Graph the points A(-1,2) B(1,8) & C(5,2)  find all possible cordinatesof the point D so the ABCD is a...?
596434
Three such points:\nD1: (-5,8)\nD2: (7,8)\nD3: (3,-4)\n\nIf you are a CBSE-board student in India, logon to www.mathguru.com more than just a wonderful site.
787261
Let the 2 points be (x1,y1)  and  (x2,y2)\n\neqn of line s given by\n\n(y-y1)/(y2-y1) = (x-x1)/(x2-x1)\n\nwhere  (y2-y1)/(x2-x1) is the slope of line
91768
Quadratic Functions have the form y = ax^2 + bx + c\n\nThe vertex will always have x = -b/(2a). You can calculate this number and plug it into your equation to solve for y. If a is positive, the graph will go up in both directions from the vertex. if a is negative it will go down. You can pick some x value different from the vertex to help you plot other points to allow a symmetric parabola graph.\n\nGood Luck!
367471
Add the x coordinates together and divide by 2.\nAdd the y coordinates together and divide by 2.\n(8+16)/2 = 12\n(3+7)/2 = 5\nMidpoint = (12,5)
686219
1,24 24,1 2,12 12,2 3,8 8,3 4,6 6,4 -1,-24 -24,-1 -2,-12   \n-12,-2 -3,-8 -8,-3 -4,-6 -6,-4 and if x and y don't have to be intergers, there are basicly an infaninte number of possibilitys, such as 1/2,48 1/3,72 1/4,96..... so on, and the negatives. I hope x and y must be intergers, or you're stuck. easiest way is to graph it, or put like y=24/x or x=24/y. Not much help, but it's right.
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
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.
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
829281
Each A= 4 points.\nEach B=3\nEach C=2 \nEach D=1\nEach F=0\n\nAdd them all together: 4+4+2+2+0+1=13\n\nYou have 6 grades (2+2+1+1=6)\n\nDivide 13 by 6= 2.1667
432121
This is a graph theory problem.\n\nRepresent each person by a point, connect every point with every other point. Color the line red if the people know each other, blue if they don't.\n\nThe question now can be stated, that there is at least 1 red triangle, or one blue triangle. \n\nLabel the points p1, p2, .. p6.\n\nLook at p1. p1 is connected to the five other points. At least 3 of the lines must be the same color say red. With out loss of generality, we can all the points connected to p1 by red lines p2, p3, and p4.  If any pair of p1, p2, and p3 are connecte with a red line, we have a red triangle formed.\n\nIf not then p2, p3, and p4  form a blue triangle.
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The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
998290
mathematically speaking   if you had two points on a line\n\nPoint Dexter is some unknown point right of  the center\nand Point Sinister is some unknown point left of the center.
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x1 >= .2(x1 + x2)\nx1 >= .2x1 + .2x2\nSubtract .2x1 from each side\n.8x1 >= .2x2\nDivide both sides by .2\n4x1 >= x2\nx2 <= 4x1\nNow, assuming x1 is on the horizontal axis (x) and x2 is the vertical axis (y), then\n1) Place a dot at (0,0).  This is the y-intercept.\n2) The slope is 4.  Starting at the point (0,0), move up 4 and to the right by 1.  Place a dot there.\n3) Connect the two points with a straight line.\n4) The graph will be shaded below the line.\nI found this by picking any point that is not on the line.\nFor instance, the point (1,0) is not on the line.\n(1,0) means x1 = 1 and x2 = 0\nx2 <= 4x1\n0 <= 4  Since this is true, you shade the side of the line where this point lies.
1112189
1. g(x)=2x-5\nso, g(f(x))=2f(x)-5\nor 2 (x^2+2x+3)-5\nor 2x^2+4x+1\n\n2. \na) put f(x)=y\nso y=2^x+1\n\nnow take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph\n\nIf x=0, then y=2 (2^0=1)\nif x=1, then y=3\nif x=2, then y=5\n\nb) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..
790996
if cuts the x axis when p(x) = 0\n\nthat is 2x-3 = or x= 3/2\n\n2nd point x-5 =0 or x = 5\n\n3rd point x+1 = 0 or x = -1
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
779402
Well, its very easy to do this. When you have 2 different graphs, you graph then on your graphing calculator, then wherever the two graphs intersect, is the solution to the system. So if there are more than 1 point, means there are more than one solutions. IF you have more than 2 graphs, then you have to find the intersection  point where all of them meet and that would be the solution for the graphs.