Step-by-step explanation:
Let's say R is the initial radius of the sphere, and r is the radius at time t.
The volume of the sphere at time t is:
V = 4/3 π r³
Taking derivative with respect to radius:
dV/dr = 4π r²
This is a maximum when r is a maximum, which is when r = R.
(dV/dr)max = 4π R²
This is 4 times the sphere's initial great circle area, but not the great circle circumference. The problem statement contains an error.
Answer: 12x-3 and 2x+4
Step-by-step explanation: i don’t know if i wanted to add them together. you j distribute that’s how i got my answer.
Step-by-step explanation:
You have found a function r(V(t)). We can see that this function is a one variable function. The variable is time.
So in this specific function we can call r(v(t)), r(t).
So:
![r(t) = \sqrt[3]{ \frac{3 \times (10 + 20t)}{4\pi} }](https://tex.z-dn.net/?f=r%28t%29%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020t%29%7D%7B4%5Cpi%7D%20%7D%20)
If α is the moment that the radius is 10 inches and since the function above gives radius in inches we have to solve the equation:
Which is the same as:
![\sqrt[3]{ \frac{3 \times (10 + 20 \alpha )}{4\pi} } = 10 \\ \frac{3 \times (10 + 20 \alpha )}{4\pi} = 1000 \\ (10 + 20 \alpha ) = \frac{4000\pi}{3} \\ 20 \alpha = \frac{(4000\pi - 30)}{3} \\ \alpha = \frac{(4000\pi - 30)}{60}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020%20%5Calpha%20%29%7D%7B4%5Cpi%7D%20%7D%20%20%3D%2010%20%5C%5C%20%20%5Cfrac%7B3%20%5Ctimes%20%2810%20%2B%2020%20%5Calpha%20%29%7D%7B4%5Cpi%7D%20%20%3D%201000%20%5C%5C%20%2810%20%2B%2020%20%5Calpha%20%29%20%3D%20%20%5Cfrac%7B4000%5Cpi%7D%7B3%7D%20%20%5C%5C%2020%20%5Calpha%20%20%3D%20%20%5Cfrac%7B%284000%5Cpi%20-%2030%29%7D%7B3%7D%20%5C%5C%20%20%5Calpha%20%20%3D%20%20%5Cfrac%7B%284000%5Cpi%20-%2030%29%7D%7B60%7D%20)
Answer:
the answer is D
Step-by-step explanation:
