function each with the same argument of seven π₯. We will therefore need to recall the following standard integrals. The integral of cos π₯ with respect to π₯ is equal to sin π₯ plus πΆ. And the integral of sec π₯ tan π₯ with respect to π₯ is equal to sec π₯ plus πΆ.
However, before we can apply these formulas to solve the given indefinite integral, we need to modify the argument of the trigonometric functions. To do this, we can perform a substitution. We let π’ equal seven π₯, which in turn implies that dπ’ by dπ₯ is equal to seven. Now dπ’ by dπ₯ is not a fraction, but we can treat it a little like one. So equivalently one-seventh dπ’ is equal to dπ₯. Performing the substitution, we obtain the indefinite integral of 32 cos π’ minus 48 sec π’ tan π’ one-seventh dπ’.
We can simplify this by splitting the integrand and taking each constant factor out the front to give 32 over seven multiplied by the indefinite integral of cos π’ dπ’ minus 48 over seven multiplied by the indefinite integral of sec π’ tan π’ dπ’. Applying the standard results, we obtain 32 over seven sin π’ plus a constant of integration πΆ one minus 48 over seven sec π’ plus a constant of integration πΆ two. We can combine the two constants of integration into a single arbitrary constant πΆ. Finally, we need to reverse the substitution by replacing π’ with seven π₯. And so we obtain our final answer, which is 32 over seven sin seven π₯ minus 48 over seven sec seven π₯ plus πΆ.
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# Step-By-Step Guide to Adding Irrational Numbers
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## What is an Irrational Number?
Before we go ahead to adding, first you have to understand what makes a number irrational. The definition of an irrational number is a
number that cannot be written as a ratio of two integers. So the number 1.25, for example, would be rational because it could be written as 5/4. The number 0.3333333 (with a repeating 3) could be written as 1/3. In fact, any terminating decimal (decimal that stops after a set number of digits) or repeating decimal (decimal in which one or several digits repeat over and over again, without terminating), is rational.
So what numbers are irrational? The main example of an irrational number is a number that contains a square root. Therefore, √2 is an irrational number, as is 2√57. (Obviously, √4 is rational, because it is equal to “2,” a rational number.) Other examples of irrational numbers are pi(∏) and e, neither of which can be represented by a ratio of two integers.
In order to add square roots, you can add only “like terms.” This may sound familiar from pre-algebra, in which you had to find “like terms” in order to add coefficients together. In pre-algebra, you looked at 3x2 and x2 as like terms because they both contained x2. You then added the coefficients - or the digits before the like terms - together to get 4x2.
The same works with square roots. For example, you can add 3√3 and 2√3 to get 5√3, in the same way that you can add 3x and 2x to get 5x. You cannot, however, add 3√3 and 2√2, in the same way that you cannot add 3x and 2x2.
You may, however, need to simplify the square roots before you can see whether they contain like terms. For example, if you were given the problem “3√2 + √8,” you might think that the terms cannot be added together. In truth, you could simply rewrite √8 to be √4 X √2, which is 2√2. So the results would be 3√2 + 2√2, which is 5√2.
The same would be true with adding other irrational numbers, such as ∏ and e. For example, if you had the problem “2∏ + 6∏,” you could easily add them together to get 8∏. If you had the problem “2∏ + 8_e_,” however, you would not be able to add the two terms together. The same would be true if you had a problem that contained both a square root and another irrational number, such as “2∏ + 2√2.”
Adding irrational numbers is actually quite simple, once you get the hang of it. The key is to find any like terms, and then add the coefficients together. Try it with the following problem, to make sure you have it right.
2√18 + 3√2 + √32 + √2 = ?
(The answer would be 6√2 + 3√2 + 4√2 + 1√2 = 14√2.)
## This post is part of the series: Math Study Guides
Confused in math class? These math study guides span various topics, from square roots to improper fractions.
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# Video: Optimization Using Derivatives
In this video, we will learn how to apply derivatives to real-world problems to optimize a function under certain constraints.
17:56
### Video Transcript
In this video, we will see one of the practical applications of differentiation to optimization problems. That is, problems where we’re asked to determine the maximum or minimum value of a given function such as time, area, or perimeter. We’ll be able to answer questions like what does the maximum area I can enclose with a given length of fencing or what is the highest point reached by a rocket following a given curve. In some examples, we’ll also see how to form the optimization function and any constraints ourselves from a worded description.
We recall, first of all, some key facts about using differentiation to find the critical points of a function. The critical points of a function are those points in its domain where its first derivative, 𝑓 prime of 𝑥, is equal to zero or is undefined. To find the 𝑥-value at a critical point, we can differentiate the function to find its gradient, or slope, function and then set this equal to zero and solve the resulting equation. We can find the value of the function itself at the critical point by substituting our 𝑥-value or values into the function.
We must also remember to confirm that our critical point is indeed a maximum or a minimum by applying the second derivative test. We recall that at a local maximum, the second derivative 𝑓 double prime of 𝑥 will be negative, whereas at a local minimum, the second derivative 𝑓 double prime of 𝑥 will be positive. We’ll now see how to apply these key principles to some optimization problems.
A rocket is launched in the air. Its height, in meters, as a function of time is given by ℎ of 𝑡 equals negative 4.9𝑡 squared plus 229𝑡 plus 234. Find the maximum height the rocket attains.
So, we’ve been given an equation for the height ℎ of this rocket in terms of 𝑡 time. And we’re asked to determine the maximum height that the rocket attains. We can follow some key steps in order to do so. First, we need to find an expression for the first derivative of our function. In this case, that’s ℎ prime of 𝑡. By applying the power rule of differentiation, we see that ℎ prime of 𝑡 is equal to negative two multiplied by 4.9𝑡 plus 229, which simplifies to negative 9.8𝑡 plus 229.
Next, we recall that at the critical points of the function, the first derivative is equal to zero. So, we’re going to take that expression we’ve found for ℎ prime of 𝑡, set it equal to zero, and then solve the resulting equation for 𝑡. We add 9.8𝑡 to both sides and then divide by 9.8 to give 𝑡 equals 229 over 9.8, which, as a decimal, is 23.36734, or 23.37 to two decimal places. Now, this is the value of 𝑡 at which our function ℎ of 𝑡 has a critical point. We don’t yet know whether it is a maximum. And we don’t yet know the height that the rocket attains at this point.
Our next step, then, is to evaluate the function ℎ of 𝑡 when 𝑡 is equal to 23.37. Substituting into our equation ℎ of 𝑡 gives ℎ of 23.37 equals negative 4.9 multiplied by 23.37 squared plus 229 multiplied by 23.37 plus 234. Which evaluates to 2909.56119, or 2909.56 correct to two decimal places. So, we believe that this is the maximum height that the rocket attains as this occurs at the only critical point of the function. But we must confirm that it is indeed a maximum.
To do so, we’ll perform the second derivative test. We’ll evaluate the second derivative ℎ double prime of 𝑡 at this critical point. Differentiating our expression for ℎ prime of 𝑡, which was negative 9.8𝑡 plus 229, we find that ℎ double prime of 𝑡 is equal to negative 9.8. And in fact, we don’t need to evaluate the second derivative when 𝑡 equals 23.37 because it is a constant. The second derivative is the same for all values of 𝑡.
We note though, that this value is negative. And therefore, by the second derivative test, our critical point is a maximum. Now, we could also have seen this if we considered that the expression we were given for ℎ in terms of 𝑡 is a quadratic expression with a negative leading coefficient. And therefore, the graph of 𝑡 against ℎ would be an inverted parabola. And we know that it will therefore have a maximum point rather than a minimum. So, we can conclude that the maximum height the rocket attains, and we have confirmed that it is indeed a maximum, is 2909.56 meters, correct to two decimal places.
Now, it’s just worth pointing out that in this question, each of the derivatives also have practical interpretations. The first derivative of our function, that’s negative 9.8𝑡 plus 229, gives the velocity of the rocket, which we see is decreasing as time increases. The second derivative, which we saw is just the constant negative 9.8, gives the acceleration of the rocket, or in fact the deceleration. Which, we see, is equal to negative 9.8 meters per second squared. And that’s the deceleration due to gravity.
In this problem, the function we needed to optimize was given to us. But it will more often be the case in optimization problems that we’ll need to form these functions ourselves from a worded description. Let’s see how this works in our next example.
Find two numbers whose sum is 156 and the sum of whose squares is the least possible.
Let’s allow these two numbers, which we don’t yet know, to be 𝑥 and 𝑦. Then, we can express the fact that their sum is 156 as 𝑥 plus 𝑦 equals 156. We want to minimize the sum of their squares, which we can call 𝑠. 𝑠 is equal to 𝑥 squared plus 𝑦 squared. In order to do so, we need to find the values of 𝑥 and 𝑦 for which the rate of change of 𝑠 with respect to either 𝑥 or 𝑦, is equal to zero. This means that we need to differentiate 𝑠 with respect to either 𝑥 or 𝑦.
First, though, we need to write 𝑠 in terms of one variable only. The choice is entirely arbitrary in this problem. We could perform a simple rearrangement of our first equation to give 𝑦 equals 156 minus 𝑥 and then substitute this expression for 𝑦 into our equation for 𝑠 to give an equation in terms of 𝑥 only. Distributing the parentheses and then simplifying gives 𝑠 is equal to two 𝑥 squared minus 312𝑥 plus 24336.
Remember, we’re looking to minimize this sum of squares, so we need to find the critical points of 𝑠. To do so, we need to find where the first derivative of 𝑠 with respect to 𝑥, that’s d𝑠 by d𝑥, is equal to zero. We can use the power rule to find this derivative. And we see that d𝑠 by d𝑥 is equal to four 𝑥 minus 312. We then set this derivative equal to zero and solve for 𝑥. We first add 312 to each side and then divide by four, giving 𝑥 equals 78.
So, we found the value of 𝑥 at which 𝑠 has a critical point. But there are two things we need to do. In a moment, we’ll confirm that this is indeed a minimum. But first, we also need to find the value of 𝑦, which we can do by substituting the value of 𝑥 into our linear equation. We see that 𝑦 is equal to 156 minus 78, which is equal to 78.
To confirm that this critical point is indeed a minimum. We need to find the second derivative of the function 𝑠 with respect to 𝑥. Differentiating d𝑠 by d𝑥 again gives d two 𝑠 by d𝑥 squared is equal to four. The second derivative of 𝑠 with respect to 𝑥 is, therefore, constant for all values of 𝑥. And more importantly, it is positive, which confirms that this critical point is indeed a minimum. The two numbers then whose sum is 156, which have the minimum sum of squares, are 78 and 78.
In this example, we saw how to set up the optimization function and any constraints ourselves from information given in the question. We’ll, now see how to do this again with a more practical example.
A wire of length 41 centimeters is used to make a rectangle. What dimensions give its maximum area?
Now, it’s important that we don’t just attempt trial and error here. We need to use a proper optimization method to find the dimensions which will give the largest possible area for this rectangle, subject to the constraint that we only have 41 centimeters of wire.
Let’s consider then a rectangle with a length of 𝑙 centimeters and a width of 𝑤 centimeters. We need to maximize its area, which for a rectangle is its length multiplied by its width. Subject to the constraint which is that the perimeter of this rectangle must be equal to 41. The perimeter of a rectangle is twice its length plus twice its width. So, we have the constraint two 𝑙 plus two 𝑤 equals 41.
Now, in order to maximize this area, we’re going to need to use differentiation to find the critical points of this function 𝐴. But before we can do that, we need to write 𝐴 in terms of one variable only. The choice of whether we use 𝑙 or 𝑤 is entirely arbitrary. So, I’ve chosen to rearrange the linear equation to give 𝑙 equals 41 minus two 𝑤 over two. Substituting this expression for 𝑙 into our area formula gives 𝐴 equals 41 minus two 𝑤 over two multiplied by 𝑤. And distributing the parentheses, we have 41𝑤 over two minus 𝑤 squared.
To find the critical points of 𝐴, we first find its first derivative, d𝐴 by d𝑤, which, using the power rule of differentiation, is equal to 41 over two minus two 𝑤. We then set this expression equal to zero and solve the resulting equation for 𝑤. We add two 𝑤 to each side and then divide by two, giving 𝑤 is equal to 41 over four. So, we’ve found the width of the rectangle at which the area has a critical point.
We also need to find the length though, which we do by substituting this value of 𝑤 back into our expression for 𝑙, giving 41 minus two times 41 over four all over two. Which also simplifies to 41 over four. However, we’re not yet finished. We know that these values of 𝑤 and 𝑙 give a critical point for the area. But we haven’t yet confirmed that it is indeed a maximum. To check this, we need to perform the second derivative test. We find d two 𝐴 by d𝑤 squared, which is equal to negative two.
Now, this is a constant for all values of 𝑤. But more specifically it is a negative constant. And as the second derivative is less than zero, this confirms that our critical point is indeed a maximum. So, we found that the length and width which give this rectangle its maximum area, subject to the given perimeter constraint, as decimals, are both 10.25 centimeters.
Now, we notice that the length and width of this rectangle are actually the same, making it in fact a square. This illustrates a general point in optimization problems where we’re looking to maximize an area with respect to a length constraint. The maximum area will be achieved when the dimensions are as similar as possible i.e. when the ratio between the dimensions is as close as possible to one to one.
In the case of a problem involving a rectangle, it will always turn out that the shape will in fact be a square. But of course, we must always go through the working out. In order to show this. In our final example, we’ll see how to maximize the sum of the volumes of two three-dimensional shapes subject to a surface area constraint.
Given that the sum of the surface areas of a sphere and a right circular cylinder is 1000𝜋 centimeters squared, and their radii are equal, find the radius of the sphere that makes the sum of their volume at its maximum value.
So, in this question, we’re asked to maximize the sum of the volumes of two three-dimensional solids, subject to a constraint about the sum of their surface areas. Let’s begin by writing down fomulae for the surface areas of both the sphere and the right circular cylinder. And as their radii are the same, we can use the same letter 𝑟 for both. For the sphere, first of all, its surface area is given by four 𝜋𝑟 squared. For the cylinder, its surface area is two 𝜋𝑟 squared plus two 𝜋𝑟ℎ, where ℎ represents the height of the cylinder.
As the sum of these surface areas is 1000𝜋 centimeters squared, we can form an equation, four 𝜋𝑟 squared plus two 𝜋𝑟 squared plus two 𝜋𝑟ℎ equals 1000𝜋. We can then combine the like terms on the left-hand side and then divide through by a 𝜋, as this is a common factor in all terms. We could also divide through by two, as all the coefficients are even, to give three 𝑟 squared plus 𝑟ℎ equals 500. We can’t do anything further with this equation at this point, as we have two unknowns 𝑟 and ℎ. So, next, we recall the formulae for the volume of a sphere and the volume of a cylinder.
The volume of a sphere is four-thirds multiplied by 𝜋 multiplied by its radius cubed. And the volume of a cylinder is 𝜋 multiplied by its radius squared multiplied by its height. So, we have that the total volume of these two solids is four-thirds 𝜋𝑟 cubed plus 𝜋𝑟 squared ℎ. We want to maximize the sum of these volumes, 𝑉 total. Now, this will be maximized when its rate of change with respect to either 𝑟 or ℎ is equal to zero, which will be when its first derivative is equal to zero. But before we can differentiate, we need to express 𝑉 total in terms of a single variable.
It’s much more straightforward to rearrange our surface area constraint to give an expression for ℎ in terms of 𝑟 than it is to give an expression for 𝑟 in terms of ℎ. We have ℎ equals 500 minus three 𝑟 squared over 𝑟. We can then substitute this expression for ℎ into our expression for the total volume so that it is in terms of 𝑟 only. We can cancel a factor of 𝑟 in the second term and then distribute the parentheses to give four-thirds 𝜋𝑟 cubed plus 500𝜋𝑟 minus three 𝜋𝑟 cubed. We have an expression for 𝑉 total in terms of 𝑟 only.
Next, we need to find the first derivative d𝑉 total by d𝑟, so we’ll create a little bit of space in order to do this. By applying the power rule of differentiation, we see that the derivative of 𝑉 total with respect to 𝑟 is equal to four-thirds 𝜋 multiplied by three 𝑟 squared plus 500𝜋 minus three 𝜋 multiplied by three 𝑟 squared, which all simplifies to 500𝜋 minus five 𝜋𝑟 squared. Next, in order to find critical points, we need to set this derivative equal to zero and solve for 𝑟.
We can divide through by five 𝜋, giving zero equals 100 minus 𝑟 squared. Adding 𝑟 squared to both sides gives 𝑟 squared equals 100. And we then find 𝑟 by square rooting. We only need to take the positive square root as the radius of a solid must be a positive value. So, we see that 𝑟 is equal to 10. We now know that the combined volume of these two solids has a critical point when the radius is equal to 10. But we must now confirm that it is a maximum.
We perform the second derivative test. Differentiating our expression for d𝑉 total by d𝑟 again, with respect to 𝑟, gives negative 10𝜋𝑟. And evaluating this when 𝑟 is equal to 10 gives negative 100𝜋. This is negative, which confirms that the critical point is indeed a maximum. So, we found that the radius of the sphere and also the radius of the right circular cylinder which maximizes the sum of their volumes, subject to the given surface area constraint, is 10 centimeters.
Let’s summarize then the key points that we’ve seen in this video. Firstly, the key principles of differentiation can be applied to optimization problems. That’s problems where we want to find the maximum or minimum value of a function. We know that the critical points of a function occur when its first derivative is equal to zero or is undefined. Once we found a critical point, we must always confirm that it is indeed a maximum or minimum by performing the second derivative test. And we’ve seen that it may be necessary for us to form both the optimization function and any constraints ourselves from a written description.
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### Transcript of Power Functions and Polynomials · PDF file polynomial. The graph has 2 horizontal intercepts,...
• This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2017, and contains content remixed with permission from College Algebra © Stitz & Zeager 2013. This material is licensed under a Creative Commons CC-BY-SA license.
Power Functions and Polynomials Power Functions & Polynomials A square is cut out of cardboard, with each side having length L. If we wanted to write a function for the area of the square, with L as the input and the area as output, you may recall that the area of a rectangle can be found by multiplying the length times the width. Since our shape is a square, the length & the width are the same, giving the formula:
2)( LLLLA =⋅= Likewise, if we wanted a function for the volume of a cube with each side having some length L, you may recall volume of a rectangular box can be found by multiplying length by width by height, which are all equal for a cube, giving the formula:
3)( LLLLLV =⋅⋅= These two functions are examples of power functions, functions that are some power of the variable.
Power Function A power function is a function that can be represented in the form
pxxf =)( Where the base is a variable and the exponent, p, is a number.
Example 1
Which of our toolkit functions are power functions? The constant and identity functions are power functions, since they can be written as
0)( xxf = and 1)( xxf = respectively. The quadratic and cubic functions are both power functions with whole number powers:
2)( xxf = and 3)( xxf = . The reciprocal and reciprocal squared functions are both power functions with negative whole number powers since they can be written as 1)( −= xxf and 2)( −= xxf . The square and cube root functions are both power functions with fractional powers since they can be written as 21)( xxf = or 31)( xxf = .
• Chapter 3 160
Try it Now 1. What point(s) do the toolkit power functions have in common? Characteristics of Power Functions Shown to the right are the graphs of
642 )(and,)(,)( xxfxxfxxf === , all even whole number powers. Notice that all these graphs have a fairly similar shape, very similar to the quadratic toolkit, but as the power increases the graphs flatten somewhat near the origin, and become steeper away from the origin. To describe the behavior as numbers become larger and larger, we use the idea of infinity. The symbol for positive infinity is ∞ , and ∞− for negative infinity. When we say that “x approaches infinity”, which can be symbolically written as ∞→x , we are describing a behavior – we are saying that x is getting large in the positive direction. With the even power functions, as the x becomes large in either the positive or negative direction, the output values become very large positive numbers. Equivalently, we could describe this by saying that as x approaches positive or negative infinity, the f(x) values approach positive infinity. In symbolic form, we could write: as ±∞→x , ∞→)(xf . Shown here are the graphs of
753 )(and,)(,)( xxfxxfxxf === , all odd whole number powers. Notice all these graphs look similar to the cubic toolkit, but again as the power increases the graphs flatten near the origin and become steeper away from the origin. For these odd power functions, as x approaches negative infinity, f(x) approaches negative infinity. As x approaches positive infinity, f(x) approaches positive infinity. In symbolic form we write: as −∞→x , −∞→)(xf and as ∞→x , ∞→)(xf .
Long Run Behavior The behavior of the graph of a function as the input takes on large negative values,
−∞→x , and large positive values, ∞→x , is referred to as the long run behavior of the function.
• 3.1 Power and Polynomial Functions 161
Example 2 Describe the long run behavior of the graph of 8)( xxf = . Since 8)( xxf = has a whole, even power, we would expect this function to behave somewhat like the quadratic function. As the input gets large positive or negative, we would expect the output to grow without bound in the positive direction. In symbolic form, as ±∞→x , ∞→)(xf .
Example 3
Describe the long run behavior of the graph of 9)( xxf −= Since this function has a whole odd power, we would expect it to behave somewhat like the cubic function. The negative in front of the 9x will cause a vertical reflection, so as the inputs grow large positive, the outputs will grow large in the negative direction, and as the inputs grow large negative, the outputs will grow large in the positive direction. In symbolic form, for the long run behavior we would write: as ∞→x , −∞→)(xf and as −∞→x , ∞→)(xf . You may use words or symbols to describe the long run behavior of these functions.
Try it Now 2. Describe in words and symbols the long run behavior of 4)( xxf −=
Treatment of the rational and radical forms of power functions will be saved for later. Polynomials An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. If we wanted to write a formula for the area covered by the oil slick, we could do so by composing two functions together. The first is a formula for the radius, r, of the spill, which depends on the number of weeks, w, that have passed. Hopefully you recognized that this relationship is linear:
wwr 824)( += We can combine this with the formula for the area, A, of a circle:
2)( rrA π=
• Chapter 3 162
Composing these functions gives a formula for the area in terms of weeks: 2)824()824())(()( wwAwrAwA +=+== π
Multiplying this out gives the formula
264384576)( wwwA πππ ++= This formula is an example of a polynomial. A polynomial is simply the sum of terms each consisting of a vertically stretched or compressed power function with non-negative whole number power.
Terminology of Polynomial Functions
A polynomial is function that can be written as nn xaxaxaaxf ++++= 2
210)( Each of the ai constants are called coefficients and can be positive, negative, or zero, and be whole numbers, decimals, or fractions. A term of the polynomial is any one piece of the sum, that is any ii xa . Each individual term is a transformed power function. The degree of the polynomial is the highest power of the variable that occurs in the polynomial. The leading term is the term containing the highest power of the variable: the term with the highest degree. The leading coefficient is the coefficient of the leading term. Because of the definition of the “leading” term we often rearrange polynomials so that the powers are descending.
01 2
2.....)( axaxaxaxf n
n ++++=
• 3.1 Power and Polynomial Functions 163
Example 4 Identify the degree, leading term, and leading coefficient of these polynomials: a) 32 423)( xxxf −+= b) ttttg 725)( 35 +−= c) 26)( 3 −−= ppph a) For the function f(x), the degree is 3, the highest power on x. The leading term is the term containing that power, 34x− . The leading coefficient is the coefficient of that term, -4. b) For g(t), the degree is 5, the leading term is 55t , and the leading coefficient is 5. c) For h(p), the degree is 3, the leading term is 3p− , so the leading coefficient is -1.
Long Run Behavior of Polynomials For any polynomial, the long run behavior of the polynomial will match the long run behavior of the leading term.
Example 5
What can we determine about the long run behavior and degree of the equation for the polynomial graphed here? Since the output grows large and positive as the inputs grow large and positive, we describe the long run behavior symbolically by writing: as ∞→x ,
∞→)(xf . Similarly, as −∞→x , −∞→)(xf . In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function which has not been reflected, so the degree of the polynomial creating this graph must be odd, and the leading coefficient would be positive.
Try it Now 3. Given the function )5)(1)(2(2.0)( −+−= xxxxf use your algebra skills to write the
function in standard polynomial form (as a sum of terms) and determine the leading term, degree, and long run behavior of the function.
• Chapter 3 164
Short Run Behavior Characteristics of the graph such as vertical and horizontal intercepts and the places the graph changes direction are part of the short run behavior of the polynomial. Like with all functions, the vertical intercept is where the graph crosses the vertical axis, and occurs when the input value is zero. Since a polynomial is a function, there can only be one vertical intercept, which occurs at the point ),0( 0a . The horizontal intercepts occur at the input values that correspond with an output value of zero. It is possible to have more than one horizontal intercept. Horizontal intercepts are also called zeros, or roots of the function. Example 6
Given the polynomial function )4)(1)(2()( −+−= xxxxf , written in factored form for your convenience, determine the vertical and horiz
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Question Video: Solving Multistep Problems Involving Right Triangle Trigonometry to Find an Unknown Length | Nagwa Question Video: Solving Multistep Problems Involving Right Triangle Trigonometry to Find an Unknown Length | Nagwa
# Question Video: Solving Multistep Problems Involving Right Triangle Trigonometry to Find an Unknown Length Mathematics • Third Year of Preparatory School
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In the given figures, ๐ด๐ต = 6, ๐ด๐ท = 3, and ๐โ ๐ด๐ถ๐ต = 33ยฐ. Calculate the length of ๐ต๐ถ. Give your answer to two decimal places.
04:32
### Video Transcript
In the given figures ๐ด๐ต equals six, ๐ด๐ท equals three, and the measure of angle ๐ด๐ถ๐ต is 33 degrees. Calculate the length of ๐ต๐ถ. Give your answer to two decimal places.
First, letโs transfer the written information in the question onto the diagram. Now letโs have a closer look at this diagram. It consists of two right-angled triangles which share a common side, ๐ต๐ท. In the triangle on the left, weโve have been given the lengths of two sides. In the triangle on the right, weโve been given the measure of one angle.
The side we have been asked to calculate, ๐ต๐ถ, is in the second triangle in which we currently only have one piece of information. In order to find this side, weโre going to need at least two pieces of information in the triangle to which it belongs. So weโre going to need to use the fact that the triangles share the side ๐ต๐ท.
Letโs begin in triangle ๐ด๐ต๐ท. As weโve already said, this is a right-angled triangle. And we know the lengths of two of its sides. Therefore, we can apply the Pythagorean theorem in order to calculate the length of the third side. The Pythagorean theorem tells us that in this triangle, ๐ต๐ท squared plus three squared is equal to six squared. And this is an equation that we can solve in order to find the length of ๐ต๐ท.
Evaluating three squared and six squared tells us that ๐ต๐ท squared plus nine is equal to 36. Subtracting nine from both sides of this equation gives ๐ต๐ท squared is equal to 27. Finally for this stage, square rooting both sides of the equation tells us that ๐ต๐ท is equal to the square root of 27.
Now I could evaluate this as a decimal but it would be a value that needed rounding and therefore an inexact value. So Iโm going to keep it as a surd for now. So now we know the length of ๐ต๐ท. Letโs turn our attention to the second triangle, the triangle on the right. Apart from the right angle, we now have one known angle and one known side. And weโre looking to calculate the length of a second side in this triangle.
We can use trigonometry in order to do this. The first step is to label the three sides of the triangle in relation to the angle of 33 degrees. So ๐ต๐ถ is the hypotenuse, ๐ถ๐ท is the adjacent, and ๐ต๐ท is the opposite. Remember itโs ๐ต๐ถ that weโre looking to calculate. We know the length of the opposite side, ๐ต๐ท. And weโre looking to calculate the hypotenuse.
So if we recall the acronym SOHCAHTOA, this tells us that we need to use the sine ratio. The sine ratio is defined as sine of an angle ๐ is equal to the opposite divided by the hypotenuse. Now letโs substitute the values for this triangle. We have that sine of 33 degrees is equal to root 27 over ๐ต๐ถ.
This is why I kept that value as a square root rather than evaluating as a decimal, because I knew I was going to use it in the next stage of the calculation. Now we want to solve this equation in order to find the value of ๐ต๐ถ. So Iโm going to multiply by ๐ต๐ถ, first of all, as itโs currently in the denominator of a fraction.
This gives me ๐ต๐ถ multiplied by sin 33 degrees is equal to root 27. Next, I need to divide both sides of the equation by sine of 33 degrees, which is just a number. So we have that ๐ต๐ถ is equal to root 27 over sin 33 degrees.
Now I can evaluate this using my calculator. And this is actually the first stage in this question that Iโve needed to use a calculator. This gives a value of 9.54054 and the decimal then continues. The question has asked us to give our answer to two decimal places. So we need to round this value.
So we have that the length of ๐ต๐ถ to two decimal places is 9.54. And there are no units with this as we werenโt given any units in the original question.
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Home » Math Theory » Geometry » Properties of Angles
Properties of Angles
Before working through the information below you may wish to review this lesson on measuring angles as well as taking a look at the lesson on adding and subtracting angles.
This lesson will provide information and guidance on:
After reviewing the lessons above you will be ready to read through the information below on angles and their relationships with your children. Discuss these as you go and, when you are ready, try the angle relationships worksheet.
Useful Terms
Parallel Lines – lines that are equidistant from each other and never intersect.
Transversal – a line that intersects two or more other lines.
Adjacent Angles – angles that share a common side and that have a common vertex.
Complementary Angles
Complementary Angles are those which add together to make 90°.
The examples above all show two angles that are complementary. Notice that the angles do not have to be adjacent to be complementary. If they are adjacent then they form a right angle.
Supplementary Angles
Supplementary Angles add together to make 180°
The two angles shown above are supplementary to each other. They add together to give 180°. They can be said supplement each other. Note that, as with complementary angles, they do not need to be adjacent to each other.
Opposite Angles
When to lines intersect they create four angles. Each angle is opposite to another and form a pair of what are called opposite angles.
Opposite angles are sometimes called vertical angles or vertically opposite angles.
Corresponding and Alternate Angles
The example below shows two parallel lines and a transversal (a line that cross two or more other lines). This results in eight angles. Each of these angles has a corresponding angle. Looking at the two intersections, the angles that are in the same relative (or corresponding) positions are called corresponding angles.
Since the two lines are parallel, the corresponding angles are equal.
As Shown below, there are also two pairs of alternate interior angles and two pairs of alternate exterior angles. Notice how the interior angles are in between the two parallel lines and the exterior angles are to the outside.
Since the two lines are parallel, the alternate angles shown above are equal.
The Sum of Interior Angles
The sum of the interior angles in a triangle is 180°.
The sum of the interior angles in a quadrilateral is 360°.
Try the 180° In a Triangle Experiment which is a 2-page (be careful with the scissors) activity to demonstrate that the sum of the interior angles in a triangle is 180°.
Angle Relationship Worksheet
Have your children try the worksheet below that has questions on angle relationships. After completing it your children will be ready to review the lesson on finding missing angles.
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# Direct Variation
When two variables change in proportion it is called as direct variation. In direct variation one variable is constant times of other. If one variable increases other will increase, if one decrease other will also decease. This means that the variables change in a same ratio which is called as constant of variation.
Direct variation is the simplest type of variation and in practical life we can find many situations which can be co-related with direct variation.
If two variables A and B are so related that when A increases ( or decreases ) in a given ratio, B also increases ( or, decreases ) in the same ratio, then A is said to vary directly as B ( or, A is said to vary as B ).
This is symbolically written as, A ∝ B (read as, ‘A varies as B’ ).
Suppose a train moving at a uniform speed travels d km. in t minutes. Now, consider the following table:
d (km) 24 12 48 36 t (min) 30 15 60 45
Like in a math examination if for one problem solving we can score 10 numbers, so five problems solving we can get 50 numbers. This can be explained with a direct variation equation. If T denotes total numbers scored, N denotes numbers of problem solved and K denotes numbers can be scored for solving a problem, then the direct variation equation for this situation will be T = KN.
As the numbers for a problem can be scored is fixed, it is a constant = K = $$\frac{T}{N}$$ = 10
For solving 5 problems total numbers scored T = KN = 10 x 5 = 50.
From the above example we can understand that the ratio of two variables is a constant K and T, N are the variables which changes in proportion with value of constant.
Direct variation can be by a linear equation Y = KX where K is a constant. When the value of constant is higher, the change of variable Y is significantly for small change of X. But when the value of K is very small, Y changes very less with change of X. For this case K is equivalent to the ratio of change of two variables. So $$\frac{σY}{σX}$$ = K when K is very small.
Now we will solve some problems on direct variation:
1. If P varies directly as Q and the value of P is 60 and Q is 40, what is the equation that describes this direct variation of P and Q?
Solution:
As P varies directly with Q, ratio of P and Q is constant for any value of P and Q.
So constant K = $$\frac{P}{Q}$$ = $$\frac{60}{40}$$ = $$\frac{3}{2}$$
So the equation that describes the direct variation of P and Q is P = $$\frac{3}{2}$$Q.
2. If a car runs at a constant speed and takes 3 hrs to run a distance of 180 km, what time it will take to run 100 km?
Solution:
If T is the time taken to cover the distance and S is the distance and V is the speed of the car, the direct variation equation is S= VT where V is constant.
For the case given in the problem,
180 = V × 3 or V = $$\frac{180}{3}$$ = 60
So speed of the car is 60kmph and it is constant.
For 100 km distance
S = VT or 100 = 60 × T
T = $$\frac{100}{60}$$ = $$\frac{5}{3}$$ hrs = 1 hr 40 mins.
So it will take 1 hr 40 mins time.
3. In X is in direct variation with square of Y and when X is 4, Y is 3. What is the value of X when Y is 6?
Solution:
From the given problem direct variation equation can be expressed as
X = KY2
For the given case
4 = K × 32
or, 4 = 9K
or, K = $$\frac{4}{9}$$
So when Y is 6,
X = $$\frac{4}{9}$$ × 6 = $$\frac{8}{3}$$
So the value of X is $$\frac{8}{3}$$.
Variation
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