jmc

When adding the numbers in base $12$, we start by adding the rightmost digits as we do in normal addition. Since $4+9$ yields a residue of $1$ upon division by $12$, we write down a $1$ as the rightmost digit of the sum, and carry-over a $1$. The remaining two digits do not yield and carry-overs, so we can add them as normal. Carrying out this addition, we find that:\begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & \stackrel{}{7} & \stackrel{1}{0} & \stackrel{}{4}_{12} \\ &+ & & 1 & 5 & 9_{12} \\ \cline{2-6} && & 8 & 6 & 1_{12} \\ \end{array} .Thus, our answer is $\boxed{861_{12}}$.