Homework 1 - PreCalc Review-solutions

Homework 1 - PreCalc Review-solutions - Version 028 –...

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Unformatted text preview: Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Welcome to Quest. This homework over- laps #2 and #3. Use it as a precalculus review and diagnostic. Good luck with M408K. 001 10.0 points Rewrite the expression radicalBig a + √ b. 1. ( a + b 1 / 2 ) 1 / 2 correct 2. a 1 / 2 + b 1 / 4 3. a + b 4. a + b 1 / 2 5. a + b 1 / 4 Explanation: Since √ x = x 1 / 2 , we see that radicalBig a + √ b = ( a + b 1 / 2 ) 1 / 2 . 002 10.0 points Rationalize the numerator of √ x + 4 − √ x − 3 x . 1. 7 x ( √ x + 4 + √ x − 3) correct 2. 1 x ( √ x + 4 + √ x − 3) 3. x √ x + 4 + √ x − 3 4. 7 x ( √ x + 4 − √ x − 3) 5. 7 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) . 003 10.0 points Simplify the expression parenleftBig xy − 3 √ z parenrightBig 6 ÷ parenleftBig y 2 x 1 / 3 z − 3 parenrightBig 12 as much as possible, leaving no negative ex- ponents and no radicals. 1. x 10 y 42 z 39 correct 2. x 2 y 6 z 33 3. x 10 y 6 z 33 4. x 10 y 42 z 39 5. x 10 z 39 y 42 Explanation: By the Laws of Exponents parenleftBig xy − 3 √ z parenrightBig 6 = x 6 y 18 z 3 , while parenleftBig y 2 x 1 / 3 z − 3 parenrightBig 12 = y 24 z 36 x 4 . Version 028 – Homework 1 - PreCalc Review – Helleloid – (58250) 2 Consequently, the given expression can be rewritten as x 6 y 18 z 3 × x 4 y 24 z 36 = x 10 y 42 z 39 . 004 10.0 points Simplify the expression f ( x ) = 5 + 15 x − 4 4 + 60 parenleftBig x x 2 − 16 parenrightBig as much as possible. 1. f ( x ) = x − 4 x − 16 2. f ( x ) = 5 4 parenleftBig x + 4 x + 16 parenrightBig correct 3. f ( x ) = 5 4 parenleftBig x + 4 2 x + 16 parenrightBig 4. f ( x ) = x + 4 x − 16 5. f ( x ) = 5 4 parenleftBig x − 4 x + 16 parenrightBig Explanation: After bringing the numerator to a common denominator it becomes 5 x − 20 + 15 x − 4 = 5 x − 5 x − 4 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes 4 x 2 − 64 + 60 x x 2 − 16 = 4( x − 1)( x + 16) x 2 − 16 . Consequently, f ( x ) = 5 + 15 x − 4 4 + 60 parenleftBig x x 2 − 16 parenrightBig = 5 x − 5 4( x − 1)( x + 16) parenleftBig x 2 − 16 x − 4 parenrightBig . On the other hand, x 2 − 16 = ( x + 4)( x − 4) . Thus, finally, we see that f ( x ) = 5 4 parenleftbigg x + 4 x + 16 parenrightbigg ....
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This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.

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Homework 1 - PreCalc Review-solutions - Version 028 –...

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