Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Review Exercises - Page 498: 59

Answer

$10y^{3}\sqrt y$

Work Step by Step

$\sqrt (10y^{7})=\sqrt (10\times y^{6}\times y)=\sqrt 10\times\sqrt y^{6}\times\sqrt y=10y^{3}\sqrt y$ We know that $\sqrt 100=10$, because $10^{2}=100$. We know that $\sqrt y^{6}=y^{3}$, because $(y^{3})^{2}=y^{3\times2}=y^{6}$.
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