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differentiated would be p(t) = .5^t/30 ?

you get
\[30k=\ln(.5)\] or
\[k=\frac{\ln(.5)}{30}\]

ok i believe i have confused youi
lets go slow

allright i guess I cant solve it, maybe they jst want the formulas.

that is weird

then you get \(k=\frac{\ln(\frac{1}{2})}{30}\) and leave it at taht

solve via
\[30k=\ln(\frac{1}{2})\]
\[k=\frac{\ln(\frac{1}{2})}{30}\]

which is very strange to me, but if you say so then i believe you

to answer the next question
replact \(t\) by \(75\), that is, compute
\[p(75)\]

yeaa i go to city college in new york and they arent allowing any calcs for calc or precal

hold on

they are not asking for the derivative, they are asking for the "differential equation"

the answer to the first question is
\[\frac{dP}{dt}=kP\]

this is the same as saying the instantaneous rate of change is proportional to the current amount

the solution to
\[\frac{dP}{dt}=-kP\] is
\[P(t)=P_0e^{-kt}\] which is what we solved above

i guess i should have said in your case \[\frac{dP}{dt}=-kP\] and \[P(0)=200\]

we cannot do 3 or 4 without a calculator, but i can show you how to do them if you like

wouldnt a be kt ? from the original decay equation?

ok nvm i got it ! thanks :)

yw