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For a simply connected convex polyhedron , that is, shapes that are topologically equivalent to a sphere, F - E + V = 2 where F = faces, E = edges and V = vertices. For shapes that are topologically equivalent to a sphere with one hole in it (torus or doughnut shape), F - E + V = 0 For shapes that are topologically equivalent to a sphere with two holes in it F - E + V = -2 For shapes that are topologically equivalent to a sphere with three holes in it F - E + V = -4 and so on.
There cannot be such shapes.The Euler characteristic for each shape requires Faces + Vertices = Edges + 2Therefore, for 2 shapes, F + V = E + 4The equation fails in this case.There cannot be such shapes.The Euler characteristic for each shape requires Faces + Vertices = Edges + 2Therefore, for 2 shapes, F + V = E + 4The equation fails in this case.There cannot be such shapes.The Euler characteristic for each shape requires Faces + Vertices = Edges + 2Therefore, for 2 shapes, F + V = E + 4The equation fails in this case.There cannot be such shapes.The Euler characteristic for each shape requires Faces + Vertices = Edges + 2Therefore, for 2 shapes, F + V = E + 4The equation fails in this case.
thats easy here r some rectangles and squares
Triangles ar e 2 demensional shapes so they don't have volume, but you calculate the area by the base ttimes the height divided by 2
ellipsis. ...