course Mth 163 I still cannot get the Q_A form to work. Only the Query works. Also, I hope you got my email about Test 1. I went ahead and did the best I could on this Assignment. ???S???????|
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11:00:37 Query last asst before test 1, problem 1.Give your solution to x ^ 3 / 17 = 58
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RESPONSE --> For finding the solution to the problem, first both sides must be multiplyed by 17 to get x^3 = 58 * 17. Next,the problem solver must take the 1/3 power of both sides, getting a solution such as...... (x^3)^(1/3) = (58 * 17)^(1/3)=x=9.95 confidence assessment: 3
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11:24:40 ** The solution to x^3 / 17 = 58 is found by first multiplying both sides by 17 to get x^3 = 58 * 17 then taking the 1/3 power of both sides, obtaining (x^3)^(1/3) = (58 * 17)^(1/3) or x = 9.95, approx.. COMMON ERROR: If you interpret the equation as x^(3/17) = 58 you will get solution x = 58^(17/3) = 9834643694. However this is not the solution to the given equation To interpret x ^ 3 / 17 you have to follow the order of operations. This means that x is first cubed (exponentiation precedes multiplication or division) then divided by 17. If you introduce the grouping x^(3/17) you are changing the meaning of the expression, causing 3 to be divided by 17 before exponentiation. **
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RESPONSE --> Answering the problem step by step like you asked me to do helps me review the problem better as I am giving the answer rather than taking it straight off the paper. self critique assessment: 3
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11:37:17 Give your solution to (3 x) ^ -2 = 19
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RESPONSE --> For the problem, (3x)^-2 = 19, the first step is to be solved by taking the 1/2 power of both sides. ((3x)^-2)^(-1/2)) = 19^(-1/2)=3x = 19^(-1/2), therefore, x = [ 19^(-1/2) ] / 3 = .0765 confidence assessment: 2
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11:40:30 ** (3x)^-2 = 19 is solved by taking the -1/2 power of both sides, or the negative of the result: ((3x)^-2)^(-1/2)) = 19^(-1/2) gives us 3x = 19^(-1/2) so that x = [ 19^(-1/2) ] / 3 = .0765 or -.0765. **
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RESPONSE --> I see where I messed up...I didn't put the negative sign in. I do that a lot, unfortunately. I need to pay more attention to the problem. self critique assessment: 2
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11:53:18 Give your solution to 4 x ^ -.5 = 7
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RESPONSE --> For solving the equation, first both sides must be multiplied by 1/4, then I get x ^ -.5 = 7 / 4 Then both sides should be raised to the - Power------ For example,(x^ -.5)^-2 = (7/4)^-2 therefore the solution is x = .327 confidence assessment: 3
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11:59:46 ** to solve the equation we first multiply both sides by 1/4 to get x ^ -.5 = 7 / 4. Then we raise both sides to the -2 power: (x^-.5)^-2 = (7/4)^-2 so x = .327 approx **
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RESPONSE --> I was more confident about this question. self critique assessment: 3
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12:11:43 Give your solution to 14 x ^ (2/3) = 39
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RESPONSE --> First, I multiply both sides by 1/14 which equals x^(2/3) = 39/1 Then I once again raise both sides, this time to the 3/2 power to get a solution of x = (39/14)^(3/2) = 4.65 confidence assessment: 2
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12:24:40 ** We first multiply both sides by 1/14 to get x^(2/3) = 39/14. The we raise both sides to the 3/2 power to get x = (39/14)^(3/2) = 4.65. **
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RESPONSE --> I got it. self critique assessment: 0
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12:31:16 Give your solution to 5 ( 3 x / 8) ^ (-3/2) = 9
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RESPONSE --> x = 8/3 * (9/5)^(-2/3) = 1.80 confidence assessment: 2
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12:40:45 ** multiplying both sides by 1/5 we get (3x/8)^(-3/2) = 9/5. Raising both sides to the -2/3 power we have 3x / 8 = (9/5)^(-2/3). Multiplying both sides by 8/3 we obtain x = 8/3 * (9/5)^(-2/3) = 1.80 **
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RESPONSE --> Oh crap. I got carried away and forgot to put my steps in. Sorry. self critique assessment: 0
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12:47:39 Query problem 2. a(n+1) = a(n) + .5 n, a(0) = 2 What are a(1), a(2), a(3), a(4) and a(5)?
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RESPONSE --> Substituting n = 0, I got a(0+1) = a(0) + .5 * 0 Which is simplifyed to get: a(1) = a(0) Substituting a(0) = 2 from the information given I got a(1) = 2 Substituting n = 1 I got a(1+1) = a(1) + .5 * 1 which is simplifyed to get a(2) = a(1) + .5 Substituting a(1) = 2 from the earlier step I got a(2) = 2.5 Substituting n = 2 I got a(2+1) = a(2) + .5 * 2 which is simplifyed to get: a(3) = a(2) + 1 Substituting a(2) = 2.5 from the earlier step I got a(3) = 2.5 + 1 = 3.5 Substituting n = 3 I got a(3+1) = a(3) + .5 * 3 which is simplifyed to get: a(4) = a(3) + 1.5 Substituting a(3) = 3.5 from the earlier step I got a(4) = 3.5 + 1.5 = 5 confidence assessment: 3
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12:57:14 ** Substituting n = 0 we get a(0+1) = a(0) + .5 * 0 which we simplify to get a(1) = a(0). Substituting a(0) = 2 from the given information we get a(1) = 2. Substituting n = 1 we get a(1+1) = a(1) + .5 * 1 which we simplify to get a(2) = a(1) + .5. Substituting a(1) = 2 from the previous step we get a(2) = 2.5. Substituting n = 2 we get a(2+1) = a(2) + .5 * 2 which we simplify to get a(3) = a(2) + 1. Substituting a(2) = 2.5 from the previous step we get a(3) = 2.5 + 1 = 3.5. Substituting n = 3 we get a(3+1) = a(3) + .5 * 3 which we simplify to get a(4) = a(3) + 1.5. Substituting a(3) = 3.5 from the previous step we get a(4) = 3.5 + 1.5 = 5. Substituting n = 4 we get a(4+1) = a(4) + .5 * 4 which we simplify to get a(5) = a(4) + 2. Substituting a(4) = 5 from the previous step we get a(5) = 5 + 2 = 7. **
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RESPONSE --> That was a long process. I think I kept up with it pretty well though. self critique assessment: 3
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13:30:54 What is your quadratic function and what is its value for n = 4? Does it fit the sequence exactly?
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RESPONSE --> Points (1,2), (3,3.5) and (7,5) were substituted into the problem y = a x^2 + b x + c to get the follwing three equations: 2 = a * 1^2 + b * 1 + c 3.5 = a * 3^2 + b * 3 + c 7 = a * 5^2 + b * 5 + c Then solving the resulting system for a, b and c, I find a = .25, b = -.25 and c = 2, giving one the equation of: 0.25?^2 - 0.25? + 2 confidence assessment: 2
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13:45:08 ** Using points (1,2), (3,3.5) and (7,5) we substitute into the form y = a x^2 + b x + c to obtain the three equations 2 = a * 1^2 + b * 1 + c 3.5 = a * 3^2 + b * 3 + c 7 = a * 5^2 + b * 5 + c. Solving the resulting system for a, b and c we obtain a = .25, b = -.25 and c = 2, giving us the equation 0.25?^2 - 0.25? + 2. **
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RESPONSE --> Correct! self critique assessment: 3
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13:57:10 Query problem 3. f(x) = .3 x^2 - 4x + 7, evaluate at x = 0, .4, .8, 1.2, 1.6 and 2.0.
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RESPONSE --> Points:(0, 7), (4, 5.448), (.8, 3.992), (1.2, 2.632), (1.6, 1.368), (2, .2) The Y values are 7, 5.448, 3.992, 2.632, 1.368, 0.2 Differences between the Y values would be: -1.552, -1.456, -1.36, -1.264, -1.168 confidence assessment: 1
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14:35:26 ** We obtain the points (0, 7) (4, 5.448) (.8, 3.992) (1.2, 2.632) (1.6, 1.368) (2, .2) y values are 7, 5.448, 3.992, 2.632, 1.368, 0.2. Differences are 7-5.448 = -1.552, 3.992 - 2.632 = -1.456, etc. The sequence of differences is -1.552, -1.456, -1.36, -1.264, -1.168. The rate of change of the original sequence is proportional to this sequence of differences. The differences of the sequence of differences (i.e., the second differences) are .096, .096, .096, .096, .096.. These differences are constant, meaning that the sequence of differences is linear.. This constant sequence is proportional to the rate of change of the sequence of differences. The differences are associated with the midpoints of the intervals over which they occur. Therefore the difference -1.552, which occurs between x = 0 and x = .4, is associated with x = .2; the difference -1.456 occuring between x = .4 and x = .8 is associated with x = .6, etc.. The table of differences vs. midpoints is } 0.2, -1.552 -.6, -1.456 1, -1.36 1.4, -1.264 1.8, -1.168 This table yields a graph whose slope is easily found to be constant at .24, with y intercept -1.6. The function that models these differences is therefore y = 2.4 x - 1.6. **
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RESPONSE --> I could get the basis of the problem, but after that I was confused. After looking over it though and writing the entire problem down step by step the way you worked it, i understand it now and I can use it to practice problems like this in the future. self critique assessment: 3
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14:50:13 Query problem 4. f(x) = a x^2 + b x + c What symbolic expression stands for the average slope between x = h and x = k?
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RESPONSE --> Average slope(rise/Run) = [ f(k) - f(h) ] / (k - h) = ( a k^2 + b k + c - ( a h^2 + b h + c) ) / ( k - h) After simplifying, the average slope would be: slope(rise/Run) = a ( k + h) + b confidence assessment: 3
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15:22:30 ** The average slope is rise / run = [ f(k) - f(h) ] / (k - h) = ( a k^2 + b k + c - ( a h^2 + b h + c) ) / ( k - h). We simplify this to get ave slope = ( a ( k^2 - h^2) + b ( k - h) ) / ( k - h), which we write as ave slope = ( a ( k-h) ( k+h) ) + b ( k - h) ) / (k - h). k - h is a factor of the numerator so we have the final form ave slope = a ( k + h) + b. **
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RESPONSE --> I could've went in to detail more.