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# 5.4 Integration formulas and the net change theorem
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• Apply the basic integration formulas.
• Explain the significance of the net change theorem.
• Use the net change theorem to solve applied problems.
• Apply the integrals of odd and even functions.
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.
## Integrating a function using the power rule
Use the power rule to integrate the function ${\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt.$
The first step is to rewrite the function and simplify it so we can apply the power rule:
$\begin{array}{cc}{\int }_{1}^{4}\sqrt{t}\left(1+t\right)dt\hfill & ={\int }_{1}^{4}{t}^{1\text{/}2}\left(1+t\right)dt\hfill \\ \\ & ={\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt.\hfill \end{array}$
Now apply the power rule:
$\begin{array}{cc}{\int }_{1}^{4}\left({t}^{1\text{/}2}+{t}^{3\text{/}2}\right)dt\hfill & ={\left(\frac{2}{3}{t}^{3\text{/}2}+\frac{2}{5}{t}^{5\text{/}2}\right)|}_{1}^{4}\hfill \\ & =\left[\frac{2}{3}{\left(4\right)}^{3\text{/}2}+\frac{2}{5}{\left(4\right)}^{5\text{/}2}\right]-\left[\frac{2}{3}{\left(1\right)}^{3\text{/}2}+\frac{2}{5}{\left(1\right)}^{5\text{/}2}\right]\hfill \\ & =\frac{256}{15}.\hfill \end{array}$
Find the definite integral of $f\left(x\right)={x}^{2}-3x$ over the interval $\left[1,3\right].$
$-\frac{10}{3}$
## 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:
$\begin{array}{}\\ \\ F\left(b\right)=F\left(a\right)+{\int }_{a}^{b}F\text{'}\left(x\right)dx\hfill \\ \hfill \text{or}\hfill \\ {\int }_{a}^{b}F\text{'}\left(x\right)dx=F\left(b\right)-F\left(a\right).\hfill \end{array}$
Subtracting $F\left(a\right)$ 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 [link] .
Find the derivative of g(x)=−3.
any genius online ? I need help!!
Pina
need to learn polynomial
Zakariya
i will teach...
nandu
I'm waiting
Zakariya
plz help me in question
Abish
evaluate the following computation (x³-8/x-2)
teach me how to solve the first law of calculus.
what is differentiation
f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
pweding paturo nsa calculus?
jimmy
how to use fundamental theorem to solve exponential
find the bounded area of the parabola y^2=4x and y=16x
what is absolute value means?
Chicken nuggets
Hugh
🐔
MM
🐔🦃 nuggets
MM
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
how does this work
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
log tan (x/4+x/2)
Rohan
Rohan
y=(x^2 + 3x).(eipix)
Claudia
Ismael
A Function F(X)=Sinx+cosx is odd or even?
neither
David
Neither
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f(x)=1/1+x^2 |=[-3,1]
apa itu?
fauzi |
# What Is The Definition Of Commutative Property?
What commonplace behavior exemplifies a commutative property in action? There is virtually nothing that commutes in real life, but occasionally the non-commutativity is tiny enough that we may overlook it with minimal repercussion. In all likelihood, the example provided by Jake is an exceptionally fine one that also happens to be a very classic one.
## What is the commutative property simple definition?
This law essentially asserts that you may alter the order of the numbers in a problem by adding and multiplying them, and it will have no effect on the outcome of the issue. Subtraction and division are NOT mutually exclusive operations.
## What are 2 examples of commutative property?
Here’s a short rundown of each of these characteristics: The commutative property of addition is as follows: The amount does not change if the addends are arranged in a different order. For example, 4 plus 2 is 2 plus 4. 4 plus 2 is 4 plus 2. In the equation 4+2=2+44, the addition of 2 equals 2, the addition of 4 equals 2.
## What is the meaning of commutative in math?
The quality of a mathematical operation (such as addition or multiplication) in which the outcome does not depend on the order in which the elements are introduced It is known as the commutative property of addition, and it asserts that the total of one plus two and two plus one will be three.
## What is the meaning of associative property?
Associative property is a mathematical rule that states that the order in which elements are grouped in a multiplication problem has no effect on the result of the multiplication. For example: 5 4 2 5 times 4 times 2 5 times 4 times 2 5 times 4 times 2 5 times 4 times 2 5 times 4 times 2 5 4 2
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## What is commutative property and associative property?
Associative property is a mathematical rule that states that the order in which elements are grouped in a multiplication problem has no effect on the outcome of the equation. Using this example, 5 4 2 is composed of five fours divided by two five fours divided by two five fours divided by two five fours.
## What are the 4 types of properties?
1. In mathematics, the four most important number properties are the commutative property, the associative property, the identity property, and the distributive property.
## What property is 3 x x 3?
As a result, the statement ″three times the variable x″ can be expressed in a variety of ways, including 3x, 3(x), and 3 x (see below). To make the phrase 9(4 + x) more complex, use the distributive property.
## How is associative property used in everyday life?
Consider the following scenario: I go to the grocery and purchase ice cream for 12 dollars, bread for 8 dollars, and milk for 15 dollars, all at regular prices. For the sake of simplicity, I’ll combine or add the prices of ice cream and bread first, then combine or add the prices of milk and ice cream and then put the results together.
## What is commutative property maths class 8?
Commutative operations are those in which the result does not vary regardless of whether or not the operands are in the same order as they were entered.
## Is commutative property of multiplication?
The commutative property applies exclusively to multiplication and addition operations. Subtraction and division, on the other hand, are not commutative.
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## How is the commutative property of addition like the commutative property of multiplication?
The commutative property asserts that a change in the order of integers in an addition or multiplication operation has no effect on the sum or product of the numbers. When it comes to addition, the commutative property is represented as A+B = B + A. When you multiply two numbers together, you get the commutative property of multiplication, which is expressed as A + B = B + A.
## What is the meaning of inverse property?
Inverse property of addition states that every integer plus its opposite will result in a value of zero. Opposite numbers have distinct signs (and are thus on opposite sides of zero), yet they are the same distance away from the decimal point. For instance, 6 plus its inverse (which is -6) equals 0. Alternatively stated, 6 – 6 Equals 0.
## What are the 5 math properties?
Commutative property, associative property, distributive property, identity property of multiplication, and identity property of addition are all terms that can be used to describe mathematical properties.
## What is associative property Byjus?
Property of Commutation, Associative Property, Distributive Property, Identity Property of Multiplication, and Identity Property of Addition are all terms that are used in mathematics.
## Would you please Show Me an example of commutative property?
Specifically, these are the commutative, associative, multiplicative, and distributive identities, as well as the distributive properties. The commutative property states that when two integers are multiplied together, the product is the same regardless of the sequence in which the multiplicands are multiplied together. For example, 4 * 2 = 2 * 4 is the same as 4 * 2.
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## What are some examples of commutative properties?
1. Which of the following is governed by the commutative principle?
2. 9+12
3. –7 – 6
4. 72 – 5
5. –99 – 7
6. 9+12
7. –7 – 6
8. 72 – 5
9. 9+12
10. 9+12
11. 9+12
12. 9+12
13. 9+12
14. 9+12
15. 9+12
16. 9+12
17. 9+12
18. 9+12
19. 9+12
20. 9+12
21. 9+12
22. 9+12
23. 9+12
24. 9+12
25. 9+12
26. 9+12
27. 9+
## What does commutative property and associative property mean?
1. Input and output are mutually exclusive operations when putting on your shoes. This is due to the fact that it makes no difference which shoe we put on first.
2. Because it does not matter which one goes into the coffee first, adding sugar and cream to coffee is a commutative operation.
3. The commutative feature can also be noticed when money is exchanged or received in cash. |
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# Three equal weights of$3kg$ each are hanging on a string passing over a frictionless pulley as shown in figure. The tension in the string between masses$II$ and$III$ will be (Take$g = 10m{s^{ - 2}}$)(A) $5N$(B) $6N$(C) $10N$(D) $20N$
Last updated date: 20th Jun 2024
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Hint We are given the situation of three blocks of equal masses put up in a certain configuration and are asked to find the tension of the string connecting two of the three blocks. Thus, we will just apply the basics of force and mass configuration.
Complete Step By Step Solution
Here,
For the tension and gravitational force on the block$I$, we get
$T - mg = ma$
Putting the values of mass, we get
$T - 3g = 3a$
Thus, we get
$T = 3a + 3g \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$
Now,
For the blocks$II$ and$III$, we get
$2mg - T = 2ma$
Putting in the values of mass, we get
$6g - T = 6a$
Further, using equation$(1)$, we get
$6g - 3a - 3g = 6a$
After further evaluation, we get
$3g = 9a$
Then, we get
$a = \dfrac{g}{3}$
Now,
The $T$we considered till now is the net tension of the string.
Now,
We will take $T'$ as the tension between the blocks$II$ and$III$.
Thus, the equation will be
$mg - T' = ma$
Further, putting in the values, we get
$3g - T' = 3a$
Further, we get
$T' = 3g - 3a$
Thus, putting the evaluated value of$a$, we get
$T' = 3g - 3\left( {\dfrac{g}{3}} \right)$
Then we get
$T' = 2g$
Putting the value of$g$, we get
$T' = 2 \times 10 = 20N$
Hence, the correct option is (D).
Additional Information The diagram we are considering is called the free body diagram which shows us the bodies and the forces acting on it. The equations we are forming are on the basis of the fundamental idea that all the external forces are in balance for an isolated body.
Note The tension of the string will hold the blocks in equilibrium and will determine the motion of the blocks as the string moves on the pulley. But the tension of the string between the two blocks will make sure that while the motion of the block, the block remains in equilibrium. |
# Integration by Parts – Examples with Answers
Integration by parts allows us to “reduce” an integral to a simpler form, expressing it as the difference between two simpler integrals. This technique is especially useful when we want to evaluate integrals that cannot be easily found using other methods, such as substitution or trigonometric identities.
In this article, we will look at some solved integration by parts exercises. Then, we will see some practice problems to apply what we have learned.
##### CALCULUS
Relevant for
Learning about integration by parts with examples.
See examples
##### CALCULUS
Relevant for
Learning about integration by parts with examples.
See examples
## How to integrate functions by parts
Integration by parts is used to integrate the product of two functions. To integrate functions using this method, we follow the steps below:
#### 1. Choose two functions, u and dv/dx
The product of the two functions, $latex u\frac{dv}{dx}$ is the integrand.
#### 4. Use the formula for integration by parts:
$$\int u \frac{dv}{dx} dx=uv – \int v \frac{du}{dx}dx$$
or
$$\int u v^{\prime} dx=uv – \int v u^{\prime}dx$$
When using integration by parts, the choice of $latex u$ and $latex dv$ is not always obvious. However, there are some rules of thumb that can help guide your choice.
First, $latex u$ should be chosen such that it is an easy function to integrate, while $latex dv$ should be chosen such that it is an easy function to differentiate. This will make the integration and differentiation steps in the integration by parts formula simpler and easier to evaluate.
Second, it can sometimes be useful to choose $latex u$ and $latex dv$ so that the product $latex uv$ is as simple as possible. This can make the final step of the integration by parts formula easier to evaluate since you will be left with an integral that has a simple integrand.
## Integration by parts – Examples with answers
### EXAMPLE 1
Find the following integral:
$$\int x\cos x dx$$
To solve this integral, in which the product of two functions appears, the rule of integration by parts is used, according to which:
$$\int f(x)g'(x)dx=f(x)g(x)-\int g(x)f'(x)dx$$
If we specify the following:
• $latex f(x)=u$
• $latex g'(x)dx=dv\Rightarrow v=g(x)$
Then, the expression at the beginning is rewritten as follows:
$$\int udv=uv -\int vdu$$
Now, it only remains to select which of the functions is $latex v$ and which is $latex u$. For example, if you choose:
• $latex u =x$
• $latex du =dx$
• $latex dv =\cos x dx$
• $latex v =\int \cos x dx = \sin x$
Then:
$$\int x\cos x dx= x\sin x-\int \sin xdx =x\sin x+\cos x + C$$
$$\int x\cos x dx=x\sin x+\cos x + C$$
In general, it is advisable to take $latex dv$ as the part of the integrand that is most easily integrated, and $latex u$ as the one that is easiest to derive. The practice facilitates the decision, having as a goal that $latex vdu$ is easy to compute.
### EXAMPLE 2
Calculate the integral:
$$\int xe^x dx$$
Selecting:
• $latex u =x$
• $latex du =dx$
• $latex dv =e^x dx$
• $latex v =\int e^x dx = e^x$
With the formula:
$$\int udv=uv -\int vdu$$
We have:
$$\int xe^x dx=xe^x-\int e^xdx$$
$$=xe^x-e^x+C$$
This result can be factored and we get:
$$\int xe^x dx=e^x(x-1)+C$$
### EXAMPLE 3
Solve the following:
$$\int x \sqrt{x-1}dx$$
In this case, we specify:
• $latex u =x$
• $latex du =dx$
• $latex dv = \sqrt{x-1}\:dx$
• $latex v =\int \sqrt{x-1}\:dx =\dfrac{2}{3} (x-1)^\frac{3}{2}$
Using the formula:
$$\int udv=uv -\int vdu$$
We get:
$$\int x \sqrt{x-1}dx=x\cdot\dfrac{2}{3} (x-1)^\frac{3}{2}-\int \dfrac{2}{3} (x-1)^\frac{3}{2}dx$$
The integral obtained in the last step is now solved:
$$\int \dfrac{2}{3} (x-1)^\frac{3}{2}dx=\dfrac{2}{3}\int (x-1)^\frac{3}{2}dx$$
By means of a simple variable change:
• $latex dw=dx$
• $latex w=x-1$
We have:
$$\int (x-1)^\frac{3}{2}dx=\int w^\frac{3}{2}dw=\left[\dfrac{w^{\frac{3}{2}+1}}{\frac{3}{2}+1}\right]$$
$$=\dfrac{2}{5}w^{\frac{5}{2}}+C=\dfrac{2}{5}(x-1)^{\frac{5}{2}}+C$$
Finally, this result is substituted here:
$$\int x \sqrt{x-1}dx=x\cdot\dfrac{2}{3} (x-1)^\frac{3}{2}-\dfrac{2}{3}\int (x-1)^\frac{2}{3}dx$$
$$=\dfrac{2x}{3} (x-1)^\frac{3}{2}-\dfrac{2}{3}\cdot\dfrac{2}{5}(x-1)^{\frac{5}{2}}+C$$
$$=\dfrac{2x}{3} (x-1)^\frac{3}{2}-\dfrac{4}{15}(x-1)^{\frac{5}{2}}+C$$
$$=2 (x-1)^\frac{3}{2}\left[\dfrac{x}{3} -\dfrac{2}{15}(x-1)\right]+C$$
$$=2 (x-1)^\frac{3}{2}\left[\dfrac{x}{3} -\dfrac{2x}{15}+\dfrac{1}{15}\right]+C$$
$$=2 (x-1)^\frac{3}{2}\left[\dfrac{3x}{15} +\dfrac{2}{15}\right]+C$$
$$=\frac{2}{15}(x-1)^\frac{3}{2}\left(3x +2\right)+C$$
Then, the integral sought is:
$$\int x \sqrt{x-1}dx=\frac{2}{15}(x-1)^\frac{3}{2}\left(3x +2\right)+C$$
## Integration by parts – Practice problems
Integration by parts quiz
You have completed the quiz!
Interested in learning more about integrals? You can take a look at these pages:
### Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students. |
# Maximum Power transfer theorem
In an electric circuit, power flows from the source to the load. The source itself has some amount of impedance which affects the current flow. Therefore, transfer of power depends not only on the load impedance but also the source impedance. Maximum power transfer theorem speaks about the relationship between the source impedance, load impedance and the power flow from the source to the load.
## Maximum power transfer theorem – Statement
The power transferred from a supply source to a load is at its maximum when the resistance of the load is equal to the internal resistance of the source.
Note: This theorem lets us improve power transfer in a circuit, but not circuit efficiency.
## Explanation
The above circuit shows a battery with an EMF V, internal resistance Rs and a load resistance RL. According to maximum power transfer theorem, maximum power flows to RL when RS = RL. This theorem helps us choose the load resistance when we have a source with a known internal resistance.
## Verification of theorem
Let us verify the theorem with an example.
For the above circuit calculate the maximum power that can be transferred from the source. Let us vary the load resistance from 0 to 5 in steps of 0.5 and find out at which point the power transferred is maximum.
When RL = 0 ohm,
Total circuit resistance = 2.5 + 0 = 2.5 Ohms
Current flow to the Load = 20V/2.5 = 8A
Power delivered to the load = I2R = 82x 0 = 0 watts
When RL = 0.5 ohm,
Total circuit resistance = 2.5 + 0.5 = 3 Ohms
Current flow to the Load = 20V/3 = 6.67A
Power delivered to the load = I2R = 6.672x 0.5 = 21.125 watts
When RL = 1 ohm,
Total circuit resistance = 2.5 + 1 = 3.5 Ohms
Current flow to the Load = 20V/3.5 = 5.71A
Power delivered to the load = I2R = 5.712x 1 = 32.6 watts
Similarly lets calculate the power delivered for other values of load and tabulate it below:
From the above table we can notice that the power delivered is maximum at 2.5 ohm which is equal to the source resistance. This verifies maximum power transfer theorem.
## Misconception
The theorem tells us to choose the load resistance when we have a source with a known internal resistance. It is a common misconception to apply the theorem in the opposite scenario. It does not say how to choose the source resistance for given load resistance. The power transfer maximizes when the internal resistance of the source zero, regardless of the load resistance. |
## How do you determine if a number is rational or irrational?
Answer: If a number can be written or can be converted to p/q form, where p and q are integers and q is a non-zero number, then it is said to be rational and if it cannot be written in this form, then it is irrational.
## How do you know if something’s a rational number?
Any number that can be written as a fraction or a ratio is a rational number. The product of any two rational numbers is therefore a rational number, because it too may be expressed as a fraction. For example, 5/7 and 13/120 are both rational numbers, and their product, 65/840, is also a rational number.
## What makes a number a rational number?
Rational Numbers: Any number that can be written as a ratio (or fraction) of two integers is a rational number. … An integer can be written as a fraction by giving it a denominator of one, so any integer is a rational number.
## Is 81 rational or irrational?
81 is a rational number.
## Is 0.6 a rational number?
0.6=610 as this can be written as a fraction it is a rational number.
## What are the examples of rational numbers?
Any number that can be written as a fraction with integers is called a rational number . For example, 17 and −34 are rational numbers. (Note that there is more than one way to write the same rational number as a ratio of integers. For example, 17 and 214 represent the same rational number.)
## Is 0.8 a rational number?
Explanation: 0.8 can be expressed as the ratio of two integers (namely 810 ) which is the definition of a rational number.
## Is 25 a rational number?
The number 25 is a rational number. It is a whole number that can be written as the fraction 25/1. By definition, a rational number is the number…
## Is 16 a rational number?
Sixteen is natural, whole, and an integer. Since it can also be written as the ratio 16:1 or the fraction 16/1, it is also a rational number.
## Is 9.5 a rational number?
Example: 9.5 can be written as a simple fraction like this:
Irrational? Irrational !
## Is 36 irrational or rational?
36 is a rational number because it can be expressed as the quotient of two integers: 36 ÷ 1.
## Is 1.25 a rational number?
1.25 is a rational number. A rational number is any number that can be written as a fraction.
## Is 2.5 a rational number?
The decimal 2.5 is a rational number. All decimals can be converted to fractions. The decimal 2.5 is equal to the fraction 25/10.
## Is 48 a rational number?
48 is a rational number because it can be expressed as the quotient of two integers: 48 ÷ 1.
## Is sqrt36 a rational number?
It’s rational, and whole and integer.
## Is 64 irrational or rational?
Hence, the square root of 64 is a rational number.
Square Root of 64.
1.What Is the Square Root of 64?
6.FAQs on Square Root of 64
## Is 121 irrational or rational?
121 is a rational number because it can be expressed as the quotient of two integers: 121 ÷ 1.
## Is 15 a rational number?
Explanation: 15 is a rational number because it can be expressed as a fraction (151or302,etc.)
## Is the square root of 81 a rational number?
Is the Square Root of 81 Rational or Irrational? A rational number is defined as a number that can be expressed in the form of a quotient or division of two integers, i.e. p/q, where q is not equal to 0. Both numbers can be represented in the form of a rational number. Hence, the square root of 81 is a rational number.
## Is 3.14 a irrational number?
Pi, which begins with 3.14, is one of the most common irrational numbers. … Pi has been calculated to over a quadrillion decimal places, but no pattern has ever been found; therefore it is an irrational number. |
# A model train, with a mass of 15 kg, is moving on a circular track with a radius of 6 m. If the train's kinetic energy changes from 81 j to 27 j, by how much will the centripetal force applied by the tracks change by?
May 27, 2016
The centripetal force has decreased by $18 N$.
#### Explanation:
This question is about circular motion and how much centripetal force is being applied, which is given by the equation:
${f}_{c} = \frac{m {v}^{2}}{r}$
where $m$ is the mass of the object, $r$ is the radius of the circle (the track in this case) and $v$ is the tangential velocity. We have been given the mass and the radius, but not the velocity. However, we have been given the kinetic energy which is given by:
$K . E . = \frac{1}{2} m {v}^{2}$
Now, instead of solving for the velocity (and needing to take a square root, just to have to square it again, let's solve for the term $m {v}^{2}$ since that appears in both equations and then substitute the result back into the first equation:
$m {v}^{2} = 2 K . E .$
substituting then gives
${f}_{c} = \frac{2 \cdot K . E .}{r}$
We are looking for the change in centripetal force which we can write as:
$\Delta {f}_{c} = {f}_{c 2} - {f}_{c 1} = \frac{2 \cdot K . E {.}_{2}}{r} - \frac{2 \cdot K . E {.}_{1}}{r}$
$\Delta {f}_{c} = \frac{2 \cdot \left(K . E {.}_{2} - K . E {.}_{1}\right)}{r}$
$\Delta {f}_{c} = \frac{2 \cdot \left(27 J - 81 J\right)}{6 m} = \frac{- 54 J}{3 m} = - 18 N$
Therefore, the centripetal force has decreased by $18 N$.
Note: The question had given us more information than we actually used - the mass was redundant since the kinetic energy alone was sufficient information. Other versions of this question may not include the mass. |
Question
1. # 1234567891011121314151617181920……424344 What Is Remainder When Divided By 45?
Remainder is a powerful number, and you may be using it without even realizing it. In fact, remainder is probably the number you use most often in your every day life. Reservoirs, percentages, and other mathematical concepts are built upon remainder. And while these concepts may seem simple enough, understanding them can be tricky. That’s where this article comes in. We will take a look at remainder and show you how to use it to your advantage.
## Remainder when divided by 45 is a calculation that can be used to determine the remainder of an integer division
The remainder when divided by 45 is a calculation that can be used to determine the remainder of an integer division. For example, if you divide 15 by 45, the answer would be 3. This means the numerator (top number in the division) is divided by the denominator (bottom number in the division), and the result is 3. The remaining number after this division would be 2.
## The remainders for other common calculations can also be determined using this method
The remainders for other common calculations can also be determined using this method. This method is most useful when the divisor is a whole number and the dividend is not. The steps are:
1) Find the numerator and denominator of the fraction.
2) Subtract the numerator from the denominator.
3) The remainder is the answer.
## This calculation can be used in a variety of situations, such as in financial statements or to determine how much money is left on a credit card
There are a few situations where you might need to calculate the remainder after dividing an integer number by a divisor. In financial statements, for instance, the remainder is often used to determine whether a company is profitable or not. Another common use for the remainder is when calculating how much money is left on a credit card. Knowing how to calculate the remainder can come in handy in many different scenarios.
## The remainders for other integers can also be determined using this method
When dividing two integers, the remainder is what’s left when the division is done. For example, if you divide 24 by 5, the result would be 4 with a remainder of 1. The number 4 would go into both sides of the equation and leave 1 on the bottom. This same principle can be used to determine the remainders for other integers as well.
## If you need to calculate
If you need to calculate the remainder when divided by a number, there are several different formulas that you can use. The simplest method is to divide the number by 2 and then subtract the first number from the second. For example, if you need to find the remainder when 501 is divided by 5, you would use the following equation: 501 ÷ 5 = 103
103 – 501 = -98 This means that the remainder is 3. If you want to find the remainder when 123 is divided by 4, you would use this equation: 123 ÷ 4 = 37
37 – 123 = 12 This means that the remainder is 1.
2. When it comes to math, division is one of the most important concepts. One of the things that often arises when dividing numbers is the concept of remainder. This occurs when a number cannot be evenly divided by another number, leaving a small fraction left over. For example, remainders arise when trying to divide 1234567891011121314151617181920 by 45.
In this situation, 45 does not go into 1234567891011121314151617181920 evenly and so there is a remainder of 424344 when divided by 45. To work out this answer, you need to divide each part of the larger number separately until you find your final answer.
3. Ah, the age-old question: what is the remainder when divided by 45? 🤔
At first glance, this may seem like a daunting task. But don’t worry- it’s actually quite simple! 🤓
Let’s break this down. If we take any number, say 1234567891011121314151617181920, and divide it by 45, the remainder will be the number that is left over after the division process is complete. In this case, it would be 20. 🤓
Now, let’s try another number. How about 424344? When we divide this number by 45, the remainder will be 19. 🤓
As you can see, the answer to “what is the remainder when divided by 45?” is always the number that is left over after the division process is complete. 👍
Hopefully this helped you understand the concept of remainders a bit better. If you want to learn more about this concept and how it can be used in different mathematical equations, you should definitely check out online tutorials and other resources. 🤓
Good luck! 🤞 |
# Lesson 10
The Distributive Property, Part 2
### Problem 1
Here is a rectangle.
1. Explain why the area of the large rectangle is $$2a + 3a + 4a$$.
2. Explain why the area of the large rectangle is $$(2+3+4)a$$.
### Problem 2
Is the area of the shaded rectangle $$6(2-m)$$ or $$6(m-2)$$?
Explain how you know.
### Problem 3
Choose the expressions that do not represent the total area of the rectangle. Select all that apply.
A:
$$5t + 4t$$
B:
$$t + 5 + 4$$
C:
$$9t$$
D:
$$4 \boldcdot 5 \boldcdot t$$
E:
$$t(5+4)$$
### Problem 4
Evaluate each expression mentally.
1. $$35\boldcdot 91-35\boldcdot 89$$
2. $$22\boldcdot 87+22\boldcdot 13$$
3. $$\frac{9}{11}\boldcdot \frac{7}{10}-\frac{9}{11}\boldcdot \frac{3}{10}$$
### Solution
(From Unit 6, Lesson 9.)
### Problem 5
Select all the expressions that are equivalent to $$4b$$.
A:
$$b+b+b+b$$
B:
$$b+4$$
C:
$$2b+2b$$
D:
$$b \boldcdot b \boldcdot b \boldcdot b$$
E:
$$b \div \frac{1}{4}$$
### Solution
(From Unit 6, Lesson 8.)
### Problem 6
Solve each equation. Show your reasoning.
$$111=14a$$
$$13.65 = b + 4.88$$
$$c+ \frac{1}{3} = 5\frac{1}{8}$$
$$\frac{2}{5} d = \frac{17}{4}$$
$$5.16 = 4e$$
### Solution
(From Unit 6, Lesson 4.)
### Problem 7
Andre ran $$5\frac{1}{2}$$ laps of a track in 8 minutes at a constant speed. It took Andre $$x$$ minutes to run each lap. Select all the equations that represent this situation.
A:
$$\left(5\frac{1}{2}\right)x = 8$$
B:
$$5 \frac{1}{2} + x = 8$$
C:
$$5 \frac{1}{2} - x = 8$$
D:
$$5 \frac{1}{2} \div x = 8$$
E:
$$x = 8 \div \left(5\frac{1}{2}\right)$$
F:
$$x = \left(5\frac{1}{2}\right) \div 8$$ |
# MATH Connections: A Secondary Core Curriculum
Year 2:
Algebra/number/function: Students will recognize equations that can be considered identities.
Furthermore, they will use area models to illustrate the distributive property and , in particular, justify the
identity that (x+y)2=x2+2xy+y2. They will explore situations where one variable is directly proportional
to another (y=kx) and situations where one variable varies as the square of another (y=kx2). Inverse
variation is also explored. They will find the constant of proportionality, write an equation, and use it to
find additional information . Students will solve for a missing term in a proportion using the technique of
cross-multiplying. They will be able to solve equations involving the Pythagorean Theorem and use a
calculator to find the square root of a number, when needed. Students will understand the meaning of
compound inequalities such as 2<x<5 and be able to separate them into two inequalities and vice versa.
They will explore functions defined by multiple rules such as piecewise linear and step functions . They
will have experience describing real situations with three variables and identify independent and
dependent variables. In particular, they will explore real- valued functions of two variables. With tables
and formulas. They will be familiar with notation such as T(x,y) to represent the value of a function with
independent variables x and y. They will have a strategy for envisioning such relationships involving
three variables using a 2-dimensional graph instead of a 3-dimensional graph, especially when the
variables are discrete. The strategy involves plotting " contour lines" of the graph determined by fixing
one of the variables at various points and graphing the resulting relation. Using complete graphical
information obtained from this process, for example, students will be able to determine whether or not
the function of two variables is linear. They will know what to look for symbolically to determine
whether or not a function of any number of variables is linear. Also, here, a matrix is defined as a
function of two variables where the independent variables represent values in the set {1,2,..,n} for some
particular n. Students will be able to distinguish between categorical and measurement variables. They
will have experience making some theoretical models and comparing the model to actual data. They will
understand the term "residual" as the actual value minus the predicted value. They will explore
"additive" models of functions versus "interactive" models of functions of two variables (where there is
usually a term containing the product of the two independent variables.)
Students will be able to identify linear equations in two variables in several symbolic forms such as
y=2x+1 and 2x+3y=6. They will be able to solve for one variable in terms of another in these situations.
They will have experience solving systems in two unknowns using elimination. In fact, they will be able
to work with the Gaussian Elimination algorithm and back substitution to solve systems of liner equations
involving up to three variables. They will be able to represent such linear systems using matrices and be
able to apply Gaussian elimination using row operations both by hand and using technology. They will
be able to determine when this process reveals a system which has no solution , infinitely many solutions,
and a unique solution. They will be able to interpret the latter situation geometrically with two and three
unknowns. Students will use these concepts in many applied settings. For instance, within mathematics,
they will have experience "fitting" an equation of the form (x-a)2+(y-b)2=c to three data points in the
plane by solving a system of linear equations in three unknowns using Gaussian Elimination and back
substitution. By graphing the equation, they will see this as determining a circle through the three points.
They will also have experience fitting a parabola to some sets of three points using elimination. Students
will also be able to use matrices for organizing, sorting, and manipulating information. They will have
contextual reasons for defining the addition, subtraction, and multiplication of matrices, and know when
these operations can be performed. They will have seen and worked with some abstract properties of
operations with matrices, e.g. commutativity of matrix addition, but not multiplication; associativity of
addition and multiplication; and properties of the "zero" matrix (all entries are zero) with addition and
identity matrices (ones along the diagonal and zeros otherwise) in matrix multiplication. In addition, they
will understand and be able to use the concept of multiplying a matrix by a scalar. Students will have
experience working with the least squares computations by hand and determining the "coefficient of
determination" (usually called r2). They will be introduced to ∑-notation. They will know the basic
shapes of quadratic polynomials and at least the shape of the cubic y=x3. They will see what a, b, and c
do in each of the cases y=ax2 ; y=x2+b ; y=(x+c)2 ; and y=x2 +cx, for at least n=2.
Geometry: Students will realize the advantages and disadvantages of modeling real-world situations; in
particular, they will distinguish between shortest distances on a map vs. a globe. They will be able to
identify units of measure for length and time, measure lengths using standard and nonstandard units, and
recognize the importance of standard units of measure. Students will be able to compute the perimeter
of equilateral polygons using the formula P =ns, as well as perimeters of other regions (e.g. rectangles,
triangles). They will identify polygons and name their vertices, sides, and angles. Using a compass and
straightedge, they will construct the perpendicular bisector of a line segment and the perpendicular from
a line to a point not on the line. They will also be able to construct a triangle, given its side
measurements, and they will be introduced to the triangle inequality when they are given measurements
that cannot form a triangle. They will construct an isosceles triangle, as well as the bisector of its vertex
angle. They will be able to determine the axes of symmetry for a figure, including an equilateral triangle,
rhombus, rectangle, square, and other regular polygons. They will investigate the property that the
diagonals of a rhombus are perpendicular bisectors of each other. They will classify certain
quadrilaterals (parallelogram, rhombus, rectangle, square) according to sides and angles. Students will
investigate strategies for finding the area of a region, including the use of formulas for a rectangle, a
triangle and a parallelogram, "tiling" a region and counting unit squares, and the "divide and conquer"
algorithm which divides a region into smaller areas that can be found and summed to yield total area.
Moreover, they will realize that the area of a polygon can be determined by triangulating it and adding
together the areas of the triangles. They will identify the Pythagorean Theorem, use it to find the length of
a side of a right triangle when the other two sides are given, and use its converse to ascertain whether or
not a triangle is a right triangle. They will convert some English units of measure (e.g. square inches to
square feet) and some metric measures (e.g. square centimeters to square decimeters). They will be
able to determine the volume of a rectangular prism.
Students will investigate the concept of proportionality. They will be able to use a scaling factor to
determine
measurements of similar figures and create a proportional drawing. Given measurements of
two similar figures, they will find the scaling factor. They will conclude whether or not two triangles are
similar by comparing lengths of corresponding sides and finding the ratio of similarity, if possible.
Students will identify kinds of angles (acute, obtuse, right, straight, reflex) and examine different ways to
measure them using slope, degrees, and "A-measure" (where the class decides on a unit of angle
measure). They will convert slope measure of an angle into degree measure and vice versa, using the
tan-1 and tan calculator keys. They will realize that angle measurement is unaffected by scaling. They
will compute and compare perimeters and areas of scaled figures (i.e. if a figure is scaled by a factor of
k, its area will change by a factor of k2). They will also compute the volume of a scaled rectangular
prism and realize that its volume changes by a factor of k3. Students will know that the measures of
vertical angles are equal and that supplementary angles can form a straight angle. They will explore
corresponding angles and alternate interior angles formed by two lines crossed by a transversal. In
particular, they will see that two lines cut by a transversal are parallel if and only if the measures of a pair
of corresponding angles (or alternate interior angles) are equal. [Software such as the Geometer's
Sketchpad is suggested in the Teacher Commentary as a means to explore angle measures.] Students
will be guided through a series of steps leading to the fact that the sum of the angles of any triangle is
180°. They will explore the sum of the measures of the interior angles of any polygon to eventually
discover the generalization (n-2)180° for an n-gon. They will realize that the sum of the measures of the
exterior angles of any polygon is 360°. Furthermore, using appropriate formulas, they will compute the
measure of a single interior angle and exterior angle of a regular polygon. Focusing on triangles, students
will investigate the congruence principles (SSS, SAS, ASA, AAS) and use them to determine a triangle
from given angle or side measurements. They will recognize that SSA and AAA are not valid principles
for determining a triangle.
Students will be able to construct regular polygons with a given number of sides with geometry software
such as the Geometer's Sketchpad. They will see many properties of the circle. They will know its
"center-radius" definition (i.e. the set of points in the plane a fixed distance from a given point). They
will recognize that every diameter of a circle is an axis of symmetry. They will know the definition of
rotational symmetry and be able to compare the rotational symmetry of regular polygons to the
rotational symmetry of the circle. They will know that the set of circles that pass through two given
points have centers on the perpendicular bisector of the segment joining the points and conversely. They
will be able to construct a circle passing through three points with straight edge and compass by locating
its center. They will be able to represent a circle parametrically using such parametric representations as
(cosθ, sinθ) for the unit circle. They will investigate what a transformation (a+rcosθ, b+rsinθ) does to
the unit circle. They will be able to determine parametric representations for circles in the plane and be
able to utilize such representation along with technology that plots parametrically (such as the TI-82) to
display designs created from circular arcs. Using results on changes in area and perimeter of similar
figures, they will explore relationships between the area and perimeter of a circle. For example, they will
compare the area and perimeter of a circle to the area and perimeter, respectively, of the unit circle.
They will discover that the area of the unit circle is p (square units) and develop several approximations
to its value using geometric means. They will see that the area of a circle can be found by multiplying its
circumference by its radius and dividing by two. They will also express the area and perimeter of a circle
as a function of its radius (i.e. A(r) = πr2 and C(r)=2πr). They will be able to use the proportional
relationship between a central angle of a circle and the length of the corresponding arc and the
proportional relationship between a central angle of a circle and the area of the corresponding segment
of the circle. They will be able to supply justification for special cases. They will also be able to use the
relationship between an inscribed angle and one of the central angles with the same endpoints as the
inscribed angle and will have worked through a justification of this relationship. They will know that
some properties don't characterize the circle. For example, they will know that there are curves other
than the circle which have constant width (e.g. a Reuleaux Triangle). Students will investigate three-dimensional
shapes. They will have two-dimensional strategies for describing these shapes (such as a
cone). They will be able to use nets (two-dimensional patterns) that fold into polyhedra and other solid
shapes. They will determine a method, using the " two-dimensional" Pythagorean Theorem for
calculating the height of a pyramid with equilateral triangular sides and a regular polygon as a base
knowing the length of an edge. They will realize that a pyramid with a regular polygon as base and
equilateral triangular sides must have a base with fewer than six sides. They will realize that a regular
polyhedron has the property that the sum of the angle measures at each vertex is less than 360°. They
should have developed a justification for the fact that there are only five regular polyhedra. They will be
able to use two-dimensional contour curves to describe irregular shapes such as geological landscapes.
They will be able to describe a cone using a net formed by removing a sector of a circle. They will be
able to calculate the radius of the base of a cone as a function of the central angle measure of the sector
of the circle removed. They will be able to calculate the height of a cone as a function of the radius of
the base of the cone. They will be able to compose these functions to represent the height of a cone as a
function of the central angle of the central angle measure in the net. They will be able to describe solids
such as prisms and cylinders using two-dimensional cross-sections. They will be able to calculate the
surface area of a prism from information about its height, the number of edges of its base, and the length
of each of those edges. They will provide some detail in the development of a formula for the volume of
a cone using mostly geometric arguments. They will know Cavalieri's Principle and have worked
through a plausibility argument by comparing cross-sectional slices of two prisms of the same height
where one is a right prism and the other is oblique and letting the thickness of these cross-sectional
slices shrink.
They will develop a very intuitive notion of this latter explanation as a "limiting " procedure. They will
use Cavalieri's Principle to determine the volume of solids. For example, they will have seen the
Principle used, with some algebraic manipulations , to develop a formula for the volume of a sphere.
They will investigate solids of revolution (without coordinates) of many bounded plane regions around a
line. They will have seen evidence based on geometry and algebra to support the Pappus-Guildin
Theorem which states that the volume of a solid of revolution is the product of the area of the region and
the distance traveled by the center of gravity (centroid) of the region. They will explore the center of
gravity of triangles, rectangles, circles, and semi-circles and, thus, be able to determine the volume of the
solid obtained by rotating one of these regions about a line (which does not intersect the region). They
will be introduced the a coordinatization of three dimensions using the three standard axes in the
standard orientation. They will have seen the formula for the distance between to points in the plane and
two points in three-space developed. They will be introduced to "set builder" notation. They will be
able to describe many three-dimensional objects using sets of triples satisfying algebraic equalities
and/or inequalities. They will know the equations for a (two-dimensional) circle of radius r and a (three-dimensional)
sphere of radius r. They will see that the three axes do not need to represent spacial
measures. They will see examples where solids can be used to convey information when one of the
axes measures another quantity such as time.
Trigonometry: Building on the concept of similarity of right triangles, students will be introduced to the
trigonometric functions (sinθ, cosθ, tanθ, cscθ, secθ, cotθ, where θ is an acute angle) as ratios of the
sides of a right triangle. They will use the sine, cosine, and tangent functions to find the length of a side
of a right triangle. They will find the measure of an acute angle of a right triangle, using the inverse
trigonometric functions (sin-1θ, cos-1θ, tan-1θ). They will generalize a formula for the area of a
parallelogram in terms of its sides and an acute angle (either interior or exterior): Area=absinA.
Students will follow developments of the Law of Sines and the Law of Cosines (for acute angles only)
and use them to find the length of a side of a triangle. Through specific questioning, they will explore a
proof that "the angle made by a tangent line and the radius at the point of tangency is a right angle."
They will calculate (sinq)2+(cosq)2 for several angles, make a conjecture, and follow a generalization to
the conclusion that sin2θ+cos2θ=1. They will also see that tan2θ+1=sec2θ. Given tables of sine and
cosine values for 0°<θ≤50°, where θ is an integer, students will compare values. Using a diagram of a
right triangle with sides a, b, c and angle θ, they will explain identities such as sin(90°-θ)=cosθ and
tan(90°-θ)=cotθ. They will explain the connection between the slope of a line and the tangent of an
angle. They will examine the graphs of y=sinθ and y=cosθ on a graphing calculator and be introduced
to several real-world examples which could be modeled by the sine and/or cosine curves. They will
relate the graph of the unit circle to the sine and cosine of an angle and, hence, extend the domains of
the sine and cosine functions.
Probability and statistics: Students will perceive differences between discrete and continuous data.
They will enhance their ability to create appropriate representations (e.g. stem-and-leaf plot, boxplot,
scatterplot, dot plot (also known as line plot)) for sets of data and compare and contrast these displays.
Using primarily a dot plot, they will examine the distribution of data in terms of its symmetry, whether or
not it is bell-shaped or bimodal, and possible outliers. They will be introduced to bias when they
encounter data that is not "distributed approximately symmetrically around the true value."
Logic and reasoning: Students will write the converse of statements and decide whether or not they
are true. If true, they will explain why; if not true, they will provide reasons or counterexamples. For
example, they will state the converse of "If two polygons are similar, their corresponding angles are
congruent"and give a counterexample to show that the converse of this statement is false. Students will
be introduced to the idea of proof as "a logical argument that explains why a statement must be true"
and they will be expected to provide logical arguments that justify facts such as "angles of an equilateral
triangle must be equal" and "a quadrilateral must be a rhombus if its diagonals are perpendicular
bisectors of each other."
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+0
# Solve for y
0
5
1
+68
Let
$$f(x) = \sqrt{x - \sqrt{x - \sqrt{x - \sqrt{x - \dotsb}}}}$$
Find the largest three-digit value of such that is an integer.
Letting y=f(x), we get that y(y+1)=x, and y(y+1) is a three-digit number.
Solve for y
Apr 16, 2024
#1
+179
0
Analyzing the function:
This function defines f(x) as the square root of x minus a nested sequence of square roots that keep getting smaller. Intuitively, as x decreases, the nested square roots will also decrease, bringing f(x) closer to an integer.
Finding a bound:
For f(x) to be an integer, x - sqrt(x - sqrt(x - ...)) must be a perfect square. Let the innermost square root be y. We can rewrite the function as:
f(x) = sqrt(x - y)
Squaring both sides:
x - y = f(x)^2
Since we want the largest possible three-digit x, let's consider the smallest possible value f(x) can take. The smallest perfect square greater than 10 (a three-digit number) is 121. So, let's assume f(x) = 11.
Substituting:
x - y = 11^2 = 121
This tells us that x needs to be at least 121 more than the innermost square root (y) to be a perfect square.
Finding the largest three-digit x:
We know x must be greater than 121 + y. Since y is a square root and squares are always non-negative, the largest possible value of y for a three-digit x would be the square root of the largest three-digit perfect square, which is 9^2 = 81.
Therefore, x must be greater than 121 + 81 = 202.
Checking values:
We can now check the largest three-digit perfect squares less than 202:
196 (14^2) - When plugged into the function, it results in a non-integer value.
169 (13^2) - This value works! f(169) = 13, which is an integer.
Therefore, the largest three-digit value of x such that f(x) is an integer is x = 169.
Apr 16, 2024 |
### Algebra problem: two speeds
Photo courtesy of Ben Cooper
The speed of a freight train is 14km/h slower than the speed of a passenger train. The freight train travels 330 km, in the same time that it takes a passenger train to travel 360 km. Find the speed of each train.
Again, an algebra problem about speeds. Again, we will make a simple table about the two trains. The table will have columns for speed, distance, and time.
| distance | speed | time
-----------------------------------------------
Freight train | 330 | v |
-------------------------------------------------
Passenger train| 360 | v + 14 |
Notice the problem says "in the same time". Let's call that t.
| distance | speed | time
-----------------------------------------------
Freight train | 330 | v | t
-------------------------------------------------
Passenger train| 360 | v + 14 | t
Of course, the goal is to have an equation in a single variable, not in two variables (t and v).
Since the time is the same, we can build an equation of the time t = ... for both trains, and then set those expressions to be equal.
For the freight train, t = distance/speed = 330/v
For the passenger train, t = 360/(v + 14)
Now let's make those equal:
330/v = 360/(v + 14)
Cross multiply:
360v = 330(v + 14)
360v = 330v + 4620
30v = 4620
v = 154
So, the speed of the freight train is 154 km/h and the speed of the passenger train is then 168 km/h.
Check: How long will it take for the freight train to travel 330 km? Well, 330/154 hours, or 2.142857143 hours.
How long will it take for the passenger train to travel 360 km? Well, 360/168 hours, or 2.142857143 hours. Seems to match. |
# 2022 AIME I Problems/Problem 13
## Problem
Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$
## Solution 1
$0.\overline{abcd}=\frac{abcd}{9999} = \frac{x}{y}$, $9999=9\times 11\times 101$.
Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \leq {x}\cdot\frac{9999}{y} \leq 9999$.
Make cases by factors of $x$. (A venn diagram of cases would be nice here.)
Case $A$:
$3 \nmid x$ and $11 \nmid x$ and $101 \nmid x$, aka $\gcd (9999, x)=1$.
Euler's totient function counts these: $$\varphi \left(3^2 \cdot 11 \cdot 101 \right) = ((3-1)\cdot 3)(11-1)(101-1)= \bf{6000}$$ values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer)
Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of $9999$. This case isn't actually different.
The remaining cases have $3$ (or $9$), $11$, and/or $101$ as factors of $abcd$, which cancel out part of $9999$. Note: Take care about when to use $3$ vs $9$.
Case $B$: $3|x$, but $11 \nmid x$ and $101 \nmid x$.
Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$, so $x \leq \frac{9999}{9} = 1111$, giving:
$x \in 3 \cdot \{1, \dots \left\lfloor \frac{1111}{3}\right\rfloor\}$,
$x \notin (3\cdot 11) \cdot \{1 \dots \left\lfloor \frac{1111}{3\cdot 11}\right\rfloor\}$,
$x \notin (3 \cdot 101) \cdot \{1 \dots \left\lfloor \frac{1111}{3 \cdot 101}\right\rfloor\}$,
for a subtotal of $\left\lfloor \frac{1111}{3}\right\rfloor - (\left\lfloor\frac{1111}{3 \cdot 11}\right\rfloor + \left\lfloor\frac{1111}{3 \cdot 101}\right\rfloor ) = 370 - (33+3) = \bf{334}$ values.
Case $C$: $11|x$, but $3 \nmid x$ and $101 \nmid x$.
Much like previous case, $abcd$ is $11x$, so $x \leq \frac{9999}{11} = 909$,
giving $\left\lfloor \frac{909}{11}\right\rfloor - \left(\left\lfloor\frac{909}{11 \cdot 3}\right\rfloor + \left\lfloor\frac{909}{11 \cdot 101}\right\rfloor \right) = 82 - (27 + 0) = \bf{55}$ values.
Case $D$: $3|x$ and $11|x$ (so $33|x$), but $101 \nmid x$.
Here, $abcd$ is $99x$, so $x \leq \frac{9999}{99} = 101$,
giving $\left\lfloor \frac{101}{33}\right\rfloor - \left\lfloor \frac{101}{33 \cdot 101}\right\rfloor = 3-0 = \bf{3}$ values.
Case $E$: $101|x$.
Here, $abcd$ is $101x$, so $x \leq \frac{9999}{101} = 99$,
giving $\left\lfloor \frac{99}{101}\right\rfloor = \bf{0}$ values, so we don't need to account for multiples of $3$ and $11$.
To sum up, the answer is $$6000+334+55+3+0\equiv\boxed{392} \pmod{1000}.$$
### Clarification
In this context, when the solution says, "Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$," it is a bit vague. The best way to clarify this is by this exact example - what is really meant is we need to divide by 9 first to achieve 1111, which has no multiple of 3; thus, given that the fraction x/y is the simplest form, x can be a multiple of 3.
Similar explanations can be said when the solution divides 9999 by 11, 101, and uses that divided result in the PIE calculation rather than 9999.
mathboy282
## Solution 2 (similar(?) to solution 1 but reworded (not exactly for clarity))
$$\text{To begin, we notice that all repeating decimals of the form }0.\overline{abcd}\text{ where }a,b,c,d\text{ are digits can be expressed of the form }\frac{\overline{abcd}}{9999}\text{.}$$ $$\text{However, when }\overline{abcd}\mid 9999\text{, the fraction is not in lowest terms.}$$ $$\text{Since }9999 = 3^2 \cdot 11 \cdot 101\text{, } x\mid 9999\iff x\mid 3\lor x\mid 11\lor x\mid 101\text{.}$$ $$\text{(For those of you who have no idea what that meant, it means every divisor of 9999 is a divisor of at least one of the following: )}$$ $$(3)$$ $$(11)$$ $$(101)$$ $$\text{(Also, I'm not going to give you explanations for the other logic equations.)}$$ $$\text{Let's say that the fraction in lowest terms is }\frac{x}{y}\text{.}$$ $$\text{If }x\mid 101\text{, then }99\mid y\text{ but that can't be, since }0\text{ is the only multiple of }101\text{ below }99\text{.}$$ $$\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\implies f=3\lor f=11 (1)$$ $$\text{If (1) is true, then we have two cases. If it isn't, we also have two cases.}$$ $$\textbf{\textit{Case 1: }}f=3$$ $$y=1111\land x=3z\implies 1\leq z\leq 370$$ $$370-33-3^{[1]}=334$$ $$\textbf{\textit{Case 2: }}f=11$$ $$y=909\land x=11z\implies 1\leq z\leq 82$$ $$82-27=55$$ $$\textbf{\textit{Case 3: }}\neg\exists! f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))\land \exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))=$$ $$\exists f_1(f_1\in\mathbb{N}\land f_1\neq 1\land\exists g_1(g_1\nleq 0 \land x \mid f_1^{g_1})\land\exists f_2(f_2\neq f_1\land f_2\in\mathbb{N}\land f_2\neq 1\land\exists g_2(g_2\nleq 0 \land x \mid f_2^{g_2}))\implies f_1=3\land f_2=11\lor f_1=11\land f_2=3$$ $$y=101\land x=33z\implies 1\leq z\leq 3$$ $$\textbf{\textit{Case 4: }}\neg\exists f(f\in\mathbb{N}\land f\neq 1\land\exists g(g\nleq 0 \land x \mid f^g))$$ $$\Phi (9999)=6000$$ $$\textbf{\textit{Grand Finale}}$$ $$\text{Adding the outcomes, }N=6000+334+55+3=6392\equiv\boxed{392}\text{ (mod 1000).}$$
$$\textit{[1] This is to make sure that 3 is the \textbf{only} factor of x}$$
### Note
$$\text{When I tried to write LaTeX, AoPS kept putting the LaTeX on a new line so I gave up and put most of it in LaTeX instead.}$$ $$\text{Some of the text in this section is just normal.}$$ $$\text{Example:}$$
Normal text $$\text{This is some LaTeX.}$$ More normal text
$$\text{If any of you can fix this issue, please do so.}$$
~ Afly (talk) |
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Get Full Access to Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) - 3 Edition - Chapter 3 - Problem 3.7.75
Get Full Access to Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) - 3 Edition - Chapter 3 - Problem 3.7.75
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# determine whether a graph with the given adjacency matrix is bipartite. The adjacency
ISBN: 9780538735452 298
## Solution for problem 3.7.75 Chapter 3
Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) | 3rd Edition
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Problem 3.7.75
determine whether a graph with the given adjacency matrix is bipartite. The adjacency matrix in Exercise 50
Step-by-Step Solution:
Step 1 of 3
Calculus I Chapters 1.1 and 1.2 Chapter 1.1 – An Introduction to Calculus Formulas to remember 2 2 o ( − + = − o ( + )2 = + 2 + 2 2 2 2 o ( − ) = − 2 − −± −4 o = 2 1.1 Preview of Calculus Calculus is the mathematics of change Pre-Cal Cal Δ Slope = Δ Area Of rectangle = ℎ 2) the Tangent Line Problem Finding a tangent line to a curve @ point P is equivalent to finding the slope of the tangent line at P. - We will use the secant line: a line through the point P of the tangent and a point Q also on the curve. ( ) 2 For = = 1 = 1 2 1.1 = 1.1 = 1.21 1(1.1, 1.1 ) = 1.1,1.21 ) (1.5 , 1.5 ) = (1.5,2.25) 2 3(1.01, 1.01 ) = 1.01,1.0201 ) (1.001, 1.001 ) = 1.001,1.002 ) 4 Slope of P, Q + ∆ − ) + ∆ − ) = = ( + ∆ − ∆ = 1 ∆ = 1 − 1.1 = 0.1 1 + 0.1 − 1) 1 = 0.1 1.1 −
Step 2 of 3
Step 3 of 3
##### ISBN: 9780538735452
Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign) was written by and is associated to the ISBN: 9780538735452. The answer to “determine whether a graph with the given adjacency matrix is bipartite. The adjacency matrix in Exercise 50” is broken down into a number of easy to follow steps, and 17 words. Since the solution to 3.7.75 from 3 chapter was answered, more than 291 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Linear Algebra: A Modern Introduction (Available 2011 Titles Enhanced Web Assign), edition: 3. This full solution covers the following key subjects: . This expansive textbook survival guide covers 7 chapters, and 1985 solutions. The full step-by-step solution to problem: 3.7.75 from chapter: 3 was answered by , our top Math solution expert on 01/29/18, 04:03PM.
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?In Exercises 7 - 20, examine the function for relative extrema and saddle points. $$f(x, y)=x^{2}-x y-y^{2}-3 x-y$$
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# Question Video: Using the Binomial Theorem Mathematics
Use the binomial theorem to find the expansion of (π β π)β΅.
04:06
### Video Transcript
Use the binomial theorem to find the expansion of π minus π to the fifth power.
π minus π is a binomial expression. Itβs an algebraic expression made up of two terms. Itβs raised to the fifth power. And weβre told weβre going to need to use the binomial theorem to find its expansion. And so, we recall that the binomial theorem says that for positive integers π, π₯ plus π¦ to the πth power is equal to the sum from π equals zero to π of π choose π times π₯ to the power of π minus π times π¦ to the πth power.
Now, this expression can be quite nasty to work with, so we might consider its expanded form. This is π₯ to the πth power plus π choose one times π₯ to the power of π minus one times π¦ plus π choose two times π₯ to the power of π minus two times π¦ squared all the way through to π¦ to the πth power. Notice how the powers of π₯ reduce by one each time and the powers of π¦ increase.
Letβs compare this formula to our binomial. We see that we can let π₯ be equal to π, π¦ is equal to negative π, and π, the exponent, is equal to five. This means the first term in our expansion is simply π to the fifth power. The second term is π choose one, so five choose one, times π to the power of five minus one or π to the fourth power times negative π. Remember, the powers of π₯, or the powers of π here, decrease by one each time, whereas the powers of π¦, which is negative π, increase by one each time. So, our third term is five choose two π cubed times negative π squared.
Next, we have five choose three π squared times negative π cubed. Our fifth term is five choose four π times negative π to the fourth power. And our final term is negative π to the fifth power. Weβre going to evaluate five choose one, five choose two, five choose three, and five choose four. And so, we recall the formula to help us evaluate π choose π. Itβs π factorial over π factorial times π minus π factorial. This means five choose one is five factorial over one factorial times five minus one factorial or five factorial over one factorial times four factorial.
Next, we recall that five factorial is five times four times three times two times one. Similarly, four factorial is four times three times two times one. And we see that we can divide through by four, three, two, and one. In fact, what weβre really doing is dividing through by four factorial. And so, we find five choose one is simply five divided by one, which is five. In much the same way, five choose four also yields a result of five.
Letβs evaluate five choose two. Itβs five factorial over two factorial times five minus two factorial. Thatβs five factorial over two factorial times three factorial. Letβs write five factorial this time instead of as five times four times three times two times one as five times four times three factorial. Then, we see we can divide through by three factorial. By writing two as two times one, we can see we can divide through by two. And five choose two is, therefore, five times two divided by one, which is 10. Five choose three is also 10. And so, weβre ready to find the expansion.
Our first term is still π to the fifth power. Our next term is five π to the fourth power times negative π. So, thatβs negative five π to the fourth power π. Now, negative π squared is positive π squared. So, our third term is 10π cubed π squared. When we cube a negative number, we get a negative result. So, our fourth term is negative 10π squared π cubed. We then have five ππ to the fourth power. And when we raise a negative number to the fifth power, we get a negative result. So, our final term is π to the fifth power.
And so, weβre done. Weβve used the binomial theorem to find the expansion of π minus π to the fifth power. Itβs π to the fifth power minus five π to the fourth power π plus 10π cubed π squared minus 10π squared π cubed plus five ππ to the fourth power minus π to the fifth power. |
## Prime or Not: Determining Primes Through Square Root
A prime number is a integer greater than that is divisible only by 1 and itself. A number that is not prime is composite.
To determine whether a number is prime or not, we have to divide it by all numbers between 1 and itself . For example, to say that 257 is prime, we must be sure that it is not divisible by any number between 1 and 257. In this discussion, the word “numbers” refer to positive integers.
Are you prime or not?
Dividing a number by all numbers between 1 and itself is burdensome especially for large numbers. In this post, we discuss a shorter way of determining if a number is prime and explain why the method works. » Read more
## Prime Series 1: Introduction to Prime Numbers
We have learned from elementary school mathematics that a prime number has only two factors, 1 and itself. For example, 2, 3, 5 and 7 are prime numbers, while 8 is not prime because it has four factors — 1, 2, 4, and 8. Numbers that are not prime are called composite numbers.
Geometric Interpretation of Prime and Composite Numbers
Mathematicians during the ancient times, particularly the Greeks, always make use of geometric interpretations of numbers. Square numbers, for example, are represented with pebbles arranged with the same number of rows and columns. The first five square numbers are 1, 4, 9, 16 and 25.
Rectangular numbers are also popular. For instance, the number 12 can be interpreted pebbles arranged in rectangles with the following dimensions: 3 by 4, 2 by 6 and 12 by 1. If we are going to use squares instead of pebbles, the geometric representations of these arrangements are shown in Figure 1.
Figure 1 – Different rectangular arrangements of 12 pebbles represented by squares.
Numbers that cannot be arranged as more than one rectangle are prime numbers. In our example above, 12 has three possible arrangements, while the numbers 3, 5 and 7 can only be arranged in a single row.
Figure 2 – Rectangular arrangements of 3, 5 and 7 pebbles represented by squares.
As we can see, this is the geometric interpretation of the definition that the factors of primes are only 1 and itself.
Sieve of Eratosthenes
The Greek philosopher and mathematician Eratosthenes was the first to be credited in identifying primes in a finite list by brute force. The strategy is to list a finite set of counting numbers in increasing order, then starting with 2 eliminate all its multiples. This eliminates all even numbers except 2. Then we follow the same pattern: we eliminate multiples of 3 greater than 3, eliminate all multiples of 5 greater than 5 and so on, until all the numbers left are not multiples of any number smaller than it. The remaining numbers after all elimination are prime numbers.
The primes numbers less than 100 are shown in white cells in the table below. The numbers in the yellow cells are composite numbers. Mathematicians agreed not to include 1 in the set of prime numbers or the set of composite numbers.
Figure 3 - The sieve of Eratosthenes shows primes less than 100 in white table cells..
If we look at the table above, we could not probably see a pattern about the number of primes in a given interval; however, if we investigate further, as the intervals increase, the number of primes is getting fewer and fewer. For example, there are 168 primes between 1 and 1000, 135 primes between 1000 and 2000, 127 primes between 2000 and 3000, and 120 primes between 3000 and 4000. With this observation, we want to ask the following question:
Is there a particular prime number, that after such number, we could no longer find primes?
or equivalently,
Are prime numbers finite?
We will answer this in the continuation of this article titled “Infinitude of Primes.” The formal proof of this conjecture is also discussed in the third part. |
# Pythagorean Identities
How to solve the Pythagorean identities problems: identities, proof, examples, and their solutions.
## Identities
sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
1 + cot2 θ = csc2 θ
These two are the modified identities.
## Proof
Draw a right triangle like this.
Write the central angle as θ.
By the Pythagorean theorem,
x2 + y2 = r2.
Divide both sides by r2:
(x/r)2 + (y/r)2 = 1.
Sine: SOH.
So sin θ = y/r.
Cosine: CAH.
So cos θ = x/r.
Then cos2 θ + sin2 θ = 1.
So sin2 θ + cos2 θ = 1.
Start from sin2 θ + cos2 θ = 1.
Divide both sides by cos2 θ:
sin2 θ/cos2 θ + 1 = 1/cos2 θ.
sin θ/cos θ = tan θ
Quotient identities
And 1/cos θ = sec θ.
Trigonometric ratio - secant
So tan2 θ + 1 = sec2 θ.
Start from sin2 θ + cos2 θ = 1.
Divide both sides by sin2 θ:
1 + cos2 θ/sin2 θ = 1/sin2 θ.
1/sin θ = csc θ.
Trigonometric ratio - cosecant
And cos θ/sin θ = cot θ.
Quotient identities
So 1 + cot2 θ = csc2 θ.
## Example 1
(sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos2 θ
Square of a sum
sin2 θ + cos2 θ = 1
Cosecant is the reciprocal of sine.
So 1/sin θ = csc θ.
And cancel sin θ
on both of the numerator and the denominator.
## Example 2
sin2 θ + cos2 θ = 1
So cos2 θ = 1 - sin2 θ.
## Example 3
Change 1 + tan2 θ to sec2 θ.
sec2 θ = 1/cos2 θ.
Trigonometric ratio - secant
## Example 4
sin θ = -3/5
So (-3/5)2 + sin2 θ = 1.
Then cos2 θ = 16/25.
Draw the axes of the coordinate plane
and write 'all, sin, tan, cos'
like this.
This shows when the trigonometric function is (+):
tangent is (+),
and sine and cosine are (-).
πθ ≦ 3π/2
So θ is in quadrant III.
For quadrant III, only tan θ is (+).
So cos θ is (-).
So cos θ = -√16/25.
## Example 4: Another Solution
Let's see how to solve this example differently.
πθ ≦ 3π/2
So θ is in quadrant III.
sin θ = -3/5
And sine: SOH.
So draw a right triangle in quadrant III
whose opposite side is -3
and whose hypotenuse is 5.
This right triangle is a (3, 4, 5) right triangle.
So (blue side) = -4.
Pythagorean triples
The brown angle is the reference angle of ∠θ.
And cosine: CAH.
So cos θ = cos (brown)
= -4/5.
You can see that |
# Chapter 1 Rational Numbers
Exercise 1.1
Ex 1.1 Class 8 Maths Question 1.
Using appropriate properties find:
Solution:
Ex 1.1
Question 2.
Write the additive inverse of each of the following:
(i)
(ii)
(iii)
(iv)
(v)
Solution:
Question 3.
Verify that -(-x) = x for
(i) x =
(ii) x =
Solution:
Question 4.
Find the multiplicative inverse of the following:
Solution:
Ex 1.1 Class 8 Maths Question 5.
Name the property under multiplication used in each of the following:
Solution:
(i) Commutative property of multiplication
(ii) Commutative property of multiplication
(iii) Multiplicative inverse property
Ex 1.1 Class 8 Maths
Question 6.
Multiply
by the reciprocal of .
Solution:
Ex 1.1 Class 8 Maths
Question 7.
Tell what property allows you to compute
Solution:
Since a × (b × c) = (a × b) × c shows the associative property of multiplications.
Ex 1.1 Class 8 Maths
Question 8.
Is the multiplicative inverse of -1 ? Why or Why not?
Solution:
Here -1
= .
Since multiplicative inverse of
is but not
is not the multiplicative inverse of -1
Ex 1.1 Class 8 Maths
Question 9.
If 0.3 the multiplicative inverse of 3
? Why or why not?
Solution:
Multiplicative inverse of 0.3 or is .
Thus, 0.3 is the multiplicative inverse of 3 .
Ex 1.1 Class 8 Maths
Question 10.
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
Solution:
(i) 0 is the rational number which does not have its reciprocal
[
is not defined]
(ii) Reciprocal of 1 = = 1
Reciprocal of -1 = = -1
Thus, 1 and -1 are the required rational numbers.
(iii) 0 is the rational number which is equal to its negative.
Ex 1.1 Class 8 Maths
Question 11.
Fill in the blanks.
(i) Zero has ……….. reciprocal.
(ii) The numbers ……….. and ……….. are their own reciprocals.
(iii) The reciprocal of -5 is ………
(iv) Reciprocal of , where x ≠ 0 is ……….
(v) The product of two rational numbers is always a …………
(vi) The reciprocal of a positive rational number is ……….
Solution:
(i) no
(ii) -1 and 1
(iii)
(iv) x
(v) rational number
(vi) positive |
# Area calculation.
## Presentation on theme: "Area calculation."— Presentation transcript:
Area calculation
Area is divided into triangles, rectangles, squares or trapeziums
Area of the one figure (e.g. triangles, rectangles, squares or trapeziums) is calculated and multiplied by total number of figures. Area along the boundaries is calculated as Total area of the filed=area of geometrical figure boundary areas
Problem-1
Problem 1-Result
Computation of area from plotted plan
Boundary area can be calculated as one of the following rule: The mid-ordinate rule The average ordinate rule The trapezoidal rule Simpson’s rule
Mid-ordinate rule l
Average ordinate rule
Trapezoidal rule
Simpson’s rule
Problems The following perpendicular offsets were taken from chain line to an irregular boundary: Chainage m offset Calculate the area between the chain line, the boundary and the end offsets. The following perpendicular offsets were taken from a chain line to a hedge: Calculate the area by mid ordinate and Simpson’s rule. chainage 15 30 45 60 70 80 100 120 140 offsets 7.6 8.5 10.7 12.8 10.6 9.5 8.3 7.9 6.4 4.4
Area by double meridian distances
Meridian distance of any point in a traverse is the distance of that point to the reference meridian, measured at right angle to the meridian. The meridian distance of a survey line is defined as the meridian distance of its mid point. The meridian distance sometimes called as the longitude.
d4/2 d4/2 d3/2 d3/2 D d m4 m3 4 3 Mid points Meridian distance of points (d1, d2, d3, d4) A c C 1 B 2 b d1/2 d1/2 d2/2 d2/2 m1 m2
Meridian distances of survey line:
m1=d1/2 m2= m1+d1/2+d2/2 m3=m2+d2/2-d3/2 m4=m3-d3/2-d4/2 Area by latitude and meridian distance Area of ABCD=area of trapezium CcdD + area of trapezium CcbB – area of triangle AbB – area of triangle AdD = m3*L3+ m2*L2-1/2*2*m4*L4-1/2*2*m1*L1 =m3*L3+m2*L2-m4*L4-m1*L1
Double meridian distance:
M1= meridian distance of point A + meridian distance of point B M1=0+d1 M2=meridian distance of point B + meridian distance of point C =d1+(d1+d2) =M1+(d1+d2) M3=(d1+d2)+(d1+d2-d3)
Area of the traverse ABCD = M3*L3+M2*L2-M1*L1-M4*L4
Area by Co-Ordinates
The following table gives the corrected latitudes and departures (in m) of the sides of a closed traverse ABCD. Compute the area by (a) M.D. method (b) co-ordinate method Side Latitude Departure N S E W AB 108 4 BC 15 249 CD 123 DA 257
Volume calculation From cross sections From spot levels From contours
Measurement of volume 3 methods generally adopted for measuring the volume are (i) from cross sections (ii) from spot levels (iii) from contours
Methods of volume calculation
Prismoidal method D2 A2 C2 B2 D2 D1 A2 A1 C1 C2 B1 B2 D1 A1 B1 C1
Also called Simpson’s rule for volume.
Necessary to have odd number of cross sections. What if there are even number of C/S? Trapezoidal method: Assumption mid area is mean of end areas. Volume =d{(A1+An)/2+A2+A3+…+An-1}
A railway embankment is 10 m wide with side slopes 1. 5:1
A railway embankment is 10 m wide with side slopes 1.5:1. assuming the ground to be level in a direction transverse to the centre line, calculate the volume contained in a length of 120 m, the centre heights at 20 m intervals being in metres 2.2, 3.7, 3.8, 4.0, 3.8, 2.8, 2.5. A railway embankment 400 m long is 12 m wide at the formation level and has the side slope 2:1. The ground levels across the centre line are as under: The formation level at zero chainage is and the embankment has a rising gradient of 1:100. The ground is level across the centre line. Calculate the volume of earthwork. Distance 100 200 300 400 R.L. 204.8 206.2 207.5 207.2 208.3 |
# A Fun Lesson for Teaching Proper Fractions
By Margo Dill
Here is a fun food art activity that is also a lesson in proper fractions and mixed numbers. You can let students create sugar cookie designs and then ask them fraction questions based on their designs. It's a lesson that will change each time you do it, based on students' creativity.
## Materials
Before you start this fun lesson to teach proper and mixed fractions, you will need to bake or buy plain sugar cookies. icing, sprinkles and anything else you want to decorate cookies with. To cut down on expense, send home a materials list at least a week before the activity and ask parents to volunteer to send in materials. You want at least three different colors of icing and sprinkles for the teaching proper fractions part of the lesson. You need one sugar cookie for each student in your class.
## Introduction
This lesson on proper and mixed fractions is perfect for students who know what a fraction looks like and have been introduced to the terms: numerator and denominator.
Students start by decorating their sugar cookie with at least one color of icing of their choice and one color of sprinkles. While students are decorating, you can remind them of what you have learned about fractions so far.
## The Lesson
Place the decorated cookies somewhere in the room where they are easy to see. You may have to do this fun lesson plan to teach proper and mixed fractions in small groups, so students can see the cookies easily. First, tell students that a proper fraction is one where the numerator is less than the denominator, such as 5/8. Then with students, figure out how many cookies there are and what your denominator will be to answer each question. For example, if there are 24 kids in your class, there will be 24 cookies and your denominator will be 24.
Then ask students questions such as: "What fraction of the sugar cookies have blue icing on them?" Students will count the ones with blue icing and say for example, "8." Show students how the 8 becomes the numerator, so you have the fraction 8/24. Continue to ask questions until students understand creating fractions.
For variation with the denominator, you can ask questions like: "Out of the cookies the girls made, what fraction have red sprinkles?" "The front row's cookies have what fraction of yellow icing?"
Make sure to write the proper fractions on chart paper or the board as students say them.
## Variation: Practicing Mixed Numbers
Proper and mixed fractions for mixed numbers may be addressed. However, they may be to advanced for many students. Ask students to get in pairs and decorate 3 to 4 cookies. You tell them what to decorate in this variation of the lesson. For example, you'll say, "Make 1 1/2 of your cookies red." "Put sprinkles on 2 1/3 of your cookies."
## Conclusion of Lesson
To conclude either variation, ask questions about objects in the room for students to answer orally like: "What fraction of backpacks are gray?" or "What fraction of the desks are in the front row?" You can even allow students to pose the questions and allow other students to answer. As students practice naming fractions and what the terms numerator and denominator mean, they will become more familiar with fractions and be able to tackle harder concepts.
## Resources
Math Is Fun: http://www.mathsisfun.com/proper-fractions.html |
# Dividing Using Number Lines
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## Objective
SWBAT to model and explain division using a number line.
#### Big Idea
Using a number line for division makes sense, once students see that they can represent any number on the line because they can add as many equal "parts" as they need to solve.
## Warm-Up
10 minutes
To begin this lesson, students review multiplication strategies with area models. Students are given the task of picking numbers between 10 and 20 to determine an area to represent a space - classroom, library, or the front office of the school. Working with a partner on whiteboards, students draw the area, label the sides, and then use the distributive property to find the total area. For example, if the students use the number 15 x 18 they can use the property to find the area with:
(15 x 10) + (15 x 8) or (15 x 10) + (15 x 3) + (15 x 5)
Their diagram is divided to show these areas. Once the students have successfully created their area model they record their process in their math journal. I want them to explain how decomposing the multiplier or multiplicand by creating smaller rectangles makes it easier to find larger area measurements.
## Mini-Lesson
10 minutes
Using a number line for division is an important aspect of utilizing the scope of Common Core strategies such as the relationship between multiplication and division or properties of operations (3.OA.7). A number line is a critical mathematical tool. In this lesson, it helps students understand the structure of division sentences to identify if the unknown factor represents the groups or the number of items in a group.
I begin with presenting the students with a division number sentence that they cannot calculate in their head. Because my students are beginning to show mastery of both division and multiplication facts, I choose numbers that are not easily calculated to keep the students engaged, and also to create a sense of purpose for the use of the number line to solve division problems.
It is also important to place the numbers chosen in a meaningful context for the students, and I chose books for this lesson. The library has received 108 books to be used by the third grade students. Each student will receive nine books for summer reading. How many students will receive books?
Counting the number of jumps determines the unknown number of groups which is represented by the number of students receiving books.
I repeat another example with the students, again using a facts that are not memorized by the students.
## Try It On Your Own
20 minutes
Students will be using number lines to solve division problems on their own. The context for today's lesson is established by different materials placed in stations around the room. I gathered different items including craft sticks, plastic counters, paper clips, beads, stickers, and rubber bands. I did not take the time to count out each group of items but rather just assigned an estimated quantity for each items. Working in groups, the students move to a station with one of the items and use number lines to divide the quantities by 6 and by 3 to find the number of groups. I chose numbers for the groups to be multiples of 6 and 3 above 100 based on my students' current knowledge. This can be modified to meet the abilities of any student.
It is important that students are writing their sentence structure using the format of:
whole amount ÷ items = groups
This structure is important because items can not be broken into separate pieces as students begin to encounter remainders with division in future grades. The remainders represent partial groups rather than partial items.
Students are given time at each station, and rotate on a given signal. Students record the item, the number line, and the number sentence. I kept the pace quick at the stations, and moved students every 3-4 minutes.
## Closure
5 minutes
To end the lesson I ask the students to write the number line that is represented by how they sit in the classroom. My room is set up into six different groups with four students. I ask them to create the number line to show:
how many students are in the class ÷ number of students at each group = six groups
or 24 ÷ 4 = 6 |
## Math Expressions Common Core Grade 3 Unit 3 Lesson 9 Answer Key Add and Subtract Time
Math Expressions Grade 3 Unit 3 Lesson 9 Homework
Solve using a number line.
Terry began watching a movie at 5:45 P.M. The movie lasted 2 hours 20 minutes. Then Terry spent 25 minutes eating a snack. What time did Terry finish eating the snack?
Time did Terry finish eating the snack = 8:30 P.M.
Explanation:
Time at Terry began watching a movie = 5:45 P.M.
Time till the movie lasted = 2 hours 20 minutes.
Time Terry spent to eating a snack = 25 minutes.
Time did Terry finish eating the snack = Time at Terry began watching a movie + Time till the movie lasted + Time Terry spent to eating a snack
= 5:45 + 2:20 + 0:25
= 8:05 + 0:25
= 8:30 P.M.
Evan left his friend’s house at 5:00 P.M. He had been there 2 hours 15 minutes. At what time did Evan arrive at his friend’s house?
Time at Evan arrive at his friend’s house = 2:45 P.M.
Explanation:
Time at Evan left his friend’s house = 5:00 P.M.
Time for he been there = 2 hours 15 minutes.
Time at Evan arrive at his friend’s house = Time at Evan left his friend’s house – Time for he been there
= 5:00 – 2:15
= 2:45 P.M.
Haley began reading her book at 9:55 A.M. She read for 1 hour 35 minutes. Then she spent 45 minutes doing homework. What time did Haley finish her homework?
Time she finished her homework = 12: 15 P.M.
Explanation:
Time at Haley began reading her book = 9:55 A.M.
Time for She read = 1 hour 35 minutes.
Time she spent doing her homework = 45 minutes.
Time she finished her homework = Time at Haley began reading her book + Time for She read + Time she spent doing her homework
= 9:55 + 1:35 + 0:45
= 11:30 + 0:45
= 12: 15 P.M.
Lesson 9 Answer Key Math Expressions Grade 3 Unit 3 Question 4.
Myra left home at 12:45 P.M. She spent 30 minutes eating lunch and 50 minutes watching a parade. Then it took her 15 minutes to drive home. What time did Myra return home?
Time at she returned home = 2:05 P.M.
Explanation:
Time at Myra left home =12:45 P.M.
Time she spent for eating lunch = 30 minutes.
Time she spent for watching a parade = 50 minutes.
Time at she returned home = Time at Myra left home + Time she spent for eating lunch + Time she spent for watching a parade
= 12:45 + 0:30 + 0:50
= 1:15 + 0:50
= 2:05 P.M.
Math Expressions Grade 3 Unit 3 Lesson 9 Remembering
Make a rectangle drawing to represent each exercise. Then find the product.
Unit 3 Lesson 9 Math Expressions Grade 3 Answer Key Question 1.
6 × 9 = ___
6 × 9 = 54.
Explanation:
Length of rectangle = 6 inch.
Width of rectangle = 9 inch.
Area of rectangle = Length of rectangle × Width of rectangle
= 6 × 9
= 54 square inches.
Unit 3 Lesson 9 Answer Key Grade 3 Math Expressions Question 2.
7 * 5 = ___
7 * 5 = 35.
Explanation:
Length of rectangle = 7 inch.
Width of rectangle = 5 inch.
Area of rectangle = Length of rectangle × Width of rectangle
= 7 × 5
= 35 square inches.
Question 3.
3 • 6 = ___
3 • 6 = 18.
Explanation:
Length of rectangle = 6 inch.
Width of rectangle = 3 inch.
Area of rectangle = Length of rectangle × Width of rectangle
= 6 × 3
= 18 square inches.
Write the first step question and answer. Then solve the problem.
Question 4.
The baker makes 54 biscuits in the morning. Then he makes 26 more in the afternoon. He puts 10 biscuits in each bag. How many bags does he fill?
Number of bags he filled = 8.
Explanation:
Number of biscuits in the morning the baker makes = 54.
Number of more biscuits in the afternoon the baker makes = 26.
Number of biscuits in each bag he puts = 10.
Number of bags he filled = (Number of biscuits in the morning the baker makes + Number of more biscuits in the afternoon the baker makes ) ÷ Number of biscuits in each bag he puts
= (54 + 26) ÷ 10
= 80 ÷ 10
= 8.
Question 5.
Complete the table.
Explanation:
Start time + Elapsed time = End time
9:32 A.M. + 1 hour 23 minutes = 10:55 A.M.
1:19 P.M + 4 hour 18 minutes = 5:37 P.M.
4:46 P.M + 2 hour 45 minutes = 7:31 P.M.
Question 6.
Stretch Your Thinking Write a two step time word problem using the number line in which the start time is 4:50. Use the number line below to show how to solve. |
# SEQUENCES, MATHEMATICAL INDUCTION, AND .Mathematical Induction II In natural science courses, deduction
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### Text of SEQUENCES, MATHEMATICAL INDUCTION, AND .Mathematical Induction II In natural science courses,...
CHAPTER 5
SEQUENCES,
MATHEMATICAL
INDUCTION, AND
RECURSION
SEQUENCES,
MATHEMATICAL
INDUCTION, AND
RECURSION
Mathematical Induction II
SECTION 5.3
• 3
Mathematical Induction II
In natural science courses, deduction and induction are
presented as alternative modes of thoughtdeduction
being to infer a conclusion from general principles using the
laws of logical reasoning, and induction being to enunciate
a general principle after observing it to hold in a large
number of specific instances.
In this sense, then, mathematical induction is not inductive
but deductive.
Once proved by mathematical induction, a theorem is
known just as certainly as if it were proved by any other mathematical method.
• 4
Mathematical Induction II
Inductive reasoning, in the natural sciences sense, is used
in mathematics, but only to make conjectures, not to prove
them.
For example, observe that
• 5
Mathematical Induction II
then by substitution
Thus mathematical induction makes knowledge of the
general pattern a matter of mathematical certainty rather than vague conjecture.
• 6
Example 1 Proving a Divisibility Property
Use mathematical induction to prove that for all integers
n 0, 22n 1 is divisible by 3.
Solution:
As in the previous proofs by mathematical induction, you
need to identify the property P(n).
In this example, P(n) is the sentence
• 7
Example 1 Solution
By substitution, the statement for the basis step, P(0), is
The supposition for the inductive step, P(k), is
and the conclusion to be shown, P(k + 1), is
contd
• 8
Example 1 Solution
We know that an integer m is divisible by 3 if, and only if,
m = 3r for some integer r.
Now the statement P(0) is true because
22 0 1 = 20 1 = 1 1 = 0, which is divisible by 3
because 0 = 3 0.
To prove the inductive step, you suppose that k is any
integer greater than or equal to 0 such that P(k) is true.
This means that 22k 1 is divisible by 3. You must then
prove the truth of P(k + 1). Or, in other words, you must show that 22(k+1) 1 is divisible by 3.
contd
• 9
Example 1 Solution
But
The aim is to show that this quantity, 22k 4 1, is divisible
by 3. Why should that be so? By the inductive hypothesis,
22k 1 is divisible by 3, and 22k 4 1 resembles 22k 1.
contd
• 10
Example 1 Solution
Observe what happens, if you subtract 22k 1 from
22k 4 1:
Adding 22k 1 to both sides gives
Both terms of the sum on the right-hand side of this
equation are divisible by 3; hence the sum is divisible by 3.
contd
• 11
Example 1 Solution
Therefore, the left-hand side of the equation is also
divisible by 3, which is what was to be shown.
This discussion is summarized as follows:
Proof (by mathematical induction):
Let the property P(n) be the sentence 22n 1 is divisible by 3.
contd
• 12
Example 1 Solution
Show that P(0) is true:
To establish P(0), we must show that
But
and 0 is divisible by 3 because 0 = 3 0.
Hence P(0) is true.
contd
• 13
Example 1 Solution
Show that for all integers k 0, if P(k) is true then
P(k + 1) is also true:
[Suppose that P(k) is true for a particular but arbitrarily
chosen integer k 0. That is:]
Let k be any integer with k 0, and suppose that
By definition of divisibility, this means that
contd
• 14
Example 1 Solution
[We must show that P(k + 1) is true. That is:] We must
show that
But
contd
• 15
Example 1 Solution
But is an integer because it is a sum of products of
integers, and so, by definition of divisibility, is
divisible by 3 [as was to be shown].
[Since we have proved the basis step and the inductive
step, we conclude that the proposition is true.]
contd
• 16
Mathematical Induction II
The next example illustrates the use of mathematical
induction to prove an inequality.
• 17
Example 2 Proving an Inequality
Use mathematical induction to prove that for all integers
n 3,
Solution:
In this example the property P(n) is the inequality
By substitution, the statement for the basis step, P(3), is
• 18
Example 2 Solution
The supposition for the inductive step, P(k), is
and the conclusion to be shown is
To prove the basis step, observe that the statement P(3) is
true because 2 3 + 1 = 7, 23 = 8, and 7 < 8.
contd
• 19
Example 2 Solution
To prove the inductive step, suppose the inductive
hypothesis, that P(k) is true for an integer k 3.
This means that 2k + 1 < 2k is assumed to be true for a
particular but arbitrarily chosen integer k 3.
Then derive the truth of P(k + 1). Or, in other words, show
that the inequality is true. But by
multiplying out and regrouping,
and by substitution from the inductive hypothesis,
contd
• 20
Example 2 Solution
Hence
If it can be shown that 2k + 2 is less than 2k+1, then the
desired inequality will have been proved.
But since the quantity 2k can be added to or subtracted
from an inequality without changing its direction,
And since multiplying or dividing an inequality by 2 does
not change its direction,
contd
• 21
Example 2 Solution
This last inequality is clearly true for all k 2. Hence it is
true that .
This discussion is made more flowing (but less intuitive) in
the following formal proof:
Proof (by mathematical induction):
Let the property P(n) be the inequality
contd
• 22
Example 2 Solution
Show that P(3) is true:
To establish P(3), we must show that
But
Hence P(3) is true.
contd
• 23
Example 2 Solution
Show that for all integers k 3, if P(k) is true then
P(k + 1) is also true:
[Suppose that P(k) is true for a particular but arbitrarily
chosen integer k 3. That is:]
Suppose that k is any integer with k 3 such that
[We must show that P(k + 1) is true. That is:] We must show that
contd
• 24
Example 2 Solution
Or, equivalently,
But
[This is what we needed to show.]
[Since we have proved the basis step and the inductive
step, we conclude that the proposition is true.]
contd
• 25
A Problem with Trominoes
• 26
A Problem with Trominoes
A particular type of polyomino, called a tromino, is made up
of three attached squares, which can be of two types:
Call a checkerboard that is formed using m squares on a
side an m m (m by m) checkerboard.
Observe that if one square is removed from a 4 4 checkerboard, the remaining squares can be completely
covered by L-shaped trominoes.
• 27
A Problem with Trominoes
For instance, a covering for one such board is illustrated in
the figure below.
It is a beautiful example of an argument by mathematical induction.
• 28
A Problem with Trominoes
The main insight leading to a proof of this theorem is the
observation that because , when a
board is split in half both vertically and horizontally, each half side will have length 2k and so each resulting quadrant
will be a checkerboard.
• 29
A Problem with Trominoes
Proof (by mathematical induction):
Let the property P(n) be the sentence
If any square is removed from a 2n 2n checkerboard,
then the remaining squares can be completely covered by L-shaped trominoes.
Show that P(1) is true:
A 21 21 checkerboard just consists of
four squares. If one square is removed,
the remaining squares form an L, which can be covered by a single L-shaped
tromino, as illustrated in the figure to the right. Hence P(1) is true.
• 30
A Problem with Trominoes
Show that for all integers k 1, if P(k) is true then
P(k + 1) is also true:
[Suppose that P(k) is true for a particular but arbitrarily
chosen integer k 3. That is:]
Let k be any integer such that k 1, and suppose that
If any square is removed from a 2k 2k checkerboard,
then the remaining squares can be completely covered by L-shaped trominoes.
• 31
A Problem with Trominoes
P(k) is the inductive hypothesis.
[We must show that P(k + 1) is true. That is:] We must
show that
If any square is removed from a 2k+1 2k+1 checkerboard,
then the remaining squares can be completely
covered by L-shaped trominoes.
• 32
A Problem with Trominoes
Consider a 2k+1 2k+1 checkerboard with one square
removed. Divide it into four equal quadrants: Each will
consist of a 2k 2k checkerboard.
In one of the quadrants, one square will have been
removed
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# Class 11 NCERT Solutions- Chapter 2 Relation And Functions – Exercise 2.2
• Last Updated : 03 Jan, 2021
### Problem 1: Let A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.
Solution:
Given, A = {1, 2, 3,…,14}.
Here, the relation R from A to A is given by, R = {(x, y): 3x – y = 0, where x, y ∈ A}
So, relation R = {(1,3), (2,6), (3,9), (4,12)}
Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.
So, Domain of R = {1, 2, 3, 4}
Now, Here the complete set A is the Codomain of relation R.
So, Co-Domain of R = {1, 2, 3, 4,….,14}
Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.
So, Range of R = {3, 6, 9, 12}
### Problem 2: Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range.
Solution:
Here, the relation R is given by, R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈N}
Now, As we know that the natural numbers less than 4 are 1, 2 and 3.
So, relation R = {(1,6), (2,7), (3,8)}
Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.
So, Domain of R = {1, 2, 3}
Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.
So, Range of R = {6, 7, 8}
### Problem 3: A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Solution:
Given, A = {1, 2, 3, 5} and B = {4, 6, 9}
Here, the relation from A to B is given by, R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
So, relation R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}
### Problem 4: Fig.2.7 shows a relationship between the sets P and Q. Write this relation –
(i) in set-builder form
(ii) roster form.
What is its domain and range?
Solution:
From the given figure, we can see that –
P = {5, 6, 7} and Q = {3, 4, 5}
Now, The relation between sets P and Q –
(i) In set-builder form
R = {(x, y): y = x – 2; x ∈ P} ‘or’ R = {(x, y): y = x – 2 for x = 5, 6, 7}
(ii) In roster form
R = {(5,3), (6,4), (7,5)}
Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.
So, Domain of R = {5, 6, 7} = P.
Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.
So, Range of R = {3, 4, 5} = Q.
### Problem 5: Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by –
{(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form.
(ii) Find the domain of R.
(iii) Find the range of R.
Solution:
Given, A = {1, 2, 3, 4, 6}
Here, the relation R on A is given by, R = {(a, b): a , b ∈ A, b is exactly divisible by a}
(i) The relation R in roster form will be –
R = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}
(ii) We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.
So, Domain of R = {1, 2, 3, 4, 6}
(iii) We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.
So, Range of R = {1, 2, 3, 4, 6}
### Problem 6: Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
Here, the relation R is given by, R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
So, relation R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}
Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.
So, Domain of R = {0, 1, 2, 3, 4, 5}
Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.
So, Range of R = {5, 6, 7, 8, 9, 10}
### Problem 7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Solution:
Here, the relation R is given by, R = {(x, x3) : x is a prime number less than 10}
Now, As we know that the prime numbers less than 10 are 2, 3, 5 and 7.
So, relation R = {(2,8), (3,27), (5,125), (7,343)}
### Problem 8: Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution:
Given, A = {x, y, z} and B = {1, 2}.
Now, number of elements in set A, n(A) = 3
and number of elements in set B, n(B) = 2
So, n(A × B) = n(A) × n(B) = 6.
We know that, the number of relations from A to B = 2n(A × B) = 26 = 64.
‘OR’
Given, A = {x, y, z} and B = {1, 2}.
Now, A × B = {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)}
Here, number of elements in A × B, n(A × B) = 6
So, the number of subsets of A × B = 26 = 64
Thus, the number of relations from A to B are 64.
### Problem 9: Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution:
Here, the relation R is given by, R = {(a, b): a, b ∈ Z, a – b is an integer}
As we know that the difference between any two integers is always an integer.
So, Domain of R = Z and Range of R = Z.
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What is the partial derivative, how do you compute it, and what does it mean? %PDF-1.4 Asymptotes and Other Things to Look For; 6 Applications of the Derivative. âxây2, which is taking the derivative of f ï¬rst with respect to y twice, and then diï¬erentiating with respect to x, etc. Now, the fact that weâre using $$s$$ and $$t$$ here instead of the âstandardâ $$x$$ and $$y$$ shouldnât be a problem. ��J���� 䀠l��\��p��ӯ��1_\_��i�F�w��y�Ua�fR[[\�~_�E%�4�%�z�_.DY��r�����ߒ�~^XU��4T�lv��ߦ-4S�Jڂ��9�mF��v�o"�Hq2{�Ö���64�M[�l�6����Uq�g&��@��F���IY0��H2am��Ĥ.�ޯo�� �X���>d. Letâs do the derivatives with respect to $$x$$ and $$y$$ first. When working these examples always keep in mind that we need to pay very close attention to which variable we are differentiating with respect to. Solution: Given function: f (x,y) = 3x + 4y To find âf/âx, keep y as constant and differentiate the function: Therefore, âf/âx = 3 Similarly, to find âf/ây, keep x as constant and differentiate the function: Therefore, âf/ây = 4 Example 2: Find the partial derivative of f(x,y) = x2y + sin x + cos y. << /S /GoTo /D (section.3) >> Here is the rate of change of the function at $$\left( {a,b} \right)$$ if we hold $$y$$ fixed and allow $$x$$ to vary. Now that we have the brief discussion on limits out of the way we can proceed into taking derivatives of functions of more than one variable. Learn more about livescript So, if you can do Calculus I derivatives you shouldnât have too much difficulty in doing basic partial derivatives. Weâll do the same thing for this function as we did in the previous part. Letâs look at some examples. In this case we do have a quotient, however, since the $$x$$âs and $$y$$âs only appear in the numerator and the $$z$$âs only appear in the denominator this really isnât a quotient rule problem. Theorem â 2f âxây and â f âyâx are called mixed partial derivatives. Partial derivative notation: if z= f(x;y) then f x= @f @x = @z @x = @ xf= @ xz; f y = @f @y = @z @y = @ yf= @ yz Example. This means the third term will differentiate to zero since it contains only $$x$$âs while the $$x$$âs in the first term and the $$z$$âs in the second term will be treated as multiplicative constants. Optimization; 2. For instance, one variable could be changing faster than the other variable(s) in the function. << /S /GoTo /D (subsection.3.1) >> Thus, the only thing to do is take the derivative of the x^2 factor (which is where that 2x came from). ... your example doesn't make sense. The more standard notation is to just continue to use $$\left( {x,y} \right)$$. Weâll start by looking at the case of holding $$y$$ fixed and allowing $$x$$ to vary. We will now hold $$x$$ fixed and allow $$y$$ to vary. >> Product rule Example 1. Remember that since we are assuming $$z = z\left( {x,y} \right)$$ then any product of $$x$$âs and $$z$$âs will be a product and so will need the product rule! Here is the partial derivative with respect to $$y$$. Given the function $$z = f\left( {x,y} \right)$$ the following are all equivalent notations. This first term contains both $$x$$âs and $$y$$âs and so when we differentiate with respect to $$x$$ the $$y$$ will be thought of as a multiplicative constant and so the first term will be differentiated just as the third term will be differentiated. Since we are treating y as a constant, sin(y) also counts as a constant. share | cite | improve this answer | follow | answered Sep 21 '15 at 17:26. Note that these two partial derivatives are sometimes called the first order partial derivatives. Since we can think of the two partial derivatives above as derivatives of single variable functions it shouldnât be too surprising that the definition of each is very similar to the definition of the derivative for single variable functions. Therefore, since $$x$$âs are considered to be constants for this derivative, the cosine in the front will also be thought of as a multiplicative constant. Derivative of a ⦠Before we work any examples letâs get the formal definition of the partial derivative out of the way as well as some alternate notation. Letâs start with the function $$f\left( {x,y} \right) = 2{x^2}{y^3}$$ and letâs determine the rate at which the function is changing at a point, $$\left( {a,b} \right)$$, if we hold $$y$$ fixed and allow $$x$$ to vary and if we hold $$x$$ fixed and allow $$y$$ to vary. 16 0 obj << Letâs now differentiate with respect to $$y$$. For example, the derivative of f with respect to x is denoted fx. Newton's Method; 4. If you can remember this youâll find that doing partial derivatives are not much more difficult that doing derivatives of functions of a single variable as we did in Calculus I. Letâs first take the derivative with respect to $$x$$ and remember that as we do so all the $$y$$âs will be treated as constants. In this case we call $$h'\left( b \right)$$ the partial derivative of $$f\left( {x,y} \right)$$ with respect to $$y$$ at $$\left( {a,b} \right)$$ and we denote it as follows. 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, $$f\left( {x,y} \right) = {x^4} + 6\sqrt y - 10$$, $$w = {x^2}y - 10{y^2}{z^3} + 43x - 7\tan \left( {4y} \right)$$, $$\displaystyle h\left( {s,t} \right) = {t^7}\ln \left( {{s^2}} \right) + \frac{9}{{{t^3}}} - \sqrt[7]{{{s^4}}}$$, $$\displaystyle f\left( {x,y} \right) = \cos \left( {\frac{4}{x}} \right){{\bf{e}}^{{x^2}y - 5{y^3}}}$$, $$\displaystyle z = \frac{{9u}}{{{u^2} + 5v}}$$, $$\displaystyle g\left( {x,y,z} \right) = \frac{{x\sin \left( y \right)}}{{{z^2}}}$$, $$z = \sqrt {{x^2} + \ln \left( {5x - 3{y^2}} \right)}$$, $${x^3}{z^2} - 5x{y^5}z = {x^2} + {y^3}$$, $${x^2}\sin \left( {2y - 5z} \right) = 1 + y\cos \left( {6zx} \right)$$. By using this website, you agree to our Cookie Policy. To denote the speciï¬c derivative, we use subscripts. Thereâs quite a bit of work to these. However, if you had a good background in Calculus I chain rule this shouldnât be all that difficult of a problem. endobj We will see an easier way to do implicit differentiation in a later section. Notice that the second and the third term differentiate to zero in this case. If there is more demand for mobile phone, it will lead to more demand for phone line too. To compute $${f_x}\left( {x,y} \right)$$ all we need to do is treat all the $$y$$âs as constants (or numbers) and then differentiate the $$x$$âs as weâve always done. 8 0 obj Before we actually start taking derivatives of functions of more than one variable letâs recall an important interpretation of derivatives of functions of one variable. Now, solve for $$\frac{{\partial z}}{{\partial x}}$$. x��ZKs����W 7�bL���k�����8e�l` �XK� Since uâ has two parameters, partial derivatives come into play. Example of Complementary goods are mobile phones and phone lines. 2. 1. The second derivative test; 4. Donât forget to do the chain rule on each of the trig functions and when we are differentiating the inside function on the cosine we will need to also use the product rule. Differentiation is the action of computing a derivative. Letâs start out by differentiating with respect to $$x$$. We first will differentiate both sides with respect to $$x$$ and remember to add on a $$\frac{{\partial z}}{{\partial x}}$$ whenever we differentiate a $$z$$ from the chain rule. Likewise, to compute $${f_y}\left( {x,y} \right)$$ we will treat all the $$x$$âs as constants and then differentiate the $$y$$âs as we are used to doing. If you recall the Calculus I definition of the limit these should look familiar as they are very close to the Calculus I definition with a (possibly) obvious change. Since there isnât too much to this one, we will simply give the derivatives. The function f can be reinterpreted as a family of functions of one variable indexed by the other variables: With this function weâve got three first order derivatives to compute. So, there are some examples of partial derivatives. stream 13 0 obj Itâs a constant and we know that constants always differentiate to zero. We will now look at finding partial derivatives for more complex functions. In both these cases the $$z$$âs are constants and so the denominator in this is a constant and so we donât really need to worry too much about it. Notice as well that it will be completely possible for the function to be changing differently depending on how we allow one or more of the variables to change. The plane through (1,1,1) and parallel to the yz-plane is x = 1. Letâs do the partial derivative with respect to $$x$$ first. This one will be slightly easier than the first one. The partial derivative with respect to $$x$$ is. Free derivative applications calculator - find derivative application solutions step-by-step This website uses cookies to ensure you get the best experience. Here are the derivatives for these two cases. The partial derivative of z with respect to x measures the instanta-neous change in the function as x changes while HOLDING y constant. /Filter /FlateDecode Use partial derivatives to find a linear fit for a given experimental data. Two goods are said to be substitute goods if an increase in the demand for either result in a decrease for the other. The Mean Value Theorem; 7 Integration. Gummy bears Gummy bears. Note that the notation for partial derivatives is different than that for derivatives of functions of a single variable. Now letâs take a quick look at some of the possible alternate notations for partial derivatives. /Length 2592 z= f(x;y) = ln 3 p 2 x2 3xy + 3cos(2 + 3 y) 3 + 18 2 Find f x(x;y), f y(x;y), f(3; 2), f x(3; 2), f y(3; 2) For w= f(x;y;z) there are three partial derivatives f x(x;y;z), f y(x;y;z), f z(x;y;z) Example. (Partial Derivatives) Example 1: Determine the partial derivative of the function: f (x,y) = 3x + 4y. Email. We have just looked at some examples of determining partial derivatives of a function from the Partial Derivatives Examples 1 and Partial Derivatives Examples 2 page. It should be clear why the third term differentiated to zero. Here is the derivative with respect to $$y$$. This is important because we are going to treat all other variables as constants and then proceed with the derivative as if it was a function of a single variable. We can do this in a similar way. Solution: Given function is f(x, y) = tan(xy) + sin x. Since we are holding $$x$$ fixed it must be fixed at $$x = a$$ and so we can define a new function of $$y$$ and then differentiate this as weâve always done with functions of one variable. Concavity and inflection points; 5. Here is the derivative with respect to $$z$$. Free partial derivative calculator - partial differentiation solver step-by-step This website uses cookies to ensure you get the best experience. From that standpoint, they have many of the same applications as total derivatives in single-variable calculus: directional derivatives, linear approximations, Taylor polynomials, local extrema, computation of total derivatives via chain rule, etc. Now, this is a function of a single variable and at this point all that we are asking is to determine the rate of change of $$g\left( x \right)$$ at $$x = a$$. Google Classroom Facebook Twitter. By ⦠We also canât forget about the quotient rule. We will be looking at the chain rule for some more complicated expressions for multivariable functions in a later section. For the fractional notation for the partial derivative notice the difference between the partial derivative and the ordinary derivative from single variable calculus. Also, the $$y$$âs in that term will be treated as multiplicative constants. Partial derivatives are the basic operation of multivariable calculus. Here are the formal definitions of the two partial derivatives we looked at above. Here, a change in x is reflected in uâ in two ways: as an operand of the addition and as an operand of the square operator. Now, in the case of differentiation with respect to $$z$$ we can avoid the quotient rule with a quick rewrite of the function. f(x) â f â² (x) = df dx f(x, y) â fx(x, y) = âf âx & fy(x, y) = âf ây Okay, now letâs work some examples. Combined Calculus tutorial videos. PARTIAL DERIVATIVES 379 The plane through (1,1,1) and parallel to the Jtz-plane is y = l. The slope of the tangent line to the resulting curve is dzldx = 6x = 6. For example Partial derivative is used in marginal Demand to obtain condition for determining whether two goods are substitute or complementary. Related Rates; 3. It will work the same way. the PARTIAL DERIVATIVE. We will need to develop ways, and notations, for dealing with all of these cases. In this last part we are just going to do a somewhat messy chain rule problem. The first step is to differentiate both sides with respect to $$x$$. In other words, what do we do if we only want one of the variables to change, or if we want more than one of them to change? For example,w=xsin(y+ 3z). Also, donât forget how to differentiate exponential functions. the second derivative is negative when the function is concave down. Now letâs solve for $$\frac{{\partial z}}{{\partial x}}$$. Now, we canât forget the product rule with derivatives. With functions of a single variable we could denote the derivative with a single prime. Now, as this quick example has shown taking derivatives of functions of more than one variable is done in pretty much the same manner as taking derivatives of a single variable. However, the First Derivative Test has wider application. Ontario Tech University is the brand name used to refer to the University of Ontario Institute of Technology. endobj Since we are interested in the rate of change of the function at $$\left( {a,b} \right)$$ and are holding $$y$$ fixed this means that we are going to always have $$y = b$$ (if we didnât have this then eventually $$y$$ would have to change in order to get to the pointâ¦). Since only one of the terms involve $$z$$âs this will be the only non-zero term in the derivative. However, with partial derivatives we will always need to remember the variable that we are differentiating with respect to and so we will subscript the variable that we differentiated with respect to. Partial derivatives are computed similarly to the two variable case. Since we are differentiating with respect to $$x$$ we will treat all $$y$$âs and all $$z$$âs as constants. Solution: The partial derivatives change, so the derivative becomesâfâx(2,3)=4âfây(2,3)=6Df(2,3)=[46].The equation for the tangent plane, i.e., the linear approximation, becomesz=L(x,y)=f(2,3)+âfâx(2,3)(xâ2)+âfây(2,3)(yâ3)=13+4(xâ2)+6(yâ3) 12 0 obj For the same f, calculate âfâx(1,2).Solution: From example 1, we know that âfâx(x,y)=2y3x. We will just need to be careful to remember which variable we are differentiating with respect to. Definition of Partial Derivatives Let f(x,y) be a function with two variables. Letâs start off this discussion with a fairly simple function. The So, the partial derivatives from above will more commonly be written as. In this case we donât have a product rule to worry about since the only place that the $$y$$ shows up is in the exponential. Partial derivative and gradient (articles) Introduction to partial derivatives. Here is the rewrite as well as the derivative with respect to $$z$$. To calculate the derivative of this function, we have to calculate partial derivative with respect to x of uâ(x, uâ). This is the currently selected item. We will spend a significant amount of time finding relative and absolute extrema of functions of multiple variables. Letâs start with finding $$\frac{{\partial z}}{{\partial x}}$$. First letâs find $$\frac{{\partial z}}{{\partial x}}$$. If looked at the point (2,3), what changes? This is also the reason that the second term differentiated to zero. We will be looking at higher order derivatives in a later section. For the fractional notation for the partial derivative notice the difference between the partial derivative and the ordinary derivative from single variable calculus. A function f(x,y) of two variables has two ï¬rst order partials âf âx, âf ây. Application of Partial Derivative in Engineering: In image processing edge detection algorithm is used which uses partial derivatives to improve edge detection. Now, we do need to be careful however to not use the quotient rule when it doesnât need to be used. The final step is to solve for $$\frac{{dy}}{{dx}}$$. Two examples; 2. Refer to the above examples. Note as well that we usually donât use the $$\left( {a,b} \right)$$ notation for partial derivatives as that implies we are working with a specific point which we usually are not doing. If we have a function in terms of three variables $$x$$, $$y$$, and $$z$$ we will assume that $$z$$ is in fact a function of $$x$$ and $$y$$. Before getting into implicit differentiation for multiple variable functions letâs first remember how implicit differentiation works for functions of one variable. We went ahead and put the derivative back into the âoriginalâ form just so we could say that we did. In practice you probably donât really need to do that. In this case both the cosine and the exponential contain $$x$$âs and so weâve really got a product of two functions involving $$x$$âs and so weâll need to product rule this up. partial derivative coding in matlab . Because we are going to only allow one of the variables to change taking the derivative will now become a fairly simple process. Let's find the partial derivatives of z = f(x, y) = x^2 sin(y). Partial Derivatives Examples 3. In the case of the derivative with respect to $$v$$ recall that $$u$$âs are constant and so when we differentiate the numerator we will get zero! This is an important interpretation of derivatives and we are not going to want to lose it with functions of more than one variable. 1. Finally, letâs get the derivative with respect to $$z$$. In fact, if weâre going to allow more than one of the variables to change there are then going to be an infinite amount of ways for them to change. Here are the two derivatives. 5 0 obj endobj Here is the partial derivative with respect to $$x$$. Do not forget the chain rule for functions of one variable. Hopefully you will agree that as long as we can remember to treat the other variables as constants these work in exactly the same manner that derivatives of functions of one variable do. Likewise, whenever we differentiate $$z$$âs with respect to $$y$$ we will add on a $$\frac{{\partial z}}{{\partial y}}$$. The partial derivative notation is used to specify the derivative of a function of more than one variable with respect to one of its variables. In this manner we can ï¬nd nth-order partial derivatives of a function. 1. Examples of the application of the product rule (open by selection) Here are some examples of applying the product rule. Question 1: Determine the partial derivative of a function f x and f y: if f(x, y) is given by f(x, y) = tan(xy) + sin x. Differentiation. Okay, now letâs work some examples. There is one final topic that we need to take a quick look at in this section, implicit differentiation. talk about a derivative; instead, we talk about a derivative with respect to avariable. Letâs take a quick look at a couple of implicit differentiation problems. Remember that since we are differentiating with respect to $$x$$ here we are going to treat all $$y$$âs as constants. That means that terms that only involve $$y$$âs will be treated as constants and hence will differentiate to zero. Now, letâs take the derivative with respect to $$y$$. In this case we treat all $$x$$âs as constants and so the first term involves only $$x$$âs and so will differentiate to zero, just as the third term will. We will find the equation of tangent planes to surfaces and we will revisit on of the more important applications of derivatives from earlier Calculus classes. Concavityâs connection to the second derivative gives us another test; the Second Derivative Test. In this section we are going to concentrate exclusively on only changing one of the variables at a time, while the remaining variable(s) are held fixed. ... For a function with the variable x and several further variables the partial derivative to x is noted as follows. The gradient. One Bernard Baruch Way (55 Lexington Ave. at 24th St) New York, NY 10010 646-312-1000 Examples of how to use âpartial derivativeâ in a sentence from the Cambridge Dictionary Labs Here is the derivative with respect to $$x$$. Now, letâs do it the other way. endobj The first derivative test; 3. Sometimes the second derivative test helps us determine what type of extrema reside at a particular critical point. Similarly, we would hold x constant if we wanted to evaluate the eâect of a change in y on z. We call this a partial derivative. 9 0 obj With this one weâll not put in the detail of the first two. This function has two independent variables, x and y, so we will compute two partial derivatives, one with respect to each variable. In other words, $$z = z\left( {x,y} \right)$$. Doing this will give us a function involving only $$x$$âs and we can define a new function as follows. In other words, we want to compute $$g'\left( a \right)$$ and since this is a function of a single variable we already know how to do that. Given below are some of the examples on Partial Derivatives. Linear Approximations; 5. f(x;y;z) = p z2 + y x+ 2cos(3x 2y) Find f x(x;y;z), f y(x;y;z), f z(x;y;z), We will deal with allowing multiple variables to change in a later section. Solution: Now, find out fx first keeping y as constant fx = âf/âx = (2x) y + cos x + 0 = 2xy + cos x When we keep y as constant cos y becomes a con⦠We will shortly be seeing some alternate notation for partial derivatives as well. More demand for mobile phone, it will lead to more demand for either result in a section! X and several further variables the partial derivative with respect to \ z\. Derivative, how do you compute it, and what does it mean extrema functions. ShouldnâT have too much difficulty in doing basic partial derivatives let f ( x, y } \right ) )... Z } } { { \partial z } } { { \partial z }... There is one final topic that we did in the function a little to help us with differentiation! = f ( x, y ) by differentiating with respect to \ ( z = f\left ( {,. Website uses cookies to ensure you get the best experience the derivative with respect to (... Simple process, for dealing with all of these cases start by looking at the case of \. A function with the differentiation process at a couple of implicit differentiation works in exactly the same for! Of multiple variables while HOLDING y constant the case of HOLDING \ ( =... The way as well as the derivative with respect to \ ( z\.. One final topic that we need to be substitute goods if an increase in the part! Sentence from the Cambridge Dictionary Labs partial derivative with respect to \ ( y\ ) you a. Or Complementary only thing to do implicit differentiation problems test helps us what. More demand for phone line too, how do you compute it, what... Of partial derivatives ) be a function with the variable x and several further the! Relative and absolute extrema of functions of a function with two variables is more for... Of two variables has two parameters, partial derivatives come into play f with respect to x 2x! Fractional notation for the fractional notation for partial derivatives to compute one we. We know that constants always differentiate to zero become a fairly simple function first is! A new function as we did this problem because implicit differentiation problems quotient rule when it doesnât need be... Derivatives with respect to x measures the instanta-neous change in the detail of the on... Answer | follow | answered Sep 21 '15 at 17:26 z = f ( x y. ) \ ) as partial derivative application examples formal definition of the product rule ( open by selection ) here are the operation! Into the âoriginalâ form just so we could denote the derivative with respect to x is 2x sin ( )! Did in the detail of the two variable case find a linear fit for a function with two has... Functions in a decrease for the fractional notation for the other variable ( s ) the! Test helps us determine what type of extrema reside at a particular critical point derivative test helps us determine type... Called the first step is to solve for \ ( x\ ) partial., we canât forget the product rule with derivatives with allowing multiple variables to change the., how do you compute it, and notations, for dealing with all of these cases examples on derivatives. Faster than the first one a constant shortly be seeing some alternate notation in marginal demand to condition... See an easier way to do that, partial derivatives we looked at above than other! Use \ ( x\ ) differentiation process thus, the derivative Cookie Policy and \! Of time finding relative and absolute extrema of functions of multiple variables do a somewhat messy chain rule shouldnât... Rule problem variable is that there is more than one variable that domains. Agree to our Cookie Policy for partial derivatives of functions of one variable decrease the. With a single prime the variables to change in a decrease for the fractional notation for partial derivatives from... Computed similarly to the second derivative test function \ ( x\ ) and \ ( {... A linear fit for a function with the differentiation process best experience step-by-step... We are differentiating with respect to \ ( \frac { { \partial z } } \ ) derivative single. = tan ( xy ) + sin x several further variables the derivative... Notations, for dealing with all of these cases f\left ( partial derivative application examples,! Called the first order partial derivatives to improve edge detection algorithm is used which uses partial derivatives '15! Derivative notice the difference between the partial derivative with respect to x measures the change. To differentiate both sides with respect to x is 2x sin ( y ) = (! Partial derivative in Engineering: in image processing edge detection algorithm is used which uses partial derivatives for more functions! As we did this problem because implicit differentiation works in exactly the same thing for this function we... The previous part a single prime applying the product rule will work the same manner with of. Parameters, partial derivatives let f ( x, y } \right ) \.... Cookie Policy at in this manner we can ï¬nd nth-order partial derivatives to compute.kastatic.org. | answered Sep 21 '15 at 17:26 not use the quotient rule when it doesnât need to used! That terms that only involve \ ( y\ ) somewhat messy chain for. That means that terms that only involve \ ( y\ ) fixed and allowing \ ( )... Do you compute it, and what does it mean with functions of one variable demand to obtain condition determining... That there is one final topic that we need to do implicit differentiation in! Involving only \ ( y\ ) âs in that term will be treated as constants and hence will differentiate zero! The final step is to solve for \ ( z\ ) at higher order derivatives to find a fit. ; the second derivative test helps us determine what type of extrema reside at couple... First two Ontario Tech University is the brand name used to refer the! New function as follows did in the previous part the same thing for this function weâve three... Difference between the partial derivative with respect to \ ( y\ ) first ) \ the. Since we are not going to only allow one of the possible alternate notations partial. Of Technology find a linear fit for a function involving only \ ( x\ ) first should be why! Way as well as the derivative with respect to \ ( \frac { { dx } } \.! Marginal demand to obtain condition for determining whether two goods are mobile phones and phone.. Somewhat messy chain rule problem careful however to not use the quotient rule when doesnât., \ ( \frac { { \partial z } } { { partial derivative application examples x }. Solver step-by-step this website uses cookies to ensure you get the best experience f ( x, y of... We know that constants always differentiate to zero phones and phone lines need... WeâLl start by looking at the case of HOLDING \ ( y\ ): in image processing detection. What is the partial derivative is used which uses partial derivatives are mobile phones and phone lines changing than. Best experience rule with derivatives of Complementary partial derivative application examples are said to be substitute goods if an increase in the a. Manner with functions of multiple variables now hold \ ( \frac { { \partial z } } ). \Partial z } } { { \partial x } } { { dy } } \ ) for,! To \ ( y\ ) to vary how implicit differentiation for multiple variable functions letâs first remember implicit... Spend a significant amount of time finding relative and absolute extrema of functions one. Derivative is used which uses partial derivatives for more complex functions that for derivatives of a problem changing faster the! The derivatives complex functions to obtain condition for determining whether two goods are to. Goods are said to be careful however to not use the quotient when... Is used which uses partial derivatives of a single partial derivative application examples spend a significant of... Solution: given function is f ( x, y } \right \! For multiple variable functions letâs first remember how implicit differentiation for multiple variable functions letâs first how. You agree to our Cookie Policy term differentiate to zero look for ; 6 Applications of the possible notations. Did this problem because implicit differentiation works for functions of multiple variables relative and extrema! ÂS and \ ( z\ ) compute it, and notations, dealing... Higher order derivatives in a later section from ) variable we could say that we did by differentiating with to... For dealing with all of these cases ahead and put the derivative with to! Can have derivatives of a production function variable ( s ) in demand! As it does with functions of a function on partial derivatives come play! Labs partial derivative to x is noted as follows looking at the case HOLDING... Function is f ( x, y ) = tan ( xy ) + sin x what! DonâT really need to do implicit differentiation for multiple variable functions letâs first remember implicit... Definitions of the partial derivative out of the product rule ( open by selection ) are., for dealing with all of these cases changes while HOLDING y constant we! Function \ ( z\ ) âs and we know that constants always to. Be all that difficult of a problem 2x sin ( y ) = x^2 sin ( y ) Labs! To compute, donât forget how to differentiate exponential functions answer | follow | Sep. Always differentiate to zero = f\left ( { x, y ) of two variables has two ï¬rst partials. |
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Calculus I - Notes
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## Critical Points
Critical points will show up throughout a majority of this chapter so we first need to define them and work a few examples before getting into the sections that actually use them.
Definition
We say that is a critical point of the function f(x) if exists and if either of the following are true.
Note that we require that exists in order for to actually be a critical point. This is an important, and often overlooked, point.
The main point of this section is to work some examples finding critical points. So, let’s work some examples.
Example 1 Determine all the critical points for the function. Solution We first need the derivative of the function in order to find the critical points and so let’s get that and notice that we’ll factor it as much as possible to make our life easier when we go to find the critical points. Now, our derivative is a polynomial and so will exist everywhere. Therefore the only critical points will be those values of x which make the derivative zero. So, we must solve. Because this is the factored form of the derivative it’s pretty easy to identify the three critical points. They are,
Polynomials are usually fairly simple functions to find critical points for provided the degree doesn’t get so large that we have trouble finding the roots of the derivative.
Most of the more “interesting” functions for finding critical points aren’t polynomials however. So let’s take a look at some functions that require a little more effort on our part.
Example 2 Determine all the critical points for the function. Solution To find the derivative it’s probably easiest to do a little simplification before we actually differentiate. Let’s multiply the root through the parenthesis and simplify as much as possible. This will allow us to avoid using the product rule when taking the derivative. Now differentiate. We will need to be careful with this problem. When faced with a negative exponent it is often best to eliminate the minus sign in the exponent as we did above. This isn’t really required but it can make our life easier on occasion if we do that. Notice as well that eliminating the negative exponent in the second term allows us to correctly identify why is a critical point for this function. Once we move the second term to the denominator we can clearly see that the derivative doesn’t exist at and so this will be a critical point. If you don’t get rid of the negative exponent in the second term many people will incorrectly state that is a critical point because the derivative is zero at . While this may seem like a silly point, after all in each case is identified as a critical point, it is sometimes important to know why a point is a critical point. In fact, in a couple of sections we’ll see a fact that only works for critical points in which the derivative is zero. So, we’ve found one critical point (where the derivative doesn’t exist), but we now need to determine where the derivative is zero (provided it is of course…). To help with this it’s usually best to combine the two terms into a single rational expression. So, getting a common denominator and combining gives us, Notice that we still have as a critical point. Doing this kind of combining should never lose critical points, it’s only being done to help us find them. As we can see it’s now become much easier to quickly determine where the derivative will be zero. Recall that a rational expression will only be zero if its numerator is zero (and provided the denominator isn’t also zero at that point of course). So, in this case we can see that the numerator will be zero if and so there are two critical points for this function.
Example 3 Determine all the critical points for the function. Solution We’ll leave it to you to verify that using the quotient rule we get that the derivative is, Notice that we factored a “-1” out of the numerator to help a little with finding the critical points. This negative out in front will not affect the derivative whether or not the derivative is zero or not exist but will make our work a little easier. Now, we have two issues to deal with. First the derivative will not exist if there is division by zero in the denominator. So we need to solve, We didn’t bother squaring this since if this is zero, then zero squared is still zero and if it isn’t zero then squaring it won’t make it zero. So, we can see from this that the derivative will not exist at and . However, these are NOT critical points since the function will also not exist at these points. Recall that in order for a point to be a critical point the function must actually exist at that point. At this point we need to be careful. The numerator doesn’t factor, but that doesn’t mean that there aren’t any critical points where the derivative is zero. We can use the quadratic formula on the numerator to determine if the fraction as a whole is ever zero. So, we get two critical points. Also, these are not “nice” integers or fractions. This will happen on occasion. Don’t get too locked into answers always being “nice”. Often they aren’t. Note as well that we only use real numbers for critical points. So, if upon solving the quadratic in the numerator, we had gotten complex number these would not have been considered critical points. Summarizing, we have two critical points. They are, Again, remember that while the derivative doesn’t exist at and neither does the function and so these two points are not critical points for this function.
So far all the examples have not had any trig functions, exponential functions, etc. in them. We shouldn’t expect that to always be the case. So, let’s take a look at some examples that don’t just involve powers of x.
Example 4 Determine all the critical points for the function. Solution First get the derivative and don’t forget to use the chain rule on the second term. Now, this will exist everywhere and so there won’t be any critical points for which the derivative doesn’t exist. The only critical points will come from points that make the derivative zero. We will need to solve, Solving this equation gives the following. Don’t forget the 2π n on these! There will be problems down the road in which we will miss solutions without this! Also make sure that it gets put on at this stage! Now divide by 3 to get all the critical points for this function.
Notice that in the previous example we got an infinite number of critical points. That will happen on occasion so don’t worry about it when it happens.
Example 5 Determine all the critical points for the function. Solution Here’s the derivative for this function. Now, this looks unpleasant, however with a little factoring we can clean things up a little as follows, This function will exist everywhere and so no critical points will come from that. Determining where this is zero is easier than it looks. We know that exponentials are never zero and so the only way the derivative will be zero is if, We will have two critical points for this function.
Example 6 Determine all the critical points for the function. Solution Before getting the derivative let’s notice that since we can’t take the log of a negative number or zero we will only be able to look at . The derivative is then, Now, this derivative will not exist if x is a negative number or if , but then again neither will the function and so these are not critical points. Remember that the function will only exist if and nicely enough the derivative will also only exist if and so the only thing we need to worry about is where the derivative is zero. First note that, despite appearances, the derivative will not be zero for . As noted above the derivative doesn’t exist at because of the natural logarithm and so the derivative can’t be zero there! So, the derivative will only be zero if, Recall that we can solve this by exponentiating both sides. There is a single critical point for this function.
Let’s work one more problem to make a point.
Example 7 Determine all the critical points for the function. Solution Note that this function is not much different from the function used in Example 5. In this case the derivative is, This function will never be zero for any real value of x. The exponential is never zero of course and the polynomial will only be zero if x is complex and recall that we only want real values of x for critical points. Therefore, this function will not have any critical points.
It is important to note that not all functions will have critical points! In this course most of the functions that we will be looking at do have critical points. That is only because those problems make for more interesting examples. Do not let this fact lead you to always expect that a function will have critical points. Sometimes they don’t as this final example has shown.
Rates of Change Previous Section Next Section Minimum and Maximum Values Derivatives Previous Chapter Next Chapter Integrals
[Notes]
© 2003 - 2018 Paul Dawkins |
# AP Statistics Curriculum 2007 Distrib Multinomial
(Difference between revisions)
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Revision as of 07:40, 6 March 2008 (view source)IvoDinov (Talk | contribs) (→SOCR Multinomial Examples)← Older edit Revision as of 07:42, 6 March 2008 (view source)IvoDinov (Talk | contribs) Newer edit → Line 67: Line 67: Of course, we can compute this number exactly as: Of course, we can compute this number exactly as: - : $P(A) =$ + : $P(A) = {10! \over 3!\times 3! \times 2! \times 2! } \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 = .....$ However, we can also find a pretty close empirically-driven estimate using the [[SOCR_EduMaterials_Activities_DiceExperiment | SOCR Dice Experiment]]. However, we can also find a pretty close empirically-driven estimate using the [[SOCR_EduMaterials_Activities_DiceExperiment | SOCR Dice Experiment]].
## General Advance-Placement (AP) Statistics Curriculum - Multinomial Random Variables and Experiments
The multinomial experiments (and multinomial distribtuions) directly extend the their bi-nomial counterparts.
### Multinomial experiments
A multinomial experiment is an experiment that has the following properties:
• The experiment consists of k repeated trials.
• Each trial has a discrete number of possible outcomes.
• On any given trial, the probability that a particular outcome will occur is constant.
• The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.
#### Examples of Multinomial experiments
• Suppose we have an urn containing 9 marbles. Two are red, three are green, and four are blue (2+3+4=9). We randomly select 5 marbles from the urn, with replacement. What is the probability (P(A)) of the event A={selecting 2 green marbles and 3 blue marbles}?
• To solve this problem, we apply the multinomial formula. We know the following:
• The experiment consists of 5 trials, so k = 5.
• The 5 trials produce 0 red, 2 green marbles, and 3 blue marbles; so r1 = rred = 0, r2 = rgreen = 2, and r3 = rblue = 3.
• For any particular trial, the probability of drawing a red, green, or blue marble is 2/9, 3/9, and 5/9, respectively. Hence, p1 = pred = 2 / 9, p2 = pgreen = 1 / 3, and p3 = pblue = 5 / 9.
Plugging these values into the multinomial formula we get the probability of the event of interest to be:
$P(A) = {5\choose 0, 2, 3}p_1^{r_1}p_2^{r_2}p_3^{r_3}$
$P(A) = {5! \over 0!\times 2! \times 3! }\times (2/9)^0 \times (1/3)^2\times (5/9)^3=0.19052.$
Thus, if we draw 5 marbles with replacement from the urn, the probability of drawing no red , 2 green, and 3 blue marbles is 0.19052.
### Synergies between Binomial and Multinomial processes/probabilities/coefficients
${n\choose i}=\frac{n!}{k!(n-k)!}$
${n\choose i_1,i_2,\cdots, i_k}= \frac{n!}{i_1! i_2! \cdots i_k!}$
• The Binomial vs. Multinomial Formulas
$(a+b)^n = \sum_{i=1}^n{{n\choose i}a^1 \times b^{n-i}}$
$(a_1+a_2+\cdots +a_k)^n = \sum_{i_1+i_2\cdots +i_k=n}^n{ {n\choose i_1,i_2,\cdots, i_k} a_1^{i_1} \times a_2^{i_2} \times \cdots \times a_k^{i_k}}$
$p=P(X=r)={n\choose r}p^r(1-p)^{n-r}, \forall 0\leq r \leq n$
$p=P(X_1=r_1 \cap X_1=r_1 \cap \cdots \cap X_k=r_k | r_1+r_2+\cdots+r_k=n)={n\choose i_1,i_2,\cdots, i_k}p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}, \forall r_1+r_2+\cdots+r_k=n$
### Example
Suppose we study N independent trials with results falling in one of k possible categories labeled 1,2,cdots,k. Let pi be the probability of a trial resulting in the ith category, where $p_1+p_2+\cdots++p_k =1$. Let Ni be the number of trials resulting in the ith category, where $N_1+N_2+\cdots++N_k = N$.
For instance, suppose we have 9 people arriving at a meeting according to the following information:
P(by Air) = 0.4, P(by Bus) = 0.2, P(by Automobile) = 0.3, P(by Train) = 0.1
• Compute the following probabilities
P(3 by Air, 3 by Bus, 1 by Auto, 2 by Train) = ?
P(2 by air) = ?
### SOCR Multinomial Examples
Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (one is the most likely and six is the least likely outcome).
x 1 2 3 4 5 6 P(X=x) 0.286 0.238 0.19 0.143 0.095 0.048
P(A)=?
Of course, we can compute this number exactly as:
$P(A) = {10! \over 3!\times 3! \times 2! \times 2! } \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 = .....$
However, we can also find a pretty close empirically-driven estimate using the SOCR Dice Experiment.
For instance, running the SOCR Dice Experiment 1,000 times with number of dice n=10, and the loading probabilities listed above, we get an output like the one shown below.
Now, we can actually count how many of these 1,000 trials generated the event A as an outcome. In this one experiment of 1,000 trials there were 8 outcomes of the type {3 ones, 3 twos, 2 threes and 2 fours}. Therefore, the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above
$P(A) \approx {8 \over 1,000}=0.008$ .
### References
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# Free Perimeter and Area Subjective Test 02 Practice Test - 7th grade
Given a square of side 10m and a rectangle of length 11m and breadth 9m. Which one has more area?
[1 MARK]
#### SOLUTION
Solution :
Area of square of side A isA2
Area of given square = 10 × (10)
= 100 m2
Area of a rectangle whose length and breadth are A m and B m respectively is A × B
Area of given rectangle = 11 × 9
= 99 m2
100 is greater than 99.
Hence, Square has greater area compared to the given rectangle.
Find the area of the circle whose diameter is 10m. Express your answer in cm2. [2 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Given that
The diameter of a circle is 10m.
Area of the circle whose radius is R is
π×(R)2
Radius of the given circle = 102
= 5m
Area of the given circle = π×(5)2
= 78.5m2
= 785000cm2
Area of the required circle is 785000cm2
A circle has a circumference of 70 meters. Find its area. [2 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Given that
The circumference of a circle = 70 m
Let the radius of the circle be 'R' m
Circumference of given circle is 2πR
2πR = 70
R = 35π
Area = π×(35π)2
= 389.77m2
The area of the circle is 389.77m2.
Is the figure below a polygon? [2 MARKS]
#### SOLUTION
Solution :
Reason: 1 Mark
This is a closed figure, but this does not contain any straight lines. Hence, it is not a polygon.
Given the side length as 10m and base length as 9m and height as 5m of a parallelogram. Find the area and its perimeter. [2 MARKS]
#### SOLUTION
Solution :
Process : 1 Mark
Result : 1 Mark
Perimeter of the parallelogram = 2( side + base )
= 2(10+9) =38 m
Area of the parallelogram = base × (height)
= 9 × (5) = 45 m2
Find the missing values: [3 MARKS]
BaseHeightArea of triangle15 cm87 cm231.4 mm1256 mm222 cm170.5 cm2
#### SOLUTION
Solution : Blanks : 1 Mark each
We know that the area of triangle= 12×base×height
In first row, base=15 cm and area = 87cm2
87=12×15×height height=87×215=11.6 cm
In second row, height=31.4 mm and area =1256 mm2
1256=12×base×31.4 base=1256×231.4=80 mm
In third row, base=22 cm and area =170.5 cm2
170.5=12×22×height
height=170.5×222=15.5 cm
Thus, the missing values are:
BaseHeightArea of triangle15 cm11.6 cm87 cm280 mm31.4 mm1256 mm222 cm15.5 cm170.5 cm2
A square of side 8 m is cut into two equal rectangles along its length. The rectangle is bent in the shape of a circle. Find the radius of the circle. [3 MARKS]
#### SOLUTION
Solution :
Steps: 2 Marks
Given side of a square = 8 cm
If it is cut into two equal parts the along the length, then:
the rectangles formed will have the length and breadth as 8 m and 4 m respectively
The perimeter of the rectangle will be 2×(8+4) = 24m
The perimeter of the rectangle will be equal to the circumference of the circle.
The circumference of the circle whose radius is R m = 2πR
2πR = 24
R = 12πm
R = 3.81 m
The radius of the circle is 3.81 m.
The side of a square is equal to twice the radius of the circle. Which of them will have greater area?
[3 MARKS]
#### SOLUTION
Solution :
Formula: 2 Marks
Let the side of given square be A cm
Area of the given square = A2
Radius of the given circle = A2 cm
Area of the circle = (π)(A2)2
= π4×(A2)
= 0.785A2
A2>0.785A2
So square has more area compared with circle
The area of the given square and rectangle are same. If the length of each side of the square is 4m, and the ratio of length and breadth of the rectangle is 2. Find the length of the rectangle?
[3 MARKS]
#### SOLUTION
Solution :
Steps: 2 Marks
Given that
The length of each side of a square = 4 m
Area of the given square =
(4)2
= 16m2
As per the question, the ratio of length and breadth of the rectangle is 2
Area of the rectangle is given = 16 m2
The area of the rectangle whose length is L m and breadth is B m is L × B m2
L = 2B ( as per question)
2B × B = 16
B = 2 × 2m
So, L = 2(2 × 2m)
= 4 ×2m
So, the length of the rectangle is 4
×2m.
A parallelogram of area 48 cm2 is divided into two congruent triangles. If the height of the triangle is 4 cm, find the length of base of the triangle.
[4 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Steps: 1 Mark
Application: 1 Mark
Given that
Area of a parallelogram=48 cm2
The parallelogram is divided into two congruent triangles.
Area of parallelogram = 2(Area of triangle)
Area of triangle = 48 cm22=24cm2
Area of triangle = 12×base×height
24 cm2=12×base×4cm
base = 12 cm
So, measure of base of the triangle is 12 cm.
A parallelogram has sides of length 12 cm and 8 cm. If the distance between the 12 cm sides is 5 cm. Find the distance between 8 cm sides. [4 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Steps: 2 Marks
According to the question, if base = 12 cm and height = 5 cm
So, area = base × height = 12 × 5 = 60 cm2
Now, if base = 8 cm and now we know that area of a parallelogram is 60 cm2
So, Distance between 8 cm sides = 608 = 7.5 cm
So, the distance between the 8 cm sides is 7.5 cm.
Given two squares of the same length. One square has a triangle in it with its vertices coinciding with three vertices of the square. The other square contains a circle touching all mid points of the sides of the square.Which of these two will have a greater area and what is the difference in the areas if the length of the given square is 4m?
[4 MARKS]
#### SOLUTION
Solution :
Steps: 3 Marks
The given square has a length of 4 m
Area of the Triangle = 0.5 × 4 × 4 = 8 m2
Area of the given circle = π
× 22 =12.56 m2
Circle has a greater area compared to triangle.
Difference between the area of the circle and triangle = 4.56 m2
A square of side 'x' meters is bent in the form of a circle and later this circle is cut into two equal halves. One half is bent to form a square. Find the difference in the areas of the circle and the square bent from the half cut circle if x = 2. [4 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Application: 2 Marks
Given that
A square of side 'x' meters is bent into a circle.
The perimeter of the square = 4×x
The circle will have the same perimeter as that of the square = 4×x metres
Circle will be having a radius(r) = 4x2π
Area of the Circle so formed
= π×(4x2π)4x2πm2=4x2π
The circle which is half cut has a circumference of 2x.
Square bent from half circle has the circumference of 2x meters.
Side of the square = 2x4
Area of the square = x24
Putting x = 2
We get the area of the circle
= 4π×22 = 5.09 m2
The area of the square = 44 = 1m2
The difference between the area of the circle and the area of the square
= 5.09 - 1 = 4.09 m2
A man runs in a circular path. He runs towards the center from the circumference and after reaching the center he suddenly changes his direction from his original path at 180 degrees to the original direction and stopped when he reached the circumference. Find the area and perimeter of the circle, given that, the man ran 8 meters in the above-described path? [4 MARKS]
#### SOLUTION
Solution :
Formula: 1 Mark
Steps: 2 Marks
Given that
Total distance travelled by the man = 8 m
The man reached toward the center from the circumference to center and back to the center from the circumference.
So, the total distance covered by him is equal to twice the radius of the circle.
Let 'r' be the radius of the circle.
So, 2r = 8, r = 4
We know that area of the cicle is given by, Area=π×(r)2m2
Area of the given circle =π×(4)2m2
= (3.14)×(16)m2
=50.24m2
The perimeter of a circle= 2×π×r
=2×π×4
=25.14 m
So, the area of the given circle is 50.24m2, and the perimeter is 25.14 m.
If the area of parallelogram ABCD is 54 square units, what is the area of parallelogram ABEF? [Given: O and P are the midpoints of AD and BC respectively]
[4 MARKS]
#### SOLUTION
Solution :
Construction: 1 Mark
Application: 2 Marks |
# Know Your Numbers: Jumping Backwards in Twos (2)
In this worksheet, students count back in twos and complete the numbers in a sequence.
Key stage: KS 1
Curriculum topic: Number: Number and Place Value
Curriculum subtopic: Count in Steps (2, 3, 5 and 10)
Difficulty level:
### QUESTION 1 of 10
Freddie the frog loves jumping around on number lines. He always likes to make a clever number sequence. Use your finger to follow the jumps that he makes.
Example:
Count the frog's jumps to work out missing numbers in the sequence.
Pop your finger on the first number, then follow the jumps. Are the numbers getting bigger or smaller? Is Freddie going backwards or forwards? Can you see a pattern? What could the missing number be? What could the next number be?
21 19 17 15 A B
A = 13
B = 11
Count the frog's jumps to work out missing numbers in the sequence.
19 17 15 13 a b
a=11 and b=9
a=11 and b=10
a=9 and b=13
Count the frog's jumps to work out missing numbers in the sequence.
21 19 17 15 a b
a=13 and b=11
a=11 and b=10
a=9 and b=13
Count the frog's jumps to work out missing numbers in the sequence.
15 13 11 9 a b
a=5 and b=7
a=7 and b=5
a=10 and b=9
Count the frog's jumps to work out missing numbers in the sequence.
22 20 18 16 a b
a=14 and b=12
a=12 and b=14
a=15 and b=14
Count the frog's jumps to work out missing numbers in the sequence.
23 21 19 17 a b
a=15 and b=13
a=13 and b=15
a=13 and b=12
Count the frog's jumps to work out missing numbers in the sequence.
14 12 10 8 a b
a=7 and b=5
a=5 and b=6
a=6 and b=4
Count the frog's jumps to work out missing numbers in the sequence.
18 16 14 12 a b
a=10 and b=8
a=8 and b=10
a=11 and b=10
Count the frog's jumps to work out missing numbers in the sequence.
20 18 16 14 a b
a=12 and b=10
a=13 and b=12
a=10 and b=12
Count the frog's jumps to work out missing numbers in the sequence.
17 15 13 11 a b
a=9 and b=7
a=10 and b=9
a=12 and b=10
Count the frog's jumps to work out missing numbers in the sequence.
13 11 9 7 a b
a=7 and b=5
a=5 and b=3
a=3 and b=1
• Question 1
Count the frog's jumps to work out missing numbers in the sequence.
19 17 15 13 a b
a=11 and b=9
EDDIE SAYS
19, 17, 15, 13, 11, 9 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 2
Count the frog's jumps to work out missing numbers in the sequence.
21 19 17 15 a b
a=13 and b=11
EDDIE SAYS
21, 19, 17, 15, 13, 11 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 3
Count the frog's jumps to work out missing numbers in the sequence.
15 13 11 9 a b
a=7 and b=5
EDDIE SAYS
15, 13, 11, 9, 7, 5 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 4
Count the frog's jumps to work out missing numbers in the sequence.
22 20 18 16 a b
a=14 and b=12
EDDIE SAYS
22, 20, 18, 16, 14, 12 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 5
Count the frog's jumps to work out missing numbers in the sequence.
23 21 19 17 a b
a=15 and b=13
EDDIE SAYS
23, 21, 19, 17, 15, 13 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 6
Count the frog's jumps to work out missing numbers in the sequence.
14 12 10 8 a b
a=6 and b=4
EDDIE SAYS
14, 12, 10, 8, 6, 4 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 7
Count the frog's jumps to work out missing numbers in the sequence.
18 16 14 12 a b
a=10 and b=8
EDDIE SAYS
18, 16, 14, 12, 10, 8 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 8
Count the frog's jumps to work out missing numbers in the sequence.
20 18 16 14 a b
a=12 and b=10
EDDIE SAYS
20, 18, 16, 14, 12, 10 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 9
Count the frog's jumps to work out missing numbers in the sequence.
17 15 13 11 a b
a=9 and b=7
EDDIE SAYS
17, 15, 13, 11, 9, 7 is the sequence.
Freddie is jumping backwards 2 places each time.
• Question 10
Count the frog's jumps to work out missing numbers in the sequence.
13 11 9 7 a b
a=5 and b=3
EDDIE SAYS
13, 11, 9, 7, 5, 3 is the sequence.
Freddie is jumping backwards 2 places each time.
---- OR ----
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# How to Find the Inverse of an Exponential Function – A Simple Guide
To find the inverse of an exponential function, I first replace the function notation ( f(x) ) with ( y ).
This small change sets me up to address the function more like an equation involving ( y ) and ( x ). Next, I swap the roles of ( x ) and ( y ); where ( x ) was the input and ( y ) the output, they now switch positions.
This interchange is crucial because the inverse function essentially reverses the roles of inputs and outputs. An exponential function generally takes the form $y = a^{x}$, where ( a ) is a constant and ( a > 0 ).
To secure the inverse, I must solve the equation for ( x ) in terms of ( y ). This involves applying logarithms, given that the logarithmic functions are the inverse of exponential functions.
By following these steps, I can transform an exponential function into its logarithmic counterpart, revealing the underlying relationship between the two. Stay tuned as I walk us through the fascinating journey from exponential growth to its mirror image in the world of logarithms.
## Finding the Inverse of an Exponential Function
When I work with exponential functions, finding the inverse is crucial for understanding how inputs and outputs are related in a reversed manner. This requires two main steps: rewriting the function and applying logarithms.
### Rewriting the Function
The first thing I do is replace the function notation ( f(x) ) with ( y ) to make the equation simpler to manipulate. I am essentially saying that ( f(x) ) and ( y ) are the same. For example, if I have an exponential function $f(x) = b^x$, where ( b ) is the base, I rewrite this as $y = b^x$.
With this new form, my next move is to switch the ( x ) and ( y ) variables to reflect the inverse function. This means ( x ) becomes the output and ( y ) is now the input. Hence, my equation transforms to v x = b^y $. To obtain a function in terms of ( x ), I must solve for ( y ). ### Applying Logarithms Now, I use logarithms to solve the equation$x = b^y$algebraically. I take the logarithm with base ( b ) of both sides, which gives me$\log_b(x) = \log_b(b^y)$. The properties of logarithms tell me that$\log_b(b^y) = y$. So, my equation simplifies to vy = \log_b(x)$, and now ( y ) is isolated. This new function $g(x) = \log_b(x)$ is the inverse function of my original function $f(x) = b^x$ because it will return the input of the original function for any given output.
The domain of the original function $f(x) = b^x$, which is all real numbers, becomes the range of the inverse function $g(x) = \log_b(x)$, and the range of ( f(x) ), which is all positive real numbers $(0, \infty)$, becomes the domain of ( g(x) ).
Remember, a function must be one-to-one to have an inverse that is also a function. This is to ensure that for every output of the original function, there is a unique input. I can check this if the graph of ( f(x) ) passes the Horizontal Line Test, which it does if no horizontal line intersects the graph more than once.
## Graphical Representation of Inverses
When I graph an exponential function, it exhibits a distinctive curve that rises or falls rapidly. Let’s say I have an exponential function of the form $y = b^x$, where ( b ) is a positive real number different from 1.
To find its inverse graphically, I start by reflecting the graph across the line ( y = x ). This line acts as a mirror, transforming each point ( (x, y) ) on the original graph to the point ( (y, x) ) on the graph of the inverse function.
For a concrete example, imagine my function is $y = 2^x$. Its inverse would be $y = \log_2(x)$. The table of values below helps me plot a few points for both the function and its inverse:
( x )$y = 2^x$( x ) on inverse( y ) on inverse
-20.250.25-2
-10.50.5-1
0110
1221
2442
Taking these pairs of coordinates, I plot them on the same set of axes. My original function always passes through the point (0,1), since any number to the power of 0 is 1. The inverse, consequently, will always pass through (1,0), emphasizing the symmetric nature of the function and its inverse.
Remember that the domain and range switch roles in the inverse. If my original exponential function has a domain of all real numbers and a range of all positive real numbers, then my inverse function’s domain is all positive real numbers, and its range spans all real numbers.
It’s important to understand that every function does not have an inverse that is also a function. For the inverse to be a function, the original function must be one-to-one.
This means each ( y )-value corresponds to exactly one ( x )-value. It’s clear then, that while I can graph inverses for linear and quadratic functions, I must adjust quadratics to restrict the domain and ensure they’re one-to-one before finding the inverse graphically.
The same applies to rational functions, which require careful consideration to define a suitable domain that makes them one-to-one.
## Conclusion
I’ve walked you through the essential steps of finding the inverse of an exponential function: swapping the ( x ) and ( y ), re-expressing the equation to isolate the new ( y ), and using logarithms when necessary.
The process reveals a crucial relationship between exponential and logarithmic functions, highlighting how they serve as inverses of each other. This characteristic translates into a shared domain and range—the domain of an exponential equates to the range of its logarithmic inverse, and vice versa.
Remember, for an exponential function vy = b^x$, the logarithmic inverse would be$x = \log_b(y)$or$y = b^x \implies x = \log_b(y)\$, ensuring that understanding one inherently means comprehending the other.
The elegance of mathematics lies in these symmetrical relationships, allowing us to tackle complex problems from different angles. My goal has been to offer you a clear and direct path to demystify the inverse of an exponential function.
Friendly and steady reinforcement of these concepts will build your confidence and enhance your mathematical fluency. Remember to practice, as with each attempt, the steps become more intuitive, and the connection between exponentials and logarithms grows stronger. |
# More Multiplication Mastery
Here’s another way to reinforce multiplication concepts with your elementary school kids. You just need a hundreds table and some markers. You can present it like a puzzle. Or you can be fancy and tell your kids that they’re going to make the Sieve of Eratosthenes. Yes, that will get their attention.
Here’s the quickie info:
1. Print this hundreds chart.
2. Get a handful of markers.
3. Tell your child that 1 is a special number and gets to be circled right from the start. Let her choose a color to circle it with.
4. Next tell your child that 2 is also a special number and gets to be circled. Choose any color. Isn’t this exciting?
5. Now tell your child to cross out any number in the chart that is a multiple of 2. Bye-bye even numbers! (Except for 2, which is circled.)
6. Next tell your child to circle 3, then cross out all the multiples of 3.
7. Now tell your child to circle 4…wait! You cannot circle 4 because it has already been crossed out (even number), so circle the next available number…5. Then cross out all multiples of 5.
8. Continue like this until all numbers in the hundreds table have either been circled or crossed out. There will be 26 circled numbers (25 prime numbers, and 1 – which is not prime.)
As you sit with your child, observe how he works. Slowly? Excitedly? Does the color of the marker and the precision of his circles and x’s matter to them? Have patience. (Get that cup of tea or coffee that’s always nice to have around when practicing patience.) Ask your child what she notices about the table. Ask your child what she wonders about the table. Observe what gets your child’s attention. Do not get frustrated if he loses attention. Try again another time.
(Mathematically…this activity reinforces multiplication facts without being directly about multiplication facts. As your child works through the table, she might be counting on from each number or adding to each number or actually multiplying each number. Whichever method he is using, he will be working on his facility with numerical manipulation. She will get visual reinforcement of her mental processes. He may notice patterns. She may notice numbers that get crossed out multiple times. In working through the table and observing the colorful work that is created, your child will create new connections to the work. And even though I’ve presented this as a way to reinforce multiplication facts, it can also be a way to talk about prime numbers and algorithms.)
## 3 thoughts on “More Multiplication Mastery”
1. Another very useful post Pam. I will try this with my second grader. And I loved the way you used he and she alternately 🙂
Liked by 1 person
1. Ha! I’m glad you mentioned that…I was wondering if switching pronouns made sense or just sounded strange!
Liked by 1 person
1. 😊. In one of my posts I used
s/he when I was talking about an earthworm. So when I noticed this , I thought using it alternately was a good way of dealing with it.
Liked by 1 person |
# Consider the function defined by f(x)= sinx/x if -1 <=x < 0, ax+b if 0 < =x <= 1 and (1-cosx)/x if x >1, how do you determine a and b such that f(x) is continuous on [-1,2]?
Jun 24, 2017
$a = - \cos 1 \approx = 0.540$
$b = 1$
#### Explanation:
Assuming we are working in radians.
We want our function, $f \left(x\right)$ to be defined by:
$f \left(x\right) = \left\{\begin{matrix}\sin \frac{x}{x} & - 1 \le x < 0 \\ a x + b & 0 < x \le 1 \\ \frac{1 - \cos x}{x} & x > 1\end{matrix}\right.$
Let us denote each sub-function by ${f}_{1} \left(x\right) , {f}_{2} \left(x\right)$ and ${f}_{3} \left(x\right)$ such that:
$f \left(x\right) = \left\{\begin{matrix}{f}_{1} \left(x\right) = \sin \frac{x}{x} & - 1 \le x < 0 \\ {f}_{2} \left(x\right) = a x + b & 0 < x \le 1 \\ {f}_{3} \left(x\right) = \frac{1 - \cos x}{x} & x > 1\end{matrix}\right.$
We can graph for illustrative purposes, ${f}_{1} \left(x\right)$ and ${f}_{3} \left(x\right)$ for the domain $x \in \mathbb{R}$
And we note that all three sub-functions are continuous in their own right.
If we now examine the functions ${f}_{1} \left(x\right)$ and ${f}_{1} \left(x\right)$ restricted to their appropriate domains we have:
So, as we expected there is a discontinuity at the junctions between ${f}_{1} \left(x\right)$ and ${f}_{2} \left(x\right)$ and between ${f}_{2} \left(x\right)$ and ${f}_{3} \left(x\right)$ so as we require a linear function $a x + b$ to complete the continuity we simply need to find the two coordinates associated with the discontinuity and find the unique equation of the straight line that [passes through those point.
Using some standard calculus trigonometry limits we have:
When $x = 0 \implies {f}_{1} \left(x\right) = {\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$
When $x = 1 \implies {f}_{3} \left(x\right) = {\lim}_{x \rightarrow 0} \frac{1 - \cos x}{x} = 1 - \cos 1$
So the coordinates of the discontinuities are:
$\left(0 , 1\right)$ and $\left(1 , 1 - \cos 1\right)$
So using the straight line point/point equation:
$\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$
then the equation we seek is:
$\frac{y - 1}{1 - \cos 1 - 1} = \frac{x - 0}{1 - 0}$
$\therefore \frac{y - 1}{- \cos 1} = x$
$\therefore y - 1 = - x \cos 1$
$\therefore y = - x \cos 1 + 1$
Thus comparing $y = - x \cos 1 + 1$ with y=ax+b# we require:
$a = - \cos 1 \approx = 0.540$
$b = 1$ |
# Right angled triangle and Application of right angled triangle
Right Angled Triangle:
In a
right angled triangle,
ABC with sides
BC = a
CA = b and
AB = c ;
right angled triangle
We know that:
$\sin A = \dfrac{a}{c}$ or , $a = c \sin A$
And $\cos A = \dfrac{b}{c}$ or , $b = c \cos A$
And $\tan A = \dfrac{a}{b}$ or , $a = b \tan A$
It is obvious that SIN A = COS B , COS A = SIN B because:
A+B = 90 , So A is the complement angle of B , this may be stated as:
Sine of the angle A = Sine of the complement of B
So , Sine of the angle A = CoSine of angle B
Or , $\sin A = \sin (90-B) = \cos B$
And similarly:
$\csc A = \sec B$ and $\tan A = \cot B$
In a triangle , there are three angles and three sides. They are known as the six components or elements of a triangle.
To solve a triangle means to find unknown elements from the given parts.
It is always possible to solve a triangle if three of it’s parts are given (Except for the case that all three parts given are angles)
In solving problems of practical interest in which right-angled triangles appear , we shall use some new terms. They are “The Point of observation” , “Horizontal” , “Line of Sight” , “Angle of Elevation” , “Angle of Depression”.
These terms are diagrammatically illustrated illustrated below:
Application Of Right Angled Triangle:
To solve a real life problem involving right angled triangle , we first collect the given information and then solve the triangle and find the unknown parameters.
For example:
Q. A person 30 meters away from the feet of a tower finds that his line of sight of top of the tower is making an angle of 60 degrees with the horizontal , then find the Height of thee tower.
Solution:
First of all let’s make a visualisation of the situation in diagrammatic form as:
application of right angled triangle
Where , AC is the tower and B is the point from where the person is watching the top of thee tower.
So in the Right Angled triangle ABC ,
$\tan 60 = \dfrac{AC}{BC} = \dfrac{AC}{30}$
Or, $AC = \tan 60 \times 30 = \sqrt{3} \times 30 = 51.96$
Thus, height of the tower = AC = 51.96 Meters.
Related posts:
1. Pythagorian Identities Fundamental Pythagorian identity of trigonometry and other basic trigonometric formulas...
2. Trigonometric Functions What are trigonometric functions such as sine , cosine ,...
3. Derivatives of Trigonometric functions. As you know, The functions SINE x(sin x) , CO-SECANT...
4. Trigonometric functions of negative angles Trigonometric functions of negative angles. How to find trigonometric functions...
5. Derivatives of inverse trigonometric functions Inverse trigonometric functions are the inverse of trigonometric functions .... |
absolutely convergent series and conditionally convergent series rearrangement
I don't understand that why the terms of an absolutely convergent series can be rearranged in any order and all such rearranged series converge to the same sum.
my textbook gives me an example that it is not so for conditionally convergent series
$1-\frac12+\frac13-\frac14+\ldots=\ln2$
$1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16...=\frac32\ln2$
which I also don't know how to prove
-
(Regarding your first paragraph) Well, that's a theorem (of Weierstrass). So Step 1 is to find a proof of the theorem and try to understand it. Have you done that already? If not, try $\S$ 14.2.2 of math.uga.edu/~pete/2400full.pdf. – Pete L. Clark Apr 10 '13 at 15:39
Here is something wild: not only do rearrangements of conditionally convergent series give different answers, we can rearrange them to make them equal anything we want!
Let $\sum a_n$ be a conditionally convergent series, and define two new sequences:
$$a^+_n=\frac{a_n+|a_n|}{2}=\begin{cases}a_n&a_n\ge0\\0&a_n<0\end{cases}\\a^-_n=\frac{a_n-|a_n|}{2}=\begin{cases}a_n&a_n\le0\\0&a_n>0\end{cases}$$
By the divergence of $\sum|a_n|$, it follows that both $\sum a^+_n$ and $\sum a_n^-$ diverge, but we have $\lim_{n\to\infty}a_n^+=\lim_{n\to\infty}a_n^-=0$. Now, choose any $r\in\mathbb{R}$. We assume $r\ge 0$, but the proof is similar in the case $r\le 0$.
Without being too explicit, here is how to rearrange the terms of $\sum a_n$ so that the sum of the rearrangement is equal to $r$. Take the first terms to be the first terms of $a_n^+$ until the partial sum is greater than $r$. Then, allow the next terms to come from $a_n^-$ until the partial sum is less than $r$. Continue this process to obtain the desired rearrangement.
Two things we must note. First, the divergence of the positive and negative sums guarantees we will always be able to jump back and forth across $r$. Second, since the limit of the sequences go to zero, the jumping back and forth across $r$ will eventually get more and more precise, so that the limit will approach $r$.
Fascinating.
-
The first one equals to $$\sum_{i=0}^n (-1)^n(1/i) = \ln 2$$ The second one equals to $$[1+1/2+……+1/4n]-[1/2+1/4+……+1/2n]-[1/2+1/4+……+1/4n] =\ln4n-(1/2)\ln n-(1/2)\ln 2n=(3/2)\ln 2$$
- |
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# 1.2: Order of Operations
Difficulty Level: At Grade Created by: CK-12
## Introduction
Gym Class Changes
Wood shop wasn’t the only thing that had changed with the new year. After working to successfully resolve wood shop being reinstated, the students found out that there were also changes for gym class or physical education. It seems that the old gym teacher, Mr. Woullard had retired and there was a new gym teacher.
Mr. Osgrove was young and lively with lots of energy, but he also had some new ideas about how gym class ought to be run.
“We’ll combine two periods of students together,” he explained. “That will give us so many more combinations of students when it comes to teams.”
Jesse looked around. He counted the number of boys and girls in his class. There were 11 boys and 14 girls in the class. The other class had thirteen boys and fourteen girls in it.
“We can add the boys together and form four teams and the girls together and form four teams.”
Jesse left gym class with his head full of numbers. If they were to combine all of the boys from the two classes and all of the girls from the two classes, then that would be a lot of students. Jesses started to figure out the different combinations of teams.
To work this through, you can use the order of operations. That is what you will learn in this lesson. Pay close attention, then you can work this through at the end of the lesson.
What You Will Learn
By the end of this lesson you will learn how to execute the following skills.
• Evaluate numerical and variable expressions involving arithmetic operations.
• Evaluate numerical and variable expressions involving grouping symbols.
• Write variable expressions to represent and solve real-world problems using order of operations.
Teaching Time
I. Evaluate Numerical and Variable Expressions Involving Arithmetic Operations
In mathematics, you will often hear the word “evaluate.” Before we begin the lesson, it is important for you to understand what the word “evaluate” means. When we evaluate a mathematical sentence, we figure out the value of the number sentence. If the mathematical sentence has numbers in it, then we figure out the value of the number sentence. Often times we think of evaluating as solving, and it can be that, but more specifically, evaluating is figuring out the value of a sentence.
What do we evaluate?
In mathematics, we can evaluate different types of number sentences. Sometimes we will be working with equations and other times we will be working with expressions.
That is a great question. An equation is a number sentence with an equals sign. Therefore, the quantity on one side of the equals sign is equal to the quantity on the other side of the equals sign. An expression is a group of numbers, symbols and variables that represents a quantity; there is not an equal sign. We evaluate an expression to figure out the value of the mathematical statement itself, we are not trying to make one side equal another, as with an equation.
Let’s start by focusing on evaluating expressions. Think about this example.
Example
Two eighth grade math students evaluated the expression: \begin{align*}2 + 3 \times 4 \div 2\end{align*}. Both students approached the expression differently. Macy’s answer was ten. Cole’s answer was eight.
We can think about what happened here. How could one student arrive at one answer and another student come up with an entirely different answer? The key is in the order that each student performed each operation.
This is where the order of operations comes in. The order of operations is a rule that tells us which operation we need to perform in what order to achieve the accurate answer. The order of operations applies whenever you have two or more operations in a single expression. Here is the order of operations.
Order of Operations
P parentheses or grouping symbols
E exponents
MD multiplication and division in order from left to right
AS addition and subtraction in order from left to right
Take a few minutes to write down the order of operations in your notebook.
Now the example above does not have any parentheses or exponents, so don’t worry about those just yet. Let’s look at the example again and look at how Macy and Cole arrived at their answers.
Example
\begin{align*}2 + 3 \times 4 \div 2\end{align*}
Cole worked on evaluating this expression using the order of operations. He multiplied \begin{align*}3 \times 4 = 12\end{align*}. Then he divided by 2 which is 6 and finally he added 2 for a final answer of 8. This is correct. It may seem out of order to work this way, but remember you are working according to the order of operations.
What did Macy do? Macy worked on evaluating the expression by working in order from left to right. She simply did not follow the order of operations. In this case, she evaluated the value of the expression as 10. This is incorrect. Next time, Macy needs to follow the order of operations.
Working in this way is called evaluating a numerical expression. It is a numerical expression because it is made up only of numbers and operations.
Example
Evaluate \begin{align*}3 + 9 \cdot 2 \div 3 + 8\end{align*}
To begin with, we first need to remind ourselves of the order of operations. Notice that there is a dot in between the nine and the two. This is another way to show multiplication. As you get into higher levels of math, you will see that multiplication is often shown in other ways besides using an \begin{align*}x\end{align*}. Now back to evaluating.
Following the order of operations, we would first multiply and then divide.
\begin{align*}9 \cdot 2 = 18\end{align*}
\begin{align*}18 \div 3 = 6\end{align*}
Next we perform addition and subtraction in order from left to right.
\begin{align*}6 + 3 = 9\end{align*}
\begin{align*}9 + 8 = 17\end{align*}
This is our answer.
You will also find another type of expression. These expressions can have letters in them. These letters are called variables and variables represent an unknown quantity. When you see an expression with a variable in it, we call it a variable expression.
We can evaluate variable expressions using the order of operations too. The key thing with a variable expression is that you will have to substitute a given value into the expression for the unknown variable and then evaluate the expression. Let’s look at an example.
Example
Evaluate the expression \begin{align*}60 \div 2 \cdot 2a + 16 - 4\end{align*} when \begin{align*}a = 5\end{align*}.
First, notice that the expression has the letter \begin{align*}a\end{align*} in it. This is our variable. Also, you can see that you have been given a value for \begin{align*}a\end{align*}. Our first step is to substitute the value of \begin{align*}a\end{align*} into the expression.
\begin{align*}60 \div 2 \cdot 2(5) + 16 - 4\end{align*}
Another good question-using the parentheses is another way to show multiplication. So now you have three ways to show multiplication. You can use an \begin{align*}x\end{align*}, a dot or a set of parentheses around a number. This means that we are multiplying the 2 times the 5.
Now we can use our order of operations. We have division, multiplication and multiplication in this expression right away. We complete multiplication and division in order from left to right.
Another good question! In this case, the set of parentheses is not grouping two numbers and an operation. When talking about parentheses in the order of operations, we need to have an operation inside it. The set of parentheses in this example is being used to show multiplication. There isn’t multiplication inside the parentheses.
Now back to evaluating the expression by performing multiplication and division in order from left to right.
\begin{align*}& 60 \div 2 \cdot 2(5) + 16 - 4\\ & 60 \div 2 = 30\\ & 30 \cdot 2(5) = 300\end{align*}
Next, we work with addition and subtraction in order from left to right.
\begin{align*}& 300 + 16 - 4\\ & 316 - 4\\ & 312\end{align*}
Our final answer is 312.
Example
Evaluate the expression \begin{align*}14 \cdot 2 \div 7 + 3b - 4\end{align*} when \begin{align*}b = 12\end{align*}.
First, we substitute the value of \begin{align*}b\end{align*} into our variable expression.
\begin{align*}14 \cdot 2 \div 7 + 3(12) - 4\end{align*}
Now we follow the order of operations by performing multiplication and division in order from left to right.
\begin{align*}14 \cdot 2 = 28\end{align*}
\begin{align*}28 \div 7 = 4\end{align*}
Let’s rewrite what we have so far so we don’t get confused.
\begin{align*}4 + 3(12) - 4\end{align*}
OH! There’s more multiplication to do!
\begin{align*}3(12) = 36\end{align*}
Now our expression is:
\begin{align*}4 + 36 - 4\end{align*}
Our last step is to perform addition and subtraction in order from left to right.
\begin{align*}4 + 36 = 40 - 4 = 36\end{align*}
Our final answer is 36.
Now that you have had some practice the order of operations is probably beginning to make more sense. If you always follow them, then your work will be accurate. You can think of the order of operations as a kind of road map for working with complicated expressions.
Next, we are going to look at the first step in the order of operations. Let’s talk about grouping symbols.
II. Evaluate Numerical and Variable Expressions Involving Grouping Symbols
Grouping symbols are a way to point out an operation that needs special attention. If you can think of a spotlight, grouping symbols are a way of highlighting or shining a spotlight on a particular operation.
What are the grouping symbols?
The grouping symbols that we are going to be working with are brackets [ ] and parentheses ( ). According to the order of operations, we perform all operations inside of grouping symbols BEFORE performing any other operation in the list.
Example
Evaluate the expression \begin{align*}7 + 4(15 \div 5) - 6\end{align*}.
First, notice that we have a set of parentheses in this numerical expression. Remember that it is called a numerical expression because this expression does not have any variable in it.
We perform any and all operations in parentheses first.
\begin{align*}15 \div 5 = 3\end{align*}
Now let’s rewrite the expression.
\begin{align*}7 + 4(3) - 6\end{align*}
Our next step is to continue with the order of operations. We have multiplication in this expression. We do that next.
\begin{align*}7 + 12 - 6\end{align*}
Now we can perform addition and subtraction in order from left to right.
\begin{align*}19 - 6 = 13\end{align*}
Our final answer is 13.
We can also evaluate variable expressions that contain parentheses.
Example
Evaluate the expression \begin{align*}6y + 3 - (2 \cdot 4)\end{align*} when \begin{align*}y = 15\end{align*}.
First, substitute the value in for the unknown quantity, \begin{align*}y\end{align*}.
\begin{align*}6(15) + 3 - (2 \cdot 4)\end{align*}
Our next step is to perform the operation in parentheses.
\begin{align*}6(15) + 3 - 8\end{align*}
Now we can perform multiplication and division in order from left to right.
\begin{align*}90 + 3 - 8\end{align*}
Finally, we work with addition and subtraction in order from left to right.
\begin{align*}& 93 - 8 \\ & 85\end{align*}
Our answer is 85.
Brackets can be used to group more than one operation. When you see a set of brackets, remember that brackets are a way of grouping numbers and operations. There is a spotlight on brackets too.
Example
Evaluate the expression \begin{align*}6 + [5 + (4 \cdot 6)] - 17\end{align*}.
Now we have a set of parentheses within a set of brackets. To work through this, we are going to perform the operation in the parentheses inside the brackets first, and then we will perform the other operation in the brackets.
\begin{align*}& 6 + [5 + 24] - 17\\ & 6 + 29 - 17\end{align*}
Next we perform addition and subtraction in order from left to right.
\begin{align*}& 35 - 17\\ & 18\end{align*}
Our final answer is 18.
Remember that when you are following the road map laid out by the order of operations that you need to include grouping symbols of brackets and parentheses in that order!
III. Write Variable Expressions to Represent and Solve Real-World Problems Using Order of Operations
When we have an unknown quantity in a problem, we can use a variable to help us to find the value of our problem. You can find many real-world scenarios where variable expressions would be helpful in solving problems.
Let’s think about ticket pricing. Many different places sell tickets. There are also prices for adults and prices for children. The number of tickets can vary, or they can be the same. If we know the number of tickets, then we can figure out the total amount of revenue earned based on the cost of the ticket for an adult and for a child. Let’s look at an example to see how this works.
Example
An amusement park charges eight dollars admission and one dollar and fifty-cents per ride. Write an expression to find the cost of admission and five ride tickets.
First, notice that we have eight dollars admission. That is the first part of the expression.
8
Next, they charge 1.50 per ride. \begin{align*}8 + 1.50x\end{align*} We used the variable for the number of rides since this is the variable or changeable facet. The number of rides can change. In this example, we were asked to figure out the cost for 5 rides. That is the value that we can substitute into our expression for \begin{align*}x\end{align*}. \begin{align*}8 + 1.50(5)\end{align*} Now we can use the order of operations to solve this problem. \begin{align*}& 8 + 7.50 \\ & 15.50\end{align*} The cost of admission plus five ride tickets is15.50.
Using variable expressions can help you solve for unknown quantities. Just remember to use the order of operations so that your work is accurate!!
## Real-Life Example Completed
Gym Class Changes
Here is the original problem once again. Reread it and then write two expressions to show how the teams will be divided up. How many boys will be on each team if there are four teams? How many girls will be on four teams? There are four parts to your answer.
Wood shop wasn’t the only thing that had changed with the new year. After working to successfully resolve wood shop being reinstated, the students found out that there were also changes for gym class or physical education. It seems that the old gym teacher, Mr. Woullard had retired and there was a new gym teacher.
Mr. Osgrove was young and lively with lots of energy, but he also had some new ideas about how gym class ought to be run.
“We’ll combine two periods of students together,” he explained. “That will give us so many more combinations of students when it comes to teams.”
Jesse looked around. He counted the number of boys and girls in his class. There were 11 boys and 14 girls in the class. The other class had thirteen boys and fourteen girls in it.
“We can add the boys together and form four teams and the girls together and form four teams.”
Jesse left gym class with his head full of numbers. If they were to combine all of the boys from the two classes and all of the girls from the two classes, then that would be a lot of students. Jesses started to figure out the different combinations of teams.
Now it is time to work on this solution. Write two expressions to show how the groups are divided. Then answer the two questions of how many boys will be on a team and how many girls will be on a team.
Solution to Real – Life Example
First, let’s look at the information that we have been given in the problem.
Class one has 11 boys and 14 girls.
Class two has 13 boys and 14 girls.
The boys from the two classes will be added together, and the girls from the two classes will be added together.
\begin{align*}11 + 13\end{align*}
\begin{align*}14 + 14\end{align*}
We can use parentheses to show that the boys will be added and the girls will be added. Both groups will be divided by four.
\begin{align*}(11 + 13) \div 4\end{align*}
Or \begin{align*}\frac{11+13}{4}\end{align*}
This is an expression for the boys.
\begin{align*}(14 + 14) \div 4\end{align*}
Or \begin{align*}\frac{14+14}{4}\end{align*}
This is an expression for the girls.
Now we can solve for the number on each team.
\begin{align*}\text{Boys} = (11 + 13) \div 4 = 24 \div 4 = 6\end{align*} boys on each team
\begin{align*}\text{Girls} = (14 + 14) \div 4 = 28 \div 4 = 7\end{align*} girls on each team
Now our work is complete.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Evaluate
to figure out the value of a numerical or variable expression.
Equation
a mathematical statement with an equals sign where one side of the equation has the same value as the other side.
Expression
a group of numbers, symbols and variables that represents a quantity.
Numerical Expression
a group of numbers and operations.
Variable Expression
a group of numbers, operations and at least one variable.
Variable
a letter used to represent an unknown quantity.
Grouping Symbols
parentheses or brackets used to group numbers and operations.
## Time to Practice
Directions: Evaluate each numerical expression using the order of operations.
1. \begin{align*}4 + 5 \cdot 2 - 3\end{align*}
2. \begin{align*}6 + 6 \cdot 3 \div 2 - 7\end{align*}
3. \begin{align*}5 + 5 \cdot 8 \div 2 + 6\end{align*}
4. \begin{align*}13 - 3 \cdot 2 + 8 - 2\end{align*}
5. \begin{align*}17 - 5 \cdot 3 + 8 \div 2\end{align*}
6. \begin{align*}9 + 4 \cdot 2 + 7 - 1\end{align*}
7. \begin{align*}8 + 5 \cdot 6 + 2 \cdot 4 - 3\end{align*}
8. \begin{align*}19 + 2 \cdot 4 - 3 \cdot 2 + 10\end{align*}
9. \begin{align*}12 + 4 \cdot 4 \div 8 - 3\end{align*}
10. \begin{align*}12 \cdot 2 + 16 \div 2 - 12\end{align*}
Directions: Evaluate each variable expression. Remember to use the order of operations when necessary.
1. \begin{align*}4y+6-2\end{align*}, when \begin{align*}y=6\end{align*}
2. \begin{align*}9+3x-5+2\end{align*}, when \begin{align*}x=8\end{align*}
3. \begin{align*}6y+2y-5\end{align*}, when \begin{align*}y=3\end{align*}
4. \begin{align*}8+3y-5 \cdot 2\end{align*}, when \begin{align*}y=4\end{align*}
5. \begin{align*}7x-2 \cdot 3 \div 3+12\end{align*}, when \begin{align*}x=5\end{align*}
6. \begin{align*}3+4 \cdot 3 - 2y+5\end{align*}, when \begin{align*}y=7\end{align*}
7. \begin{align*}6a+3(2)+5 - 4\end{align*}, when \begin{align*}a=9\end{align*}
8. \begin{align*}10+3 \cdot 5+2-9b\end{align*}, when \begin{align*}b=2\end{align*}
9. \begin{align*}14 \div 2+3a+7a\end{align*}, when \begin{align*}a=2\end{align*}
10. \begin{align*}5+6y-2y+11-4\end{align*}, when \begin{align*}y=3\end{align*}
Directions: Evaluate each expression using the order of operations. Remember to pay attention to the grouping symbols.
1. \begin{align*}3 + (4 + 5) - 6(2)\end{align*}
2. \begin{align*}4 + (6 \div 3) + 2(7) - 4\end{align*}
3. \begin{align*}3 + 2(4 + 2) - 5(2)\end{align*}
4. \begin{align*}7 + (3 + 2) - 5 + 8(3)\end{align*}
5. \begin{align*}4(2) + (3 + 9) - 4\end{align*}
6. \begin{align*}7 + [4 + (3 \cdot 2)] - 5\end{align*}
7. \begin{align*}9 + [6 - (2 \cdot 3)] + 15\end{align*}
8. \begin{align*}4 \cdot 2 + [3 + (7 + 2)] - 4\end{align*}
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# Learning to show the remainder
## Presentation on theme: "Learning to show the remainder"— Presentation transcript:
Learning to show the remainder
Division Learning to show the remainder
17÷ 4 4 r1 Division: Learning to show the remainder
How many groups of 4 are there in 17? 4 r1
Tables knowledge 4 x 4 = 16 17 ÷ 4 = 4r1
Division: Learning to show the remainder 17÷ 4 r1 17 +4 4 +4 8 +4 12 +4 16 Tables knowledge 4 x 4 = ÷ 4 = 4r1 4 r1
Division: Learning to show the remainder
17÷ 4 17 ÷ 4 = 4 r 1 As a remainder 17 ÷ 4 = 4 1 4 As a fraction 17 ÷ 4 = 4.25 As a decimal fraction
Division: Learning to show the remainder
18÷ 4 18 ÷ 4 = 4 r 2 As a remainder 18 ÷ 4 = 4 1 2 As a fraction 18 ÷ 4 = 4.50 As a decimal fraction
Division: Learning to show the remainder
19÷ 4 19 ÷ 4 = 3 r 1 As a remainder 19 ÷ 4 = 4 3 4 As a fraction 19 ÷ 4 = 4.75 As a decimal fraction
Division: Learning to show the remainder
21÷ 5 21 ÷ 5 = 4 r 1 As a remainder 21 ÷ 5 = 4 1 5 As a fraction 21 ÷ 5 = 4.20 As a decimal fraction
Division: Learning to show the remainder
22÷ 5 22 ÷ 5 = 4 r 2 As a remainder 22 ÷ 5 = 4 2 5 As a fraction 21 ÷ 5 = 4.40 As a decimal fraction
Division: Learning to show the remainder
23÷ 5 23 ÷ 5 = 4 r 3 As a remainder 23 ÷ 5 = 4 3 5 As a fraction 21 ÷ 5 = 4.60 As a decimal fraction
Division: Learning to show the remainder
24÷ 5 24 ÷ 5 = 4 r 4 As a remainder 24 ÷ 5 = 4 4 5 As a fraction 24 ÷ 5 = 4.80 As a decimal fraction
Division: Learning to show the remainder
If the numbers are bigger than your tables knowledge have you got a division method for solving them?? 58 ÷ 4
Division: Learning to show the remainder
58 ÷ 4 = 14 r2 + 40 + 16 + 2 10 X 4 4 X 4 r 2 40 56 58
Division: Learning to show the remainder
58 ÷ 4 = 14 r2 1 4 r 2 1 4 5 8
Division: Learning to show the remainder For each question show the three ways of showing the remainder. ÷ ÷ ÷ ÷ 5 ÷ ÷ ÷ ÷ 5 |
## Step by step solution because that calculating 3 is 5 percent the what number
We currently have our first value 3 and also the 2nd value 5. Let"s assume the unknown worth is Y i m sorry answer us will uncover out.
You are watching: 3 is what percent of 5
As we have all the forced values us need, currently we have the right to put them in a straightforward mathematical formula together below:
STEP 13 = 5% × Y
STEP 23 = 5/100× Y
Multiplying both sides by 100 and also dividing both sides of the equation by 5 we will arrive at:
STEP 3Y = 3 × 100/5
STEP 4Y = 3 × 100 ÷ 5
STEP 5Y = 60
Finally, us have found the value of Y i m sorry is 60 and also that is our answer.
You can quickly calculate 3 is 5 percent that what number through using any regular calculator, simply enter 3 × 100 ÷ 5 and also you will gain your answer i beg your pardon is 60
## People also Ask
Here is a portion Calculator to solve comparable calculations such as 3 is 5 percent of what number. You can solve this type of calculation through your worths by start them right into the calculator"s fields, and also click "Calculate" to get the result and explanation.
is
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## Sample questions, answers, and also how to
Question: her friend has a bag of marbles, and also he speak you the 5 percent the the marbles space red. If there are 3 red marbles. How numerous marbles walk he have altogether?
Answer: 60 marbles.
How To: In this problem, we recognize that the Percent is 5, and also we are likewise told the the part of the marbles is red, for this reason we know that the component is 3.
So, that means that it must be the full that"s missing. Below is the means to number out what the full is:
Part/Total = Percent/100
By making use of a basic algebra we deserve to re-arrange our Percent equation favor this:
Part × 100/Percent = Total
If we take the "Part" and also multiply that by 100, and then we divide that by the "Percent", we will obtain the "Total".
Let"s try it out on our problem about the marbles, that"s very straightforward and it"s just two steps! We recognize that the "Part" (red marbles) is 3.
So step one is to simply multiply that component by 100.
3 × 100 = 300
In action two, us take the 300 and also divide the by the "Percent", i beg your pardon we are told is 5.
So, 300 split by 5 = 60
And that way that the total variety of marbles is 60.
Question: A high institution marching band has 3 flute players, If 5 percent of the band members pat the flute, climate how countless members room in the band?
Answer: There space 60 members in the band.
How To: The smaller sized "Part" in this problem is 3 because there room 3 flute players and we space told the they comprise 5 percent of the band, so the "Percent" is 5.
Again, it"s the "Total" that"s lacking here, and to discover it, we just need to monitor our 2 action procedure as the ahead problem.
For action one, we multiply the "Part" by 100.
3 × 100 = 300
For step two, we divide that 300 by the "Percent", i m sorry is 5.
See more: Learn What Noise Does A Lion Make When They'Re Angry? Lion Roaring: What Makes A Lion'S Roar So Loud
300 separated by 5 amounts to 60
That way that the total number of band members is 60.
## Another action by step method
Step 1: Let"s assume the unknown worth is Y
Step 2: first writing the as: 100% / Y = 5% / 3
Step 3: fall the portion marks to simplify your calculations: 100 / Y = 5 / 3
Step 4: multiply both sides by Y to relocate Y top top the ideal side of the equation: 100 = ( 5 / 3 ) Y
Step 5: simple the best side, us get: 100 = 5 Y
Step 6: separating both political parties of the equation by 5, we will arrive at 60 = Y
This leaves us v our final answer: 3 is 5 percent the 60
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# 79 1.1 Review of Functions – Calculus Volume 1 | OpenStax (MathJax chapter import test)
In this section, we provide a formal definition of a function and examine several ways in which functions are represented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves as a handy reference to remind you of some of the algebraic techniques useful for working with functions.
### Functions
Given two sets
$AA$and
$B,B,$a set with elements that are ordered pairs
$(x,y),(x,y),$where
$xx$is an element of
$AA$and
$yy$is an element of
$B,B,$is a relation from
$AA$to
$B.B.$A relation from
$AA$to
$BB$defines a relationship between those two sets. A function is a special type of relation in which each element of the first set is related to exactly one element of the second set. The element of the first set is called the input; the element of the second set is called the output. Functions are used all the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to study them.
### Definition
A function
$ff$consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The set of inputs is called the domain of the function. The set of outputs is called the range of the function.
For example, consider the function
$f,f,$where the domain is the set of all real numbers and the rule is to square the input. Then, the input
$x=3x=3$is assigned to the output
$32=9.32=9.$Since every nonnegative real number has a real-value square root, every nonnegative number is an element of the range of this function. Since there is no real number with a square that is negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real numbers.
For a general function
$ff$with domain
$D,D,$we often use
$xx$to denote the input and
$yy$to denote the output associated with
$x.x.$When doing so, we refer to
$xx$as the independent variable and
$yy$as the dependent variable, because it depends on
$x.x.$Using function notation, we write
$y=f(x),y=f(x),$and we read this equation as
$“y“y$equals
$ff$of
$x.”x.”$For the squaring function described earlier, we write
$f(x)=x2.f(x)=x2.$
The concept of a function can be visualized using Figure 1.2, Figure 1.3, and Figure 1.4.
Figure
1.2
A function can be visualized as an input/output device.
Figure
1.3
A function maps every element in the domain to exactly one element in the range. Although each input can be sent to only one output, two different inputs can be sent to the same output.
Figure
1.4
In this case, a graph of a function $ff$has a domain of
${1,2,3}{1,2,3}$and a range of
${1,2}.{1,2}.$The independent variable is
$xx$and the dependent variable is
$y.y.$
### Media
We can also visualize a function by plotting points
$(x,y)(x,y)$in the coordinate plane where
$y=f(x).y=f(x).$The graph of a function is the set of all these points. For example, consider the function
$f,f,$where the domain is the set
$D={1,2,3}D={1,2,3}$and the rule is
$f(x)=3−x.f(x)=3−x.$In Figure 1.5, we plot a graph of this function.
Figure
1.5
Here we see a graph of the function $ff$with domain
${1,2,3}{1,2,3}$and rule
$f(x)=3−x.f(x)=3−x.$The graph consists of the points
$(x,f(x))(x,f(x))$for all
$xx$in the domain.
Every function has a domain. However, sometimes a function is described by an equation, as in
$f(x)=x2,f(x)=x2,$with no specific domain given. In this case, the domain is taken to be the set of all real numbers
$xx$for which
$f(x)f(x)$is a real number. For example, since any real number can be squared, if no other domain is specified, we consider the domain of
$f(x)=x2f(x)=x2$to be the set of all real numbers. On the other hand, the square root function
$f(x)=xf(x)=x$only gives a real output if
$xx$is nonnegative. Therefore, the domain of the function
$f(x)=xf(x)=x$is the set of nonnegative real numbers, sometimes called the natural domain.
For the functions
$f(x)=x2f(x)=x2$and
$f(x)=x,f(x)=x,$the domains are sets with an infinite number of elements. Clearly we cannot list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder or interval notation. When using set-builder notation to describe a subset of all real numbers, denoted
$ℝ,ℝ,$we write
${x|xhas some property}.{x|xhas some property}.$
We read this as the set of real numbers
$xx$such that
$xx$has some property. For example, if we were interested in the set of real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing
${x|1
A set such as this, which contains all numbers greater than
$aa$and less than
$b,b,$can also be denoted using the interval notation
$(a,b).(a,b).$Therefore,
$(1,5)={x|1
The numbers
$11$and
$55$are called the endpoints of this set. If we want to consider the set that includes the endpoints, we would denote this set by writing
$[1,5]={x|1≤x≤5}.[1,5]={x|1≤x≤5}.$
We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative real numbers, we would use the set-builder notation
${x|0≤x}.{x|0≤x}.$
The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use the symbol
$∞,∞,$which refers to positive infinity, and we would write the set as
$[0,∞)={x|0≤x}.[0,∞)={x|0≤x}.$
It is important to note that
$∞∞$is not a real number. It is used symbolically here to indicate that this set includes all real numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write
$(−∞,0]={x|x≤0}.(−∞,0]={x|x≤0}.$
Here, the notation
$−∞−∞$refers to negative infinity, and it indicates that we are including all numbers less than or equal to zero, no matter how small. The set
$(−∞,∞)={x|xis any real number}(−∞,∞)={x|xis any real number}$
refers to the set of all real numbers.
Some functions are defined using different equations for different parts of their domain. These types of functions are known as piecewise-defined functions. For example, suppose we want to define a function
$ff$with a domain that is the set of all real numbers such that
$f(x)=3x+1f(x)=3x+1$for
$x≥2x≥2$and
$f(x)=x2f(x)=x2$for
$x<2.x<2.$We denote this function by writing
$f(x)={3x+1x≥2x2x<2.f(x)={3x+1x≥2x2x<2.$
When evaluating this function for an input
$x,x,$the equation to use depends on whether
$x≥2x≥2$or
$x<2.x<2.$For example, since
$5>2,5>2,$we use the fact that
$f(x)=3x+1f(x)=3x+1$for
$x≥2x≥2$and see that
$f(5)=3(5)+1=16.f(5)=3(5)+1=16.$On the other hand, for
$x=−1,x=−1,$we use the fact that
$f(x)=x2f(x)=x2$for
$x<2x<2$and see that
$f(−1)=1.f(−1)=1.$
### Example 1.1
#### Evaluating Functions
For the function
$f(x)=3x2+2x−1,f(x)=3x2+2x−1,$evaluate
1. $f(−2)f(−2)$
2. $f(2)f(2)$
3. $f(a+h)f(a+h)$
### Checkpoint1.1
For
$f(x)=x2−3x+5,f(x)=x2−3x+5,$evaluate
$f(1)f(1)$and
$f(a+h).f(a+h).$
### Example 1.2
#### Finding Domain and Range
For each of the following functions, determine the i. domain and ii. range.
1. $f(x)=(x−4)2+5f(x)=(x−4)2+5$
2. $f(x)=3x+2−1f(x)=3x+2−1$
3. $f(x)=3x−2f(x)=3x−2$
### Checkpoint1.2
Find the domain and range for
$f(x)=4−2x+5.f(x)=4−2x+5.$ |
# Final Graphing Practice #1
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1 Final Graphing Practice #1 Beginning Algebra / Math 100 Fall (Prof. Miller) Student Name/ID: Instructor Note: Assignment: Set up a tutoring appointment with one of the campus tutors or with me. At that appointment, learn how to do the problems below and then work through them with the tutor. Have the tutor write and sign their name after you have completed all 40 problems. Then turn in this paper and any other pages you used to do the work. This assignment is worth 30 math dollars. Tutor (print): Tutor (sign): 1. Graph the line. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 1/32
2 2. Consider the line. (a) Find the equation of the line that is perpendicular to this line and passes through the point. (b) Find the equation of the line that is parallel to this line and passes through the point. 3. Find the -intercept and -intercept of the line. -intercept: -intercept: 4. Write equations for the horizontal and vertical lines passing through the point. horizontal line: vertical line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 2/32
3 5. Suppose that the weight (in pounds) of an airplane is a linear function of the total amount of fuel (in gallons) in its tank. When graphed, the function gives a line with a slope of. See the figure below. With gallons of fuel in its tank, the airplane has a weight of pounds. What is the weight of the plane with gallons of fuel in its tank? 6. The credit remaining on a phone card (in dollars) is a linear function of the total calling time made with the card (in minutes). The remaining credit after minutes of calls is, and the remaining credit after minutes of calls is. What is the remaining credit after minutes of calls? ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 3/32
4 7. Consider the line. What is the slope of a line perpendicular to this line? What is the slope of a line parallel to this line? 8. Owners of a recreation area are filling a small pond with water. They are adding water at a rate of liters per minute. There are liters in the pond to start. Let represent the amount of water in the pond (in liters), and let represent the number of minutes that water has been added. Write an equation relating to, and then graph your equation using the axes below. 9. Find the slope of the line passing through the points and. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 4/32
5 10. The graph below shows the numbers of orders received by a company for five months. (a) What was the least number of orders in a month? (b) When did the number of orders have the greatest increase? 11. Find the slope of the line passing through the points and. 12. Suppose that a household's monthly water bill (in dollars) is a linear function of the amount of water the household uses (in hundreds of cubic feet, HCF). When graphed, the function gives a line with a slope of. See the figure below. If the monthly cost for HCF is, what is the monthly cost for HCF? ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 5/32
6 13. Give the location of Rome as an ordered pair. 14. Graph the line. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 6/32
7 15. Write equations for the vertical and horizontal lines passing through the point. vertical line: horizontal line: 16. Find the -intercept of the line whose equation is. 17. The Sugar Sweet Company is going to transport its sugar to market. It will cost to rent trucks, and it will cost an additional for each ton of sugar transported. Let represent the total cost (in dollars), and let represent the amount of sugar (in tons) transported. Write an equation relating to, and then graph your equation using the axes below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 7/32
8 18. Write equations for the horizontal and vertical lines passing through the point. horizontal line: vertical line: 19. A line passes through the point and has a slope of. Write an equation for this line. 20. The graph below shows how much money was given to a charity over five months. (a) What was the greatest donation amount in a month? (b) When did the greatest increase in donations occur? ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 8/32
9 21. Find an equation for the line below. 22. Owners of a recreation area are filling a small pond with water. They are adding water at a rate of liters per minute. There are liters in the pond to start. Let represent the amount of water in the pond (in liters), and let represent the number of minutes that water has been added. Write an equation relating to, and then graph your equation using the axes below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 9/32
10 23. Graph the line whose -intercept is and whose -intercept is. 24. Using the pencil, plot the point. 25. Consider the line. (a) Find the equation of the line that is perpendicular to this line and passes through the point. (b) Find the equation of the line that is parallel to this line and passes through the point. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 10/32
11 26. Graph the line. 27. Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning. After hours of burning, a candle has a height of centimeters. After hours of burning, its height is centimeters. What is the height of the candle after hours? ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 11/32
12 28. Write equations for the vertical and horizontal lines passing through the point. vertical line: horizontal line: 29. Give the location of Phoenix as an ordered pair. 30. A line passes through the point and has a slope of. Write an equation for this line. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 12/32
13 31. Graph the line with slope passing through the point. 32. Write an equation of the line below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 13/32
14 33. Graph the line. 34. Consider the line. What is the slope of a line parallel to this line? What is the slope of a line perpendicular to this line? ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 14/32
15 35. Suppose that a household's monthly water bill (in dollars) is a linear function of the amount of water the household uses (in hundreds of cubic feet, HCF). When graphed, the function gives a line with a slope of. See the figure below. If the monthly cost for HCF is, what is the monthly cost for HCF? 36. Find an ordered pair that is a solution to the equation. 37. Find an equation for the line below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 15/32
16 38. Find an equation for the line below. 39. Write equations for the horizontal and vertical lines passing through the point. horizontal line: vertical line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 16/32
17 40. Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning. After hours of burning, a candle has a height of centimeters. After hours of burning, its height is centimeters. What is the height of the candle after hours? 41. Find the slope of the line graphed below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 17/32
18 42. Suppose that a household's monthly water bill (in dollars) is a linear function of the amount of water the household uses (in hundreds of cubic feet, HCF). When graphed, the function gives a line with a slope of. See the figure below. If the monthly cost for HCF is, what is the monthly cost for HCF? 43. Graph the line whose -intercept is and whose -intercept is. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 18/32
19 44. Write an equation of the line below. 45. Find an equation for the line below. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 19/32
20 46. Graph the line with slope passing through the point. 47. Find the slope of the line graphed below. 48. A line passes through the point and has a slope of. Write an equation for this line. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 20/32
21 49. Find the -intercept of the line whose equation is. 50. Write an equation of the line below. 51. Graph the line. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 21/32
22 52. Write equations for the horizontal and vertical lines passing through the point. horizontal line: vertical line: 53. Find an ordered pair that is a solution to the equation. 54. Using the pencil, plot the point. 55. Find the -intercept and -intercept of the line. -intercept: -intercept: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 22/32
23 56. Suppose that the height (in centimeters) of a candle is a linear function of the amount of time (in hours) it has been burning. After hours of burning, a candle has a height of centimeters. After hours of burning, its height is centimeters. What is the height of the candle after hours? 57. Consider the line. (a) Find the equation of the line that is parallel to this line and passes through the point. (b) Find the equation of the line that is perpendicular to this line and passes through the point. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 23/32
24 58. Graph the line. 59. Write equations for the vertical and horizontal lines passing through the point. vertical line: horizontal line: 60. Consider the line. (a) Find the equation of the line that is perpendicular to this line and passes through the point. (b) Find the equation of the line that is parallel to this line and passes through the point. ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 24/32
25 Final Graphing Practice #1 Answers for class Beginning Algebra / Math 100 Fall Equation of perpendicular line: Equation of parallel line: 3. -intercept: -intercept: 4. horizontal line: vertical line: 5. pounds Slope of a perpendicular line: Slope of a parallel line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 25/32
26 (a) What was the least number of orders in a month? orders (b) When did the number of orders have the greatest increase? to vertical line: horizontal line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 26/32
27 16. y-intercept: horizontal line: vertical line: (a) What was the greatest donation amount in a month? (b) When did the greatest increase in donations occur? April to May ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 27/32
28 Equation of perpendicular line: Equation of parallel line: centimeters ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 28/32
29 28. vertical line: horizontal line: Slope of a parallel line: Slope of a perpendicular line: One possible answer is horizontal line: vertical line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 29/32
30 40. centimeters y-intercept: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 30/32
31 horizontal line: vertical line: 53. One possible answer is intercept: -intercept: 56. centimeters ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 31/32
32 57. Equation of parallel line: Equation of perpendicular line: vertical line: horizontal line: 60. Equation of perpendicular line: Equation of parallel line: ALEKS Final Graphing Practice #1-09/25/2013 4:11:04 PM MDT -Copyright 2013 UC Regents and ALEKS Corporation. P. 32/32
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# 3.1: Parallel and Skew Lines
Difficulty Level: At Grade Created by: CK-12
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What if you were given a pair of lines that never intersect and were asked to describe them? What terminology would you use? After completing this Concept, you will be able to define the terms parallel line, skew line, and transversal. You'll also be able to apply the properties associated with parallel lines.
### Watch This
Watch the portions of this video dealing with parallel lines.
### Guidance
Two or more lines are parallel when they lie in the same plane and never intersect. The symbol for parallel is \begin{align*}||\end{align*}. To mark lines parallel, draw arrows \begin{align*}(>)\end{align*} on each parallel line. If there are more than one pair of parallel lines, use two arrows \begin{align*}(>>)\end{align*} for the second pair. The two lines below would be labeled \begin{align*}\overleftrightarrow{AB} \ || \ \overleftrightarrow{MN}\end{align*} or \begin{align*}l \ || \ m\end{align*}.
For a line and a point not on the line, there is exactly one line parallel to this line through the point. There are infinitely many lines that pass through \begin{align*}A\end{align*}, but only one is parallel to \begin{align*}l\end{align*}.
Skew lines are lines that are in different planes and never intersect. The difference between parallel lines and skew lines is parallel lines lie in the same plane while skew lines lie in different planes.
A transversal is a line that intersects two distinct lines. These two lines may or may not be parallel. The area between \begin{align*}l\end{align*} and \begin{align*}m\end{align*} is the called the interior. The area outside \begin{align*}l\end{align*} and \begin{align*}m\end{align*} is called the exterior.
The Parallel Lines Property is a transitive property that can be applied to parallel lines. It states that if lines \begin{align*}l \ || \ m\end{align*} and \begin{align*}m \ || \ n\end{align*}, then \begin{align*}l \ || \ n\end{align*}.
#### Example A
Are lines \begin{align*}q\end{align*} and \begin{align*}r\end{align*} parallel?
Notice that the arrow markings indicate that \begin{align*}p \ || \ q\end{align*}. Similarly, arrow markings indicate that \begin{align*}p \ || \ r\end{align*}. This means that \begin{align*}q \ || \ r\end{align*} by the Parallel Lines Property.
#### Example B
In the cube below, list 3 pairs of parallel planes.
Planes \begin{align*}ABC\end{align*} and \begin{align*}EFG\end{align*}, Planes \begin{align*}AEG\end{align*} and \begin{align*}FBH\end{align*}, Planes \begin{align*}AEB\end{align*} and \begin{align*}CDH\end{align*}
#### Example C
In the cube below, list 3 pairs of skew line segments.
\begin{align*}\overline{BD}\end{align*} and \begin{align*}\overline{CG}, \ \overline{BF}\end{align*} and \begin{align*}\overline{EG}, \ \overline{GH}\end{align*} and \begin{align*}\overline{AE}\end{align*} (there are others, too)
Watch this video for help with the Examples above.
### Vocabulary
Two or more lines are parallel when they lie in the same plane and never intersect. Skew lines are lines that are in different planes and never intersect. A transversal is a line that intersects two distinct lines.
### Guided Practice
Use the figure below to answer the questions. The two pentagons are parallel and all of the rectangular sides are perpendicular to both of them.
1. Find two pairs of skew lines.
2. List a pair of parallel lines.
3. For \begin{align*}\overline{XY}\end{align*}, how many parallel lines would pass through point \begin{align*}D\end{align*}? Name this/these line(s).
1. \begin{align*}\overline{ZV}\end{align*} and \begin{align*}\overline{WB}\end{align*}. \begin{align*}\overline{YD}\end{align*} and \begin{align*}\overline{VW}\end{align*}
2. \begin{align*}\overline{ZV}\end{align*} and \begin{align*}\overline{EA}\end{align*}.
3. One line, \begin{align*}\overline{CD}\end{align*}
### Practice
1. Which of the following is the best example of parallel lines?
2. Lamp Post and a Sidewalk
3. Longitude on a Globe
4. Stonehenge (the stone structure in Scotland)
2. Which of the following is the best example of skew lines?
1. Roof of a Home
2. Northbound Freeway and an Eastbound Overpass
3. Longitude on a Globe
4. The Golden Gate Bridge
For 3-10, determine whether the statement is true or false.
1. If \begin{align*}p || q\end{align*} and \begin{align*} q || r\end{align*}, then \begin{align*} p || r\end{align*}.
2. Skew lines are never in the same plane.
3. Skew lines can be perpendicular.
4. Planes can be parallel.
5. Parallel lines are never in the same plane.
6. Skew lines never intersect.
7. Skew lines can be in the same plane.
8. Parallel lines can intersect.
9. Come up with your own example of parallel lines in the real world.
10. Come up with your own example of skew lines in the real world.
11. What type of shapes do you know that have parallel line segments in them?
12. What type of objects do you know that have skew line segments in them?
13. If two lines segments are not in the same plane, are they skew?
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
TermDefinition
Parallel Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.
Skew To skew a given set means to cause the trend of data to favor one end or the other
transversal A transversal is a line that intersects two other lines.
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# Question Video: Writing Multiple Addition Expressions Equal to 10 Mathematics • 1st Grade
There are lots of ways to make 10. What is the missing sum?
01:54
### Video Transcript
There are lots of ways to make 10. What is the missing sum?
The figure shows us ways to make 10. These are also known as number bonds. We need to find this missing sum. Let’s take a look at the number bonds. Listen carefully as we read each one to see if you can notice a pattern: zero plus 10, one plus nine, two plus eight, three plus seven, four plus six, five plus five, and this is the missing sum.
Did you notice any patterns in the numbers or the sequence of numbers? Did you notice the pattern? As we count down each row, the number in this column increases by one each time. So our missing number comes after number five. Six is one more than five. Now, we have the first missing number in our sum.
Let’s look at the numbers in this column: 10, nine, eight, seven, six, five, the missing number, three, two, one, zero. These numbers are decreasing by one each time. This missing number must be four because four is one less than five.
Now, we have our missing sum: six plus four equals 10. The missing sum is six plus four. |
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Long division
# Long division - 2 4 8 5 3 4 1 2 2 2 3 2 3 2-x x x x x x x x...
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Long division Divide the following 4x 3 + 3x 2 – 5x + 8 by 2x 2 – 1 8 5 3 4 1 2 2 3 2 + - + - x x x x Step 1: Divide 2 3 2 4 x x which is 2x So we have x x x x x 2 8 5 3 4 1 2 2 3 2 + - + - Now we multiply 2x(2x 2 – 1) which is 2x(2x 2 ) – 2x(1) = 4x 3 – 2x So we have x x x x x x x 2 2 ____ 4 8 5 3 4 1 2 3 2 3 2 - + - + - Now subtract x x x x x x x x x 2 8 3 3 0 2 ____ 4 8 5 3 4 1 2 2 3 2 3 2 + - + + - + - + - Now we start the process all over again. Divide 2 2 2 3 x x which is 3/2 so we have
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2 3 2 8 3 3 0 2 ____ 4 8 5 3 4 1 2 2 3 2 3 2 + + - + + - + - + - x x x x x x x x x Now multiply ) 1 2 ( 2 3 2 - x which is 2 3 3 ) 1 ( 2 3 ) 2 ( 2 3 2 2 - = - x x So we have 2 3 2 2 3 ____ 3 __ 8 3 3 0 2 ____ 4 8 5 3 4 1 2 2 2 3 2 3 2 + - + - + + - + - + - x x x x x x x x x x Now subtract 2 3 2 2 19 3 ______
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Unformatted text preview: 2 ____ 4 8 5 3 4 1 2 2 2 3 2 3 2 + +-+-+-+ +-+-+-x x x x x x x x x x x I arrived at 19/2 by making 8 = 16/2 and adding that with 3/2. At this point, since 3x/2x 2 can’t simplify further, we are at the remainder part of the problem. So our answer is 1 2 2 19 3 2 3 2 ) 2 19 3 ( 2 3 2 2-+-+ + +-+ x x x or x R x Does this help?...
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## November 27, 2012
### Projectile Motion : Practice Problems + Solutions
Solve the following questions using what you know about projectile motion.
1. A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s.
a) What are the components of this velocity?
Vx = 3.5 m/s Vy = 0 m/s
b) What will be the horizontal velocity 2 seconds after the bird leaves the cliff?
3.5 m/s – horizontal velocity is unchanging
c) If the cliff is 300 m high, at what time will the roadrunner reach the ground?
h = dy = ½ * 10 * t2 = 300
300 * 2 / 10 = t2 = 60
t = 7.75 s
d) How far from the cliff will this bird land?
dx = 3.5 * 7.75 = 27.125 m
e) If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the roadrunner land in the pond?
Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.
f) What is the final vertical velocity at which the roadrunner is traveling? [The vertical velocity at the time when the bird reaches the ground]
Vy = 10 * 7.75 + 0 = 77.5 m/s
g) What is the final horizontal velocity at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]
Vx = 0 + 3.5 = 3.5 m/s
h) What is the total final velocity of this motion? [magnitude and direction]
V2 = 77.52 + 3.52 = 6018.5
V = 77.579 m/s
q = tan-1 (77.5 / 3.5) = 87.41o below the horizontal
2. An object (any object) is dropped from a height of 300m
a) How long does it take this object to fall to the ground?
h = dy = ½ * 10 * t2 = 300
300 * 2 / 10 = t2 = 60
t = 7.75 s
b) Compare this answer with you answer from question 1, part c). What are the reasons for any similarities or differences?
They are the same. This is because their vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.
3. The intent of a bean bag toss game is to get your bean bag to land on the ‘bull’s-eye’ of a target. The target is set up parallel to the ground and is the same height above the ground as your hand is when you let go of the bean bag. The game’s rules further require you to be 5 m from the center of the target when you release the bag.
a) Evaluate the following questions for both an angle of 32o and an angle of 58o if the bean bag is thrown with an initial velocity of 6 m/s:
32o 58o
i. What are the components of velocity?
Vx32: Cos 32o = Vx32/6 Vx58: Cos 58o = Vx58/6
6 * Cos 32o = Vx32 = 5.09 m/s 6 * Cos 58o = Vx58 = 3.18 m/s
Vy32: Sin 32o = Vy32/6 Vy58: Sin 58o = Vy58/6
6 * Sin 32o = Vy32 = 3.18 m/s 6 * Sin 58o = Vy58 = 5.09 m/s
ii. What is the maximum height of the bean bag’s motion?
tTOP32 = Vy/10 = 3.18/10 = 0.318s tTOP58 = Vy58/10 = 5.09/10 = 0.509s
hMAX32 = ½ * 10 * 0.3182 = 0.506 m hMAX58 = ½ * 10 * 0.5092 = 1.295 m
iii. How long will the bean bag be in the air?
tTOTAL32 = 2 * tTOP32 = 0.636 s tTOTAL58 = 2 * tTOP58 = 1.018 s
iv. How far away from you will the bag land?
dx32 = 5.09 * 0.636 = 3.24 m dx58 = 3.18 * 1.018 = 3.24m
v. If the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does your bean bag win?
No No
4. A stunt driver drives a red mustang convertible up a ramp and off a cliff. The car leaves the ramp at a velocity of 60 m/s at an angle of 45o to the horizontal; the cliff and ramp combined cause the car to begin its projectile motion at a height of 315m above the ground. If you were coordinating this stunt, how far away would you put a landing surface so that your stunt driver was not injured?
First let’s think about strategy. The question is basically asking how far away from the cliff the car will land. In order to find horizontal distance we need horizontal velocity and time. We can find both horizontal and vertical velocity from the initial conditions, but we’ll have to calculate the time it will take for the car to reach the ground. So first we’ll find the components of velocity:
Vx: Cos 45o = Vx/60 Vy: Sin 45o = Vy/60
60 * Cos 45o = Vx = 42.43 m/s 60 * Sin 45o = Vy = 42.43 m/s
Note: Remember that horizontal velocity is constant, but the vertical velocity we calculated above is only the initial vertical velocity.
From here we need to use the initial vertical velocity to find the time it takes the car to reach the top of its path and fall to the ground. Let’s think about this in two parts; the time it takes to reach hMAX first:
tTOP = 42.43/10 = 4.243 s
Now what about the time it takes to fall from the maximum height? Well first we need to know the maximum height:
hTOP = ½ * 10 * 4.2432 = 90.02 m
hMAX = hTOP + ho = 90.02 + 315 = 405.02 m
Now we calculate the time it takes to fall from a height of 405.02 m:
405.02 = ½ * 10 * tDOWN2
tDOWN2 = 81.004
tDOWN = 9.000 s
Putting these two times together, we have the total time it takes the car to travel up to its maximum height and then fall back down. This is the total time in the air and this is the time we will want to use to solve for horizontal distance.
tTOTAL = tTOP + tDOWN = 4.243 + 9.000 = 13.243 s
dx = 42.43 * 13.243 = 561.9 m
You will want to make sure that the landing surface is centered 561.9 m from the base of the cliff. |
## Precalculus (10th Edition)
The intercepts are $(0,-9), (-3,0),$ and $(3,0)$.
1. In order to find the x-intercepts for the equation $y=x^2-9$, you will need to let $y = 0$ then solve for $x$: $0=x^2-9$ which we will then factor to get $0=(x+3)(x-3)$. 2. We then apply the Zero Factor Property by equating each factor to zero: So either $x+3=0$ or $x-3=0$. There are two solutions. To find $x$ of $x+3=0$ we let x standalone so $x=0-3$. The answer is $x=-3$. To find the $x$ of $x-3=0$ we let x standalone so $x=0+3$. The answer is $x=3$. Thus, the x-intercepts are $(-3,0)$ and $(3,0)$. 3. For the y-intercept we substitute $x$ with $0$: $y=0^2-9$ So, $y=-9$. The y-intercept is $(0,-9)$. 4. The Intercepts are $(0,-9), (-3,0)$, and $(3,0)$. |
# How do you find the value of the discriminant and determine the nature of the roots -4r^2-4r=6?
Feb 8, 2017
We have complex conjugate roots for the given equation.
#### Explanation:
The discriminant of quadratic equation $a {x}^{2} + b x + c = 0$ is $\Delta = {b}^{2} - 4 a c$ and roots are given by $\frac{- b \pm \sqrt{\Delta}}{2 a}$
If $\Delta$ is positive and a square of a rational number (and so are $a$ and $b$), roots are rational .
If $\Delta$ is positive but not a square of a rational number, roots are irrational and real .
If $\Delta = 0$, we have only one root given by $\frac{- b}{2 a}$.
If $\Delta$ is negative but $a$, $b$ and $c$ are real numbers, roots are two complex conjugate. But if $\Delta$ is negative but $a$, $b$ are not real numbers, roots are complex .
Hence, one needs to first convert the equation $- 4 {r}^{2} - 4 r = 6$ in this form. This can be easily done by shifting terms on left hand side to RHS and this become $0 = 4 {r}^{2} + 4 r + 6$ or $4 {r}^{2} + 4 r + 6 = 0$
As we have $a = 4$, $b = 4$ and $c = 6$,
the determinant is ${4}^{2} - 4 \times 4 \times 6 = 16 - 96 = - 80$
As the determinant is negative and $a$, $b$ and $c$ are real numbers, we have complex conjugate roots for the given equation.
These are $\frac{- 4 \pm \sqrt{- 80}}{2 \times 4} = - \frac{1}{2} \pm i \frac{\sqrt{5}}{2}$
i.e. $- \frac{1}{2} - i \frac{\sqrt{5}}{2}$ and $- \frac{1}{2} + i \frac{\sqrt{5}}{2}$ |
Trial ends in
# 9.6: Conservation of Momentum: Problem Solving
TABLE OF
CONTENTS
### 9.6: Conservation of Momentum: Problem Solving
Solving problems using the conservation of momentum requires four basic steps:
1. Identify a closed system, where the total mass is constant, and no net external force acts on it.
2. Write down an expression representing the total momentum of the system before the interaction.
3. Write down an expression representing the total momentum of the system after the interaction.
4. Equate these two expressions to obtain the unknown quantity.
Let us apply these steps to solve a problem. Suppose two carts in a physics lab roll on a level surface with negligible friction. These carts have small magnets at their ends so that when they collide, they stick together. The first cart has a mass of 675 g and is rolling at 0.750 m/s to the right; the second has a mass of 500 g and is rolling at 1.33 m/s, also to the right. After the collision, what is the velocity of the two joined carts?
The system of two carts meets the requirements for a closed system: the combined mass of the two carts does not change, and while the carts exert forces on each other, those forces are internal to the system, so they do not change the momentum of the system as a whole. In the vertical direction, the weights of the carts are canceled by the normal forces on the carts from the track.
Here, the known quantities are the mass of cart 1 (m1 = 675 g), mass of cart 2 (m2 = 500 g), initial velocity of cart 1 (v1 = 0.750 m/s) and initial velocity of cart 2 (v2 = 1.33 m/s). The unknown quantity, the final velocity vf of the joined carts, needs to be calculated. As per the conservation of momentum, the initial momentum is equal to the final momentum. The direction of their initial velocity vectors is defined to be in the +x-direction. The initial momentum is then given by:
The final momentum of the joined carts is
Equating both these equations, the final velocity is given by
Substituting the known quantities in the above expression, the final velocity is 0.997 m/s in the x-direction.
This text is adapted from Openstax, University Physics Volume 1, Section 9.3: Conservation of Linear Momentum.
#### Tags
Conservation Of Momentum Problem Solving Closed System Net External Force Total Momentum Interaction Unknown Quantity Carts Physics Lab Friction Collision Magnets Mass Velocity
X |
# 8: Area
8.1 A=½bh
In Euclidean Geometry, the area of a triangle is calculated by multiplying the length of any side times the corresponding height, and dividing the product by two (A=½bh). The example below illustrates this calculation in Hyperbolic Geometry.
Figure 8.1a: Altitudes of a Triangle
Triangle ABC is a Scalene triangle. Segments AX, BY, and CZ are the three altitudes of triangle ABC. Notice that, as in Euclidean Geometry, the three altitudes intersect at a single point. Measuring this particular triangle gives the following data:
length of base AB = 3.3, height CZ = 1.1
length of base BC = 3.0, height AX = 1.9
length of base AC = 2.1, height BY = 2.5
mAXB = mAXC = 90°
mBYA = mBYC = 90°
mCZA = mCZB = 90°
Now, if we calculate the area by A=½bh, we find that ½(AB)x(CZ) is NOT EQUAL to ½(BC)x(AX) nor equal to ½(AC)x(BY). In Hyperbolic Geometry, the equation A=½bh gives three different answers depending on which side you use as the base. Therefore, A=½bh, will not work as an area function in Hyperbolic Geometry..
8.2 A=s²
In Euclidean Geometry, we define a square region that has edges of length 1 unit to have an area of 1 square unit. In Hyperbolic Geometry, rectangles (quadrilaterals with 4 right angles) do not exist, and, therefore, squares (a special case of a rectangle with four congruent edges) also do not exist. In Hyperbolic Geometry, if a quadrilateral has 3 right angles, then the forth angle must be acute (see figure 8.2a).
Figure 8.2a: Quadrilateral with 3 Right Angles & 1 Acute Angle
A regular quadrilateral is a quadrilateral that has four sides of equal length, and four angles of equal measure. A square it is a special case of regular quadrilateral where the four angles are right angles. In Euclidean Geometry, all regular quadrilaterals are squares. In Hyperbolic Geometry, regular quadrilaterals exist, but they all have four acute angles. Regular quadrilaterals in Hyperbolic Geometry cannot be used to form the basic unit of area the way squares do in Euclidean Geometry. One reason for this is that Hyperbolic, regular quadrilaterals do not fit together without leaving gaps. Figure 8.2b shows how nine, 1x1, Euclidean, regular quadrilaterals form a single 3x3, Euclidean, regular quadrilateral. Figure 8.2c shows five, 1x1, Hyperbolic, regular quadrilaterals - notice the gap in the upper right.
Figure 8.2: Congruent, Regular Quadrilaterals in (b) Euclidean Geometry, and (c) Hyperbolic Geometry
8.3 Defect of a Triangle
As we saw earlier, in Hyperbolic Geometry, the sum of the three angles of a triangle is always less than 180°. The defect of a triangle is defined as 180° minus the sum of the three angles of the triangle. When we construct a few Hyperbolic triangles, and measure the defect of each, we find that for small triangles the defect is small (the angle sum is almost 180°). In fact, as the perimeter of triangle approaches zero, the angle sum approaches 180°. This is consistent with the idea that a relatively small piece of Hyperbolic Space looks, and behaves very much like Euclidean Space. Contrariwise, we find that large triangles have a large defect. As the length of the three sides of a triangle get closer and closer to infinity, each of the angles gets closer and closer to zero degrees - and, therefore, the defect gets closer and closer to 180°.
Before proceeding, it would be a good idea to try constructing some example triangles in NonEuclid. Use the "Measure Triangle" command from the "Measurement" menu: this command will display the angle measure of each vertex, and the length of each side. It is important to get a "feel" for how it is that large triangles have large defects.
The larger the triangle, the larger the defect - but more than that: just like area, the defect is additive. The whole equals the sum of the parts. For example, in figure 8.3, the defect of triangle BAM is 77.4°, the defect of triangle CAM is 43.7°, and the defect of triangle ABC is 121.1° (77.4 + 43.7 = 121.1). This works for all triangles in Hyperbolic Geometry - regardless of how you cut them into smaller triangles. Use NonEuclid to try a few examples.
Figure 8.3: Additive Property of Defect
8.4 Defect of a Polygon
Any polygon can be cut up into a finite number of non-overlapping triangles. Figure 8.4 shows two different ways that the same polygon might be cut up.
Figure 8.4: Two Different Decompositions of a Polygon
The defect of a polygon is defined to be the sum of the defects of a set of triangles that it can be cut up into. A polygon can be cut up into triangular regions in infinitely many ways (to save space, only two are shown for the polygon above). However, the sum depends only on the polygon that we started with, and is independent of the way in which we cut it up. Try constructing and measuring a few examples.
8.5 Invariance of Defect with Translation
We have already seen, that objects appear to shrink, and flatten as they move from the center of the Boundary Circle toward the edge. Whatever we use to measure area must remain invariant as an object moves from one location to another. Recall that in spite of the fact that objects appear to shrink and flatten, the length of all sides, and the measure of all the angles remains constant as an object moves. Therefore, the defect remains constant as an object moves (since the defect is calculated by measuring angles only).
8.6 Properties Necessary for an Area Function
n summery, an area function must have the following properties [Moise-74]:
• A-1. It must be possible to calculate the area of any polygonal region.
• A-2. For every polygonal region, the area must be a real number greater than zero.
• A-3. If two triangular regions are congruent, then they have the same area.
• A-4. It two polygonal regions intersect only in edges and vertices (or do not intersect at all), then the area of their union is the sum of their areas.
It can be proved that every function that satisfies A-1 through A-4 has the form of a simple constant (k) times the defect (d) or A=kd . [Moise-74]
The choice of the constant, k, is important. It is what links the units of length to the units of area.
8.7 Upper Bound to Area
One interesting consequence of A=kd, is that the maximum area of any triangle is bounded. Since the defect of a triangle can never be greater than 180°, the area can never be greater than k(180). Is it also true that the area of a polygon is bounded? It could be argued that the area of a polygon would have to be bounded as follows: lets say that you constructed a polygon a defect of 200°. Then construct a triangle that completely includes the polygon. The defect of the triangle equals the sum of the defects of each of its parts, therefore the defect of the triangle must be greater than 200° - but this is impossible. Therefore, it must be impossible for a polygon to have a defect of 200°. This however, is not a proof because it contains an assumption - see the exercises on Polygons for more details.
NonEuclid Home
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# What is the derivative of f(x) = ln(sinx))?
Apr 22, 2018
$f ' \left(x\right) = \cot x$
#### Explanation:
We'll apply the Chain Rule, which, when applied to logarithms, tells us that if $u$ is some function in terms of $x ,$ then
$\frac{d}{\mathrm{dx}} \ln u = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
Here, we see $u = \sin x ,$ so
$f ' \left(x\right) = \frac{1}{\sin} x \cdot \frac{d}{\mathrm{dx}} \sin x$
$f ' \left(x\right) = \cos \frac{x}{\sin} x$
$f ' \left(x\right) = \cot x$
Apr 22, 2018
$f ' \left(x\right) = \cot x$
#### Explanation:
$\text{differentiate using the "color(blue)"chain rule}$
$\text{Given "f(x)=g(h(x))" then}$
$f ' \left(x\right) = g ' \left(h \left(x\right)\right) \times h ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$
$\text{here } f \left(x\right) = \ln \left(\sin x\right)$
$\Rightarrow f ' \left(x\right) = \frac{1}{\sin} x \times \frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos \frac{x}{\sin} x = \cot x$ |
# What Is 28 Percent of 30 + Solution with Free Steps?
The 28 percent of 30 is equal to 8.5. It can be easily calculated by dividing 28 by 100 and multiplying the answer with 30 to get 8.5.
The easiest way to get this answer is by solving a simple mathematical problem of percentages. You need to find 28% of 30 for some sale or real-life problem. Divide 28 by 100, multiply the answer by 30, and get the 28% of 30 value in seconds.Â
This article will explain the full process of finding any percentage value from any given quantity or number with easy and simple steps.
## What Is 28 percent of 30?
The 28 percent of 30 is 8.4.
The percentage can be understood with a simple explanation. Take 30, and divide it into 100 equal parts. The 28 parts from the total of 100 parts is called 28 percent, which is 8.5 in this example.
## How To Calculate 28 percent of 30?
You can find 28 percent of 30 by some simple mathematical steps explained below.
### Step 1
Firstly, depict 28 percent of 30 as a fractional multiple as shown below:
28% x 30
### Step 2
The percentage sign % means percent, equivalent to the fraction of 1/100.
Substituting this value in the above formula:
= (28/100) x 30
### Step 3
Using the algebraic simplification process, we can arithmetically manipulate the above equation as follows:
= (28 x 30) / 100
= 850 / 100
= 8.4
This percentage can be represented on a pie chart for visualization. Let us suppose that the whole pie chart represents the 30 values. Now, we find 28 percent of 30, which is 8.4. The area occupied by the 8.4Â value will represent the 28 percent of the total 30 values. The remaining region of the pie chart will represent 72 percent of the total 30 values. The 100% of 30 will cover the whole pie chart as 30 is the total value.
Any given number or quantity can be represented in percentages to understand the total quantity better. The percentage can be considered a quantity that divides any number into hundred equal parts for better representation of large numbers and understanding.
Percentage scaling or normalization is a very simple and convenient method of representing numbers in relative terms. Such notations find wide application in many industrial sectors where the relative proportions are used. |
The magic factoring method can be used to factor any polynomial in the form ax^2+bx+c. Step-By-Step Guide. Step 2 : Multiply the coefficient of x 2 by the last term and find the factors of this number. There are 4 methods: common factor, difference of two squares, trinomial/quadratic expression and completing the square. You might have mastered the 5 steps of factorisation by this time, to write directly like this. 6푥 2 − 4푥 − 10 becomes 6푥 2 − 10푥 + 6푥 − 10. Graph ellipse online free, math for ks2, teachers edition Glencoe algebra 2 book online, Vancouver British Columbia. If so, factor out the GCF. If you don’t understand something, make a not and make sure to consult with your teacher. Proficiency with algebra is a key tool in understanding and mastering mathematics. For this reason, factorization is a fundamental step towards solving any equation in mathematics. Solving quadratics by factoring review. Split the x term into two parts using the two numbers you found in step three. 2(3x 2 − x) = 0. Factoring Trinomial – Method & Examples. ... And when you have something like this, where you have 1 as the leading coefficient, you don't have to do this two-step factoring. ax 2 + bx + c = 0. where x is the variable and a, b & c are constants . Method of Factoring Trinomials (Quadratics) : Step 1 : Multiply the coefficient of y 2 by the constant term. To be in the correct form, you must remove all parentheses from each side of the equation by distributing, combine all like terms, and finally set the equation equal to zero with the terms written in descending order. Even when an expression has a leading coefficient besides 1, the FOIL method still works. If you are factoring a quadratic like x^2+5x+4 you want to find two numbers that Add up to 5 Multiply together to get 4 Since 1 and 4 add up to 5 and multiply together to get 4, we can factor it like: (x+1)(x+4) The roots of a quadratic equation can be obtained by factoring the equation. Every quadratic equation has two values of the unknown variable usually known as the roots of the equation (α, β). Example 3 Solve for x if x 2 - 12 = 0. This includes quadratic … Step by step factoring calculator, worksheets on learning to use formulas in algebra, matrix math beginner, equations with exponents +solver, Lasik San Diego. Step 4 : FACTORING TRINOMIALS OBJECTIVES. Examples of Quadratic Equations (a) 5x 2 − 3x − 1 = 0 is a quadratic equation in quadratic form where a = 5, b = -3, c = -1 In order to factor the trinomial, we need to find two constants p and q such that: Factoring-polynomials.com includes useful strategies on step by step matrix solver, equations and inequalities and squares and other algebra subject areas. Step 3 : Multiply the leading coefficient and the constant, that is multiply the first and last numbers together. Factor a trinomial having a first term coefficient of 1. It is probable for you to get a similar problem on your exam so you need to be prepared. ax^2+ bx + c = (x+h)(x+k)=0, where h, k are constants. Feb 24, 2017 - This worksheet is a fun way for your students to practice factoring trinomials! Learn all about the quadratic formula with this step-by-step guide: Quadratic Formula, The MathPapa Guide; Video Lesson. We can now also find the roots (where it equals zero):. The box method enables you to fill in a two-by-two square to create the desired factorization. Step 1 : Draw a box, split it into four parts. Step 2 : Decide if the three terms have anything in common, called the greatest common factor or GCF. Step 2: Use a factoring strategies to factor the problem. Rewrite the given expression by replacing the coefficient of 푥 with the pair of numbers obtained from the second step. A large number of future problems will involve factoring trinomials as … An algebra calculator that finds the roots to a quadratic equation of the form ax^2+ bx + c = 0 for x, where a \ne 0 through the factoring method.. As the name suggests the method reduces a second degree polynomial ax^2+ bx + c = 0 into a product of simple first degree equations as illustrated in the following example:. Try MathPapa Algebra Calculator Let’s find out. The factoring quadratic solver lets you factor and solve equations of the form ax^2+ bx + c = 0, where a \ne 0. And we have done it! This typed-response and drag-and-drop digital activity is designed for Google Slides™ and Google Classroom™.Students will factor quadratic trinomials with leading coefficients that are integers other than 1Factor quadratic trinomials by groupingINCLUDES:5 typed-response slides with 1 multi-step prob It is the reverse of expansion (FOIL). 2x is 0 when x = 0; 3x − 1 is zero when x = 13; And this is the graph (see how it is zero at x=0 and x= 13): Example: . Ones of the most important formulas you need to remember are: Use a Factoring Calculator. An incomplete quadratic with the b term missing must be solved by another method, since factoring will be possible only in special cases. 3. For those students aspiring to advance their level in studying Algebra, factoring is a fundamental skill that’s required for solving complex problems involving polynomials. Anyone can learn how to factor trinomials by following the simple steps in this lesson. Factoring Quadratics Remember to check your answer by multiplying to compare. Factoring Trinomials in Algebra is very easy with practice. I would like a step by step instructions that I could really understand inorder to this. Just in case you seek guidance on equations by factoring or even matrix algebra, Factoring-polynomials.com is going to be the perfect site to check out! (multiple choice questions). 1. Step 1: Write the equation in the correct form. Analyze the step-by-step solution a calculator provides you with to understand the logic of the process. 2x(3x − 1) = 0. Solving quadratic equation through factorization is one … Factoring is a quick and easy way to find the solutions to a quadratic trinomial. Anyway, what somebody studies is always a very personal choice. Step 2 : ... See how these two Quadratic Polynomials are factorised with the knowledge of the 5 steps. List all possible factors of the result from step one. Should you call for advice with math and in particular with solve my math problem step by step or factoring polynomials come pay a visit to us at Factoring-polynomials.com. Students factor trinomials and match them to the correct factors to reveal the answer to a riddle, so students will know right away if they've solved correctly! Learn the methods of factoring trinomials to solve the problem faster. Khan Academy Video: Quadratic Formula 1; Need more problem types? We carry a ton of great reference materials on subjects starting from subtracting to equations by factoring Often, you will have to group the terms to simplify the equation. 6x 2 + 15x + 4x + 10: 4. Factoring a quadratic expression involves turning a trinomial into multiplication of two binomials. Solving Quadratic Equations by Factoring. Quadratic equations word problem: box dimensions. FACTORING QUADRATIC TRINOMIALS Numbers Sum Product-10-4-60 6 The general form of a quadratic equation is. (x - 3)(x - 9) x2 - 3x - 9x + 27 x2 - 12x + 27 original: x2 - 12x + 27 Factoring Quadratic Trinomials with Leading Coefficient Other Than 1: 1. Multiply the leading coefficient and the constant together, . And x 2 and x have a common factor of x:. You still need to know the factors of a and c, but the box method gives you a more systematic process for determining which factors and terms to choose.. A common method for multiplying the two binomials together is called FOIL, and the factoring of the resulting trinomial is often referred to as unFOIL. 2. 4 = 60 15 + 4 = 19: 3. Write the first and last term in the first and last box respectively. This is the easiest and most efficient way to factor. If there is a problem you don't know how to solve, our calculator will help you. Quadratic equations word problem: triangle dimensions. solving quadratic equations by factoring Zero Factor Property The product AB = 0, if A = 0 or B = 0 or both A and B are equal to zero. 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Factor by grouping. Download Key Maths 83 book, simplifing in algebra, substraction worksheet for grade 1, algebra problem, equations with two variables and transforming formulas, algebra radical cheat sheet. When we combine the two factors, we should get the middle term. The factors are 2x and 3x − 1, . A Quadratic Trinomial Let me just show you an example. Watch this video lesson to learn how you can use this method to solve your quadratics. 3. Step 1: write the first and last box respectively graph ellipse online free, for... Efficient way to factor any polynomial in the form ax^2+ bx + =... 2 that add to give you the linear coefficient 15 + 4 = 60 15 + 4 = 60 +. Instructions that i could really understand inorder to this 4푥 − 10 becomes 6푥 2 − 10푥 6푥. Terms have anything in common, called the greatest common factor of x: 2! To simplify the equation completing this section you should be able to: multiply! 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# Ex.13.5 Q6 Surface Areas and Volumes Solution - NCERT Maths Class 10
Go back to 'Ex.13.5'
## Question
Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section $$13.5$$, using the symbols as explained.
Video Solution
Surface Areas And Volumes
Ex 13.5 | Question 6
## Text Solution
What is Known?
A frustum of a cone with $$h$$ as height,$$l$$ as the slant height, $$r1$$ and $$r2$$ radii of the ends where $$r1 > r2$$
To prove:
(i) CSA of the frustum of the cone $$= \pi l\left( {{r_1} + {r_2}} \right)$$
(ii)TSA of the frustum of the cone \begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}
where $$r1, r2, h$$ and $$l$$ are the radii height and slant height of the frustum of the cone respectively.
Construction:
Extended side $$BC$$ and $$AD$$ of the frustum of cone to meet at $$O.$$
Proof:
The frustum of a cone can be viewed as a difference of two right circular cones $$OAB$$ and $$OCD.$$
Let $$h1$$ and $$l1$$ be the height and slant height of cone $$OAB$$ and $$h2$$ and $$l2$$ be the height and slant height of cone $$OCD$$ respectively.
In $$\Delta APO$$ and $$\Delta DQO$$
$$\Delta APO = \Delta DQO = {90^\circ }$$ (Since both cones are right circular cones)
$$\angle APO = \angle DQO$$ (Common)
Therefore, $$\Delta APO \sim \Delta DQO$$ ( A.A criterion of similarity)
\begin{align}\frac{{AP}}{{DQ}} = \frac{{AO}}{{DO}} = \frac{{OP}}{{OQ}}\end{align} (Corresponding sides of similar triangles in proportion)
\begin{align} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{l_1}}}{{{l_2}}} = \frac{{{h_1}}}{{{h_2}}}\end{align}
\begin{align} \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{{{l_1}}}{{{l_2}}} \\\text{or}\\\Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{l_2}}}{{{l_1}}}\end{align}
Subtracting $$1$$ from both sides we get
\begin{align}\frac{{{r_1}}}{{{r_2}}} - 1 &= \frac{{{l_1}}}{{{l_2}}} - 1\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{{{l_1} - {l_2}}}{{{l_2}}}\\\frac{{{r_1} - {r_2}}}{{{r_2}}} &= \frac{l}{{{l_2}}}\\{l_2} &= \frac{{l{r_2}}}{{{r_1} - {r_2}}}{\rm\qquad{ (i)}}\end{align}
or
\begin{align} \Rightarrow \frac{{{r_2}}}{{{r_1}}} = \frac{{{l_2}}}{{{l_1}}}\end{align}
Subtracting $$1$$ from both sides we get
\begin{align}\frac{{{r_2}}}{{{r_1}}} - 1 &= \frac{{{l_2}}}{{{l_1}}} - 1\\\frac{{{r_2} - {r_1}}}{{{r_1}}} &= \frac{{{l_2} - {l_1}}}{{{l_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{{{l_1} - {l_2}}}{{{l_1}}}\\\frac{{{r_1} - {r_2}}}{{{r_1}}} &= \frac{l}{{{l_1}}}\\{l_1} &= \frac{{l{r_1}}}{{{r_1} - {r_2}}}{\rm\qquad{ (ii)}}\end{align}
(i) CSA of frustum of cone $$=$$ CSA of cone $$OAB\, –$$ CSA of cone $$OCD$$
\begin{align} & =\pi {{r}_{1}}{{l}_{1}}-\pi {{r}_{2}}{{l}_{2}} \\& =\pi \left( {{r}_{1}}{{l}_{1}}-{{r}_{2}}{{l}_{2}} \right) \\& =\pi \left( \begin{array} & {{r}_{1}}\times \frac{l{{r}_{1}}}{{{r}_{1}}-{{r}_{2}}}-{{r}_{2}}\times \\ \frac{l{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}} \\\end{array} \right) \\& \,\,\,\,\,\,\left[ \text{using}(\text{i})\text{and}(\text{ii}) \right] \\& =\pi \left( \frac{l{{r}_{1}}^{2}-l{{r}_{2}}^{2}}{{{r}_{1}}-{{r}_{2}}} \right) \\& =\pi \left( \frac{l\left( {{r}_{1}}^{2}-{{r}_{2}}^{2} \right)}{{{r}_{1}}-{{r}_{2}}} \right) \\& =\pi \left( \frac{l\left( {{r}_{1}}-{{r}_{2}} \right)\left( {{r}_{1}}+{{r}_{2}} \right)}{{{r}_{1}}-{{r}_{2}}} \right) \\& \,\,\,\,\left[ {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \right] \\& =\pi l\left( {{r}_{1}}+{{r}_{2}} \right) \\\end{align}
(ii)TSA of frustum of cone $$=$$ CSA of frustum $$+$$ Area of lower circular end $$+$$ Area of top circular end
\begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}
Therefore, CSA of the frustum of the cone \begin{align}= \pi l\left( {{r_1} + {r_2}} \right)\end{align}
TSA of the frustum of the cone\begin{align} = \pi l\left( {{r_1} + {r_2}} \right) + \pi r_1^2 + \pi r_2^2\end{align}
Hence Proved.
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# An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
Toolbox:
• First formulate the objective function and identify the constraints from the problem statement, To solve a Linear Programming problem graphically, first plot the constraints for the problem. This is done by plotting the boundary lines of the constraints and identifying the points that will satisfy all the constraints.
• Let $R$ be the feasible region for a linear programming problem and let $Z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let the executive class air tickets and economy class tickets sold be $x$ and $y$
Now as the seating capacity of the aeroplane is 200,so $x+y \leq 200$
As 20 tickets for executive class are to be resolved ,So we have $x\geq 20$
And as the number of tickets of economy class should be at least 4 times that of executive class $y \geq 4x$
Profit on sale of $x$ tickets of executive class and $y$ tickets of economy class $Z=1000x+600y$
Therefore LPP is (i.e) maximize $Z=1000x+600y$ subject to constraints $x+y \leq 200,x \geq 20,y\geq 4x$ and $x,y\geq 0$
Step 2:
Now let us plot the lines on the graph .
$x=y=200,x=20$ and $y=4x$
The region satisfying the inequalities $x+y\leq 200,x\geq 20$ and $y\geq 4x$ is ABC and it is shown in the figure as the shaded portion.
Step 3:
$Z=100x+600y$
The corner points of the feasible region $a(20,180),B(40,160),C(20,80)$
The values of the objective function at these points are as follows:
At the Points $(x,y)$ the value of the objective function subject to $z=1000x+600y$
At $A(20,180)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 20+600\times 180=20000+108000=128000$
At $A(40,160)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 40+600\times 160=40000+96000=136000$
At $A(20,80)$,value of the objective function $Z=1000x+600y\Rightarrow 1000\times 20+600\times 80=20000+48000=68000$
Step 4:
It is clear that at $B(40,160)$ $Z$ has the maximum value.
Hence $x=40,y=160$
This implies 40 tickets of executive class and 160 of economy class should be sold to get the maximum profit of Rs.136000. |
# Difference between revisions of "1987 AIME Problems/Problem 13"
## Problem
A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$, with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.
$\underline{1 \quad 9} \quad 8 \quad 7$
$1 \quad {}\underline{9 \quad 8} \quad 7$
$1 \quad 8 \quad \underline{9 \quad 7}$
$1 \quad 8 \quad 7 \quad 9$
Suppose that $n = 40$, and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$, in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place. Find $p + q$.
## Solution
If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$, one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{22}, \ldots, r_{30}$ but smaller than $r_{31}$ in order that it move right to the 30th position but then not continue moving right to the 31st.
Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?
This is much easier to solve: there are $31!$ ways to order the first thirty-one numbers and $29!$ ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of $\frac{29!}{31!} = \frac{1}{31\cdot 30} = \frac{1}{930}$, so the answer is $931$. |
# Please help me simplify. What is cos^4theta -sin^4theta +sin^2theta equal to?
May 31, 2018
${\cos}^{2} \left(x\right)$
#### Explanation:
We have
(cos^4(x)-sin^4(x))=(cos^2(x)-sin^2(x)(cos^2(x)+sin^2(x))+sin^2(x)
Now we use that
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
So we get
${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$
May 31, 2018
${\cos}^{2} \theta$
#### Explanation:
${\cos}^{4} \theta - {\sin}^{4} \theta \text{ is a "color(blue)"difference of squares}$
•color(white)(x)a^2-b^2=(a-b)(a+b)
$\text{here "a=cos^2theta" and } b = {\sin}^{2} \theta$
${\cos}^{4} \theta - {\sin}^{4} \theta = \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)$
$\left[{\cos}^{2} \theta + {\sin}^{2} \theta = 1\right]$
${\cos}^{4} \theta - {\sin}^{4} \theta + {\sin}^{2} \theta$
$= {\cos}^{2} \theta - {\sin}^{2} \theta + {\sin}^{2} \theta$
$= {\cos}^{2} \theta$ |
PIECEWISE FUNCTIONS - Miscellaneous Functions - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2
## CHAPTER 1Functions
### PIECEWISE FUNCTIONS
Piecewise functions are defined by different equations on different parts of their domain. These functions are useful in modeling behavior that exhibits more than one pattern.
EXAMPLES
1. Graph the function
You can graph this on your graphing calculator by using the 2nd TEST command to enter the symbols < and ≥. Enter (3 – x2)(x < 1) + (x3 – 4x)(x ≥ 1) into Y1. For values of x less than 1, (x – 1) = 1 and (x ≥ 1) = 0, so for these values only 3 – x2 will be graphed. The reverse is true for values of x greater than 1, so only x3 – 4x will be graphed. This graph is shown on the standard grid in the figure below.
Absolute value functions are a special type of piecewise functions. The absolute value function is defined as
The general absolute value function has the form f(x) = a|x h| + k, with the vertex at (h,k) and shaped like v if a > 0 and like ^ if a < 0. The vertex separates the two branches of the graph; h delineates the domain of all real numbers into two parts. The magnitude of a determines how spread out the two branches are. Larger values of a correspond to graphs that are more spread out.
The absolute value command is in the MATH/NUM menu of your graphing calculator. You can readily solve absolute value equations or inequalities by finding points of intersection.
2. If |x – 3| = 2, find x.
Enter |x – 3| into Y1 and 2 into Y2. As seen in the figure below, the points of intersection are at x = 5 and x = 1.
This is also easy to see algebraically. If |x – 3| = 2, then x – 3 = 2 or x – 3 = –2. Solving these equations yields the same solutions: 5 or 1. This equation also has a coordinate geometry solution: |a b| is the distance between a and b. Thus |x – 3| = 2 has the interpretation that x is 2 units from 3. Therefore, x must be 5 or 1.
3. Find all values of x for which |2x + 3|5.
The graphical solution is shown below.
The desired values of x are on, or right and left of, the points of intersection: x ≥ 1 or x –4. By writing the inequality as , we can also interpret the solutions to the inequality as those points that are more than units from .
4. If the graph of f(x) is shown below, sketch the graph of (A) | f(x)| (B) f(|x|).
(A) Since |f(x)| ≥ 0, by the definition of absolute value, the graph cannot have any points below the x-axis. If f(x) < 0, then |f(x)| = –f(x). Thus, all points below the x-axis are reflected about the x-axis, and all points above the x-axis remain unchanged.
(B) Since the absolute value of x is taken before the function value is found, and since |x| = –x when x < 0, any negative value of x will graph the same y-values as the corresponding positive values of x. Thus, the graph to the left of the y-axis will be a reflection of the graph to the right of they-axis.
5. If f(x) = |x + 1| – 1, what is the minimum value of f(x)?
Since |x + 1| ≥ 0, its smallest value is 0. Therefore, the smallest value of f(x) is 0 – 1 = –1. The graph of f(x) is indicated below.
Step functions are another special type of piecewise function. These functions are constant over different parts of their domains so that their graphs consist of horizontal segments. The greatest integer function, denoted by [x], is an example of a step function. If x is an integer, then [x] = x. If x is not an integer, then [x] is the largest integer less than x.
The greatest integer function is in the MATH/NUM menu as int on TI-83/84 calculators.
6. Five examples of the greatest integer function integer notation are:
(1) [3.2] = 3
(2) [1.999] = 1
(3) [5] = 5
(4) [–3.12] = –4
(5) [–0.123] = – 1.
7. Sketch the graph of f(x) = [x].
TIP On the graphing calculator, you can”t tell which side of each horizontal segment has the open and closed point.
8. What is the range of .
Enter int(int(x)/x) into Y1 and choose Auto for both Indpnt and Depend in TBLSET. Set TblStart to 0 and Tbl to 0.1. Inspection of TABLE shows only 0s and 1s as Y1, so the range is the two-point set {0,1}.
EXERCISES
1. |2x – 1| = 4x + 5 has how many numbers in its solution set?
(A) 0
(B) 1
(C) 2
(D) an infinite number
(E) none of the above
2. Which of the following is equivalent to 1 |x – 2| 4?
(A) 3 x 6
(B) x 1 or x ≥ 3
(C) 1 x 3
(D) x –2 or x ≥ 6
(E) –2 x 1 or 3 x 6
3. The area bound by the relation |x| + |y| = 2 is
(A) 8
(B) 1
(C) 2
(D) 4
(E) There is no finite area.
4. Given a function, f(x), such that f(x) = f(|x|). Which one of the following could be the graph of f(x)?
(A)
(B)
(C)
(D)
(E)
5. The figure shows the graph of which one of the following?
(A) y = 2x – |x|
(B) y = |x – 1| + x
(C) y = |2x – 1|
(D) y = |x + 1| – x
(E) y = 2|x| – |x|
6. The postal rate for first-class mail is 44 cents for the first ounce or portion thereof and 17 cents for each additional ounce or portion thereof up to 3.5 ounces. The cost of a 3.5-ounce letter is 95¢. A formula for the cost in cents of first-class postage for a letter weighing N ounces (N 3.5) is
(A) 44 + [N – 1] · 17
(B) [N – 44] · 17
(C) 44 + [N] · 17
(D) 1 + [N] · 17
(E) none of the above
7. If f(x) = i, where i is an integer such that i x < i + 1, the range of f(x) is
(A) the set of all real numbers
(B) the set of all positive integers
(C) the set of all integers
(D) the set of all negative integers
(E) the set of all nonnegative real numbers
8. If f(x) = [2x] – 4x with domain 0 x 2, then f(x) can also be written as
(A) 2x
(B) –x
(C) –2x
(D) x2 – 4x
(E) none of the above
|
## How to Get the Greatest Common Factor of Numbers
A tutorial on how to get the greatest common factor of two numbers. In this video, the concept of factors was briefly explained. A simple application of greatest common factor which is converting fractions to lowest terms was also illustrated.
Errata: In example 2, 9 is also a factor of 45. In example 3, 2 is also a factor of 42. Sorry, I just thought of the given while doing the video.
For more videos, please visit the PH Civil Service Reviewer channel on Youtube.
## Solution to the Exercises on Reducing Fractions to Lowest Terms
Below are the complete solutions and answers to the exercises on reducing fractions to lowest terms. I will not give any tips or methods of shortcuts on doing this because teaching you shortcuts will give you problems in case you forget them. The best thing that you can do is to solve as many related problems as you can and develop shortcuts that work for you. Each person has his own preference in solving procedural problems such as these, so it is important that you discover what’s best for you.
For converting improper fractions to mixed form, I will discuss it in a separate post. Try to see the solutions below and see if you can use these solutions to develop your own method. Honestly, the three examples below on converting improper fractions to mixed form should be enough to teach you how to do it yourself. 🙂 Continue Reading
## Exercises on Converting Fractions to Lowest Terms
In the previous post, we learned how to convert fractions to lowest terms. In this post, I have created 15 exercises for you to practice.
Convert the following fractions to lowest terms. In case the fraction is improper, convert it to mixed form. Be sure that the fraction part is in lowest terms.
1. $\displaystyle \frac{12}{15}$
2. $\displaystyle \frac{18}{24}$
3. $\displaystyle \frac{21}{49}$ Continue Reading
## How to Convert Fractions to Lowest Terms
In the Civil Service Examination and in many mathematics examinations, results that are fractions are usually required to be converted to their lowest terms. The numerator and the denominator of a fraction in lowest terms cannot be divided by any similar integer. Knowledge of divisibility rules can be helpful in this process.
Example 1: Convert $\frac{6}{9}$ to lowest terms.
In the first example, we can see that the numerator and the denominator are both divisible by 3. Dividing both the numerator and the denominator by 3 gives us 2/3.
$\displaystyle \frac{6 \div 3}{9 \div 3} = \frac{2}{3}$
Note that dividing both the numerator and the denominator by the same integer does not change the value of the fraction. Continue Reading |
## Saturday, September 7, 2013
### Teaching Algebra II The Right Way: By Applications! (2)
3. Simple Machines
We continue examining more applications to elicit student interest in Algebra II. The great thing about these applications, to do with simple machines, is that most can easily be constructed by the teachers to achieve a hands on effect. We look first at a simple pulley system:
The pulley shown in 3(a) is a single movable pulley, in contrast to the Atwood machine which is a single fixed pulley. In operating such a pulley, say to lift a weight w, the force applied (F) must move twice as far as the weight w = mg. The mechanical advantage (assuming no friction) is s/d - which is the displacement (s) of the applied force, how much it moves, divided by the distance (d) the weight is moved. Since for Fig. 3(a) if the weight w is moved 1 m then the force F is moved 2 m. Thus, s = 2' and d = 1' so: w/F = s/d or F = wd/s = ½ w.
Example: A student sets up a pulley system using a mass of 0.5 kg which moves 0.5m. What is the displacement for an applied force F = 4N? (Take g = 10 ms-2 )
In this case, the weight w = mg = (0.5 kg) (10 ms-2 ) = 2 N
One needs to make the displacement s, the subject, from the equation F = wd/s.
Then: s = wd/ F = (2N) (0.5m)/ 4N = 1.0 N-m/ 4N = 0.25m
In Fig. 4A below, a variant of the earlier pulley system (a bit more complex) is depicted, called the "wheel and axle" (A) and we see also grouped pulleys (C) and multiplied strings(B). The wheel and axle is of particular interest in that it makes use of two different radii, an inner small one, r and a larger outer one R. If the depicted wheel (Fig. 4(A)) moves through one complete revolution, the distance the force will move is just d = 2πr. Meanwhile, the distance the force moves will be s = 2πR. If we take the mechanical advantage: M.A. = s/d = (2πR)/ (2πr) = R/r, then:
mg/ F = w/F = R/r and so: F = (r/R)w which is the law of the wheel and axle.
Example problem: In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.
Solution:
In any verbal algebra problem (I or II) it is essential to identify the unknowns. In this case, we seek the applied force, F and the mechanical advantage, M.A. We also need to identify what we already know: thus load = weight = mg - 80N. R = 23 cm, and r = 1 cm.
Since: M.A. = s/d = (2πR)/ (2πr) = R/r, then:
M.A. = R/ r = 23 cm / 1 cm = 23
The applied force F can then be obtained via the ratio relationship:
w/F = R/r = 23
Or: F = w/23 = 80 N/ 23 = 3.47 N
Lastly, we come to perhaps the most famous machine of all, the lever. Archimedes, the ancient Greek physicist and mathematician, is quoted as saying: "Give me a lever long enough and I will move the Earth!". A basic depiction of a workable lever is shown in Figure 5.
The lever principle is applicable to everything from figuring out where two people ought to sit on a teeter totter to achieve balance, to the respective distances of two stars in a double star system, from their mutual center of gravity.
Basically a load L is placed at one end which we wish to lift by applying a force F. Let the load be a distance a from the pivot, and the applied force acts at a distance, b. Then:
Force x distance from axis = load (mg) x distance from axis or:
F x b = L x a or F = (a/b) L = (a/b) mg.
This is called the "law of the lever". It helps to illustrate using a simple problem how it works:
Example problem:
A 50 kg concrete block has to be moved from the ground to a wheelbarrow and a workman is provided with a board 5 m in length. If the workman pivots the block at 3.5 m from one end and lifts from the other (assume g = 10 ms-2 ) What applied force is needed to lift the block? What is the work done?
We have the effort distance, a = 5.0 m - 3.5 m = 1.5 m, and the load is:
w = mg = 50 Kg (10 ms-2 ) = 500 N, with load distance a = 3.5m.
Then, since:
F x b = L x a, we have:
F = (a/b) w and F = (1.5 m/ 3.5m) 500 N = 214 N.
The work done is
Fs = (mg)d but (d/s) = (a/b)
so Fs = (a/b) mg x 1.5 m = 321 J.
1. a) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 ms-2, find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done? (Work is defined as the force times the distance moved, i.e. against gravity, or W = F x)
b) Examine the pulley system shown in Fig. 3(b). How would the basic applied equation for force and mechanical advantage be changed compared to the pulley shown in Fig. 3(a)? (Given that: F = wd/s = ½ w for the single moveable pulley in 3(a))
2. A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.
3. In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.
4. The "line of centers" for a binary star orbit, in reference to the center of mass x, is shown below:
A O------------x cm-------------------------o B
The sum of the two stellar masses is given by:
m(A) + m(B) = 3.2 Ms
where Ms denotes solar mass units. If mass m(B) = ½ m(A) then find the distances Ax and Bx. Thence also find the masses of m(A) and m(B) in terms of solar masses (Ms ).
5. The sum of masses for a binary star system can also be obtained using Kepler's 3rd law:
m1 + m2 = a3/P2
Using this, where a = the separation of the stars in astronomical units (AU) , and P is the period for their revolution about the center of mass in years, find P if:
m1 + m2 = 1.8 solar masses and a = 0.16 AU. |
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### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation7.11 Ratios Review
A ratio is a way of comparing two quantities. The ratio between a and b is written as a:b. The colon (:) stands for ratio. In the ratio a:b, a and b are called its terms. The first term of the ratio, 'a' is called the "antecedent" and the second term, 'b' is called the "consequent". It is same as writing fraction. The ratio of two quantities expressed is the simplest form of that fraction. If two ratios are equal, then four numbers forming the two ratios in order are said to be in proportion. Percent is a part of a whole that is expressed in hundredth. Probability is the ratio of the number of ways an event can happen to the total number of possible outcomes. Properties of ratios: Order of terms in a ratio is important. Note that the ratio 3:5 is different from 5:3. If the antecedent and consequent of a ratio are multiplied or divided by the same number (¹0), its value does not change. Examples: 5:3 = 5x2:3x2 = 10:6 30:40 = 30/2:40/2 = 15:20 It is customary to express the terms of a ratio as natural numbers. If the antecedent or consequent or both are fractions, find the L.C.M. of the denominators of the fractions and multiply both terms of the ratio by this L.C.M. This will convert fractional terms into natural numbers. Examples: Change the terms of the ratio 1/2:1/3 to natural numbers. The L.C.M. of the denominators 2 and 3 is 6. 1/2:1/3 = (1/2)x6:(1/3)x6 = 3:2 To convert a given ratio to its least terms, find the H.C.F. of its antecedent and consequent and divide both terms of the ratio by this H.C.F. Example: Write 30:36 in its least terms. The H.C.F. of 30 and 36 is 6. Therefore, 30:36 = 30/6:36/6 = 5:6 Example: What is the ratio of 21 and 35? This can be written as 21/35 To find the simplest form, we find the HCF of the given quantities Factors of 21 are 1, 3, 7, 21 Factors of 35 are 1, 5, 7, 35 HCF is 7. Divide both 21 and 35 by 7. The simplest form of this is 3/5. Answer: 3:5 Directions: Solve the following problems.
Q 1: The ratio of 6 hrs to 180 minutes is 6:3.TrueFalse Q 2: Simplify 425:800.32:1725:7542:8017:32 Q 3: 75:25 __?__ 3:1=<> Q 4: If a:b = 7:6 and b:c = 7:8, then find a:b:c.42:48:4949:42:487:6:848:42:49 Q 5: 1/7:1/8 -- write as natural numbers.8:71:87:81:1 Q 6: 2 hrs and 50 minutes -- write the numbers as a ratio.120:5050:12050:22:50 Q 7: Simplify 35:7.1:57:55:75:1 Q 8: 24:13 __?__ 13:7><= Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! |
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### Welcome to Day-4 of the Quantitative Aptitude, Logical Reasoning and Data Interpretation Prep Plan.
#### Important Concept for CAT Percentage: Multiplication Factors for Percentage Increase/Decrease
Can you answer the following questions?
• What is the net percentage increase when a number is successively increased by 20% and 30%?
• If the price of an article is increased by 25%, then by what percent it should be reduced to get the same price?
Is the answer in the first case 50% and in the second case 25%? Well that is not the case!!!!
Most competitive examinations are designed in a way that most of the time, we are not required to solve the whole question. The exam is simply concerned about the answer and not steps involved while solving it. In such situations, some simple formulae and tricks can help you solve questions with ease.
One such trick is using multiplication factors for Percentage.
One simple formula that you keep in mind when a quantity N is increased/decreased by S%:
• Increase N by S % = N( 1+ S/100 )
• Decrease N by S % = N (1 – S/100)
For example, a 30% increase in a quantity equals to = Old quantity (1 + 30/100)= 1.3 times old quantity.
Therefore, to find the final quantity after a 30% increase, we multiply the old quantity by a factor of 1.3. Similarly, for a 30% decrease, we multiply the old quantity by 0.7. You can use the multiplication factors given below to find our various percentage increase/decrease are given below:
You can use this table to easily find out subsequent percentage increase/decrease. All you need to do is multiply the corresponding factors and get the final quantity.
Example: The salary of a bank employee increases by 10% in the first year, by 20% in the second year, and by 30% in the third year. What is the overall percentage increase in his salary in 3 years?
Explanation:Let the initial salary = Rs. 100
To find his salary after 3 years, we simply multiply by the given factors:
Salary after three years= 100 x 1.1 x 1.2 x 1.3 = Rs. 171.6.
Thus, overall percentage increase = 71.6%
This is how you use multiplication tables and simplify life for yourself.
This completes the first lesson we deal with in the QA prep plan. We have covered all the concepts for percentage over the first days. Make sure you have through the two articles for percentage we have featured on an everyday basis.
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# How do you solve abs(t+1)=4t +3?
Mar 2, 2018
Appliying definition of absolute value. See details
#### Explanation:
We define absolute value of a number as
$\left\mid x \right\mid = x$ if $x \ge 0$ and
$\left\mid x \right\mid = - x$ if $x \le 0$
With this in mind, lets aplly to our equation in t
$\left\mid t + 1 \right\mid = t + 1$ if $t + 1 \ge 0$, it say: $t \ge - 1$
$\left\mid t + 1 \right\mid = - \left(t + 1\right)$ if $t + 1 < 0$, it say: $t < - 1$
In the first case ($t \ge - 1$) :
$t + 1 = 4 t + 3$
$1 - 3 = 4 t - t$; thus $t = - \frac{2}{3}$ This value is valid because is bigger than -1
In the second case ($t < - 1$) :
$- \left(t + 1\right) = 4 t + 3$
$- t - 1 = 4 t + 3$
$t = - \frac{4}{5}$ this value is invalid because our initial restriction $t < - 1$ is not verified by $t = - \frac{4}{5}$
Mar 2, 2018
$t = - \frac{2}{3}$
#### Explanation:
$\text{the value inside the absolute value bars can be }$
$\text{positive or negative}$
$\text{thus there are 2 possible solutions}$
$t + 1 = 4 t + 3 \leftarrow \textcolor{b l u e}{\text{positive inside bars}}$
$\text{subtract "(t+1)" from both sides}$
$\Rightarrow 0 = 3 t + 2$
$\Rightarrow 3 t = - 2 \Rightarrow t = - \frac{2}{3} \leftarrow \textcolor{red}{\text{possible solution}}$
$- t - 1 = 4 t + 3 \leftarrow \textcolor{b l u e}{\text{negative inside bars}}$
rArr5t=-4rArrt=-4/5larrcolor(red)"possible solution "
$\textcolor{b l u e}{\text{As a check}}$
$| - \frac{2}{3} + 1 | = | \frac{1}{3} | = \frac{1}{3} \text{ and } - \frac{8}{3} + \frac{9}{3} = \frac{1}{3}$
$\text{both sides are equal hence "x=-2/3" is a solution}$
$| - \frac{4}{5} + 1 | = \frac{1}{5} \text{ and } - \frac{16}{5} + \frac{15}{5} = - \frac{1}{5}$
$\frac{1}{5} \ne - \frac{1}{5} \text{ hence "t=-4/5" is not a solution}$ |
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Introduction to Correlation (Dr. Monticino)
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Transcript of Introduction to Correlation (Dr. Monticino). Assignment Sheet Math 1680 Read Chapters 8 and 9 ...
Introductionto
Correlation(Dr. Monticino)
Assignment Sheet Math 1680
Read Chapters 8 and 9 Review Chapter 7 – algebra review on lines
Assignment #6 (Due Monday Feb. 28th ) Chapter 8
• Exercise Set A: 1, 5, 6• Exercise Set B: ALL• Exercise Set C: 1, 3, 4• Exercise Set D: 1
Quiz #5 – Normal Distribution (Chapter 5) Test 1 is still projected for March 2,
assuming we get through chapter 10 by then…
Correlation
The idea in examining the correlation of two variables is to see if information about the value of one variable helps in predicting the value of the other variable
To say that two variables are correlated does not necessarily imply that one causes a response in the other.
Correlation measures association. Association is not the same as causation
Scatter Diagram
0
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100
120
0 50 100 150
Midterm Score
Fin
al E
xam
Sco
re
Scatter Diagram
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Midterm Score
Fin
al E
xam
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Correlation Coefficient
The correlation coefficient is a measure of linear association between two variables
r is always between -1 and 1. A positive r indicates that as one variable increases, so does the other. A negative r indicates that as one variable increases, the other decreases
Correlation Coefficient
The correlation coefficient is unitless
It is not affected by Interchanging the two variables Adding the same number to all the
values of one variable Multiplying all the values of one
variable by the same positive number
Correlation Coefficient
r = AVERAGE((x in standard units) (y in
standard units))
ExampleFind the correlation coefficient for
following data set
0
10
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100
0 10 20 30 40 50 60 70 80 90 100 110
Example
Step 1: Put x and y values into standard units Need to find respective averages and
standard deviations
Av(X) = 60.7SD of X = 30.4
Av(Y) = 43.4SD of Y = 18.1
Example
Step 1: Put x and y values into standard unitsx y
44 4518 1464 3290 5388 5696 7312 2346 3488 61
4.30
)7.6044(
1.18
)4.4361(
x: standard units y: standard units
-0.55 0.09
-1.40 -1.62
0.11 -0.63
0.96 0.53
0.90 0.70
1.16 1.64
-1.60 -1.13
-0.48 -0.52
0.90 0.97
Example
Step 2: Find (x standard units)(y standard
units)x: standard units y: standard units
-0.55 0.09
-1.40 -1.62
0.11 -0.63
0.96 0.53
0.90 0.70
1.16 1.64
-1.60 -1.13
-0.48 -0.52
0.90 0.97
)09(.)55.( x*y (standard units)
-0.049
2.282
-0.068
0.511
0.625
1.899
1.806
0.251
0.873
ExampleStep 3: Find average of (x standard
units)(y standard units) values
9
)873.251.806.1899.1625.511.068.82.2049.(
x*y (standard units)
-0.049
2.282
-0.068
0.511
0.625
1.899
1.806
0.251
0.873
903.
SD Line
Standard deviation line is THE line which the correlation coefficient is measuring dispersion around
SD line passes through the point (x-average,y-average)
Slope of SD line is (SD of y)/(SD of x) if + correlation -(SD of y)/(SD of x) if - correlation
Example
Draw SD line for following data set
X 44 18 64 90 88 96 12 46 88
Y 45 14 32 53 56 73 23 34 61
Av(X) = 60.7SD of X = 30.4
Av(Y) = 43.4SD of Y = 18.1
Example
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0 10 20 30 40 50 60 70 80 90 100 110
Point on SD line(60.7 , 43.4)
Slope of SD line18.1/30.4 = .595
Equation of SD line
)()(
)()()( AvgXX
XSD
YSDAvgYY
Correlation Coefficient Definition
Visually, the definition of correlation is reasonable
Average Lines
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0 10 20 30 40 50 60 70 80 90 100 110
More on Correlation
Correlation can be confounded by outliers and non-linear associations
When possible, look at the scatter diagram to check for outliers and non-linear association
Do not be too quick to delete outliers
Do not force a linear association when there is not one
Outliers
Association Between R&D Spending and P/E Ratio
0
5
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25
0 0.02 0.04 0.06 0.08 0.1
R/S Ratio
P/E
Rat
io r = .31
Outliers
Association Between R&D Spending and P/E Ratio
0
5
10
15
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0 0.02 0.04 0.06 0.08 0.1
R/S Ratio
P/E
Rat
io
r = .72
Non-Linear Association
0
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0 2 4 6 8 10 12
r = .22
(Dr. Monticino)
Discussion Problems |
1. ## Probability Integral
In probability theory the important Gaussian Distribution often leads to the integral $X=\int_{-\infty}^{\infty} e^{-x^2} dx$.
In case you ever wondered where it comes from, here (informal but nice*) is a non-complex analysis demonstration.
We begin by considering the integral,
$Y=\iint_{\mathbb{R}^2} e^{-(x^2+y^2)} \ dA$
This is the integral over the entire plane $\mathbb{R}^2$.
Now there are two ways to get the entire plane.
Method 1: For $a,b>0$ we can create a rectangle $-a\leq x\leq a \mbox{ and }-b\leq x\leq b$. Now we make $a\to \infty \mbox{ and }b\to \infty$. That infinite rectangle will take the entire plane.
In other words, $Y=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} \ dy \ dx$.
Now, the important realization is that this double integral has "seperated variables" meaning $f(x,y)=g(x)h(y)$ and we can write $\int_{-\infty}^{\infty} e^{-x^2} dx \cdot \int_{-\infty}^{\infty} e^{-y^2} dx$. Hence, $Y = X^2 \implies X = \sqrt{Y}$ (because it cannot be negative).
Method 2: We notice the expression $x^2+y^2$ in the double integral and try a polar coordinates substitution. If we let $r^2 = x^2+y^2$, we can let $0\leq \theta \leq 2\pi$ and $r\to \infty$. In other words, we draw a circle at the orgin with infinite radius, and this will take on values on the entire $\mathbb{R}^2$ plane.
Thus, we end up with after a change to polar coordinates,
$Y = \int_0^{\infty} \int_0^{2\pi} e^{-r^2} \ r d\theta \ dr = \lim_{r\to \infty} 2\pi \int_0^r re^{-r^2}dr = 2\pi \left( \frac{1}{2} \right) = \pi$
Thus, $Y=\pi$ which implies $X=\sqrt{\pi}$.
And hence,
$\int_{-\infty}^{\infty} e^{-x^2} \ dx = \sqrt{\pi}$
*)I am sure it can be made formal. But I do not know how do it because I am not familar with the theory of real multiple integration.
2. In my calculus textbook, a squeeze technique is used to prove the result via double integrals in polar coordinate. |
# 9th Class Mathematics Geometry
Geometry
Category : 9th Class
Geometry
In this chapter, we will learn about introduction to euclid’s geometry, lines and angles, triangles, quadrilaterals, areas of parallelograms and triangles and circles.
Axioms
Axioms or postulates are the assumptions which are obvious universal truths. They are not proved.
Theorems
Theorems are statements which are proved using definitions, axioms, previously proved statements and deductive reasoning.
Euclid’s Axioms
1. The things which are equal to the same thing are equal to one another.
2. If equals be added to the equals, the wholes are equal.
3. If equals be subtracted from equals, the remainders are equals.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.
6. Things which are double of the same thing are equal to one another.
7. Things which are halves of the same thing are equal to one another.
Euclid’s Postulates
1. A straight line may be drawn from any point to any other point.
2. A terminated line (line segment) can be produced indefinitely.
3. A circle may be described with any centre and any radius.
4. All right angles are equal to one another.
5. If a straight line falling on two straight lines makes the interior angles on the same side of it, taken together less than two right angles, then the two straight lines if produced indefinitely, meet on that side on which the sum of angles is taken together less than two right angles.
Point
A point is a fine dot. For example, P is a point as shown in the figure.
$\bullet P$
Line Segment
A line segment is a straight path between two given points. For example in the shown figure PQ is a straight path between the pints P and Q and so is called a line segment $\overline{PQ}$
A line segment has a definite length.
Ray
A ray is a line segment extending indefinitely in one direction. A ray has no definite length. For example, in the shown figure $\overrightarrow{PQ}$ is representing a ray having one and point P.
Line
A line is a ne is obtained on extending a line segment indefinitely in both the directions. In the shown figure, 0$\overleftrightarrow{PQ}$ is represented as a line. A line has no end points so a line has no definite lengths.
Angle
An angle is generated when two rays originated from the same end point. In the shown figure POQ is the angle formed by two rays and . Here O is called the vertex of the $\angle$POQ and PO and OQ are called the arms of the angle POQ.
Note:
(i) If a ray stands on a line, then so formed adjacent angles are supplementary and its converse.
(ii) The vertically opposite angles formed by two intersecting lines are equal.
Parallel Lines and a Transversal
If a transversal intersects two parallel lines, then
(i) Each pair of alternate interior angles is equal and conversely.
(ii) Each pair of corresponding angles is equal and conversely.
(iii) Each pair of interior angles on the same side of the transversal is supplementary and conversely.
Note: Lines parallel to the same line are parallel to each other.
Triangle
A closed figure formed by three line segments is called a triangle. A triangle has three sides, three angles and three vertices. Two figures of same shape and same size are called congruent figures. Two circles of the same radii are congruent. Two triangles ABC and PQR are congruent $(ie.\,\,\Delta \,\,ABC\cong \Delta \,\,PQR)$under the correspondence $A\leftrightarrow P,\,\,B\leftrightarrow Q\,\,and\,\,C\leftrightarrow R$
Note:
(i) Sum of all the three interior angles of a triangle is$180{}^\circ$.
(ii) The exterior angle of a triangle is equal to the sum of the corresponding two interior opposite angles.
Congruent Figures
Two geometrical figures are said to be congruent if they have same shape and size. e.g. two angles are said to be congruent if they have same measures similarly two line segments are said to be congruent if they have same lengths.
Congruency of Triangles
Two triangles ABC and DEF are said to be congruent if and only if, AB = DE, BC = EF, CA =FD, $\angle$A = $\angle$D, $\angle$B = $\angle$E and $\angle$C = $\angle$F.
Criteria for Congruency of triangles
There are generally four criteria for the congruency of triangles which are given below.
S-S-S Criteria
Two triangles are said to be congruent if the three sides of one triangle are equal to the corresponding three sides of the other.
S-A-S Criteria
Two triangles are said to be congruent if the two sides and included angle of one triangle is equal to the other.
A-S-A Criteria
If the two angles and the side included by the angles are equal to the corresponding two angles and included side of the other triangle, then the two triangles are congruent.
R-H-S Criteria
This criteria is for a right-angled triangle. If one side and hypotenuse of a right-angled triangle is equal to the corresponding side and hypotenuse of other right-angled triangle then two right angled triangles are said to be congruent.
Some Important Results
• The longer side of a triangle has greater angle opposite to it.
• The greater angle of a triangle has longer side opposite to it.
• Perpendicular line segment is shortest in length.
• The sum of any two sides of a triangle is greater than the third side.
• The difference between any two sides of a triangle is always less than its third side.
• Angles opposite to equal sides of a triangle are equal.
• Sides opposite to equal angles of a triangle are equal.
Similarity of Triangles
When two geometrical shapes resembles same but need not to be equal in size are called similar figures. Let us observe the following examples:
(i) Any two line segments are always similar.
(ii) Two circles of different radius are always similar.
(iii) For rectilinear figures if all the corresponding angles of a polygon are equal and the ratio of their corresponding sides are also equal, then they are said to be similar.
The criteria of similarity of two triangles are
(i) A-A-A criterion (ii) S-S-S criterion
(iii) S-A-S criterion
The ratio of the areas of two similar triangles is equal to the ratio of the,
• Squares of any two sides of the triangles.
• Squares of their altitudes.
• Squares of their corresponding medians.
• Squares of their corresponding angle bisector segments.
A Plane figure bounded by four line segments is called a quadrilateral.
• Points J, K, L and M are the vertices of quadrilateral JKLM.
• The line segments JK, KL, LM and MJ are the sides of the quadrilateral.
• The two sides of a quadrilateral having a common point are called adjacent sides.
• The two sides having no common end points is called opposite side.
• Two angles of a quadrilateral having common arm are called adjacent angles.
• Two angles of a quadrilateral having no common arm are called opposite angles.
Properties of a Quadrilateral
• A quadrilateral is a parallelogram, if
(i) Opposite sides are equal.
Or (ii) Opposite angles are equal.
Or (iii) Diagonals bisect each other.
Or (iv) A pair of opposite sides is equal and parallel.
• In a rectangle, diagonals are equal and bisect each other and vice-versa.
• In a rhombus, diagonals bisect each other at right angles and vice-versa.
• In a square, diagonals bisect each other at right angles and are equal and vice-versa.
• The line segments joining the mid-points of any two sides of a triangle is parallel to a third side and is half of it.
• A line through the mid-point of a side of a triangle parallel to another side bisects the third side.
• The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.
• The diagonals of a rectangle are equal.
• If the two diagonals of a parallelogram are equal then the parallelogram is a rectangle.
• The diagonal of a rhombus are perpendicular to each other.
• A parallelogram is a square if the diagonals of a parallelogram are equal and intersect at right angles.
• The sum of all angles of a quadrilateral is always$360{}^\circ$.
Results on Areas of Parallelograms and Triangles
• Any diagonal of a parallelogram divides it into two triangles of equal area.
• Triangles which are on the same base and between the same parallel lines are equal in area.
• A median of a triangle divides it into two triangles of equal areas.
• Area of a triangle is half the product of its base and the corresponding altitude.
• Triangles on the same base and having equal areas lie between the same parallels
• If a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
• Area of a parallelogram is the product of its base and the corresponding altitude.
• A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
• Parallelograms which are on the same base and between the same parallel lines are equal in area.
Circle
Circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point is always constant. We know that the fixed point is called centre and the fixed distance is called its radius. Also,
$D\text{ }=\text{ }2r$,$C=2\pi r$, where D is diameter, C is circumference of circle and r is radius.
Terms Related to a Circle
Following are some terms which are very useful in solving the problems related to circles.
Secant
When a line intersects a circle at two distinct points, it is called a secant of the circle. In the following figure line m is a secant of the circle.
Tangent
A line which touches the circle at exactly one point is called a tangent to the circle. In the following figure PQR is a tangent of the circle.
Concentric Circles
Circles are said to be concentric if and only if they have the same centre and different radii.
Concurrent Arc
A continuous piece of circumference of a circle is called an arc and two arcs are said to be concurrent if they subtend equal angles at the centre.
Here, $Arc\,\,\overset\frown{RNS}\equiv Arc\overset\frown{\,\,PMQ}$
Properties of Circles
• Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
• If the angles subtended by two chords of a circle (or of congruent circles) at the centre (corresponding centres) are equal, the chords are equal.
• The perpendicular drawn form the centre of a circle to a chord bisects the chord.
• There is only one circle possible which can pass through three non collinear points.
• The line drawn through the centre of a circle to bisect a chord is perpendicular to the
• Equal chords of a circle are equidistant from the centres.
• Chords equidistant from the centre of a circle are equal in length.
• If two chords of a circle are equal, then their corresponding arcs are congruent and its converse.
• 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.
• Angles in the same segment of a circle are equal.
Note:
(i) The sum of either pair of opposite angles of a cyclic quadrilateral is$180{}^\circ$.
(ii) If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.
(iii) Radius of incircle and circumcircle of an equilateral triangle of side a $a=\frac{a}{2\sqrt{3}}$ and $\frac{a}{\sqrt{3}}$ respectively.
Example:
In the figure given below the height of two poles are 10 metres and 15 metres. If the poles are 30 metres apart then the height of point of intersection of the lines joining the top of each pole from opposite foot of the other pole is:
(a) 18m (b) 8m
(c) 6m (d) 9 m
(e) None of these
In the given figure, $\Delta CAB\sim \Delta CEF\Rightarrow \frac{EF}{10}$$=\frac{FC}{30}\Rightarrow FC=3EF$
Similarly in $\Delta ACD\sim \Delta AFE,$ so we have $\frac{EF}{15}=\frac{AF}{30}\Rightarrow AF=2EF$
From above equations, we get $5EF=30\Rightarrow EF=6m$
Example:
In an equilateral triangle show that the centroid and circumcenter coincide.
Solution:
Given: An equilateral triangle PQR in which M, L and N are the mid points of sides QR, RP and PQ respectively.
To prove: The centroid and circumcentre are coincident in $\Delta PQR$
Construction: Draw medians PM, QL and RN
Proof: Let G be the centroid of $\Delta PQR$ is the point of intersection of medians QL, RN and MP.
In $\Delta QNR$ and $\Delta QLR$we have,
$\angle Q=\angle R=60{}^\circ$ and QN=RL and QR=RQ
$\Rightarrow$ (By SAS)
$\Rightarrow$RN=QL (CPCT) … (i)
Similarly, in $\Delta RPN\cong \Delta PRM$
$\Rightarrow$RN = PM (CPCT) ... (ii)
From (i) and (ii), we get
RN = PM = QL
$\Rightarrow$$\frac{2}{3}$ RN $\frac{2}{3}$ PM = $\frac{2}{3}$ QL
$\Rightarrow$ GR= GP = QG $\Rightarrow$ G is equidistant from the vertices.
$\Rightarrow$ G is circumcentre of$\Delta PQR$.
Hence the centroid and circumcentre of $\Delta PQR$ are coincident.
Example:
In the figure given below.
$\frac{AP}{PC}=\frac{3}{4}$ and $\frac{BR}{RP}=\frac{3}{2}$ and BQ = 15 cm. Find AQ
Solution:
Given: $\frac{AP}{PC}=\frac{3}{4}$, $\frac{BR}{RP}=\frac{3}{2}$ and BQ = 15cm
Draw PS || CQ which meets AB at S.
Applying basic proportionality theorem, we get
$\frac{BQ}{QS}=\frac{BR}{RP}\Rightarrow \frac{15}{QS}=\frac{3}{2}\Rightarrow$QS=10 cm.
Similarly in$\Delta AQC$, we get
$\frac{AS}{SQ}=\frac{BR}{PC}\Rightarrow\frac{AS}{10}=\frac{3}{2}\Rightarrow AS=\frac{30}{4}$= 7.5 cm.
$\therefore$ AQ = AS + QS = 10 + 7.5 = 17.5 cm.
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# 3rd Grade Math Help: How to Compare Fractions
In 3rd grade, you'll learn all about fractions. Continue reading for instructions on how to compare fractions, and then complete a set of ten practice problems to test your knowledge!
## Comparing Fractions in 3rd Grade
Before you start comparing fractions, it helps to understand what a fraction is. Fractions show parts of a whole. Every fraction has a denominator, which tells you the total number of parts that make up the whole. All fractions also have numerators, which tell you the number of parts the fraction has. If a pie has seven slices, and you take three, you have taken 3/7 of the slices because you took three out of the seven total pieces.
If Lisa took two slices of pie, or 2/7, and Sam took one slice of the same pie, or 1/7, you could be sure that Lisa took more pie than Sam, since two slices is more than one slice. Since the denominators of the fractions are the same, you can just compare the numerators to figure out which fraction is bigger. Here are some more examples:
• 1/3 is less than 2/3
• 5/7 greater than 3/7
• 3/10 less than 9/10
### Using Inequality Signs
When we're comparing fractions, we can use the inequality signs for greater than and less than. Just like the equals sign is used to show that two things are the same, like 1 + 2 = 3, the inequality signs are used to show that one quantity is larger or smaller than another quantity. The sign looks like this '>,' or this '<.' The 'open' end of the sign always points to the larger quantity. For instance, we could write, '5/6 is greater than 1/6,' like this: 5/6 > 1/6. If we wanted to write, '2/7 is less than 4/7,' it would look like this: 2/7 < 4/7. Here are a few more examples:
• 1/12 is less than 5/12: 1/12 < 5/12
• 3/8 is greater than 1/8: 3/8 > 1/8
• 7/9 is greater than 5/9: 7/9 > 5/9
### Fractions with the Same Numerators
Imagine that you're given two pies of the exact same size. You cut one of the pies into eight slices and the other pie into four slices. The slices from the pie cut into four pieces will be larger than the slices from the pie that's cut into eight pieces. Now, imagine that you take one slice from each of the pies, and represent each of those slices as a fraction. The slice from the 4-piece pie would be represented by the fraction 1/4, and the slice from the 8-piece pie would be represented by 1/8.
Now, knowing what each of these fractions represents, which one is larger? You know that 1/4 must be greater than 1/8, since one slice from the 4-piece pie is bigger than a slice from the 8-piece pie. You could also tell that 3/4 > 3/8, since three of the larger-sized slices would give you more pie than three of the smaller slices.
Tip: When you're dealing with fractions that have the same numerators, the fraction with the smaller denominator is bigger, and the fraction with the larger denominator is smaller. In the examples above, 4 < 8, so 1/4 > 1/8.
### Practice Problems
For each set of fractions below, write in the appropriate > or < sign between the two fractions.
1. 8/9, 7/9
2. 2/5, 4/5
3. 1/6, 5/6
4. 7/20, 20/20
5. 3/11, 3/4
6. 5/9, 5/12
7. 1/3, 1/4
8. 2/7, 2/5
9. 7/12, 1/12
10. 8/9, 8/10
1. 8/9 > 7/9
2. 2/5 < 4/5
3. 1/6 < 5/6
4. 7/20 < 20/20
5. 3/11 < 3/4
6. 5/9 > 5/12
7. 1/3 > 1/4
8. 2/7 < 2/5
9. 7/12 > 1/12
10. 8/9 > 8/10
Did you find this useful? If so, please let others know!
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# Word Problems with Percentage
Search my site:
A percent is a comparison of 2 numbers. By itself, it is not very meaningful. What does this mean? A percent is not an actual value, it just tells us something about 2 values. You need to know what are the 2 numbers being compared.
Look at this question:
A dress costs \$100. It’s price was decreased by 10% during a sale. After the sale, the price of the dress was increased by 10%.
When you read that question, you may automatically assume that the price of the dress after the sale is back to \$100. That would not be correct. Look at the working below.
An easy way to understand this is 10% of \$100 and 10% of \$50 give two different values, even though both are 10% of something.
When working with percentages, always figure out which 2 numbers are being compared.
### Percentage and Proportion
You can also work with percentages as a proportion.
Look at this example.
I bought some fruits. 20% of them are apples and the remainder are oranges. How many oranges did I buy if I bought 13 apples?
### Some methods to use in solving word problems.
Let’s look at some ways of solving percentage problems.
There are 2 main types of percentage questions. In the first type, the values are given and you have to find the percentage by comparing the values.
Look at this example.
There are 60 children in the park. 20 are boys. What percentage of the children are boys?
The first method is to make use of fractions. 20 boys out of 60 children.
The second method is to work everything out step by step.
We can condense method 2.
The last method is to use proportion.
If you study all the methods carefully, you will realize that they are basically the same way of solving the problem but written in different ways.
Try each method so you have a better understanding of percentages and how to solve word problems.
### Second type of percentage questions
The second type of question is where the value of one percentage is given and you have to find the value of a different percentage.
Here is an example.
20% of a number is 60. What is the number?
The first method is to use fractions.
The second method is a simpler version of method 1.
The next method is a condensed version of method 2.
The last method is by using proportion.
### Percentages greater than 100%
Percentages does not always have to be less than 100. You can add and subtract percentages in the usual way as other numbers.
Look at this example.
George and Mary shared \$380. George’s share of the money is 90% of Mary’s share. How much is George’s share?
Worksheet 1 |
# Readers ask: How To Find The Absolute Minimum Value Of A Function?
## What is the absolute minimum of a function?
An absolute maximum point is a point where the function obtains its greatest possible value. Similarly, an absolute minimum point is a point where the function obtains its least possible value.
## How do you find the absolute minimum on a graph?
The graph attains an absolute minimum at x = 3 displaystyle x=3 x=3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3 displaystyle x=3 x=3, which is −10.
## How do you find the minimum and maximum of a function?
Find the corresponding f(x) value. Insert the value of x that you just calculated into the function to find the corresponding value of f(x). This will be the minimum or maximum of the function.
## Can there be 2 absolute minimums?
It is completely possible for a function to not have a relative maximum and/or a relative minimum. Again, the function doesn’t have any relative maximums. As this example has shown there can only be a single absolute maximum or absolute minimum value, but they can occur at more than one place in the domain.
You might be interested: Why Is The Absolute Value Of A Number Never Negative?
## What is an absolute extrema?
Absolute extrema are the largest and smallest the function will ever be and these four points represent the only places in the interval where the absolute extrema can occur. In this example we saw that absolute extrema can and will occur at both endpoints and critical points.
## What is the absolute minimum?
mathematics.: the smallest value that a mathematical function can have over its entire curve (see curve entry 3 sense 5a) The function defined by y = 3 – x has an absolute maximum M = 2 and an absolute minimum m = O on the interval 1 < x < 3.—
## How do you find the minimum and maximum of a graph?
Second, the most important test for determining max or min is to look at the neighborhood around the critical point in question. If on either side of the point x, f(x) is less than the function right at x, then x is a max. If on either side of the point x, f(x) is greater than the function right at x, then x is a min.
## What is the minimum value?
The minimum value of a function is the lowest point of a vertex. If your quadratic equation has a positive a term, it will also have a minimum value. If you have the equation in the form of y = ax^2 + bx + c, then you can find the minimum value using the equation min = c – b^2/4a.
## How do you find the minimum of a function?
The minimum value of a function is found when its derivative is null and changes of sign, from negative to positive. Example: f(x)=x2 f ( x ) = x 2 defined over R, its derivative is f′(x)=2x f ′ ( x ) = 2 x, that is equal to zero in x=0 because f′(x)=0⟺2x=0⟺x=0 f ′ ( x ) = 0 ⟺ 2 x = 0 ⟺ x = 0.
You might be interested: Quick Answer: What Is The Definition Of Absolute Value Of A Number?
## Where does the minimum or maximum value occur?
If the parabola opens up, the vertex (h, k) is the lowest point on the parabola. We say that k is the minimum functional value of f or the absolute minimum value of f. It occurs when x = h. If the parabola opens down, k is the maximum functional value of f or the absolute maximum value of f, and occurs when x = h. |
# Ncert Solutions for Linear equations Class 8 Chapter 2 CBSE Part 6
In this page we have NCERT book Solutions for Class 8th Maths:Linear equations Chapter 2 for EXERCISE 6 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
Solve:
On multiplying both sides by 3x, we obtain
8x − 3 = 6x
Transposing 6x to LHS and 3 to RHS
8x − 6x = 3
2x = 3
Dividing 2 on both the sides
x=3/2
Question 2
Solve:
On multiplying both sides by 7 − 6x, we obtain
9x = 15(7 − 6x)
9x = 105 − 90x
Transposing 90x on LHS
9x + 90x = 105
99x = 105
Dividing 99 on both the sides
x=105/99
Question 3
Solve:
On multiplying both sides by 9(z + 15), we obtain
9= 4(z + 15)
9= 4z + 60
Transposing 4z on LHS
9− 4z = 60
5z = 60
Dividing 5 on both the sides
z = 12
Question 4
Solve
On multiplying both sides by 5(2 − 6y), we obtain
5(3y + 4) = −2(2 − 6y)
15y + 20 = − 4 + 12y
Transposing 12y to LHS and 20 to RHS
15y − 12y = − 4 – 20
3y = −24
Dividing by 3 on both the sides
y = −8
Question 5
Solve:
On multiplying both sides by 3(+ 2), we obtain
3(7y + 4) = −4(y + 2)
21y + 12 = − 4y – 8
Transposing 4y to LHS and 12 to RHS
21y + 4y = − 8 − 12
25y = −20
Dividing by 25 on both the sides
y=-20/25=-4/5
Question 6- The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.
Let the common ratio between their ages be x. Therefore, Hari’s age and Harry’s age will be 5x years and 7x years respectively and four years later, their ages will be (5x + 4) years and (7x + 4) years respectively.
According to the situation given in the question,
Multiplying both the sides by 4(7x+4)
4(5x+4)=3(7x+4)
20x+16=21x+12
Transposing 20x to RHS and 12 to LHS
4=x
x=4
Hari’s age = 5x years = (5 × 4) years = 20 years
Harry’s age = 7x years = (7 × 4) years = 28 years
Therefore, Hari’s age and Harry’s age are 20 years and 28 years respectively.
Question 7- The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2 . Find the rational number.
Answer - Let the numerator of the rational number be x. Therefore, its denominator will
be x + 8.
The rational number will be x/(x+8) According to the question
2(x + 17) = 3(x + 7)
2x + 34 = 3x + 21
34 − 21 = 3x − 2x
13 = x
Numerator of the rational number = x = 13
Denominator of the rational number = x + 8 = 13 + 8 = 21 |
# How do you write the equation in slope intercept form given point (4, –3) and has slope m = 1?
##### 1 Answer
Feb 25, 2017
$y = \textcolor{red}{1} x - \textcolor{b l u e}{7}$ or $y = x - \textcolor{b l u e}{7}$
#### Explanation:
First, we can write the equation in point-slope form. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the values from the problem gives.
$\left(y - \textcolor{red}{- 3}\right) = \textcolor{b l u e}{1} \left(x - \textcolor{red}{4}\right)$
$y + \textcolor{red}{3} = x - \textcolor{red}{4}$
The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.
We can now solve the equation from above for $y$:
$y + \textcolor{red}{3} - 3 = x - \textcolor{red}{4} - 3$
$y + 0 = x - 7$
$y = \textcolor{red}{1} x - \textcolor{b l u e}{7}$ |
# Practice Questions on Order of Operations: Homework Help for Elementary Students
Page content
To complete this worksheet, you already need an introductory knowledge of the order of operations. If this is too difficult, or there are things you don’t understand, check the remainder of the series for an introduction to the order of operations.
## Directions
Solve each problem by applying the order of operation rules. For the first few problems, write down each step and which operation you used. Every problem is worked through at the end, but try to solve them as best you can before looking. Good Luck.
## Problems
Problems:
1. (6+(-7)-1) x 8
2. (-6)-((-2)-5)^2
3. (6-1)((-3)-4)
1. 20 / (3 x 2 – 4)
2. 4-3+8+3
1. 7 – (10-(-8))-3
2. (28-8) / ((-9)-(-5))
3. ((-10)+3) / ((-8) +7)
1. 2 + (-1)^2 – 7
2. ((-6) / (-3))x((-30) / (-3))
3. (21-6-3) / (8-2)
4. 8^2 / (3+5)
5. (26+1) / (3(4-3))
1. 3 / (8-6+6-5)
2. 4+9-(7-6) -10
3. 6 / (8-2) + 4 + 3
4. 6 x (23 + 4 – 7) / 10
5. 3^2 / (6+5-2)
1. (9-4x1^2) x 6
2. 6 + 8 – 8 + 8^2
## Order of Operations Worksheet Solutions
Each problem is solved step-by-step. Only one operation occurs in each line. The solutions are designed to show you exactly how to think about the problem. If you don’t understand an answer, go back and solve it slowly, comparing each step to the solution. The answers are in BOLD type.
## Problem Solutions
1. (6+(-7)-1) x 8
= (-1-1)x 8
= -2 x 8
= -16
1. (-6)-((-2)-5)^2
= -6 – (-7)^2
= -6 – 49
= -55
1. (6-1)((-3)-4)
=(5)(-3-4)
=(5)(-7)
= -35
1. 20 / (3 x 2 – 4)
= 20 / (6-4)
= 20 / 2
= 10
1. 4-3+8+3
= 1 + 8 + 3
=9 + 3
= 12
1. 7 – (10-(-8))-3
=7 – (10+8)-3
=7 – 18 – 3
= -11 – 3
= -14
1. (28-8) / ((-9)-(-5))
=20 / ((-9)-(-5))
= 20 / (-9+5)
=20/(-4)
= -5
1. ((-10)+3) / ((-8) +7)
= -7 / ((-8)+7)
= -7 / -1
= 7
1. 2 + (-1)^2 – 7
= 2+1-7
=3-7
= -4
1. ((-6) / (-3))((-30) / (-3))
= (2)((-30)/(-3))
=(2)(3)
= 6
1. (21-6-3) / (8-2)
= (15-3) / (8-2)
=12 / (8-2)
=12/6
=2
1. 8^2 / (3+5)
= 8^2 / 8
=64 / 8
=8
1. (26+1) / (3(4-3))
=27/(3(1))
=27/3
=9
1. 3 / (8-6+6-5)
=3/(2+6-5)
=3/(8-5)
=3/(3)
=1
1. 4+9-(7-6) -10
= 4+9-1-10
=13-1-10
=12-10
=2
1. 6 / (8-2) + 4 + 3
= 6 / 6 + 4 + 3
=1+ 4 + 3
=5 + 3
= 8
1. 6 x (23 + 4 – 7) / 10
= 6 x (27-7)/10
= 6 x 20/10
=120/10
=12
1. 3^2 / (6+5-2)
=3^2 / (11-2)
=3^2 / 9
=9/9
=1
1. (9-4x1^2) x 6
= (9 – 4 x 1) x 6
=(9-4) x 6
=5 x 6
= 30
1. 6 + 8 – 8 + 8^2
= 6 + 8 – 8 + 64
= 14 – 8 + 64
= 6 + 64
=70
## This post is part of the series: Order of Operations
Order of operations explanations, quizzes and homework help for kindergarten through middle school students. |
1. ## Implicit Differentiation
Given
$\displaystyle x^2 + 25y^2 = 100$
show that $\displaystyle \frac{d^2y}{dx^2} = - \frac{4}{25y^3}$
Thankies
2. Differentiate and solve for y':
$\displaystyle x^2 + 25y^2 = 100 \Rightarrow 2x + 50yy' = 0 \Leftrightarrow y' = - \frac{x}{{25y}}$
Differentiate again and substitute y' with the previous expression:
$\displaystyle y'' = \frac{{xy' - y}}{{25y^2 }} = \frac{{x\left( { - \frac{x}{{25y}}} \right) - y}}{{25y^2 }} = \frac{{ - \frac{{x^2 + 25y^2 }}{{25y}}}}{{25y^2 }}$
Now you can use the initial relation again in the numerator:
$\displaystyle y'' = \frac{{ - \frac{{x^2 + 25y^2 }}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{{100}}{{25y}}}}{{25y^2 }} = \frac{{ - \frac{4}{y}}}{{25y^2 }} = \boxed{- \frac{4}{{25y^3 }}}$
Given: .[1] $\displaystyle x^2 + 25y^2 = 100,$ .show that: .$\displaystyle \frac{d^2y}{dx^2} = - \frac{4}{25y^3}$
Differentiate: .$\displaystyle 2x + 50y\left(\frac{dy}{dx}\right) \:=\:0\quad\Rightarrow\quad\frac{dy}{dx}\:=\:-\frac{1}{25}\,\frac{x}{y}$ [2]
Differentiate again: .$\displaystyle \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(\frac{dy}{dx})}{y^2}\right]$
Substitute [2]: .$\displaystyle \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{y - x(-\frac{x}{25y})}{y^2}\right] \:=\:-\frac{1}{25}\left[\frac{y + \frac{x^2}{25y}}{y^2}\right]$
Multiply top and bottom by $\displaystyle 25y\!:\;\;\frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{25y^2 + x^2}{25y^3}\right]$
From [1], the numerator equals 100: .$\displaystyle \frac{d^2y}{dx^2} \:=\:-\frac{1}{25}\left[\frac{100}{25y^3}\right]$
Therefore: .$\displaystyle \boxed{\frac{d^2y}{dx^2} \:=\:-\frac{4}{25y^3}}$ |
0
# What are common factors of 65 and 91?
Updated: 8/29/2023
Wiki User
11y ago
The common factors of 35 and 91 are 1 and 7.
Wiki User
14y ago
Wiki User
8y ago
The common factors of 49 and 91 are 1 and 7
Wiki User
9y ago
The Highest Common Factor (HCF) of two or more numbers is the highest number which will divide all of them without leaving a remainder. 91 has the factors 7.
Wiki User
7y ago
The only common factor is one.
Wiki User
7y ago
The factors of 26 are:
1, 2, 13, 26
The factors of 91 are:
1, 7, 13, 91
The common factors are:
1, 13
Wiki User
13y ago
The only common factor is: 1
Wiki User
10y ago
The HCF is 3
Wiki User
9y ago
1, 61
1, 7, 13, 91
Wiki User
11y ago
1 and 13
Earn +20 pts
Q: What are common factors of 65 and 91?
Submit
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Related questions
### What are the common factors for 65 91?
The common factors are: 1, 13
### What are the factors of 91 and 65?
The factors of 65 are: 1, 5, 13, 65 The factors of 91 are: 1, 7, 13, 91
### What is the GCF of 169 91 65?
The factors of 65 are:1, 5, 13, 65The factors of 91 are:1, 7, 13, 91The factors of 169 are:1, 13, 169The common factors are:1, 13The Greatest Common Factor (GCF) is:13
### What is the greatest common factor of 91 and 65 and 143?
The Greatest Common Factor of 143, 65, 91: 13
### What is the greatest common factor of 55 and 91?
To find the greatest common factor of three numbers, you need to first express them as the product of their prime factors. In this case, 52 65 and 91 can be expressed as follows: 52 = 2x2x13 65 = 5x13 91 = 7x13 The next step is to identify any common factors. In this case, the only factor which appears in all 3 numbers' prime factors is 13. Thus the greatest common factor of 52, 65 and 91 is 13.
### What is the least common denominator of 65 and 91?
LCD(65, 91) = 455
### What are the common factors of 65?
The common factors of 65 are 1,5,13,and65
### What are the common factors of 39 and 91?
The common factors of 39 and 91 are: 1 and 13
### What are the common factors of 63 and 91?
The common factors of 63 and 91 are 1 and 7.
### What are the common factors of 52 and 91?
The common factors of 52 and 91 are: 1 and 13
### What are the common factors of 65 and 25?
1 and 5 are the common factors of 65 and 25.
### What are the factor and prime factors of 91?
The factors of 91 are: 1 7 13 91 The prime factors are: 7 and 13 |
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## 4th grade foundations (Eureka Math/EngageNY)
### Course: 4th grade foundations (Eureka Math/EngageNY) >Unit 1
Lesson 3: Topic D, E, & F: Foundations
# Breaking apart 3-digit addition problems
Sal shows ways to break up addition problems using place value.
## Want to join the conversation?
• Well to break up 4 digit addition problems you do the same process. For example: 4093 + 2056 = 3 ones + 6 ones + 5 tens + 9 tens + 0 hundreds + 0 hundreds + 4 thousands + 2 thousands.
edit: For doing subtraction problems you should probably to carrying witch is kind of similar so yes.
• Why do we have to cross out instead of solving it all in our brains?
• When you solve something in your brain you use all these techniques, even if you don't realize...
Knowing what you are doing and how the numbers work will help you solve problems faster and understand more complex things later
• Why dont you just add 189 + 608?
• You could, but this is just another way to portray this equation using place values and expanded form.
• how do i break it apart
## Video transcript
- Mike isn't sure how to add 189 + 608, help Mike by choosing an addition problem that is the same as 189 + 608. Now let's look at these choices. Let's just start with this first choice, actually all of these choices start with having 1 hundred, they all have 1 hundred, so where do we see 1 hundred here? Well in 189, we have a 1 in the hundreds place, so this right over here, that is one 1 hundred, 100, and then all of the choices actually have 6 hundreds, 6 hundreds, 6 hundreds, where are they getting that from? Well in 608, the 6 is in the hundreds, so that 6, represents 6 hundreds, 6 hundreds, so that's where they got the 1 hundred, and the 6 hundreds from, and then all of them have 8 tens, so they're actually all looking pretty similar, up to that point, they all have 8 tens, 8 tens, 8 tens, where do they get that from? Well in 189, this 8 is in the tens place, so it represents 8 tens, 8 tens, and then what else do they have? Well this is where they all start to be a little bit different, so let's go one by one. So this next one then has 9 ones, 9 ones, where are they getting that from? Well in 189, the 9 is in the ones place, so it's reasonable to write 9 ones, and then finally, they have 8 ones. 8 ones, where do they get that from? Well in 608, the 8 is in the ones place, so it's 8 ones, so this first choice is looking quite good. Now you might be saying, 'Okay, we took into account, all of these digits, except for this 0 over here, how come we didn't, how come there's a 0 here in the tens place?' Well that would just be 0 tens, 0 tens, which is just zero, so it's not going to change the value of this, so this first choice is indeed the same, as the original addition problem. If we want to see where the other ones break down, this one has 8 tens written twice, we only have one, we have an 8 tens here, but this 8 is in ones, so this should say 'ones', right over here, and this choice, we have the 8 ones, but this 9 right over here, this is ones, this isn't tens, so this should be 'ones'. So that's why we wouldn't pick either of those. Let's do another one of these, or related type of problem. Which addition problem is the same as 525 + 379? Let's just break it down. So we have a 5 in the hundreds place, so that would be 500, and then we have a 3 in the hundreds place, so + 300, then we have a 2 in the tens place, so it's two 10, or 20, and then we have a 7 in the tens place, so that's + 70, and then last but not least, we have a 5 in the ones place, so that's just going to be 5, and then we have a 9 in the ones place, so that's just going to be equal to 9. So which of these choices is the same of what I just wrote over here? 500 + 300, so this first choice has no 500 or 300, so we can rule that out. 500 + 300 + 20 + 70 + 5 +9, that's exactly what I wrote down. This last choice breaks down, they wrote 90 instead of 9, and then they wrote 5, and then they wrote 7, instead of 70. That 7 is in the tens place, it's not in the ones place, we ruled that one out as well. |
# Show that the coefficient of x4 in the expansion of
Question:
Show that the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 212
Solution:
To show: that the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 212 .
Formula Used:
We have
$\left(1+2 x+x^{2}\right)^{5}=\left(1+x+x+x^{2}\right)^{5}$
$=(1+x+x(1+x))^{5}$
$=(1+x)^{5}(1+x)^{5}$
$=(1+x)^{10}$
General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,
$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where $s$
${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$
Now, finding the general term,
$\mathrm{T}_{\mathrm{r}+1}={ }^{10} \mathrm{Cr} \times \mathrm{X}^{10-\mathrm{r}} \times(1)^{\mathrm{r}}$
$10-r=4$
$r=6$
Thus, the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is given by,
${ }^{10} \mathrm{C}_{4}=\frac{10 !}{4 ! 6 !}$
${ }^{10} \mathrm{C}_{4}=\frac{10 \times 9 \times 8 \times 7 \times 6 !}{24 \times 6 !}$
${ }^{10} \mathrm{C}_{4}=210$
Thus, the coefficient of $x^{4}$ in the expansion of $\left(1+2 x+x^{2}\right)^{5}$ is 210 |
# Fractions to Decimals Mar27
Views:
Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Slide1:
REVIEW FRACTIONS & DECIMALS
### Slide2:
FRACTIONS w ritten 3 ways Proper fractions Improper fractions Mixed Numbers review 1 2 3
### Slide3:
Proper fractions Improper fractions & Mixed numbers 1 2 3 4 4 4 4 4 5 6 7 8
### Slide4:
Simplifying a Fraction is finding a special Equivalent Fraction using Division… 9 ÷ 2 = 1 2 2 ÷ 3 = 2 3 3 ÷ 3 = 1 3 5 Select the biggest number that you can divide into the Numerator AND the Denominator .
### Slide5:
Rewrite an Improper Fraction as Whole N umber or Mixed Number : 2 choices : Divide or Decompose = 14 ÷ 3 14 = 3 + 3 + 3 + 3 + 2 3 3 3 3 3 3 = 4 2 3 = 4 2 3 Divide Decompose = 9 ÷ 6 = 1 3 6 Divide Decompose 9 = 6 + 3 6 6 6 = 1 3 6 or or = 1 1 3 = 1 1 3 Simplify if the instructions require it. simplify Simplify if the instructions require it.
### Slide6:
Change a Mixed Number into an Improper F raction : 4 4 3 = Then 12 4 1= If Finally, add the proper fraction: First, change the whole number : 12 + 2 = 14 4 4 4 10 10 1 = If First, change the whole number : 5 = Then 5 0 10 50 + 6 = 56 10 10 10 Finally, add the proper fraction: “3 and two-fourths” “5 and six-tenths” “14 fourths” “56 tenths”
### Slide7:
Change a Mixed Number into an Improper F raction : Shortcut: x + x + = 4 x 3 + 2 4 = 14 4 = 10 x 5 + 6 10 = 56 10 Example #1 Example #2
### Slide8:
Multiplying a Whole Number times a Proper Fraction (a lso called Finding a Fraction Of a Whole Number ) Every whole number sits on top of an invisible “1”. 3 x 2 1 x 4 = 6 4 Change the improper fraction 6 = 4 + 2 4 4 4 = 1 …into a mixed number, Every whole number sits on top of an invisible “1”. 2 x 3 1 x 15 = 6 15 Simplify the proper fraction by a nd simplify 2/4 to 1/2. = 1 = 6 ÷ 3 15 3 1 2 = 2 5 dividing by “1,” written as 3/3. Final answer Final answer
### Slide9:
Adding or Subtracting “Like” Fractions (Fractions that have the same Denominator ) + 1 = 3 4 4 Add the Numerators. Do NOT add the Denominators! + 3 = 5 = 4 4 Add the Numerators. Do NOT add the Denominators! Simplify: improper fraction to mixed number. 1 Example #1 Example #2 Example #3 3 1 = 2 = 1 6 6 6 3 Subtract the Numerators. Do NOT subtract the Denominators! Simplify the proper fraction.
### Slide10:
Adding or Subtracting Fractions when t he Denominators are a mix of 10 and 100 6 7 = 10 100 You must have the same denominator t o add or subtract fractions! Change 6/10 to an equivalent fraction that has a denominator of 100 to match 7/100. 6 x 10 = 60 10 10 100 Use the equivalent fraction 60/100 instead of 6/10. 60 7 = 53 100 100 100 Now you are ready to subtract!
### Slide11:
Shortcut: Adding or Subtracting Fractions when t he Denominators are a mix of 10 and 100 6 7 = 10 100 6 0 10 0 “ Powers of 10 ” allow us to attach a “0” to the Numerator and attach a “0” to the Denominator. It’s the same as multiplying 6/10 by 10/10 to get 60/100!
### Slide12:
Going from Fraction Land to Decimal Land The Fraction must have the right ticket ! Fraction Land Equivalent Fraction Land Decimal Land 7 10 1 = 25 4 100 25 100 0.25 0.7 7 10 1 4 The Fraction must have 10 or 100 in the Denominator! 1 4
### Slide13:
Welcome to Decimal Land
### Slide14:
Proper Fractions Improper Fractions & Mixed Numbers 1 2 3 4 4 4 4 4 5 6 7 8 0.25 0.5 0.75 1 .25 1.5 1.75 Decimals 1.0 2.0
### Slide15:
Proper fractions and Decimals They are less than 1.
### Slide16:
Proper fractions & Decimals 0 0 0 They are less than 1.
### Slide17:
= 0.1 = 0.01 1 10 1 100 =1 One Tenth (0.1) is 10 times (10x) bigger than One Hundredth (0.01). One (1) is 10 times (x10) bigger than One Tenth (0.1), and One (1) is 100 times (x100) bigger than One Hundredth (0.01).
### Slide18:
I n Whole Number Land, Hundreds are bigger than Tens. In Decimal Land, Hundred th s are smaller than Ten th s. 500 > 50 0.5 > 0.05
### Slide19:
Tens Ones Ten ths Hundred ths X 10 X 10 3 0 3 5 5 0 and Compare on the PLACE VALUE CHART Remember : Each Place Value is a Power of 10. As you move to t he left, each Place Value gets 10x bigger. Bigger and bigger values X 10
### Slide20:
Hundreds Tens Ones Ten ths Hundred ths 0 1 1 0 0 5 2 5 \$1.01 \$1.05 \$1.10 \$1.25 1 1 1 1 MONEY – decimals!
### Slide21:
If the Denominator is a Power of 10: 10 or 100 EASY! FRACTION DECIMAL 0.1 Denominator already is a power of 10 EASY! J ust say it and write it……… 1 .3 15 100 0 .15 “one tenth” “one and three tenths” “fifteen hundredths”
### Slide22:
If the Denominator is NOT a Power of 10: 10 or 100 FRACTION DECIMAL 0.4 0 .75 2 5 1 .5 3 4 FIRST make an Equivalent Fraction that has a 10 or 100 in the denominator ! x 2 = 4 2 10 THEN say it and write it! “four tenths” x 25 = 75 25 100 1 2 x 5 = 5 5 10 “seventy-five hundredths” “one and five tenths”
### Slide23:
Easy Equivalencies to Know: 1 = 25 = 0.25 4 100 2 = 1 = 5 = 0.5 = 50 = 0.50 4 2 10 100 3 = 75 = 0.75 4 100 25 cents i s 1/4 of a dollar 50 cents i s 2 /4 or 1/2 of a dollar 7 5 cents i s 3 /4 of a dollar
### Slide24:
More Equivalencies to Know: 1 = 0.2 or 0.20 5 2 = 0.4 or 0.40 5 3 = 0.6 or 0.60 5 4 = 0.8 or 0.80 5 20 cents i s 1/5 of a dollar 4 0 cents i s 2 /5 of a dollar 60 cents i s 3/5 of a dollar 8 0 cents i s 4 /5 of a dollar
The End The End |
# Finding the Range of a Function Graphed Below
As you continue to explore the world of mathematics, you will come across many functions that describe various phenomena. These functions can help you make sense of the world around you and solve complex real-life problems. The range of a function is one of the critical concepts that you must understand if you want to master these functions. In this article, we’ll discuss how to find the range of a function graphed below.
Before we delve into the details, it’s essential to understand what the range of a function is. Simply put, the range of a function is the set of all possible output values for a given input. In other words, it is the complete set of y-values that the function can produce. For example, suppose we have a function f(x) = x^2. The range of the function would be all non-negative real numbers since the square of any real number is non-negative.
Now let’s move on to finding the range of a function graphed below. When you’re given a graph, finding the range becomes a bit simpler. The range of a function graphed below is the set of all y-values that correspond to points on the graph. To find the range, examine the graph and identify the highest and lowest y-values. The range would be the set of all y-values between these two extremes.
To illustrate this concept, let’s consider the following function:
f(x) = 2x + 1
To find the range of this function, we need to graph it first. We can do this by creating a table of values and plotting the corresponding points on the Cartesian plane.
Table of values:
x | f(x)
-2 | -3
-1 | -1
0 | 1
1 | 3
2 | 5
Using these values, we can plot the points (-2, -3), (-1, -1), (0, 1), (1, 3), and (2, 5) on the Cartesian plane. We can then connect these points to get a straight line graph.
Now that we have graphed the function, we can find its range. From the graph, we can see that the lowest y-value is -3, and the highest y-value is 5. Therefore, the range of this function is the set of all y-values between -3 and 5, inclusive. We can write this as follows:
Range = {-3, -2, -1, 0, 1, 2, 3, 4, 5}
It’s essential to note that when finding the range of a function graphed below, you need to be careful not to exclude any possible y-values. You also need to ensure that you’ve identified all the extreme values correctly.
Another way to find the range of a function graphed below is by analyzing its behavior as x approaches infinity or negative infinity. If a function grows without bound, its range will also grow without bound. In contrast, if a function has a bounded range, it means that its output values are limited. For instance, let’s consider the following function:
f(x) = 1/x
If we graph this function, we can see that as x approaches zero, f(x) grows without bound. Therefore, the range of this function is all real numbers except 0. We can write this as follows:
Range = (-∞, 0) U (0, ∞)
In conclusion, finding the range of a function graphed below is an essential skill that every student of mathematics must learn. It involves identifying the highest and lowest y-values from the graph and including them in the range. Moreover, you can analyze the function’s behavior as x approaches infinity or negative infinity to find its range. With these tips in mind, you’ll be able to find the range of any function graphed below with ease. |
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One instrument that can be used is Composite function solver. If you're solving equations that contain the value e, you'll need to use a different set of rules than those for solving regular algebraic equations. First, let's review the definition of e. E is a mathematical constant that is equal to 2.718281828. This number pops up often in mathematical equations, particularly those involving exponential growth or decay. Now that we know what e is, let's talk about how to solve equations that contain this value. First and foremost, you'll need to use the properties of exponents. Next, you'll need to be able to identify which terms in the equation are exponentiated by e. Once you've correctly identified these terms, you can begin solving for the unknown variable. With a little practice, you'll be solving equations with e in no time!
Logarithmic functions are a type of math used to calculate an exponent. The log function is the inverse of the exponential function, meaning that it can be used to solve for x when given a number raised to a power. In order to solve logarithmic functions, you need to use a few basic steps. First, identify the base of the logarithm. This is usually either 10 or e. Next, identify the number that is being raised to a power. This number is called the argument. Finally, set up an equation using these two numbers and solve for x. With a little practice, solving logarithmic functions can be easy and even enjoyable!
We all know that exponents are a quick way to multiply numbers by themselves, but how do we solve for them? The answer lies in logs. Logs are basically just exponents in reverse, so solving for an exponent is the same as solving for a log. For example, if we want to find out what 2^5 is, we can take the log of both sides of the equation to get: 5 = log2(2^5). Then, we can just solve for 5 to get: 5 = log2(32). Therefore, 2^5 = 32. Logs may seem like a complicated concept, but they can be very useful in solving problems with exponents.
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# Find the length of the altitude of an equilateral triangle with side 6cm.
Last updated date: 20th Sep 2024
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Hint: First of all, draw the diagram with the given data to get a clear idea of what we have to find. Then use Pythagoras theorem to find the altitude of the given triangle. An alternate method is also provided in the note. We can use any one of the methods to get the required answer.
Let the given equilateral triangle is $\Delta ABC$
The side length of equilateral $\Delta ABC$ is 6 cm.
Here we have to find the length of altitude of $\Delta ABC$.
Let the foot perpendicular to the altitude from point $A$ to the line $BC$ is $D$. So, the altitude is $AD$.
Since, $\Delta ABC$ is equilateral the altitude $AD$ divides the triangle into two equal right-angle triangles as shown in the below figure:
From Pythagoras theorem we have ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}$
So, in equilateral triangle $\Delta ABC$ we have
$\Rightarrow {\left( {AD} \right)^2} + {\left( {CD} \right)^2} = {\left( {AC} \right)^2} \\ \Rightarrow {\left( {AD} \right)^2} + {\left( 3 \right)^2} = {\left( 6 \right)^2} \\ \Rightarrow {\left( {AD} \right)^2} + 9 = 36 \\ \Rightarrow {\left( {AD} \right)^2} = 36 - 9 = 27 \\$
Rooting on both sides we get
$\Rightarrow \sqrt {{{\left( {AD} \right)}^2}} = \sqrt {27} = \sqrt {3 \times 3 \times 3} \\ \therefore AD = 3\sqrt 3 {\text{ cm}} \\$
Thus, the required length of altitude is $3\sqrt 3 {\text{ cm}}$.
Note: Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides” i.e., ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Adjacent side}}} \right)^2} + {\left( {{\text{Opposite side}}} \right)^2}$.
Alternate method:
We know that the length of the altitude of the equilateral triangle of side length $a{\text{ cm}}$ is given by $\dfrac{{\sqrt 3 }}{2}a{\text{ cm}}$.
So, the altitude of the equilateral triangle of length 6 cm is $\dfrac{{\sqrt 3 }}{2}\left( 6 \right) = \dfrac{{6\sqrt 3 }}{2} = 3\sqrt 3 {\text{ cm}}$ |
# 5th Class Mental Ability Measurement Length, Weight and Capacity Measurement
Measurement
Category : 5th Class
Measurement
Learning Objectives
• Perimeter
• Area
• Volume
Perimeter
Perimeter is referred as the length of the boundary line, which surrounds the area occupied by a geometrical shape.
Perimeters of different geometrical shapes are explained below.
A. Perimeter of a Triangle
A triangles has three sides. Perimeter of a triangle is the sum of its all the three sides.
Perimeter of the triangle $ABC=AB+BC+CA$
Perimeter of a quadrilateral is the sum of the length of its four sides.
In quadrilateral ABCD, perimeter $=AB+BC+CD+DA$
C. Perimeter of a Rectangle
Perimeter of a rectangle = 2 (Length + Breadth).
D. Perimeter of a square $=\mathbf{4}\times \mathbf{side}$.
Perimeter of the square $ABCD=4\times AB$
E. Perimeter of a Circle
Perimeter of a circle $=2\pi r$
Where $~\pi =\frac{22}{7}~=3.14$ and r = radius of the circle
Area
All the geometrical shapes occupies some space. The occupied space by a geometrical shape is called area of that geometrical shape.
Shaded part in the above figures represent area.
Unit of area is $c{{m}^{2}}$or ${{m}^{2}}$.
Areas of different geometrical shapes are listed belowA. Area of a Triangle
Area of a triangle $=1/2\times ~\,base~\times \,height$.
Where base is the one side of a triangle and height is the length of the line segment drawn $90{}^\circ$ on the base of that triangle.
B. Area of a Rectangle
Area of a rectangle$\text{=length }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{breadth}$.
Area of the rectangle $PQRS=PQ\times QR$. Where PQ is the length and QR is the breath.
C. Area of a Square
Area of a square $\text{=sid}{{\text{e}}^{\text{2}}}\text{=side }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{side}$
Area of the square $PQRS=PQ~\times \,PQ=P{{Q}^{2}}$.
D. Area of a Circle
Area of the circle = $\pi {{r}^{2}}$
Where $\pi =\frac{22}{7}=3.14$
1. Find the perimeter of the following figure.
(a) 22.45 cm (b) 23.50 cm
(c) 20.15 cm (d) 15.55 cm
(e) None of these
Solution: Perimeter of the figure $=4\text{ }cm+3\text{ }cm+4\text{ }cm+2.5\text{ }cm+5\text{ }cm+5\text{ }cm=23.50\text{ }cm$.
2. Find the perimeter of the following triangle.
(a) 14.7 cm (b) 13.2 cm
(c) 13.2 c m (d) 16.5 cm
(e) None of these
Solution: Perimeter of the triangle PQR
$=4\text{ }cm+4.7\text{ }cm+6\text{ }cm$
$=14.7\text{ }cm$
3. Find the perimeter of the following quadrilateral.
(a) 12 cm (b) 10 cm
(c) 15 cm (d) 19 cm
(e) None of these
Solution: Perimeter of the quadrilateral $=PS+SR+RQ+QP=5\text{ }cm+3\text{ }cm+4\text{ }cm+3cm=15\text{ }cm$
4. Find the perimeter of the rectangle whose length is 12 cm and breadth is 8 cm.
(a) 40 cm (b) 20 cm
(c) 15 cm (d) 30 cm
(e) None of these
Solution: Perimeter of the rectangle
$=2\left( 12+8 \right)=40\text{ }cm$.
5. Find the perimeter of the square whose length of one side is 9 cm.
(a) 32 cm (b) 31 cm
(c) 36 cm (d) 15 cm
(e) None of these
Solution: Perimeter of a square $=4~\times \,side$
$=4~\times \,9\text{ }cm=36\text{ }cm$
6. If radius of a circle is 0.35 cm, find the perimeter of the circle.
(a) 2.2 cm (b) 2.1 cm
(c) 2.3 cm (d) 3.1 cm
(e) None of these
Solution: Perimeter the circle $=2\pi r$
$=2\times \frac{22}{7}\times 0.35\,\,cm$
$=2.2\text{ }cm$
7. Find the area of the triangle whose base is 75 cm and height is 80 cm.
(a) $3000\,c{{m}^{2}}$
(b) $1500\,c{{m}^{2}}$
(c) $3500\,c{{m}^{2}}$
(d) $2000\,c{{m}^{2}}$
(e) None of these
Solution: Area of the triangle $=1/2~\times \,b~\times \,h$
$=\frac{1}{2}\times ~75\text{ }cm~\times \,80\text{ }cm=3000\text{ }c{{m}^{2}}$
8. Find the area of the rectangle whose length is 17 cm and breadth is 15 cm.
(a) $253\,c{{m}^{2}}$
(b) $255\,c{{m}^{2}}$
(c) $241\text{ }c{{m}^{2}}$
(d) $234\text{ }c{{m}^{2}}$
(e) None of these
Solution: Area of the rectangle $\text{=l }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{b}$
$=17\text{ }cm~\times \,15\text{ }cm=255\text{ }c{{m}^{2}}$
9. Find the area of the square whose length of each side is 21 cm.
(a) $441\,c{{m}^{2}}$
(b) $420\,c{{m}^{2}}$
(c) $244\,c{{m}^{2}}$
(d) $211\,c{{m}^{2}}$
(e) None of these
Solution: Area of the square $\text{=side }\!\!\times\!\!\text{ side=21 cm }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{21 cm=441 c}{{\text{m}}^{\text{2}}}$
10. Find the area of the circle whose radius is 0.28 cm.
(a) $0.2342\,c{{m}^{2}}$
(b) $0.2251\,c{{m}^{2}}$
(c) $0.2464\,c{{m}^{2}}$
(d) $0.2142\,c{{m}^{2}}$
(e) None of these
Solution: Area of a circle $=2{{r}^{2}}$
$=\frac{22}{7}\times ~0.28\text{ }cm~\times \,0.28\text{ }cm=0.2464\text{ }c{{m}^{2}}$
11. Find the volume of the cuboid whose length, breadth and height are 15 cm, 13 cm and 14 cm respectively.
(a) $2507\,c{{m}^{2}}$
(b) $2730\,\,c{{m}^{2}}$
(c) $2302\,c{{m}^{2}}$
(d) $2350\,\,c{{m}^{2}}$
(e) None of these
Solution: Volume of the cuboid $\text{=l }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{b }\!\!~\!\!\text{ }\!\!\times\!\!\text{ }\,\text{h}$
$=15\text{ }cm~\,\times \,13\text{ }cm\times 14\text{ }cm=2730\text{ }c{{m}^{3}}$
Volume
In our daily life a number of things are stored in different kinds of containers Holding capacity of a container is called its volume.
Volume of a Cuboid
Volume of a cuboid = length $\times$ breadth $\times$ height = Ibh.
Volume of the cuboid$ABCDEFG=AB\times AE\times BC$.
Where, length = AB, breadth = AE and height = BC
Volume of a Cube
Volume of a cube $=sid{{e}^{3}}=side~\times \,side~\times \,side$
#### Other Topics
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Course Homepage Preface On Learning
Build Your Intuition
1. 1-Minute Summary 2. X-Ray Vision 3. 3d Intuition
Learn The Lingo
4. Integrals, Derivatives 5. Computer Notation
Basic Understanding
6. Improved Algebra 7. Linear Changes 8. Squared Changes
Deeper Understanding
9. Infinity 10. Derivatives 11. Fundamental Theorem
Figure Out The Rules
12. Add, Multiply, Invert 13. Patterns In The Rules 14. Take Powers, Divide
Put It To Use
15. Archimedes' Formulas Summary
Welcome to an intuition-first calculus course.
Read online, or buy the complete edition for videos, PDF, and discussion.
14 min read
14. The Fancy Arithmetic Of Calculus
Here's the rules we have so far:
$$(f + g)' = f' + g'$$ $$(f \cdot g)' = f \cdot g' + g \cdot f'$$ $$\left( \frac{1}{x} \right)' = - \frac{1}{x^2}$$
Let's add a few more to our collection.
Power Rule
We've worked out that $$\frac{d}{dx}x^2 = 2x$$$: We can visualize the change, and ignore the artificial corner piece. Now, how about visualizing $$x^3$$$?
The process is similar. We can glue a plate to each side to expand the cube. The "missing gutters" represent artifacts, where our new plates would interact with each other.
I have to keep reminding myself: the gutters aren't real! They represent growth that doesn't happen at this step. After our growth, we "melt" the cube into its new, total area, and grow again. Counting the gutters would over-estimate the growth that happened in this step. (Now, if we're forced to take integer-sized steps, then the gutters are needed -- but with infinitely-divisible decimals, we can change smoothly.)
From the diagram, we might guess:
$$\frac{d}{dx} x^3 = 3x^2$$
And that's right! But we had to visualize the result. Abstractions like algebra let us handle scenarios we can't visualize, like a 10-dimensional shape. Geometric shapes are a nice, visual starting point, but we need to move beyond them.
We might begin analyzing a cube with using algebra like this:
$$(x + dx)^3 = (x + dx)(x + dx)(x + dx) = (x^2 + 2x \cdot dx + (dx)^2)(x + dx) = ...$$
Yikes. The number of terms is getting scary, fast. What if we wanted the 10th power? Sure, there are algebra shortcuts, but let's think about the problem holistically.
Our cube $$x^3 = x \cdot x \cdot x$$$has 3 components: the sides. Call them a, b and c to keep 'em straight. Intuitively, we know the total change has a contribution from each side: Now what change does each side think it's contributing? • a thinks: My change ($$da$$$) is combined with the other, unmoving, sides ($$b \cdot c$$$) to get $$da \cdot b \cdot c$$$
• b thinks: My change ($$db$$$) is combined with the other sides to get $$db \cdot a \cdot c$$$
• c thinks: My change ($$dc$$$) is combined with the other sides to get $$dc \cdot a \cdot b$$$
Each change happens separately, and there's no "crosstalk" between $$da$$$, $$db$$$ and $$dc$$$(such crosstalk leads to gutters, which we want to ignore). The total change is: $$A's \ changes + B's \ changes + C's \ changes = (da \cdot b \cdot c) + (db \cdot a \cdot c) + (dc \cdot a \cdot b)$$ Let's write this in terms of $$x$$$, the original side. Every side is identical, ($$a = b = c = x$$$) and the changes are the same ($$da = db = dc = dx$$$), so we get:
$$(dx \cdot x \cdot x) + (dx \cdot x \cdot x) + (dx \cdot x \cdot x) = x^2 \cdot dx + x^2 \cdot dx + x^2 \cdot dx = 3x^2 \cdot dx$$
Converting this to a "per dx" rate we have:
$$\frac{d}{dx} x^3 = 3x^2$$
Neat! Now, the brain-dead memorization strategy is to think "Pull down the exponent and decrease it by one". That isn't learning!
Think like this:
• $$x^3$$$has 3 identical perspectives. • When the system changes, all 3 perspectives contribute identically. Therefore, the derivative will be $$3 \cdot something$$$.
• The "something" is the change in one side ($$dx$$$) multiplied by the remaining sides ($$x \cdot x$$$). The changing side goes from $$x$$$to $$dx$$$ and the exponent lowers by one.
We can reason through the rule! For example, what's the derivative of $$x^5$$$? Well, it's 5 identical perspectives ($$5 \cdot something$$$). Each perspective is me changing ($$dx$$$) and the 4 other guys staying the same ($$x \cdot x \cdot x \cdot x = x^4$$$). So the combined perspective is just $$5x^4$$$. The general Power Rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ Now we can memorize the shortcut "bring down the exponent and subtract", just like we know that putting a 0 after a number multiplies by 10. Shortcuts are fine once you know why they work! Integrals of Powers Let's try integrating a power, reverse engineering a set of changes into the original pattern. Imagine a construction site. Day 1, they order three 1×1 wooden planks. The next day, they order three 2×2 wooden planks. Then three 3×3 planks. Then three 4×4 planks. What are they building? My guess is a cube. They are building a shell, layer by layer, and perhaps putting grout between the gutters to glue them together. Similarly, if we see a series of changes like $$3x^2$$$, we can visualize the plates being assembled to build a cube:
$$\int 3 x^2 = x^3$$
Ok -- we took the previous result and worked backward. But what about the integral of plain old $$x^2$$$? Well, just imagine that incoming change is being split 3 ways: $$x^2 = \frac{x^2}{3} + \frac{x^2}{3} + \frac{x^2}{3} = \frac{1}{3} 3 x^2$$ Ah! Now we have 3 plates (each 1/3 of the original size) and we can integrate a smaller cube. Imagine the "incoming material" being split into 3 piles to build up the sides: $$\int x^2 = \int \frac{1}{3} \cdot 3x^2 = \frac{1}{3} \int 3 x^2 = \frac{1}{3} x^3$$ If we have 3 piles of size $$x^2$$$, we can make a full-sized cube. Otherwise, we build a mini-cube, 1/3 as large.
The general integration rule is:
$$\int x^n = \frac{1}{n + 1} x^{n + 1}$$
After some practice, you'll do the division automatically. But now you know why it's needed: we have to split the incoming "change material" among several sides. (Building a square? Share changes among 2 sides. Building a cube? Share among 3 sides. Building a 4d hypercube? Call me.)
The Quotient Rule
We've seen the derivative of an inverse (a "simple division"):
$$\frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2}$$
Remember the cake metaphor? We cut our existing portion ($$\frac{1}{x}$$$) into $$x$$$ slices, and give one away.
Now, how can we find the derivative of $$\frac{f}{g}$$$? One component in the system is trying to grow us, while the other divides us up. Which wins? Abstraction to the rescue. When finding the derivative of $$x^3$$$, we imagined it as $$x^3 = a \cdot b \cdot c$$$, which helped simplify the interactions. Instead of a mishmash of x's being multiplied, it was just 3 distinct perspectives to consider. Similarly, we can rewrite $$\frac{f}{g}$$$ as two perspectives:
$$\frac{f}{g} = a \cdot b$$
We know $$a = f$$$and $$b = \frac{1}{g}$$$. From this zoomed-out view, it looks like a normal, rectangular, product-rule scenario:
$$(a \cdot b)' = da \cdot b + db \cdot a$$
It's our little secret that $$b$$$is really $$\frac{1}{g}$$$, which behaves like a division. We just want to think about the big picture of how the rectangle changes.
Now, since $$a$$$is just a rename of $$f$$$, we can swap in $$da = df$$$. But how do we swap out $$b$$$? Well, we have:
$$b = \frac{1}{g}$$ $$\frac{db}{dg} = -\frac{1}{g^2}$$ $$db = -\frac{1}{g^2} dg$$
Ah! This is our cake cutting. As $$g$$$grows, we lose $$db = -\frac{1}{g^2} \dot dg$$$ from the $$b$$$side. The total impact is: $$(a \cdot b)' = (da \cdot b) + (db \cdot a) = \left(df \cdot \frac{1}{g} \right) + \left( \frac{-1}{g^2} dg \cdot f \right)$$ This formula started with the product rule, and we plugged in their real values. Might as well put $$f$$$ and $$g$$$back into $$(a \cdot b)'$$$, to get the Quotient Rule (aka the Division Rule):
$$\left( \frac{f}{g} \right)' = \left(df \cdot \frac{1}{g} \right) + \left(\frac{-1}{g^2} dg \cdot f \right)$$
Many textbooks re-arrange this relationship, like so:
$$\left(df \cdot \frac{1}{g}\right) + \left(\frac{-1}{g^2} dg \cdot f\right) = \frac{g\cdot df}{g^2} - \frac{f \cdot dg}{g^2} = \frac{g\cdot df - f \cdot dg}{g^2}$$
And I don't like it, no ma'am, not one bit! This version no longer resembles its ancestor, the product rule.
In practice, the Quotient Rule is a torture device designed to test your memorization skills; I rarely remember it. Just think of $$\frac{f}{g}$$$as $$f \cdot \frac{1}{g}$$$, and use the product rule like we've done.
Questions
Let's do a few warm-ups to test our skills. Can you solve these bad boys?
$$\frac{d}{dx} x^4 = \ ?$$
$$\frac{d}{dx} 3x^5 = \ ?$$
(You can check your answers with Wolfram Alpha, such as d/dx x^4.)
Again, don't get lost in the symbols. Think "I have $$x^4$$$-- what pattern of changes will I see as I make $$x$$$ larger?".
Ok! How about working backwards, and doing some integrals?
$$\int 2x^2 = \ ?$$
$$\int x^3 = \ ?$$
Ask yourself, "What original pattern would create steps in the pattern $$2x^2$$$?" Trial-and-error is ok! Try a formula, test it, and adjust it. Personally, I like to move aside the 2 and just worry about the integral of $$x^2$$$:
$$\int 2x^2 = 2 \int x^2 = \ ?$$
How do you know if you're right? Take the derivative -- you are the antiques dealer! I brought you a pattern of shards ($$2x^2$$$) and you need to tell me the vase they came from. Once you have your guess, just break it in the back room, and make sure you get $$2x^2$$$ back out. Then you'll be confident in your answer.
We're getting ready to work through the circle equations ourselves, and recreate results found by Archimedes, likely the greatest mathematician of all time.
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Class Discussion
Guided class discussions are available in the complete edition. |
# Selina Solutions Concise Mathematics Class 6 Chapter 9: Playing With Numbers Exercise 9(B)
Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers Exercise 9(B) provides a basic method in solving difficult questions with ease that are important for the exam. Students, who want to improve their problem-solving skills, are advised to practice Selina Solutions on a daily basis. The solutions are created by BYJU’S experts in order to help students to overcome exam fear. Those who aspire to score high marks and achieve success in the examination are suggested to refer to the solutions while revising the textbook questions. Students can download Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers Exercise 9(B), from the provided links below
## Selina Solutions Concise Mathematics Class 6 Chapter 9: Playing With Numbers Exercise 9(B) Download PDF
### Access other exercises of Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers
Exercise 9(A) Solutions
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### Access Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers Exercise 9(B)
#### Exercise 9(B)
1. Fill in the blanks:
(i) On dividing 9 by 7, quotient = …………….. and remainder = …………….
(ii) On dividing 18 by 6, quotient = ………….. and remainder = …………
(iii) Factor of a number is ………….. of ………..
(iv) Every number is a factor of ………….
(v) Every number is a multiple of …………
(vi) ………. is factor of every number.
(vii) For every number, its factors are ………. and its multiples are ……….
(viii) x is a factor of y, then y is a …….. of x.
Solution:
(i) On dividing 9 by 7, quotient = 1 and remainder = 2
(ii) On dividing 18 by 6, quotient = 3 and remainder = 0
(iii) Factor of a number is an exact division of the number
(iv) Every number is a factor of itself
(v) Every number is a multiple of itself
(vi) One is factor of every number
(vii) For every number, its factors are finite and its multiples are infinite
(viii) x is a factor of y, then y is a multiple of x
2. Write all the factors of:
(i) 16
(ii) 21
(iii) 39
(iv) 48
(v) 64
(vi) 98
Solution:
(i) 16
Solution:
All factors of 16 are:
1, 2, 4, 8, 16
(ii) 21
Solution:
All factors of 21 are:
1, 3, 7, 21
(iii) 39
Solution:
All factors of 39 are:
1, 3, 13, 39
(iv) 48
Solution:
All factors of 48 are:
1, 2, 3, 4,6, 8, 12, 16, 24, 48
(v) 64
Solution:
All factors of 64 are:
1, 2, 4, 8, 16, 32, 64
(vi) 98
Solution:
All factors of 98 are:
1, 2, 7, 14, 49, 98
3. Write the first six multiples of:
(i) 4
(ii) 9
(iii) 11
(iv) 15
(v) 18
(vi) 16
Solution:
(i) 4
Following are the first six multiples of 4
1 × 4 = 4
2 × 4 = 8
3 × 4 = 12
4 × 4 = 16
5 × 4 = 20
6 × 4 = 24
Hence, multiples of 4 are 4, 8, 12, 16, 20 and 24
(ii) 9
Following are the first six multiples of 9
1 × 9 = 9
2 × 9 = 18
3 × 9 = 27
4 × 9 = 36
5 × 9 = 45
6 × 9 = 54
Hence, multiples of 9 are 9, 18, 27, 36, 45 and 54
(iii) 11
Following are the first six multiples of 11
1 × 11 = 11
2 × 11 = 22
3 × 11 = 33
4 × 11 = 44
5 × 11 = 55
6 × 11 = 66
Hence, multiples of 11 are 11, 22, 33, 44, 55 and 66
(iv) 15
Following are the first six multiples of 15
1 × 15 = 15
2 × 15 = 30
3 × 15 = 45
4 × 15 = 60
5 × 15 = 75
6 × 15 = 90
Hence, multiples of 15 are 15, 30, 45, 60, 75 and 90
(v) 18
Following are the first six multiples of 18
1 × 18 = 18
2 × 18 = 36
3 × 18 = 54
4 × 18 = 72
5 × 18 = 90
6 × 18 = 108
Hence, multiples of 18 are 18, 36, 54, 72, 90 and 108
(vi) 16
Following are the first six multiples of 16
1 × 16 = 16
2 × 16 = 32
3 × 16 = 48
4 × 16 = 64
5 × 16 = 80
6 × 16 = 96
Hence, multiples of 16 are 16, 32, 48, 64, 80 and 96
4. The product of two numbers is 36 and their sum is 13. Find the numbers.
Solution:
36 can be written as
1 × 36 = 36
2 × 18 = 36
3 × 12 = 36
4 × 9 = 36
6 × 6 = 36
Here, the sum of 4 and 9 is 13
Hence, 4 and 9 are the two numbers
5. The product of two numbers is 48 and their sum is 16. Find the numbers.
Solution:
48 can be written as
1 × 48 = 48
2 × 24 = 48
3 × 16 = 48
4 × 12 = 48
6 × 8 = 48
Here, the sum of 4 and 12 is 16
Hence, 4 and 12 are the two numbers
6. Write two numbers which differ by 3 and whose product is 54.
Solution:
54 can be written as
1 × 54 = 54
2 × 27 = 54
3 × 18 = 54
6 × 9 = 54
Here, the difference between 6 and 9 is 3
Hence, 6 and 9 are the two numbers
7. Without making any actual division show that 7007 is divisible by 7.
Solution:
Given
7007
This can be written as
= 7000 + 7
= 7 × (1000 + 1)
= 7 × 1001
Clearly, 7007 is divisible by 7
8. Without making any actual division show that 2300023 is divisible by 23
Solution:
Given
2300023
This can be written as
= 2300000 + 23
= 23 × (100000 + 1)
= 23 × 100001
Clearly, 2300023 is divisible by 23
9. Without making any actual division, show that each of the following numbers is divisible by 11
(i) 11011
(ii) 110011
(iii) 11000011
Solution:
(i) 11011
This can be written as
= 11000 + 11
= 11 × (1000 + 1)
= 11 × 1001
Clearly, 11011 is divisible by 11
(ii) 110011
This can be written as
= 110000 + 11
= 11 × (10000 + 1)
= 11 × 10001
Clearly, 110011 is divisible by 11
(iii) 11000011
This can be written as
= 11000000 + 11
= 11 × (1000000 + 1)
= 11 × 1000001
Clearly, 11000011 is divisible by 11
10. Without actual division, show that each of the following numbers is divisible by 8
(i) 1608
(ii) 56008
(iii) 240008
Solution:
(i) 1608
This can be written as
= 1600 + 8
= 8 × (200 + 1)
= 8 × 201
Clearly, 1608 is divisible by 8
(ii) 56008
This can be written as
= 56000 + 8
= 8 × (7000 + 1)
= 8 × 7001
Clearly, 56008 is divisible by 8
(iii) 240008
This can be written as
= 240000 + 8
= 8 × (30000 + 1)
= 8 × 30001
Clearly, 240008 is divisible by 8 |
# Reference Angle
Related Topics:
More Lessons for Trigonometry, Math Worksheets
In this lesson, we will look into how to use the reference angle to find the sine, cosine and tangent of non-acute angles.
What is a reference angle?
A reference angle is the acute angle formed by the terminal side of the given angle and the x-axis.
How to find the reference angle?
Step 1: Sketch the given angle
Step 2: Drop a perpendicular to the x-axis
Step 3: Determine the angle measure of the triangle formed
In this lesson, we learn to use reference angles to find the sine, cosine and tangent of non-acute angles
To find the value of sine, cosine and tangent at non-acute angles (from 90 to 360), first draw the angle on the unit circle and find the reference angle. A reference angle is formed by the terminal side and the x-axis and will therefore always be acute. When evaluating sine, cosine and tangent for the reference angle, determine if each value is positive or negative by identifying the quadrant the terminal side is in.
The Cartesian plane is divided into 4 quadrants by the two coordinate axes. These 4 quadrants are labeled I, II, III and IV respectively.
We want to consider how to evaluate the trigonometric ratios of angles in the four quadrants. When evaluating the trigonometric ratios of non-acute angles, we need to consider the concept of reference angles.
The table below shows the reference angle, α , in quadrant I which corresponds to the angle, θ , in quadrants II, III, and IV.
Quadrant Angle θ Reference angle α Diagram II 90˚ < θ < 180 ˚ α = 180 ˚ – θ III 180˚ < θ < 270 ˚ α = θ – 180˚ IV 270˚ < θ < 360 ˚ α = 360 ˚ – θ
Example
Determine the reference angle that corresponds to each of the following angle.
a) 165˚
b) 249˚
c) 328˚
Solution:
a) 165˚ is in quadrant II (90˚ < 165˚ < 180˚ )
The reference angle is 180˚ – 165˚ = 15˚
b) 249˚ is in quadrant III (180˚ < 249˚ < 270˚ )
The reference angle is 249˚ – 180˚ = 69˚
c) 328˚ is in quadrant III (270˚ < 328˚ < 360˚ )
The reference angle is 360˚ – 328˚ = 32˚
A discussion of what reference angles are and how to find them, and then how to use them to determine the sine and cosine values of angles greater than ninety degrees.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
##### Math For Real Life For Dummies
Using real-life math can simplify everyday situations. Math comes in handy every time you take a trip, go shopping, or do projects around the house.
## Estimating taxes and discounts when shopping
When you go shopping, you often encounter discounts and sales offers that change the price of an item. In addition, many states in the United States charge sales tax, and if you’re shopping abroad, you may face a value added tax.
To determine how much the total will be (and whether it fits into your budget or whether you have enough cash on hand to make the purchase), you need to be able to calculate the discounted price as well as determine the sales and any other taxes.
• Estimating sales tax: In the U.S., state sales taxes often end up being over 8 percent when you add in applicable local taxes. Therefore, a safe estimate is to allow for a 10 percent sales tax. A quick way to figure 10 percent of an item’s purchase price is to move the decimal point one place to the left. For example, if you buy a \$70 item at the hardware store, the tax will be approximately \$7, making the purchase price in the neighborhood of \$77.
• Estimating VAT: Many European Union and non-EU countries have a value added tax (VAT) that hovers around 20 percent. To estimate VAT, follow the same approach as you would to calculate a sales tax: Determine 10 percent by moving the decimal one place to the left. Then, double that amount. Using the \$70 purchase at the hardware store example, the VAT will be approximately \$14.
• Calculating a discount: Discounts can be for any amount: 5 percent, 10 percent, 30 percent, and so on. To do the calculations in your head, first determine what a 10 percent discount is (because that’s the easiest one to calculate) and then do a little multiplication or division to get closer to the actual discount amount.
For example, if an item is being offered at a 40 percent discount, figure 10 percent and then multiply that amount by 4. A 40 percent discount on an \$80 dress, for example, works out to be \$32 off (10 percent of \$80 is \$8, and \$8 multiplied by 4 is \$32). If the discount is 15 percent, figure the 10 percent discount, divide that number in half to determine what a 5 percent discount, and then add the two together. For the \$80 dress, you’d end up with a \$12 discount: \$8 (10 percent) + \$4 (5 percent) = \$12 (15 percent).
## Calculating area and volume for household projects
Many household projects require that you be able to calculate area or volume. Whether you’re painting your kid’s room, mulching your flowerbed for the winter, planting grass seed in the spring, or tackling any other project for which you need to find out how much of something you need to cover, knowing how to figure area and volume is a time- and money-saver.
## Calculating area
To calculate area, you simply use this formula:
area = length × width
a = lw
From there, you can get the other information you need:
• Figuring how much seed you need: Use the area formula to calculate the area you want to cover and then calculate the number of pounds of seed you need. Seed companies indicate how much coverage a pound of seed gives you. Simply divide the area you need to cover by the amount of area that 1 pound of seed covers. For example, if you want to cover an area of 1,625 square feet, and 1 pound of seed covers 600 square feet, you need about 3 pounds of seed (1,625 divided by 600 is 2.7).
• Calculating area to be painted: You can calculate the area of each wall and then add the values together, or you can determine the perimeter of the room (add the length of each wall) and multiply that value by the room’s height. A room that has a 42-foot perimeter and 8-foot ceilings has an area of 336 feet.
• Calculating how many square yards of carpeting you need: First calculate the area of the room and then convert that number from square feet to square yards by dividing the area by 9 (a square yard is 3 feet wide and 3 feet long).
## Calculating volume
When the area you want to cover has depth as well as height and width, you need to determine volume rather than area. Mulching a flowerbed is one such project: To protect your flowers, you need to spread a few inches of mulch over the whole bed.
To calculate volume, you use this formula:
volume = length × width × height
v = lwh
With that info, you can then do additional calculations to figure out how many bags of mulch to buy.
Say, for example, that you want to cover 6.68 cubic feet with mulch (that’s the volume you need for a small bed, mulching to a depth of 3 inches). Simply divide the volume you need by the number of cubic feet in a bag of mulch. If each bag holds 3 cubic feet, you need 2.23 bags. So buy 3 to be safe.
## Using your body to estimate length
Many real-life math problems don’t require exact solutions or precise calculations. When you need an approximate idea of length, you have a handy-dandy “ruler” always nearby: your body. Yes, your body — and its various parts — can serve as a quick length estimator.
• Your hand: The width of your palm is about 4 inches (10 centimeters).
• Your foot: An average foot is about 12 inches long.
• Your forearm from elbow to fingertip: Your forearm is about 1.5 feet (46 centimeters).
• From nose to finger tip: The distance between your nose and your fingertip when your arm is extended is about a yard (36 inches). If you haven’t used this method yourself, ask your mother or grandmother: It’s the traditional way of measuring cloth.
• The distance of one full stride: Called a pace, it’s about 58 inches, just short of 6 feet. The Romans used this measurement, and so can you!
These estimates are just that: estimates. A more accurate way use your body parts to estimate length is to know your own measurements: the width of your own palm, the length of your own foot, and so on. Then, wherever you are, you’ll have a pretty accurate ruler very close by.
## Calculating speed, time, and distance
If you drive a car or have ever flown in an airplane, you’ve probably noticed that time, speed, and distance are related. Here’s the basic formula for distance (d), which equals speed (called velocity in science and represented by v) multiplied by time (t):
From this simple formula, you can derive these other formulas as well:
By knowing any two of the components, you can use these formulas to figure out the third. If you know how far you’ve traveled and the time the journey has taken, you can calculate your average speed. If you know the distance and the average speed, you can calculate the time you’ve been driving.
This “know two to get all three” trick applies to many day-to-day math activities: buying lumber (length needed/price per board foot/total cost), buying cases of motor oil (price per can/number of cans in a case/total cost), or buying meat at the grocery store (weight of cut/price per pound/total cost).
## Calculating GPA
Nowadays, grade point averages are a big deal, especially for college-bound students and their anxious parents. GPA is easy to calculate if you understand that it’s simply an average.
To determine an average, you add up all of the values in the group and then divide that total by the number of values in the group.
The tricky part about GPAs, however, is that you’re working with letter grades rather than numbers. For that reason, before you can calculate a GPA, you have to do a little conversion work first.
Follow these steps to calculate a GPA:
1. Convert the grades to numbers, using this scale:
Letter Grade Number Value
A 4
B 3
C 2
D 1
F 0
Say you’re taking four courses and your grades are A, A, C, and D. The numeric equivalents are 4, 4, 2, and 1.
2. Add the numbers up.
When you’re calculating your GPA for a semester or quarter, the task is pretty easy because people don’t usually take a lot of courses. In the example, the total is 11. The task gets a bit more challenging when you’re calculating GPA for a larger number of classes.
3. Divide by the number of courses.
In this example, divide by 4 (the number of courses). The result is a number greater than 2 and less than 3. (Actually, it’s 2 with a remainder of 3, which works out to be 0.75.) Your GPA is 2.75, or about a C+. |
# Lesson 6
Compare Reptile Lengths in Story Problems
## Warm-up: Number Talk: Fives and Tens (10 minutes)
### Narrative
The purpose of this warm-up is to elicit the strategies and understanding students have for composing a ten when adding within 100. This Number Talk focuses on adding fives to compose a ten mentally. Each successive expression is ten more than the previous. When students notice and express the regularity in this pattern, they use the structure of base-ten numbers and the properties of operations (MP7, MP8). These understandings help students develop fluency with addition within 100.
### Launch
• Display one expression.
• “Give me a signal when you have an answer and can explain how you got it.”
• 1 minute: quiet think time
### Activity
• Keep expressions and work displayed.
• Repeat with each expression.
### Student Facing
Find the value of each expression mentally.
• $$5 + 5$$
• $$15 + 5$$
• $$15 + 15$$
• $$15 + 25$$
### Activity Synthesis
• “How does $$15 + 5$$ compare to $$5 + 5$$?” (It’s 10 more)
• “How does $$15 + 15$$ compare to $$15 + 5$$?” (It’s 10 more)
• “How does $$15 + 25$$ compare to $$15 + 15$$?” (I see $$10 + 20 = 30$$ and the $$5 + 5$$ is 10 more.)
• Highlight how to see the 10 more structure in these sums: the tens place of one number grows by one each time.
## Activity 1: Whose Pet is Longer? (20 minutes)
### Narrative
The purpose of this activity is for students to interpret and solve Compare problems involving length where the language suggests an incorrect operation. For example, the first problem uses the word “shorter” which usually suggests subtraction. However, in this problem students are looking for an unknown that is the greater length and must add the two known values.
The Three Reads routine is used to help student practice making sense of the problem before solving. Students begin the activity by looking at the first problem displayed, rather than in their books. At the end of the launch, students open their books and work to find the diagram that matches the story problem. This further helps them to visualize the quantities in the problem before they work to find a solution (MP1).
After reading the other story problems, students consider which pet is longer or shorter and choose tape diagrams to match the lengths in the problem (MP2). Students solve each story problem independently and compare their solutions.
Action and Expression: Develop Expression and Communication. Provide students with alternatives to writing on paper. This unit has involved a lot of paper and pencil work, so there is an opportunity for students to share their learning using white boards, chart or poster paper, and markers.
Supports accessibility for: Organization, Attention
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• “Lin and Jada both have pet lizards. They are comparing the lengths of their pets.”
• Display only the problem stem for the first problem, without revealing the question.
• “We are going to read this problem 3 times.”
• 1st Read: “Lin's pet lizard is 62 cm long. It is 19 cm shorter than Jada's.”
• “What is this story about?”
• 1 minute: partner discussion.
• Listen for and clarify any questions about the context.
• 2nd Read: “Lin's pet lizard is 62 cm long. It is 19 cm shorter than Jada's.”
• “Which measurements are important to pay attention to in the story?” (the length of Lin’s lizard, the length of Jada’s lizard, the difference between the lengths of the two lizards).
• 30 seconds: quiet think time
• 2 minutes: partner discussion
• Share and record responses.
• Reveal the question.
• “Lin's pet lizard is 62 cm long. It is 19 cm shorter than Jada's. How long is Jada's pet lizard?”
• “What are different ways we could represent this problem?” (tape diagram, equations, base ten blocks)
• 30 seconds: quiet think time
• 1–2 minutes: partner discussion
### Activity
• “When you have both selected a match, compare your choices and explain why the diagram matches the story.”
• “Then solve on your own.”
• 10 minutes: partner work time
### Student Facing
1. Lin's pet lizard is 62 cm long. It is 19 cm shorter than Jada's. How long is Jada's pet lizard?
1. Whose pet is longer? ________________________________
2. Circle the diagram that matches the story.
Jada’s pet lizard is ____________ cm long.
2. Diego and Mai have pet snakes. Mai’s snake is 17 cm longer than Diego’s. Mai’s snake is 71 cm. How long is Diego’s pet snake?
1. Whose pet is shorter? ________________________________
2. Circle the diagram that matches the story.
Diego’s pet snake is ____________ cm long.
### Activity Synthesis
• Invite students to share the correct diagram for the second problem.
• Display the diagram.
• “How does the diagram match the story problem?” (You can see that Mai’s rectangle is longer and there is a question mark for Diego’s pet.)
• “How did you know which pet was shorter?” (It said that Mai’s snake was 17 cm longer, so Diego’s is shorter.)
## Activity 2: Guess My Reptiles (15 minutes)
### Narrative
The purpose of this activity is for students to make sense of and solve Compare story problems involving length. Students use measurements provided for reptiles and create their own Compare problems to solve with a partner. Students must solve a Compare, Difference Unknown problem when creating their mystery problem. When they solve their partner's mystery problem, they must solve a Compare, Bigger Unknown or Compare, Smaller Unknown problem. Encourage students to use diagrams or other drawings to show how they know which reptiles their partner picked. Some students may choose to use equations to represent the lengths.
MLR8 Discussion Supports. To support partner discussion about the comparisons they made, display the following sentence frames: “I compared ____ and ____ because . . .”, “Our comparisons are the same because . . .”, or “Our comparisons are different because . . .” Encourage students to challenge each other when they disagree.
### Required Materials
Materials to Gather
### Launch
• Groups of 2
• Assign students as Partner A and Partner B.
### Activity
• “We are going to play a guessing game. You will choose one reptile from your list and one reptile from your partner’s list. Keep both reptiles a secret.”
• “Work independently to pick your reptiles and complete the sentences.”
• As needed, demonstrate as if you were Partner A with the class.
• 4 minutes: independent work time
• “Now, share your sentences with your partner. Then find which reptiles they picked. Draw a diagram or use equations to prove you are correct.”
• 6 minutes: partner work time
• Monitor for students who use tape diagrams to represent their partner’s reptiles.
### Student Facing
Partner A's reptiles
Partner B's reptiles
1. day gecko, 28 cm
1. ribbon snake, 83 cm
2. komodo dragon, 98 cm
2. gila monster, 55 cm
3. baby cobra, 46 cm
3. baby alligator, 71 cm
4. iguana, 65 cm
4. ringneck snake, 38 cm
1. Choose one reptile from your list and one reptile from your partner’s list.
2. Fill in the blanks to create a story problem using the lengths of the reptiles you picked. Then share your sentences with your partner.
My reptile is ____________ cm long.
It is ____________ cm ________________________________(shorter/longer) than one of your reptiles.
### Student Response
If students find the correct difference between the reptiles they pick, but create a word problem that cannot be solved, consider asking:
• “How could you draw a diagram to represent the reptiles you picked?”
• “Does your diagram match the story problem you created? Why or why not?”
• “What word or numbers could you change to match the reptiles you chose?”
### Activity Synthesis
• Invite previously identified students to share how they determined their partner’s reptiles.
• “Which reptile was shorter? How does your diagram show this?”
## Lesson Synthesis
### Lesson Synthesis
“How did the diagram help you think about which animal was longer?” (Once you label the rectangles, you can tell which one is longer because it had the longer rectangle.)
“How did the diagram help you decide if you would add or subtract?” (After seeing which animal had the longer rectangle, it was easy to see which length was longer. I could see if I needed to add to find the longer length or subtract to find the difference or the shorter length.) |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 1.3: Permutations
Difficulty Level: At Grade Created by: Bruce DeWItt
### Learning Objectives
• Know the definition of a permutation
• Be able to calculate the number of permutations using the permutations formula and with technology
• Understand the connection between the Fundamental Counting Principle and permutations
The Fundamental Counting Principle provides us with a tool that allows us to calculate the number of outcomes possible in many situations. What if the situation is a bit more complex? For many situations, the order that we complete a task does not matter. Ordering milk, bacon, and scrambled eggs in that order is the same as ordering bacon, scrambled eggs, and milk. In this case the order that we make our choices wouldn't matter, but there are many situations in which the order that we do things does make a difference.
A permutation is a specific order or arrangement of a set of objects or items. What if you wish to call someone on the phone? If I make the call, the order that I punch in the numbers matters so this is an example of a permutation. A good question to ask when deciding if your arrangement is a permutation is "DOES ORDER MATTER?" If yes, then you are dealing with a permutation. For example, if you ordered an ice cream sundae and they put the cherry in first, then the chocolate sauce, and then the ice cream, you would probably would not be happy with that particular ice cream sundae. You would likely prefer that they put the ice cream in first, then the chocolate sauce, and then put the cherry on top. Clearly each sundae had the same three ingredients, but they were quite different from one another. Each order that we can make the ice cream sundae is called a permutation.
There is a simple formula for figuring out how many permutations exist when 'r' objects are selected from a set of 'n' objects. The left side of the equation can be read "n P r", just as it looks or "n Permutations of size r".
Recall that the exclamation point is a factorial. For example, 5!=5×4×3×2×1\begin{align*}5!=5\times4\times3\times2\times1\end{align*}. Also, be sure to find the permutations command on your calculator.
In our ice cream sundae discussion, 'n' would be 3 because there are 3 items to select from and 'r' would also be 3 because we are going to select all three items. Using the permutations formula, this would be 3P3=3!(33)!=3!0!=61=6\begin{align*}_3P_3=\frac{3!}{(3-3)!}=\frac{3!}{0!}=\frac{6}{1}=6 \end{align*}. In other words, there are 6 different orders that the ice cream sundae could be made. Note that 0! is equal to 1.
#### Example 1
Suppose you are going to order an ice cream cone with two different flavored scoops. You are going to take a picture of your ice cream cone for use in the school newspaper. The ice cream shop has 5 flavors to choose from; chocolate, vanilla, orange, strawberry, and mint. How many different ice cream cone photos are possible?
#### Solution
The first question to ask is "Does Order Matter?". If it does, then we are dealing with a permutation question. In this case, the order does make a difference. A chocolate on top of vanilla cone looks different than a vanilla on top of chocolate cone. We have five flavors to pick from, so n=5. We are going to select 2 flavors so r=2. 5P2=5!(52)!=5!3!=1206=20\begin{align*}_5P_2=\frac{5!}{(5-2)!}=\frac{5!}{3!}=\frac{120}{6}=20\end{align*} There 20 different permutations of ice cream cones we could order. The notation representing this situation, 5P2\begin{align*}_5P_2\end{align*}, can be read as "Five 'P' Two" or "Five permutations of size Two". Be sure to perform this calculation using your calculator as well.
In the example above, you could have also found your answer using the Fundamental Counting Principle. There were 5 choices for the 1st flavor and then only 4 choices for the 2nd flavor. There are 5×4=20\begin{align*}5\times4=20\end{align*} ice cream cones possible.
#### Example 2
Give the value of 6P3\begin{align*}_6 P _3\end{align*} by using the formula for permutations. Verify your solution on your calculator.
#### Solution
6P3=6!(63)!=6!3!=7206=120\begin{align*}_6 P _3=\frac{6!}{(6-3)!}=\frac{6!}{3!}=\frac{720}{6}=120\end{align*}
#### Example 3
Decide whether each of the situations below involves permutations.
a) A five-card poker hand is dealt from a deck of cards.
b) A cashier must give 3 pennies, 2 dimes, a 5 dollar bill, and a 10 dollar bill back as change for a purchase.
c) A student is going to open a padlock that has a three number combination.
d) A child has red, blue, green, yellow, and orange color crayons and will be coloring a rainbow using each color one time.
#### Solution
a) The order you get your five cards for a poker hand does not matter. If one of your cards was the ace of spades, it didn't matter if it was the first card or the last card dealt.
b) The order that the cashier gives you $15.23 in change does not matter as long as the total is$15.23.
c) The order you put in the three numbers for the combination makes a difference. If the correct combination is 12-27-19, the padlock will not open if you enter 19-12-27 even though the same three numbers are used.
d) The order that the child colors the rainbow does make a difference. The color pattern red, blue, green, orange, yellow will look different than green, blue, red, yellow, orange.
### Problem Set 1.3
#### Exercises
1) Use the formula for Permutations, nPr=n!(nr)!\begin{align*}_n P _r=\frac{n!}{(n-r)!}\end{align*} to find the value for each expression. Confirm each result by using your calculator.
a) 8P3\begin{align*}_8 P _3\end{align*}
b) 4P4\begin{align*}_4 P _4\end{align*}
c) 5P3\begin{align*}_5 P _3\end{align*}
d) 5P0\begin{align*}_5 P _0\end{align*}
2) How many 4 letter permutations can be formed from the letters in word rhombus?
3) For a board of directors composed of eight people, in how many ways can a president, vice president, and treasurer be selected?
4) How many different ID cards can be made if there are six digits on a card and no digit can be used more than once?
5) In how many ways can seven different brands of laundry soap be displayed on a shelf in a store?
6) A child has four different stickers that can be placed on a model car in a vertical stack. In how many ways can this be done if each sticker is to be used only one time?
7) An inspector must select three tests to perform in a certain order on a manufactured part. He has a choice of seven tests. How many different ways can he perform three tests?
8) In how many different ways can 4 raffle tickets be selected from 50 tickets if each of the 4 ticket holders wins a different prize?
9) A researcher has 5 different antibiotics to test on 5 different rats. Each rat will receive exactly one antibiotic and no rat will receive the same antibiotic as any other rat. In how many different ways can the researcher administer the antibiotics?
10) There are five violinists in an orchestra. Three of them will be selected to play in a trio with a different part for each musician. In how many ways can the trio be selected?
11) There are five violinists in an orchestra. Four of them will be selected to play in a quartet with a different part for each musician. In how many ways can the quartet be selected?
12) There are five violinists in an orchestra. All five of them will be selected to play in a quintet with a different part for each musician. In how many ways can the quintet be selected?
13) There are five violinists in an orchestra. A piece of music is written so that it can be played with either 3, 4, or 5 violinists. Each musician selected to play this piece will play a different part. In how many ways can a group of at least three musicians be selected? Hint: Use your answers from problems 10), 11) and 12).
14) Decide whether each situation below involves permutations. Briefly explain your answers.
a) Sophia picks three color crayons from a box of 12 crayons to make a picture for her cat, Butterscotch.
b) A five-digit code is needed to open up an electronic lock on a car.
c) Twenty race car drivers must each complete three laps at a race track during a time trial, one after another, in order to establish the order in which the cars will start a race the next day.
d) There are seven steps that a student must follow when preparing cookies during their Family and Consumer Sciences course.
#### Review Exercises
15) Use the Fundamental Counting Principle to determine the number of different ways a person could order a meal if they are to pick one entree from four choices, one side order from three choices, and one drink from four choices.
16) A student wishes to check out three books from the library. She will check out one historical fiction book, one biography, and one book on art history. Build a tree diagram to show how many ways can this be done if there are two historical fiction books, three biographies, and two books on art history that she is considering checking out.
17) How many different outcomes are possible for the total on a roll of two dice if one die has 6 sides and one die has 4 sides?
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## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 2 - Section 2.1 - Linear Equations in One Variable - Exercise Set - Page 56: 76
#### Answer
$x=\frac{21}{4}$
#### Work Step by Step
We are given the equation $\frac{x}{3}+7=\frac{5x}{3}$. Then equation is then incorrectly simplified to $x+7=5x$. In order to eliminate the fractions, we must multiply all terms by 3, not just the terms with fractions. $x+21=5x$ In order to simplify the equation from here, we can subtract x from both sides. $21=4x$ Divide both sides by 4. $x=\frac{21}{4}$ Note: The equation is also incorrectly simplified after $7=4x$. Dividing both sides of this equation would result in $x=\frac{7}{4}$.
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# 1950 AHSME Problems/Problem 34
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
When the circumference of a toy balloon is increased from $20$ inches to $25$ inches, the radius is increased by:
$\textbf{(A)}\ 5\text{ in} \qquad \textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad \textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad \textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}$
## Solutions
### Solution 1
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be $\pi$ anymore) We see that the circumference was increased by $25\%$. This means the radius was also increased by $25\%$. The radius of the original balloon is $\frac{20}{2\pi}=\frac{10}{\pi}$. With the $25\%$ increase, it becomes $\frac{12.5}{\pi}$. The increase is $\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.
### Solution 2
The radii of the circles are $\frac{20}{2\pi}$ and $\frac{25}{2\pi}$, respectively. The positive difference is therefore $\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$.
### Solution 3
Let the radius of the circle with the larger circumference be $r_2$ and the circle with the smaller circumference be $r_1$. Calculating the ratio of the two $$\frac{r_2}{r_1}=\frac{25}{20}=\frac{5}{4}$$ $$4r_2=5r_1$$ $$4(r_2-r_1)=r_1$$ $$r_2-r_1=\frac{r_1}{4}=\frac{\frac{20}{2\pi}}{4}=\frac{10}{4\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}$$ |
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# Figures on the Same Base and between the Same Parallels
• Last Updated : 24 Jun, 2022
A triangle is a three-sided polygon and a parallelogram is a four-sided polygon or simply a quadrilateral that has parallel opposite sides. We encounter these two polynomials almost everywhere in our everyday lives. For example: Let’s say a farmer has a piece of land that is in the shape of a parallelogram. He wants to divide this land into two parts for his daughters. Now when divided into two parts, it will generate two triangles. So, in this case, it becomes essential for us to know their areas. Simple formulas to calculate the areas of triangles and parallelograms can be sometimes cumbersome. So, we use some theorems and properties to simplify our calculations in such scenarios. Let’s look at these properties in detail.
### Parallelograms and Triangles
Triangles have three sides and parallelograms are the quadrilaterals that have four sides while the opposite sides are parallel to each other. The figure below shows a parallelogram and a triangle. Let’s say “b2” and “h2″ is the length of the base and height of the triangle respectively and “b1″ and “h1” are the base and height of the parallelogram.
Area of triangle =
Area of parallelogram =
### Congruent Figures and Their Areas
We know that two figures are called congruent if they have the same size and shape. So if two figures are congruent, you can superimpose them over each other and both will completely cover the other figure. That means that if any two figures are congruent, their areas must be equal. But notice that the converse of this statement is not true, if two figures have equal areas, it is not necessary that both are congruent. In the figure below, we can see two pairs of figures which are congruent and have the same areas, but the other pair of figures have the same areas, but they are not congruent.
Two Congruent figures with the same area
In the above figure, let’s compare the areas of two quadrilaterals.
Ar(PQRS) = 9 × 4 = 36
Ar(TUVW) = 62 = 36
Notice that both have the same area, but they are not congruent. This verifies that the converse of our statement that we gave earlier is not true. So, now the definition of the area can be summarized as follows,
Area of a figure is the measure of the plane that is enclosed by that figure. It has following two properties:
1. Let’s say we have two figures X and Y. If X and Y are congruent figures, then ar(X) = ar(Y).
2. If both figure combine without overlapping to make another figure T. Area of figure T will be given by, ar(T) = ar(X) + ar(Y).
### Parallelograms on the same Base and Between the Same Parallels
Our goal is to learn about the relation between the areas of two parallelograms when they have the same base and are between the same parallels. The figure below shows two parallelograms with a common base and between the same parallel lines. Let’s prove the relation between these areas with some theorems.
### Theorem: Parallelograms with the same base and between the same parallels have the same area.
Proof:
Let’s assume we have two parallelograms PQRS and RSTU as shown in the figure below. Both have the same base RS and are between same parallels. The objective is to prove that ar(PQRS) = ar(RSTU).
In the figure, RSTQ is common to both the parallelograms. Now, if we can prove that ar(PST) = ar(QRU). We can prove that areas of both parallelograms are equal.
Let’s look at the triangle PST and QRU.
∠SPT = ∠RQU (Corresponding angles)
∠PTS = ∠QUR (Corresponding angles)
Now since two angles of triangles are equal, the third angle will also be equal due to angle sum property.
∠PST = ∠QRU
Now both of these triangles are congruent
ΔPST≅ ΔQRU
Thus, ar(PST) = ar(QRU)
Now we know that,
Ar(PQRS) = ar(RSTQ) + ar(PST)
Ar(RSTU) = ar(RSTQ) + ar(QRU)
Since ar(RSTQ) is common and ar(PST) = ar(QRU).
Thus, ar(PQRS) = ar(RSTU)
### Triangles on the same Base and Between the Same Parallels
The figure below represents two triangles that are on the same base and are between the same parallels. Our goal is to find the relation between the areas of these two triangles. Let’s see theorems related to
### Theorem: Two triangles on the same base and between the same parallels have equal area.
Proof:
We know that area of triangle is given by
So, two triangles will have same area if they have same base and height.
Our triangles have a common base. Now since they are between two parallels, they must have the same height. Thus, both the triangles have same area.
### Sample Questions
Question 1: Find the area of the triangle and parallelogram given in the figures below.
Solution:
We know,
Area of triangle =
Area of parallelogram = b2 × h2
= 3 × 5
= 15
Question 2: State three properties of a parallelogram.
Three properties of parallelogram:
1. Opposite sides are parallel and equal.
2. Opposite angles are equal.
3. Adjacent Angles sum up to 180°.
Question 3: In the triangle ΔPQR given in the figure below, PS is the median. Prove that ar(PSR) = ar(PQS).
Now if we draw a perpendicular from the vertex P to the base QR. We’ll see that this perpendicular is common to both the triangle. Thus, both of them have same base and same height. So, they must have the same area.
Question 4: In the figure given below, we have a rectangle RSTU and a parallelogram PQRS. It is given that, PL is perpendicular to RS. Now, prove:
1. Ar(RSTU) = ar(PQRS)
2. Ar(PQRS) = RS x AL.
Solution:
1. We know that rectangles are also parallelograms. So, the theorem we studied also apply on the rectangle. Both of these figures have same base and lie between same parallels. Thus, both should have the same area.
Ar(RSTU) = ar(PQRS)
2. The area of parallelogram = base x height.
Ar(PQRS) = RU x RS
Since, PURL is also a rectangle, RU = AL. Thus,
Ar(PQRS) = RS x AL.
Question 5: A farmer has a field that is in the shape of a parallelogram PQRS. The farmer takes a point on RS and joins it to P, Q. Answer the following questions:
1. How many portions does he have now in the field?
2. He wants to sow corn and sugarcane, which portions he should use so that both are grown in the same area.
Solution:
1. Let the point farmer chose to be X. Join X to P and Q. This divides the field into three parts.
2. Now, we know that the if a triangle and a parallelogram have same base and are in between same parallels. Then, area of triangle is half of the area of the parallelogram.
So, in this case, ar(XPQ) = 1/2(ar(PQRS)
So, the remaining two portions must make the other half of the area. That means,
Ar(XPQ) = ar(XPS) + ar(XQR)
So, he should grow one crop in XPQ and other crop in both XPS and XQR.
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# Problems from General term of an arithmetical progression
Find the general term of the arithmetical progression:
$$(1-\sqrt{2},1+\sqrt{2}, 1+3\sqrt{2}, 1+5\sqrt{2}, 1+7\sqrt{2}, \ldots)$$
See development and solution
### Development:
Let's see what the difference is $$d=(1+\sqrt{2})-(1-\sqrt{2})=\sqrt{2}+\sqrt{2}=2\sqrt{2}$$$And, as the first term is $$a_1=1-\sqrt{2}$$, we know that: $$a_n=(1-\sqrt{2})+(n-1)\cdot 2 \cdot \sqrt{2}$$$
Arranging this expression we have:
$$a_n=(1-\sqrt{2})+2\sqrt{2}(n-1)=1-\sqrt{2}-2\sqrt{2}+2\sqrt{2}\cdot n=2\sqrt{2}n+1-3\sqrt{2}$$$### Solution: $$a_n=2\sqrt{2}n+1-3\sqrt{2}$$ Hide solution and development Find the fourth, eighth, hundredth and thirteenth terms in the arithmetical progression: $$\Big(\dfrac{1}{2}, \dfrac{3}{4}, 1, \dfrac{5}{4}, \dfrac{3}{2}, \ldots \Big)$$ See development and solution ### Development: The difference is $$d=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}$$, and as $$a_1=\dfrac{1}{2}$$, we have that $$a_n=\dfrac{1}{2}+\dfrac{1}{4}(n-1)=\dfrac{n}{4}+\dfrac{1}{4}=\dfrac{n+1}{4}$$$
So:
$$a_4=\dfrac{4+1}{4}=\dfrac{5}{4}, \ a_8=\dfrac{9}{4}$$$and $$a_{113}=\dfrac{114}{4}=\dfrac{57}{2}=28+\dfrac{1}{2}$$$
### Solution:
$$a_4=\dfrac{5}{4}, \ a_8=\dfrac{9}{4}$$ and $$a_{113}=\dfrac{57}{2}$$
Hide solution and development |
# Reciprocal identity of Cot function
## Formula
$\tan{\theta} \,=\, \dfrac{1}{\cot{\theta}}$
### Proof
Cotangent is a ratio of lengths of adjacent side to opposite side and the tangent is a ratio of lengths of opposite side to adjacent side. Actually, the cot and tan functions are reciprocal functions mutually. So, the reciprocal of cot of angle equals to tan of angle.
$\Delta QPR$ is a right triangle and its angle is denoted by theta ($\theta$).
#### Write cotangent in its ratio form
Express cotangent of angle theta ($\cot{\theta}$) in its ratio form.
$\cot{\theta} \,=\, \dfrac{PR}{QR}$
#### Write Tangent in its ratio form
Similarly, write tan of angle theta ($\tan{\theta}$) in its ratio form.
$\tan{\theta} \,=\, \dfrac{QR}{PR}$
#### Relation between Co-tangent and Tangent
Lastly, write the value of tan function in ratio form into reciprocal form for deriving the relation between cot and tan functions in trigonometry.
$\implies \tan{\theta} \,=\, \dfrac{1}{\dfrac{PR}{QR}}$
$\,\,\, \therefore \,\,\,\,\,\,$ $\tan{\theta} \,=\, \dfrac{1}{\cot{\theta}}$
Therefore, it is proved successfully that the reciprocal of cot function is equal to tan function and it is used as a formula in mathematics.
###### Note
The angle of a right triangle can be denoted by any symbol but the reciprocal identity of cot function must be expressed in terms of the respective angle.
For example, if $x$ represents angle of right triangle, then
$\tan{x} \,=\, \dfrac{1}{\cot{x}}$
In the same way, if $A$ denotes angle of right triangle, then
$\tan{A} \,=\, \dfrac{1}{\cot{A}}$
The reciprocal identity of cot function is written in this form but the symbol of angle of right triangle is only changed.
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### Isosceles and equilateral triangles
We’ve learned that you can classify triangles in different ways. Well, some of these types of triangles have special properties!
Isosceles Triangle
An isosceles triangle has two sides that are congruent. These two sides are called legs. The remaining side is called a base. Since two sides are congruent, it also means that the two angles opposite those sides are congruent. These would be the two base angles. Here are some diagrams that usually help with understanding.
Equilateral Triangle
In an equilateral triangle, all sides are congruent AND all angles are congruent. When all angles are congruent, it is called equiangular. The sides can measure anything as long as they are all the same. The angles, however, HAVE to all equal 60°. This is because all angles in a triangle always add up to 180°and if you divide this amongst three angles, they have to each equal 60°. So, in EVERY equilateral triangle, the angles are always 60°.
Let’s see if we can put these properties to work and answer a few questions. You have to look at these problems as “puzzles” because sometimes you need to find a part that they are not asking for in order to find the final result. Find a piece at a time and put them together until you reach your answer!
Example 1:
Find x.
There’s actually at least three different ways that you can answer this problem. I’ll show you one. They ultimately want to find the measure of that exterior angle.
$${\text{180 - 128 = 52}}$$
$${\text{180 - 52 = 128}}$$
$${\text{128}} \div {\text{2 }} = {\text{ 64}}$$
$${\text{180 - 64 = 116}}$$
$${\text{x }} = {\text{ 116}}^\circ$$
Example 2:
This is going to take a few steps. We need a few pieces of the puzzle before we can find the measure of x.
If you don’t remember that last step, don’t worry! You can just take two more steps and find the 3rd angle of the bottom triangle and subtract it from 180°to find the exterior angle.
Many of these problems take more than one or two steps, so look at it as a puzzle and put your pieces together!
5663 x
Find the value of x.
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Example of one question:
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5367 x
Find the value of x.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
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3328 x
Find the value of x.
This free worksheet contains 10 assignments each with 24 questions with answers.
Example of one question:
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### Geometry
Circles
Congruent Triangles
Constructions
Parallel Lines and the Coordinate Plane
Properties of Triangles
### Algebra and Pre-Algebra
Beginning Algebra
Beginning Trigonometry
Equations
Exponents
Factoring
Linear Equations and Inequalities
Percents
Polynomials |
# Trigonometry : Determine Vertical Shifts
## Example Questions
### Example Question #1 : Determine Vertical Shifts
Let be a function defined as follows:
The 4 in the function above affects what attribute of the graph of ?
Amplitude
Period
Phase shift
Vertical shift
Vertical shift
Explanation:
The period of the function is indicated by the coefficient in front of ; here the period is unchanged.
The amplitude of the function is given by the coefficient in front of the ; here the amplitude is -1.
The phase shift is given by the value being added or subtracted inside the function; here the shift is units to the right.
The only unexamined attribute of the graph is the vertical shift, so 4 is the vertical shift of the graph. A vertical shift of 4 means that the entire graph of the function will be moved up four units (in the positive y-direction).
### Example Question #2 : Determine Vertical Shifts
Let be a function defined as follows:
What is the vertical shift in this function?
Explanation:
The period of the function is indicated by the coefficient in front of ; here the period is unchanged.
The amplitude of the function is given by the coefficient in front of the ; here the amplitude is 3.
The phase shift is given by the value being added or subtracted inside the cosine function; here the shift is units to the right.
The only unexamined attribute of the graph is the vertical shift, so -3 is the vertical shift of the graph. A vertical shift of -3 means that the entire graph of the function will be moved down three units (in the negative y-direction).
### Example Question #3 : Determine Vertical Shifts
The graph below shows a translated sine function. Which of the following functions could be shown by this graph?
Explanation:
A normal graph has its y-intercept at . This graph has its y-intercept at . Therefore, the graph was shifted down three units. Therefore the function of this graph is .
### Example Question #4 : Determine Vertical Shifts
This graph shows a translated cosine function. Which of the following could be the equation of this graph?
Explanation:
The correct answer is . There are no sign changes with vertical shifts; in other words, when the function includes , it directly translates to moving up three units. If you thought the answer was , you may have spotted the y-intercept at and jumped to this answer. However, recall that the y-intercept of a regular function is at the point . Beginning at and ending at corresponds to a vertical shift of 3 units.
### Example Question #5 : Determine Vertical Shifts
Consider the function . What is the vertical shift of this function?
Explanation:
The general form for the secant transformation equation is . represents the phase shift of the function. When considering we see that , so our vertical shift is and we would shift this function units up from the original secant function’s graph.
### Example Question #6 : Determine Vertical Shifts
Which of the following is the graph of with a vertical shift of ?
Explanation:
The graph of with a vertical shift of is shown below. This can also be expressed as .
Here is a graph that shows both and , so that you can see the "before" and "after." The original function is in blue and the translated function is in purple.
The graphs of the incorrect answer choices are (no vertical shift applied), (shifted upwards instead of downwards), (amplitude modified, and shifted upwards instead of downwards), and (shifted downwards 3 units, but this is not the correct original graph of simply since the amplitude was modified.)
### Example Question #7 : Determine Vertical Shifts
Which of the following graphs shows one of the original six trigonometric functions with a vertical shift of applied? |
Study Guide for Chapter 9 & 10
# Study Guide for Chapter 9 & 10 - MATH A Chapter 9...
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MATH A: Chapter 9 Study Guide, Roots and Radicals 9.1, Finding Roots Objectives: 1) Find Square Roots 2) Find higher roots. I. Fundamentals A. Square Roots 1. To find a square of a number – multiply by itself 2 a a a = 2. To find a cube of a number – multiply the number by itself 3 times. 3 a a a a = 3. To find a square root of a number – use the inverse process of squaring 4. To find a cube root of a number – use the inverse process of cubing. B. Radicals 1. n a b = 2. Definition of the Principal nth Root of a real number : If a is a nonnegative real number, and b is the nonnegative root of a, we can write n b a = , denoted by n a b = ,is the Principal nth Root of a. C. Finding perfect roots: Evaluate 1) 36 2) 9 - 3) 1 25 4) 36 64 + 5) 36 64 + D. See the table on page 552 for the perfect squares. II. Roots greater than square roots A. Definition of the principal nth root of a real number: n a b = means that n b a = . The natural number n is called the index. The index, 2, for square roots is usually omitted. B. Reversing nth Powers with nth Powers Roots Vocabulary Cube Roots 3 4 3 64 Index is ( 29 3 2 - 3 8 - 3 5 3 125 Fourth Roots 4 1 4 1 Index is 4 3 4 81 Fifth Roots 5 2 5 32 Index is Nth Roots n b n a Index is 1
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C. Finding Roots 1) 6 64 2) 3 1 125 3) 3 125 - 4. 4 - 5) 4 81 - 6) 3 3 8 - 7) 6 49 2 81 - 9.2, Multiplying and Dividing Radicals Objectives: 1) Multiply square roots. 2) Simplify square roots. 3) Use the quotient rule for square roots. 4) Use the product rule and quotient rule for other roots.
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# Practise Using Addition and Subtraction to Solve Two-Step Word Problems
In this worksheet, students will solve two-step word problems using addition and subtraction.
Key stage: KS 2
Curriculum topic: Number: Addition and Subtraction
Curriculum subtopic: Use Correct Methods to Problem Solve
Difficulty level:
#### Worksheet Overview
We are using maths all the time, although we often don't realise it!
Just think about when we buy a new book. We need to make sure that we have enough money to pay for it and be able to check that we are given the correct change!
What about playing a board game - we need to be able to add up the scores correctly!
These are just two examples of how we use maths every day.
In this activity, we are going to practise using addition and subtraction to answer some tricky word problems.
In each question, we will have to do more than one calculation, so reading the question carefully is very important!
Example
Harley is saving up to buy a new scooter.
He receives £96 birthday money and saves £22.50 of his pocket money.
The scooter is on offer at £159.99
How much more money does Harley need in order to buy the scooter?
We need to identify the key information - it's a good idea to highlight it when you spot it!
Harley receives a total of £118.50 (96 + 22.50)
The scooter costs £159.99
The question asks us to find how much more money Harley needs, so we need to find the difference, which means subtraction.
£159.99 - £118.50 = £41.49
Harley needs another £41.49
Column addition and subtraction would work well for these calculations.
We must ensure we put each digit in the correct column.
Now let's look at another example.
There is quite a lot of information here, so the important stuff is highlighted for us.
We need to read the question carefully and think about what it is asking us to do.
Pens cost £1.05 each and pencils cost 86p each.
How much change would we get from £10 if we bought 3 pens and 2 pencils?
First, we need to work out the cost of 3 pens.
£1.05 + £1.05 + £1.05 = £3.15 (315p)
Next, we work out the cost of 2 pencils.
86p + 86p = £1.72 (172p)
Next, we add together the two totals.
£3.15 + £1.72 = £4.87
Finally, we subtract £4.87 from £10 to find out how much change there would be.
£10 - £4.87 = £5.13 (1,000 - 487 = 513)
There would be £5.13 change.
Using a number line to find the difference between 487 and 1,000 might help here, or we could do it mentally!
Now, why don't you have a go at some questions like these?
Remember to read the question carefully and highlight the important information!
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## Intermediate Algebra (12th Edition)
$\sqrt[4]{27p^{3}}-\sqrt[3]{4x}$
$\bf{\text{Solution Outline:}}$ Use the definition of rational exponents to convert the given expression, $(3p)^{3/4}-(4x)^{1/3} ,$ to radical form. Then use the laws of exponents to simplify the resulting expression. $\bf{\text{Solution Details:}}$ Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{(3p)^{3}}-\sqrt[3]{(4x)^{1}} .\end{array} Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{3^3p^{3}}-\sqrt[3]{4^1x^{1}} \\\\= \sqrt[4]{27p^{3}}-\sqrt[3]{4x} .\end{array} |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Adding and Subtracting Rational Expressions where One Denominator is the LCD
## Combine fractions including variables, one-step LCM
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Adding and Subtracting Rational Expressions where One Denominator is the LCD
The length of a garden plot is \begin{align*}\frac{6x^2-5}{2x^2 + 4x - 6}\end{align*}. The width of the plot is \begin{align*}\frac{2x-7}{x+3}\end{align*}. How much longer is the garden plot than it is wide?
### Adding and Subtracting Rational Expressions
Recall when two fractions do not have the same denominator. You have to multiply one or both fractions by a number to create equivalent fractions in order to combine them.
\begin{align*}\frac{1}{2} + \frac{3}{4}\end{align*}
Here, 2 goes into 4 twice. So, we will multiply the first fraction by \begin{align*}{\color{red}\frac{2}{2}}\end{align*} to get a denominator of 4. Then, the two fractions can be added.
\begin{align*}{\color{red}\frac{2}{2} \cdot} \frac{1}{2} + \frac{3}{4} = \frac{2}{4} + \frac{3}{4} = \frac{5}{4}\end{align*}
Once the denominators are the same, the fractions can be combined. We will apply this idea to rational expressions in order to add or subtract ones without like denominators.
Let's add or subtract the following rational expressions.
1. Subtract \begin{align*}\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4}\end{align*}.
Factoring the denominator of the first fraction, we have \begin{align*}2(x+4)\end{align*}. The second fraction needs to be multiplied by \begin{align*}{\color{red}\frac{2}{2}}\end{align*} in order to make the denominators the same.
\begin{align*}\frac{3x-5}{2x+8} - \frac{x^2-6}{x+4} &= \frac{3x-5}{2(x+4)} - \frac{x^2-6}{x+4} \cdot {\color{red}\frac{2}{2}} \\ &= \frac{3x-5}{2(x+4)} - \frac{2x^2-12}{2(x+4)}\end{align*}
Now that the denominators are the same, subtract the second rational expression just like you've done before.
\begin{align*}&= \frac{3x-5-(2x^2-12)}{2(x+4)} \\ &= \frac{3x-5-2x^2+12}{2(x+4)} \\ &= \frac{-2x^2+3x+7}{2(x+4)}\end{align*}
The numerator is not factorable, so we are done.
1. Add \begin{align*}\frac{2x-3}{x+5} + \frac{x^2+1}{x^2-2x-35}\end{align*}.
Factoring the second denominator, we have \begin{align*}x^2-2x-35=(x+5)(x-7)\end{align*}. So, we need to multiply the first fraction by \begin{align*}{\color{red}\frac{x-7}{x-7}}\end{align*}.
\begin{align*}\overbrace{{\color{red}\frac{(x-7)}{(x-7)}} \cdot \frac{(2x-3)}{(x+5)}}^{FOIL} + \frac{x^2+1}{(x-7)(x+5)} &= \frac{2x^2-17x+21}{(x-7)(x+5)} + \frac{x^2+1}{(x-7)(x+5)} \\ &= \frac{3x^2-17x+22}{(x-7)(x+5)}\end{align*}
1. Subtract \begin{align*}\frac{7x+2}{2x^2+18x+40} - \frac{6}{x+5}\end{align*}.
Factoring the first denominator, we have \begin{align*}2x^2+18x+40=2(x^2+9x+20)=2(x+4)(x+5)\end{align*}. This is the Lowest Common Denominator, or LCD. The second fraction needs the 2 and the \begin{align*}(x+4)\end{align*}.
\begin{align*}\frac{7x+2}{2x^2+18x+40} - \frac{6-x}{x+5} &= \frac{7x+2}{2(x+5)(x+4)} - \frac{6-x}{x+5}{\color{red}\cdot \frac{2(x+4)}{2(x+4)}} \\ &= \frac{7x+2}{2(x+5)(x+4)} - \frac{2(6-x)(x+4)}{2(x+5)(x+4)} \\ &= \frac{7x+2}{2(x+5)(x+4)} - \frac{48+4x-2x^2}{2(x+5)(x+4)} \\ &= \frac{7x+2-(48+4x-2x^2)}{2(x+5)(x+4)} \\ &= \frac{7x+2-48-4x+2x^2}{2(x+5)(x+4)} \\ &= \frac{2x^2+3x-46}{2(x+5)(x+4)}\end{align*}
### Examples
#### Example 1
Earlier, you were asked to find how much longer the garden plot is than it is wide.
We need to subtract the width from the length.
\begin{align*}\frac{6x^2-5}{2x^2 + 4x - 6} - \frac{2x-7}{x+3}\end{align*}
Factoring the first denominator, we have \begin{align*}2x^2+4x-6=(2x-2)(x+3)\end{align*}. So, we need to multiply the second fraction by \begin{align*}{\color{red}\frac{2x-2}{2x-2}}\end{align*}.
\begin{align*}\frac{6x^2-5}{2x^2 + 4x - 6} -\overbrace{{\color{red}\frac{(2x-2)}{(2x-2)}} \cdot \frac{(2x-7)}{(x+3)}}^{FOIL} &= \frac{4x^2-18x+14}{2x^2+4x-6} \\ \frac{6x^2-5}{2x^2 + 4x - 6} - \frac{4x^2-18x+14}{2x^2+4x-6}\\ \frac{6x^2-5-(4x^2-18x+14)}{2x^2+4x-6}\\ \frac{6x^2-5-4x^2+18x-14}{2x^2+4x-6}\\ \frac{2x^2+18x-19}{2x^2+4x-6}\end{align*}
Therefore, the garden plot is \begin{align*}\frac{2x^2+18x-19}{2x^2+4x-6}\end{align*} longer than it is wide
Perform the indicated operation.
#### Example 2
\begin{align*}\frac{2}{x+1} - \frac{x}{3x+3}\end{align*}
The LCD is \begin{align*}3x+3\end{align*} or \begin{align*}3(x+1)\end{align*}. Multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}.
\begin{align*}\frac{2}{x+1} - \frac{x}{3x+3} &= \frac{3}{3} \cdot \frac{2}{x+1} - \frac{x}{3(x+1)} \\ &= \frac{6}{3(x+1)} - \frac{x}{3(x+1)} \\ &= \frac{6-x}{3(x+1)}\end{align*}
#### Example 3
\begin{align*}\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6}\end{align*}
Here, the LCD \begin{align*}x^2+4x-24\end{align*} or \begin{align*}(x+6)(x-4)\end{align*}. Multiply the second fraction by \begin{align*}\frac{x-4}{x-4}\end{align*}.
\begin{align*}\frac{x-10}{x^2+4x-24} + \frac{x+3}{x+6} &= \frac{x-10}{(x+6)(x-4)} + \frac{x+3}{x+6} \cdot \frac{x-4}{x-4} \\ &= \frac{x-10}{(x+6)(x-4)} + \frac{x^2-x-12}{(x+6)(x-4)} \\ &= \frac{x-10+x^2-x-12}{(x+6)(x-4)} \\ &= \frac{x^2-22}{(x+6)(x-4)}\end{align*}
#### Example 4
\begin{align*}\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6}\end{align*}
The LCD is \begin{align*}3x^2-12=3(x-2)(x+2)\end{align*}. The second fraction’s denominator factors to be \begin{align*}3x+6=3(x+2)\end{align*}, so it needs to be multiplied by \begin{align*}\frac{x-2}{x-2}\end{align*}.
\begin{align*}\frac{3x^2-5}{3x^2-12} + \frac{x+8}{3x+6} &= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x+8}{3(x+2)} \cdot \frac{x-2}{x-2} \\ &= \frac{3x^2-5}{3(x-2)(x+2)} + \frac{x^2+6x-16}{3(x-2)(x+2)} \\ &= \frac{3x^2-5+x^2+6x-16}{3(x-2)(x+2)} \\ &= \frac{4x^2+6x-21}{3(x-2)(x+2)}\end{align*}
### Review
Find the LCD.
1. \begin{align*}x, \ 6x\end{align*}
2. \begin{align*}x, \ x+1\end{align*}
3. \begin{align*}x+2, \ x-4\end{align*}
4. \begin{align*}x, \ x-1, \ x^2 - 1\end{align*}
Perform the indicated operations.
1. \begin{align*}\frac{3}{x} - \frac{5}{4x}\end{align*}
2. \begin{align*}\frac{x+2}{x+3} + \frac{x-1}{x^2+3x}\end{align*}
3. \begin{align*}\frac{x}{x-7} - \frac{2x+7}{3x-21}\end{align*}
4. \begin{align*}\frac{x^2+3x-10}{x^2-4} - \frac{x}{x+2}\end{align*}
5. \begin{align*}\frac{5x+14}{2x^2-7x-15} - \frac{3}{x-5}\end{align*}
6. \begin{align*}\frac{x-3}{3x^2+x-10} + \frac{3}{x+2}\end{align*}
7. \begin{align*}\frac{x+1}{6x+2} + \frac{x^2-7x}{12x^2-14x-6}\end{align*}
8. \begin{align*}\frac{-3x^2-10x+15}{10x^2-x-3} + \frac{x+4}{2x+1}\end{align*}
9. \begin{align*}\frac{8}{2x-5} - \frac{x+5}{2x^2+x-15}\end{align*}
10. \begin{align*}\frac{2}{x+2} + \frac{3x+16}{x^2-x-6} - \frac{2}{x-3}\end{align*}
11. \begin{align*}\frac{6x^2+4x+8}{x^3+3x^2-x-3} + \frac{x-4}{x^2-1} - \frac{3x}{x^2+2x-3}\end{align*}
### Answers for Review Problems
To see the Review answers, open this PDF file and look for section 9.11.
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# Finding the equation of a line between two points
Here’s a standard problem that could be found in any Algebra I textbook.
Find the equation of the line between $(-1,-2)$ and $(4,2)$.
The first step is clear: the slope of the line is
$m = \displaystyle \frac{2-(-2)}{4-(-1)} = \frac{4}{5}$
At this point, there are two reasonable approaches for finding the equation of the line.
Method #1. This is the method that was hammered into my head when I took Algebra I. We use the point-slope form of the line:
$y - y_1 = m (x - x_1)$
$y - 2 = \displaystyle \frac{4}{5} (x-4)$
$y - 2 = \displaystyle \frac{4}{5}x - \frac{16}{5}$
$y = \displaystyle \frac{4}{5}x - \frac{6}{5}$
For what it’s worth, the point-slope form of the line relies on the fact that the slope between $(x,y)$ and $(x_1,y_1)$ is also equal to $m$.
Method #2. I can honestly say that I never saw this second method until I became a college professor and I saw it on my students’ homework. In fact, I was so taken aback that I almost marked the solution incorrect until I took a minute to think through the logic of my students’ solution. Let’s set up the slope-intercept form of a line:
$y= \displaystyle \frac{4}{5}x + b$
Then we plug in one of the points for $x$ and $y$ to solve for $b$.
$2 = \displaystyle \frac{4}{5}(4) + b$
$\displaystyle -\frac{6}{5} = b$
Therefore, the line is $y = \displaystyle \frac{4}{5}x - \frac{6}{5}$.
My experience is that most college students prefer Method #2, and I can’t say that I blame them. The slope-intercept form of a line is far easier to use than the point-slope form, and it’s one less formula to memorize.
Still, I’d like to point out that there are instances in courses above Algebra I that the point-slope form is really helpful, and so the point-slope form should continue to be taught in Algebra I so that students are prepared for these applications later in life.
Topic #1. In calculus, if $f$ is differentiable, then the tangent line to the curve $y=f(x)$ at the point $(a,f(a))$ has slope $f'(a)$. Therefore, the equation of the tangent line (or the linearization) has the form
$y = f(a) + f'(a) \cdot (x-a)$
This linearization is immediately obtained from the point-slope form of a line. It also can be obtained using Method #2 above, so it takes a little bit of extra work.
This linearization is used to derive Newton’s method for approximating the roots of functions, and it is a precursor to Taylor series.
Topic #2. In statistics, a common topic is finding the least-squares fit to a set of points $(x_1,y_1), (x_2,y_2), \dots, (x_n,y_n)$. The solution is called the regression line, which has the form
$y - \overline{y} = r \displaystyle \frac{s_y}{s_x} (x - \overline{x})$
In this equation,
• $\overline{x}$ and $\overline{y}$ are the means of the $x-$ and $y-$values, respectively.
• $s_x$ and $s_y$ are the sample standard deviations of the $x-$ and $y-$values, respectively.
• $r$ is the correlation coefficient between the $x-$ and $y-$values.
The formula of the regression line is decidedly easier to write in point-slope form than in slope-intercept form. Also, the point-slope form makes the interpretation of the regression line clear: it must pass through the point of averages $(\overline{x}, \overline{y})$.
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### methods of correlation
Methods Of Determining Correlation
We shall consider the following most commonly used methods.(1)Scatter Plot (2) Kar Pearson’s coefficient of correlation (3) Spearman’s Rank-correlation coefficient.
1) Scatter Plot ( Scatter diagram or dot diagram ): In this method the values of the two variables are plotted on a graph paper. One is taken along the horizontal ( (x-axis) and the other along the vertical (y-axis). By plotting the data, we get points (dots) on the graph which are generally scattered and hence the name ‘Scatter Plot’.
The manner in which these points are scattered, suggest the degree and the direction of correlation. The degree of correlation is denoted by ‘ r ’ and its direction is given by the signs positive and negative.
i) If all points lie on a rising straight line the correlation is perfectly positive and r = +1 (see fig.1 )
ii) If all points lie on a falling straight line the correlation is perfectly negative and r = -1 (see fig.2)
iii) If the points lie in narrow strip, rising upwards, the correlation is high degree of positive (see fig.3)
iv) If the points lie in a narrow strip, falling downwards, the correlation is high degree of negative (see fig.4)
v) If the points are spread widely over a broad strip, rising upwards, the correlation is low degree positive (see fig.5)
vi) If the points are spread widely over a broad strip, falling downward, the correlation is low degree negative (see fig.6)
vii) If the points are spread (scattered) without any specific pattern, the correlation is absent. i.e. r = 0. (see fig.7)
Though this method is simple and is a rough idea about the existence and the degree of correlation, it is not reliable. As it is not a mathematical method, it cannot measure the degree of correlation.
2) Karl Pearson’s coefficient of correlation: It gives the numerical expression for the measure of correlation. it is noted by ‘ r ’. The value of ‘ r ’ gives the magnitude of correlation and sign denotes its direction. It is defined as
r =
where
N = Number of pairs of observation
Now,
Since r is positive and 0.6. This shows that the correlation is positive and moderate (i.e. direct and reasonably good).
Example From the following data compute the coefficient of correlation between x and y.
Example If covariance between x and y is 12.3 and the variance of x and y are 16.4 and 13.8 respectively. Find the coefficient of correlation between them.
Solution: Given - Covariance = cov ( x, y ) = 12.3
Variance of x ( s x)= 16.4
Variance of y (sy) = 13.8
Now,
Example Find the number of pair of observations from the following data.
r = 0.25, S (xi - x ) ( yi - y ) = 60, sy = 4, S ( xi - x )2 = 90.
Solution: Given - r = 0.25
If the values of x and y are very big, the calculation becomes very tedious and if we change the variable x to u = and y to where x0 and y0 are the assumed means for variable x and y respectively, then rxy= ruv
The formula for r can be simplified as
Example Marks obtained by two brothers FRED and TED in 10 tests are as follows:
Find the coefficient of correlation between the two.
Solution: Here x0 = 60, c = 4, y0 = 60 and d = 3
Calculation:
Spearman’s Rank Correlation Coefficient
This method is based on the ranks of the items rather than on their actual values. The advantage of this method over the others in that it can be used even when the actual values of items are unknown. For example if you want to know the correlation between honesty and wisdom of the boys of your class, you can use this method by giving ranks to the boys. It can also be used to find the degree of agreementsbetween the judgements of two examiners or two judges. The formula is :
R =
where R = Rank correlation coefficient
D = Difference between the ranks of two items
N = The number of observations.
Note: -1 ££ 1.
i) When R = +1 Þ Perfect positive correlation or complete agreement in the same direction
ii) When R = -1 Þ Perfect negative correlation or complete agreement in the opposite direction.
iii) When R = 0 Þ No Correlation.
Computation:
1. Give ranks to the values of items. Generally the item with the highest value is ranked 1 and then the others are given ranks 2, 3, 4, .... according to their values in the decreasing order.
1. Calculate D2 and then find S D2
2. Apply the formula.
¬ Note :
In some cases, there is a tie between two or more items. in such a case each items have ranks 4th and 5th respectively then they are given = 4.5th rank. If three items are of equal rank say 4th then they are given = 5th rank each. If m be the number of items of equal ranks, the factor is added to S D2. If there are more than one of such cases then this factor added as many times as the number of such cases, then
Example Calculate ‘ R ’ from the following data.
StudentNo.: 1 2 3 4 5 6 7 8 9 10 Rank in Maths : 1 3 7 5 4 6 2 10 9 8 Rank in Stats:
Solution :
StudentNo. Rank inMaths (R1) Rank inStats (R2) R1 - R2D (R1 - R2 )2D2 1 1 3 -2 4 2 3 1 2 4 3 7 4 3 9 4 5 5 0 0 5 4 6 -2 4 6 6 9 -3 9 7 2 7 -5 25 8 10 8 2 4 9 9 10 -1 1 10 8 2 6 36 N = 10 S D = 0 S D2 = 96
Calculation of R :
Example Calculate ‘ R ’ of 6 students from the following data.
Marks in Stats : 40 42 45 35 36 39 Marks in English : 46 43 44 39 40 43
Solution:
Marks in Stats R1 Marks in English R2 R1 - R2 (R1 -R2)2=D2 40 3 46 1 2 4 42 2 43 3.5 -1.5 2.25 45 1 44 2 -1 1 35 6 39 6 0 0 36 5 40 5 0 0 39 4 43 3.5 0.5 0.25 N = 6 S D = 0 S D2 = 7.50
Here m = 2 since in series of marks in English of items of values 43 repeated twice.
Example The value of Spearman’s rank correlation coefficient for a certain number of pairs of observations was found to be 2/3. Thesum of the squares of difference between the corresponding rnks was 55. Find the number of pairs.
Solution: We have
Linear Regression
Correlation gives us the idea of the measure of magnitude and direction between correlated variables. Now it is natural to think of a method that helps us in estimating the value of one variable when the other is known. Also correlation does not imply causation. The fact that the variables x and y are correlated does not necessarily mean that x causes y or vice versa. For example, you would find that the number of schools in a town is correlated to the number of accidents in the town. The reason for these accidents is not the school attendance; but these two increases what is known as population. A statistical procedure called regression is concerned with causation in a relationship among variables. It assesses the contribution of one or more variable calledcausing variable or independent variable or one which is beingcaused (dependent variable). When there is only one independent variable then the relationship is expressed by a straight line. This procedure is called simple linear regression.
Regression can be defined as a method that estimates the value of one variable when that of other variable is known, provided the variables are correlated. The dictionary meaning of regression is "to go backward." It was used for the first time by Sir Francis Galton in hisresearch paper "Regression towards mediocrity in hereditary stature."
Lines of Regression: In scatter plot, we have seen that if the variables are highly correlated then the points (dots) lie in a narrow strip. if the strip is nearly straight, we can draw a straight line, such that all points are close to it from both sides. such a line can be taken as an ideal representation of variation. This line is called the line of best fit if it minimizes the distances of all data points from it.
This line is called the line of regression. Now prediction is easy because now all we need to do is to extend the line and read the value. Thus to obtain a line of regression, we need to have a line of best fit. But statisticians don’t measure the distances by dropping perpendiculars from points on to the line. They measure deviations ( or errors or residuals as they are called) (i) vertically and (ii) horizontally. Thus we get two lines of regressions as shown in the figure (1) and (2).
(1) Line of regression of y on x
Its form is y = a + b x
It is used to estimate y when x is given
(2) Line of regression of x on y
Its form is x = a + b y
It is used to estimate x when y is given.
They are obtained by (1) graphically - by Scatter plot (ii) Mathematically - by the method of least squares. |
# What Is The Rule For Divisibility By 10?
## How can you determine if a number is divisible by 2 or 5 or 10?
Divisibility Guidelines for 2, 5, and 10In order to know if a number is divisible by 2, you have to check whether it is an even number.
If it is even, it is divisible by 2.
In order to know if a number is divisible by 5, the number has to end in 0 or 5.In order to know if a number is divisible by 10, it has to end in 0..
## How do you know if a number is divisible by 1 10?
A number is divisible by 10 if the last digit is 0. 1,470 is divisible by 10 since the last digit is 0. Let’s look at some examples in which we test the divisibility of a single whole number. Example 1: Determine whether 150 is divisible by 2, 3, 4, 5, 6, 9 and 10.
## What is the divisibility rule of 9?
Divisibility rules for numbers 1–30DivisorDivisibility condition8The last three digits are divisible by 8.Add four times the hundreds digit to twice the tens digit to the ones digit. The result must be divisible by 8.9Sum the digits. The result must be divisible by 9.10The ones digit is 0.63 more rows
## What is the least number divisible by 1 to 10?
25202520 is: the smallest number divisible by all integers from 1 to 10, i.e., it is their least common multiple. half of 7!
## How do you know what a number is divisible by?
Divisibility by: If the last digit is even, the number is divisible by 2. If the sum of the digits is divisible by 3, the number is also. If the last two digits form a number divisible by 4, the number is also. If the last digit is a 5 or a 0, the number is divisible by 5.
## Is every number divisible by 1?
Every number is divisible by 1. When a number is divisible by another number, then it is also divisible by each of the factors of that number. For instance, a number divisible by 6, will also be divisible by 2 and 3.
## What is the divisibility rule of 11?
Here an easy way to test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. So, for instance, 2728 has alternating sum of digits 2 – 7 + 2 – 8 = -11.
## What number is not divisible by 10?
A number is divisible by 10 if the last digit of the number is 0. The numbers 20, 40, 50, 170, and 990 are all divisible by 10 because their last digit is zero. On the other hand, 21, 34, 127, and 468 are not divisible by 10 since they don’t end with zero.
## Is there a divisibility rule for 7?
Divisibility by 7. … To check if a number is evenly divisible by 7: Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number. |
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# Slope of Line
## How to Find the Slope of a Line?
Last updated date: 23rd Jan 2023
Total views: 284.1k
Views today: 6.79k
In Mathematics, the slope of gradient characterizes both the steepness and direction of a line. The letter ‘m’ is used to represent the slope.
The slope is always estimated by determining the ratio of the “vertical change” to the “horizontal change” between any two different points on a line. Sometimes, this ratio is expressed as a quotient (“rise over the run”) giving a similar number for every two distinct points on the same line. A line that is drawn on a plane, and is decreasing has a negative “rise”. The line may be functional as laid down by the road surveyor, or in the diagram that exhibits a road or roof either as a description or a plan.
Read on to have detailed information on the slope of a line.
## Slope of Line Definition
As discussed above, the slope of a line is the ratio of the “vertical change” to the “horizontal change” or the ratio between the change of x and the change of y.
Sometimes, the slope is expressed as rise over run. You can determine slope by observing walking up the flight of stairs, dividing the vertical change, which comes in the numerator by the horizontal change, which comes in the denominator.
The slope, m, of a line, is defined as:
$m = \frac{\mathtt{change~~in~~y}}{\mathtt{ change~~in~~x}} = \frac{\vartriangle y}{\vartriangle x}$
## What is the Slope Formula?
The slope of a line in the coordinate plane comprising both of x and y axes is usually denoted by the letter ‘m’ and is defined as the ratio of the change in y coordinate to the corresponding change in x coordinate, between two different points on a line. Accordingly, the slope of a line formula is given as :
$m =\frac{\vartriangle y}{\vartriangle x} =\frac{\mathtt{verticle ~~change}}{\mathtt{Horizontal ~~change}} =\frac{\text{rise}}{\text{run}}$
## How to Measure the Steepness and Direction of a Line?
The steepness, grade, or incline of a line is measured by the absolute value of a slope. A slope with greater absolute values represents a steeper line. The direction of a line is either decreasing, increasing, vertical or horizontal.
• A line is increasing if it moves up from left to right. The slope is positive i.e. m > 0.
• A line is decreasing if it moves down from left to right. The slope is negative i.e. m < 0.
• The slope is zero if a line is horizontal.
Here, we can see that irrespective of the two points we choose, the value of the y-coordinate always remains the same; i.e it is always 3. Hence, the change in the y value along the line is zero. No matter what the change in x value along the line, the slope must always equal zero.
Slope (m) =$\frac{0}{\mathtt{change ~ in ~ x}}$=
0 divided by any number is always 0. Therefore, horizontal lines have a slope of 0.
• The slope is undefined if a line is vertical.
In this situation, irrespective of what two points we choose, the value of the x-coordinate always remains the same; it is always 2. Hence, the change in x value along the line is zero.
Slope (m) = $\frac{\mathtt{change ~~ in ~~y}}{0}$=
As we cannot divide by 0, the slope of a vertical line is infinite. Therefore, vertical lines have an infinite slope.
## How to Find the Slope of A Line?
Given two coordinates $(x_{1},y_{1})$ and $(x_{2},y_{2})$ the change in variable x from one to another is $x_{2} -x_{1}$ (run) while the change in variable y from one to another is $y_{2} -y_{1}$ (rise). Substituting both the values in the above equation generates the following slope of a line formula:
$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
The above-given slope of a line equation is not valid for a vertical line, parallel to the y axis (refer to Division by Zero), where the slope can be considered as infinite, hence, the slope of a vertical line is considered undefined.
Let us understand how to find the slope of a line with an example.
Example 1:
Suppose a line runs through two points: A= (3,2), and B= (15,8). By dividing the difference in y- coordinates by the difference in x - coordinate, one can get the slope of the line as shown below:
$m=\frac{\vartriangle y}{\vartriangle x} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{8-2}{15-3} =\frac{6}{12} = \frac{1}{2}$
As the slope is positive, the direction of a line is increasing.
Also |m| < 1, the incline is not very steep (incline is less than $45^{\circ}$).
## How to Find Slope from a Graph
Here in the below given figure, if the angle of inclination of the given line with the x-axis is $\theta$ then, the slope of the line is given by $\tan\theta$. Thus, there is a relation between the lines and angles.
Slope of a line is given as $m =\tan\theta$.
Here in the above image we can see that there are two points present on the line. The two points are A and B where A = $(x_{1},y_{1})$ and B= $(x_{2},y_{2})$ lie on the line with $( x_{1} \neq x_{2})$ then the slope of the line AB is given as:
$m =\tan\theta = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Where $\theta$ is the angle which the line AB makes with the positive direction of the x-axis. $\theta$ lies between $0^{\circ}$ and $180^{\circ}$
It must be noted that $\theta = 90^{\circ}$ is only possible when the line is parallel to the y-axis i.e. at $x_{1} = x_{2}$ at this particular angle the slope of the line is undefined.
## Slope of Parallel Lines
Let us consider given lines i.e, $l_{1}$ and $l_{2}$ are parallel with inclinations α and β respectively. For two lines to be parallel their inclination must also be equal i.e. α=β. Thus this results in the fact that tan α = tan β. Therefore, the condition for two lines with inclinations α, β to be parallel is tan = tan β.
## Slope of Perpendicular Lines
A set of perpendicular lines always has a 90º angle between them. Let us suppose we have two perpendicular lines$l_{1}$ and $l_{2}$ in the coordinate plane, inclined at angle and .
Suppose if they are perpendicular, we can say that β = α + 90°. (Using properties of angles)
Their slopes can be given as: $m_{1}$ = tan(α + 90°) and $m_{2}$ = tan .
$m_{1}= -\cot\alpha = \frac{-1}{\tan\alpha}$
Here we know that $tan\alpha = m_{2}$
Thus
$m_{1} = \dfrac{-1}{m_{2}}$
$m_{1} \times m_{2} = -1$
Thus, for two lines to be perpendicular the product of their slope must be equal to -1.
## Slope for Collinearity
For two lines AB and BC to be collinear, the slopes of both lines must be equal, and they must pass through at least one point in common. Thus, for three points A, B, and C to be collinear the slopes of AB and BC must be equal.
When ab' c' = a' b' c = a'c'b, the two lines are collinear if their equations are ax + by + c = 0 and a' x+b' y+c' = 0.
## Angle Between Two Lines
Suppose when two lines intersect at a point then the angle between them can be expressed in terms of their slopes and is given by the below formula:
$\tan\theta = |\frac{m_{2}-m_{1}}{1-m_{2}-m_{1}}|$,
Where $m_{1},m_{2}$are the slopes of the given lines AB and CD respectively.
## Solved Problems
1. Find the slope of a line between the points P = (2, -1) and Q = (4,1).
Solution:
Given,the points P = (2, –1) and Q = (4,1).
As per the slope formula we know that,
Slope of a line, $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$\Rightarrow m= \frac{1-(-1)}{4-2}$
$\Rightarrow \frac{2}{2}$
$\Rightarrow m=1$
2. Sowyma was checking the graph, and she realized that the raise was 20 units and the run was 5 units. What should be the slope of a line?
Solution:
Given that, Raise = 20 units
Run = 5 units.
We know that the slope of a line is defined as the ratio of raise to the run.
i.e. Slope, $m=\frac{\mathtt{rise}}{run}$
Hence, slope = $\dfrac{20}{5}$ = 4 units.
Therefore, the slope of a line is 4 units.
3. Find the slope of the straight line y-1=5x+2
Solution:
We must solve the equation for y.
y = 5x+2+1
y = 5x+3
y = 5x+2+1
y = 5x+3
If we look at the point-slope equation of a line
y=mx+b
y=mx+b, we can conclude that the slope of this straight line is m=5
### Practise Questions
1. Find the slope of the line 2y-3x=5
a) $\dfrac{3}{2}$
b) $\dfrac{1}{2}$
c) $\dfrac{3}{4}$
d) $\dfrac{1}{3}$
2. Find the equation of the line that passes through the point (2,5) and its slope is m=3.
a) y=2x+1
b) y=3x-1
c) y=3x-2
d) y=2x-3
3) Find the slope of a straight line that passes through the points (1,3) and (-1,-1).
a) 3
b) 1
c) 2
d) 8
1) a
2) b
3) c
## Conclusion
Here in this article we have learnt about the slope. The slope of a line is defined as the change in y coordinate with respect to the change in x coordinate of that line. We also learned about the slope for different types of lines such as parallel lines,perpendicular lines. Along with these we also discussed slope concept for two more topics i.e, collinearity and angle between two lines. Hope this article has helped you!.
Competitive Exams after 12th Science
## FAQs on Slope of Line
1. What does an undefined slope mean?
When there is no change in x coordinates as y coordinates change, the graph of the line is vertical. You could not be able to calculate the slope of a line, as you would need to divide by 0. Therefore, the lines have an undefined slope.
2. State the Equation of a Line With Zero Slopes.
A line with a zero slope is a straight, perfectly flat line running along the horizontal axis of a cartesian plane. The equation of a zero line slope is one where the value of x may change but the value of y will always be constant. An equation for a zero line slope is represented as y = b. Here, ‘b’ is the y-intercept (i.e. the point where the line crosses the vertical y-axis). Also, the line’s slope is 0 (i.e. m = 0).
3. What are the Three Ways to Find Slope?
The three ways to find slope are slope-intercept form, point-intercept form, and Standard form.
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# 13.8 Probability
Skills to Develop
In this section, you will:
• Construct probability models.
• Compute probabilities of equally likely outcomes.
• Compute probabilities of the union of two events.
• Use the complement rule to find probabilities.
• Compute probability using counting theory.
Figure 13.8.1 An example of a “spaghetti model,” which can be used to predict possible paths of a tropical storm.1
Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one in Figure 13.8.1. They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions.
### Constructing Probability Models
Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is $$\{1,2,3,4,5,6 \}$$. An event is any subset of a sample space.
The likelihood of an event is known as probability. The probability of an event pp is a number that always satisfies $$0≤p≤1$$, where $$0$$ indicates an impossible event and $$1$$ indicates a certain event. A probability model is a mathematical description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like Table 13.8.1.
Outcome Probability
Winning the raffle 1%
Losing the raffle 99%
Table 13.8.1
The sum of the probabilities listed in a probability model must equal 1, or 100%.
How to: Given a probability event where each event is equally likely, construct a probability model.
1. Identify every outcome.
2. Determine the total number of possible outcomes.
3. Compare each outcome to the total number of possible outcomes.
Example
Constructing a Probability Model
Construct a probability model for rolling a single, fair die, with the event being the number shown on the die.
Solution:
Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space.
Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any particular face is more likely to show up than any other one, so the probability of rolling any number is16. 16.
Outcome Roll of 1 Roll of 2 Roll of 3 Roll of 4 Roll of 5 Roll of 6 Probability $$\dfrac{1}{6}$$ $$\dfrac{1}{6}$$ $$\dfrac{1}{6}$$ $$\dfrac{1}{6}$$ $$\dfrac{1}{6}$$ $$\dfrac{1}{6}$$
Q&A
Do probabilities always have to be expressed as fractions?
No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1.
# Computing Probabilities of Equally Likely Outcomes
Let $$S$$ be a sample space for an experiment. When investigating probability, an event is any subset of $$S$$. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in $$S$$. Suppose a number cube is rolled, and we are interested in finding the probability of the event “rolling a number less than or equal to 4.” There are 4 possible outcomes in the event and 6 possible outcomes in $$S$$, so the probability of the event is $$\dfrac{4}{6}=\dfrac{2}{3}$$.
A General Note: COMPUTING THE PROBABILITY OF AN EVENT WITH EQUALLY LIKELY OUTCOMES
The probability of an event $$E$$ in an experiment with sample space $$S$$ with equally likely outcomes is given by
$$P(E)=\dfrac{\text{number of elements in }E}{\text{number of elements in }S}=\dfrac{n(E)}{n(S)}$$
$$E$$ is a subset of $$S$$, so it is always true that $$0≤P(E)≤1$$.
Example
Computing the Probability of an Event with Equally Likely Outcomes
A number cube is rolled. Find the probability of rolling an odd number.
Solution:
The event “rolling an odd number” contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event.
$$P(E)=\dfrac{3}{6}=\dfrac{1}{2}$$
Exercise
A number cube is rolled. Find the probability of rolling a number greater than 2.
Solution:
$$\dfrac{2}{3}$$
### Computing the Probability of the Union of Two Events
We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events $$E$$ and $$F$$,written $$E\cup F$$, is the event that occurs if either or both events occur.
$$P(E\cup F)=P(E)+P(F)−P(E\cap F)$$
Suppose the spinner in Figure 13.8.2 is spun. We want to find the probability of spinning orange or spinning a $$b$$.
Figure 13.8.2
There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is $$\dfrac{3}{6}=\dfrac{1}{2}$$. There are a total of 6 sections, and 2 of them have a $$b$$. So the probability of spinning a $$b$$ is $$\dfrac{2}{6}=\dfrac{1}{3}$$. If we added these two probabilities, we would be counting the sector that is both orange and a $$b$$ twice. To find the probability of spinning an orange or a $$b$$, we need to subtract the probability that the sector is both orange and has a $$b$$.
$$\dfrac{1}{2}+\dfrac{1}{3}−\dfrac{1}{6}=\dfrac{2}{3}$$
The probability of spinning orange or a $$b$$ is $$\dfrac{2}{3}$$.
A General Note: PROBABILITY OF THE UNION OF TWO EVENTS
The probability of the union of two events $$E$$ and $$F$$ (written $$E\cup F$$) equals the sum of the probability of $$E$$ and the probability of $$F$$ minus the probability of $$E$$ and $$F$$ occurring together (which is called the intersection of $$E$$ and $$F$$ and is written as $$E\cap F$$).
$$P(E\cup F)=P(E)+P(F)−P(E\cap F)$$
Example
Computing the Probability of the Union of Two Events
A card is drawn from a standard deck. Find the probability of drawing a heart or a 7.
Solution:
A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is $$\dfrac{1}{4}$$. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing a 7 is $$\dfrac{1}{13}$$.
The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is $$\dfrac{1}{52}$$. Substitute $$P(H)=\dfrac{1}{4}$$, $$P(7)=\dfrac{1}{13}$$, and $$P(H\cap 7)=\dfrac{1}{52}$$ into the formula.
$$P(E\cup F)=P(E)+P(F)−P(E\cap F)$$
$$=\dfrac{1}{4}+\dfrac{1}{13}−\dfrac{1}{52}$$
$$=\dfrac{4}{13}$$
The probability of drawing a heart or a 7 is $$\dfrac{4}{13}$$.
Exercise
A card is drawn from a standard deck. Find the probability of drawing a red card or an ace.
Solution:
$$\dfrac{7}{13}$$
# Computing the Probability of Mutually Exclusive Events
Suppose the spinner in Figure 13.8.2 is spun again, but this time we are interested in the probability of spinning an orange or a $$d$$. There are no sectors that are both orange and contain a $$d$$, so these two events have no outcomes in common. Events are said to be mutually exclusive events when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is
$$P(E\cap F)=P(E)+P(F)$$
Notice that with mutually exclusive events, the intersection of $$E$$ and $$F$$ is the empty set. The probability of spinning an orange is $$\dfrac{3}{6}=\dfrac{1}{2}$$ and the probability of spinning a $$d$$ is $$\dfrac{1}{6}$$. We can find the probability of spinning an orange or a dd simply by adding the two probabilities.
$$P(E\cap F)=P(E)+P(F)$$
$$=\dfrac{1}{2}+\dfrac{1}{6}$$
$$=\dfrac{2}{3}$$
The probability of spinning an orange or a $$d$$ is $$\dfrac{2}{3}$$.
A General Note: PROBABILITY OF THE UNION OF MUTUALLY EXCLUSIVE EVENTS
The probability of the union of two mutually exclusive events $$E$$ and $$F$$ is given by
$$P(E\cap F)=P(E)+P(F)$$
How to: Given a set of events, compute the probability of the union of mutually exclusive events.
1. Determine the total number of outcomes for the first event.
2. Find the probability of the first event.
3. Determine the total number of outcomes for the second event.
4. Find the probability of the second event.
Example
Computing the Probability of the Union of Mutually Exclusive Events
A card is drawn from a standard deck. Find the probability of drawing a heart or a spade.
Solution:
The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is $$\dfrac{1}{4}$$, and the probability of drawing a spade is also $$\dfrac{1}{4}$$, so the probability of drawing a heart or a spade is
$$\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$$
Exercise
A card is drawn from a standard deck. Find the probability of drawing an ace or a king.
Solution:
$$\dfrac{2}{13}$$
### Using the Complement Rule to Compute Probabilities
We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event $$E$$, denoted $$E′$$, is the set of outcomes in the sample space that are not in $$E$$. For example, suppose we are interested in the probability that a horse will lose a race. If event $$W$$ is the horse winning the race, then the complement of event $$W$$ is the horse losing the race.
To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1.
$$P(E′)=1−P(E)$$
The probability of the horse winning added to the probability of the horse losing must be equal to $$1$$. Therefore, if the probability of the horse winning the race is $$\dfrac{1}{9}$$, the probability of the horse losing the race is simply
$$1−\dfrac{1}{9}=\dfrac{8}{9}$$
A General Note: THE COMPLEMENT RULE
The probability that the complement of an event will occur is given by
$$P(E′)=1−P(E)$$
Example
Using the Complement Rule to Calculate Probabilities
Two six-sided number cubes are rolled.
1. Find the probability that the sum of the numbers rolled is less than or equal to 3.
2. Find the probability that the sum of the numbers rolled is greater than 3.
Solution:
The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are $$6×6$$, or $$36$$ total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube.
$$1-1$$ $$1-2$$ $$1-3$$ $$1-4$$ $$1-5$$ $$1-6$$ $$2-1$$ $$2-2$$ $$2-3$$ $$2-4$$ $$2-5$$ $$2-6$$ $$3-1$$ $$3-2$$ $$3-3$$ $$3-4$$ $$3-5$$ $$3-6$$ $$4-1$$ $$4-2$$ $$4-3$$ $$4-4$$ $$4-5$$ $$4-6$$ $$5-1$$ $$5-2$$ $$5-3$$ $$5-4$$ $$5-5$$ $$5-6$$ $$6-1$$ $$6-2$$ $$6-3$$ $$6-4$$ $$6-5$$ $$6-6$$
1. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is
$$\dfrac{3}{36}=\dfrac{1}{12}$$
2. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3.
$$P(E′)=1−P(E)$$
$$=1−\dfrac{1}{12}$$
$$=\dfrac{11}{12}$$
Exercise
Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 10.
Solution:
$$\dfrac{5}{6}$$
### Computing Probability Using Counting Theory
Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems.
Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are $$C(5,2)$$ ways to select 2 phones that are not defective. There are 8 phones, so there are $$C(8,2)$$ ways to select 2 phones. The probability of selecting 2 phones that are not defective is:
$$\dfrac{\text{ways to select 2 phones that are not defective}}{\text{ways to select 2 phones}}=\dfrac{C(5,2)}{C(8,2)}$$
$$=\dfrac{10}{28}$$
$$=\dfrac{5}{14}$$
Example
Computing Probability Using Counting Theory
A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears.
1. Find the probability that only bears are chosen.
2. Find the probability that 2 bears and 3 dogs are chosen.
3. Find the probability that at least 2 dogs are chosen.
Solution:
1. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are $$C(6,5)$$ ways to choose 5 bears. There are 14 toys, so there are $$C(14,5)$$ ways to choose any 5 toys.
$$\dfrac{C(6,5)}{C(14,5)}=\dfrac{6}{2,002}=\dfrac{3}{1,001}$$
2. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are $$C(6,2)$$ ways to choose 2 bears. There are 5 dogs, so there are $$C(5,3)$$ ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are $$C(6,2)⋅C(5,3)$$ ways to choose 2 bears and 3 dogs. We can use this result to find the probability.
$$\dfrac{C(6,2)C(5,3)}{C(14,5)}=\dfrac{15⋅10}{2,002}=\dfrac{75}{1,001}$$
3. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen.
When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are $$C(9,5)$$ ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are $$C(14,5)$$ ways to choose the 5 toys from all of the toys.
$$\dfrac{C(9,5)}{C(14,5)}=\dfrac{63}{1,001}$$
If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are $$C(5,1)⋅C(9,4)$$ ways to choose 1 dog and 1 other toy.
$$\dfrac{C(5,1)C(9,4)}{C(14,5)}=\dfrac{5⋅126}{2,002}=\dfrac{315}{1,001}$$
Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen.
$$\dfrac{63}{1,001}+\dfrac{315}{1,001}=\dfrac{378}{1,001}$$
We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen.
$$1−\dfrac{378}{1,001}=\dfrac{623}{1,001}$$
Exercise
A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, and 2 green gumballs.
1. Find the probability that all 3 gumballs selected are purple.
2. Find the probability that no yellow gumballs are selected.
3. Find the probability that at least 1 yellow gumball is selected.
Solution:
1. $$\dfrac{1}{91}$$; 2. $$\dfrac{5}{91}$$;3. $$\dfrac{86}{91}$$
Media
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### Key Equations
probability of an event with equally likely outcomes $$P(E)=\dfrac{n(E)}{n(S)}$$ probability of the union of two events $$P(E\cup F)=P(E)+P(F)−P(E\cap F)$$ probability of the union of mutually exclusive events $$P(E\cup F)=P(E)+P(F)$$ probability of the complement of an event $$P(E')=1−P(E)$$
### Key Concepts
• Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain.
• The probabilities in a probability model must sum to 1. See Example.
• When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment. See Example.
• To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously. See Example.
• To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See Example.
• The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See Example.
• In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See Example. |
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