Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.1 - Algebraic Expressions, Mathematical Models, and Real Numbers - Exercise Set - Page 18: 119

Answer

$\frac{-1}{2}$

Work Step by Step

$\frac{(5 - 6)^2 - 2 |3 - 7|}{89 - 3*5^2}=\frac{(-1)^2 - 2 |-4|}{89 - 3*25}=\frac{1 - 2 (4)}{89 - 75}=\frac{1 - 8}{14}=\frac{-7}{14}=\frac{-1}{2}$
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