To determine how far Joey is from home after walking east and then west, we can set up the problem as follows:
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Convert mixed numbers to improper fractions:
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15 2/3 meters: \[ 15 \frac{2}{3} = \frac{15 \times 3 + 2}{3} = \frac{45 + 2}{3} = \frac{47}{3} \]
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21 3/4 meters: \[ 21 \frac{3}{4} = \frac{21 \times 4 + 3}{4} = \frac{84 + 3}{4} = \frac{87}{4} \]
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Convert both fractions to a common denominator.
The least common multiple of 3 and 4 is 12.
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Convert \(\frac{47}{3}\) to a denominator of 12: \[ \frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12} \]
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Convert \(\frac{87}{4}\) to a denominator of 12: \[ \frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12} \]
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Calculate Joey's total distance walked from home.
Since he first walked east and then west, we subtract the two distances: \[ \text{Distance from home} = \text{East distance} - \text{West distance} \] \[ = \frac{188}{12} - \frac{261}{12} = \frac{188 - 261}{12} = \frac{-73}{12} \]
The negative value indicates that Joey walked 73/12 meters to the west past his starting position (home).
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Convert \(\frac{73}{12}\) back to a mixed number:
Divide 73 by 12: \[ 73 \div 12 = 6 \quad \text{(whole number)} \] \[ 73 - (12 \times 6) = 73 - 72 = 1 \quad \text{(remainder)} \] Thus, \[ \frac{73}{12} = 6 \frac{1}{12} \]
Therefore, Joey was \(6 \frac{1}{12}\) meters from home.
Joey was \(6 \frac{1}{12}\) meters from home.
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