If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

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If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

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Answer 1 · 2018-03-11T18:55:18

$y = \frac{21}{125}$ $y=21/125$

Explanation:

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

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

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

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

$y = \frac{1}{6} when x = 20 and z = 6$ $y=1/6" when "x=20" and "z=6$

$y = \frac{k x}{z^{2}} \Rightarrow k = \frac{y z^{2}}{x} = \frac{\frac{1}{6} \times 36}{20} = \frac{3}{10}$ $y=(kx)/z^2rArrk=(yz^2)/x=(1/6xx36)/20=3/10$

$"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(3x)/(10z^2))color(white)(2/2)|)))$

$"when "x=14" and "z=5" then"$

$y=(3xx14)/(10xx25)=21/125$

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If y varies directly as x and inversely as the square of z and y=1/6 when x=20 and z =6, how do you find y when x = 14 and z=5?

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Explanation:

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