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14:19:51 Query 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 --> x^2 + 4x + 4 factors as (x+2)(x-2)
x^4 - 16 is a perfect square that factors to (x+4)(x-4) =(x+2)(x-2) / (x+4)(x-4).................................................
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14:22:21 ** 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)]. **......!!!!!!!!...................................
RESPONSE --> I see I forgot that x^2 - 16 factors again to a difference of 2 squares. I also had the correct (x+2)^2 as my numerator but I think I entered it in as (x+2)(x-2) in my response. I see my mistake here.
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14:25:15 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 --> flip your denom and multiply to divide the fraction.
(x -2) /4x * 12x/(x^2 -4x +4) = x-2 * 12x / 4x (x -2)(x-2) cancelling out x-2 = 12x/4x(x-2) = 3/x-2.................................................
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14:25:36 ** [ ( 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 --> right
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14:27:31 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) + (x+4) / 3x +2 = combining like terms in numerator since there is already a common denom
=3x-1 / 3x +2.................................................
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14:27:41 ** 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). **......!!!!!!!!...................................
RESPONSE --> right
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14:31:54 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 --> LCD is x^3 (x^2 + 1)
Mulitply first fraction by x^2 + 1 and second fraction by x^3 (x^2 +1)(x -1) /x^3(x^2 + 1) + x(x^3) / x^2 +1 (x^3) = x^3 + x - x^2 -1 / x^3 (x^2 +1) + x^4/x^3(x^2 +1) = x^6 -1 / x^3 (x^2 +1).................................................
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14:34:14 ** 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, once again I have seen my mistake. I added my exponents as if I were multiplying which I know I am not supposed to do here. A sorry excuse, but it was very late last night when I did these.
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14:40:24 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 --> You find the LCM by factoring out the terms and using the polynomial of least degree. NOTE: in the book x^2 + 3x is the second expression
x-3 remains as x-3 x^2 + 3x = x(x+3) x^3 - 9x = x(x^2-9) = x(x+3)(x-3) So, the LCM is x(x+3)(x-3).................................................
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14:41:21 ** 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) **......!!!!!!!!...................................
RESPONSE --> I see that this is the answer to the problem including x^3 + 3x.
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14:45:49 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 --> LCM = (x -1)^2
3x(x-1)^2 / (x-1)^2 - x-4 /(x-1)^2 = 3x^2 - 3x -x -4 / (x-1)^2 = 3x^2 - 4x - 4 / (x-1)^2 = num factors to: (3x +2)(x-2) / (x-1)^2.................................................
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14:46:22 ** 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**......!!!!!!!!...................................
RESPONSE --> Ok, so I don't need to further factor the numerator from that point?
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14:50:36 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).......!!!!!!!!...................................
RESPONSE --> I see.
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