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 - Concept and Vocabulary Check - Page 16: 6

Answer

a. $|1-\sqrt 2|$ =$ -(1-\sqrt 2) $ = $\sqrt 2 - 1$ b. $|\pi-3|$ = $\pi-3$ c. $|x|/x$ = 1

Work Step by Step

a. The number inside the absolute value bars, $|1-\sqrt 2|$ , is negative. Because $\sqrt 2$ $\approx$ 1.414.$ (1-1.414) $ = -0.414. The absolute value of negative number is negative. Thus $|1-\sqrt 2|$ =$ -(1-\sqrt 2) $ =$\sqrt 2 - 1$ b. The number inside the absolute value bars is positive. Because $\pi\approx3.14$. Thus |3.14 - 3| $\gt$ 0. The absolute value of a positive number is the number itself. So, $|\pi-3|$ = $\pi-3$ c. If x$\gt$0, then $|x|$ = x. Thus, $|x|/x$ = $x/x$ = 1
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