Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Review Exercises - Page 798: 24

Answer

$(2,2\sqrt 3)$
1584192872

Work Step by Step

Step 1. See plot; move to $\theta=\frac{4\pi}{3}$ and use $r=-4$ to get the point. Step 2. Using the formulas $x=r\ cos\theta$ and $y=r\ sin\theta$, we have $x=-4cos\frac{4\pi}{3}=2$ and $y=-4sin\frac{4\pi}{3}=2\sqrt 3 $, giving a point $(2,2\sqrt 3)$
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