course MTH 163

?????V?????? ^??assignment #002

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002.

Precalculus I

09-08-2007

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15:54:09

`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 -->

In all three problems c is the easiest variable to eliminate.

(2a + 3b + c = 128) - (60a + 5b + c = 90) = -58a - 2b = 38

(2a + 3b + c = 128) - (200a + 10b + c = 0) = -198a - 5b = 128

confidence assessment: 3

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15:55:19

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 used a different system of calcualtions, but recieved equivalent systems.

self critique assessment: 2

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16:05:57

`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 -->

It would be easiest to eliminate the b variable. First we'll multiply the first equation by the second equation's b variable:

(58a + 2b = -38) * 7 = (406a + 14b = -266)

Then we'll multiply the second equation by the neagtive of the first equation's variable:

(198a + 7b = -128) * -2 = (-396a - 14b = 256)

From there, we solve:

(406a + 14b = -266) + (-396a - 14b = 256) = (10a = -10)

(10a = -10) = (a = -1)

We then plug -1 into the a variable of the original equation:

58(-1) + 2b = -38

-58 + 2b = -38

2b = 20

b = 10

a = -1 , b = 10

confidence assessment: 3

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16:06:27

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 overshot a little in solving for b and all, but I did it right.

self critique assessment: 2

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16:09:21

`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 -->

As in my previous answer, we plug -1 into the a variable:

58(-1) + 2b = -38

-58 + 2b = -38

2b + 20

b = 10

198(-1) + 7(10) = -128

-198 + 70 = -128

confidence assessment: 3

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16:09:35

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 understand.

self critique assessment: 2

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16:11:26

`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

confidence assessment: 3

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16:11:43

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.

self critique assessment: 2

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16:13:28

`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 -->

-2 = a(1^2) + b(1) + c

or

a(1^2) + b + c = -2

confidence assessment: 3

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16:13:51

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 should've carried it out a little more. In understand.

self critique assessment: 1

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16:15:27

`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 = a(7^2) + b(7) + c

8 = 49a + 7b + c

confidence assessment: 2

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16:15:48

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 understand.

self critique assessment: 2

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16:39:47

`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 -->

a + b + c = -2

9a + 3b + c = 5

49a + 7b + c = 8

First we eliminate our c variables:

(49a + 7b + c = 8) - (a + b + c = -2) = (48a + 6b = 10)

(49a + 7b + c = 8) - (9a + 3b + c = 5) = (40a + 4b = 3)

We now eliminate b:

(49a + 7b + c = 8) * 3 = (147a + 21b = 24)

(9a + 3b + c = 5) * -7 = (-63a - 21b = -35)

(147a + 21b = 24) - (-63a - 21b = -35) = (84a = -11)

(84a = -11) = (a = 0.131)

We can then plug a back into an original equation:

147(0.131) + 21b = 24

19.257 + 21b = 24

21b = 4.743

b = 0.226

We can then plug both a and b back into the original equation:

0.131 + 0.226 + c = -2

.357 + c = -2

c = -2.357

a = 0.131 , b = 0.226 , c = -2.357

confidence assessment: 1

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16:50:04

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 I eliminated c I recieved the system:

(48a + 6b = 10)

(40a + 4b = 3)

Here I see where I made my error. I copied the wrong equations. From here we eliminate b:

(48a + 6b = 10) * 2 = (96a + 12b = 20)

(40a + 4b = 3) * 3 = (120a + 12b = 9)

Then we subtract the second equation from the third:

(96a + 12b = 20) - (120a + 12b = 9) = (-24a = 11)

(-24a = 11) = (a = -0.458)

Then we plug a into a previous equation:

48(-0.458) + 6b = 10

-22 + 6b = 10

6b = 32

b = 5.333

We then plug both a and b into an original equation:

9(-0.458) + 3(5.333) + c = 5

-4.122 + 16 + c = 5

11.878 + c = 5

c = -6.875

I understand now.

self critique assessment: 1

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16:57:30

`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 = -0.458(x^2) + 5.333(x) + -6.875

y = -0.458(1^2) + 5.333(1) + -6.875

y = -0.458 + 5.333 + -6.875

y =4.872 + -6.875

y = -2.003

y = -0.458(3^2) + 5.333(3) + -6.875

y = -4.122 + 15.999 + -6.875

y = 11.877 + -6.875

y = 5.002

y = -0.458(5^2) + 5.333(5) + -6.875

y = -11.45 + 26.665 + -6.875

y = 15.215 + -6.875

y = 8.34

y = -0.458(7^2) + 5.333(7) + -6.875

y = -22.442 + 37.331 + -6.875

y = 14.889 + -6.875

y = 8.014

confidence assessment: 2

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16:58:32

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: 2

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"

It seems clear that you understand this. Let me know if you have questions.

course MTH 163

This send file originally contained part of the assignment 2 query as well. I had gotten a few steps into it and realized I was referring to the wrong set of data when answering the questions. I copied and pasted the assignment 1 query into the field below. Assignment 2 query will be included in the next send file. I hope this is ok.

?J?{????????R?assignment #001001. `query1

As long as you're careful about the process and don't lose any of your original information, no problem at all.

Precalculus I

09-08-2007

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03:09:17

Query Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising?

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RESPONSE -->

I found it interesting to ponder why it's better to understand arithmetic rather than doing all our work on a calculator. I also found the fishing pond hypothesis interesting; more specifcally all the variables and factors that could go into the design of such.

confidence assessment: 3

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03:10:47

Query Introductory Flow Experiment (no summary needed) Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? Support your conclusion.

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RESPONSE -->

It is changing at a faster and faster rate. Through the sequence of pictures it is clear that the surface of the water is dropping at a faster rate.

confidence assessment: 3

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03:11:40

** If you time the water at equal time intervals you should find that the depth changes by less and less with each new interval.

If you timed the depth at equal intervals of depth you should find that each interval takes longer than the one before it.

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Either way you would conclude that water flows from the hole at a decreasing rate.

The reason is that as the water depth decreases the pressure forcing the water out of the hole decreases. **

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RESPONSE -->

I understand.

self critique assessment: 2

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03:17:27

What does the graph of depth vs. clock time look like? Is it increasing or decreasing? Does the rate of increase or decrease speed up or slow down? Does your graph intercept the y axis? Does it intercept the x axis? How would you describe its overall shape?

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RESPONSE -->

The graph is decreasing and the rate of decrease slows down. My graph does intercept the y-axis at 63.7 cm.It does not intercept the x-axis, although theoritically it would if more data were given in the 'simulated data' page. The graph is curved.

confidence assessment: 2

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03:17:58

** The graph will start on the positive y axis and will decrease at a decreasing rate.

The shape of the graph is the left-hand side of a parabola that opens upward. The right-hand half of the parabola does not correspond to the flow. The lef-hand half of the parabola, which corresponds to the flow, gets less and less steep with increasing clock time, matching the fact that that the rate of decrease is slowing.

At the instant the containers empties, the water will be at the level of the hole. If depth is measured relative to the hole, then at the instant depth will reach zero. The corresponding graph point will lie on the t axis and will correspond to the vertex of the parabola. **

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RESPONSE -->

I understand.

self critique assessment: 2

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Your work looks good. Let me know if you have any questions. &#