Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 5 - Secton 5.1 - The Unit Circle - 5.1 Exercises: 10

Answer

$$P(\frac{24}{25}, -\frac{7}{25})$$

Work Step by Step

Since $([], -\frac{7}{25})$ is an $(x, y)$ coordinate we can plug it into the equation of a circle, $x^{2}+y^{2}=1$. $$(x)^{2}+(-\frac{7}{25})^{2}=1$$ $$x^{2}+\frac{-7\times-7}{25\times25}=1$$ $$x^{2}+\frac{49}{625}=1$$ $$x^{2}=1-\frac{49}{625}$$ $$x^{2}=\frac{625}{625}-\frac{49}{625}$$ $$x^{2}=\frac{576}{625}$$ $$\sqrt (x^{2})=\sqrt (\frac{576}{625})$$ $$x=±\frac{24}{25}$$ Since the coordinate needs to be in the 4th quadrant, the x value needs to be positive. Therefore, $x=\frac{24}{25}$ and $P(-\frac{24}{25}, -\frac{7}{25})$
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