Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Appendix A - A.3 - Exercise Set: 2

Answer

Volume of the sphere is exactly $36π$ mi.$^{3}$ or approximately $113\frac{1}{7}$ mi.$^{3}$ Surface area of the sphere is exactly $36π$ mi.$^{2}$ or approximately $113\frac{1}{7}$ mi.$^{2}$

Work Step by Step

Let r = $3$ mi. $V$ = $\frac{4}{3}$π$r^{3}$ $V$ = $\frac{4}{3}$ π$($3$ mi.)^{3}$ $V$ = $\frac{4}{3}$ π$($27$ mi.)^{3}$ $V$ = $36π$ mi.$^{3}$ $V$ = $36$ times$\frac{22}{7}$ mi.$^{3}$ $V$ = $\frac{792}{7}$ or $113\frac{1}{7}$ mi.$^{3}$ Volume of the sphere is exactly $36π$ mi.$^{3}$ or approximately $113\frac{1}{7}$ mi.$^{3}$ $SA$ = ${4}$ π $r^{2}$ $SA$ = ${4}$ π $(3$mi.)$^{2}$ $SA$ = ${4}$ π ($9$ mi.)$^{2}$ $SA$ = $36π$ mi.$^{2}$ $SA$ = ${36}$ times $\frac{22}{7}$ mi.$^{2}$ $SA$ = $\frac{792}{7}$ or $113\frac{1}{7}$ mi.$^{2}$ Surface area of the sphere is exactly $36π$ mi.$^{2}$ or approximately $113\frac{1}{7}$ mi.$^{2}$
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