To find the water level of the river at the end of summer, we need to add the two negative elevations:
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Initial Water Level: \( 1 \frac{1}{2} \) meters below sea level
- Converting this to an improper fraction: \[ 1 \frac{1}{2} = \frac{3}{2} \text{ meters below sea level} \]
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Summer Drop: \( 2 \frac{1}{3} \) meters below sea level
- Converting this to an improper fraction: \[ 2 \frac{1}{3} = \frac{7}{3} \text{ meters below sea level} \]
Next, we need to add these two values together. Since both are below sea level, we will combine them, keeping in mind that they are both negative values.
We need a common denominator to perform the addition. The least common multiple of 2 and 3 is 6.
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Convert both fractions to have a denominator of 6:
- For \(\frac{3}{2}\): \[ \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \]
- For \(\frac{7}{3}\): \[ \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \]
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Now, we can add the fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{9 + 14}{6} = \frac{23}{6} \]
This result, \(\frac{23}{6}\) meters, means that at the end of summer, the river is \( \frac{23}{6} \) meters below sea level.
- Converting \(\frac{23}{6}\) to a mixed number: \[ 23 \div 6 = 3 \text{ remainder } 5 \] Therefore, \[ \frac{23}{6} = 3 \frac{5}{6} \text{ meters below sea level} \]
Thus, the water level at the end of summer is \( 3 \frac{5}{6} \) meters below sea level.