Answer
G. 4
Work Step by Step
As shown on page 445, we know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$.
Therefore, $(-32)^{\frac{2}{5}}=\sqrt[5] (-32)^{2}=(\sqrt[5] -32)^{2}=(-2)^{2}=-2\times-2=4$.
We know that $\sqrt[5] -32=-2$, because $(-2)^{5}=-32$.
