Endless Paradigm

ZiNgA BuRgA Wrote:What is the 48,574th digit after the decimal point of PI?

easily done, if you know a way to jump to that particular digit.. (I don't.. thinking how). I have obtained the first 2million digits after the decimal. Now I can either sit and count or find a way to jump...

just for those interested, here are the first 4million digits of Pi

Spoiler:

feinicks Wrote:
Kaiser Wrote:10000 cows get a serious illness. they decrease exponentially as they die at a rate that in 60 weeks, only half (1/2) of the cows remain, how much cows will remain after 70 weeks?
i dont really know the answer, this was from a pre calculus quiz and i didn't knew what to do, som if somebody could explain to me i would be grateful enough to give piggies
by the way, im currently taking exponential/quadratic functions... so it should be using that...
my answer: the total no. of cows alive at end of 70 weeks= 4167 cows

how do you do it?

Kaiser Wrote:
feinicks Wrote:
Kaiser Wrote:10000 cows get a serious illness. they decrease exponentially as they die at a rate that in 60 weeks, only half (1/2) of the cows remain, how much cows will remain after 70 weeks?
i dont really know the answer, this was from a pre calculus quiz and i didn't knew what to do, som if somebody could explain to me i would be grateful enough to give piggies
by the way, im currently taking exponential/quadratic functions... so it should be using that...
my answer: the total no. of cows alive at end of 70 weeks= 4167 cows
how do you do it?

can solve it using quadratic equations..

verify if the answer is correct. if so, then I'll solve it here.

feinicks Wrote:
Kaiser Wrote:
feinicks Wrote:my answer: the total no. of cows alive at end of 70 weeks= 4167 cows
how do you do it?
can solve it using quadratic equations..

verify if the answer is correct. if so, then I'll solve it here.

That's wrong. (Lolwut? quadratic equations... for an exponential decay problem?)

The correct answer is 4454 cows.

Let t=time.
N(t)=the number of cows alive at time t.
r=rate of decay

Wee are given
N(0)=10000. (That is, there are 10000 at the start).
N(60)=5000. (At 60 weeks, there are 5000 cows).

Wee want to find N(70).

Because the number of cows decreases exponentially, it can be modeled as

N(t)= N(0)*(1-r)^t

Spoiler for What the fudge that means:

Now wee are given n(60)=5000.

Therefore N(60)=N(0)*(1-r)^60
=>  5000=10000*(1-r)^60
=>  1/2=(1-r)^60.
=>  (1-r)=0.5^(1/60)

Now substituting (1-r) into the original equation, wee have

N(t)=10000*(0.5^(1/60))^t
=>  N(70)=10000*(0.5^(1/60))^70

Stick that into a calculator/excel/whatever, and you should get

N(70)=4454.493591...

Ok, that's the simple way of doing it (which is perfectly fine by the way).

But if you want to be/feel hardcore/troublesome/annoying/whatever

The more rigorous way is with the differential equation

dN/dt=-λt

Solve that (don't ask me how) to get N(t)=N(0)*e^(-λt)

And then do the substitutions and stuff on that. If you want to do it that way then figure it out yourself. Will get you the same answer in the end.

Ok, slightly better explanation.

(+ shameless bump).

Thanks!!

Haha so that was just to help you!?!>!.

diego Wrote:Haha so that was just to help you!?!>!.

well, it was a mathematical challenge thread and i needed help so...

feinicks

feinicks

feinicks

Kaiser

feinicks

Assassinator

Assassinator

Kaiser

diego

Kaiser