To solve the problem \( 1 \frac{2}{9} + 5 \frac{2}{3} \), we can first convert each mixed number into an improper fraction.
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Convert \( 1 \frac{2}{9} \): \[ 1 \frac{2}{9} = \frac{1 \times 9 + 2}{9} = \frac{9 + 2}{9} = \frac{11}{9} \]
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Convert \( 5 \frac{2}{3} \): \[ 5 \frac{2}{3} = \frac{5 \times 3 + 2}{3} = \frac{15 + 2}{3} = \frac{17}{3} \]
Now we need to add \( \frac{11}{9} \) and \( \frac{17}{3} \). To do this, we need a common denominator. The least common multiple of 9 and 3 is 9.
Convert \( \frac{17}{3} \) to have a denominator of 9: \[ \frac{17}{3} = \frac{17 \times 3}{3 \times 3} = \frac{51}{9} \]
Now we can add: \[ \frac{11}{9} + \frac{51}{9} = \frac{11 + 51}{9} = \frac{62}{9} \]
Finally, convert \( \frac{62}{9} \) back to a mixed number: \[ 62 \div 9 = 6 \quad \text{(remainder } 8\text{)} \] Thus, \( \frac{62}{9} = 6 \frac{8}{9} \).
Therefore, the sum \( 1 \frac{2}{9} + 5 \frac{2}{3} \) is: \[ \boxed{6 \frac{8}{9}} \]