MATH SOLVE

2 months ago

Q:
# 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.?

Accepted Solution

A:

Answer:4 feetStep-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):[tex]\frac{2.5}{1}=\frac{1}{x}\\2.5x=1*1\\2.5x=1\\x=\frac{1}{2.5}\\x=0.4[/tex]So, in 1 hour, the water level will rise 0.4 feetSo, at 6pm (1 hour from 5 pm) it will rise to 3.6 + 0.4 = 4 feet