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Homework 1 - PreCalc Review-solutions

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

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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 .
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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 . 005 10.0 points Find the solution set of the inequality x 2 x 4 x x 1 , expressing your answer in interval notation. 1. ( −∞ , 2 ) (1 , 4) 2. ( −∞ , 2 ] (1 , 4) correct 3. ( 2 , 1) (4 , ) 4. [ 2 , 1) (4 , ) 5. ( −∞ , 2 ] (4 , ) Explanation: To solve the inequality x 2 x 4 x x 1 we first bring all the terms to one side: x 2 x 4 x x 1 0 , and then bring the left hand side to a common
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