The expression \( \frac{47}{3} + 121^2 \) results in \( \frac{47}{3} + 14641 \), which is a sum of two rational numbers; hence, the result is also rational.
To find the sum: \[ 121^2 = 14641 \ \frac{47}{3} + 14641 = \frac{47}{3} + \frac{14641 \times 3}{3} = \frac{47 + 43923}{3} = \frac{43970}{3} \] Thus, the sum is \( \frac{43970}{3} \), which is rational.