I do not have direct access to my webs this week, which is causing delays in my responses. I do have access to my email. If you want quicker feedback, you may use email; however continue to also submit your work by the form. The work included here will be posted by the first of next week. assignment #019 019. `query 19 College Algebra 03-13-2007 yvvHFw assignment #019 019. `query 19 College Algebra 03-13-2007
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19:08:27 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 --> x - 2y= - 5 (0,0) slope= - 1/2 -2y= - x - 5 -2y/-2= (- x - 5)/ - 2 y= - 1/2x - 5/2 y - y1=m(x-x1) y - 0= - 1/2(x - 0) y= - 1/2x or -1/2x-y=0??? confidence assessment: 2
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19:09:55 ** 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 close its + 5/2 not - 5/2...and 1/2 not - 1/2, I got the concept! yaay! self critique assessment: 2
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19:31:36 ** 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 question why do the program do that it gives me the question and asner at the sametime its crazy! but anyways.... x - 2y= - 5 (0,4) slope=1/2 - 2y= - x - 5 divide all by - 2 y= 1/2x + 5/2 slope - 2/1 = -2 y-y1=m(x-x1) y - 4= - 2(x-0) or y= - 2x+4 self critique assessment: 2
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21:01:12 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 --> (0,1) and (2,3) center=midpoint of the two points radius=distance from two points equation=(x-h)^2+(y-k)^2=r^2 c=[(0+2/2),(1+3/2)]=(2/2,4/2)=(1,2) r=d=sqrt[(2-1)^2+(3-2)^2]=sqrt(2) (x-1)^2+(y-2)=sqrt2 or 2???maybe confidence assessment: 2
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21:01:35 ** 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 --> yes.... self critique assessment: 2
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21:03:04 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-h)^2+(y-k)^2=r^2 (x-1)^2+(y-0)^2=3^2 confidence assessment: 2
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21:03:24 ** 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 --> yeah! self critique assessment: 2
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21:14:33 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) radius=1 x-int=(0,0) y-int=(0,2), (0, - 2) ii quadrant and i quadrant confidence assessment: 2
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21:15:05 ** 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 --> i was close...yes! self critique assessment: 2
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21:17:20 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 --> uhmmmm.... confidence assessment: 0
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21:33:10 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 --> (4,3)(0,1) center=[(4+0/2),(3+1/2)]=(2,2) radius=sqrt[(2-4)^2+(2-3)^2] =sqrt(4+1) =sqrt(5) equation=(x-2)^2+(y-2)^2=]sqrt(5)]^2 =x^2+y^2-4x-4y+3=0 confidence assessment: 2
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21:33:26 ** 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 --> yes....whoooah! self critique assessment: 2
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" assignment #020 020. `query 20 College Algebra 03-14-2007
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16:58:56 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 --> a curve, that seems to be nonlinear confidence assessment: 2
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17:01:05 ** 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 self critique assessment: 2
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17:16:24 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 --> (10,4) (20,11) m=(4-11)/(20-10)= - 7/10 y - 4= - 7/10(x - 11) y= - 7/10x +117/10 confidence assessment: 2
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17:17:06 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 self critique assessment: 2
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17:30:58 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 --> (20000, 54100) (50000, 135400) m=(54100 - 135400)/(50000 - 20000)= - 81300/30000 y - 54100=( - 81300/30000)(x - 20,000) y= - 81300/30000x +54200 self critique assessment: 2
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