Answer:
1st problem: b) [tex]A=2500(1.01)^{12t}[/tex]
2nd problem: c) [tex]A=2500e^{.12t}[/tex]
Step-by-step explanation:
1st problem:
The formula/equation you want to use is:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
t=number of years
A=amount he will owe in t years
P=principal (initial amount)
r=rate
n=number of times the interest is compounded per year t.
We are given:
P=2500
r=12%=.12
n=12 (since there are 12 months in a year and the interest is being compounded per month)
[tex]A=2500(1+\frac{.12}{12})^{12t}[/tex]
Time to clean up the inside of the ( ).
[tex]A=2500(1+.01)^{12t}[/tex]
[tex]A=2500(1.01)^{12t}[/tex]
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2nd Problem:
Compounded continuously problems use base as e.
[tex]A=Pe^{rt}[/tex]
P is still the principal
r is still the rate
t is still the number of years
A is still the amount.
You are given:
P=2500
r=12%=.12
Let's plug that information in:
[tex]A=2500e^{.12t}[/tex].
