What are the coordinates of the centroid C of a triangle whose vertices are F(-12,1) G(-2,1) and H(-7,-11)?
960360
To do this accurately, you need to find the midpoint of each side of the triangle. Connecting each vertex to the midpoint of the opposite side will give you the Centroid(the place where all the medians meet).  I graphed it out and it appears to be (-7,-3).
109605
Just use Cosinus Theorem: c^2=a^2+b^2-2abCos(gamma)\n1.\nc=sqrt(4+64-2*2*8*sqrt(3)/2))\nc=6.34\n\n2. Is not a right Triangle.. but anyway lets find the angle A:\n22^2=15^2+33^2-2*15*33*Cos(A)\nCos(A)=830/990\nA=ArcCos(83/99)\nA=33°1' 48.54"
915081
the standard form of the equation for a parabola with vertex (0,0) is x^2 = 4py  where p is the distance from the vertex to the focus, which is equal to the distance from the vertex to the directrix.  So if x^2 = 12y, then 4p = 12, so p = 3.  This means the focus is 3 units in the positive direction from the vertex and the directrix is 3 units away in the negative direction.  Since the vertex is (0,0), the focus is (0,0+3)  or (0,3), and the directrix is the horizontal line y=-3.
910387
The nearest vertex will be any of the four lower vertices because the ball is closer to the bottom than to the top of the cube, and the centre of a square is equidistant from all four vertices.\n\nPicture the right triangle whose vertices are:\nA = the center of the ball\nB = one of the lower vertices\nC = the center point of the bottom face of the cube.\n\nYou want to know the length of segment AB which is the hypotenuse of this triangle.  You know that AC is 6 cm long.  And BC is half the diagonal of the bottom face so it measures 1/2 * sqrt(2) * 14 = 7*sqrt(2) cm.\n\nSo AB = sqrt(AC^2 + BC^2)\n= sqrt(6^2 + (7*sqrt(2))^2)\n= sqrt(36 + 49*2)\n= sqrt(134) cm.
904256
there are 6 vertices and you form 4 triangles from the first 2, then 3, then 2, then 1 for a total of 14.\n\nNow we form other triangles that overlap and it gets really complex...\n\nI count 18 smaller triangles, for a total of 32 so far...\n\nNow middle triangles...\n\n36, for a running total of 68
787980
i believe this is a continuation of the previous prob.\n\nit can be solved graphically. \ndraw the lines x + y = 2,  y - x = 2 and y = -2 on the cartesian plane. the feasible reason will be the area enclosed by the triangle formed by these 3 lines. the vertices being (0,2) , (4, -2) and (-4, -2). it can be proved that the max value of 2x + 3y will be one of these vertices. \n\nsubstituting, u can see the maximum occurs at (0,2) whence 2x + 3y = 6
764538
For any given triangle ABC with sides a, b and c.\n\nCos C = (a^2 + b^2 - c^2)/2ab  This is the cosine rule.\n\nSin C =sqrt(1 - Cos^2 C). You may need to adjust the sign of the sign to match the triangle.
14639
Assuming your first statement is correct, then\nA. g(c) = c/231\n\nIf c = 231, g(c)=1,\nIf c = 462, g(c)=2,\netc.
329126
You have to graph them and find the corner points.  One of the corner points will be the answer.\nI found the corner points to be\n(28,52) and (40,40)\nPlug each point into g = 3x+y \nFor (28,52): g = 3(28) + 52 = 136\nFor (40,40): g = 3(40) + 40 = 160\n136 is the smaller value, so (28,52) is the answer.
883980
find the mid pointof any side by using the formula (x1+x2/2,y1+y2/2)\nusing that and the opposite vertex using the two point formula (y-y1/y2-y1)=(x-x1/x2-x1) find the equation\ni shall do one for you\nbid point of BC=(0,-2)\npoint A=(0,7)\naquation of median AD=(y+2)/(7+2)=(x-0)/(0-0)\n0(y+2)=9(x-0)\n9x=0 is th equation of median AD
555107
MY guess is that your trying to find the Hypotenuse of a triangle.  Thus A^2+B^2=C^2 or in this case H^2+W^2 = C^2, 25 + 196 = C^2, 221 = C^2, C = 14.86.  Missing side is 14.86.
456084
angle b and angle c.\nthey are both on the same line of the x- axis and collide at a central point with respect to A.
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
631670
V = pi*(r^2)*h\n121 = pi*(3^2)*h\nh = 4.28 cm
497748
In rt triangle ABC with B= 90degrees (draw figure)      \nIf C is theta, cosC=7/8(ignoring sign)\ni.e., BC = 7, CA = 8 (in simplest terms)\nusing pythagoras theorem, AB= sqrt(15)\n \nsince C=theta is in the second quadrant tan C will be negative\n\nso tan C = AB / BC \n\n  tan C= - { sqrt(15) } / 7\n\ntan(theta) = -0.5533
1127635
To solve for the centroid, treat this problem like a center of mass problem, because the centroid of an object is also the object's center of mass. To find the x coordinate of the center of mass, multiply each piece of mass by its x coordinate, add them all together, and divide the sum by the total mass. In our situation, we use y as our piece of mass, multiply it by x, and divide by the total area. Of course, we need to use integrals, and the formula looks like this:\n\nx = integral(y*x*dx) / integral(y*dx)\n\nwhere y is the difference in the two equations (2x+3-x^2).\n\nI hope you know how to do integrals, so I won't show all that work. It comes out to be (32/3) / (32/3) which is 1.\n\nThe process is similar for calculating the y coordinate, except the integrating is complicated. First, rewrite the equations into "x =" form. The total area remains the same, but you will need to break the numerator into two parts: an integral from 0 to 1 of 2 times square root y * y * dy plus an integral from 1 to 9 of (square root of y - (.5y - 1.5)) * y * dy. After evaluating, you should get (.8+532/15)/(32/3) = 3.4. The coordinates of the centroid are (1, 3.4)
939325
centeris\n(5,-1)\nradius\n=(25+1-25)^1/2\n=1