Answer
$$5.$$
Work Step by Step
Since $x=3$ and $y=-2$, then we have
$$\frac{2x^3-3y}{x y^2}=\frac{2(3)^3-3(-2)}{3 (-2)^2}\\
=\frac{2(27)+6}{3 (4)}=\frac{54+6}{12}=\frac{60}{12}=5.$$
Elementary Technical Mathematics
by Nelson, Robert; Eren, Dale;
Published by
Brooks Cole
ISBN 10:
1285199197
ISBN 13:
978-1-28519-919-1
Chapter 5 - Review - Page 232: 12
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