Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Study Summary - Practice Exercises: 9

Answer

$\dfrac{\sqrt{6xy}}{3y}$

Work Step by Step

Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{\dfrac{2x}{3y}} \\\\= \sqrt{\dfrac{2x}{3y}\cdot\dfrac{3y}{3y}} \\\\= \sqrt{\dfrac{1}{(3y)^2}\cdot6xy} \\\\= \dfrac{1}{3y}\sqrt{6xy} \\\\= \dfrac{\sqrt{6xy}}{3y} .\end{array}
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