course Mth158
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12:19:16 **** query 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).
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RESPONSE --> I enter each value into the distance equation: sqrt of ((a-0)^2+(a-0)^2= sqrt of a^2+a^2= Sqrt of 2a^2= d=a*sqrt of 2
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12:19:27 ** 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 --> ok
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12:22:56 **** query 2.1.22 (was 2.1.12). Dist (2,-3) to (4,2).
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RESPONSE --> I enter these values into the distance equation: d=sqrt of(4-2)^2+(2-(-3))^2= sqrt of 4+ 25= d= sqrt of 29
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12:24:57 ** 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 --> Is there an arithmetic error in this problem? Line 2: Is 2-4 not -2? Is -3-2 not -5? And then those quantities would be squared? Just wondering.
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12:33:31 **** 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. First I find the distance between A,B by entering the values into the distance formula: sqrt of (12-(-2))^2+(3-5)^2= 196+4= sqrt 200= 10*sqrt of 2 Then I find the distance between B,C= sqrt of (10-12)^2+(-11-3)^2= 4+196= sqrt of 200= 10*sqrt 2 Finally I find the distance between A,C= sqrt of (10-(-2))^2+(11-5)^2=144+36= sqrt of 180= 6*sqrt of 5 Enter these into the Pythagorean theorem: [d(A,B)]^2+ {d(B,C)]^2= [d(A,C)]^2= {10sqrt2}^2= 200+200= I think I messed up here
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12:34:33 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 was right on everything until the d (A,C), now I see my error
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12:37:05 **** query 2.1.46 (was 2.1.36) midpt btwn (1.2, 2.3) and (-.3, 1.1)
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RESPONSE --> I enter these values into the midpt formula and I get 1.2-3 all divided by 2 and 2.3+1.1 all divided by 2= (-1.8, 1.7)
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12:37:31 ** 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 --> Forgot to divide the first part- otherwise ok!
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12:41:45 **** 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. First I plot these points on a graph. I would find the midpoint of the segment, which in this case is (0,2). I am guessing that is the third vertex, and could also be at point (0,-2)
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12:43:47 ** 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 have no idea how you did the last part of this equation. Let me look again: Okay, I see now. In order for this triangle to be equilateral, the side with the third vertex would have to have a total length of 4 also. So you treat 0,2 as x,2 and plot the distance from the origin (vertex A). I will continue to work on this.
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12:44:57 **** What are the coordinates of the third vertex and how did you find them?
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RESPONSE --> I am guessing this refers to the previous question, so I used the midpoint formula and got 0,2 as the midpoint. I treat this as x,2 and enter the distance from 0,0 to x,2 into the distance formula to find the third vertex.
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