Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 5 - Section 5.1 - Fundamental Operations - Exercise - Page 214: 42



Work Step by Step

Since $x=-1$, $y=2$, $z=-3$ We take into account the order of operations: Parenthesis, Exponents, Multiplication/Division, Addition/Subtraction Thus, we have: $$\frac{(3x^2+2)^2-y^2}{6-3x^2y^2}=\frac{(3(-1)^2+2)^2-2^2}{6-3(-1)^22^2}\\ =\frac{(3+2)^2-4}{6-3(4)}=\frac{5^2-4}{6-12}=\frac{21}{-6}=-\frac{7}{2} .$$
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