Find the co-ordinates of the vertex and state whether it has a maximum or a minimum.?
919933
I assume this is equal to y?\n\nIf it is, it's definitely a maximum because it's a parabola that opens downward\n\nNow I'll work on the vertex
1010902
This is  basic\nslope= m=(y2-y1)/(x2-x1)=(12-7)/(8-6)=5/2\nangle=arctan(5/2)
919538
it is narrow because a is more than 1.\nThe vertex=1,-1 you take the opposite of h which=x and y=-1\ndo a t table to find the max and min \nput the vertex in and put points for x and solve for y for each one using the equation
817524
For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.\n\nThere are 2 methods:\nCalculus (using derivatives):\nf(x) = x2 - 2x - 4\nf'(x) = 2x - 2\nthe maximum/minimum will be where f'(x) = 0.\n2x - 2 = 0\nx = 1\ny = 1^2 - 2*1 - 4 = -5\n\nAlgebra:\nThe vertex can be found using the formula:\n(-b/2a , -(b^2 - 4ac)/4a)\n(-(-2)/2, -(4 - (4*1*-4))/4)\n(1,-5)\n\nCheck: x = 1, y = 1^2 - 2*1 - 4 = -5
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
692865
V = x * (4-x) * (6-x) = x^3 - 10x^2 + 24x\n\n0 < x < 4\n\nGraph the function and find the max between 0 and 4
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
690146
Optimisation problems are normally done using partial derivatives, however you're problem is discreet so it will be handled differently. First make a map of the network lets say with coloured lines to represent the individual paths from the beginning to the end of the manufacturing process and nodes where they intersect, i.e. nodes that represent each machine. Then there are multiple ways to run the algorithm. For the case of minimum number search through the nodes to see if there is one that bundles all the lines through it if not start looking for pairs that would have all the lines going through one or the other of the nodes and so on until you find a solution. Unless visually it is a simple graph you will need a computer to do the processing.\nThe problem with a cost assigned is more difficult. In this case each node has a value associated with it. In this case generate an ordering from cheapest to dearest of the possible costs of selecting any number of nodes. The test each of these in succession until you find one that corresponds to having all the lines pass through at least one of the selected nodes associated with that particular cost.\nIn knowing details of the way in which the graph would look or the cost for certain locations it may be possible to modify the algorithm so it runs quicker, but this should do it for the information given.
952205
You are missing a constraint in your problem.  If your only constraints are \n- theta must be less than or equal to pi/2 \n- you want to maximize theta\nthen theta = pi/2.\n\nSo how far should you be?  0 meters.
986056
the above is maximum when ax^2+bx+c is minimum\n\nax^2+bx+c is minium or c when x = 0 as a,b are positive\n\nso maximum value f(x) is at 0 that is A exp(-c)
526722
float ComputeEdgeCollapseCost(Vertex *u,Vertex *v)\n{\n   // if we collapse edge uv by moving u to v then how\n   // much different will the model change, i.e. the\n   // "error".\n   float edgelength = magnitude(v->position - u->position);\n   float curvature=0;\n   // find the "sides" triangles that are on the edge uv\n   List<Triangle *> sides;\n   for(i=0;i<u->face.num;i++)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         sides.Add(u->face[i]);\n      }\n   }\n   // use the triangle facing most away from the sides\n   // to determine our curvature term\n   for(i=0;i<u->face.num;i++)\n   {\n      float mincurv=1;\n      for(int j=0;j < sides.num;j++)\n      {\n         // use dot product of face normals.\n         float dotprod = \n            u->face[i]->normal ^   sides[j]->normal;\n         mincurv = min(mincurv,(1-dotprod)/2.0f);\n      }\n      curvature = max(curvature,mincurv);\n   }\n   return edgelength * curvature;\n}\n\nvoid ComputeEdgeCostAtVertex(Vertex *v)\n{\n   if(v->neighbor.num==0)\n   {\n      v->collapse=NULL;\n      v->cost=-0.01f;\n      return;\n   }\n   v->cost = 1000000;\n   v->collapse=NULL;\n   // search all neighboring edges for Òleast costÓ edge\n   for(int i=0;i < v->neighbor.num;i++)\n   {\n      float c;\n      c = ComputeEdgeCollapseCost(v,v->neighbor[i]);\n      if(c < v->cost)\n      {\n         v->collapse=v-neighbor[i];\n         v->cost=c;\n      }\n   }\n}\n\nvoid Collapse(Vertex *u,Vertex *v)\n{\n   // Collapse the edge uv by moving vertex u onto v\n   if(!v)\n   {\n      // u is a vertex all by itself so just delete it\n      delete u;\n      return;\n   }\n   int i;\n   List<Vertex *>tmp;\n   // make tmp a list of all the neighbors of u\n   for(i=0;i<u->neighbor.num;i++)\n   {\n      tmp.Add(u->neighbor[i]);\n   }\n   // delete triangles on edge uv:\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      if(u->face[i]->HasVertex(v))\n      {\n         delete(u->face[i]);\n      }\n   }\n   // update remaining triangles to have v instead of u\n   for(i=u->face.num-1;i>=0;i--)\n   {\n      u->face[i]->ReplaceVertex(u,v);\n   }\n   delete u;\n   // recompute the edge collapse costs in neighborhood\n   for(i=0;i<tmp.num;i++)\n   {\n      ComputeEdgeCostAtVertex(tmp[i]);\n   }\n}
1057389
Look at the coefficient of the x^2 term.  If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum.  You have 2x^2, so you will have a minimum value.\n\nNext find the vertex.  The x-coordinate of the vertex lies at -b/2a.  So -8/(2)(2) = -8/4 = -2.  Plug this value in to get the y-coordinate of the vertex.  2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.\n\nSo you have a minimum value of -3.
1148398
the shortest distance between two points
185652
since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
727910
P = NL where N is the number of sides the polygon has\n\nSo if L = 6 cm \nthen P = 6N cm
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
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