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 yearsA=amount he will owe in t yearsP=principal (initial amount)r=raten=number of times the interest is compounded per year t.We are given:P=2500r=12%=.12n=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]----------------------------------------------------2nd Problem:Compounded continuously problems use base as e.[tex]A=Pe^{rt}[/tex]P is still the principalr is still the ratet is still the number of yearsA is still the amount.You are given:P=2500r=12%=.12Let's plug that information in:[tex]A=2500e^{.12t}[/tex].