Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 9 - Roots and Radicals - Chapter 9 Review Problem Set - Page 430: 3



Work Step by Step

Recall the following rule: $(a^{b})^{c}=a^{bc}$ Thus, we obtain: $\sqrt {1600}=\sqrt {40\times40}=\sqrt {40^{2}}=(40^{2})^{1/2}=40^{1}=40$
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