Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
Step-by-step explanation:
Since we have given that
Height = 9 inches
Diameter = 6 inches
Radius = 3 inches
So, 
Volume of cone is given by

By differentiating with respect to time t, we get that

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie 
and h = 2 inches deep.

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.
Answer:
44
Step-by-step explanation:
This is just another way of asking "what is the volume of a sphere of 6cm"
The volume of a sphere is

The R is 3cm in this problem since the diameter is twice the radius. For a 3cm radius sphere this is 113.1cm^3
Answer:

Step-by-step explanation:
We require 2 equations with the repeating digits (63) placed after the decimal point.
let x = 0.636363..... (1) multiply both sides by 100
100x = 63.6363... (2)
Subtract (1) from (2) thus eliminating the repeating digits
99x = 63 ( divide both sides by 99 )
x =
=
← in simplest form