There cannot be a 3-d shape all of whose faces are regular hexagons.
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that claim above is not true, because for example a classic soccerball consists of hexagons.
It's called, not altogether surprisingly, a hexagonal prism. If the bases were hexagons and the other faces were triangles, it would be a hexagonal antiprism.
If by faces you mean sides, then lots. Parallelograms and squares come to mind but I'm sure there are more (hexagons for would be another).
A hexahedron is a three-dimensional figure with six identical faces - in other words, a cube. So a hexahedron is not made up of hexagons at all, but of squares. However, if you were to balance a cube on one of the vertices, the horizontal plane cutting the cube in half would make a cut in the shape of a regular hexagon. Four such regular hexagons can be found in the cube.
I would mostly say yes if it was a cube but I don't know
A standard sphere, like a ball, has no flat faces; it is a three-dimensional shape with a continuous curved surface. Therefore, it can be said to have zero faces. However, if you're referring to a polyhedral ball, such as a soccer ball, it is made up of multiple flat faces, typically hexagons and pentagons. In that case, the number of faces would depend on the specific design of the ball.
If by "vertex," you mean "apex," than any pyramid would fit the description.
The 3d shape that have five faces would be a triangular prism **************** or a pyramid
Since a cube has 6 faces, you would be looking for a solid shape with only 2 faces and no such solid exists.
the other faces on a pentagonal prism would be rectangular
Its a Octagon * * * * * No. An octagon is a shape with 8 sides. 8 faces implies a 3-dimensional object, which would have to be an octahedron.
A cuboid would fit the given description which has 6 faces, 12 edges and 8 vertices
The shape would be impossible. The faces and vertices have to add up to two more than the edges.