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.

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22:46:45 query 2.4.30 (was 2.3.24). Slope 4/3, point (-3,2) Give the three points you found on this line and explain how you obtained them.

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RESPONSE --> (-3,2)(1,6)(4,10) I used the rise over run method of going up 4 over 3

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22:47:24 STUDENT SOLUTION: (-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get ((-3+3), (2+4)), which simplifies to (0,6) (-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get ((-3-3), (2-4)) which simplifies to (-6,-2) From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get ((0+3), (6+4)), which simplifies to (3,10). The three points I obtained are (-6,-2), (0,6), (3,10).

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RESPONSE --> ok

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22:52:57 query 2.4.36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.

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RESPONSE --> y=mx+b therefore y=2-1/2--1 which is y=2-1/2=1=1/3 slope i used the slope intercept form

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22:53:41 STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3. Point-slope form gives us y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get y-1=1/3(x+1), which can be solved for y to obtain y = 1/3 x + 4/3.

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RESPONSE --> i dont get this

The equation of a straight line through (x1, y1) and (x2, y2) is y - y1 = m (x - x1), where m = (y2-y1)/(x2-x1).

You got the slope, which is 1/3.

But you didn't get the equation of the line.

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22:59:29 **** query 2.4.46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?

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RESPONSE --> y-y1=m(x-x1) y-4=m*x*2

I'm not sure where you got the 2. Can you explain that step?

If you are using (0, 4) as your point (x1, y1), then x1 = 0 while y1 = 4.

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22:59:50 STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1. The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to y=x+4.

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RESPONSE --> ok

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23:01:12 **** query 2.4.56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?

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RESPONSE --> y=mx+b slope is 2 whats in front of x y intercept is 1/2 what is where b is

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�󘩝�E䘱��㲸������Ŕ�� assignment #019 ��������{�����g逷����ẕ���� College Algebra 02-26-2006

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23:10:36 **** query 2.5.22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE --> 0-2*0=-5 y=-2.5+1/2x y=-2.5

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23:11:11 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **

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RESPONSE --> ok

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23:14:54 **** query 2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE --> y=.5x+4

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23:15:05 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **

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RESPONSE --> ok

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23:17:27 **** query 2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?

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RESPONSE -->

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23:17:52 ** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1. The equation (x-h)^2 + (y-k)^2 = r^2 becomes (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). **

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RESPONSE --> it would not let me type

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23:21:04 **** query 2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?

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RESPONSE --> (x-1)squared+(y+0)squared=3squared=9

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23:21:35 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example we have (h, k) = (1, 0). We therefore have (x-1)^2 +(y - 0)^2 = 3^2. This is the requested standard form. This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get x^2 - 2x +1+y^2 = 9 x^2 - 2x + y^2 = 8. However this is not the standard form.

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RESPONSE --> ok

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23:24:05 query 2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE --> center(0,1) set xand y=0 radius is 1 square it

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23:24:15 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example the equation can be written as (x - 0)^2 + (y-1)^2 = 1 So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1. The x intercept occurs when y = 0: x^2 + (y-1)^2 = 1. I fy = 0 we get x^2 + (0-1)^2 = 1, which simplifies to x^2 +1=1, or x^2=0 so that x = 0. The x intercept is therefore (0, 0). The y intercept occurs when x = 0 so we obtain 0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that (y-1) = +-1. If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are (0,0) and (0,-2)

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RESPONSE --> ok

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23:24:49 **** query 2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE --> ok

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23:24:55 ** Starting with 2x^2+ 2y^2 +8x+7=0 we group x and y terms to get 2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain (x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that the center is (-2,0) the radius is sqrt (1/2). To get the intercepts: We use (x+2)^2 + y^2 = 1/2 If y = 0 then we have (x+2)^2 + 0^2 = 1/2 (x+2)^2 = 1/2 (x+2) = +- sqrt(1/2) x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx. x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx If x = 0 we have (0+2)^2 + y^2 = 1/2 4 + y^2 = 1/2 y^2 = 1/2 - 4 = -7/2. y^2 cannot be negative so there is no y intercept. **

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RESPONSE --> ok

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23:27:06 **** query 2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.

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RESPONSE --> x+2+y+4squared

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23:27:21 ** The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2). The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5). The equation of the circle is therefore (x-2)^2 + (y-2)^2 = (sqrt(5))^2 or (x-2)^2 + (y-2)^2 = 5. **

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RESPONSE --> ok

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\����[�I֥��w�������ҫz|�� assignment #020 ��������{�����g逷����ẕ���� College Algebra 02-26-2006

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23:29:00 **** query 2.6.8 (was 2.5.6). graph like basic stretched cubic centered around (20,20) How well does the graph appear to indicate a linear relation? Describe any significant deviation of the data from its best-fit linear approximation.

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RESPONSE --> lineear

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23:29:11 ** The graph is curved and in fact changes its concavity. The data points will lie first above the best-fit straight line, then as the straight line passes through the data set the data points will lie below this line. **

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RESPONSE --> ok

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23:28:52 query 2.5.12. x = 5, 10, ..., 25; y = 2, 4, 7, 11, 18 **** What two points did you select on the line you graphed, and what is the equation of the line through those points? **** What is the equation of the best-fit line and how well does the line fit the data? Describe any systematic deviation of the line from the best-fit line. ****

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RESPONSE --> 2 and 4

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23:29:00 STUDENT SOLUTION WITH INSTRUCTOR COMMENT: I chose the points (5,2) and (10,4) The slope between these points is slope = rise / run = (4-2)/(10-5) = 2/5 so the equation is y-4 = 2/5(x - 10), which we solve for y to get y = 2/5 x. INTRUCTOR COMMENT: This fits the first two data points, but these are not appropriate points to select. The data set curves, with increasing slope as we move to the right. You need to sketch the best-fit line, as best you can see it, and pick two points on that line. The best-fit line is not likely to pass through any of the data points, and you should never use data points to determine the equation of the best-fit line. Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that comes as close as possible on the average to the points. Extend the line slightly beyond the data set. Estimate the y coordinates of the x = 1 and x = 20 points of this line. Find the equation of the straight line through these points. The coordinates of your points should be reasonably close to (1, 5.5) and (20, 30), though because it's a little difficult to judge exactly where the line should be you are unlikely to obtain these exact results. The equation will be reasonably close to y = .8 x - 3. **

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RESPONSE --> ok

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23:29:41 query 2.5.18. Incomes 15k, 20k, ..., 70k, loan amts 40.6, 54.1, 67.7, 81.2, 94.8, 108.3, 121.9, 135.4, 149, 162.5, 176.1, 189.6 k.

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RESPONSE --> this is not in the book

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23:29:48 ERRONEOUS STUDENT SOLUTION WITH INSTRUCTOR COMMENT: Using the points (15,000, 40,600) and (20,000 , 67,700) we obtain slope = rise / run = (67,700 - 40,600) / (20,000 - 15,000) = 271/50 This gives us the equation y - 40,600= (271/50) * (x - 15,000), which we solve for y to obtain y = (271/50) x - 40,700. INSTRUCTOR COMMENT: You followed most of the correct steps to get the equation of the line from your two chosen points. However I think the x = 20,000 value is y = 54,100, not 67,700; the latter corresponds to x = 25,000. So your equation won't be likely to fit the data very well. Another reason that your equation is not likely to be a very good fit is that you used two data points, which is inappropriate; and in addition you used two data points near the beginning of the data list. If you were going to use two data points you would need to use two typical points much further apart. {]In any case to solve this problem you need to sketch the best-fit line, as best you can see it, and pick two points on that line. The best-fit line is not likely to pass through any of the data points, and you should never use data points to determine the equation of the best-fit line. Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that comes as close as possible on the average to the points. Extend the line slightly beyond the data set. Estimate the y coordinates of the x = 10,000 and x = 75,000 points of this line. Find the equation of the straight line through these points. The coordinates of your points should be reasonably close to (5000, 19000) and (75000, 277,000). It's fairly easy to locate this line, which does closely follow the data points, though due to errors in estimating you are unlikely to obtain these exact results. The equation will be reasonably close to y = 2.7 x - 700 . If we let y = 42,000 we can solve for x: 42,000 = 2.7 x - 700 so 2.7 x = 42,700 and x = 42,700 / 2.7 = 15,800 approx.. Your solution will differ slightly due to differences in your estimates of the line and the two points on the line. **

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RESPONSE --> ok

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23:30:32 **** What is the equation of the line of best fit? **** How well does the line fit the scatter diagram of the data? Describe any systematic deviation of the line from the best-fit line. **** What is your interpretation of the slope of this line? **** What loan amount would correspond to annual income of $42,000?

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RESPONSE --> 42000=12 months annual income

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Student Name: assignment #001 001. Rates"

I've inserted a couple of notes, but you need to self-critique your work after seeing the given solutions. It is your self-critiques I generally respond to.

I suggest to rework the problems in these sections and redo the queries, inserting detailed self-critiques.

Here is a copy of my note on your first assignment:

Note that I don't critique your solutions. I look at your self-critiques to identify what you do and do not understand after you have given your solution and studied the given solution.

You should therefore always include a detailed self-critique when your solution doesn't completely agree with the given solution. Without the self-critique I can't tell what you did and did not understand about the solution and I have no way of helping you address anything you might not understand.

In a good self-critique you need identify the specific things you do and do not understand in the given solution, and either demonstrate your understanding or ask specific questions about what you don�t understand. It doesn't accomplish the intended learning goals to simply say that you understand.

I suggest that you copy this document into Notepad, insert self-critiques after the given solution on those problems where your solution differed from the given solution, and submit the resulting document so I can help you address anything you might not be clear on.