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# HW2 sol - Math 2312 Homework 2 Summer 2010 Due Wednesday...

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Math 2312 Homework 2 Summer 2010 Due: Wednesday, June 9, 2010 at the beginning of class 1. Write the equation for the line passing through 4 3 , 8 7 and a) Parallel to 0 3 5 y x Solutions: Solving this equation for y, x y x y y x 3 5 5 3 0 3 5 We see that the slope is -5/3, so that any line parallel to this line will also have a slope of -5/3. Since we now have a point and a slope for the desired line, we can apply the point-slope form: 24 53 3 5 24 18 24 35 3 5 4 3 24 35 3 5 8 7 3 5 3 5 4 3 ) 8 7 ( 3 5 4 3 ) ( 1 1 x y x y x y x y x y x x m y y The last equation being the slope-intercept form. b) ( 7 pts) Perpendicular to 0 3 5 y x Solution: We see that the slope is -5/3, so that any line perpendicular to this line will have a slope of 3/5 (the negative of the reciprocal). Since we now have a point and a slope for the desired line, we can apply the point-slope form:

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40 9 5 3 40 30 40 21 5 3 4 3 40 21 5 3 8 7 5 3 5 3 4 3 ) 8 7 ( 5 3 4 3 ) ( 1 1 x y x y x y x y x y x x m y y 2. a) Given x x x f 2 ) ( 2 Evaluate: ) 2 ( f 3 1 f ) 2 ( x f Solution: Replacing every occurrence of x with ( ): () 2 () () 2 f We compute f
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HW2 sol - Math 2312 Homework 2 Summer 2010 Due Wednesday...

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