a certain poly hedron has 10 edges and 4 faces. determine the number of vertices on this polyhedron?
362454
The number of vertices in this polyhedron is exactly 8.\nAccording to Euler's Theorem, (number of vertices)-(number of edges)+(number of faces)=2 for all polyhedrons.
293695
It is a triangular prism. It has 5 faces, 9 edges, and last but not least 6 vertices. There is your answer you have been waiting for this whole time!
701332
p=72, each side=6\nyou can divide the hexagon into 6 equilateral triangles with side=6\nA of each triangle=(1/2)*6*6sqrt(3)/2=9*sqrt(3)\nfor all six, the area is 54*sqrt(3)=93.5
667352
octagons have 8\npolygons with "x" sides will have "x" axes of symmetry
416560
You should be able to picture it. Or at least, find a D+D player and look at a 12-sided die.\n\nOne half of this is a pentagon, surrounded on each of the 5 edges by other equal-sized pentagons.  The pentagon has 5 edges, and from each of the five corners is another edge which reaches to the 'equator' of the thing.  The bottom half is identical, with anoth 10 edges just like the flipside. Finally, around the 'equator', is another 5 edges.\n\n25 edges.  Don't need formulas, just good 3-D visualization skills.\n\nPS  It's 'polyhedron', by the way.  Polyhedra is plural.  Saying 'a polyhedra' is just like saying 'a women'.
904256
there are 6 vertices and you form 4 triangles from the first 2, then 3, then 2, then 1 for a total of 14.\n\nNow we form other triangles that overlap and it gets really complex...\n\nI count 18 smaller triangles, for a total of 32 so far...\n\nNow middle triangles...\n\n36, for a running total of 68
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.
774030
There are in infinate number of 2d shapes, but here are diagrams of some of them --> http://www.mathleague.com/help/geometry/polygons.htm
623411
Pentagon = 5, decagon = 10, quadrilateral = 4, and heptagon = 7.
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
1169447
The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
543718
6 faces, 12 edges, 8 vertices
147899
Here is a graph and some explanation:\nhttp://mathworld.wolfram.com/Tetrahedron6-Compound.html
50989
Eleven if you count the two adjacent sides of the polygon, otherwise nine.\n\nThinking outside the box answer:\nUnlimited if one does not limit themselves to the other vertices as termini.
1147660
Euler's Theorem \nA graph divides the plane into several regions called faces. Let us denote the number of the faces by F, the number of the vertices by V, the number of the edges by E. \n the given graph divides the plane into 4 regions, so we have that F=4,V=6,E=8 \n\nProblem . Evaluate F, V and E for the complete graph of 5 vertices, i.e. the pentagon and its diagonals. \n\nAnswer. F=12,V=5,E=10 \n\nA graph that can be drawn in such a way that its edges do not intersect each other (except at their endpoints) is called planar. \nAlthough these two graphs are different, they have equally many vertices and we can numerate the vertices of both graphs in such a way that any two vertices of the first graph are connected by an edge if and only if the corresponding vertices of the second graph (with the same numbers) are connected by an edge. Such graphs are called isomorphic. Actually we will not distinguish isomorphic graphs. So a graph may be planar even if some of its edges intersect in a given picture (see Figure 4). Try to make another couples of pictures describing planar graphs \n\nEuler's Theorem. For a planar graph the equality V-E+F=2 always holds true. \n\n\nRemark. Euler's Theorem is an example of topological obstruction. Problem shows that the requirement for a graph to be a planar is important. An interesting problem is to show that a complete graph is planar if and only if V<5 \n\nProof of Euler's Theorem. For the proof we need to introduce a class of special graphs. A tree G is a connected graph without cycles, i.e. any two vertices of G can be connected by a path (a sequences of edges) and there is no path whose starting and ending vertices coincide. On Figure 5 you can see an example of a tree (a), and a graph which is not a tree (b). \nObviously the trees are simplier then the other graphs. So our goal is to delete some edges until we get a tree, keeping the graph connected. Figure 6 shows how after deleting one edge the quantity V-E+F does nor change. Really, before the operation we have F=5,V=5,E=8 and after that F=4,V=5,E=7. So in both cases V-E+F=2. Since for trees we have that V-E=1 (prove it!) and F=1 the theorem is proved. \nProblem 6. There are 7 lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. Haw many islands are there in Lakeland? \n\nHint. Consider the planar graph whose vertices are the lakes, whose edges are the canals, and whose faces are the islands. So F=5. Answer: 4
287514
number of doors\n\ncoupe = 2\nsedan = 4
910387
The nearest vertex will be any of the four lower vertices because the ball is closer to the bottom than to the top of the cube, and the centre of a square is equidistant from all four vertices.\n\nPicture the right triangle whose vertices are:\nA = the center of the ball\nB = one of the lower vertices\nC = the center point of the bottom face of the cube.\n\nYou want to know the length of segment AB which is the hypotenuse of this triangle.  You know that AC is 6 cm long.  And BC is half the diagonal of the bottom face so it measures 1/2 * sqrt(2) * 14 = 7*sqrt(2) cm.\n\nSo AB = sqrt(AC^2 + BC^2)\n= sqrt(6^2 + (7*sqrt(2))^2)\n= sqrt(36 + 49*2)\n= sqrt(134) cm.