Cone Volume = (PI * radius^2 * height) / 3
3 = (PI * radius^2* x) / 3
radius^2 = 9 / (PI * x)
radius = square root (9 / (PI * x))
Step-by-step explanation:

Compound interest formula = a=P(1+r/n)^nt
P= lump sum to deposit (solving for)
A= amount accumulated over the entire time (20000)
n= number of times interest is compounded annually (1)
r= rate of interest (0.82)
T= total number of years (15)
20000=P(1+0.082/1)^1*15
20000=P(1.082)^15
20000=P(3.26143638)
20000/3.26143638=P
P=$6132.2674
Looks like your function is

Rewrite it as

Recall that for
, we have

If we replace
with
, we get

By the ratio test, the series converges if

Solving for
gives the interval of convergence,

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.