How to find inverse variation - Algebra

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Question

The number of days to construct a house varies inversely with the number of people constructing that house. If it takes 28 days to construct a house with 6 people helping out, how long will it take if 20 people are helping out?

Answer

The statement, 'The number of days to construct a house varies inversely with the number of people constructing that house' has the mathematical relationship $D=\frac{k}{P}$ , where D is the number of days, P is the number of people, and k is the variation constant. Given that the house can be completed in 28 days with 6 people, the k-value is calculated.

$28=\frac{k}{6}$

This k-value can be used to find out how many days it takes to construct a house with 20 people (P = 20).

$D=\frac{168}{P}$

$D=\frac{168}{20}$

So it will take 8.4 days to build a house with 20 people.

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Question

varies directly with , and inversely with the square root of .

If and , then .

Find if and .

Answer

The variation equation can be written as below. Direct variation will put in the numerator, while inverse variation will put $\sqrt{z}$ in the denominator. is the constant that defines the variation.

$x=k* \frac{y}{\sqrt{z}}$

To find constant of variation, , substitute the values from the first scenario given in the question.

$10=k* \frac{6}{\sqrt{64}}=k* \frac{6}{8}=k* \frac{3}{4}$

$k=10\div\frac{3}{4}}= \frac{40}{3}$

We can plug this value into our variation equation.

$x=\frac{40}{3}\cdot \frac{y}{\sqrt{z}}$

Now we can solve for given the values in the second scenario of the question.

$x=\frac{40}{3}* \frac{9}{\sqrt{144}}=\frac{40}{3}* \frac{9}{12}=10$

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Question

is a one-to-one function specified in terms of a set of $\left ( x,y \right )$ coordinates:

A = $\left { \left ( 0,0 \right ), \left ( 1,2\right ),\left ( 2,4 \right ), \left ( 3,6 \right )\right }$

Which one of the following represents the inverse of the function specified by set A?

B = $\left { \left ( 0,0 \right ), \left ( 0,1 \right ), \left ( 0,2 \right ), \left ( 0,3 \right ) \right }$

C = $\left { \left ( 0,0 \right ), \left ( 2,1 \right ), \left ( 4,2 \right ), \left ( 6,3 \right ) \right }$

D = $\left { \left ( 1,1 \right ), \left ( 2,1 \right ), \left ( -2,1 \right ), \left ( 3,1 \right ) \right }$

E = $\left { \left ( -2,1 \right ), \left ( -2,5 \right ), \left ( -2,3 \right ), \left ( -2, -2 \right ) \right }$

F = $\left { \left ( 2,0 \right ), \left ( 2,-3 \right ),\left ( 2,5 \right ), \left ( 2,-7 \right ) \right }$

Answer

The set A is an one-to-one function of the form

One can find $f^{-1}\left ( x \right )$ by interchanging the and coordinates in set A resulting in set C.

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Question

$f(x) = x^{3} -5$

Which one of the following functions represents the inverse of $f\left ( x \right )?$

A)

B) $\sqrt[3]{x^{3}-5}$

C) $\sqrt[3]{x+5}$

D) $x^{3 } + 5$

E) $x^{3} - 5$

Answer

Given $y = x^{3} - 5$

Hence $x^{3}=y+5$

Interchanging with we get:

$y^{3}=x+5$

Solving for results in $y=\sqrt[3]{x+5}$ .

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Question

Find the inverse of the following function:

$f(x) = x^{2}-3$

Answer

$f\left ( x \right ) = x^{2}-3$

which is the same as

$y = x^{2}-3$

If we solve for we get

$x^{2} = y +3$

Taking the square root of both sides gives us the following:

$x = \pm \sqrt{y+3}$

Interchanging and gives us

$y = \pm \sqrt{x + 3}$

Which is not one-to-one and therefore not a function.

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Question

Given

$f\left ( x \right ) = 2x+3$

and

$g\left ( x \right ) = \frac{x-3}{2}$ .

Find $f\left ( g\left ( x \right ) \right )$ .

Answer

Starting with $f\left ( x \right ) = 2x+3$

Replace with $g\left ( x \right )$ .

We get the following:

$2\left ( \frac{x-3}{2} \right )+3=x-3+3$

Which is equal to .

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Question

Given:

and

$g(x)=\frac{x-3}{2}$ .

Find .

Answer

Start with which is equal to

$\frac{x-3}{2}$

and then replace with . We get the following:

$\frac{(2x+3)-3}{2}$

which is equal to

$\frac{2x}{2}=x$

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Question

Which of the following is not a one-to-one function?

$f\left ( x \right ) = 2x + 5$

$f\left ( x \right ) = 2^{-x}$

$f\left ( x \right ) = x^{3} + 5$

$f\left ( x \right ) = \pm \sqrt{x+3}$

$f\left ( x \right ) = x^{5}$

Answer

Expression 4 is not even a function because for any value of , one gets two values of violating the definition of a function. If it is not a function, then it can not be an one-to-one function.

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Question

varies inversely as the square root of . If , then . Find if (nearest tenth, if applicable).

Answer

The variation equation is $A = \frac{K}{\sqrt{x}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{\sqrt{15}}$

$K = 45 \sqrt{15}$

The equation is now $A = \frac{45 \sqrt{15}}{\sqrt{x}}$ .

If , then

$A = \frac{45 \sqrt{15}}{\sqrt{30}} \approx 31.8$

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Question

varies inversely as the square of . If , then . Find if (nearest tenth, if applicable).

Answer

The variation equation is $A = \frac{K}{x^{2}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{15^{2}} = \frac{K}{225}$

$K = 45 \cdot225 = 10,125$

The equation is now $A = \frac{10,125}{x^{2}}$ .

If , then

$A = \frac{10,125}{30^{2}} = \frac{10,125}{900} \approx 11.3$

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Question

The current, in amperes, that a battery provides an electrical object is inversely proportional to the resistance, in ohms, of the object.

A battery provides 1.2 amperes of current to a flashlight whose resistance is measured at 20 ohms. How much current will the same battery supply to a flashlight whose resistance is measured at 16 ohms?

Answer

If is the current and is the resistance, then we can write the variation equation for some constant of variation :

$I = \frac{k}{R}$

or, alternatively,

To find , substitute :

$k=1.2 \cdot 20 = 24$

The equation is $I = \frac{24}{R}$ . Now substitute and solve for :

$I = \frac{24}{16} = 1.5 \textrm{ A}$

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Question

If is inversely proportional to and knowing that when , determine the proportionality constant.

Answer

The general formula for inverse proportionality for this problem is

$c=\frac{k}{d}$

Given that when , we can find by plugging them into the formula.

$3=\frac{k}{5}$

Solve for by multiplying both sides by 5

$3\cdot5=\frac{k}{5}\cdot5$

So .

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Question

The number of days needed to construct a house is inversely proportional to the number of people that help build the house. It took 28 days to build a house with 7 people. A second house is being built and it needs to be finished in 14 days. How many people are needed to make this happen?

Answer

The general formula of inverse proportionality for this problem is

$d=\frac{k}{p}$

where is the number of days, is the proportionality constant, and is number of people.

Before finding the number of people needed to build the house in 14 days, we need to find . Given that the house can be built in 28 days with 7 people, we have

$28=\frac{k}{7}$

Multiply both sides by 7 to find .

$28\cdot7=\frac{k}{7}\cdot7$

So . Thus,

$d=\frac{196}{p}$

Now we can find the how many people are needed to build the house in 14 days.

$14=\frac{196}{p}$

Solve for . First, multiply by on both sides:

$14\cdot p=\frac{196}{p}\cdot p$

Divide both sides by 14

$\frac{14p}{14}=\frac{196}{14}$

So it will take 14 people to complete the house in 14 days.

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Question

The amount of lonliness you feel varies inversely with the number of friends you have. If having 4 friends gives you a 10 on the lonliness scale, how much lonliness do you feel if you have 100 friends?

Answer

If lonliness and friends are inversely proportional, we can set up an equation to solve for some missing constant, k. To make things easier to write, let's use the variable L for the lonliness scale, and the variable F for how many friends you have. First we can just set up the equation:

$\small L = \frac{k}{F}$ . We know that having 4 friends gives you a 10 on the lonliness scale, so we can plug those values in to start solving for k:

$\small 10 = \frac{k}{4}$ to solve, multiply both sides by 4:

$\small 40 = k$ Now that we know the constant is 40, we can figure out how much lonliness, L, corresponds to having 100 friends by setting up a new formula. We are still generally using $\small L = \frac{k}{F}$ , and we know k is 40 and F is 100:

$\small L = \frac{40}{100}$

We can divide to get $\small L = 0.4$

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Question

The number of lights on in your room varies inversely with the number of monsters you think are under the bed. If you only have 1 light on in your room, you're pretty sure that there are 15 monsters under the bed. How many monsters do you suspect if you turn on 4 more lights \[for a total of 5\]?

Answer

If monsters and lights vary inversely, we can set up an equation to solve for some unknown constant k. To make things easier, we can use the variable M for monsters and L for the number of lights on:

$M = \frac{k}{L}$

In the first example, there are 15 monsters under the bed and 1 light on:

$15 = \frac{k}{1}$ It's pretty easy to see that our constant, k, is 15. To solve for M when there are 5 lights on, we can go back to our original $M = \frac{k}{L}$ and plug in 15 for k and 5 for L:

$\small M = \frac{15}{5 }$ Dividing gives us an answer of 3.

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Question

The number of math problems Ricky solves in half and hour varies inversely with the number of distractions present. Even though there were 12 distractions in his room, he was able to complete 10 problems in 30 minutes. If he needs to do 15 problems in the next half hour, how many distractions does he need to remove?

Answer

If the number of math problems varies inversely with the number of distractions, we can write an equation to solve for some constant, k. To make things easier, let's use the variable P for the number of math problems Ricky can solve, and D for the number of distractions. We can write this in a couple ways, but it will be easier later if we write it as:

$\small D = \frac{k}{P}$ . We could just as easily have written it as $\small P = \frac{k}{D}$ , and actually those equations are equivalent, or the same.

We know that Ricky can solve 10 problems with 12 distractions, so we can plug in 10=P and 12=D to solve for k:

$\small 12 = \frac{k}{10}$ to solve for k, multiply both sides by 10:

$\small 120 = k$ Now that we know k, we can figure out how many distractions Ricky can handle in order to solve 15 problems in the same amount of time. We'll do this by pluggin in 120 for k, and 15 for P:

$\small D = \frac{120}{15 }$

Dividing gives us an answer of $\small D=8$ . This means that he can only handle 8 distractions. The question is asking us how many he needs to remove. Right now he has 12, so to get only 8 he'd have to remove 4.

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Question

Find the inverse of the following algebraic function:

$f(x)=\frac{1}{x-3}$

Answer

To find the inverse, switch the placement of the and variables:

$f(x)=\frac{1}{x-3}$

$y=\frac{1}{x-3}$

$x=\frac{1}{y-3}$

Next, should be isolated, providing the inverse function:

$\frac{y-3}{1}\cdot x=\frac{1}{y-3}\cdot \frac{y-3}{1}$

$\frac{xy}{x}=\frac{1+3x}{x}$

$y=\frac{1+3x}{x}$

$y=\frac{1}{x}+3$

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Question

Find the inverse of:

Answer

The notation means a function in terms of . Replace this notation with .

Interchange the and variables.

Solve for . Subtract six on both sides.

Divide by three on both sides to isolate .

$y=\frac{x-6}{3} = \frac{1}{3}x-2$

The answer is: $y= \frac{1}{3}x-2$

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Question

If , what is $f^{-1}(x)$ ?

Answer

The notation $f^{-1}(x)$ is asking for the inverse of the function.

First, replace with .

Swap the variables.

Solve for . Subtract seven on both sides.

Divide by four on both sides.

$\frac{x -7}{4}= \frac{4y}{4}$

Simplify both sides.

The inverse function is: $y= \frac{1}{4}x-\frac{7}{4}$

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Question

Given , find $f^{-1}(x)$ .

Answer

To determine the inverse function, first replace the with .

Interchange the x and y variables.

Solve for y. Add nine on both sides.

Divide by two on both sides.

$\frac{x+9}{2}=\frac{2y}{2}$

Simplify both sides.

$y=\frac{x+9}{2} = \frac{1}{2}x+\frac{9}{2}$

The answer is: $y=\frac{1}{2}x+\frac{9}{2}$

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