what is the sum of the number of edges and the number of vertices on a cube?
462376
20\n\n 12 edges and 8 vertices
781998
27 cubes, then 81, then 243, then 729...
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}
977457
1 to 64k.. so there exists about 40 cubes (1^3, 2^3...  39^3, 40^3)..  and sum of 4 cubes. that's like. "40 choose 4" (combinatorics) so that comes out to be about 91390 combinations of 4 cubes and then you just find the cube roots of each. each should have its own cube root. and 91 thousand isn't too bad for an average computer to handle. i suggest you program a computer to do it for you. it wouldn't be hard, just basic programming language.\n\nif you cant do that, then i suggest you try to narrow down what you're looking for...
24746
I think the sum of all internal angels of a polygon is given by (n-2)*180. So, for a 40 sided polygon, it should be (40-2)*180=38*180=6840 I guess.
152387
polyhedron (pŏl'ēhē'drən) , closed solid bounded by plane faces; each face of a polyhedron is a polygon. A cube is a polyhedron bounded by six polygons (in this case squares) meeting at right angles. Although regular polygons are possible for any number of sides, there are only five possible regular polyhedrons, having congruent faces, each a regular polygon and meeting at equal angles. The five regular polyhedrons are also known as the Platonic solids, although they were known to the Greeks before the time of Plato. They are the tetrahedron, bounded by four equilateral triangles; the hexahedron, or cube, bounded by six squares; the octahedron, bounded by eight equilateral triangles; the dodecahedron, bounded by twelve regular pentagons; and the icosahedron, bounded by twenty equilateral triangles. The 18th-century Swiss mathematician Leonhard Euler showed that for any simple polyhedron, i.e., a polyhedron containing no holes, the sum of the number of vertices V and the number of faces F is equal to the number of edges E plus 2, or V+F=E+2
319863
It has to do woth the number of vertices of hexagons written inside other hexagons.  Go to \n\nhttp://mathworld.wolfram.com/HexagonalNumber.html\n\nfor more details.
876576
in a cube, there are 6 sides, in each side, there is only one number, thus called a one number cube.\n\nThe order is 1-6. number 3 comes out only once. Thus, the chance of getting a 3, is only one out of 6. making it 1/6 in math language.
173051
area is .001 X .001 = .00001 (the area per cube face) X 6 ( the number of faces on a cube)=.00006 cm squared.\nVolume is length times width times height, or .001 X .001 X .001 = .0000001 cm cubed\n\nso, .00006/.0000001 = 6/.001 = 6000.
801416
sphere none of the list\ncone one vertex and one face\ncylinder two faces-two circular edges
651972
A nonagon has 9 sides and 9 vertices.
694740
Is this for the sum of the first n cubes formula? I can tell you what your problem is right now: that two should be _inside_ the brackets. Otherwise, this formula doesn't work.\n\nNow, using the correct formula:\n(k(k+1)/2)² + (k+1)³\nWe start by actually multiplying out the stuff in the parentheses:\n((k²+k)/2)² + (k+1)³\n(k^4 + 2k³ + k²)/4 + (k³ + 3k² + 3k + 1)\nNow we give these fractions a common denominator:\n(k^4 + 2k³ + k²)/4 + (4k³ + 12k² + 12k + 4)/4\nAnd add:\n(k^4 + 6k³ + 13k² + 12k + 4)/4\nFactoring the polynomial:\n(k+1)²(k+2)²/4\n((k+1)(k+2)/2)²\nThus completing the proof.
957591
line = 1 side\nparallel lines = 2 sides\ntriangle = 3 sides\nsquare/ rectangle = 4 sides\npentagon = 5 sides\nhexagon = 6 sides\nheptagon = 7 sides\noctagon = 8 sides\nnonagon = 9 sides\ndecagon = 10 sides\n\ni hope i have helped you.... good luck
277462
Within your 3x3x3 cube, we have ten classes of rectangular prisms, including the two boundary cases of the whole cube and of the single cublet:\n1x1x1, 1x1x2, 1x2x2, 1x1x3, 1x2x3, 2x2x2, 2x2x3, 1x3x3, 2x3x3, 3x3x3.\n\nFor each prism class, we have to count all possible arrangements within the main cube. The contributes to this number are from translations, then from rotations (but only for non-symmetric prisms). If I'm not wrong, they are:\n1x1x1: 27\n1x1x2: 54\n1x2x2: 12\n1x1x3: 27\n1x2x3: 18\n2x2x2: 8\n1x3x3: 9\n2x2x3: 12\n2x3x3: 6\n3x3x3: 1\n\nTotal=174 prisms\n\nAs per your later addition, if we want to discard cubic prisms, we have: 174-27-8-1 = 138 non-cubic prisms.
301913
A cube can be negative, so the sum of two pairs of cubes can also be negative.  There is no smallest.
956378
There are 35 triangles.  Count them this way:  All triangles whose vertices are vertices of the pentagon.  There are exactly 10 of hese, one for every different set of three vertices.  Next count the triangles using exactly two adjacent vertices of the pentagon, plus one vertex inside the figure.  Each pair of adjacent vertices has three such triangles, so we get 15 more.  Now count the triangles which use two NON-adjacent vertices of the pentagon plus an inside vertex.  There is just one triangle for each such pair, and there are five pairs, so 5 more triangles.  Now count the triangles using one pentagon vertex and two inside vertices.  There is only one for each vertex, so 5 more.  There are NO triangles using only inside vertices, so we are done, and it adds up to 35.\n\nI copied and pasted this directly from http://mathforum.org/library/drmath/view/54888.html
947383
cubes have all sides with the same length. Volume = length x width x height. So for a cube = length^3