41 cts Abird lies \( 21 \frac{9}{10} \) feet above sea level Atish swims \( 12 \frac{1}{5} \) foet below sea level Wite an expression that compares the locations of the bird and the fish Then write an expression that compares detances from sea lovel Which is farther fom sea lovel? Which expression compares the locations of the Nid and the fish? A. \( 21 \frac{9}{10}>-12 \frac{1}{5} \) B. \( 21 \frac{9}{10}<12 \frac{1}{5} \) C. \( 21 \frac{9}{10}=-12 \frac{1}{5} \) B. \( 21 \frac{9}{10}>12 \frac{1}{5} \)

The correct expression that compares the locations of the bird and the fish is A: \( 21 \frac{9}{10} > -12 \frac{1}{5} \). This indicates that the bird, which is \( 21 \frac{9}{10} \) feet above sea level, is higher than the fish, which is below sea level at \( -12 \frac{1}{5} \) feet. To determine the distances from sea level, we can express both measurements as positive values. The distance of the bird from sea level is \( 21 \frac{9}{10} \) feet, while the distance of the fish is \( 12 \frac{1}{5} \) feet below sea level, when expressed positively it becomes \( 12 \frac{1}{5} \) feet above sea level. Since \( 21 \frac{9}{10} > 12 \frac{1}{5} \), the bird is farther from sea level!