suppose that the length x (in centimeters) of an individual certain species of fish is given by x=50-40e^-0.05t, where t is its age in months.
a. Find its length after one year. (2 decimal places)
b. how long to the nearest whole month will it be the length of 40 cm?
a. x=50-40e^-0.05(12) = 28.05 cm
b. 40 = 50-40e^-0.05t
40e^-0.05t = 50 - 40 = 10
e^-0.05t = 10/40 = 0.25
-0.05t = ln(0.25)
-0.05t = -1.3863
t = -1.3863/-0.05 = 28 months
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