1. WATER At 2 P.M., the level of the water in the pool was 10 feet. At 6 P.M., the level of water was 2 feet. Find the constant rate of change of the water.
r = (2-10)/4 = -8/4 = -2Ft/h.
The water decreased at a rate of 2 ft. per hour.
92-1
In 4 hours the water decreased 8 feet.
How much did it decrease per hour?
To find the constant rate of change of the water level, we need to calculate the change in water level and divide it by the change in time.
Change in water level = Final water level - Initial water level
= 2 feet - 10 feet
= -8 feet (since the water level decreased)
Change in time = 6 PM - 2 PM
= 4 hours
Constant rate of change of water = Change in water level / Change in time
= -8 feet / 4 hours
= -2 feet per hour
Therefore, the constant rate of change of the water is -2 feet per hour.
To find the constant rate of change of the water, we need to determine the change in the water level and divide it by the change in time.
Given that the water level was 10 feet at 2 P.M. and decreased to 2 feet at 6 P.M., the change in the water level is:
10 feet - 2 feet = 8 feet
The change in time is 4 hours (from 2 P.M. to 6 P.M.).
To find the constant rate of change, we divide the change in the water level by the change in time:
8 feet / 4 hours = 2 feet per hour
Therefore, the constant rate of change of the water is 2 feet per hour.