## Algebra and Trigonometry 10th Edition

For $n=1$: $A=1828.49$ For $n=2$: $A=1830.29$ For $n=4$: $A=1831.19$ For $n=12$: $A=1831.80$ For $n=365$: $A=1832.09$ For continuous compounding: $A=1832.10$
$A=P(1+\frac{r}{n})^{nt}$ (n compounds per year) $A=Pe^{rt}$ (continuous compounding) $P=1500,~~n=1,~~t=10,~~r=0.02=2$%: $A=1500(1+\frac{0.02}{1})^{1\times10}=1500(1.02)^{10}=1828.49$ $P=1500,~~n=2,~~t=10,~~r=0.02=2$%: $A=1500(1+\frac{0.02}{2})^{2\times10}=1500(1.01)^{20}=1830.29$ $P=1500,~~n=4,~~t=10,~~r=0.02=2$%: $A=1500(1+\frac{0.02}{4})^{4\times10}=1500(1.005)^{40}=1831.19$ $P=1500,~~n=12,~~t=10,~~r=0.02=2$%: $A=1500(1+\frac{0.02}{12})^{12\times10}=1500(1+\frac{0.01}{6})^{120}=1831.80$ $P=1500,~~n=365,~~t=10,~~r=0.02=2$%: $A=1500(1+\frac{0.02}{365})^{365\times10}=1500(1+\frac{0.02}{365})^{3650}=1832.09$ $P=1500,~~n=continuous,~~t=10,~~r=0.02=2$%: $A=1500e^{0.02\times10}=1832.10$