Identifying Variable Relationships - Algebra II

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Question

varies directly with the square root of . If , then . What is the value of if ?

Answer

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

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

If , then ; therefore, the equation becomes

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

or

.

Divide by 5 to get , making the equation

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

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

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Question

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

Answer

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

1. Solve for when and .

$y=Kx^{3}$

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

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

2. Use your equation to solve for when .

$y=0.489x^{3}$

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

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Question

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

Answer

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

1. Solve for when and .

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

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

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

2. Use the equation you found to solve for when .

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

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

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Question

varies directly with $x^{2}$ . If , what is if ?

Answer

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:

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

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

$y=3x^{2}$

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

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Question

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

Answer

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:

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

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

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

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

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

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Question

varies directly with $\sqrt[3]{x}$ and when . What is when ?

Answer

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:

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

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

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

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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Question

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

Answer

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

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

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

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

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

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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