Assignment 9

course Mth 163

hþy¯Ã^ã÷ÑŸ¨äo³²¥é}–Bšþå´“úassignment #009

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009.

Precalculus I

02-17-2008

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19:39:42

`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 -->

For x = x1 the coordinate would be (x1, 1.1(x1)+.8). For x = x2 the coordinate would be (x2, 1.1(x2)+.8)

The rise is 1.1(x2)+.8 - 1.1(x1)+.8 = 1.1(x2)-1.1(x1). The run is x2-x1

The slope is found by: 1.1(x2-x1)/(x2-x1) which equals 1.1

confidence assessment: 3

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19:43:37

`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 coordinate for x=x1 is (x1, 3.4(x1)+7). The coordinate for x=x2 is (x2, 3.4(x2)+7).

The rise is 3.4(x2) - 3.4(x1). The run is x2-x1.

The slope is as follows: (3.4(x2)-3.4(x1)/(x2-x1) which is 3.4(x2-x1)/(x2-x1) = 3.4

confidence assessment: 3

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Good responses. Let me know if you have questions. &#