Question

y varies directly as x and inversely as the square of z. y = 16 when x = 72 and z = 6. Find ywhen x = 2 and z= 6.y =(Simplify your answer. Type an integer or a simplified fraction.)

Answer 1

`\begin{gathered} y\text{ }\propto x\text{ } \\ y\text{ }\propto\text{ }(1)/(z^2) \end{gathered}`
`\begin{gathered} y\propto x\propto(1)/(z^2) \\ y\text{ }\propto(x)/(z^2) \\ y\text{ = k }*\text{ }(x)/(z^2) \end{gathered}`

Where k is a constant

when y = 16 , x = 72 , 6

`\begin{gathered} 16\text{ = }(k*72)/(6^2) \\ 16\text{ }*36\text{ = 72k} \\ k\text{ = }(576)/(72) \\ k=\text{ 8} \end{gathered}`
`\begin{gathered} y\text{ = }(8x)/(z^2) \\ \text{when x =2 , z = 6} \\ y\text{ =}\frac{8\text{ }*2}{6^2}\text{ =}(16)/(36)\text{ = }(4)/(9) \\ \end{gathered}`

The value of y = 4/9 or 0.44