Answer
$p=\dfrac{9}{8}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use $
p=k\cdot\dfrac{z^2}{r}
$ and solve for the value of $k$ with the given $
p,z,
$ and $
r
$ values. Then use the equation of variation to solve for the value of the unknown variable.
$\bf{\text{Solution Details:}}$
Since $p$ varies directly as the square of $z$ and inversely as $r,$ then $
p=k\cdot\dfrac{z^2}{r}
.$ Substituting the given values, $
p=\dfrac{32}{5},z=4,
$ and $
r=10
,$ then the value of $k$ is
\begin{array}{l}\require{cancel}
p=k\cdot\dfrac{z^2}{r}
\\\\
\dfrac{32}{5}=k\cdot\dfrac{4^2}{10}
\\\\
\dfrac{32}{5}=k\cdot\dfrac{16}{10}
\\\\
\dfrac{10}{16}\left(\dfrac{32}{5}\right)=\left(k\cdot\dfrac{16}{10}\right)\dfrac{10}{16}
\\\\
\dfrac{10}{\cancel{16}}\left(\dfrac{\cancel{32}^2}{5}\right)=k
\\\\
\dfrac{10}{\cancel{16}}\left(\dfrac{\cancel{32}^2}{5}\right)=k
\\\\
\dfrac{\cancel{10}^2}{1}\left(\dfrac{2}{\cancel5}\right)=k
\\\\
2(2)=k
\\\\
k=4
.\end{array}
Hence, the equation of variation is given by
\begin{array}{l}\require{cancel}
p=4\cdot\dfrac{z^2}{r}
.\end{array}
If $z=3,$ and $r=32,$ then
\begin{array}{l}\require{cancel}
p=4\cdot\dfrac{3^2}{32}
\\\\
p=4\cdot\dfrac{9}{32}
\\\\
p=\cancel4\cdot\dfrac{9}{\cancel{32}^8}
\\\\
p=\dfrac{9}{8}
.\end{array}