course Mth 163
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20:30:57 `questionNumber 20000 `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 --> Use elimination- c then eliminate another variable- end up w/ one variable. Use c because it has simplest coefficient. Subtract equation 1 from equation 2 then subtract equation 1 from 3. 60a+5b+c=90 -2a+3b+c=128= 58a+2b=-38 200a+10b+c=0 -2a+3b+c=128= 198a+7b=-128 confidence assessment: 2
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20:31:37 `questionNumber 20000 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 totally understand self critique assessment: 3
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20:37:54 `questionNumber 20000 `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 --> 58 a + 2 b = -38 198 a + 7 b = -128 Eliminate variable with smaller coefficients=b Multiply 1st equation by-7=-38(-7)=58a(-7)+2b(-7) Multiply 2nd equation by 2=-128(2)=198a(2)+7b(2) 1st equ- 266=-406a-14b + -256=396a+14b= 10=-10a -1=a confidence assessment: 2
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20:38:32 `questionNumber 20000 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 got it. self critique assessment: 3
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20:39:58 `questionNumber 20000 `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 --> 198(-1)+7b=-128 7b=-128+198 7b/7=70/7 b=10 confidence assessment: 2
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20:40:16 `questionNumber 20000 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 --> Got it. self critique assessment: 3
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20:42:52 `questionNumber 20000 `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 --> a=-8+b=10 2(-1)+3(10)+c=128 c=128+2-30 c=100 Verify all 3 numbers a=-1, b=10, and c=100 by plugging into another equation. 60(-1)+5(10)+100=90 -60+50+100=90 90=90 confidence assessment: 3
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20:43:16 `questionNumber 20000 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 understand this very well. self critique assessment: 3
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20:45:06 `questionNumber 20000 `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 --> y=ax^2+bx+c just substitute the (x,y) value (1,-2) -2=a*1^2+b*1+c confidence assessment: 3
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20:45:39 `questionNumber 20000 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 got it self critique assessment: 3
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20:50:04 `questionNumber 20000 `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 --> y=ax^2+bx+c x and y values (3,5) 5=a*3^2+b*3+c 5=a*9+3b+c 5=9a+3b+c y=ax^2+bx+c 2nd x and y values (7,8) 8=a*7^2+b*7+c 8=49a+7b+c confidence assessment: 2
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20:51:15 `questionNumber 20000 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 --> I understand the answer. self critique assessment: 2
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20:56:56 `questionNumber 20000 `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 --> y=a*x^2+b*x+c Use points (1,-2), (3,5), and (7,8) and substitute into formula. (1,-2) -2=a+b+c (3,5) 5=9a+3b+c (7,8) 8=49a+7b+c 5=9a+3b+c -(-2)=a+b+c 7=8a+2b 8=49a+7b+c -(-2)=a+b+c 10=48a+6b confidence assessment: 2
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21:12:19 `questionNumber 20000 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 --> When reading my notes on my paper I forgot to finish the last half of my equation. I will do so now. 7(6)=8a(6)+2b(6) 42=48a+12b +(-20)=-96a+-12b 22/-48=-48a/-48 a=-.4583 10(-2)=48a(-2)+6b(-2) -20=-96a+(-12b) Use equation from a and b=7=8a+2b Plug in a = 7=8(-.4583)+2b 7=-3.667+2b 7+3.667=2b 10.667/2=2b/2 5.3335=b Then plug a & b into one of the original equations -2=a+b+c -2=(-.4583)+(5.3335)+c -2+(.4583)-(5.3335)=c -6.8752=c A=-.4583 B=5.333 C=-6.8752 self critique assessment: 2
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21:16:44 `questionNumber 20000 `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 = a x^2 + b x + c y=(-.4583)x^2+(5.333)x+(-6.8752) y=(-.4583)*1^2+(5.333)(1)+(-6.8752) y=-2.00 y=(-.4583)*3^2+(5.333)(3)+(-6.8752) y=4.9991 y=(-.4583)*5^2+(5.333)(5)+(-6.8752) y=8.3323 y=(-.4583)*7^2+(5.333)(7)+(-6.8752) y=7.9991 confidence assessment: 2
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21:17:06 `questionNumber 20000 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 --> I understand self critique assessment: 3
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