Question

The number of bacteria in a culture is growing at a rate of 3000e^2/5of time at t=0 the number of present bacteria 7500 find the number present at the t= 5

Answer 1

Answer:

Best Answer

== Find the formula for b ==

db/dt = 3000 e^(2/5 t)

db = 3000 e^(2/5 t) dt

b = ∫(3000 e^(2/5 t) dt)

b = (5/2) 3000 e^(2/5 t) + C

b = 7500 e^(2/5 t) + C

== Solve for C ==

b = 7500 e^(2/5 t) + C

7500 = 7500 e^(2/5 (0)) + C

7500 = 7500 + C

C = 0

== Fill C into the formula for b ==

b = 7500 e^(2/5 t) + C

b = 7500 e^(2/5 t) + (0)

b = 7500 e^(2/5 t)

== Find b at t = 5 ==

b = 7500 e^(2/5 t)

b = 7500 e^(2/5 (5))

b = 7500 e^2

b ≈ 55417.9207