A triangle has three verices A(11,1), B(3,6),and C(3,-4).?
351601
If I'm correct, it y=5/-8 +17/8
671114
1. d; 2. d; 3=a; 4. c;  5.a; 6.a 7.c; 8.d.
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
272561
1 b = 5\n2 c = 7\n3 c = -8\n4 b = 9
893652
with 1q, 2 ways (0,1 n)\nwith 3d, 1 ways (0 n)\nwith 2d, 3 ways (0,1,2 n)\nwith 1d, 5 ways (0,1,2,3,4 n)\nwith 0d, 7 ways (0,1,2,3,4,5,6 n)\n\nTotal of 18 ways
445500
(18)(-3)=(-11)r\n-54=(-11)r\n54/11=r
1074757
The problem is that 30, 31, and 63 are not the sides of a valid triangle.  For any triangle with sides a,b,c, you must have\na + b > c\nb + c > a\nc + a > b\nBut you have a + b < c, so the triangle isn't valid.
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
60270
5+3(18/6) = 5+3(3) = 5+9 = 14\n\n[2(4)+17]*9 = (8+17)*9 = 25*9 = 225
638842
a heptagon\n3: triangle\n4: quadrilateral\n5: pentagaon\n6: haxagon\n7: heptagon\n8: octagon\n9: nonagon\n10: decagon\n11: undecagon\n12: dodecagon\n...\nn: n-gon
57033
the answer is 3 because a-b|a-b & a-b|a so a-b|b
1008994
a*a(cos^2 B-cos^2 C)\n+b*b(cos^2 C-cos^2A)\n+c*c(cos^2 A-cos^2 B)\n\nare A,B, C angles in a triangular?\n\nif it is,  you may use the relation  A+B+C=180\nand cosine rule in a triangular.
737238
the 3 x 3 matrix A times the 3 x 3 matix B will yield a 3 x 3 matrix AB\n\nAB1,1 = (A1,1)(B1,1) + (A1,2)(B2,1) + (A1,3)(B3,1)\nAB1,1 = (-3)(-1) + (6)(0) + (2)(-5) = -7\n\nAB1,2 = (A1,1)(B1,2) + (A1,2)(B2,2) + (A1,3)(B3,2)\nAB1,2 = (-3)(8) + (6)(-3) + (2)(0) = -42\n\nAB2,1 = (A2,1)(B1,1) + (A2,2)(B2,1) + (A2,3)(B3,1)\nAB2,1 = (-4)(-1) + (8)(0) + (1)(-5) = -1\n\netc.
1010584
Just think....\n\nA triangle (3 sides) always totals 180 degrees\nA quadrilateral (4 sides) always totals 360 degrees\nA pentagon (5) will total 540 degrees\n\nSo, the angles, A, will always equal 180 degrees per side, for each side (N) after the first 2.\n\nIn other words,   A = 180 * (N-2)\n\nif A is 32040, then 32040 = 180 * (N-2)\n\nN - 2 = 178\n\nN = 180
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
905618
(A) 3+2(x-1)\n\n3+2x+2\n2x+5\n\n(B) 5-2(x+3)\n\n5-2x-6\n-2x-1\n\n(C) 1+2(x-1)\n\n1+2x-2\n2x-1\n\n(D) 6-3(w-2)\n\n6-3w+6\n-3w+12
1184743
Heron's formula for triangles:\na = √s(s-a)(s-b)(s-c)\ns is the semiperimeter or 1/2 (4+4+4)\ns = 6\na = √6(2)(2)(2)\na = √3×2²×2²\na = 2×2×√3\na = 4×1.7\na = 6.8\n\n\nJose, you calculated the area of the right triangle, not the entire equilateral triangle you started with.