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.?

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

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²

fwms@lil_ben06

step-by-step explanation: