course Mth 158 I am having trouble sending this SEND file. I sent it on 9-9 and again today (9-10). It is not showing up in my portfolio. ºüCë•—õ¬ËÎ÷£Ílò¾šóŽ½|E´Ôassignment #008
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14:50:01 **** query R.8.12. Simplify the cube root of 54
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RESPONSE --> This would be the cube root of 27 * cube root of 2 = 3 * cube root of 2.
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14:57:53 **** query R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ).
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RESPONSE --> ok The answer to 18 is (3xy^2)(1/3)/ [(27x^3 * 3x *y^2)]^1/3= (3xy^2)(^1/3)/ [(3x)^3 * (3xy^2)]^1/3= 1/3x * the cube root of 3xy^2/3xy^2.
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15:01:33 The cube root of (3 x y^2 / (81 x^4 y^2) ) is (3 x y^2 / (81 x^4 y^2) ) ^ (1/3) = (1 / (27 x^3) ) ^(1/3) = 1 / ( (27)^(1/3) * ^x^3^(1/3) ) = 1 / ( (3^3)^(1/3) * (x^3)^(1/3) ) = 1 / ( 3^(3 * 1/3) * x^(3 * 1/3) ) = 1 / (3 * x) = 1 / (3x).
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RESPONSE --> ok I wasn't sure if you could simplify the 3/81 first.
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15:07:32 **** query R.8.30. Simplify 2 sqrt(12) - 3 sqrt(27).
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RESPONSE --> =2 sqrt (4*3) - 3 sqrt( 9 *3)= 4 sqrt 3 - 9 sqrt 3 = -5sqrt 3.
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15:09:56 2 sqrt(12) - 3 sqrt(27) = 2 sqrt( 2*2*3) - 3 sqrt(3*3*3) = 2 sqrt(2^2 * 3) - 3 sqrt(3^3) = 2 sqrt(2^2) sqrt^3) - 3 sqrt(3^2) sqrt(3) = 2 * 2 - 3 * 3 sqrt(3) = 4 - 9 sqrt(3).
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RESPONSE --> Would this not be -5sqrt3?
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15:26:24 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 --> 2sqrt (6) * 3sqrt (6) + 3 * 3sqrt (6)= 6sqrt(6)^2 + 9sqrt (6)= (6*6) +9sqrt (6)= 36+ 9sqrt (6).
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15:26:33 ** (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 --> ok
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15:28:24 **** query R.8. Expand (sqrt(x) + sqrt(5) )^2
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RESPONSE --> (sqrt x)^2 + (sqrt 5)^2= x+5
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15:34:00 (sqrt(x) + sqrt(5) )^2 = (sqrt(x) + sqrt(5) ) * (sqrt(x) + sqrt(5) ) = sqrt(x) * (sqrt(x) + sqrt(5) ) + sqrt(5) * (sqrt(x) + sqrt(5) ) = sqrt(x) * sqrt(x) + sqrt(x) * sqrt(5) + sqrt(5) * sqrt(x) + sqrt(5) * sqrt(5) = x + 2 sqrt(x) sqrt(5) + 5.
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RESPONSE --> I'm sorry. I tried to simplify it.
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15:37:11 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 don't have this problem in my book. I would multiply both numerator and denominator by sqrt 2 to = 3sqrt2/2
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15:37:17 ** 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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15:43:42 **** query R.8.48. Rationalize denominator of sqrt(2) / (sqrt(7) + 2)
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RESPONSE --> =sqrt 2 *(sqrt7-2)/(sqrt7 +2) *( sqrt 7 - 2)= sqrt14 - 2sqrt2/11
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16:04:08 To rationalize the denominator sqrt(7) + 2 we multiply both numerator and denominator by sqrt(7) - 2. We obtain ( sqrt(2) / (sqrt(7) + 2) ) * (sqrt(7) - 2) / (sqrt(7) - 2) = sqrt(2) * (sqrt(7) - 2) / ( (sqrt(7) + 2) * ( sqrt(7) - 2) ) = sqrt(2) * (sqrt(7) - 2) / (sqrt(7) * sqrt(7) - 4) = sqrt(2) * (sqrt(7) - 2 ) / (7 - 4) = sqrt(2) * (sqrt(7) - 2 ) / 3.
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RESPONSE --> I know how I messed up the denominator. I had a positive 4 instead of a negative. I keep getting confused as to whether or not I should multiply the square roots or not.
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16:09:30 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 --> I would multiply ^3 *^1/6 and simplify to ^1/2 which would equal sqrt of x.
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16:09:42 ** Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). **
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RESPONSE --> ok
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16:10:55 **** query R.8.60. Simplify 25^(3/2).
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RESPONSE --> sqrt25 ^3 = 5^3 = 125
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16:11:02 25^(3/2) = (5^2)^(3/2) = 5^(2 * 3/2) = 5^(2 * 3/2) = 5^3.
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RESPONSE --> ok
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16:23:30 **** query R.8.72. Simplify and express with only positive exponents: (xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4).
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RESPONSE --> =x^1/4(y^1/4) (xy)/x^6/4(y^3/4)= (y)^1/4+^4/4-^-3/4/(x)^6/4 -^1/4 -^4/4= y^1/2/x^1/4
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16:23:58 (xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4) = x^(1/4) * y^(1/4) * (x^2)^(1/2) * y^2 ^ (1/2) / ( (x^2)^(3/4) * y^(3/4) ) = x^(1/4) * y^(1/4) * x^(2 * 1/2) * y^(2 * 1/2) / ( (x^(2 * 3/4) * y^(3/4) ) = x^(1/4) y^(1/4) * x^1 * y^1 / (x^(3/2) y^(3/4) ) = x^(1 + 1/4) y^(1 + 1/4) / (x^(3/2) y^(3/4) ) = x^(5/4) y^(5/4) / (x^(3/2) y^(3/4) ) = x^(5/4 - 3/2) y^(5/4 - 3/4) = x^(5/4 - 6/4) y^(2/4) = x^(-1/4) y^(1/2) = y^(1/2) / x^(1/4).
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RESPONSE --> ok
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16:46:41 **** query R.8.84. Express with positive exponents: ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2).
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RESPONSE --> =(9-x^2)^1/2 + x/(9-x^2)^1/2 / 9-x^2=Working on numerator first: (9-x^2)^1/2(9-x^2)^1/2 +x^2/(9-x^2)^1/2= 9 -x^2 + x^2/9-x^2= [9/(9 - x^2)^1/2 ] /9 - x^2 multiply by 1/9 - x^2= 9/(9 - x^2)^1/2 (9 - x^2)
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16:53:14 **** query R.8.108. v = sqrt(64 h + v0^2); find v for init vel 0 height 4 ft; for init vel 0 and ht 16 ft; for init vel 4 ft / s and height 2 ft.
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RESPONSE --> a. sqrt 64 * 4 = 8 *2 = 16 b. sqrt64 * 16 = 8 * 4=24 c. sqrt 64 * 2 + 4 = 8 +2sqrt2
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16:54:58 If initial velocity is 0 and height is 4 ft then we substitute v0 = 0 and h = 4 to obtain v = sqrt(64 * 4 + 0^2) = sqrt(256) =16.+vbcrlf+vbcrlf+If initial velocity is 0 and height is 16 ft then we substitute v0 = 0 and h = 4 to obtain v = sqrt(64 * 16 + 0^2) = sqrt(1024) = 32. Note that 4 times the height results in only double the velocity.+vbcrlf+vbcrlf+If initial velocity is 4 ft / s and height is 2 ft then we substitute v0 = 4 and h = 2 to obtain v = sqrt(64 * 2 + 4^2) = sqrt(144) =12.
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RESPONSE --> Didn't square the 4 on the last one.
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16:56:58 Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result?
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RESPONSE --> cube root of 24 = cube root of 8 *3= 2(3)^1/3
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16:57:13 ** (24)^(1/3) = (8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) **
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RESPONSE --> ok
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17:16:20 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)^1/3(y)^1/3 * (125)^1/3 (x^3)^1/3 / (8)^1/3 (x^3)^1/3 (y^4)^1/3 = 5 * x(^2/3 + 1- 1) / y(^4/3 - 1/3)= 5*x^2/3 / y
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17:17:01 ** (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 --> Left off my 2. Got the general idea.
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17:20:37 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 --> Sqrt of 4 is 2 * sqrt of (x+4)^2 is x+4 =2*(x+4 )= 2x + 8
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17:21:27 ** 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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17:22:30 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This is very confusing material to have to test on all at once. I have had considerable trouble with this section.
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