Y is directly proportional to x and inversely proportional to the square of z and y = 40 when x = 80 and z = 4, how do you find y when x = 7 and z = 16?

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Y is directly proportional to x and inversely proportional to the square of z and y = 40 when x = 80 and z = 4, how do you find y when x = 7 and z = 16?

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Answer 1 · 2015-07-16T16:18:04

$y = \frac{7}{32}$ $y=7/32$ when $x = 7$ $x=7$ and $z = 16$ $z=16$

Explanation:

$y$ $y$ being directly proportional to $x$ $x$ and inversely proportional to the square of $z$ $z$ means there is a constant $k$ $k$ such that $y = k \frac{x}{z^{2}} = \frac{k x}{z^{2}}$ $y=kx/z^2=(kx)/z^2$ .

Since $y = 40$ $y=40$ when $x = 80$ $x=80$ and $z = 4$ $z=4$ , it follows that $40 = \frac{80 k}{4^{2}} = 5 k$ $40=(80k)/4^2=5k$ which implies $k = 8$ $k=8$ .

Therefore, $y = \frac{8 x}{z^{2}}$ $y=(8x)/z^2$ .

Hence, when $x = 7$ $x=7$ and $z = 16$ $z=16$ , $y = \frac{56}{16^{2}} = \frac{7}{2 \cdot 16} = \frac{7}{32}$ $y=56/16^2=7/(2*16)=7/32$ .

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Y is directly proportional to x and inversely proportional to the square of z and y = 40 when x = 80 and z = 4, how do you find y when x = 7 and z = 16?

1 Answer

Explanation:

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