Explanation Given Information: The provided inequality equation is, ( x − 1 ) 2 + ( y + 2 ) 2 > 9 . Graph: Consider the inequality equation, ( x − 1 ) 2 + ( y + 2 ) 2 > 9 . The related equation to inequality is, ( x − 1 ) 2 + ( y + 2 ) 2 = 9 , which is circle of radius 3 with center at ( 1 , − 2 ) . First graph the related equation, ( x − 1 ) 2 + ( y + 2 ) 2 = 9 using the dashed curve as the points satisfying the equation ( x − 1 ) 2 + ( y + 2 ) 2 = 9 are not part of the solution to the given original inequality equation, ( x − 1 ) 2 + ( y + 2 ) 2 > 9 . The dashed curve is shown below, Two regions are made by the dashed curve, the inner region and the outer region of the circle. Now, select a test point from each region and test the point in original inequality equation, ( x − 1 ) 2 + ( y + 2 ) 2 > 9 . Test point, “Inside”: ( 0 , 0 ) ( x − 1 ) 2 + ( y + 2 ) 2 > 9 ( 0 − 1 ) 2 + ( 0 + 2 ) 2 ≥ ? 9 1 + 4 ≥ ? 9 5 ≥ ? 9 Which is false

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ISBN: 9781266148941

Author: Miller

Publisher: MCG CUSTOM

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Chapter 13.5, Problem 15PE

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To graph: The solution set of the inequality, (x−1)2+(y+2)2>9.

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Chapter 13.5, Problem 14PE

Chapter 13.5, Problem 16PE

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BEGINNING+INTERM.ALG.(LL) >CUSTOM PKG.<

Ch. 13.1 - Find the distance between the points ( − 4 , − 2 )...Ch. 13.1 - Prob. 2SP Ch. 13.1 - Prob. 3SP Ch. 13.1 - Prob. 4SP Ch. 13.1 - Prob. 5SP Ch. 13.1 - Prob. 6SP Ch. 13.1 - Prob. 7SP Ch. 13.1 - Prob. 8SP Ch. 13.1 - Prob. 1PE Ch. 13.1 - Prob. 2PE

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