Elementary Technical Mathematics

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

Chapter 5 - Review - Page 232: 12



Work Step by Step

Since $x=3$ and $y=-2$, then we have $$\frac{2x^3-3y}{x y^2}=\frac{2(3)^3-3(-2)}{3 (-2)^2}\\ =\frac{2(27)+6}{3 (4)}=\frac{54+6}{12}=\frac{60}{12}=5.$$
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