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20:02:21 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 --> working within the paren first: 4(x^2 +8x + 16) = 4x^2 + 32x + 64 =
sqrt (4x^2 + 32x + 64).................................................
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20:04:08 ** 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 | **......!!!!!!!!...................................
RESPONSE --> I see. I worked the problem within the paren first. This makes sense.
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20:06:19 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) is the same as 24 to the first power cube root.
cube root of 24 = cbrt 8 * 3 = 2*cbrt(3).................................................
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20:06:50 ** (24)^(1/3) =
(8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) **......!!!!!!!!...................................
RESPONSE --> can this be done either way? It is the same answer, just written differently.
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20:13:25 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^(2/3)y^(1/3) * 5x / [8^(1/3)*xy(y^1/3)]
x^(2/3)5x / 2xy(y^(1/3)) 5x^(5/3) / 2xy.................................................
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20:15:43 ** (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) **......!!!!!!!!...................................
RESPONSE --> I see I forgot to cancel out my y^(1/3).
Why did you factor out 5x (x^(2/3)) ? Could it not have been left as is?.................................................
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20:17:37 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 --> the sqrt of 12 = 4*3 and the sqrt of 27 = 9*3
2 sqrt(4*3) - 3 sqrt (9*3) 4 sqrt (3) - 9 sqrt (3) = -5sqrt (3).................................................
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20:17:57 ** 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. **......!!!!!!!!...................................
RESPONSE --> right
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20:22:15 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 --> use the distributive property
2 * sqrt (6) * 3 *sqrt (6) + 3 * 3 sqrt (6) = 2*3 sqrt (6) * sqrt (6) + 9 sqrt (6) 6 sqrt (36^2) = 6 * 6 + 9 sqrt (6) = 36 + 9 sqrt (6).................................................
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20:22:47 09-22-2005 20:22:47 ** (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). **......!!!!!!!!...................................
NOTES -------> right
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20:22:48 ** (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). **......!!!!!!!!...................................
RESPONSE -->
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20:24:04 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 --> 3 * sqrt (2) / sqrt (2) * sqrt (2) =
3*sqrt(2) / 2.................................................
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20:24:15 ** 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.......!!!!!!!!...................................
RESPONSE --> right
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20:27:43 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 --> you multiply by the conjugate of sqrt (7) - sqrt (2)
sqrt (3) * sqrt (7) + sqrt (2) / sqrt (7) - sqrt (2) * sqrt (7)+sqrt (2)= denom works out to be sqrt (7)^2 - sqrt (2)^2 = 7-2 sqrt (3) * sqrt (7)* sqrt (2) / 5.................................................
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20:28:59 ** 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.......!!!!!!!!...................................
RESPONSE --> I thought the problem was entered as sqrt (7) +sqrt (2) so then wouldn't you end up with the sqrt of (2)^2 = 2
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20:34:34 Extra Question: What steps did you follow to simplify (-8)^(-5/3) and what is your result?
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RESPONSE --> hmm, not sure but here we go
cube root of (-8) = -2 (-2)^-5= 1/(2)^(5) = 1/32.................................................
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20:36:12 ** (-8)^(-5/3) = [ (-8)^(1/3) ] ^-5. Since -8^(1/3) is -2 we get
[-2]^-5 = 1 / (-2)^5 = -1/32. **......!!!!!!!!...................................
RESPONSE --> a sign error on my part.
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20:38:15 query R.8.64. What steps did you follow to simplify (8/27)^(-2/3) and what is your result?
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RESPONSE --> (27/8) ^(2/3) =
(cubert (27) )^(2) / (cubert (8)) ^(2) = 3^(2) / 2^(2) = 9 / 4.................................................
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20:39:04 ** Starting with
(8/27)^(-2/3) we can write as (8^(-2/3)/27^(-2/3)). Writing with positive exponents this becomes (27^(2/3)/8^(2/3)) 27^(2/3) = [ 27^(1/3) ] ^2 = 3^2 = 9 and 8^(2/3) = [ 8^(1/3) ] ^2 = 2^2 = 4 so the result is (27^(2/3)/8^(2/3)) = 9/4. **......!!!!!!!!...................................
RESPONSE --> Again, I did mine taking the root then raising to the power. Do you accept either method?
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20:41:21 Extra Question: What steps did you follow to simplify 6^(5/4) / 6^(1/4) and what is your result?
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RESPONSE --> 6^ (5/4 - 1/4) = 6^(4/4) = 6
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20:41:31 ** Use the laws of exponents (mostly x^a / x^b = x^(a-b) as follows:
6^(5/4) / 6^(1/4) = 6^(5/4 - 1/4) = 6^1 = 6. **......!!!!!!!!...................................
RESPONSE --> right
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20:43:26 Extra Question: What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?
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RESPONSE --> x^(3*1/6) = x^(1/2)
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20:43:34 ** Express radicals as exponents and use the laws of exponents.
(x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). **......!!!!!!!!...................................
RESPONSE --> right
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20:50:42 Extra Question: What steps did you follow to simplify (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 and what is your result, assuming that x is positive and expressing your result with only positive exponents?
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RESPONSE --> [x^(1/2) / y^(2)]^4 * [x^(2/3) / y^(1/3)]^3 =
[x^(1/2)]^4 / [ y^(2)] ^4 * [x^(2/3)]^3 / [y^(1/3)]^3 = x^2 / y^2 * x^2 / y = x^4 / y^3.................................................
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20:56:01 ** (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 =
x^(1/2 * 4) / y^(2* 4) * y^(1/3 * 3) / x^(-2/3 * 3)= x^2 / y^8 * y / x^(-2) = x^2 * x^2 / y^7 = x^4 / y^7. **......!!!!!!!!...................................
RESPONSE --> once again, a simple mistake. Shouldn't the final y be y^8 since you are adding y^8 and y?
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20:58:08 query R.8.96. Factor 8 x^(1/3) - 4 x^(-2/3), x <> 0.
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RESPONSE --> 4x^(1/3) (2 -x^(-1/2) ) = 4x^(1/3) *[2 - (1/x^2)]
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21:01:33 ** To factor 8x^(1/3)- 4x^(-2/3) we first need to write the expression without negative exponents. To accomplish this we multiply through by x^(2/3) / x^(2/3), obtaining
(8 x^(1/3 + 2/3) - 4x^(-2/3 + 2/3) / x^(2/3) = (8 x - 4) / x^(2/3). We then factor 2 out of the numerator to obtain 4 ( 2x - 1) / x^(2/3). Other correct forms include: ( 4x^(1/3) ) ( 2 - ( 1/x) ) 8 x^(1/3) - 4 / x^(2/3). **......!!!!!!!!...................................
RESPONSE --> I see.
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21:02:26 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found the writing as a single quotient to be a little difficult. Even though this was review, I don't remember doing that before so it was surprising for me.
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¡÷àÑ×äåßz™cœ‰ÔºC¤¾ÕŠ assignment #009 ‰{kÇ…±ò³¦µ_”Ê… ª’Î÷ϯaò~é¨Ùû College Algebra 09-22-2005
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21:03:21 Query 9 is presently blank. Query 8 covers the former Query 9 as well.
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RESPONSE --> ok
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