quiz-1-110114-solutions

# quiz-1-110114-solutions - former is much easier to evaluate...

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MATH 190–02 Quiz #1 Solutions Show all work for problems 3–5; use the back of the sheet if necessary. 1. (4 points) Answer the following form-conversion problems. (a) (2 points) Express the interval ( - 2 , 5] in terms of inequalities. The parenthesis indicates a non-inclusive inequality; the square bracket indicates an in- clusive inequality, so this interval can be ddescribed as - 2 < x 5. (b) (2 points) Express the inequality x > - 1 in interval notation. Since this interval does not include its lower bound, we would use a parenthesis for the left boundary. Since there is no bound on the right, we use the peculiar inﬁnite-boundary notation to give interval ( - 1 , ) or ( - 1 , + ). 2. (2 points) Evaluate the expression 4 - 3 / 2 . Since this has a negative exponent, we might start by rewriting it as 1 4 3 / 2 . This can be de- composed as 1 (4 1 / 2 ) 3 = 1 4 3 , or as 1 [ (4) 3 ] 1 / 2 = 1 4 3 . Both will give you the same answer, but the
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Unformatted text preview: former is much easier to evaluate: 1 √ 4 3 = 1 2 3 = 1 8 1 √ 4 3 = 1 √ 64 = 1 8 3. (3 points) Simplify the rational expression x x +2-3 2 x-1 . x x + 2-3 2 x-1 = x (2 x-1) ( x + 2)(2 x-1)-3( x + 2) ( x + 2)(2 x-1) = x (2 x-1)-3( x + 2) ( x + 2)(2 x-1) = 2 x 2-4 x-6 2 x 2 + 3 x-2 4. (3 points) Expand the polynomial 2 x 3-( x 2-x )( x + 2) . 2 x 3-( x 2-x )( x + 2) = 2 x 3-( x 3 + 2 x 2-x 2-2 x ) = 2 x 3-x 3-2 x 2 + x 2 + 2 x = x 3-x 2 + 2 x 5. (3 points) Using any method you like, solve the equation x 2-3 x = 6 . We start by writing the entire quadratic on a single side of the equation, as x 2-3 x-6 = 0. Then, using the quadratic formula: x =-(-3) ± p (-3) 2-4 · 1(-6) 2 = 3 ± √ 33 2 Page 1 of 1 Friday, January 14, 2011...
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