Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Test - Page 470: 11

Answer

$\dfrac{3\sqrt[]{y}}{y}$

Work Step by Step

Multiplying both the numerator and the denominator by a factor that will make the denominator a perfect power of the radical, the rationalized-denominator form of the given expression, $ \sqrt[]{\dfrac{9}{y}} ,$ is \begin{array}{l}\require{cancel} \sqrt[]{\dfrac{9}{y}\cdot\dfrac{y}{y}} \\\\= \sqrt[]{\dfrac{9}{y^2}\cdot y} \\\\= \sqrt[]{\left( \dfrac{3}{y} \right)^2\cdot y} \\\\= \dfrac{3}{y}\sqrt[]{y} \\\\= \dfrac{3\sqrt[]{y}}{y} .\end{array}
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