Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 8 - Section 8.1 - Finding Roots - Exercise Set - Page 574: 21



Work Step by Step

The given expression, $ \sqrt{144+25} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt{169} \\\\= \sqrt{(13)^2} \\\\= 13 .\end{array}
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