course Mth 163 Sorry this is soo late. ϶㋁Tvb䞄
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09:11:35 `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 --> First we look at which variable is the easiest to eliminate. In this case, c would be the easiest to remove first. So, we subtract the equations: 200a + 10b + c = 0 2a + 3b + c = 128 We get 198a + 7b = -128 for the difference. Then we subtract the next ones: 60a + 5b + c = 90 2a + 3b +c = 128 We arrive at the new equation 58a + 2b = -38 confidence assessment: 2
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09:13:47 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 --> The c variable was easy to eliminate because it was just 1 on all three equations. self critique assessment: 2
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09:21:15 `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 --> The next easiest variable to eliminate would be b since it's numbers are smaller than a's. To eliminate it we'd have to multiply both equations by the opposite numbers for b. 7 * (58a + 2b = -38) = 406a + 14b + -266 -2 (198a + 7b = -128) = -398a - 14b = 256 So now we subtract these two new equationis: 406a +14b = -266 -398a - 14b = 256 Now we get to the equation just with the variable a: 8a = -10 confidence assessment: 1
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09:24:31 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 multiplied the equations differently, the first by a positive 7 and the secoond by a negative 2. I see what a difference it makes by making the 7 negative and 2 positive. self critique assessment: 1
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09:28:24 `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 --> I used the first equation since the numbers were smaller to solve for b. With a = -1 we can substitute it in the equation: 58a + 2b = -38 58(-1) + 2b = -38 -58 + 2b = -38 Now we have to add 58 on both sides of the equation to get the equation just for b: 58 + -58 +2b = -38 + 58 2b = 20 b = 10 If you substitute -1 for a in the second equation, b will also equal 10. confidence assessment: 2
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09:28:55 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 --> I arrived at the same answer. self critique assessment: 3
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09:31:39 `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 --> I will use the first equation to find the value of variable c: 2a + 3b + c = 128 2(-1) + 3(10) + c = 128 -2 + 30 + c = 128 28 + c = 128 -28 + 28 + c = 128 - 28 c = 100 confidence assessment: 2
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09:32:30 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 arrived at the same answer and also substituted the values in to the other equations to check myself. self critique assessment: 3
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09:38:13 `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 the point (1, -2) into the equation y = a x ^2 + b x + c we will get : -2 = a (1)^2 + b(1) + c We could take that down to: -2 = a^2 + b + c confidence assessment: 2
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09:39:22 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 wasn't sure about a * 1 ^2, I didn't know if the power of 2 was just for the x variable or for the a * x raised to the power of 2. self critique assessment: 2
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09:42:05 `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 --> First, for the points (3,5) into the form would be: 5 = a*3^2 + b*3 + c 5 = 9a + 3b + c For the points (7,8) we get: 8 + a*7^2 + b*7 + c 8 = 49a +7b + c confidence assessment: 3
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09:42:23 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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09:56:39 `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 --> From the following problems we got these following equations when we substituted the x and y values into the form y = a*x^2 + b*x + c: (1, -2) : -2 = a + b + c (3, 5) : 5 = 9a + 3b + c (7, 8) : 8 = 49a + 7b + c Next we will eliminate the variables until we find the values for the variables. We'll subtract the third and first equations, then the second and first equations: 8 = 49a + 7b + c -2 = a + b + c We get 10 = 48a + 6b 5 = 9a + 3b +c -2 = a + b + c We get 7 = 8a + 2b We eliminate another variable: Since 6 is a multiple of 2 we can just multiply the second equation by -3: -3 ( 7 = 8a + 2b) = -21 = -24a - 6b Now we can subtract the two equations: 10 = 48a + 6b -21 = -24a - 6b -11 = 24a a = -11 / 24 a = -0.5 confidence assessment: 1
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09:57:57 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 didn't go and find the values of all the variables. I rounded the value of a to 0.5, I did get the same value in decimal form as there. I'm sure that if I had solved for b and c I would've obtained the same values. self critique assessment: 2
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10:09:05 `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 --> Using the values from the previous problem we get: y = 0.4583*x^2 + b*-1.9997 + -0.4588 If we substitutes 1, 3, 5, and 7 into it we'll find: y = 0.4583*1^2 + 1*-1.9997 + -0.4588 y = 0.4583 + -1.9997 = - 0.4588 y = -2.0002 y = 0.4583*3^2 + 3*-1.9997 + -0.4588 y = 4.1247 + -5.9991 = -0.4588 y = -2.3332 y = 0.4583*5^2 + 5*-1.9997 + -0.4588 y = 11.4575 + -9.9985 + -0.4588 y = 1.0002 y = 0.4583*7^2 + 7*-1.9997 + -0.4588 Y = 22.4567 + -13.9979 + -0.4588 Y = 8 confidence assessment: 2
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10:10:03 02-29-2008 10:10:03 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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NOTES -------> I didn't substitute the right numbers into the equations. Not sure if I used the wrong values for the a, b, c or what I did.
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10:10:04 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 --> self critique assessment: 2
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