Answer:
The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.
Step-by-step explanation:
The area of a sphere is given by the following formula:

In which A is the area, measured in cm², and r is the radius, measured in cm.
Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.
This means that 
Determine the rate of change in surface area when r = 20 cm.
This is
when
. So

Applying implicit differentiation.
We have two variables, A and r, so:



The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.
Answer: i think 0.8.
Step-by-step explanation:
Answer:
D. 3/2
Step-by-step explanation:
The scale factor is 1/2 from OP / NQ = 1/2. Because of that, MQ is equal to half of MP and QP is equal to half of MP. Therefore, QP = MQ = 2. The slope of MO is rise over run which is 6 / 4 or 3 / 2.
Answer:
no solution
Step-by-step explanation:
Given the 2 equations
y =
x + 4 → (1)
-
x + y = - 5 → (2)
Substitute y =
x + 4 into (2)
-
x +
x + 4 = - 5 , that is
4 = - 5 ← not possible
This indicates the system has no solution