course Mth 163 ˜á¿²§Ž©êµÆè·©»â´Dàw£assignment #013
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19:52:55 Query last asst before test 1, problem 1.Give your solution to x ^ 3 / 17 = 58
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RESPONSE --> x^3 / 17 = 58 x^3 = 986 x = 986^(1/3) x = 9.95 confidence assessment: 3
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19:53:07 ** 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 --> ok self critique assessment: 3
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19:59:06 Give your solution to (3 x) ^ -2 = 19
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RESPONSE --> (3x)^-2 = 19 1/9x^2 = 19
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19:59:47 ** (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 --> ((3x)^-2)^(-1/2)) = 19^(-1/2) gives us 3x = 19^(-1/2) so that x = [ 19^(-1/2) ] / 3 = .0765 or -.0765. self critique assessment: 3
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20:02:03 Give your solution to 4 x ^ -.5 = 7
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RESPONSE --> 4x^-.5 = 7 x^-.5 = 7/4 = 1.75 x = 1.75^-2 x = 1/3.06 or .327 confidence assessment: 2
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20:02:12 ** 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 --> ok self critique assessment: 3
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20:03:39 Give your solution to 14 x ^ (2/3) = 39
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RESPONSE --> 14x^(2/3) = 39 x^(2/3) = 39/14 = 2.79 x = 2.79^(3/2) x = 4.66 confidence assessment: 3
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20:03:47 ** 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 --> ok self critique assessment: 3
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20:10:42 Give your solution to 5 ( 3 x / 8) ^ (-3/2) = 9
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RESPONSE --> 5(3x/8)^(-3/2) = 9 3x/8^(-3/2) = 9/5 = 1.8 3x/8 = 1.8^(-2/3) = .676 .375x = .676 x = 1.8 confidence assessment: 2
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20:10:52 ** 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 --> ok self critique assessment: 3
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20:11:41 ** 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 --> 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. self critique assessment: 3
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20:12:45 ** 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 --> 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. self critique assessment: 3
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20:21:53 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 --> f(0) = .3(0^2) - 4(0) + 7 f(0) = 7 f(.4) = .3(.4^2) - 4(.4) +7 f(.4) = .048 - 1.6 +7 f(.4) = 5.448 f(.8) = .3(.8^2) - 4(.8) + 7 f(.8) = .192 - 3.2 +7 f(.8) = 3.992 f(1.2) = .3(1.2^2) - 4(1.2) +7 = .432 - 4.8 + 7 = 2.632 f(1.6) = .3(1.6^2) - 4(1.6) + 7 = .768 - 6.4 + 7 = 1.368 f(2) = .3(2^2) - 4(2) +7 = 1.2 - 8 +7 = 0.2 confidence assessment: 3
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20:22:23 ** 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 --> (0, 7) (4, 5.448) (.8, 3.992) (1.2, 2.632) (1.6, 1.368) (2, .2) self critique assessment: 3
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20:23:08 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 --> not sure confidence assessment: 0
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20:23:24 ** 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 --> 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. self critique assessment: 3
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