No, F + V = E + 2That's Euler's polyhedron formula (or Theorem). For a normal 3-d polyhedron to exist it must conform to that equation.
A polyhedron must have at least 4 faces, at least 4 vertices and at least 6 edges.
eighthedron eighthedron
Such a polyhedron cannot exist. According to the Euler characteristics, V + F - E = 2, where V = vertices, F = faces, E = edges. This would require that the polyhedron had only two faces.
12 vertices A prism with an n-sided base will have 2n vertices, n + 2 faces, and 3n edges.
There is not a specific formula fro vertices and edges. The Euler characteristic links the number of vertices, edges AND faces as follows: E + 2 = V + F for a simply connected polyhedron.
It has 6 vertices.
There is no such convex polyhedron in normal geometries because it does not satisfy the Euler characteristic. That requires that Faces + Vertices = Edges + 2
If the object is a convex polyhedron, then, by Euler's characteristics, it should have 23 faces.
A 20-sided polyhedron is called an icosahedron. It has 20 equilateral triangle faces, 30 edges, and 12 vertices. It is a regular convex polyhedron with rotational symmetry.
The number of edges and vertices ina polyhedron will depend on the polyhedron one selects either to study, build or etc...
A polyhedron has 30 edges and 12 vertices. How many faces does it have
A polyhedron must have at least 4 faces, at least 4 vertices and at least 6 edges.
This polyhedron has 7 vertices and 12 edges.
The numbers in the question do not satisfy the Euler characteristic so there cannot be such a [convex] polyhedron.
For a simply connected polyhedron,Faces + Vertices = Edges + 2
Edges join two vertices. Three or more edges meet at a vertex (it could be two if not in a polyhedron).
I believe you intend to talk about a polyhedron if it is a convex polyhedron, there is a relation : F + V * E = 2 (you can experiment with current polyhedrons) the relation is not satisfied by your numbers