A truncated cube is a convex polyhedron with 36 edges and 24 vertices. A truncated tetrahedron is a convex polyhedron with 18 edges and 12 vertices.

How do the number of faces of a truncated cube and a truncated tetrahedron compare?

The truncated cube has 6 more faces than the truncated tetrahedron.

The truncated cube has 8 more faces than the truncated tetrahedron.

The truncated cube has 12 more faces than the truncated tetrahedron.

The truncated cube has 18 more faces than the truncated tetrahedron.

The truncated cube has 6 more faces than the truncated tetrahedron.

Step-by-step explanation:

==>Given:

truncated cube=>

edges (E) = 36

vertices (V) = 24

faces (F) = ? (unknown)

According to Euler's polyhedron formula, Vertices (V) - Edges (E) + Faces (F) = 2, let's find how many faces the truncated cube has

=> 24 - 36 + F = 2

- 12 + F = 2

F = 2 + 12

F = 14

truncated tetrahedron=>

edges (E) = 18

vertices (V) = 12

faces (F) = ? (unknown)

Using V - E + F = 2, find F

=>12 - 18 + F = 2

- 6 + F = 2

F = 2 + 6

F = 8

Comparing the faces of the truncated cube (14) and the truncated tetrahedron (8), the truncated cube has 6 more faces than the truncated tetrahedron (14 - 8 = 6)

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead