course mth 163 \J~v̜кRdخassignment #002
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19:18:35 `q001. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. NOTE SOLN IS -1, 10, 100.
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RESPONSE --> The variable that is easiest to eliminate is c. So to eliminate it from the first two equations, multiply one by -1: -1 (2a + 3b + c = 128) -2a - 3b - c = -128 60a + 5b + c = 90 Add together to get : 58a + 2b = -38 Then eliminate c the same way with the first and third equations: -1 (2a + 3b + c = 128) -2a - 3b - c = -128 200a + 10b + c = 0 Add together and you get: 198a + 7b = -128 So your two final equations with remaining variables are: 58a + 2b = -38 and 198a + 7b = -128 confidence assessment: 2
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19:20:05 The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become 'new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might have obtain this system, or one equivalent to it, using a slightly different sequence of calculations.
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RESPONSE --> I used a dfferent method but achieved the same results. self critique assessment: 3
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19:26:19 `q002. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 , which can be obtained from the system in the preceding problem, by eliminating the easiest variable.
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RESPONSE --> b is the easiest variable to eliminate. In order to do so, multiply the equation 58a + 2b = -38 by 7 to get: 406a + 14b = -266 Take the second equation and multiply by -2: -396a - 14b = 256 Add the two equations: 406a + 14b = -266 -396a - 14b = 256 10a = -10 (divide by 10) a = -1 Substitute back in (58 * -1) + 2b = -38 -58 + 2b = -38 (add 58 to both sides) 2b = 20 b = 10 confidence assessment: 3
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19:26:54 Neither variable is as easy to eliminate as in the last problem, but the coefficients of b a significantly smaller than those of a. So we will eliminate b. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the program multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128). Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1.
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RESPONSE --> I went a step further and found b as well. self critique assessment: 3
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19:33:34 `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.
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RESPONSE --> Substitute back in (58 * -1) + 2b = -38 -58 + 2b = -38 (add 58 to both sides) 2b = 20 b = 10 Validation: (58 * -1) + (2 * 10) = -38 -58 + 20 = -38 -38 = -38 it is validated confidence assessment: 3
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19:33:52 You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10.
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RESPONSE --> self critique assessment: 3
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19:42:06 `q004. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.
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RESPONSE --> To find c substitute a = -1 and b = 10 into the first equation: (2 * -1) + (3 * 10) + c = 128 -2 + 30 + c = 128 28 + c = 128 c = 100 Validate by using the second equation: (60 * -1) + (5 * 10) + 100 = 90 -60 + 50 + 100 = 90 -10 + 100 = 90 90 = 90 confidence assessment: 3
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19:42:15 Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which was some simple arithmetic gives us c = 100. Substituting these values into the second equation we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will begin obtain an identity. This would completely validate our solution.
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RESPONSE --> self critique assessment: 3
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19:46:10 `q005. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case for the graph you sketched in the preceding assignment, then what equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?
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RESPONSE --> If we substitute into the formula: -2 = a * 1^2 + b* 1 + c We get: -2 = a + b + c confidence assessment: 1
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19:46:39 We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2.
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RESPONSE --> I thought there was more to that problem, but there was not and I had the right answer: self critique assessment: 3
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20:08:24 `q006. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what equations do we get if we substitute the x and y values corresponding to the point (3, 5), then (7, 8) into the form y = a x^2 + b x + c?
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RESPONSE --> Substitute (3, 5) into equation: 5 = a * 3^2 + b * 3 + c 5 = 9a + 3b +c Substitute (7, 8) into equation: 8 = a * 7^2 + b * 7 + c 8 = 49a + 7b + c confidence assessment: 3
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20:08:39 Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 7.
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RESPONSE --> self critique assessment: 3
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20:58:03 `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then what system of equations do we get if we substitute the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c? What is the solution of this system?
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RESPONSE --> As we saw in the preceeding questions, the equations we get when substituting the points are: (1, -2) : a + b + c = -2 (3, 5): 9a + 3b + c = 5 (7, 8): 49a + 7b + c = 8 Add equations together: Multiply first equation by -1 -a - b - c = 2 9a + 3b + c = 5 This gives us: 8a + 2b = 7 Add the first and third equations -a - b - c = 2 49a + 7b + c =8 This gives us: 48a + 6b = 10 Add these two together (after multiplying the first equation by -3): 48a + 6b = 10 -24a - 6b = -21 24a = -11 a= -11/24 I don't know if I am supposed to get a fraction.... confidence assessment: 1
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21:03:03 The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different system, you should show the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.
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RESPONSE --> I wasn't sure if a fraction was right or not. I see that it was and that the -11/24 = a was correct. I understand how to find b and c. Like I mentioned before, I just didn't realize that a fraction was correct. self critique assessment: 2
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21:26:44 `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c. What function do you get? What do you get when you substitute x = 1, 3, 5 and 7 into this function?
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RESPONSE --> When we substitute the values in we get: y = (-.45833)x^2 + 5.3333x -6.8745 x = 1: y = (-.45833 * 1^2) + (5.333 * 1) -6.8745 y = -.45833+5.3333 - 6.8745 y = -1.9995 (in our points it is -2) x = 3: y = (-.45833 * 3^2) + (5.3333 * 3) -6.8745 y = -4.12497 + 15.9999 - 6.8745 y = 5.00043 x = 5: y = (-.45833 * 5^2) + (5.3333 * 5) -6.8745 y = -11.45825 + 26.6665 - 6.8745 y = 8.33375 x = 7: y = (-.45833 * 7^2) + 5.3333 * 7 - 6.8745 y = -22.45817 + 37.3331 - 6.8745 y = 8.0043 confidence assessment: 3
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21:27:06 Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values gives us the points (1, -2), (3, 5) and (7, 8) we used to obtain the formula, plus the point (5, 8.33333).
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RESPONSE --> I just did not put them in coordinate form. self critique assessment: 3
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