course Mth163 {Á΄¿W®žz¬Â²ó¸¡ïÅæéy¬æìassignment #002
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12:13:07 `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 --> 2a + 3b + c = 128 60a + 5b + c = 90 -58a-2b=38 2a + 3b + c = 128 200a + 10 b + c = 0 -198a-7b=128 7(-58a-2b=38) 2(-198a-7b=128) (-406a-14b=266) (-396a-14b=256) -10a=10 a=-1 58-2b=38 -2b=-20 b=10 -2+30+c=128 30+c=130 c=100 confidence assessment: 2
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12:13:52 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 answered the whole thing and checked my work by solving the two equations. self critique assessment: 3
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12:14:58 `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 --> a=-1 b=10 i answered this in the last question, so my work is on it. confidence assessment: 3
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12:16:03 `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 --> b=10 my work is in question 1 confidence assessment: 3
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12:16:22 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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12:16:51 `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 --> c=100 my work is in question 1 confidence assessment: 3
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12:17:07 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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12:18:41 `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 --> -2=a(1^2)+b(1)+c confidence assessment: 3
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12:19:08 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 didnt solve the problem down to simplest form, but i understand how to do so. self critique assessment: 2
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12:20:57 `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 --> 5=a(3^2)+b(3)+c 9a+3b+c=5 8=a(7^2)+b(7)+c 49+7b+c=8 confidence assessment:
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12:21:13 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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12:35:08 `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 --> a+b+c=-2 9a+3b+c=5 49a+7b+c=8 a=-11/24 b=-16/3 c=-91/24 confidence assessment: 2
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12:36:30 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 missed C. I think i just plugged in the wrong numbers and went wrong somewhere with my addition and subtraction. self critique assessment: 2
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12:48:44 `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 --> y=-.45833x^2+-5.33333x+3.79166 when x=1 y=-2 when x=3 y=-16.3333 when x=5 y=-34.33324 when x=7 y=-55.99982 confidence assessment: 2
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12:49:42 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 --> we have a different value for c. i dont understand how you got your value of c. i worked the problem 2 times and got the same answer. self critique assessment: 2
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