course phy121 Hello. assignment #007 007. `query 7 College Algebra 08-26-2008
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11:18:43 08-26-2008 11:18:43 ** 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 -------> Wups... (x^2 + 4x + 4)/(x^4 - 16) =((x + 2)^2)/((x^2 - 4)(x^2 + 4)) = ((x + 2)^2)/((x + 2)(x - 2)(x^2 + 4)) = (x + 2)/((x - 2)(x^2 + 4))
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11:21:41 Query 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)] = [(x - 2)/(4x)] * [(12x)/(x^2 - 4x + 4)] = (12x(x - 2))/(4x(x^2 - 4x + 4)) = (3(x - 2))/((x - 2)^2) = 3/(x - 2)
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11:21:48 08-26-2008 11:21:48 ** [ ( 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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11:23:07 Query 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 --> (2x - 5)/(3x + 2) + (x + 4)(3x + 2) = (2x - 5 + x + 4)/(3x + 2) = (3x - 1)/(3x + 2)
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11:23:14 08-26-2008 11:23:14 ** 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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11:25:43 Query 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 --> (x - 1)/x^3 + x/(x^2 + 1) = ((x - 1)(x^2 + 1))/(x^3(x^2 + 1)) + (x(x^3))/(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))
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11:26:02 08-26-2008 11:26:02 ** 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 -------> yea me
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11:27:28 Query 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 --> x - 3, x^2 + 3x, x^3 - 9x x - 3 = (x - 3) x^2 + 3x = x(x + 3) x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3) LCM: x(x - 3)(x + 3)
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11:27:55 08-26-2008 11:27:55 ** 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 -------> Oh, the problem was x^2 + 3x, but I do understand how to factor x^3 + 3x.
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11:30:16 Query 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 --> 3x/(x - 1) - (x - 4)/(x^2 - 2x + 1) = 3x/(x - 1) - (x - 4)/(x - 1)^2 = 3x/(x - 1) * (x -1)/(x - 1) - (x - 4)/(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 = ((3x - 4)(x - 1))/(x - 1)^2
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11:31:57 08-26-2008 11:31:57 ** 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'm not sure what I was thinking with that last line.
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11:32:27 08-26-2008 11:32:27 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 -------> ok
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