course Mth 163 assignment #013013. `query 13
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12:18:51 `questionNumber 130000 Query last asst before test 1, problem 1.Give your solution to x ^ 3 / 17 = 58
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RESPONSE --> Plug in and solve for x. x=9.947 confidence assessment: 2
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12:19:13 `questionNumber 130000 ** 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 --> I see self critique assessment: 2
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12:20:28 `questionNumber 130000 Give your solution to (3 x) ^ -2 = 19
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RESPONSE --> remove the -1/2 from each side to get x=[19^(-1/2)]/3=- or + .0765 confidence assessment: 2
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12:20:39 `questionNumber 130000 ** (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 get it. self critique assessment: 2
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12:21:41 `questionNumber 130000 Give your solution to 4 x ^ -.5 = 7
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RESPONSE --> multiply by 1/4 and get x^-.5^-2=7/4^-2 x=.3269 confidence assessment: 2
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12:22:03 `questionNumber 130000 ** 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 --> Answer as rounded up but we both got similar answers self critique assessment: 2
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12:22:55 `questionNumber 130000 Give your solution to 14 x ^ (2/3) = 39
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RESPONSE --> Again, multiply sides by 1/14 and the solution is x=(39/14)^(3/2)=4.65 confidence assessment: 2
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12:23:06 `questionNumber 130000 ** 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: 2
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12:24:58 `questionNumber 130000 Give your solution to 5 ( 3 x / 8) ^ (-3/2) = 9
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RESPONSE --> Again we multiply both sides, this time by 1/5 this gives 9/5. then raise sides to -2/3 rd power to get 3x/8=9/5^-2/3. Multiply again 8/3 and get x=1.8 confidence assessment: 2
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12:25:20 `questionNumber 130000 ** 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 --> This one was a little harder b/c there were more steps but we got the same answer. self critique assessment: 2
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12:27:25 `questionNumber 130000 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 --> Substitute values in on paper and solutions are a(1)=2 a(2)=2.5 a(3)=3.5 a(4)=5 a(5)=7 confidence assessment: 2
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12:27:36 `questionNumber 130000 ** 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 --> I got it. self critique assessment: 2
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12:30:17 `questionNumber 130000 What is your quadratic function and what is its value for n = 4? Does it fit the sequence exactly?
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RESPONSE --> Substitute numbers into quadratic form, getting the equations, solving for a,b,and c then putting into the equation and solving for 0.25x^2-0.25x+2 I believe it fits confidence assessment: 2
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12:30:26 `questionNumber 130000 ** 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·x^2 - 0.25·x + 2. **
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RESPONSE --> I got it. self critique assessment: 2
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12:32:43 `questionNumber 130000 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 --> after solving we get the points (0,7) (4, 5.45) (.8, 3.99) (1.2, 2.632) (1.6, 1.368) (2,.2) confidence assessment: 2
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12:33:05 `questionNumber 130000 ** 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 just solved for the points but I see how the other values were obtained. self critique assessment: 2
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12:36:20 `questionNumber 130000 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 --> Slope= rise/run average slope = a (k+h) +b confidence assessment: 2
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12:36:30 `questionNumber 130000 ** 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 got it. self critique assessment:
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