## Intermediate Algebra (6th Edition)

$9^{-\frac{1}{2}}=\frac{1}{(9)^{\frac{1}{2}}}=\frac{1}{\sqrt 9}=\frac{1}{3}$ $\frac{1}{3}$ is greater than 0, so it is a positive number We know that $\sqrt 9=3$, because $3^{2}=9$