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Mathematics 17 Online
OpenStudy (anonymous):

how do you find the recursive formula for 200.00, 216.00, 233.28, 251.94? please explain

OpenStudy (amistre64):

it looks moenyical

OpenStudy (amistre64):

moneyical

OpenStudy (amistre64):

maybe its interest?

OpenStudy (amistre64):

looks to be compounded

OpenStudy (anonymous):

.....what? i dunno if this helps but i think its supposed to be a shifted sequence...

OpenStudy (amistre64):

251.94 = 200(1+r)^4 A/P = (1+r)^t trt(A/P)-1 = r id try for kicks: (251.94/200)^(1/4) -1 = r or about 5.94% maybe?

OpenStudy (anonymous):

...so is it just geometric?

OpenStudy (amistre64):

i think its almost geometric; i still think its some form of compunded interest

OpenStudy (anonymous):

oh...ok....thanks!

OpenStudy (amistre64):

(216/200)-1 = .08 for the first year (233.28/216)-1 = .0799999 for the next one (251.94/233.28)-1 = .07999 for the last set

OpenStudy (amistre64):

A = 200(1.08)^t

OpenStudy (amistre64):

just about ....

OpenStudy (anonymous):

ok! thanks!!!! haha....

OpenStudy (amistre64):

i must be tired, that works fine

OpenStudy (anonymous):

haha im tired too....*sigh* just got back home and now i have to finish like all my homework

OpenStudy (amistre64):

good luck with it :)

OpenStudy (anonymous):

thanks again!

OpenStudy (phi):

to make a recursion you world write f(0)= 200 f(n+1)= f(n)*1.08

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