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. \( 21 \frac{9}{10}>12 \frac{1}{5} \)

1. The bird is located at \[ 21\frac{9}{10} \quad (\text{feet above sea level}) \] and the fish is at \[ -12\frac{1}{5} \quad (\text{feet, as it is below sea level}). \] 2. To compare their locations (taking into account the signs, where above sea level is positive and below sea level is negative), we have: \[ 21\frac{9}{10} > -12\frac{1}{5} \] This shows that the bird is higher than the fish relative to sea level. 3. To compare the distances of the bird and the fish from sea level, we consider their absolute distances (ignoring the sign): - The bird’s distance: \[ 21\frac{9}{10} \quad (\text{feet}) \] - The fish’s distance: \[ 12\frac{1}{5} \quad (\text{feet}) \] The comparison of these distances is: \[ 21\frac{9}{10} > 12\frac{1}{5} \] This tells us that the bird is farther from sea level than the fish. 4. Therefore, the correct expression that compares the locations of the bird and the fish is: \[ \boxed{21\frac{9}{10} > -12\frac{1}{5}} \] and the bird is the one that is farther from sea level.