Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapter 9 Review Problem Set: 19

Answer

$-\dfrac{\sqrt{6}}{3}$

Work Step by Step

Using the properties of radicals, the given expression, $ \dfrac{-3\sqrt{2}}{\sqrt{27}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{-3\sqrt{2}}{\sqrt{9\cdot3}} \\\\= \dfrac{-3\sqrt{2}}{\sqrt{(3)^2\cdot3}} \\\\= \dfrac{-3\sqrt{2}}{3\sqrt{3}} \\\\= \dfrac{-\cancel{3}\sqrt{2}}{\cancel{3}\sqrt{3}} \\\\= \dfrac{-\sqrt{2}}{\sqrt{3}} \\\\= \dfrac{-\sqrt{2}}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\\\= \dfrac{-\sqrt{6}}{3} \\\\= -\dfrac{\sqrt{6}}{3} .\end{array}
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