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Homework 1

# Homework 1 - vu(tv2894 Homework 1(Quest miner(55096 This...

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vu (tv2894) – Homework 1 (Quest) – miner – (55096) 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) correct 2. 7 x x + 4 x 3 3. x x + 4 + x 3 4. 7 x ( x + 4 x 3) 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 - 3 z parenrightBig 6 ÷ parenleftBig y 2 x 1 / 2 z - 2 parenrightBig 4 as much as possible, leaving no negative ex- ponents and no radicals. 1. x 8 y 26 z 11 2. x 8 y 10 z 5 3. x 8 z 11 y 26 4. x 8 y 26 z 11 correct 5. x 4 y 10 z 5 Explanation: By the Laws of Exponents parenleftBig xy - 3 z parenrightBig 6 = x 6 y 18 z 3 , while parenleftBig y 2 x 1 / 2 z - 2 parenrightBig 4 = y 8 z 8 x 2 . Consequently, the given expression can be rewritten as x 6 y 18 z 3 × x 2 y 8 z 8 = x 8 y 26 z 11 . 003 10.0 points Simplify the rational expression 8 x 10 x + 20 80 10 x 2 + 20 x + 4 x as much as possible. 1. 4 5 parenleftBig x + 2 x + 5 parenrightBig 2. 4 5 x ( x + 4) 3. 4 5 parenleftBig x + 5 x + 2 parenrightBig correct 4. 8 5 x parenleftBig 2 x + 5 x + 4 parenrightBig 5. 8 5 x parenleftBig x + 5 x + 2 parenrightBig

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vu (tv2894) – Homework 1 (Quest) – miner – (55096) 2 Explanation: Factoring and bringing to a common de- nominator we get 4 braceleftbigg 2 x 10( x + 2) 20 10 x ( x + 2) + 1 x bracerightbigg = 4 braceleftbigg 2 x 2 20 + 10( x + 2) 10 x ( x + 2) bracerightbigg = 8 x 10 x parenleftbigg x + 5 x + 2 parenrightbigg . Thus after simplification the given rational expression becomes 4 5 parenleftbigg x + 5 x + 2 parenrightbigg . 004 10.0 points Write the expression ( x + 2) 1 / 6 1 4 x ( x + 2) - 5 / 6 as a single fraction containing only positive exponents. 1. 3 x + 8 4( x + 2) 5 / 6 correct 2. 3 x + 4 ( x + 2) 6 / 5 3. 4 x + 8 ( x + 2) 5 / 6 4. 4 x + 8 ( x + 2) 6 / 5 5. 3 x + 4 4( x + 2) 5 / 6 6. 3 x + 8 4( x + 2) 6 / 5 Explanation: Bringing the expression to a common de- nominator, we see that ( x + 2) 1 / 6 1 4 x ( x + 2) - 5 / 6 = 4( x + 2) x 4( x + 2) 5 / 6 = 3 x + 8 4( x + 2) 5 / 6 . 005 10.0 points Simplify the expression f ( x ) = 3 + 15 x 4 1 15 parenleftBig x x 2 16 parenrightBig as much as possible. 1. f ( x ) = 3( x + 4) x 16 correct 2. f ( x ) = 3( x + 4) 2 x 16 3. f ( x ) = x 4 x + 16 4. f ( x ) = x 4 2 x + 16 5. f ( x ) = x + 4 x + 16 6. f ( x ) = 3( x 4) x 16 Explanation: After bringing the numerator to a common denominator it becomes 3 x 12 + 15 x 4 = 3 x + 3 x 4 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes x 2 16 15 x x 2 16 = ( x + 1)( x 16) x 2 16 . Consequently, f ( x ) = 3 + 15 x 4 1 15 parenleftBig x x 2 16 parenrightBig = 3 x + 3 ( x + 1)( x 16) parenleftBig x 2 16 x 4 parenrightBig .
vu (tv2894) – Homework 1 (Quest) – miner – (55096) 3 On the other hand, x 2 16 = ( x + 4)( x 4) .

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