2.1 and Review-solutions

# 2.1 and Review-solutions - jordan(lj5655 2.1 and Review...

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jordan (lj5655) – 2.1 and Review – Arledge – (55100) 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 +4 - x - 3 x . 1. 1 x ( x +4+ x - 3) 2. 7 x x - x - 3 3. x x x - 3 4. 7 x ( x x - 3) correct 5. 7 x ( x - x - 3) Explanation: By the di±erence oF squares, ( x - x - 3)( x x - 3) =( x +4) 2 - ( x - 3) 2 =7 . Thus, aFter multiplying both the numerator and the denominator in the given expression by x x - 3 , we obtain 7 x ( x x - 3) . 002 10.0 points SimpliFy the expression ± xy - 3 z ² 4 ÷ ± y 2 x 1 / 2 z - 3 ² 4 as much as possible, leaving no negative ex- ponents and no radicals. 1. x 2 y 4 z 10 2. x 6 y 4 z 10 3. x 6 y 20 z 14 correct 4. x 6 y 20 z 14 5. x 6 z 14 y 20 Explanation: By the Laws oF Exponents ± xy - 3 z ² 4 = x 4 y 12 z 2 , while ± y 2 x 1 / 2 z - 3 ² 4 = y 8 z 12 x 2 . Consequently, the given expression can be rewritten as x 4 y 12 z 2 × x 2 y 8 z 12 = x 6 y 20 z 14 . 003 10.0 points SimpliFy the rational expression 4 x 10 x +20 - 40 10 x 2 x + 2 x as much as possible. 1. 2 5 ± x +2 x +5 ² 2. 2 5 x ( x 3. 2 5 ± x x ² correct 4. 4 5 x ± x x ² 5. 4 5 x ± 2 x x ²

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jordan (lj5655) – 2.1 and Review – Arledge – (55100) 2 Explanation: Factoring and bringing to a common de- nominator we get 2 ± 2 x 10( x +2) - 20 10 x ( x + 1 x ² =2 ± 2 x 2 - 20 + 10( x 10 x ( x ² = 4 x 10 x ³ x +5 x +2 ´ . Thus after simpli±cation the given rational expression becomes 2 5 ³ x x ´ . 004 10.0 points Write the expression ( x +3) 1 / 9 - 1 4 x ( x - 8 / 9 as a single fraction containing only positive exponents. 1. 4 x +12 ( x 9 / 8 2. 3 x +4 4( x 8 / 9 3. 3 x 4( x 9 / 8 4. 4 x ( x 8 / 9 5. 3 x 4( x 8 / 9 correct 6. 3 x ( x 9 / 8 Explanation: Bringing the expression to a common de- nominator, we see that ( x 1 / 9 - 1 4 x ( x - 8 / 9 = 4( x - x 4( x 8 / 9 = 3 x 4( x 8 / 9 . 005 10.0 points Simplify the expression f ( x )= 3+ 15 x - 4 1 - 15 µ x x 2 - 16 as much as possible. 1. f ( x 3( x - 4) x - 16 2. f ( x x - 4 x +16 3. f ( x x - 4 2 x 4. f ( x 3( x +4) 2 x - 16 5. f ( x 3( x x - 16 correct 6. f ( x x x 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 acommondenom ina to randfac r ingi tbe - comes x 2 - 16 - 15 x x 2 - 16 = ( x +1)( x - 16) x 2 - 16 . Consequently, f ( x 15 x - 4 1 - 15 µ x x 2 - 16 = 3 x ( x x - 16) µ x 2 - 16 x - 4 .
jordan (lj5655) – 2.1 and Review – Arledge – (55100) 3 On the other hand, x 2 - 16 = ( x +4)( x - 4) . Thus, fnally, we see that f ( x )= 3( x +4) x - 16 . 006 10.0 points Find the solution set o± the absolute value inequality | 6 x - 5 |≥ 4 , expressing your answer in interval notation. 1. solution set = ± -∞ , 1 6 ² ³ ´ 3 2 , µ cor- rect 2. solution set = ± -∞ , 1 6 ² ³ ´ - 1 6 , µ 3. solution set = ± -∞ , - 3 2 ² ³ ´ 3 2 , µ 4. solution set = ± 1 6 , 3 2 µ 5. solution set = ± - 3 2 , - 1 6 µ 6. solution set = ± -∞ , - 3 2 ² ³ ´ - 1 6 , µ Explanation:

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## This note was uploaded on 12/05/2010 for the course M 408N taught by Professor Gualdini during the Spring '10 term at University of Texas.

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2.1 and Review-solutions - jordan(lj5655 2.1 and Review...

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