Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 5 - Trigonometric Functions - Chapter 5 Test Prep - Review Exercises - Page 559: 5

Answer

$9360^{\circ}$

Work Step by Step

Recall the conversions $360^{\circ}=1 \ \text{revolutions}$ and $1 \ min =60 \ second$ The angular speed can be computed as: $650 {\dfrac{\text {revolutions}}{\text{min}}} \times {\dfrac{360^{\circ}}{1 \ \text{revolution}}} \times \dfrac{1}{60} \dfrac{min}{sec}=\dfrac{3900^{\circ}}{sec}$ Therefore the number of degrees are: $=\dfrac{3900^{\circ}}{sec} \times 2.4 \sec =9360^{\circ}$
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