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 83: 78

Answer

$\dfrac {3}{8r}$

Work Step by Step

$\dfrac {3r^{3}-9r^{2}}{r^{2}-9}:\dfrac {8r^{3}}{r+3}=\dfrac {3r^{3}-9r^{2}}{r^{2}-9}\times \dfrac {r+3}{8r^{3}}=\dfrac {3r^{2}\left( r-3\right) }{\left( r-3\right) \left( r+3\right) }\times \dfrac {r+3}{8r^{3}}=\dfrac {3}{8r}$
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