# HW 1 Calc - garcia(jjg2564 – HW 01 – gualdani –(56410 1 This print-out should have 20 questions Multiple-choice questions may continue on the

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Unformatted text preview: garcia (jjg2564) – HW 01 – gualdani – (56410) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Rationalize the numerator of √ x + 4 − √ x − 3 x . 1. 7 x ( √ x + 4 − √ x − 3) 2. x √ x + 4 + √ x − 3 3. 7 x √ x + 4 − √ x − 3 4. 7 x ( √ x + 4 + √ x − 3) correct 5. 1 x ( √ x + 4 + √ x − 3) Explanation: By the difference of squares, ( √ x + 4 − √ x − 3)( √ x + 4 + √ x − 3) = ( √ x + 4) 2 − ( √ x − 3) 2 = 7 . Thus, after multiplying both the numerator and the denominator in the given expression by √ x + 4 + √ x − 3 , we obtain 7 x ( √ x + 4 + √ x − 3) . 002 10.0 points Simplify the expression parenleftBig xy- 2 √ z parenrightBig 8 ÷ parenleftBig y 2 x 1 / 3 z- 2 parenrightBig 12 as much as possible, leaving no negative ex- ponents and no radicals. 1. x 12 z 28 y 40 2. x 12 y 40 z 28 correct 3. x 4 y 8 z 20 4. x 12 y 8 z 20 5. x 12 y 40 z 28 Explanation: By the Laws of Exponents parenleftBig xy- 2 √ z parenrightBig 8 = x 8 y 16 z 4 , while parenleftBig y 2 x 1 / 3 z- 2 parenrightBig 12 = y 24 z 24 x 4 . Consequently, the given expression can be rewritten as x 8 y 16 z 4 × x 4 y 24 z 24 = x 12 y 40 z 28 . 003 10.0 points Simplify the rational expression 4 x 6 x + 12 − 24 6 x 2 + 12 x + 2 x as much as possible. 1. 2 3 parenleftBig x + 3 x + 2 parenrightBig correct 2. 4 3 x parenleftBig x + 3 x + 2 parenrightBig 3. 2 3 parenleftBig x + 2 x + 3 parenrightBig 4. 4 3 x parenleftBig 2 x + 3 x + 4 parenrightBig 5. 2 3 x ( x + 2) garcia (jjg2564) – HW 01 – gualdani – (56410) 2 Explanation: Factoring and bringing to a common de- nominator we get 2 braceleftbigg 2 x 6( x + 2) − 12 6 x ( x + 2) + 1 x bracerightbigg = 2 braceleftbigg 2 x 2 − 12 + 6( x + 2) 6 x ( x + 2) bracerightbigg = 4 x 6 x parenleftbigg x + 3 x + 2 parenrightbigg . Thus after simplification the given rational expression becomes 2 3 parenleftbigg x + 3 x + 2 parenrightbigg . 004 10.0 points Write the expression ( x + 2) 1 / 5 − 1 3 x ( x + 2)- 4 / 5 as a single fraction containing only positive exponents. 1. 3 x + 6 ( x + 2) 4 / 5 2. 2 x + 6 3( x + 2) 4 / 5 correct 3. 3 x + 6 ( x + 2) 5 / 4 4. 2 x + 3 3( x + 2) 4 / 5 5. 2 x + 3 ( x + 2) 5 / 4 6. 2 x + 6 3( x + 2) 5 / 4 Explanation: Bringing the expression to a common de- nominator, we see that ( x + 2) 1 / 5 − 1 3 x ( x + 2)- 4 / 5 = 3( x + 2) − x 3( x + 2) 4 / 5 = 2 x + 6 3( x + 2) 4 / 5 . 005 10.0 points Simplify the expression f ( x ) = 5 + 20 x − 3 4 − 32 parenleftBig x x 2 − 9 parenrightBig as much as possible. 1. f ( x ) = x − 3 2 x + 9 2. f ( x ) = x + 3 x + 9 3. f ( x ) = x − 3 x + 9 4. f ( x ) = 5 4 parenleftBig x − 3 x − 9 parenrightBig 5. f ( x ) = 5 4 parenleftBig x + 3 2 x − 9 parenrightBig 6. f ( x ) = 5 4 parenleftBig x + 3 x − 9 parenrightBig correct Explanation: After bringing the numerator to a common...
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## This note was uploaded on 10/08/2010 for the course MATH 408K taught by Professor Gualdani during the Fall '09 term at University of Texas at Austin.

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HW 1 Calc - garcia(jjg2564 – HW 01 – gualdani –(56410 1 This print-out should have 20 questions Multiple-choice questions may continue on the

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