Complete question is;
A 10 cm thick grindstone is initially 200 cm in diameter, and it is wearing away at a rate of 50 cm/hr. At what rate is its diameter decreasing?
Answer:
Diameter is decreasing at the rate of 5/(2πr) cm/hr
Step-by-step explanation:
We are told the stone is wearing away at a rate of 50 cm/hr. This means the volume is decreasing. Thus;
dV/dt = -50 cm/hr
Now, a grindstone is in the shape of a cylinder. Thus, volume of grindstone is;
V = πr²h
dV/dr = 2πrh
Now,to find the rate at which the diameter is decreasing, we'll write;
dr/dt = (dV/dt)/(dV/dr)
dr/dt = -50/(2πrh)
We are given;
Diameter = 200 cm
Radius; r = 200/2 = 100 cm
Thickness; h = 10 cm
Thus;
dr/dt = -50/(2π × r × 10)
dr/dt = -5/(2πr) cm/hr
-3/5 × -4/9
= 4/15
The answer is option a. 4/15
Answer:
3.076 × 10¹⁵
Step-by-step explanation:
40C10 × 10!
= 3.076 × 10¹⁵
I believe that the answer to your question in b = 0.
