# If Y varies inversely as x^2. If y=2 when x=10, how do you find y when x=5?

Hence y varies inversely as x^2 then there is a constant $k$ such as

$y = \frac{k}{x} ^ 2$

Hence when $x = 10$ then $y = 2$ the constant $k$ is

$k = y \cdot {x}^{2} = 2 \cdot {10}^{2} = 200$

So when x=5 then

$y = \frac{k}{x} ^ 2 = \frac{200}{5} ^ 2 = \frac{200}{25} = 8$