PLEASE HELP A cylinder has a 21.02 in. diameter. The linear speed of a point on the cylinder's surface is 18.12 ft/sec. What is the angular speed of the cylinder, in revolutions per hour?

11860.22 rev/hour

Step-by-step explanation:

The diameter if a cylinder, d = 21.02 inches

Radius, r = 10.51  inches

or

r = 0.875 feet

The linear speed of a point on the cylinder's surface is 18.12 ft/sec.

We need to find the angular speed of the cylinder.

The relation between the linear speed and the angular speed is given by :

Where

is angular speed

or

So, the required angular speed of the cylinder is 11860.22 rev/hour.

d. 8

step-by-step explanation:

the given radical expression is:

we can rewrite this as:

we express as square numbers to get;

the exponents will now cancel out to give.

the correct answer is d

answer: the correct answer is -32

step-by-step explanation: