course Mth 158 I do not currently own a graphing calculator. If I will need one, please let me know.
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14:19:18 **** query 2.4.10 (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 --> center = (1,2) radius = sqrt2 equation: (x-5/2)^2 + (y - 2)^2 = 9/4
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14:22:30 ** 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 --> I did #9. I had the correct solution for 10.
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14:27:34 **** query 2.4.14 / 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)^2 + (y - 0)^2 = 3^2 x^2 - 2x + 1 + y^2 = 9 x^2 + y^2 -2x -8 = 0
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14:33:35 query 2.4.22 / 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 --> radius = 1 center = (0,1) These were taken from the standard form of the equation. x^2 is the same as (x-0)^2. The graph of the circle showed it in the first and second quadrants with x-intercept (0,0) and y-intercepts (0,0) and (0,2)
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14:36:11 ** 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 had the k as 1 instead of -1.
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14:44:44 **** query 2.4.32 / 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 --> I put the equation in standard form: 2x^2 + 8x +2y^2 = -7 2(x^2 + 4x + y^2) = -7 x^2 + 4x + y^2 = -7/2 x^2 + 4x + 4 + y^2 = -7/2+4 (x + 2)^2 + (y + 0)^2 = 1/2 center is (-2,0) and radius is sqrt 1/2 x-intercept = -2+/-sqrt 1/2 No y-intercept. The graph was in the quadrants II and III.
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14:45:27 ** 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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14:49:51 **** query 2.4.40 / 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 --> Midpoint of the diameter would be the center. (2,2) using the distance formula, the radius = sqrt5 The standard equation would be (x-2)^2 + (y-2)^2 = 5 The general equation would be x^2 + y^2 -4x - 4y + 1= 0
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14:50:45 ** 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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Kdl~Ȗӹۂ՚ assignment #020 020. `query 20 College Algebra 09-24-2008
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14:52:46 **** query 4.2.8 / 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 --> It does not appear to me to be linear. It curves.
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14:52:59 ** 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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15:04:13 query 4.2.22 / 3.1.90. Sales S vs. advertising expenditures A. 335 339 337 343 341 350 351 vs. 20, 22, 22.5, 24, 24, 27, 28.3 in thousands of dollars.Does the given table describe a function? Why or why not?What two points on your straight line did you pick and what is the resulting equation?What is the meaning of the slope of this line?Give your equation as a function and give the domain of the function.What is the predicted sales if the expenditures is $25,000?
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RESPONSE --> It does describe a function because there is a direct correlation between sales and advertising.My two points were (28.3, 351) ; (20, 335) the equation is: y = 1.928x + 296.438 The slope shows that for every thousand dollars spent on advertising, sales increase $1,928. Predicted sales for 25,000 is 344,638.
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15:06:10 The table does not describe a function because ordered pairs that have the same first element and a different second element. Specifically 24,000 is paired with both 343,000 and 341,000. I picked the points (20000,335000) (27000,350000). INSTRUCTOR COMMENT: These are data points, not points on the best-fit straight line graph. You should have sketched your line then picked two points on the line, and the line will almost never pass through data points. STUDENT SOLUTION CONTINUED: The slope between these points is slope = (350000-335000)/(27000-20000) = 15000/7000 = 15/7 = 2.143 approximately. Our equation, using this slope and the first chosen point, is therefore y-335000=2.143(x-20000) y- 335 = 2.143x-42857.143 y= 2.143x+29214.857 equation of the line Expressed as a function we have f(x) = 2.143x+292142.857. Predicted sales for expenditure $25000 will be f(25000) = 2.143(25000) + 292142.857 = 53575 + 292142.857 = 345717.857 We therefore have predicted sales f(25000)= $345,717.86 INSTRUCTOR COMMENT: Excellent solution, except for the fact that you used data points and not points on the best-fit line.
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RESPONSE --> I guessed at whether or not it was a function. I am not familiar with functions...yet.
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15:08:41 query extra problem (was 2.5.12). Sketch a graph of y vs. x for x = 5, 10, 15, 20, 25; y = 2, 4, 7, 11, 18. Fit a good straight line to the data and pick two points on this line. Use these points to find an estimated equation for your line. **** 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 -->
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15:09:51 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 -->
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15:22:08 query extra problem (was 2.5.18). Individuals with incomes 15k, 20k, 25k, 30, 35, 40, 45k, 50k, 55k, 60k, 65k, 70k (meaning $15,000, $20,000, etc; 'k' means 'thousand') have respective loan amounts 40.6, 54.1, 67.7, 81.2, 94.8, 108.3, 121.9, 135.4, 149, 162.5, 176.1, 189.6 k. Sketch a graph of loan amount vs. income, fit a good straight line to the data and use two points on your line to estimate the equation of the line.
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RESPONSE --> I used the first and last points of (15, 40.6) and )70, 189.6). The equation is came up with is y = 2.71x - .05
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15:23:32 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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15:27:07 **** 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 --> the line is non-linear. The slope is positive which means that the more money you make, the greater your debt. Annual income of$42,000 would mean a loan amount of approx. $113,000
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