Answer
$\frac{4}{9}$
Work Step by Step
We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer).
Therefore, $3^{-2}+3^{-1}=\frac{1}{3^{2}}+\frac{1}{3}=\frac{1}{9}+\frac{1}{3}=\frac{1}{9}+\frac{1\times3}{3\times3}=\frac{1}{9}+\frac{3}{9}=\frac{1+3}{9}=\frac{4}{9}$.
