A person places $641 in an investment account earning an annual rate of 5.8%, compounded continuously. Using the formula V = Pe^{rt}V=Pe rt , where V is the value of the account in t years, P is the principal initially invested, e is the base of a natural logarithm, and r is the rate of interest, determine the amount of money, to the nearest cent, in the account after 2 years.

r=5.8\%=0.058

r=5.8%=0.058

Move decimal over two places

P=641

P=641

Given as the pricipal

t=2

t=2

Given as the time

V=Pe^{rt}

V=Pe

rt

V=641e^{0.058( 2)}

V=641e

0.058(2)

Plug in

V=641e^{0.116}

V=641e

0.116

Multiply

V=719.8404\approx 719.84

V=719.8404≈719.84 - answer

Use calculator and round to nearest cent

Step-by-step explanation:

555.97

Step-by-step explanation:

Move decimal over two places

r = 5.2% = 0.052

Given as the principal

P = 207

Given as the time

t = 19

V=Pe^rt

Plug in

V = 207e^0.052(19)

Multiply

V = 207e^0.988

Use calculator and round to nearest cent

V = 555.9725 ≈ 555.97

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step-by-step explanation:

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