interest problem

[dimava]

I tried doing this, but I can't seem to make a formula that will work.

The formula for the simple interest earned on an investment is I=prt, where I is the interest earned, p is the principal, r is the interest rate, and t is the time in years. Assume that $5000 is invested at the annual interest rate of 8% and that the interest is added to the principal at the end of each year.

a. Find the amount of interest that will be earned each year for five years. I came up with the formula of I = 5000(.08)(t) but that doesnt work how it should.

b. State the domain and range of the relation. I'm not too sure about what would domain and range would be in this specific problem.

c. Is this relation a function? Why or Why Not? I say yes it is a function, because logically there shouldn't be 2 items that begin with the same number.

Thanks

Dimava

[jemidiah]

a. I = 5000*.08*1;I = 5000*.08*2; ... I = 5000*.08*5
b. Well, domain and range should both be infinity (if pr <> 0, if so, then domain is infinity, range is 0)
c. Yes, it is a function (y = prx, as p & r are constants)

[dimava]

well a. can't be that, because the interest of year 2 is 5000+interest of year1 * 8.08

[jemidiah]

As I understood your question, it was asking for the interest every year for five years (so you'd have five answers, one for each year). If I have misconstrued it, then please correct me.

[dimava]

and that the interest is added to the principal at the end of each year

so year 2 would have the interest of year 1 + the 8% on top of that

[jemidiah]

Oh, of course. How could I miss that? Anyway, if you just want the interest then use this formula: 5000 * (1.08 ^ (year, 1-5)) - 5000. Skip the - 5000 if you just want the total of interest and principal.

(btw, this is a form of the compound interest formula [I=P(1+R)^t; I = Interest, P = Principal, R = Rate, & t = Time])

[dimava]

eh, that doesnt work out exactly, here would be the answers (which i did by applying what they wanted into the calc.)

after 1st year = 5400
after 2nd year = 5832
after 3rd year = 6298.56
after 4th year = 6802.4448
after 5th year = 7346.640384

because the interest is multiplied by the sum of the previous year

the 5000 are only for the 1st year, they dont want the interest for 5 years, of 5000 per year, they want the interest to be added to the previous year (which has interest already)

EDIT: NVM thanks for the formula it works

[jemidiah]

You're welcome

[nishantp]

If you have a TI-83, it's got tons of financial functions built in to it

[dimava]

I have an 83+ where are the financial functions/

[jemidiah]

Hit the 'Apps' button (it should be blue) and hit '1:Finance...'

[alkatran]

Cost = 5000*1.08^t


Beginning cost*Interest(+1 for the main cost)^time

this makes more sense when its 2, and not 1.08, it doubles every year.