Question

Please help me. This is real confusing. ​

Answer 1

Answer:

1st problem: b) $A=2500(1.01)^{12t}$

2nd problem:  c) $A=2500e^{.12t}$

Step-by-step explanation:

1st problem:

The formula/equation you want to use is:

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

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)

$A=2500(1+\frac{.12}{12})^{12t}$

Time to clean up the inside of the ( ).

$A=2500(1+.01)^{12t}$

$A=2500(1.01)^{12t}$

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2nd Problem:

Compounded continuously problems use base as e.

$A=Pe^{rt}$

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:

$A=2500e^{.12t}$.