course Mth 158 ????w?o??T??assignment #007
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14:24:51 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 --> First I factored the numerator- (x +2)(x+2). Then the denominator- (x-4)(x+4). This reduces to (x+2)(x+2)/ (x+4)(x-4). confidence assessment: 1
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14:26:42 ** 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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RESPONSE --> I factored the denominator as a difference of squares. That is where I went wrong. I understand how I should've factored it, and how the terms would've reduced. self critique assessment: 2
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14:35:02 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 --> First, I factored (x^2 - 4x + 4) to (x-2)(x+2). Then I multiplied the numerator of the whole fraction by the reciprocal of the denominator. [(x-2)/4x] * 12x/ (x-2)(x+2). (x-2) cancels out and you are left with 12x/ 4x (x+2). I simplified this as 3x/ (x+2) confidence assessment: 1
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14:35:30 ** [ ( 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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RESPONSE --> I got it. self critique assessment: 3
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14:38:07 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 --> I added 2x - 5 + x + 4 which equals 2x^2 - 1. Since the denominators in both fractions are the same the final sum is 2x^2 - 1 / 3x + 2 confidence assessment: 2
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14:39:02 ** 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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RESPONSE --> I messed up in the numerator. I added exponents instead of coefficents. I understand where I went wrong. self critique assessment: 2
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14:44:03 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(x^2 +1) * x^2 +1/ x. The x^2+1 cancels giving you x -1 / 2x. confidence assessment: 1
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14:46:11 ** 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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RESPONSE --> I completely messed that up. My mind was still on dividing fractions. I see how to find the common denominator and how to simplify everything to find the solution. self critique assessment: 2
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14:57:42 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-3 x^3 + 3x = x(x^2 +3) x^3 - 9x = x (x^2 - 9) The LCM is x(x-3) confidence assessment: 1
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14:59:18 ** 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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RESPONSE --> I didn't realize that (x^2 -9) was a difference of squares, I see that now. I understand that the factors of all three polynomials are needed in the LCM. self critique assessment: 2
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15:02:59 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 --> First I factored (x^2 - 2x + 1) as (x - 1) (x -1). I found the common denominator to be (x-1)(x-1) so I multiplied 3x/ x- 1 * x- 1 and got 3x^2 - 3x / x-1(x-1). Then I combined the fractions to get 3x^2 - 4x - 4/ (x-1)(x-1) confidence assessment: 1
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15:03:34 ** 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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RESPONSE --> That is what I did and found the correct solution. self critique assessment: 3
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15:04:34 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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RESPONSE --> Makes sense to me, but it isn' my question. self critique assessment: 3
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