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anonymous
 5 years ago
how do i set this up?! a bacteria culture initially contains 100 dells and grows at a rate proportional to its size. after an hour the population has incresed to 420.
a)find an expression for the number of bacteria after t hours.
b)find the number of bacteria after 3 hours
c)find the rate of growth after 3 hours
d)when will the population reach 10,000
anonymous
 5 years ago
how do i set this up?! a bacteria culture initially contains 100 dells and grows at a rate proportional to its size. after an hour the population has incresed to 420. a)find an expression for the number of bacteria after t hours. b)find the number of bacteria after 3 hours c)find the rate of growth after 3 hours d)when will the population reach 10,000

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have trouble converting it to numbers. here what i have so far and i working on more. i use the P(t)=P(0)e^kt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i set the equation P(3)=100e^3k=420 i solved and got k=ln42/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is a [100e^[(ln42/3)(t)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b) i just plug in 3 for t to the equation in a) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for d) i put equation a) equal to 10,000 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im stuck on c) though. how do you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess that you can derive the expression you got before to obtain dP/dt. And then you sustitute the value of 3 on it. .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got it down to 100(42^t/3) i dont how to dervie?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think your initial conditions are wrong. You wrote P(3) = 420, and on the problem says P(1) = 420

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[420 = 100e^{k(1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok crap .yeah your right! good catch

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[P = 100e^{\ln(4.2)t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you derive you get: \[\frac{d}{dt}100e^{(\ln 4.2)k}= 100(\ln4.2)e^{(\ln 4.2)k}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry its t instead of k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to get waht you got for the derviative right now. its not coming out for me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0instead of ln4.2 i get 4.2k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember that \[\frac{d}{dx}e^{ax} = ae^x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry i missed an "a" \[\frac{d}{dx}e^{ax}=ae^{ax}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh. ok. got it now. thanks
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