To find out how much longer the first fish is than the second fish, we need to subtract the length of the second fish from the length of the first fish.
The lengths of the fish are:
- First fish: \( 6 \frac{1}{2} \) feet
- Second fish: \( 1 \frac{2}{3} \) feet
First, we convert these mixed numbers into improper fractions.
For the first fish: \[ 6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{6 \times 2 + 1}{2} = \frac{12 + 1}{2} = \frac{13}{2} \]
For the second fish: \[ 1 \frac{2}{3} = 1 + \frac{2}{3} = \frac{1 \times 3 + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3} \]
Now we have:
- First fish: \( \frac{13}{2} \)
- Second fish: \( \frac{5}{3} \)
Next, we need to subtract \( \frac{5}{3} \) from \( \frac{13}{2} \). To do this, we need a common denominator. The least common multiple of 2 and 3 is 6.
Now we convert each fraction: \[ \frac{13}{2} = \frac{13 \times 3}{2 \times 3} = \frac{39}{6} \] \[ \frac{5}{3} = \frac{5 \times 2}{3 \times 2} = \frac{10}{6} \]
Now we can perform the subtraction: \[ \frac{39}{6} - \frac{10}{6} = \frac{39 - 10}{6} = \frac{29}{6} \]
Now we can convert \( \frac{29}{6} \) back to a mixed number: \[ 29 \div 6 = 4 \quad \text{(with a remainder of 5)} \] This means: \[ \frac{29}{6} = 4 \frac{5}{6} \]
Thus, the first fish is \[ \boxed{4 \frac{5}{6}} \] feet longer than the second fish.
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