s-test1-1510w10

# s-test1-1510w10 - Math 1510 Test#1 solutions(1(8pts Let A...

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Math 1510, Test #1, October 19, 2010, solutions (1) (8pts) Let A = [ - 2 , 3) and B = [ - 4 , 1]. Find the indicated set. Graph the indicated set. (a) A B = [ - 2 , 1) (b) A B = [ - 4 , 3] (2) (8pts) Simplify the expression and remove any negative exponents. ( a 3 b - 2 ) 2 ( a - 3 b 2 ) 2 · ( b 3 a - 2 ) 3 ( b - 3 a 2 ) 3 = a 6 b - 4 a - 6 b 4 · b 9 a - 6 b - 9 a 6 = b 10 (3) (12pts) Factor the expressions completely. (a) x 2 + x - 6 = ( x + 3)( x - 2) (b) x 4 + x 2 - 6 Solution. Use (a) to factor ( x 2 ) 2 + x 2 - 6 = ( x 2 + 3)( x 2 - 2). This is ( x 2 + 3)( x - 2)( x + 2). (c) x 5 / 2 + x 3 / 2 - 6 x 1 / 2 = x 1 / 2 ( x 2 + x - 6) = x 1 / 2 ( x + 3)( x - 2). (4) (10pts) Add the fractions and simplify. x x + 2 + 1 x - 2 = x ( x - 2) ( x + 2)( x - 2) + x + 2 ( x + 2)( x - 2) = x 2 - 2 x + x + 2 x 2 - 4 = x 2 - x + 2 x 2 - 4 (5) (12pts) Solve the equation. (a) x 2 + 3 x - 4 = 0 Solution. Using the quadratic formula or factorization, one gets solutions x 1 = - 4 and x 2 = 1. (b) 1 + 3

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## This note was uploaded on 02/13/2011 for the course CSE/MATH 1510 taught by Professor Mixprofs during the Winter '09 term at York University.

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s-test1-1510w10 - Math 1510 Test#1 solutions(1(8pts Let A...

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