course Mth 158 That last one confused me. ~}y¤UءvсԚassignment #016
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11:31:42 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).
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RESPONSE --> distance d=(P1,P2) P1=(a,a) P2=(0,0) d=sqrt[(x2-x1)^2+(y2-y1)^2] d=sqrt[(0-a)^2+(0-a)^2] d=sqrt(a^2+a^2) d=sqrt(2a^2) d=sqrt(2) * sqrt(a^2) d=sqrt(2) a confidence assessment: 2.
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11:32:42 ** 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 self critique assessment: 2
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11:41:35 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[(2)^2+(5)^2] d=sqrt(4+25) d=sqrt(29) confidence assessment: 2
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11:44:48 ** 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 --> ok i used (2, - 3) as (x2-x1)^2 and (4,2) as (y4-y2)^2 to get my answer sqrt(29) it seems to me you used them backwards to what i got???? self critique assessment: 2
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11:55:32 2.1.30 (was 2.1.20). (-2, 5), (12,3), (10, -11) A , B, C.
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RESPONSE --> when i did this problem in and with q_a 13 i am finding that you use the numbers differently. A=(-2,5) B=(12,3) C=(10,-11) d=sqrt[(x2-x1)^2+(y2-y1)^2] d(A,B)=sqrt(200) d=sqrt[(12-(-2))^2+(3-5)^2] d(B,C)=sqrt(200) d=sqrt[(10-12)^2+(-11-3)^2] d(A,C)=sqrt(400) or 20 d=sqrt[(10-(-2))^2+(-11-5)^2] pythagorean thoerem that triangle ABC is a right triangle....it's not [d(A,C)]^2+[d(B,C)]^2=[d(A,B)]^2 (20)^2+[sqrt(200)]^2=(sqrt(200))^2 400+200=200
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11:55:51 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 don't know! self critique assessment: 2
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12:01:07 2.1.46 (was 2.1.36) midpt btwn (1.2, 2.3) and (-.3, 1.1)
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RESPONSE --> M=(x,y)=[(x1+x2)/(2),(y1+y2)/(2)] M=[(1.2+(-.3))/(2),(2.3+1.1)/(2)] M=[(.9)/(2),(3.4)/(2)] M=(.45,1.7) confidence assessment: 2
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12:03:53 ** 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 --> ok so i got the y but not the x!? isn't the 3 you have supposed to be (-0.3) in ((1.2-3)/2)??? unless i wrote the numbers down wrong... self critique assessment: 2
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16:30:44 2.1.50 (was 2.1.40). Third vertex of equil triangle with vertices (0, 0) and (0, 4).
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RESPONSE --> ok confidence assessment: 0
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16:38:38 ** 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 --> ok self critique assessment: 2
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16:38:54 **** What are the coordinates of the third vertex and how did you find them?
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RESPONSE --> ok confidence assessment: 0
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