An investment of $4000 grows at the rate of 320e^(0.08t) dollars per year after t years.?

Question

An investment of $4000 grows at the rate of 320e^(0.08t) dollars per year after t years. Its value after 10 years is approximately

(A) $4902 (B) $8902 (C) $7122 (D) $12,902 (E) None of these

Answer 1

the growth per year after 10 years =

320 e^(0.08*10) = 320e^(0.8) = 712.17

The growth for 10 years = 712.17 * (10) = 7121.7 = $7122

Answer 2

i've got faith the equation is: A = P (a million + (r/n))^nt the place... A is the quantity of money in the financial enterprise account P is the quantity of money you began with r = annual pastime fee (as a decimal) n = style of cases pastime is calculated monthly t = style of years

Answer 3

$7122