#$&* course mth 158 7/3/13 9:55pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * The cube root of 54 is expressed as 54^(1/3). The number 54 factors into 2 * 3 * 3 * 3, i.e., 2 * 3^3. Thus 54^(1/3) = (2 * 3^3) ^(1/3) = 2^(1/3) * (3^3)^(1/3) = 2^(1/3) * 3^(3 * 1/3) = 2^(1/3) * 3^1 = 3 * 2^(1/3), i.e., 3 * cube root of 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ( I will type it out the cube roots this way next time) ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.8.18. Simplify the cube root of (3 x y^2 / (81 x^4 y^2) ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: cube root of (3 x y^2 / (81 x^4 y^2) ) = [3 / (81 x^3) ]^1/3 = x (3/81)^1/3 = x [3 / (27^3 * 3) ]^1/3 = 3x (3/3)^1/3 = 3x confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 did not initially simplify the 3/81. I still could have gotten the correct answer, but I forgot to leave the 3 (that factored out of the cube of 27) and x in the denominator when I pulled them out. This is why I failed to have the 3x in the denominator. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.8.30. Simplify 2 sqrt(12) - 3 sqrt(27). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 sqrt(12) - 3 sqrt(27) = 2 sqrt(4 * 3) - 3 sqrt(9 * 3) = 4 sqrt(3) - 9 sqrt(3) = -5 sqrt (3) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 2 sqrt(12) - 3 sqrt(27) = 2 sqrt( 2*2*3) - 3 sqrt(3*3*3) = 2 sqrt(2^2 * 3) - 3 sqrt(3^3) = 2 sqrt(2^2) sqrt^3) - 3 sqrt(3^2) sqrt(3) = (2 * 2 - 3 * 3) sqrt(3) = (4 - 9) sqrt(3) = -5 sqrt(3) Extra Question: What is the simplified form of (2 sqrt(6) + 3) ( 3 sqrt(6)) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [2 sqrt(6) + 3]* [ 3 sqrt(6)] = [2 sqrt(6) * 3 sqrt(6)] + 3 * 3 sqrt(6) = 6 * [sqrt (6) ]^2 + 9 sqrt (6) = 36 + 9 sqrt(6) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.8. Expand (sqrt(x) + sqrt(5) )^2 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your 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)] = x + sqrt(5x) + sqrt (5x) + 5 = x + 2 sqrt(5x) + 5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (sqrt(x) + sqrt(5) )^2 = (sqrt(x) + sqrt(5) ) * (sqrt(x) + sqrt(5) ) = sqrt(x) * (sqrt(x) + sqrt(5) ) + sqrt(5) * (sqrt(x) + sqrt(5) ) = sqrt(x) * sqrt(x) + sqrt(x) * sqrt(5) + sqrt(5) * sqrt(x) + sqrt(5) * sqrt(5) = x + 2 sqrt(x) sqrt(5) + 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??? You can still combine the sqrt(x) and the sqrt(5) into sqrt(5x) can you not? I thought that was one of the radical properties? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: Starting with 3/sqrt(2) we multiply numerator and denominator by sqrt(2) to get (3*sqrt(2))/(sqrt(2)*sqrt(2)) = (3 sqrt(2) ) /2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.8.48. Rationalize denominator of sqrt(2) / (sqrt(7) + 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt(2) / (sqrt(7) + 2) = [sqrt(2) / (sqrt(7) + 2] * [ (sqrt (7) - 2 ) / sqrt (7) -2) ] = [ sqrt (2) * (sqrt (7) - 2 ] / (7 - 4) = [ sqrt (2) * (sqrt (7) - 2 ] / 3 confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK 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: (x^3)^(1/6) = x^(3/6) = x^ (1/2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * R.8.60. Simplify 25^(3/2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 25^(3/2) = [sqrt (25)]^3 = 5^3 = 125 confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (xy)^(1/4) (x^2 y^2) ^(1/2) / (x^2 y)^(3/4) = (xy)^(1/4) (xy) / (x^2 y)^(3/4) = [ x^2 * y^(1/4 - 1) ] / [x^2 y^(3/4) ] = y^(1/4 - 1) / y^(3/4) = y^(-3/4) / y^(3/4) = 1 / y^(3/4) * y^(3/4) = 1 / y^(3/2) confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). STUDENT QUESTION I wrote the entire given solution on paper to see how to solve, but I am still confused when it gets to the = x^(1 + 1/4) y^(1 + 1/4) / (x^(3/2) y^(3/4) How do you get 1 + ¼? Does the 1 come from the xy on the right of the numerator? INSTRUCTOR RESPONSE The numerator of the expression x^(1/4) y^(1/4) * x^1 * y^1 / (x^(3/2) y^(3/4) ) contains two factors which are powers of x. The two are x^(1/4) and x^1 (the latter could be written just as x, but to apply the laws of exponents it's not a bad idea to write the exponent explicitly). When you multiply these two factors, the laws of exponent tell you that you get x^(1/4 + 1) = x^(5/4). The same thing happens with the y factors. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I started the problem wrong from the very beginning. If I had distributed the exponents as in the given solution, I would have had a much simpler time in determining the proper steps to take in order to reach the answer. I find it easy to mix up the operations used when distributing exponents in different situations, and that also trips me up. I did go back through and work the problem following the steps of the given solution and find it makes much more sense than what I was trying to do. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * R.8.84. Express with positive exponents: ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2), defined for -3 < x < 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: [ (9 - x^2)^(1/2) + x^2 (9 - x^2)^(-1/2) ] / (9 - x^2) = [ (9 - x^2)^(1/2) / (9 - x^2) ] + x^2 / [ (9 - x^2) (9 - x^2)^(1/2) ] = [ (9 - x^2)^(1/2) / (9 - x^2) ] + x^2 / (9 - x^2) = [ (9 - x^2)^(1/2) + x^2 ] / (9 - x^2) = [ 9^(1/2) - x + x^2 ] / (9 - x^2) = (3 - x + x^2) / (9 - x^2) confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) = ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) / (9 - x^2) = [ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ] = (9 - x^2) ^(-1/2) + x^2 / (9 - x^2)^(3/2) = 1 / (9 - x^2)^(1/2) + x^2 / (9 - x^2)^(3/2). In the third step the exponent ^1 on the (9 - x^2) expressions wasn't necessary, but was included to explicitly show the exponents and the application of the laws of exponents. The first term in the 4th step is obtained as follows: (9 - x^2) (1/2) / (9 - x^2)^1 = (9 - x^2) ^ (1/2 - 1) = (9 - x^2)^(-1/2). EXPANDED EXPLANATION OF STEPS ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) = ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) / (9 - x^2) In the above step we have replace (9 - x^2) ^ (-1/2) in the numerator by (9 - x^2)^(1/2) in the denominator, following the rule that a^-b = 1 / (a^b) with a = (9 - x^2) and b = 1/2. ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) / (9 - x^2) = [ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ] The above step is just the distributive law of multiplication over addition, in which we multiply through the expression ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) by 1 / (9 - x^2). The brackets [ ] have been added to clarify the two terms in the resulting expression, but the expression has the same meaning without them. [ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ] = (9 - x^2) ^(-1/2) + x^2 / (9 - x^2)^(3/2) (9 - x^2) ^ (1/2) / (9 - x^2)^2 = (9 - x^2)^-1/2, by the laws of exponents; and (9 - x^2)^(1/2) * (9 - x^2) = (9 - x^2) ^(3/2) by the laws of exponents. (9 - x^2) ^(-1/2) + x^2 / (9 - x^2)^(3/2) = 1 / (9 - x^2)^(1/2) + x^2 / (9 - x^2)^(3/2) (9 - x^2) ^(-1/2) has been replaced by 1 / (9 - x^2) ^(1/2), using a^-b = 1 / a^b. All the exponents in the final expression are positive. It would also be possible to factor out 1 / (9 - x^2)^(1/2), though this wasn't requested and isn't necessary in the problem as stated. The result would be 1 / (9 - x^2)^(1/2) * ( 1 + x^2 / (9 - x^2) ). This could be further simplified to 1 / (9 - x^2)^(1/2) * ( 9 / (9 - x^2) ) , which is equal to 9 / (9 - x^2)^(3/2) You aren't expected to be able to read these expressions. You are expected to be able to write them out in standard form; having done so you should understand. However these expressions are fairly challenging, so some of the expressions will be depicted here ( (9 - x^2) ^(1/2) + x^2 ( 9 - x^2) ^(-1/2) ) / (9 - x^2) would be depicted in standard notation as (9 - x^2) ^(-1/2) + x^2 / (9 - x^2)^(3/2) would be depicted in standard notation as 1 / (9 - x^2)^(1/2) + x^2 / (9 - x^2)^(3/2) would be depicted in standard notation as &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I get lost in this problem around the 3rd step when it transitions from ( (9 - x^2) ^(1/2) + x^2 / ( 9 - x^2) ^(1/2) ) / (9 - x^2) = [ (9 - x^2) (1/2) / (9 - x^2)^1 ] + [ x^2 / ( (9 - x^2)^(1/2) * (9 - x^2)^ 1) ]. I don't understand how it goes from one to the other. ???
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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): OK ------------------------------------------------ Self-critique Rating: OK Extra Question: What is the simplified form of (24)^(1/3) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (24)^(1/3) = cubed root of 24 = cubed root of (8 * 3) = cubed root of (2^3 * 3) = 2 * cubed root of (3) also equal to 2 (3)^(1/3) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * (24)^(1/3) = (8 * 3)^(1/3) = 8^(1/3) * 3^(1/3) = 2 * 3^(1/3) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Extra Question: What is the simplified form of (x^2 y)^(1/3) * (125 x^3)^(1/3) / ( 8 x^3 y^4)^(1/3) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x^2 y)^(1/3) * (125 x^3)^(1/3) / ( 8 x^3 y^4)^(1/3) = [ (x^2 y)^(1/3) / 1 ] * [ 5x / 2xy * y^(1/3) ] = [ (x^2 y)^(1/3) / 1 ] * [ 5 / 2y * y^(1/3) ] = [ (x^2)^(1/3) (y)^(1/3) / 1 ] * [ 5 / 2y * y^(1/3) ] = [ (x^2)^(1/3) / 1 ] * ( 5 / 2y ) = [ 5x^(2/3) ] / 2y confidence rating #$&*: 2y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Extra question. What is the simplified form of sqrt( 4 ( x+4)^2 ) and how did you get this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt( 4 ( x+4)^2 ) = sqrt ( 2^2 (x + 4)^2 ) = 2 (x + 4) confidence rating #$&*: 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 did the math correctly, but I forgot the rule that requires absolute value ------------------------------------------------ Self-critique Rating: OK * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: * Add comments on any surprises or insights you experienced as a result of this assignment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!