## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 6

#### Answer

$2y^{5}$

#### Work Step by Step

$\sqrt (4y^{10})=\sqrt 4\times\sqrt(y^{10})=2y^{5}$ We know that $\sqrt 4=2$, because $2^{2}=4$. Also, we know that $\sqrt(y^{10})=y^{5}$, because $(y^{5})^{2}=y^{5\times2}=y^{10}$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.