To find out how far Joey is from home after his walk, we can start by determining the net distance he walked away from home and then the distance he returned towards home.
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Distance walked away from home: \[ 21 \frac{3}{4} \text{ meters} \]
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Distance walked back towards home: \[ 15 \frac{2}{3} \text{ meters} \]
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Convert both mixed numbers to improper fractions:
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For \( 21 \frac{3}{4} \): \[ 21 \frac{3}{4} = \frac{21 \times 4 + 3}{4} = \frac{84 + 3}{4} = \frac{87}{4} \]
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For \( 15 \frac{2}{3} \): \[ 15 \frac{2}{3} = \frac{15 \times 3 + 2}{3} = \frac{45 + 2}{3} = \frac{47}{3} \]
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Calculate the total distance from home: When Joey walked away initially, he was \( \frac{87}{4} \) meters away, and then he walked back towards home \( \frac{47}{3} \) meters. We can express the distance still away from home as: \[ \text{Distance from home} = \text{Distance away} - \text{Distance back} = \frac{87}{4} - \frac{47}{3} \]
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Find a common denominator: The least common multiple of 4 and 3 is 12. We can convert both fractions:
- Convert \( \frac{87}{4} \): \[ \frac{87}{4} = \frac{87 \times 3}{4 \times 3} = \frac{261}{12} \]
- Convert \( \frac{47}{3} \): \[ \frac{47}{3} = \frac{47 \times 4}{3 \times 4} = \frac{188}{12} \]
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Now subtract the two fractions: \[ \frac{261}{12} - \frac{188}{12} = \frac{261 - 188}{12} = \frac{73}{12} \]
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Convert \( \frac{73}{12} \) to a mixed number: Divide 73 by 12:
- \( 73 \div 12 = 6 \) (whole part)
- Remainder: \( 73 - (12 \times 6) = 73 - 72 = 1 \)
Hence, \[ \frac{73}{12} = 6 \frac{1}{12} \]
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Final answer: Joey is \( \boxed{6 \frac{1}{12}} \) meters away from home.