course Mth 163 «X}Îÿ«øßÕ¹˜šÙë\Èñ³ññÓ´í|assignment #002
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10:57:30 `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 first step would be to divide up the three equations and i distinguished them by saying that the first two are A and B while the third equation is C. So we take B equation (60a +5b +c=90) and subtract that from the A equation (2a +3b+c=128) which comes to the solution of 58a + 2b= -38 To get the second part, we take the C equation (200a+10b+c=0) and subtract it from the B equation (60a +5b +c=90) which comes to the answer of 140a+5b=-90. confidence assessment: 3
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11:05:32 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 was a bit lost in exactly how to get the third equation. I get that we are to use the answer of the two equations that we solved but then what do we do is lost on me. I tried to use the example that you gave us but I couldn't see how that answer came about. self critique assessment: 2
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11:55:35 `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 --> To solve this equation we subtract both the equations where the end equation is 406a+2b(7b) = -266 subtracted from 396a - 7b(2b)= -25b. This results in 10a=-10 so that a=-1. confidence assessment: 2
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11:56:35 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 understood this equation fairly well compared to the first equation because I was confused there. self critique assessment: 2
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12:03:10 09-07-2008 12:03:10 `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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NOTES ------->
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12:05:15 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 --> Here it is essential to remember that the a factors need to be subtracted from the equation for just the b factor to be solved. self critique assessment: 2
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12:35:53 `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 --> For this equation, I used the first equation 2a+3b+c=128 which gets us 2(-1) +3(10)+c = 128 , -2 +30 +c = 128. The equation would then give us 28 + c = 128. We subtract 28 from both sides and the result is c = 100. confidence assessment: 3
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12:37:26 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 --> I understood and got the answer that is provided for the second equation as well. self critique assessment: 2
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12:43:19 `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 --> I would think that the equation here is -2=a(1) +b(1) + c which could later equal each other. confidence assessment: 1
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12:43:32 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 --> ok self critique assessment: 3
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12:46:52 `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 --> When we use (3,5) points, we get an equation of 9a+3b+c =5 and with the points (7,8), we get the equation of 49a+7b+c=8. confidence assessment: 2
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12:48:18 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 --> ok, I think that the answer on the answer box is wrong because the second equation should have an answer of 8 instead of 7. self critique assessment: 2
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13:25:05 `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 --> To begin with the first three equations are as follows A--> 1a+1b+c=-2 B--> 9a+3b+c=5 C--> 49a+7b+c=8. The first thing that we do here is take the B equation - A equation which gives us 8a+2b+c=7.
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13:25:41 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 --> ok self critique assessment: 2
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14:34:59 `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 --> The following functions I got : the first time I put the numbers y = -2. The second was y=5.002, y=31.24 and y=33.662 confidence assessment: 2
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14:35:24 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 --> ok self critique assessment: 2
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