155k views
3 votes
Water is being pumped into a 12-foot-tall cylindrical tank at a constant rate. • The depth of the water is increasing linearly. • At 1:30 p.M., the water depth was 2.6 feet. • It is now 3:30 p.M., and the depth of the water is 3.8 feet. What will the depth of the water be at 4:00 p.M?

1 Answer

4 votes

Answer:

The depth will be 4.1 ft

Explanation:

Here, we have a linear model to work with

Between 1:30 pm and 3:30 pm, there was a difference of 3.8-2.6 = 1.2 ft on the depth

What we can see there is that in 2 hours, there was a difference of 1.2 ft

So the filling rate per hour will be 1.2ft/2 hours = 0.6 ft/hr

Now, we want to know the depth of the water at 4:00 pm

The difference between 3:30 and 4 pm is 30 minutes = 0.5 h

so by addition;

3.8 ft + 0.5(0.6)

will give the depth of the water at 4 pm

That will be;

3.8 + 0.3 = 4.1 ft

We used 3.8 ft as it is the depth at 3:30 pm

Then we need to add what is topping this after 30 minutes of filling using the constant filling rate and the time

User Sirish Renukumar
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories