Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 2 - Section 2.5 - Linear Functions and Models - 2.5 Exercises - Page 197: 47

Answer

The change in horizontal distance is equal to 3.16 mi

Work Step by Step

Let $f(x)$ be the linear function that gives the value of vertical distance, where $x$ represents the horizontal distance. $$f(x) = ax = -\frac {6}{100}x$$ Substituting the value: $$-1000 \space ft = -\frac {6}{100}x$$ $$(-1000 \space ft)(-\frac{100} 6 ) = x$$ $$x = \frac{50000}{3} \space ft$$ - Since $1 \space mi = 5280 \space ft$ $$x = \frac{50000}{3} \space ft \times \frac{1 \space mi}{5280 \space ft} =3.16 \space mi$$
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