The surface area of the sphere is:
A = 4 * pi * r ^ 2
Deriving we have:
A '= 8 * pi * r * r'
We clear the speed:
r '= A' / (8 * pi * r)
Substituting values:
r '= 24 / (8 * pi * 3)
r '= 1 / pi inch / sec
Then, the volume of the sphere is:
V = (4/3) * (pi) * (r ^ 3)
Deriving we have:
V '= (3) * (4/3) * (pi) * (r ^ 2) * (r')
Substituting values:
V '= (3) * (4/3) * (pi) * (3 ^ 2) * (1 / pi)
Rewriting:
V '= (4) * (9)
V '= 36 inch3 / sec
Answer:
the volume of the ice is changing at 36 inch3 / sec at that moment
Answer:
The answer I got is 3. I attached above my solutions. I hope that helps :))
(a-b)^2 = a^2-2ab+b^2
(8-5i)^2 = 8^2-2(8)(5i)+(5i)^2
= 64-80i+25i^2
i^2=-1
So
= 64-80i+25(-1)
=64-25-80i
= <em><u>39 - 80i</u></em>
which is your answer :)
Answer is A or your first option.
