Answer:
A. .325L
Step-by-step explanation:
All you have to do is had .125 with .20 to get your answer
Answer:
A
Step-by-step explanation:
Any number that have the | | symbol will automatically be positive
For example
| -5.8 | = 5.8
Answer:
The answer is B.
Step-by-step explanation:
Let us go through each of the points one by one:
The label A represents the radius of the base of the cone.
The label B represents the height of the cone.
The label C represents the origin of the base of the cone.
The label D represents the vertex of the cone (where the cone ends).
So it is choice B that represents the height of the cone.
<em>P.S: it is tempting to pick label D to represent the height, but since label A already points to a line that is the height of the cone, we don't pick Label D.</em>

a. The gradient is


b. The gradient at point P(1, 2) is

c. The derivative of
at P in the direction of
is

It looks like

so that

Then

