course Mth 158 雇^assignment #007 007. `query 7 College Algebra 02-27-2008
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12:11:51 R.7.10 (was R.7.6). Show how you reduced (x^2 + 4 x + 4) / (x^4 - 16) to lowest terms.
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RESPONSE --> 10) (x^2 + 4x + 4) / (x^4 - 16) = ((x + 2)^2) / ((x + 2) (x - 2)) Cancel common factors that are in both the numerator and denominator, in this case (x + 2). = (x + 2) / (x - 2) confidence assessment: 3
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12:19:58 02-27-2008 12:19:58 ** We factor the denominator to get first (x^2-4)(x^2+4), then (x-2)(x+2)(x^2+4). The numerator factors as (x+2)^2. So the fraction is (x+2)(x+2)/[(x-2)(x+2)(x^2+4)], which reduces to (x+2)/[(x-2)(x^2+4)]. **
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NOTES -------> I am not quite sure where the (x^2+4) is coming into this problem. x^2 - 4 factors to (x+2)(x-2), right? I am slightly confused by your answer.
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12:29:20 R.7.28 (was R.7.24). Show how you simplified[ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ].
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RESPONSE --> [ (x - 2) / (4x) ] / [ (x^2 - 4x + 4) / (12x) ] Based on the rule that (a / b) / (c / d) = (a / b) * (d / c) = (ad / bc) = [ (x - 2) / (4x) ] * [ (12x) / (x^2 - 4x + 4) ] Now reduce ll rational expressions to lowest terms. = [ (x - 2) / (4x) } * [ (12x) / (x - 2)^2 ] Multiply the rational expressions. = [ (x - 2)(12x) ] / [ (4x)(x-2)^2 ] Use Cancellation Property: = (12x) / (4x)(x - 2) Simplify = 3 / (x - 2) confidence assessment: 3
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12:29:31 02-27-2008 12:29:31 ** [ ( x - 2) / (4x) ] / [ (x^2 - 4 x + 4) / (12 x) ] = (x-2) * / 4x * 12 x / (x^2 - 4x + 4) = (x-2) * 12 x / [ 4x ( x^2 - 4x + 4) ] = 12 x (x-2) / [4x ( x-2) ( x-2) ] = 3/(x - 2) **
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NOTES -------> ok
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12:35:17 R.7.40 (was R.7.36). Show how you found and simplified the sum (2x - 5) / (3x + 2) + ( x + 4) / (3x + 2).
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RESPONSE --> Because both rational expressions have equal denominators the answer is the sum of the numerators over the equal denominator. [ (2 x - 5) / (3x + 2) ] + [ (x + 4) / (3x + 2) ] a / b + c / b = (a + c) / b = [ (2x - 5) + (x + 4) ] / (3x + 2) = (2x + x - 5 + 4) / (3x + 2) = (3x - 1) / (3x + 2) confidence assessment: 3
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12:35:26 02-27-2008 12:35:26 ** We have two like terms so we write (2x-5)/(3x+2) + (x+4)/(3x+2) = [(2x-5)+(x+4)]/(3x+2). Simplifying the numerator we have (3x-1)/(3x+2). **
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NOTES -------> ok
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12:45:09 R.7.52 (was R.7.48). Show how you found and simplified the expression(x - 1) / x^3 + x / (x^2 + 1).
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RESPONSE --> Because these two rational expressions haveunequal denominators, in order to add them we must use the following formula. (a / b) + (c / d) = [ (ad / bd) + (bc / bd) ] = (ad + bc) / bd [ (x - 1) / (x^3) ] + [ x / (x^2 + 1) ] = {[ (x - 1) / (x^3) ] * [ (x^2 + 1) / (x^2 + 1) ]} + {[ (x^3 / x^3) * (x / (x^2 + 1) ]} = [(x - 1)(x^2 + 1)] / [(x^3)(x^2 + 1)] + [(x^3)(x)] / [(x^3)(x^2 + 1)] = (x^3 + x - x^2 - 1 + x^4) / (x^3(x^2 + 1)) = (x^4 + x^3 - x^2 + x - 1) / (x^3 (x^2 + 1)) confidence assessment: 3
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12:45:37 02-27-2008 12:45:37 ** Starting with (x-1)/x^3 + x/(x^2+1) we multiply the first term by (x^2 + 1) / (x^2 + 1) and the second by x^3 / x^3 to get a common denominator: [(x-1)/(x^3) * (x^2+1)/(x^2+1)]+[(x)/(x^2+1) * (x^3)/(x^3)], which simplifies to (x-1)(x^2+1)/[ (x^3)(x^2+1)] + x^4/ [(x^3)(x^2+1)]. Since the denominator is common to both we combine numerators: (x^3+x-x^2-1+x^4) / ) / [ (x^3)(x^2+1)] . We finally simplify to get (x^4 +x^3 - x^2+x-1) / ) / [ (x^3)(x^2+1)] **
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NOTES -------> OK
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12:49:39 R.7.58 (was R.7.54). How did you find the LCM of x - 3, x^3 + 3x and x^3 - 9x, and what is your result?
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RESPONSE --> In order to determine the LCM of the three polynomials: x - 3, x^3 + 3x, and x^3 - 9x you must first factor completely x - 3 = x - 3 x^2 + 3x = x(x + 3) x^3 - 9x = x (x^2 - 9) = x(x - 3)(x + 3) Starting with the first set of factors we have (x - 3) The second set of factors x(x+ 3) is not previously in the LCM so we add it x(x - 3)(x + 3) The third set of factors are already listed in the LCM so we need not add them again. Thus LCM = x(x - 3)(x + 3) confidence assessment: 3
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12:51:45 02-27-2008 12:51:45 ** x-3, x^3+3x and x^3-9x factor into x-3, x(x^2+3) and x(x^2-9) then into (x-3) , x(x^2+3) , x(x-3)(x+3). The factors x-3, x, x^2 + 3 and x + 3 'cover' all the factors of the three polynomials, and all are needed to do so. The LCM is therefore: x(x-3)(x+3)(x^2+3) **
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NOTES -------> I understand why our answers differ - through me for a loop, but I found it - you used x^3 + 3x rather than x^2 + 3x
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13:05:52 R.7.64 (was R.7.60). Show how you found and simplified the difference3x / (x-1) - (x - 4) / (x^2 - 2x + 1).
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RESPONSE --> I'm not totally sure about my answer but this is what I got: [ 3x / (x - 1) ] - [ (x - 4) / (x^2 - 2x + 1) ] = [ 3x / (x - 1) ] - [ (x - 4) / (x - 1)^2 ] LCM = (x - 1)^2 = [ (3x * (x - 1)) / (x - 1)^2 ] - [ (x - 4) / (x - 1)^2 ] = [ (3x^2 - 3x) - (x - 4) ] / (x - 1)^2 = (3x^2 - 4x + 4) / (x - 1)^2 confidence assessment: 2
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13:07:50 02-27-2008 13:07:50 ** Starting with 3x / (x-1) - (x-4) / (x^2 - 2x +1) we factor the denominator of the second term to obtain (x - 1)^2. To get a common denominator we multiply the first expression by (x-1) / (x-1) to get 3x(x-1)/(x-1)^2 - (x-4)/(x-1)^2, which gives us (3x^2-3x-x-4) / (x-1)^2 = (3x^2 - 4x - 4) / (x-1)^2. DRV**
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NOTES -------> I had + 4, but I do understand that it is - 4. I was right other than that. I wasn't sure about having a square in the denominator, but I went with my gut and was right.
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13:09:56 02-27-2008 13:09:56 QUESTION FROM STUDENT: On the practice test I'm having problems with problem #5 I don't know where to start or how to set it up. I'm probably missing something simple and will probably feel stupid by seeing the solution. Could you help with this problem. A retailer is offering 35% off the purchase price of any pair of shoes during its annual charity sale. The sale price of the shoes pictured in the advertisement is $44.85. Find the original price of the shoes by solving the equation p-.35p = 44.85 for p. INSTRUCTOR RESPONSE: It's very easy to get ahold of the wrong idea on a problem and then have trouble shaking it, or to just fail to look at it the right way. Nothing stupid about it, just human nature. See if the following makes sense. If not let me know. p - .35 p = 44.85. Since p - .35 p = 1 p - .35 p = (1 - .35) p = .65 p we have .65 p = 44.85. Multiplying both sides by 1/.65 we get p = 44.85 / .65 = etc. (you can do the division on your calculator); you'll get something near $67).
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NOTES -------> Where are the practice tests?
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