Polyhedra with at least 3 pentagonal faces

Question

A convex polyhedron has at least three faces which are pentagons. What is the minimum number of faces the polyhedron might have?

I have a polyhedron with seven faces but I don't know whether it is possible with six:

Answer 1

At least $3$ edges meet at each vertex. Thus (denoting the numbers of vertices, edges and faces by $V$, $E$ and $F$, respectively), we have $3V\le 2E$. With $V-E+F=2$, this yields $F\ge2+\frac E3$. For $F=6$, we'd need $E\le12$. The $3$ pentagons have $15$ edges, and each pair of them can share at most one edge, so the pentagons themselves already have $12$ distinct edges, so $E=12$. But the $3$ pentagons have $15$ vertices, and each pair of them can share at most two vertices, so $V\ge15-3\cdot2=9$, contradicting $V=2+E-F=2+12-6=8$.

Thus your polyhedron has the minimal number of faces.