Answer:
The height of the pile is increasing at the rate of 
Step-by-step explanation:
Given that :
Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min
i.e 
we know that radius r is always twice the diameter d
i.e d = 2r
Given that :
the shape of a cone whose base diameter and height are always equal.
then d = h = 2r
h = 2r
r = h/2
The volume of a cone can be given by the formula:




Taking the differentiation of volume V with respect to time t; we have:


we know that:

So;we have:



The height of the pile is increasing at the rate of 
The slope would represent the cost per minute, since m is the length of the call in minutes. Logically, D) Cost per minute, is the only one that would work. The connection cost would be just added in, and you wouldn't multiply the cost of having a phone line by how many minutes you're on the phone. The length of the call is already there, it's m, so that wouldn't work either. Therefore, D, cost per minute, is the logical answer. The slope in the equation represents D, cost per minute.
64° i think because. its an iscoceles triangle so the bottom two angles are the same so you do 58+58=116 and because all angles in a triangle add to 180° you do 180-116=64°
You're answers are a bit jumbled up but here is the answer:
T for Two
B for Blue
M for Moon Roof
F for four
G for Gray
NM for No Moon Roof... so
TBM
TBNM
TGM
TGNM
FBM
FBNM
FGM
FGNM
8 Possibilities
Since 216 is the cube of 6, 64 is the cube of 4 and x^3 is the cube of x, we have

so the expression is the sum of the cubes of 6x and 4