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13:10:53 **** query 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).
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RESPONSE --> d = sqrt ((0 - a)^(2) + (0-a)^(2)) d = sqrt (a+a) d = 2a
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13:13:12 ** Using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((a-0)^2+(a-0)^2) = sqrt(a^2+a^2) = sqrt(2 a^2) = sqrt(2) * sqrt(x^2) = sqrt(2) * a. COMMON ERROR: sqrt(a^2 + a^2) = a + a = 2 a INSTRUCTOR'S CORRECTION: sqrt( x^2 + y^2 ) is not the same thing as x + y. For example sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 but 3 + 4 = 7. So you can't say that sqrt(a^2 + a^2) = a + a. **
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RESPONSE --> I see I made the common mistake you are referring to. Your explanation makes sense to me.
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13:14:52 **** query 2.1.22 (was 2.1.12). Dist (2,-3) to (4,2).
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RESPONSE --> d = sqrt [(4-2)^(2) + (2+3)^(2)] d = sqrt (4 + 25) d= sqrt (29)
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13:16:27 ** using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((2-4)^2+(-3-2)^2) = sqrt((-4)^2+(-6)^2) = sqrt(16+36) = sqrt(52) = sqrt(4) * sqrt(13) = 2 sqrt(13) **
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RESPONSE --> this is the opposite of what the problem states in the book. P1 is (2,-3) P2 is (4,2)
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13:22:08 **** query 2.1.30 (was 2.1.20). (-2, 5), (12,3), (10, -11) A , B, C.
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. d (A.B) = sqrt [(12 + 2)^(2) + (3-5)^(2)] = sqrt (200) = 10sqrt(2) d (B,C) = sqrt [(10-12)^(2) +( -11 - 3)^(2)] = sqrt (196+4) = 10sqrt(2) d (A,C) = sqrt [(10+2)^(2) + (-11-5)^(2)] = sqrt (400) = 20 test for right triangle: [d(A,B)]^2 + [d(B,C)]^2 = [d(A, C)]^2 (10sqrt(2))^2 + (10sqrt(2))^2 =? 20 40 does not = 20 so it is not a right triangle.
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13:25:46 STUDENT SOLUTION: The triangle is a right triangle if the Pythagorean Theorem holds. d(A,B)= sqrt((-2-12)^2+(5-3)^2) sqrt(196+4)sqrt(200) 10 sqrt2 d(B,C)= sqrt((12-10)^2+(3+11)^2) sqrt(4+196) sqrt200 10 sqrt2 d(A,C)= sqrt((-2-10)^2 + (5+11)^2) sqrt(144+256) sqrt(400) 20 The legs of the triangle are therefore both 10 sqrt(2) while the hypotenuse is 20. The Pythagorean Theorem therefore says that (10sqrt2)^2+(10sqrt2)^2=(20)^2 which simplifies to 10^2 (sqrt(2))^2 + 10^2 (sqrt(2))^2 = 20^2 or 100 * 2 + 100 * 2 = 400 or 200+200=400 and finally 400=400. This verifies the Pythagorean Theorem and we conclude that the triangle is a right triangle. **
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RESPONSE --> I miscalculated my pythagorean theorem. I failed to square 20 and did the math wrong on the legs as well. i see my error and copied down the correct answer.
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13:27:47 **** query 2.1.46 (was 2.1.36) midpt btwn (1.2, 2.3) and (-.3, 1.1)
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RESPONSE --> M = for x coord (1.2 + -.3)/2 , for y coord(2.3 + 1.1)/2 =.09/2, (3.4)/2 (0.45, 1.7)
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13:29:22 ** The midpoint is ( (x1 + x2) / 2, (y1 + y2) / 2) = ((1.2-3)/2) , ((2.3+1.1)/2) = (-1.8 / 2 , 3.4 / 2) = (-0.9, 1.7) **
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RESPONSE --> the book has P2 = (-0.3, 1.1), not (-3,1.1)
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13:31:36 **** query 2.1.50 (was 2.1.40). Third vertex of equil triangle with vertices (0, 0) and (0, 4).
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. the 3rd vertex is (4,0). when you figure the distance of each: A(0,0) C(0,4) and B(4,0) it equals 4.
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13:35:26 ** The point (0, 2) is the midpoint of the base of the triangle, which runs from (0,0) to (0, 4). This base has length 4, so since the triangle is equilateral all sides must have length 4. The third vertex lies to the right or left of (0, 2) at a point (x, 2) whose distance from (0,0) and also from (0, 4) is 4. The distance from (0, 0) to (x, 2) is sqrt(x^2 + 2^2) so we have sqrt(x^2 + 2^2) = 4. Squaring both sides we have x^2 + 2^2 = 16 so that x^2 = 16 - 4 = 12 and x = +-sqrt(12) = +-sqrt(4) * sqrt(3) = +-2 * sqrt(3). The third vertex can therefore lie either at (2, 2 sqrt(3)) or at (2, -2 sqrt(3)). **
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RESPONSE --> I see that this explanation is different from mine. I would not have thought to work it out this way. In my mind, since all sides must =4, I figured out the distance of (A.B) (B,C) AND (A,C). Using a trial and error method but knowing that 4 must be a point.
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13:39:55 **** What are the coordinates of the third vertex and how did you find them?
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RESPONSE --> if A=(0,0) C=(0,4) and you know the triangle is equilateral then you can derive that it must have a distance of 4 for each side. Let B=(4,0) d(A,B) = sqrt{ 0 + (-4 + 0)^(2)] = sqrt(16) = 4 d(B,C) = sqrt [(0-4)^(2) + 0 ] = sqrt (16) = 4 d(A,C) = sqrt [0+(4-0)^(2)] =sqrt (16) = 4
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Qf}x|ѫp assignment #017 {kDž_ʅ ϯa~ College Algebra 10-14-2005
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13:40:41 Note that you can't use a calculator graph to document your solutions to these problems. You have to use the analytical methods as in the given solutions. Documentation is required on tests, and while you may certainly use the calculator to see symmetry, intercepts etc., you have to support your solutions with the algebraic details of why the graph looks the way it does.
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RESPONSE --> ok
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13:44:34 query 2.2.10 (was 2.2.6). Point symmetric to (-1, -1) wrt x axis, y axis, origin. What point is symmetric to the given point with respect to each: x axis, y axis, the origin?
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RESPONSE --> for point (-1,-1) the point symmetric to x axis would be for the point (-1,-1) the point (-1,1) for the y axis: (1,-1) for the origin (1,1)
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13:45:20 ** There are three points: The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1). The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1) The point symmetric to (-1, -1) with respect to the origin is ( 1,1). **
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RESPONSE --> right
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13:48:12 **** query 2.1.19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.
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RESPONSE --> the intercept is (0,0) because that is where the graph crosses each axis. it is symmetric to the x-axis because the part above the x-axis is a mirror image of the part of the graph below the x-axis.
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13:48:22 ** The graph intercepts both axes at the same point, (0,0) The graph is symmetric to the x-axis, with every point above the x axis mirrored by its 'reflection' below the x axis. **
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RESPONSE --> right
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ASSIGNMENT 17 13:50:30 **** query 2.2.24 (was 2.2.20). basic cubic poly arb vert stretch **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.
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RESPONSE --> I am not sure on this one since it doesn't give an equation and the quality of the photo isn't that great. I think it intercepts at points (1/2, 0) (-1/2, 0) and is symmetric to the origin.
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13:51:34 STUDENT SOLUTION: origin of the graph is (-.5,0) and (.5,0) graph is symmetric to the origin INSTRUCTOR COMMENT: Check and see whether the graph passes thru the origin (0, 0), which according to my note it should (but my note could be wrong). }If so, and if it is strictly increasing except perhaps at the origin where it might level off for just an instant, in which case the only intercept is at the origin (0, 0). I believe the graph is symmetric with respect to the origin, and if so (0, 0) must be an intercept.
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RESPONSE --> I don't know if it passes through origin. The picture isn't clear enough for me to tell for certain.
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14:00:18 **** query 2.2.40 (was 2.2.36). 4x^2 + y^2 = 4 **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.
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RESPONSE --> for x let y=0 4x^2 + 0^2 = 4 4x^2 = 4 x=1 for y let x=0 4(0)^2 + y^2 = 4 y^2 = 4 y =2 to test for symmetry for the x axis, replace y by -y 4x^2 + (-y)^2 = 4 which simplifies to 4x^2 + y^2 =4 which is equivalant to the original equation. Yes, it is symmetrical to the x=axis. following the same pattern for y-axis 4(-x)^2 + y^2 = 4 4x^2 + y^2 = 4 eqauls the original eq. so it's symmetrical to the y-axis for the origin, replace both x and y with -x and -y 4(-x)^2 + (-y)^2 = 4 4x^2 = y^2 = 4 symmetircal to the origin.
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14:03:10 ** Starting with 4x^2 +y^2 = 1 we find the x intercept by letting y = 0. We get 4x^2 + 0 = 1 so 4x^2 = 1 and x^2=1/4 . Therefore x=1/2 or -1/2 and the x intercepts are (1/2,0) and ( -1/2,0). Starting with 4x^2 +y^2 = 1 we find the y intercept by letting x = 0. We get 0 +y^2 = 1 so y^2 = 1 and y= 1 or -1, giving us y intercepts (0,1) and (0,-1). To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis. Substituting we get 4 (-x)^2 + y^2 = 1. SInce (-x)^2 = x^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the y axis. To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis. Substituting we get 4 (x)^2 + (-y)^2 = 1. SInce (-y)^2 = y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the x axis. To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin. Substituting we get 4 (-x)^2 + (-y)^2 = 1. SInce (-x)^2 = x^2 and (-y)^2 - y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the origin. **
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RESPONSE --> Again, the problem I just worked from the book is different than what you have here. # 40 from the book is 4x^2 + y^2 = 4. Perhaps I overlooked this difference when reading the problem. Nonetheless, I see we are working them the same way.
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14:12:30 **** query 2.2.46 (was 2.2.42). y = (x^2-4)/(2x) **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.
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RESPONSE --> following exactly the problem typed in here we get: for x int. (x^2 +4) /2x = 0 multiply both sides by 2x x^2 + 4 = 0 x=2 (2,0) for y int y= (0^2 -4)/2*0 y= -4 (0,-4) tests for symmetry: for x: -y=(x^2 -4)/2x is not = to original equation so it's not symmetrical to x-axis. for y: y=[(-x)^2 -4]/2*-x (x^2 - 4)/-2x is not =to original equation so it's not symmetrical to y-axis. for origin: -y = [(-x)^2 - 4]/2*-x isn't = to original equation so it isn't symmetrical to the origin either.
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14:13:12 ** We do not have symmetry about the x or the y axis, but we do have symmetry about the origin: To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis. }Substituting we get y = ( (-x)^2 - 4) / (2 * (-x) ). SInce (-x)^2 = x^2 the result is y = -(x^2 - 4) / (2 x). This is not identical to the original equation so we do not have symmetry about the y axis. To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis. Substituting we get -y = (x^2-4)/(2x) , or y = -(x^2-4)/(2x). This is not identical to the original equation so we do not have symmetry about the x axis. To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin. Substituting we get -y = ((-x)^2-4)/(2(-x)) SInce (-x)^2 = x^2 the result is -y = -(x^2-4)/(2x), or multiplying both sides by -1, our result is y = (x^2-4)/(2x). This is identical to the original equation so we do have symmetry about the origin. **
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RESPONSE --> yes, I forgot I could multiply by -1.
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