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anonymous
 5 years ago
How do I show that y = Ce^kt is y=800(9)^t/4 given:
The initial size of a bacterial culture is 800
After 4h there are 7200 bacteria
anonymous
 5 years ago
How do I show that y = Ce^kt is y=800(9)^t/4 given: The initial size of a bacterial culture is 800 After 4h there are 7200 bacteria

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0plug in t = 0, that will give you y = 800 (9^0) = 800*1 =800. then after 4 hours, so t=4 , plug in that into the original equation. so y = 800 (9^(4/4))= 800 * 9^1 = 800*9 = 7200

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0t=0 is the initial size, at the beginning

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Where does the 9 come from? I'm not sure if you got the right meaning of the question. The exact question is: The initial size of a bacterial culture is 800. After 4h there are 7200 bactera. By starting with y=Ce^kt, show that the simplified equation is y=800(9)^(t/4). I think what it is saying is put the information it gives you into Ce^kt to get y=800(9)^t/4. Thanks for replying, I appreciate it :)
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