Precalculus (10th Edition)

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

Chapter 1 - Graphs - 1.3 Lines - 1.3 Assess Your Understanding - Page 32: 123

Answer

$F=\frac{9}{5}C+32$ or $C=\frac{5}{9}(F-32)$ $21.1^\circ C$

Work Step by Step

Step 1. Given $0^\circ C$ to $32^\circ F$ and $100^\circ C$ to $212^\circ F$, we have two points $(0,32), (100,212)$ Step 2. Find the slope of the line: $m=\frac{212-32}{100-0}=\frac{9}{5}$ Step 3. Using the first point, we have the equation: $F-32=\frac{9}{5}(C-0)$ or $F=\frac{9}{5}C+32$ or $C=\frac{5}{9}(F-32)$ Step 4. For $F=70^\circ F$, we have $C=\frac{5}{9}(70-32)\approx21.1^\circ C$
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