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Mathematics 15 Online
OpenStudy (anonymous):

Word problem, please help! You invested $52,000.00 at 6% for 5 years. What is your total return on this investment for the following compounding: i) Annually ii) Quarterly iii) Semiannual iv) Continuously v) When will your investment double? What do those ^ mean & how do I figure it out?? Also this came with it: A=P(1+r/n)^(nt) A=Pe^(rt)

OpenStudy (anonymous):

\[A=P(1+\frac{r}{n})^{nt}\]

OpenStudy (anonymous):

\[A=Pe ^{rt}\]

OpenStudy (anonymous):

1) once per year 2) 4 times per years 3) twice a year 4) I = Pe^(rt)

OpenStudy (anonymous):

@sourwing thank you so much! what numbers would I plug into the equation and into which of the 2 equations??

OpenStudy (anonymous):

n = number of times per year

OpenStudy (anonymous):

so sorry to bother you but i just want to make sure i do the correct work... would #2 look like this: A=52000(1+6/4)^(6*5)?

OpenStudy (anonymous):

#3. would look like: A=52000(1+6/2)^(2*5) how do I know what the I and E are in the equation I=Pe^(rt) & how would I know when my investment doubles? @sourwing

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

woohoo! :D so my last question is how do I know what the I and E are in the equation I=Pe^(rt) & how would I know when my investment doubles?

OpenStudy (anonymous):

@sourwing for #2 and #3 would R=6 or .06 when I plug it into the equation?

OpenStudy (anonymous):

set I = 2P

OpenStudy (anonymous):

thank you @sourwing but can you tell me what E in the equation A=Pe^(rt) is??

OpenStudy (anonymous):

& would 6% in the equation be 6 or .06?

OpenStudy (anonymous):

2P = P e^(rt), e is just a number. Just like pi

OpenStudy (anonymous):

6% = 0.06

OpenStudy (anonymous):

so to figure out when the investment doubles i just do i=2(52000) which is 104000

OpenStudy (anonymous):

and continuously would be A=52000e^30?

OpenStudy (anonymous):

@sourwing I promise LAST question for you, but I know you said that e is just a number like pi, but what number is it?

OpenStudy (anonymous):

nevermind, I figured out that e is 2.71...

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