Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 3 - Radian Measure and the Unit Circle - Section 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 119: 86a

Answer

The temperature on March 1 is $1.0^{\circ}F$

Work Step by Step

March 1 is day 60, so we can use $x = 60$ in the equation: $T(x)=37~sin[\frac{2\pi}{365}~(x−101)]+25$ $T(60)=37~sin[\frac{2\pi}{365}~(60−101)]+25$ $T(60)=37~sin(-0.70578)+25$ $T(60)=37~[-sin(0.70578)]+25$ $T(60)=(37)(-0.64863)+25$ $T(60) = 1.0^{\circ}F$ The temperature on March 1 is $1.0^{\circ}F$
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