An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 3 cm, and the height is 10 cm. Find the rate at which the water level is rising when the water level is 3 cm.

When the water level is 3 cm, the rate at which the water level is rising is 30.86 cm/s.

Step-by-step explanation:

We first have to relate the height H of the pyramid with the volume V.

The volume of the pyramid is V=Sh/3, being S: area of the base square, and h: height of the pyramid.

The base is changing as the pyramid is filled with water.

Then, the volume for every height is:

The side of the square will grow linearly with the height. They start at H=0 and L=0 at the bottom of the pyramid, and they end at H=10 and L=3.

So we have:

If we replace in the volume equation we have:

We know that the rate of variation of the volume in time is contant and equal to 25 cm3:

We can derive the volume equation to calculate the variation of h in time:

The rate of variation of the height in time is dh/dt=278h^(-2). The units are cm/s.

When the water level is 3 cm, the rate at which the water level is rising is:

whatare you sayig

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180