Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 5

Answer

$-\dfrac{1}{9}$

Work Step by Step

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