To find out how far Joey is from home, we need to calculate the distance he walked away from home and then subtract the distance he walked back towards home.
- Joey walked away from home: \( 21 \frac{3}{4} \) meters
- He walked back towards home: \( 15 \frac{2}{3} \) meters
First, convert these mixed numbers to improper fractions:
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For \( 21 \frac{3}{4} \): \[ 21 \frac{3}{4} = 21 + \frac{3}{4} = \frac{21 \times 4 + 3}{4} = \frac{84 + 3}{4} = \frac{87}{4} \]
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For \( 15 \frac{2}{3} \): \[ 15 \frac{2}{3} = 15 + \frac{2}{3} = \frac{15 \times 3 + 2}{3} = \frac{45 + 2}{3} = \frac{47}{3} \]
Next, we need a common denominator to subtract these two distances. The least common multiple of \( 4 \) and \( 3 \) is \( 12 \).
Convert the fractions to have a denominator of \( 12 \):
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For \( \frac{87}{4} \): \[ \frac{87}{4} \times \frac{3}{3} = \frac{261}{12} \]
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For \( \frac{47}{3} \): \[ \frac{47}{3} \times \frac{4}{4} = \frac{188}{12} \]
Now, we can subtract the distance he walked back from the distance he initially walked away: \[ \frac{261}{12} - \frac{188}{12} = \frac{261 - 188}{12} = \frac{73}{12} \]
Now, let's convert \( \frac{73}{12} \) back to a mixed number:
- Divide \( 73 \) by \( 12 \):
- \( 12 \) goes into \( 73 \) a total of \( 6 \) times (because \( 12 \times 6 = 72 \)).
- The remainder is \( 73 - 72 = 1 \).
Thus, \( \frac{73}{12} = 6 \frac{1}{12} \).
Therefore, Joey is \( 6 \frac{1}{12} \) meters away from home.
Final Answer: 6 1/12
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