College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 48: 11

Answer

$13$

Work Step by Step

As written on page 37, the principle square root of $a^{2}$ is the absolute value of $a$. $\sqrt{(-13)^{2}}$ $=\sqrt{169}$ $=13$
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