Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - Cumulative Review - Page 849: 45



Work Step by Step

The volume of the cone is:$ \frac{1}{3}\pi r^2h$, $h=r+4$ Thus our equation is: $\frac{1}{3}\pi r^2(r+4)=75\pi\\r^2(r+4)=225\\(r-5)(r^2+9r+45)=0$ The second term has a negative determinant and thus never equals 0. Thus, the only valid zero for the function is $r=5$. Hence the dimensions are $r=5,h=9$.
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