The table below shows the multiplicative inverses of the first $9$ positive residues modulo $47$.
\begin{tabular}{c|ccccccccc}
$b$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline
inverse of $\,b$ & 1 & 24 & 16 & 12 & 19 & 8 & 27 & 6 & 21
\end{tabular}
Find the multiplicative inverse of $10\pmod{47}$. Express your answer as an integer from $0$ to $46$, inclusive.