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redsakura
26.06.2019 •
Mathematics
Water is being pumped into a 12-foot-tall cylindrical tank at a constant rate. • the depth of the water is increasing linearly. • at 2: 30 p.m., the water depth was 2.6 feet. • it is now 5: 00 p.m., and the depth of the water is 3.6 feet. what will the depth (in feet) of the water be at 6: 00 p.m.?
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Ответ:
4 feet
Step-by-step explanation:
It is given that:
• At 2:30 p.m., the water depth was 2.6 feet.
• It is now 5:00 p.m., and the depth of the water is 3.6 feet.
So in 2.5 hours ( 5pm - 2:30pm) the water rose 1 feet (3.6 - 2.6).
Now we can find how much water rises in 1 hour by setting up a ratio (let x be the depth increase of water in 1 hr):
So, in 1 hour, the water level will rise 0.4 feet
So, at 6pm (1 hour from 5 pm) it will rise to 3.6 + 0.4 = 4 feet
Ответ:
X = 2/3