## Calculus with Applications (10th Edition)

$\dfrac{1}{9}$
RECALL: $a^{-m} = \dfrac{1}{a^m}, a \ne0$ Use the rule above to have: $=-\left(-\dfrac{1}{(-3)^2}\right) \\=-\left(-\dfrac{1}{9}\right) \\=\dfrac{1}{9}$