Answer
$z=5$ is a solution.
Work Step by Step
The given value into the inequality. If the resullting statement is true, then the given value is a solution.
$\begin{align*}
20&\stackrel{}{\le}\frac{10}{2z}+20\\
20&\stackrel{?}{\le}\frac{10}{2(5)}+20\\
20&\stackrel{?}{\le}\frac{10}{10}+20\\
20&\stackrel{?}{\le}1+20\\
20&\stackrel{\checkmark}{\le}21\\
\end{align*}$
Thus, the given value is a solution.
