None.Vertices is a plural term and therefore "a vertices" cannot exist. As a result "a vertices" cannot have any vertices. In fact, it cannot have anything apart from non-existence.
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More than one unique triangle exist
Yes, such a quadrilateral can exist.
Yes, they exist.
Yes, they do exist!
None.Vertices is a plural term and therefore "a vertices" cannot exist. As a result "a vertices" cannot have any vertices. In fact, it cannot have anything apart from non-existence.
it does not exist and please do not delete my answer. it truly does not exist
There is no such thing as a hexagonal solid. A regular hexagon will tessellate and so it forms a plane (2-D) surface, not a 3-D shape.A 3-D shape with six faces is a hexahedron and this could be a triangular bipyramid with 5 vertices, a parallelepiped with 8 vertices, a pentagonal pyramid with six vertices. There are also other possible shapes.
I believe that such an object cannot exist in normal 3-d space. If there are 6 vertices, the maximum number of edges is 12.
A shape with four sides and three vertices does not exist in Euclidean geometry. In Euclidean geometry, a shape must have the same number of sides as vertices. Therefore, a shape with four sides would have four vertices.
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 dodehedron does not exist. A regular dodecahedron has 20 vertices, 30 edges and 12 faces. A dodecahedron must have 12 faces, but it can have any number from 8 to 20 vertices and so 18 to 30 edges.
i think i would be five defanly five.
The Euler characteristic indicates that such a solid does not exist.
Of course lesbians exist. There are millions of them.
No. Since the graph is simple, none of the vertices connect to themselves - that is, there are no arcs that loop back on themselves. Then the two vertices with degree 6 must connect to all the other vertices. Therefore there can be no vertex with less than two arcs [ to these two vertices]. So a vertex with degree 1 cannot be part of the graph.