A cup has the shape of a right circular cone. The height of the cup is 12 cm, and the radius of the opening is 3 cm. Water is poured into the cup at a constant rate of 32 cm /sec. What is the rate at which the water level is rising when the depth of water in the cup is 5cm?

  about 6.52 cm/s

Step-by-step explanation:

The radius of the cone is 3/12 = 1/4 of its height, so the radius of the water surface at the time of interest is (5 cm)/4 = 1.25 cm.

The area of the water's surface is ...

  A = πr² = π(5/4 cm)² = 25π/16 cm²

The rate of change of depth multiplied by this area will give the rate of change of volume.

  dV/dt = (25π/16 cm²)(dh/dt)

  dh/dt = (32 cm³/s)/(25π/16 cm²) = 512/(25π) cm/s ≈ 6.52 cm/s

The water is rising at the rate of about 6.52 cm/s.

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.