Assignment 9

course Mth 163

ѮXassignment #009

009.

Precalculus I

09-26-2008

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18:58:18

09-26-2008 18:58:18

`q001. Note that this assignment has 2 questions

For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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NOTES ------->

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19:03:13

`q002. For the function y = 3.4 x + 7, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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RESPONSE -->

confidence assessment:

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assignment #009

009.

Precalculus I

09-26-2008

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19:46:18

`q001. Note that this assignment has 2 questions

For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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RESPONSE -->

To solve the equation, we have start with the two equations: y = 1.1(x1)+.8 and y= 1.1(x2)+.8

The points come to (x1, 1.1x1+.8) and (x2, 1.1x2+.8). The two coordinates make the rise of 1.1x2+.8 - 1.1x1+.8. The run would be x2-x1. Since we say that to get a slope, we have to divide the rise by the run. (1.1x2-1.1x1) / (x2-x1). The slope is 1.1

confidence assessment: 2

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19:48:43

In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 1.1 x1 + .8) and the coordinates of the x = x2 point are ( x2, 1.1 x2 + .8).

The rise between the two points is therefore

rise = (1.1 x2 + .8) - (1.1 x1 + .8) = 1.1 x2 + .8 - 1.1 x1 - .8 = 1.1 x2 - 1.1 x1.

The run is

run = x2 - x1.

The slope is therefore (1.1 x2 - 1.1 x1) / (x2 - x1) = 1.1 (x2 - x1) / (x2 - x1) = 1.1.

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RESPONSE -->

ok

self critique assessment: 3

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20:00:10

`q002. For the function y = 3.4 x + 7, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2?

What therefore is the rise between these two points, and what is the run?

What is the average slope of the graph between these two points? Be sure to simplify your result.

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RESPONSE -->

the same equation and the way would be used to acquire the answers here.

The coordinates of the points are (x1,3.4x1+.8) and the second coordinate (x2,3.4x2+.8).

The rise comes to 3.4x2-3.4x1. The run is x2-x1. The slope is 3.4.

confidence assessment: 2

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20:00:40

In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 3.4 x1 + 7) and the coordinates of the x = x2 point are ( x2, 3.4 x2 + 7).

The rise between the two points is therefore

rise = (3.4 x2 + 7) - (3.4 x1 + 7) = 3.4 x2 + 7 - 3.4 x1 - 7 = 3.4 x2 - 3.4 x1.

The run is

run = x2 - x1.

The slope is therefore (3.4 x2 - 3.4 x1) / (x2 - x1) = 3.4 (x2 - x1) / (x2 - x1) = 3.4.

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RESPONSE -->

ok, I think I understood and do this exercise well.

self critique assessment: 3

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You do appear to have picked up the right ideas from the q_a_ and the worksheets. Good job.