Respuesta :

WojtekR

We could use Euler's polyhedron formula:

[tex] F=n\qquad\qquad\text{(Faces)}\\\\E=2n\qquad\qquad\text{(Edges)}\\\\V=2n-4\qquad\qquad\text{(Vertices)}\\\\\\V-E+F=2\\\\2n-4-2n+n=2\\\\-4+n=2\qquad|+4\\\\\boxed{n=6} [/tex]