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Algebra II : Variable Relationships

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Example Question #1 : How To Find Direct Variation

varies directly with the square root of . If x = 25 , then y = 48 . What is the value of if x = 81 ?

Possible Answers:

86.4

155.52

$26\tfrac{2}{3}$

None of these answers are correct.

491.52

Correct answer:

86.4

Explanation:

If varies directly with the square root of , then for some constant of variation ,

$\small y = K \sqrt{x}$

If x = 25 , then y = 48 ; therefore, the equation becomes

$\small \small 48 = K \sqrt{25}$ ,

5K = 48 .

Divide by 5 to get K = 9.6 , making the equation

$\small \small y = 9.6 \sqrt{x}$ .

If x = 81 , then $\small \small \small y = 9.6 \sqrt{81} = 9.6 \cdot 9 = 86.4$ .

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Example Question #1 : Identifying Variable Relationships

If varies directly with $x^{3}$ and when x=5 , y=61.125 due to the effect of a constant, what is the value of when x=11 ?

Possible Answers:

y=650.859

y=134.475

y=1,331

y=43.331

y=672.375

Correct answer:

y=650.859

Explanation:

Since varies directly with $x^{3}$ , $y=Kx^{3}$ where is a constant.

1. Solve for when x=5 and y=61.125 .

$y=Kx^{3}$

$61.125=K(5^{3})$

$K=\frac{61.125}{125}=0.489$

2. Use your equation to solve for when x=11 .

$y=0.489x^{3}$

$y=0.489(11^{3})=650.859$

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Example Question #2 : Identifying Variable Relationships

If varies indirectly with $\sqrt{x}$ and when x=4 , y=12.3 due to the effect of a constant, what is the value of when x=17 ?

Possible Answers:

y=5.97

y=25.36

y=50.71

y=16.49

y=0.24

Correct answer:

y=5.97

Explanation:

Since varies indirectly with $\sqrt{x}$ , $y=\frac{K}{\sqrt{x}}$

1. Solve for when y=12.3 and x=4 .

$y=\frac{K}{\sqrt{x}}$

$12.3=\frac{K}{\sqrt{4}}$

$12.3= \frac{K}{2}$

K=24.6

2. Use the equation you found to solve for when x=17 .

$y=\frac{24.6}{\sqrt{x}}$

$y=\frac{24.6}{\sqrt{17}}=5.97$

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Example Question #1 : Variable Relationships

varies directly with $x^{2}$ . If x=5, y=75 , what is if x=10 ?

Possible Answers:

300

150

250

100

Correct answer:

300

Explanation:

1. Since varies directly with $x^{2}$ :

$y=Kx^{2}$ with K being some constant.

2. Solve for K using the x and y values given:

$75=K(5^{2})$

75=K(25)

K=3

3. Use the equation you solved for to find the value of y:

$y=3x^{2}$

$y=3(10^{2})$

y=300

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Example Question #4 : Identifying Variable Relationships

varies inversely with $x^{3}$ . If x=2, y=2 , then what is equal to when x=5 ?

Possible Answers:

0.654

0.128

0.226

0.197

Correct answer:

0.128

Explanation:

1. Since varies indirectly with $x^{3}$ :

$y=\frac{K}{x^{3}}$

2. Use the given x and y values to determine the value of K:

$2=\frac{K}{2^{3}}$

$2=\frac{K}{8}$

K=16

3. Using the equation along with the value of K, find the value of y when x=5:

$y=\frac{16}{x^{3}}$

$y=\frac{16}{5^{3}}$

y=0.128

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Example Question #5 : Identifying Variable Relationships

varies directly with $\sqrt[3]{x}$ and when x=8, y=3 . What is when x=27 ?

Possible Answers:

4.5

5.5

Correct answer:

4.5

Explanation:

1. Since varies directly with $\sqrt[3]{x}$ :

$y=K\sqrt[3]{x}$

2. Use the values given for x and y to solve for K:

$y=K\sqrt[3]{x}$

$3=K\sqrt[3]{8}$

3=K(2)

$K=\frac{3}{2}$

3. Use your new equation with the K you solved for to solve for y when x=27:

$y=(\frac{3}{2})\sqrt[3]{27}$

$y=(\frac{3}{2})(3)=4.5$

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Example Question #6 : Identifying Variable Relationships

varies inversely with $\sqrt{x}$ . When $x=4,\ y=0.5$ . What is the value of when x=100 ?

Possible Answers:

$\frac{1}{20}$

$\frac{1}{8}$

$\frac{1}{10}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{10}$

Explanation:

1. Since y varies indirectly with $\sqrt{x}$ :

$y=\frac{K}{\sqrt{x}}$

2. Solve for K using the x and y values given:

$0.5=\frac{K}{\sqrt{4}}$

$0.5=\frac{K}{2}$

K=1

3. Using the equation you created by solving for K, find y when x=100:

$y=\frac{1}{\sqrt{x}}$

$y=\frac{1}{\sqrt{100}}$

$y=\frac{1}{10}$

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Example Question #1 : Interpolations

Given the two following points, use interpolation to determine the best estimate for the value f(24.75)

(24,81) , (25,93)

Possible Answers:

91.5

82.5

Correct answer:

Explanation:

Using our two known points, we can use interpolation to determine the value at any point between them with the following formula:

$\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}$

Where (x_1,y_1) is our first given point, (x_2,y_2) is our second given point, and (x,y) is the point we want to find. We know our two given points, as well as the x value of our unknown point, so now all we must do is plug in all of our known values and solve for y, our only unknown:

$\frac{93-81}{25-24}=\frac{y-81}{24.75-24}$

$12=\frac{y-81}{0.75}$

$y-81=9\rightarrow y=f(24.75)=90$

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Example Question #2 : Interpolations

The output of a factory in units per day versus the number of employees working is plotted on the graph below, with the following data points collected:

(Workers, Units of output per day):

$\left(100,1300\right)$

$\left(75,1050\right)$

$\left(200,2300\right)$

$\left(225,2550\right)$

Assuming a linear relationship, interpolate to find how many units will be made per day if 110 workers are present.

Linear_interp

Possible Answers:

$1375\textup{ units}$

$1400\textup{ units}$

$1370\textup{ units}$

$1380\textup{ units}$

$1335\textup{ units}$

Correct answer:

$1400\textup{ units}$

Explanation:

We want to do a linear interpolation since the relationship between workers and units can be assumed to be linear. This means there is a constant slope between the points, so the slope between two known points will be equal to the slope between the point we are trying to find and some known point. This is expressed in the relation:

$\frac{y-y_0}{x-x_0}=\frac{y_1-y_0}{x_1-x_0}$ ,

where and are the points we want to find and (x_1,y_1) and (x_0,y_0) are known. We choose the known points to be those that are just to the left and right of the point we are trying to find,

(100,1300) and (200,2300) .

Plugging these into our interpolation formula and knowing x=110 , we can find , the units output per day.

$\frac{y-1300}{110-100}=\frac{2300-1300}{200-100}$ .

Simplifying and rearranging to solve for :

y=1400 .

So there are 1400 units produced when the number of workers is 110 .

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Example Question #3 : Interpolations

Given the points $\left ( 30,51\right )$ and $\left ( 20,36\right )$ , use linear interpolation to find the value of when x=26.5 .

Possible Answers:

42.38

44.60

45.75

49.99

Correct answer:

45.75

Explanation:

Use the formula for interpolation to determine the value of y:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{x-x_{1}}$

We will use (30, 51) as our x2 and y2 and (20, 36) as our x1 and y1 and we will solve for y using 26.5 for x.

$\frac{51-36}{30-20}=\frac{y-36}{26.5-20}$

$\frac{15}{10}=\frac{y-36}{6.5}$

$\frac{3}{2}(6.5)=y-36$

y=45.75

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Algebra II : Variable Relationships

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1 : How To Find Direct Variation

Example Question #1 : Identifying Variable Relationships

Example Question #2 : Identifying Variable Relationships

Example Question #1 : Variable Relationships

Example Question #4 : Identifying Variable Relationships

Example Question #5 : Identifying Variable Relationships

Example Question #6 : Identifying Variable Relationships

Example Question #1 : Interpolations

Example Question #2 : Interpolations

Example Question #3 : Interpolations

All Algebra II Resources

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