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23:16:30 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= (x+2)^2/ x^4-16= (x^2+4)(x^2-4) (x^2-4)= (x+2)(x-2) (x+2)(x+2)/(x^2+4)(x+2)(x-2)= (x+2)/(x^2+4)(x-2)
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23:16:43 ** 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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RESPONSE --> ok
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23:19:25 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)^2= 3/(x-2)
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23:19:34 ** [ ( 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
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23:21:31 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)/(3x+2)+(x+4)/(3x+2)= 3x-1/3x+2
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23:21:46 ** 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 --> ok
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23:22:53 08-31-2005 23:22:53 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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NOTES ------->
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assignment #007 C{ꃌtKޡY College Algebra 08-31-2005
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23:30:02 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 -->
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23:30:05 ** 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 -->
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23:30:07 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 -->
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23:30:08 ** [ ( 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 -->
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23:30:10 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 -->
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23:30:13 ** 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 -->
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23:31:50 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/(x^2+1)= (x-1)/x(x^2+1)+x/(x^2+1)=(x-1)/(x^2+1)
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23:32:38 ** 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 see now where I messed up.
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̜}|yi䷓ assignment #007 C{ꃌtKޡY College Algebra 09-01-2005
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21:14:43 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 -->
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21:14:44 ** 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 -->
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21:14:46 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 -->
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21:14:48 ** [ ( 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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RESPONSE -->
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21:14:50 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 -->
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21:14:55 ** 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 -->
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21:15:18 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 -->
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21:15:20 ** 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 -->
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21:22:37 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^3-9= x(x^2-9)=x(x+3)(x-3) x^3+3x= x(x^2+3) x-3= x-3 LCM=?
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21:23:14 ** 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 thought it was the answer, but second guessed myself and backed out because it looked like such a long answer. Sorry!
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21:26:06 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= 3(x-1)/x-4= 3x-3/x-4
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21:26:51 ** 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 --> I understand now- get a common denominator, NOT cross reduce.
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21:28:11 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, I see
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iҼ\Xayi assignment #008 C{ꃌtKޡY College Algebra 09-01-2005
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21:30:33 Extra question. What is the simplified form of sqrt( 4 ( x+4)^2 ) and how did you get this result?
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RESPONSE --> The simplified form of the sqrt (4(x+4)^2) is 2(x+4) You take the sqrt of 4, and then the sqrt of (x+4)^2= x+4
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21:30:43 ** sqrt(a b) = sqrt(a) * sqrt(b) and sqrt(x^2) = | x | (e.g., sqrt( 5^2 ) = sqrt(25) = 5; sqrt( (-5)^2 ) = sqrt(25) = 5. In the former case x = 5 so the result is x but in the latter x = -5 and the result is | x | ). Using these ideas we get sqrt( 4 ( x+4)^2 ) = sqrt(4) * sqrt( (x+4)^2 ) = 2 * | x+4 | **
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RESPONSE --> ok
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21:38:39 Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result?
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RESPONSE --> 24^1/3= the cube root of 24= the cube root of 3*8= 2 *cube root of 3
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21:40:09 ** (24)^(1/3) = (8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) **
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RESPONSE --> Not what I expected at all. Ok, so instead of taking thje cube root of 8 and 3, raise each one to the 1/3 power and that would equal 2*3^1/3.
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22:00:09 Extra Question: What is the simplified form of (x^2 y)^(1/3) * (125 x^3)^(1/3) / ( 8 x^3 y^4)^(1/3) and how did you get this result?
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RESPONSE --> (x^2y)*(125x^3)/(8x^3y^4) all ^1/3= x^2y*125x^3*8x^3y^4= x^2-3-1 y^1-4= 125*8x^-3y^-3 all ^1/3= 1000x^-3y^-3 = 1000*1/x^3y^3= 1000/x^3y^3 all ^ 1/3= 10/xy
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22:02:38 ** (x^2y)^(1/3) * (125x^3)^(1/3)/ ( 8 x^3y^4)^(1/3) (x^(2/3)y^(1/3)* (5x)/[ 8^(1/3) * xy(y^(1/3)] (x^(2/3)(5x) / ( 2 xy) 5( x^(5/3)) / ( 2 xy) 5x(x^(2/3)) / ( 2 xy) 5 ( x^(2/3) ) / (2 y) **
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RESPONSE --> I am so confused by this question that I will have to write it all down on paper and probably email you about it.
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22:04:59 Extra Question: What is the simplified form of 2 sqrt(12) - 3 sqrt(27) and how did you get this result?
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RESPONSE --> 2 sqrt of 12= 2 sqrt 4*3= 4 sqrt 3 3 sqrt of 27= 3 sqrt of 9*3= 9 sqrt of 3= 4 sqrt 3- 9 sqrt 3= -5 sqrt 3
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22:07:12 ** 2* sqrt(12) - 3*sqrt(27) can be written as 2* sqrt (4*3) - 3 * sqrt (9*3) by factoring out the maximum possible perfect square in each square root. This simplifies to 2* sqrt (4) sqrt(3) - 3 * sqrt (9) sqrt(3) = 2*2 sqrt 3 - 3*3 * sqrt 3 = } 4*sqrt3 - 9 * sqrt3 = -5sqrt3. **
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RESPONSE --> ok
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22:12:56 Extra Question: What is the simplified form of (2 sqrt(6) + 3) ( 3 sqrt(6)) and how did you get this result?
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RESPONSE --> Raise all parts of the equation to ^2 to get rid of the radical= 4*6+9(9*6)=1782
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22:18:51 ** (2*sqrt(6) +3)(3*sqrt(6)) expands by the Distributive Law to give (2*sqrt(6) * 3sqrt(6) + 3*3sqrt(6)), which we rewrite as (2*3)(sqrt6*sqrt6) + 9 sqrt(6) = (6*6) + 9sqrt(6) = 36 +9sqrt(6). **
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RESPONSE --> Again, a really off notion on my part. I thought I could do what I did and get the correct answer. Let me look at this- Distribute 3* sqrt 6 into the first part of the equation to get 2*sqrt 6*3 sqrt 6+ 3*3 sqrt 6= 2*3 (sqrt6)^2+9 sqrt 6 6*6 +9 sqrt6= 36 +9 sqrt6 This is an area where I must work harder, however, now that I have it on paper, I see better how the distributive property worked and what I need to do.
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22:20:48 Query R.8.42. What do you get when you rationalize the denominator of 3 / sqrt(2) and what steps did you follow to get this result?
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RESPONSE --> I multiply the top and bottom by sqrt of 2= 3* sqrt 2/ sqrt2*sqrt2= 3*sqrt2/2
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22:20:54 ** Starting with 3/sqrt(2) we multiply numerator and denominator by sqrt(2) to get (2*sqrt(2))/(sqrt(2)*sqrt(2)) = (3 sqrt(2) ) /2.
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RESPONSE --> ok
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22:26:04 Query R.8.46. What do you get when you rationalize the denominator of sqrt(3) / (sqrt(7) - sqrt(2) ) and what steps did you follow to get this result?
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RESPONSE --> Multiply the top and bottom by sqrt 7 +sqrt2= sqrt3* (sqrt 7+sqrt2)/ sqrt7-sqrt2(sqrt7+sqrt2)= sqrt3(sqrt7+sqrt2)/5
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22:28:38 ** Starting with sqrt(3)/(sqrt(7)-sqrt2) multiply both numerator and denominator by sqrt(7) + 2 to get (sqrt(3)* (sqrt(7) + 2))/ (sqrt(7) - 2)(sqrt(7) + 2). Since (a-b)(a+b) = a^2 - b^2 the denominator is (sqrt(7)+2 ) ( sqrt(7) - 2 ) = sqrt(7)^2 - 2^2 = 7 - 4 = 3 so we have sqrt(3) (sqrt(7) + 2) / 3.
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RESPONSE --> I forgot to remove the radical from the sqrt 2. Now too tired and frustrated to continue.
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22:28:50 09-01-2005 22:28:50 Extra Question: What steps did you follow to simplify (-8)^(-5/3) and what is your result?
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NOTES ------->
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