course Mth 158 FÙä³Âpû퇖¥ùSº¨òï’€Ýô†wŒàassignment #007
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17:50:57 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 --> Reduce to lowest terms (x^2+4x+4)/(x^4-16) (x+2)(x-2)/(x-2)^4 confidence assessment: 2
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17:53:32 ** 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 --> Ok I got that but didn't feel like it was right that maybe to facor out the last (x^2+4) but it did go back to the original nomial. self critique assessment: 2
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18:05:24 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 to divide is to multiply by the recipical of the expression. x-2 12x --------- * ------------- = 4x x^2-4x+4 Then simplify the the nomials. x-2 12x (x-2)(12x) ------- * ------------ = -------------------- = 4x (x-2)(x-2) (4x)(x-2)(x-2) And then cross out any alike nomials. 12x ---------------- = (4x)(x-2) Then simplify the expressions 12x ans 4x. 3x 3x ----------- = ----------- x(x-2) x^2-2x confidence assessment: 2
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18:05:54 ** [ ( 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 --> ok the x cancels out, too! self critique assessment: 2
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18:10:36 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 --> Since they have common denominator. Just add the numerators. (2x-5)+(x+4) 3x-1 -------------------- = ---------- 3x+2 3x+2 confidence assessment: 2
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18:10:56 ** 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 --> YEAH! i got it right! self critique assessment: 2
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18:22:31 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 --> Since there is no common denominator I have to find it. To get that I multiply all sides by x^2+1. (x-1)(x^2-1) + (x)(x^2-1) ---------------------------------- = (x^3)(x^2+1) And simplify 2x^3-x^2+2x ------------------- (x^3)(x^2+1) confidence assessment: 2
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18:23:06 ** 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 --> ok confusing! self critique assessment: 2
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18:26:01 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 --> LCM=(x)(x-3)(x+3)(x^2+3) simplify all expressions confidence assessment: 2
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18:28:26 ** 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 --> yeah! sorry but I like when I get answers right! self critique assessment: 3
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18:42:57 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 --> Since there is no common denominator, I have to find it. By multiplying all sides by (x^2-2x+2). Then factor out and simplify. (3x)(x^2-2x+1) - (x-4)(x^2-2x+1) ----------------------------------------------- = (x-1)(x^2-2x+1) -x^3-3x^2-12x+1 ------------------------ (x-1)(x-1)(x-1) confidence assessment: 2
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18:43:27 ** 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 --> ok self critique assessment: 1
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18:44:30 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 --> ok self critique assessment: 2
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