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16:32:35 `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.......!!!!!!!!...................................
RESPONSE --> The easiest variable to eliminate is the c as its coefficient is always 1 in a quadratic formula.
So for 2a + 3b + c = 128 and 60a +5b +c = 90 First we have to multiply the second equation by a negative 1 in order to change the sign for the second c to a negative. 60a + 5b + c = 90 -1(2a + 3b + c) = 128 60a + 5b + c = 90 -2a - 3b - c = -128 58a + 2b = -38 60a + 5b + c = 90 200a + 10b +c = 0 60a + 5b + c = 90 -1(200a + 10b + c = 0) 60a + 5b + c = 90 -120a - 10b - c = 0 -60a + - 5b = 90 So the two new equations are: 58a + 2b = -38 and -60a - 5b = 90.................................................
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16:37:06 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.......!!!!!!!!...................................
RESPONSE --> I mistakenly subtracted the incorrect equation. Instead of subtracting the the first equation from the third equation, I subtracted the third from the second equation giving me an incorrect final equation.
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16:55:55 `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.......!!!!!!!!...................................
RESPONSE --> First I would multiply the first equation 58a + 2b = -38 by 7 and then the second equation 198a + 7b = -128
So we would then have: 406a + 14b = -266 -396a - 14b = 256 This would give us 10a = -10 So a = -1 To find b I will multiply the first equation by 198 and the second equation by -58. So then we have 11484a - 396b = 7524 and 11484a + 406b = -7424 So we can get 10b = 100 b = 10 So if a = -1 and b = 10 then we can solve the equations and check our answers. 58(-1) + 2(10) = -38 -58 + 20 = -38 (checks out) 198(-1) + 7(10) = -198 + 70 = -128 (checks out).................................................
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17:00:57 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.......!!!!!!!!...................................
RESPONSE --> ok
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17:02:57 `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.......!!!!!!!!...................................
RESPONSE --> 58(-1) + 2b = -38 -58 + 2b = -38 2b = 20 b = 10
198(-1) + 7b = -128 -198 + 7b = -128 7b = 70 b = 10 So we can see that b = 10.................................................
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17:03:03 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.......!!!!!!!!...................................
RESPONSE --> ok
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17:09:45 `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.......!!!!!!!!...................................
RESPONSE --> 2(-1) + 3(10) +c = 128 -2 + 30 + c = 128 28 + c = 128 c = 100
60(-1) + 5(10) + c = 90 -60 + 50 + c = 90 -10 + c = 90 c = 100 200(-1) + 10(10) + c = 0 -200 + 100 + c = 0 -100 + c = 0 c = 100 We can see that c = 100.................................................
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17:09:57 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.......!!!!!!!!...................................
RESPONSE --> ok
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17:14:51 `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 points (1, -2) into the equation y = ax^2 + bx + c =
-2 = a1^2 + b1 + c a + b + c = -2.................................................
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17:14:57 We substitute y = -2 and x = 1 to obtain the equation
-2 = a * 1^2 + b * 1 + c, or a + b + c = -2.......!!!!!!!!...................................
RESPONSE --> ok
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17:19:02 `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 --> If we substitute the x and y values into the equation y = ax^2 + bx + c
For point (3,5) 5 = a(3^2) + b3 + c 9a + 3b + c = 5 For point (7,8) 8 = a(7^2) + b(7) + c 49a + 7b + c = 8.................................................
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17:19:14 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.......!!!!!!!!...................................
RESPONSE --> ok
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17:42:06 `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 --> We get the equations
a(1^2) + b(1) + c = -2 a(3^2) + b(3) + c = 5 a(7^2) + b(7) + c = 8 a + b + c = -2 9a + 3b + c = 5 49a + 7b + c = 8 If we subtract equation 1 from equation 3 we get 49a + 7b + c = 8 a + b + c = -2 We would multiply the second equation by -1 49a + 7b + c = 8 -a - b - c = 2 So we get 48a + 6b = 10 If we subtract equation 2 from equation 3 we get 49a + 7b + c = 8 9a + 3b + c = 5 multiply the second equation by -1 to get rid of c 49a + 7b + c = 8 -9a - 3b - c = 5 So we get 40a + 4b = 13 So now we have 48a + 6b = 10 and 40a + 4b = 13 We have to get rid of one variable in order to solve for the other. We can multiply the first equation by 2 and the second equation by -3 in order to get rid of b. 2(48a + 6b = 10) -3(40a + 4b = 13) 96a + 12b = 20 -120a - 12b = -39 -24a = -19 a = 19 / 24 or 0.79 In order to get b we have to eliminate a. 48a + 6b = 10 40a + 4b = 13 40(48a + 6b = 10) -48(40a + 4b = 13) 1920a + 240b = 400 -1920a - 192b = -624 48b = -224 b = -224 / 48 b = -4.67 0.79(1^2) + -4.67(1) + c = -2 0.79 - 4.67 + c = -2 -3.88 + c = -2 So c = -2 / -3.88 c = 2 / 3.88 c = 0.52.................................................
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17:43:55 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.......!!!!!!!!...................................
RESPONSE --> The system of 2 equations that I obtained when I eliminated c were
48a + 6b = 10 and 40a + 4b = 13.................................................
ݠԜnS|蚢Fa Student Name: assignment #002
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22:59:50 `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.......!!!!!!!!...................................
RESPONSE --> ok
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22:59:51 `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.......!!!!!!!!...................................
RESPONSE -->
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23:00:29 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:00:37 `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.......!!!!!!!!...................................
RESPONSE --> ok
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23:00:42 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:00:47 `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.......!!!!!!!!...................................
RESPONSE --> ok
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23:00:52 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:00:57 `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.......!!!!!!!!...................................
RESPONSE --> ok
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23:01:03 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:01:08 `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 --> ok
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23:01:13 We substitute y = -2 and x = 1 to obtain the equation
-2 = a * 1^2 + b * 1 + c, or a + b + c = -2.......!!!!!!!!...................................
RESPONSE --> ok
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23:01:18 `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 --> ok
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23:01:23 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:01:34 `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 --> ok
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23:01:44 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.......!!!!!!!!...................................
RESPONSE --> ok
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23:17:21 `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 you substitute the numbers into the equation you get a quadratic equation.
f(x) = 0.79x^2 + 4.67x + 0.52 When you substitute 1 into the equation you get f(1) = 0.79(1^2) + 4.67(1) + 0.52 f(1) = 0.79(1) + 4.67 + 0.52 f(1) = 0.79 + 4.67 + 0.52 f(1) = 5.98 When you substitute 3 for x you get f(3) = 0.79(3^2) + 4.67(3) + 0.52 f(3) = 0.79(9) + 4.67(3) + 0.52 f(3) = 7.11 + 14.01 + 0.52 f(3) = 21.64 When we substitute 5 into the equation you get f(5) = 0.79(5^2) + 4.67(5) + 0.52 f(5) = 0.79(25) + 4.67(5) + 0.52 f(5) = 19.75 + 23.35 + 0.52 f(5) = 43.62 When we substitute 7 into this equation you get f(7) = 0.79(7^2) + 4.67(7) + 0.52 f(7) = 0.79(49) + 4.67(7) + 0.52 f(7) = 38.71 + 32.69 + 0.52 f(7) = 71.92.................................................
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23:17:31 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).......!!!!!!!!...................................
RESPONSE --> ok
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vad٫ʙ͑I Student Name: assignment #003
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