course Mth 158 It took me two days to complete this assignment. I haven't had any college math since 1978 when I graduated from CVC (UVA/wise). I work on this class daily, but I'm concerned that the pace at which I am completing the assignments is not keeping pace with the expectations of the timetable. My teaching obligations were not fullfiled until June 6. I am willing to work hard, however, I don't think I can go any faster. Do you think I should continue it or try to pick it up this fall? ∓ݐ{rzwzŧassignment #008
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19:23:43 The cube root of 54 is expressed as 54^(1/3). The number 54 factors into 2 * 3 * 3 * 3, i.e., 2 * 3^3. Thus 54^(1/3) = (2 * 3^3) ^(1/3) = 2^(1/3) * (3^3)^(1/3) = 2^(1/3) * 3^(3 * 1/3) = 2^(1/3) * 3^1 = 3 * 2^(1/3), i.e., 3 * cube root of 2.
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RESPONSE --> I thought the cube root of 54 could be written as the cube root of 27* the cube root of two.
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19:26:34 **** query R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ).
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RESPONSE --> 1/ 27x^3
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ϡᝳȣͧye assignment #008 008. `query 8 College Algebra 06-18-2008
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19:36:02 **** query R.8.12. Simplify the cube root of 54
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RESPONSE --> the cube root of 54 = cube root of 27 * cube root of 2 or, 3 * cub root of 2.
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19:36:43 The cube root of 54 is expressed as 54^(1/3). The number 54 factors into 2 * 3 * 3 * 3, i.e., 2 * 3^3. Thus 54^(1/3) = (2 * 3^3) ^(1/3) = 2^(1/3) * (3^3)^(1/3) = 2^(1/3) * 3^(3 * 1/3) = 2^(1/3) * 3^1 = 3 * 2^(1/3), i.e., 3 * cube root of 2.
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RESPONSE --> ok
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19:51:20 **** query R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ).
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RESPONSE --> 1/ (3 *cube root of 30
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20:06:07 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 --> obviously I' misssing something. I stated 3^ 1/3 * x^1/3 *(y^)^1/3 / [ 3^1/3* 27^1/3 * y^1/3. I then cancelled the common parts of the numerator and denominator. I don't see my mistake.
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20:21:05 **** query R.8.30. Simplify 2 sqrt(12) - 3 sqrt(27).
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RESPONSE --> 2* sqrt 4 * sqrt 3 -( 3 * sqrt 9 *sqrt3) 4* sqrt 3 -( 9 * sqrt 3) -5sqrt 3
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22:15:09 ** (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 -->
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22:23:31 **** query R.8. Expand (sqrt(x) + sqrt(5) )^2
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RESPONSE --> ( sqrtx )^2 *( 5^2) x +25
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22:42:00 06-18-2008 22:42:00 (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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NOTES ------->
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22:46:00 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* sqrt2 divided by sqrt 2 *sqrt 2 3 * 2^ 1/2 / 2
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22:46:11 ** 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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23:25:41 **** query R.8.48. Rationalize denominator of sqrt(2) / (sqrt(7) + 2)
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RESPONSE --> sqrt2 *sqrt 7 -2 divided by sqrt 7 + 2 times sqrt 7 - 2 sqrt 14 - 2sqrt2 / 3
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23:26:07 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 --> ok
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23:36:18 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 --> multiply 3 by 1/6 =1/2 x^ 1/2 = sqrt x
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23:36:46 ** 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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23:38:00 **** query R.8.60. Simplify 25^(3/2).
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RESPONSE --> sqrt25^3 =125
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00:25:47 **** 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 --> I could not find an explanation in the book that was adequate enough for me to solve this problem.
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00:26:03 06-19-2008 00:26:03 (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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NOTES ------->
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00:57:30 **** 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 --> [ 81 -18x - [(x^2 /x - 2] ] /4
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01:05:02 ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) =
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RESPONSE --> [ (3 -x ) + 2x^2 - 3x ]/ 9
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01:13:34 **** 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. 16.3 ft/sec b. 32 ft/sec c. 6.9 ft/sec
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01:14:38 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 --> ok
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01:17:25 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 24 = cube rt 8 * cube rt 3 2 * cube rt 3
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01:18:09 ** (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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01:26:44 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 -->
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01:33:30 ** (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 --> ok
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01:39:34 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 --> ( x +4 ) ( x +4 ) = x^2+ 8x + 16 4 ( x^2 + 8x + 16 ) = 4x^2 + 32x + 64
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01:40:32 ** 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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01:46:29 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> For me, this was very tough. The questions relating ot algebra in calcus I did not understand at all. At times, I don't think the problems were adequately explained in the r.8 section. After being out of college since 1978, the review has been most challenging.
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