Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

Chapter 8 - Geometry - 8.8 Pythagorean Theorem - 8.8 Exercises - Page 594: 46

Answer

The distance from his current and initial position is $\approx 16.6$ miles.
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Work Step by Step

1. Find the distance between his current position and his initial position Let $x =$ the distance $ a^{2} + b^{2} = x^{2}$ $x = \sqrt(a^{2} + b^{2})$ $x = \sqrt(15^{2} + 7^{2})$ $x = \sqrt (274)$ $x = 16.5529...$ miles $x \approx 16.6$ miles
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