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# A bacteria population starts with 400 bacteria and grows at a rate of $r(t) = (450.268)e^{1.12567t}$ bacteria per hour. How many bacteria will there be after three hours?

## 11713 bacteria

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because you want to find the number of bacteria present after three hours If there are 400 bacteria initially. And the population growth at the rate defined by this function. Now let p. Be the population of bacteria in T. Hours. Then in general p. Is just the integral of R. F. T. D. T. Which is just the integral of Dysfunction. .268 times. He raised to 1.1256 70 B. T. Which gives us 450.268 times. The integral of the race to 1.1256 70. DT. Not that the integral of E. You raced to A. T. D. T. This is just the race to 80 over A. And so using this formula we have 450.268 times. He raised to 1.12567 T. All over 1.12567. And then class C. Now to find the sea notes that P. of zero is 400 and so we have 400. This is equal to 450.268 times the race too. 1.12567 time zero over 1.12567 plus C. And this is just 400. That's equal to 4 50.268 Divided by 1.12567 plus C. Or that C. Is 400 minus 450.268 over one point 12567. Now this is approximately or rather equal to 400 -400. That's just zero. And so the population of bacteria and T. Ours is just pft. This is 450.268 times E. raised to 1.1256 70 Over 1.12567 plus C. Or this is just 400 times the race too. 1.12567 T. And then plus zero. Or This is just 400 Times E. raised to 1.12567 T. and so in three hours we have P. of three. This is just 400 Times erase to one 12567 Times three. This is approximately 11,713 bacteria. And so this is the number of bacteria present in three hours.

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