SOLUTION: A culture of bacteria obeys the law f(t)= f(0)*e^0.47*t. If 500 bacteria are present initially , and there are 800 after 1 hour, how many will be present in the culture after 5 ho
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-> SOLUTION: A culture of bacteria obeys the law f(t)= f(0)*e^0.47*t. If 500 bacteria are present initially , and there are 800 after 1 hour, how many will be present in the culture after 5 ho
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Question 62632: A culture of bacteria obeys the law f(t)= f(0)*e^0.47*t. If 500 bacteria are present initially , and there are 800 after 1 hour, how many will be present in the culture after 5 hours? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A culture of bacteria obeys the law f(t)= f(0)*e^0.47*t.
:
If 500 bacteria are present initially , and there are 800 after 1 hour, how many will be present in the culture after 5 hours?
:
f(5) = 500 * e^(.47*5)
f(5) = 500 * e^2.35; find the value of e^2.35 on a good calc.
f(5) = 500 * 10.4855697
f(5) = 5248 bacteria after 5 hrs
