hw01solution - chi(ic4443 HW 01 gualdani(56410 This...

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chi (ic4443) – HW 01 – gualdani – (56410) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Rationalize the numerator oF x + 5 x 1 x . 1. 4 x ( x + 5 + x 1) 2. 6 x ( x + 5 x 1) 3. 6 x ( x + 5 + x 1) correct 4. 6 x x + 5 x 1 5. 4 x x + 5 + x 1 Explanation: By the di±erence oF squares, ( x + 5 x 1)( x + 5 + x 1) = ( x + 5) 2 ( x 1) 2 = 6 . Thus, aFter multiplying both the numerator and the denominator in the given expression by x + 5 + x 1 , we obtain 6 x ( x + 5 + x 1) . 002 10.0 points SimpliFy the expression p xy - 3 z P 8 ÷ p y 2 x 1 / 3 z - 3 P 6 as much as possible, leaving no negative ex- ponents and no radicals. 1. x 10 y 12 z 14 2. x 10 y 36 z 22 correct 3. x 10 z 22 y 36 4. x 6 y 12 z 14 5. x 10 y 36 z 22 Explanation: By the Laws oF Exponents p xy - 3 z P 8 = x 8 y 24 z 4 , while p y 2 x 1 / 3 z - 3 P 6 = y 12 z 18 x 2 . Consequently, the given expression can be rewritten as x 8 y 24 z 4 × x 2 y 12 z 18 = x 10 y 36 z 22 . 003 10.0 points SimpliFy the rational expression 6 x 10 x + 20 60 10 x 2 + 20 x + 3 x as much as possible. 1. 6 5 x p 2 x + 5 x + 4 P 2. 3 5 p x + 2 x + 5 P 3. 6 5 x p x + 5 x + 2 P 4. 3 5 x ( x + 4) 5. 3 5 p x + 5 x + 2 P correct
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chi (ic4443) – HW 01 – gualdani – (56410) 2 Explanation: Factoring and bringing to a common de- nominator we get 3 b 2 x 10( x + 2) 20 10 x ( x + 2) + 1 x B = 3 b 2 x 2 20 + 10( x + 2) 10 x ( x + 2) B = 6 x 10 x p x + 5 x + 2 P . Thus after simpli±cation the given rational expression becomes 3 5 p x + 5 x + 2 P . 004 10.0 points Write the expression ( x + 3) 1 / 3 1 4 x ( x + 3) - 2 / 3 as a single fraction containing only positive exponents. 1. 3 x + 4 4( x + 3) 2 / 3 2. 4 x + 12 ( x + 3) 3 / 2 3. 3 x + 4 ( x + 3) 3 / 2 4. 3 x + 12 4( x + 3) 2 / 3 correct 5. 4 x + 12 ( x + 3) 2 / 3 6. 3 x + 12 4( x + 3) 3 / 2 Explanation: Bringing the expression to a common de- nominator, we see that ( x + 3) 1 / 3 1 4 x ( x + 3) - 2 / 3 = 4( x + 3) x 4( x + 3) 2 / 3 = 3 x + 12 4( x + 3) 2 / 3 . 005 10.0 points Simplify the expression f ( x ) = 4 + 16 x 5 3 + 72 ± x x 2 25 ² as much as possible. 1. f ( x ) = 4 3 ± x + 5 x + 25 ² correct 2. f ( x ) = 4 3 ± x + 5 2 x + 25 ² 3. f ( x ) = x 5 2 x 25 4. f ( x ) = 4 3 ± x 5 x + 25 ² 5. f ( x ) = x 5 x 25 6. f ( x ) = x + 5 x 25 Explanation: After bringing the numerator to a common denominator it becomes 4 x 20 + 16 x 5 = 4 x 4 x 5 . Similarly, after bringing the denominator to a common denominator and factoring it be- comes 3 x 2 75 + 72 x x 2 25 = 3( x 1)( x + 25) x 2 25 . Consequently, f ( x ) = 4 + 16 x 5 3 + 72 ± x x 2 25 ² = 4 x 4 3( x 1)( x + 25) ± x 2 25 x 5 ² .
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chi (ic4443) – HW 01 – gualdani – (56410) 3 On the other hand, x 2 25 = ( x + 5)( x 5) . Thus, fnally, we see that f ( x ) = 4 3 p x + 5 x + 25 P . 006 10.0 points Find the solution set o± the absolute value inequality | 6 x 5 | ≥ 4 , expressing your answer in interval notation.
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hw01solution - chi(ic4443 HW 01 gualdani(56410 This...

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