Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.1 Composite Functions - 5.1 Assess Your Understanding - Page 255: 63


$\frac{2\sqrt {100-p}}{25}+600$

Work Step by Step

$p=-\frac{1}{4}x+100\\p+\frac{1}{4}x=100\\\frac{1}{4}x=100-p\\x=400-4p$ Hence $C(x)=\frac{\sqrt x}{25}+600=\frac{\sqrt {400-4p}}{25}+600=\frac{2\sqrt {100-p}}{25}+600$
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