Intermediate Algebra (12th Edition)

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

Chapter 7 - Review Exercises - Page 497: 24

Answer

-11

Work Step by Step

We are given that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=(\sqrt[n] a)^{m}$, if all indicated roots are real numbers. Therefore, $-121^{\frac{1}{2}}=-(121^{\frac{1}{2}})=-\sqrt[2] 121^{1}=-(\sqrt 121)=-(11)=-11$. We know that $\sqrt 121=11$, because $11^{2}=121$.
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