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: 112

Answer

$\frac{23\sqrt 5}{30}$

Work Step by Step

As $45=3^2\times5$ and $80=4^2\times5$, we have $\frac{3}{\sqrt 5}-\frac{2}{\sqrt {45}}+\frac{6}{\sqrt {80}}=\frac{3}{\sqrt 5}-\frac{2}{3\sqrt {5}}+\frac{6}{4\sqrt {5}}=\frac{1}{\sqrt 5}(3-\frac{2}{3}+\frac{3}{2})=\frac{23}{6\sqrt 5}=\frac{23\sqrt 5}{30}$
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