course MTH158 6/19/09 1:00 A.M. This was a challenging assignment for me. I can look back through the book and find the formulas and do the problems. Not sure how well I will do when I dont have the book to refer to on tests. If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ). 3x* (3)^1/3 Confidence Assessment: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I set up the problem and tried to do it alot different than you did. The way you showed here was alot simpler than in the book. I will use these to study by. Self-critique Rating:2 ********************************************* Question: * R.8.30. Simplify 2 sqrt(12) - 3 sqrt(27). 2*sqrt(4*3)-3sqrt(9*3) =4 sqrt(3) - 9 sqrt(3) =4-9*sqrt(3) Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). Extra Question: What is the simplified form of (2 sqrt(6) + 3) ( 3 sqrt(6)) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Distribute through the parentheses 2sqrt6*3sqrt6 + 9sqrt6 6 sqrt (6)^2 + 9 sqrt 6 6 * 6 which is the sqrt of (6)^2 + 9 sqrt 6 36 + 9 sqrt 6 Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8. Expand (sqrt(x) + sqrt(5) )^2 sqrt(x) * sqrt(x) + sqrt(x) * sqrt (5) + sqrt (5) * sqrt(x) + sqrt (5) * sqrt (5) (distributive property) sqrt(x^2) + sqrt (x) * sqrt (5) + sqrt (5) * sqrt (x) + sqrt (25) x + sqrt (x) * sqrt (5) + sqrt (5) * sqrt (x) + 5 x + 2 *sqrt (5X) + 5 Confidence Assessment: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I worked this problem over and over again until I came up with the correct answer. This really threw me. Im not sure why?
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8.48. Rationalize denominator of sqrt(2) / (sqrt(7) + 2) Multiply numerator and denominator by sqrt(7) -2 sqrt(2)*sqrt(7) - (2)/(sqrt(49)-4) =sqrt(2) * sqrt(7)-2/7 - 4 =sqrt(2) * sqrt(7)-2/3 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: multiplied 3/1*1/6 to come up with x^3/6 = x ^1/2 Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8.60. Simplify 25^(3/2). =125 sqrt(25)^3 5^3=125 Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 25^(3/2) = (5^2)^(3/2) = 5^(2 * 3/2) = 5^(2 * 3/2) = 5^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8.72. Simplify and express with only positive exponents: (xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4). (x^1/4 y^1/4)(xy)/((x^6/4 y^3/4)) x^1/4-6/4) (y^1/4-3/4) = xy/(x^5/4 y^1/2) =x^1/1-5/4)=-1/4 y^1/1-1/2)=1/2 =y^1/2/(x^1/4) Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: * R.8.84. Express with positive exponents: ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ((3-x + x^2(1/3 - 1/x))/9 -x^2 = (1-x+x^2(1/3-1/x))/3-x^2 Confidence Assessment: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) = &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Could you explain this problem...I have no idea what I done. I confused myself more than I helped. Self-critique Rating:1 ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: sqrt (256+0) = 16'/s sqrt (1024+0) = 32'/s sqrt (128+4^2) = (128+16) = sqrt(144) = 12'/s Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. 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. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Same as the cube root of 24 ^1 Factor 24 to 8*3 which 2 is the cube root of 8. Move the 2 outside the radicand and leave the 3 inside =2 * cuberoot(3) Confidence Assessment: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (24)^(1/3) = (8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ((x^2/3) (y^1/3) (5) (x))/2 (x) (y^4/3) x cancels (y^(1/3 - 4/3) = -y 5(x^2/3)/2y - Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (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) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: Extra question. What is the simplified form of sqrt( 4 ( x+4)^2 ) and how did you get this result? sqrt(4(x+4)(x+4)) =sqrt (2*2)(x+4)(x+4) 2 and x+4 can come outside the radicand = 2 * (x+4) Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We use two ideas in this solution: sqrt(a b) = sqrt(a) * sqrt(b) and sqrt(x^2) = | x | To understand why sqrt(x^2) = | x | and not just x consider the following: Let x = 5. Then sqrt(x^2) = sqrt( 5^2 ) = sqrt(25) = 5, so sqrt(x^2) = x. It is also clear that in this case, | x | = | 5 | = 5, so | x | = x, and we can say that sqrt(x^2) = | x |. Now let x = -5. We get sqrt(x^2) = sqrt( (-5)^2 ) = sqrt(25) = 5. In this case sqrt(x^2) = 5 but x is not equal to 5, so sqrt(x^2) is not x. However, in this case | x | = | -5 | = 5, so it is the case the sqrt(x^2) = | x |. Using these ideas we get sqrt( 4 ( x+4)^2 ) = sqrt(4) * sqrt( (x+4)^2 ) = 2 * | x+4 | ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see how you come up with the absolute value of x+4...Something I had never thought of..I just hope I can remember it Self-critique Rating: * Add comments on any surprises or insights you experienced as a result of this assignment. Alot of surprises....This has been the hardest for me to grasp so far. Actually I dont feel I have completely grasped it, Ill try to work on these more.