Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.2 - Exponents and Radicals - Exercises - Page 16: 60

Answer

$(\sqrt {x+9})^2 = x+9 $

Work Step by Step

By the formula on page 13, $(\sqrt {x+9})^2 = ((x+9)^{0.5})^2 = (x+9)^{0.5\times2} = (x+9)^{1} = x+9 $.
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