Find the value of 'p' for which the points (-5,1), (1,p) and (4,-2) are collinear?
400659
Well... a slightly advanced method is that you should fist realize that a triangle drawn through these three points will have an area = 0\n\nformula for area of a triangle is:\n        \n| a1 b1 1 |\n| a2 b2 1 |  x 1/2 = A\n| a3 b3 1 |\n\nWhere (a1,b1) ; (a2,b2) and (a3,b3) are the 3 vertices of the triangle.\n{Sorry --- no better way to represent a matrix here...}\n\nor; A = 1/2 [ a1(b2-b3) - b1(a2-a3) + (a2.b3 - a3.b2) ]\n\nHere; A = 0 ; a1=-5; b1=1 : a2=1; b2=p and a3=4; b3=-2\n\ntherefore; 0 = 1/2 [ -5(p+2) - (1-4) + (-2 - 4p)]\nor;        0 = [-5p -10 + 3 -2 -4p]\nor;        0 = -9p -9\nor;       9p = -9\n\n=> p = -1\n\nAnother method would be to find the line through two given points and pass the line through the third point... but that is slightly lengthy.... \n\nIf you have any doubt; feel free to ask me.
967083
P = -4\n\nEntering the answer into the formula to "prove" it.\n\n-4 + 6 +3(-4) = 1 + 3(-4) +1\n\n-4 + 6 -12 = 1 -12 + 1\n\n-10 = -10
843943
4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!
793059
give values of -3,-2,-1,0,1,2,3 to x and find the corresponding values of y\nyou get the ordered pairs\nnow plot
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
1118574
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
979279
take 3p common\n\n3p(2p^2-3p-2)\n= 3p(2p^2-4p+p-2)\n= 3p(2p(p-2)+1(p-2))\n= 3p((p-2)(2p+1)
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
922969
y - y value.\nx - x value.\nm - slope of line.\nc - line intercept.\n\nSlope = -½\n pt(-2,4)\n y = mx + c\n 4 = m(-2) + c\n 4 = -2m + c\n 4 = -2(-½) + c\n 4 = +1 + c\n 3 = + c\n\n y = mx + c\n y = -½x + 3
985391
Use point-slope form:\nFirst, calculate the slope: (-2 - 0) / (5 - 0) = -2 / 5.\nHere is the general point-slope form:\n(y-y1) = m*(x-x1)\n(y+2) = (-2/5)*(x-5)\ny + 2 = -2/5*x + 2\ny = -2/5*x\n\n______\n\n\nFor parametric form, there's not that much to it:\nx = 0 + 5t\ny = 0 + -2t\n\nFor clarity, here's an example with another set of points:\n(2,-3) and (4,8)\nx = 2 + 2t\ny = -3 + 11t
714579
P (X>=5) \n= 1 - [  P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)] \n\nRandam sample of 100 days from the past 10 yrs... \nC(360*10, 100)
751380
Since we're reflecting along a line parallel to the X-axis, the X coordinates of these points doesn't change.\n\nNew_Ay = Old_Ay + (4 - Old_Ay) * 2 = 3 + (4 - 3) * 2 = 5\nNew_By = Old_By + (4 - Old_By) * 2 = 5 + (4 - 5) * 2 = 3\nNew_Cy = Old_Cy + (4 - Old_Cy) * 2 = 6 + (4 - 6) * 2 = 2\n\nA = (2,5)\nB = (0,3)\nC = (5,2)
903023
M (-7, -3) is the midpoint of segment PQ and the coordinates of Q are (6, -3).\nLet the co-ordinates of P be (x,y).\nThen, (-7,-3) = ((x+6)/2, (y-3)/2)\n=>(x+6)/2 => -7 , (y-3)/2 = -3\n=>(x+6) => -14 , (y-3) = -6\n=>x => -14-6 , y = -6+3\n=>x => -20 , y = -3\n\nSo, co-ordinates of point p are (-20, -3)\n\nAll the best
269860
p(x)=3*x^4)+(4*x^2-7)=(3*x^4)+(4*x^2-7)=p*x\nTherfore(3*x^4+4*x^2-7)+(-p*x)=p*x+(-p*x)\nand3*x^4+4*x^2-7-p*x=p*x-p*x\n3*x^4+4*x^2-p*x-7=0\nor px-px
1007342
y = -1 is a horizontal line that contains the points (n, -1)\n\nhow far is (3, 4) from that line?  Subtrace the y-coordinates: 4 - (-1) = 5
898860
First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
35268
slope = (-4-(-6))/(-2-0)\nslope = -1