Yes.
to add to that
a vertex must be connected to at least 3 edges to be 3d, an edge is always connected to 2 vertexes, so the closest the two can ever be is
vetexes x 3 = edges x 2,
but when working with any platonic solid you can follow this:
vertexes x (faces / vertexes) x [edges on one side] = edges x 2
or
vertexes x [faces meeting at one vertex] = edges x 2
when working with any other polyhedron
[vertexes with x amount of faces] x (x) + [vertexes with y amount of faces] x (y) ...{and so on} = edges x 2
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A shape with more than 12 edges is a dodecahedron which has 30 edges
I don't know but I need to know =p
The shape would be impossible. The faces and vertices have to add up to two more than the edges.
Oh, isn't that a happy little question! Let's think about it together. A prism has 2 bases and the same number of edges as the number of sides on those bases, plus the number of edges connecting the corresponding vertices on the bases. So, a prism can't have seven more edges than vertices because the number of edges is determined by the number of sides on the bases and the number of vertices.
Yes, an example of this is a sphere which does not have any edges. If you had intended to ask if there are any polyhedra with less than three edges, the answer to that would be no, as the only figure constructable from three distinct lines is a triangle.