P(t) = 9,000e.119t
P'(t) = .119·9,000e.119t
P'(10) = 1,071e1.19
Sufyan M.
asked 05/03/21An investment of $9,000 which earns 11.9% per year is growing continuously
How fast will it be growing at year 10?
_______________$/year (nearest $1/year)
P(t) = 9,000e.119t
P'(t) = .119·9,000e.119t
P'(10) = 1,071e1.19
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