Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.1 Algebra Essentials - A.1 Assess Your Understanding - Page A11: 92

Answer

$6$

Work Step by Step

... recognize $36$ as $6^{2}$ $\sqrt{36}=\sqrt{6^{2}}$ ... Apply rule $\quad\sqrt[2]{a^{2}}=|a|$ $\sqrt{(6)^{2}}=|6|=6$
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