Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - Chapter R Test Prep - Review Exercises - Page 84: 96

Answer

$ \frac{1}{p^2(m+n)}$

Work Step by Step

$\frac{[p^2(m+n)^3]^{-2}}{p^{-2}(m+n)^{-5}}=p^{-4}(m+n)^{-6}\times p^2(m+n)^5=p^{-2}(m+n)^{-1}=\frac{1}{p^2(m+n)}$
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