For given set of points in 2d, finding the line with max. points on it?
958876
A simple algorithm is to first take all n(n-1)/2 pairs of points. Each pair defines a line, count how many of the remaining points lie on the line. This will be a cubic algorithm in n, the number of points.\n\nA clever algorithm using sorting takes time O( n^2 log n). See the Sources.
593080
*** pseudocode ****\n\n\n$max = $array[0];\n$min = $array[0];\n\nfor( $x=1; $x <= $size_of_array; $x++ )\n {\n     if($array[$x] < $min)\n       {\n          $min = $array[$x];\n       }\n\n     if($array[$x] > $max)\n       {\n          $max = $array[$x];\n       }\n\n}\n\n\n\n\nhave fun\n\n\nt
49129
To find the direction of maximum ascent, find the directional derivative (DD), and find the direction where it has its max.\nNote: This is the equation of a plane, so the directional derivative is independent of the point where it is taken.\n\nIf v is a unit vector, then the DD in the direction of v, is the vector dot product of v with the vector  formed from the two partial derivatives\nPartial Z with respect to x is 1.5\nPartial Z with respedt to y is -.8\n\nDD = (1.5, -.8) dot v\n\nSince v is  unit vector, you can represent it\n\n(sin t, cos t)\n\ngiving us DD = (1.5)sint - (.8)cost \n\nTo find direction the max, take the derivative with respect to t, and set to 0.\n\n0 = (1.5)cost + (.8)sint\n-(1.5)cost = .8sint\n-1.5/.8 = tan t\n\nThe solutions to that say that v has the same direction as either of the (non-unit) vectors\n(1.5, -.8) or (-1.5, .08). The first is maximum ascent, the second maximum descent.
120697
To find a point on the plane, you simply need to find any values of x, y, and z which make the equation true. The easiest way to find a point is to choose two of the variables and set them equal to zero. So in your example, I might choose to set x and z to zero. Then -y must equal 4, so y must be -4 when x and z are zero. This point is written (0,-4,0). Other points include (2,0,0), (0,0,-2). In this plane, we can pick any values for two of the variables and find the third. This wouldn't be true for planes which are parallel to one of the axes.
1148398
the shortest distance between two points
5209
You can use some math to approximate the right zoom given boundary YGeoPoint max/min values… for example given fixed map container, you want to plot N points of which you identify YGP_MAX and YGP_MIN, you also figure out the relationship where latitude degrees per pixel doubles with each zoom out.  You can observe this by inspecting return from:\nvar DegLatPerPix =  (map.getBoundsLatLon().LatMax – map.getBoundsLatLon().LatMin)/(map.getContainerSize().height);\nThis is where you can guess roughly or go further to derive a function to accomplish the job.  Let’s suppose you obtained DegLatPerPix_1 for zoom 1 your zoom follows following behaviour: zoom = 1 + log(Y/DegLatPerPix_1)/log(2), where Y represents value of degrees per pixel obtained from YGP_MAX and YGP_MIN.\n\nOR \n\nYou can simply use supplied (undocumented option) map.getZoomLevel(YGP_MAX,YGP_MIN) to return best zoom level to contain rectangle defined by YGP_MAX and YGP_MIN, optionally you can supply as third argument, container size.\ni.e. \nvar point1 = new YGeoPoint(45.213004,-123.222656);\nvar point2 = new YGeoPoint(32.101190,-83.671875);\n\nvar bestZoom = map.getZoomLevel(point1, point2);\n\nvar centerPoint = map.getCenterLatLon(); //get auto computed middle of the rectangle defined by MAX/MIN above\n\nmap.drawZoomAndCenter(centerPoint, bestZoom);\n\nCheers!
888701
ha! max is not a builtin function.\n\nbut Math.max is!\n\nalert( Math.max( 10,5,21,54,23) )\n\nworks!
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
934709
Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
152511
Which function, Subsitution, Linear COmbination, Graph?
504335
Draw two lines from any tow angles to two opposite base
774094
- Find the x-intercepts, by setting y=0 and solving the polynomial equation. You may need to factorise\n\n- Find the y-intercepts, by setting x=0 and solving.\n\n- Find stationary points by differentiating the polynomial equation wrt x and setting the differential to 0. These give you the stationary points.\n\n- Differentiate again to find out whether these turning points are maxima, minima or points of inflexion:\nIf d²y/dx²  is positive, then it is a minimum point.  \nIf d²y/dx² is negative, then it is a maximum point  \nIf d²y/dx² is zero, then it could be a max, a min or a point of inflexion.\n\n- Find asymptotes. You may get a horizontal asymptote when x approaches +ve or -ve infinity and a vertical one as y approaches +ve,-ve infinity.\n\nThis is the basic info you need to sketch a graph depending on how complex the polynomial is.
95382
Taken from about.com:\nTo find the midpoint of a line section the formula is: \n\nx = (x1 + x2)/2 \nand \ny = (y1 + y2)/2 \n\nwhere x and y are the coordinates of the midpoint, and (x1,y1) are the coordinates of the first point and (x2,y2) are the coordinates of the second point. Let's plug it in to the equation. \n\nx = (7+10)/2 = 17/2 = 8.5 \ny = (3+6)/2 = 9/2 = 4.5 \n\nSo the midpoint is (8.5,4.5)
1127952
Try to graph the points and then decide what the best function that can fit the point; however , I tried the following:\n\nSince a, b, c, d are constants then \n\ny=(a-d)/(1+(x/c)^b)+d \n\nLet a-b = k and c^b = L then\n\ny= k /(1+(x^b)/L) + d \n\ny= (k * L)/(L+(x^b)) + d\n\nLet k*L = m\n\ny= m / (L+(x^b) + d\n\ny= ( m + L*d + d*x^b ) / ( L + x^b)\n\nLet u= m+L*d\n\ny= (  u + d*x^b ) / ( L + x^b)\n\nI think the graph of such a function as smiler to the (ln) function since both the nominator and the denominator have the same order. But I couldn't simplify the function any more..... that might help a bit ...! good luck
701304
You really need to clarify your question. \n\nI could give you the min and max path weight in a undirected graph between two vertices, but in a weighted, undirected graph, an edge's weight is an edge's weight. There's no formula needed, unless I really, really misunderstood your question.
885850
y'=-10x+2\nsetting itto 0\nx=1/5\ny''=-10 negative so a max at x=1/5\nthe maximum value=-5(1/5)^2+2/5-4\n=-1/5+2/5-4\n=-19/5
826422
Find the slope of the line that connects those points.