MATH 121 - WRITTEN EXERCISES-MODULE 3

# MATH 121 - WRITTEN EXERCISES-MODULE 3 - Jimmy...

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Jimmy Morgan-MAT-121-OL-(1 Nov 09 thru 23 Jan 2010) Module 3 Written Assignment 2.1 12.) P (-4, 3), Q(2, -5) D(P, Q) = [ ] ( 29 2 2 2 2 2 ( 4) 5 3 6 ( 8) - - + - - = + - = 36 64 100 10 + = =   M= ( 29 4 2 3 ( 5) , 1, 1 2 2 - + + - = - - ÷   a.) 10 and b.) (-1, -1) 26.) (-1, 4), (-2, -1), (1, 14) d(M, N) = [ ] 2 2 2 2 1 ( 2) (4-(-1)) (1) (5) 1 25 26 - - - + = + = + = d(M, Q) = [ ] 2 2 2 2 2 1) ( 1 14) ( 3) ( 15) 9 225 234 2 26 - - + - - = - + - = + = = d (N, Q) = [ ] 2 2 2 2 1 1) (4 14)) ( 2) ( 10) 4 100 104 2 26 - - + - = - + - = + = = Yes, the three points are collinear. 52.) 4 y x = - + y = -x - 4 y = -x - 4 y = -x - 4 y = - (-2) - 4 y = - (0) - 4 y = -(-4) - 4 y = 2 - 4 y = - 0 -4 y = 4 - 4 y = -2 y = - 4 y = 0 - 2 = - x – 4 0 = - x – 4 - 4 = - x - 4 2 1 1 x - = - - 4 1 1 x - = - - 0 1 1 x - = - - -2 = x - 4 = x 0 = x x y -2 -2 -4 0 0 -4

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Jimmy Morgan-MAT-121-OL-(1 Nov 09 thru 23 Jan 2010) Module 3 Written Assignment 2.2 8.) In exercise 8, a.) Find the center radius form of the equation of each circle, and b.) Graph it. center (-3, -2), radius 6; 2 2 2 ( ) ( ) x h y k r - + - = (h, k) =(-3, -2), r = 6 [ ] 2 2 2 ( 3) ( ( 2)) 6 x y - - + - - = a. 2 2 ( 3) ( 2) 36 x y + + + = b. 22.) 2 2 12 10 25 x y x y + - + = - ( 29 2 2 ( 12 ) 10 25 x x y y - + + = -          2 2 1 ( 12) ( 6) 36 2 - = - = and 2 2 1 (10) (5) 25 2 = = ( 29 2 2 ( 12 36) 10 25 25 25 36 x x y y - + + + + = - + + ( 29 ( 29 2 2 6 5 36 x y - + + = x = 6, y = -5 (since 36 0 > , the equation represents a circle with center at (6, - 5) and radius is 6.)
Jimmy Morgan-MAT-121-OL-(1 Nov 09 thru 23 Jan 2010) Module 3 Written Assignment 2.3 18.) Decide whether each relation defines a function and give the domain and range. Yes it does define a function and it’s domain is

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MATH 121 - WRITTEN EXERCISES-MODULE 3 - Jimmy...

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