Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises - Page 90: 109

Answer

[68,86]

Work Step by Step

The formula provided in example 9: C=$\frac{5}{9}$(F-32) 1. Substitute the formula into the given range: 20$\leq$$\frac{5}{9}$(F-32)$\leq$30 2. Distribute: 20$\leq$$\frac{5}{9}$F-$\frac{160}{9}$$\leq$30 3. Add $\frac{160}{9}$ to each section of the inequality. $\frac{340}{9}$$\leq$$\frac{5}{9}$F$\leq$$\frac{430}{9}$ 4. Divide each section by $\frac{5}{9}$, which is the same as multiplying by $\frac{9}{5}$ $\frac{3060}{45}$$\leq$F$\leq$$\frac{3870}{9}$ 5. Simplify 68$\leq$F$\leq$86 The solution interval is greater than or equal to 68, and less than or equal to 86. Use [] to denote that it's 'or equal to'.
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