If y varies inversely as $x^2$ and y = 10 when x=2 , find y when x = 0.5?

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Answer 1 · 2016-07-02T21:44:32

$y =160$

Write it as an inverse proportion first

$yprop 1/x^2$

Change into an equation by multiplying by a constant.

$y = k/x^2$

Find the value of the constant

$k = x^2 xx y = 2^2 xx 10 = 40$

Use the constant in the equation

$y = 40/x^2$

Solve for y when x = 0.5

$y = 40/0.5^2 = 160$

If y varies inversely as x^2x^2 and y = 10 when x=2 , find y when x = 0.5?