jmc

We want to find the residue $c$ such that $35c\equiv 1(\mathrm{mod}47)$. Recall that, since 35 is relatively prime to 47, this inverse exists and is unique. To make use of the table we are given, we notice that $35=5\cdot 7$. We can multiply both sides of $35c\equiv 1(\mathrm{mod}47)$ by the inverse of 5 to obtain $$\begin{array}{rl}19\cdot 5\cdot 7\cdot c& \equiv 19\cdot 1(\mathrm{mod}47)⟹\\ (19\cdot 5)\cdot 7\cdot c& \equiv 19(\mathrm{mod}47)⟹\\ 1\cdot 7\cdot c& \equiv 19(\mathrm{mod}47).\end{array}$$Now we can multiply both sides by 27, the inverse of 7, to find $$\begin{array}{rl}27\cdot 7\cdot c& \equiv 27\cdot 19(\mathrm{mod}47)⟹\\ c& \equiv 513(\mathrm{mod}47).\end{array}$$Subtracting 470 from 513 does not change its residue (mod 47), so we have $c\equiv 43(\mathrm{mod}47)$. Since $0\le 43<47$, $\boxed{43}$ is the desired residue.

Remark: More generally, the approach above shows that $(ab{)}^{-1}=b^{-1}{a}^{-1}$, where $b^{-1}$ denotes the modular inverse of $b$.