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Case 3: Each face is a pentagon. We perform the same calculation as above, this time getting \(e = 5f/2\) so \(v = 2 + 3f/2\text{.}\) Then
\begin{equation*}
e = \frac{5f}{2} = \frac{k(2+3f/2)}{2}\text{,}
\end{equation*}
so
\begin{equation*}
k = \frac{10f}{4+3f}\text{.}
\end{equation*}
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