Y varies directly as x and inversely as the square of z. y=12 when x=64 and z=4. How do you find y when x=96 and z=2?

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Y varies directly as x and inversely as the square of z. y=12 when x=64 and z=4. How do you find y when x=96 and z=2?

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Answer 1 · 2017-07-29T18:15:13

$y = 72$ $y=72$

Explanation:

$the initial statement is y \propto \frac{x}{z^{2}}$ $"the initial statement is "ypropx/z^2$

$to convert to an equation multiply by k the constant$ $"to convert to an equation multiply by k the constant"$
$of variation$ $"of variation"$

$\Rightarrow y = k \times \frac{x}{z^{2}} = \frac{k x}{z^{2}}$ $rArry=kxx x/z^2=(kx)/z^2$

$to find k use the given condition$ $"to find k use the given condition"$

$y = 12 when x = 64 and z = 4$ $y=12" when "x=64" and "z=4$

$y = \frac{k x}{z^{2}} \Rightarrow k = \frac{y z^{2}}{x} = \frac{12 \times 16}{64} = 3$ $y=(kx)/z^2rArrk=(yz^2)/x=(12xx16)/64=3$

$"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(3x)/z^2)color(white)(2/2)|)))$

$"when "x=96" and "z=2$

$rArry=(3xx96)/4=72$

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Y varies directly as x and inversely as the square of z. y=12 when x=64 and z=4. How do you find y when x=96 and z=2?

1 Answer

Explanation:

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