The radius of the base of a cylinder is decreasing at a rate of 121212 kilometers per second. The height of the cylinder is fixed at 2.52.52, point, 5 kilometers. At a certain instant, the radius is 404040 kilometers. What is the rate of change of the volume of the cylinder at that instant (in cubic kilometers per second)

7536

Step-by-step explanation:

Given that:

Rate of decreasing of radius = 12 km/sec

Height of cylinder is fixed at = 2.5 km

Radius of cylinder = 40 km

To find:

The rate of change of Volume of the cylinder?

Solution:

First of all, let us have a look at the formula for volume of a cylinder.

Where is the radius and

is the height of cylinder.

As per question statement:

= 40 km (variable)

= 2.5 (constant)

As are constant:

FG = 28 meters

Step-by-step Explanation:

==>A design ABCD and its dilated photocopy EFGH.

Length of side AB = 17m

BC = 14m

DA:HE = 1:2

==>Required:

Length of side FG on the photocopy EFGH, in meters.

==>Solution:

Dilation of design ABCD would, upon transformation would produce a photocopy EFGH that is similar. Thus, ABCD ~ EFGH

Since ABCD is similar to EFGH, it means the ratio of their corresponding sides would be the same, and also have the same scale factor.

Therefore, it means DA:HE = BC:FG

Therefore, let's find length of side FG of the photocopy using the ratio1:2 that is given for DA:HE, which would also apply as the scale factor or ratio for all other corresponding sides.

Thus,

1:2 = BC:FG

1:2 = 14:FG

1/2 = 14/FG

Cross multiply to solve for FG

1×FG = 14×2

FG = 28 meters