## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 45: 8

#### Answer

$13$

#### Work Step by Step

RECALL: For non-negative real numbers a, $\sqrt{a^2}=a$. Simplify the radicand, then use the rule above: $\sqrt{169} \\=\sqrt{13^2} \\=13.$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.