College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises: 54

Answer

$\displaystyle\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$

Work Step by Step

$A = 2\pi r^{2}+2\pi rh$ $2\pi r^{2}+2\pi hr-A = 0$ We use the quadratic formula: $r=\displaystyle \frac{-(2\pi h)\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-A)}}{2(2\pi)}=\frac{-2\pi h\pm\sqrt{4\pi^{2}h^{2}+8\pi A}}{4\pi}=\frac{-2\pi h\pm2\sqrt{\pi^{2}h^{2}+2\pi A}}{4\pi}=\frac{-\pi h\pm\sqrt{\pi^{2}h^{2}+2\pi A}}{2\pi}$
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