To subtract the mixed numbers \(9 \frac{7}{10}\) and \(6 \frac{1}{4}\), we'll first convert them to improper fractions.
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Convert \(9 \frac{7}{10}\) to an improper fraction: \[ 9 \frac{7}{10} = \frac{9 \times 10 + 7}{10} = \frac{90 + 7}{10} = \frac{97}{10} \]
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Convert \(6 \frac{1}{4}\) to an improper fraction: \[ 6 \frac{1}{4} = \frac{6 \times 4 + 1}{4} = \frac{24 + 1}{4} = \frac{25}{4} \]
Now we need to subtract \( \frac{25}{4} \) from \( \frac{97}{10} \). To do this, we need a common denominator. The least common multiple of 10 and 4 is 20.
- Convert both fractions to have a denominator of 20: \[ \frac{97}{10} = \frac{97 \times 2}{10 \times 2} = \frac{194}{20} \] \[ \frac{25}{4} = \frac{25 \times 5}{4 \times 5} = \frac{125}{20} \]
Now we can subtract: \[ \frac{194}{20} - \frac{125}{20} = \frac{194 - 125}{20} = \frac{69}{20} \]
Next, we convert \( \frac{69}{20} \) back into a mixed number: \[ 69 \div 20 = 3 \quad \text{remainder } 9 \] So, \[ \frac{69}{20} = 3 \frac{9}{20} \]
Therefore, the difference between \(9 \frac{7}{10}\) and \(6 \frac{1}{4}\) is: \[ 3 \frac{9}{20} \]
So the correct answer is: 3 9/20 (3 and 9 over 20).