Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 74: 8



Work Step by Step

RECALL: For $a\gt0$ and $b\gt0$, $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ Use the rule above to obtain: $=\dfrac{\sqrt{7}}{\sqrt{36}} \\=\dfrac{\sqrt{7}}{\sqrt{6^2}}$ Since $\sqrt{6^2}=6$, then the expression above simplifies to $=\color{blue}{\dfrac{\sqrt{7}}{6}}$
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