assignment 2

course mth 163

001. Note that this assignment has 8 questionsBegin 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:

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0. NOTE SOLN IS -1, 10, 100.

D|եnO䇔

assignment #002

002.

Precalculus I

07-08-2007

vwy~Е

assignment #002

002.

Precalculus I

07-08-2007

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09:18:14

`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 -->

equation 2 - equation 1 :

60a-2a

5b-3b

c cancels out c

= -38

58a+2b=-38

equation 3-equation 1:

200a-2a

10b-3b

c cancels out c

= -128

198a+7b=-128

confidence assessment: 3

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09:18:59

`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 -->

i understood this equation elimination

confidence assessment: 3

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09:20:32

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 understand

self critique assessment: 3

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09:25:30

`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 -->

58a+2b=-38

58(-1)+2b=-38

-58+2b=-38

2b=20

b=10

confidence assessment: 3

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09:25:48

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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09:32:10

`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 -->

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 =100

-100 +c=100

c=100

confidence assessment: 3

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09:33:01

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 -->

ok

self critique assessment: 3

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09:37:33

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 -->

a+b+c=-2

self critique assessment: 3

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09:41:54

`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

5=9a + 3b+ c

8=49a+7b+c

confidence assessment: 2

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09:42:48

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 got the equation 49a+7b+c=8

self critique assessment: 3

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09:59:07

`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 -->

equation 2-1 and equation 3-2 leaves you with

eliminate c:

3=8a+4b and

3=40a+4b

eliminate b:

12=24a+16b

-12=-160a+-16

0=160a

a=0

confidence assessment: 3

.................................................

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09:59:50

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 -->

ok

self critique assessment: 0

&#

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

&#

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10:00:11

`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 -->

0

confidence assessment: 0

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10:00:31

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

self critique assessment: 3

&#

This also requires a self-critique.

&#

.................................................

Šoج醲̼

assignment #002

002.

Precalculus I

07-08-2007

D|եnO䇔

assignment #002

002.

Precalculus I

07-08-2007

vwy~Е

assignment #002

002.

Precalculus I

07-08-2007

......!!!!!!!!...................................

09:18:14

`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 -->

equation 2 - equation 1 :

60a-2a

5b-3b

c cancels out c

= -38

58a+2b=-38

equation 3-equation 1:

200a-2a

10b-3b

c cancels out c

= -128

198a+7b=-128

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:18:59

`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 -->

i understood this equation elimination

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:20:32

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 -->

i understand

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:25:30

`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 -->

58a+2b=-38

58(-1)+2b=-38

-58+2b=-38

2b=20

b=10

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:25:48

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 -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:32:10

`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 =100

-100 +c=100

c=100

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:33:01

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

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:37:33

We substitute y = -2 and x = 1 to obtain the equation

-2 = a * 1^2 + b * 1 + c, or

a + b + c = -2.

......!!!!!!!!...................................

RESPONSE -->

a+b+c=-2

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:41:54

`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?

......!!!!!!!!...................................

RESPONSE -->

5=a(3)^2 + b(3) +c

5=9a + 3b+ c

8=49a+7b+c

confidence assessment: 2

.................................................

......!!!!!!!!...................................

09:42:48

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 -->

I got the equation 49a+7b+c=8

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:59:07

`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?

......!!!!!!!!...................................

RESPONSE -->

equation 2-1 and equation 3-2 leaves you with

eliminate c:

3=8a+4b and

3=40a+4b

eliminate b:

12=24a+16b

-12=-160a+-16

0=160a

a=0

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:59:50

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

self critique assessment: 0

.................................................

......!!!!!!!!...................................

10:00:11

`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?

......!!!!!!!!...................................

RESPONSE -->

0

confidence assessment: 0

.................................................

......!!!!!!!!...................................

10:00:31

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

self critique assessment: 3

.................................................

Šoج醲̼

assignment #002

002.

Precalculus I

07-08-2007

D|եnO䇔

assignment #002

002.

Precalculus I

07-08-2007

vwy~Е

assignment #002

002.

Precalculus I

07-08-2007

......!!!!!!!!...................................

&#I believe you submitted this as part of a previous submission. Let me know if I'm wrong about that; if I'm right, then be sure to avoid this sort of redundancy. &#

You appear to understand the process; in order to be sure, however, you need to self-critique.