The water level of a river is below normal levels it's water level is at 1 and 1/2 m below sea level during the summer the river drops another two and a third meters below sea level what is the water level at the end of summer remember the elevation is below sea level account for this in your answer

Answer 1

To calculate the water level at the end of summer, we start from the initial water level and then account for the additional drop during the summer.

1. The initial water level is \(1 \frac{1}{2}\) meters below sea level. This can be converted to an improper fraction or decimal. \(1 \frac{1}{2} = \frac{3}{2}\) meters (which is 1.5 meters).

2. During the summer, the water level drops an additional \(2 \frac{1}{3}\) meters. First, convert \(2 \frac{1}{3}\) to an improper fraction:

   • \(2 \frac{1}{3} = \frac{7}{3}\) meters.

3. Now, combine the two drops in water level:

   • The total drop in water level = \( \frac{3}{2} + \frac{7}{3} \).

To add these fractions, find a common denominator. The least common multiple of 2 and 3 is 6.

Convert both fractions to have a denominator of 6:

• \( \frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6} \)
• \( \frac{7}{3} = \frac{7 \times 2}{3 \times 2} = \frac{14}{6} \)

Now, add the two fractions: \[ \frac{9}{6} + \frac{14}{6} = \frac{23}{6} \]

This is the total drop below sea level. Since the level is below sea level, we keep the value negative:

• The final water level after summer is \(-\frac{23}{6}\) meters below sea level.

To convert \(-\frac{23}{6}\) to a mixed number: \[ -\frac{23}{6} = -3 \frac{5}{6} \]

Thus, the final water level at the end of summer is 3 and \(\frac{5}{6}\) meters below sea level.