Question

y varies directly as x and inversely as the square of z. y = 16 when x = 32 and z = 4. Find ywhen x=3 and z= 6.y=(Simplify your answer. Type an integer or a simplified fraction.)

Answer 1

y varies directly as x and inversely as the square of z means:

$y=\frac{kx}{z^2}$

where k is some constant. Substituting with y = 16, x = 32, and z = 4, and solving for k:

$\begin{gathered} 16=\frac{k\cdot32}{4^2} \\ \frac{16\cdot4^2}{32}=k \\ 8=k \end{gathered}$

Substituting with k = 8, x = 3, and z = 6, and solving for y:

$\begin{gathered} y=\frac{8\cdot3}{6^2} \\ y=\frac{2}{3} \end{gathered}$