Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 9 - Section 9.3 - Cylinders and Cones - Exercises - Page 432: 27

Answer

1200$\pi$ cm$^3$

Work Step by Step

V=$\frac{1}{3}$Bh$_1$+$\frac{1}{3}$Bh$_2$ V=$\frac{1}{3}$(144$\pi$)(9)+$\frac{1}{3}$(144$\pi$)(16) V=432$\pi$+768$\pi$ V=1200$\pi$
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