course mth 163 lDǥΦlwassignment #003
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17:20:18 `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 easiest variable to eliminate is c in the first two equations. If you eliminate c, you result in 62a+8b=218 Then from the second and third equations the result is 260a+60b=90 confidence assessment:
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MFѴk̂ assignment #002 002. Precalculus I 02-11-2008
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17:30:51 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 variable which is easiest to eliminate is c. From the first and second equations the result would be 58a+2b=-38 Then from the first and third equations the resulting equation would be 198a+13b=-128 self critique assessment: 2
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18:17:16 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 --> The second step would be to eliminate the b's since they have the smallest coefficients. (-7)58a+2b=-38 (2)198a+7b=-128 -406a - 14b=266 396a+14b=-256 --------------------------- -10a=10 a=-1 self critique assessment:
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18:27:06 `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 confidence assessment: 3
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18:28:48 `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 confidence assessment: 3
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18:29:10 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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18:36:13 `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 --> The equation would be a(1)^2+b(1)+c=-2 confidence assessment:
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18:36:42 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 --> self critique assessment: 3
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18:39:56 `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 --> The equations would be 9a+3b+c=5 49a+7b+c=8 confidence assessment: 3
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18:40:36 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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19:15:23 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 subtracted the 1st and 2nd equations and resulted in 40a+4b=3
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19:18:12 `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 --> because I cannot figure out what I did wrong in the last equation, I don't know how to solve #8 confidence assessment:
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19:18:24 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 --> self critique assessment:
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