# Y varies directly as x and inversely as the square of z. y=10 when x=80 and z=4. How do you find y when x=36 and z=2?

Sep 1, 2016

$y = 18$

#### Explanation:

As $y$ varies directly as $x$, we have $y \propto x$. Also it varies inversely as square of $z$, which means $y \propto \frac{1}{z} ^ 2$.

Hence, $y \propto \frac{x}{z} ^ 2$ or

y=k×x/z^2, where $k$ is a constant.

Now when $x = 80$ and $z = 4$, $y = 10$, so

10=k×80/4^2=k×80/16=5k

Hence $k = \frac{10}{5} = 2$ and $y = 2 \frac{x}{z} ^ 2$.

So when $x = 36$ and $z = 2$,

y=2×36/2^2=72/4=18