Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 27

Answer

$7y^{2}\sqrt y$

Work Step by Step

$\frac{\sqrt (98y^{6})}{\sqrt (2y)}=\sqrt \frac{98y^{6}}{2y}=\sqrt (49y^{5})=\sqrt (49\times y^{4}\times y)=\sqrt 49\times \sqrt(y^{4})\times \sqrt(y)=7y^{2}\sqrt y$ We know that $\sqrt 49=7$, because $7^{2}=49$. We know that $\sqrt (y^{4})=y^{2}$, because $(y^{2})^{2}=y^{2\times2}=y^{4}$
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