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: 6



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 $(y^{5})^{2}=y^{5\times2}=y^{10}$.
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