The population P=P(t) at time t of a certain | vedclass

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(A) Given the differential equation: $\frac{dP}{dt} = 0.5 P - 450 = \frac{1}{2}(P - 900)$.Separating the variables,we get: $\frac{2 dP}{P - 900} = dt$

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$2 \log 18$

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(A) Given the differential equation: $\frac{dP}{dt} = 0.5 P - 450 = \frac{1}{2}(P - 900)$.Separating the variables,we get: $\frac{2 dP}{P - 900} = dt$.Integrating both sides: $2 \int \frac{dP}{P - 900} = \int dt \Rightarrow 2 \ln |P - 900| = t + C$.Using the initial condition $P(0) = 850$: $2 \ln |850 - 900| = 0 + C \Rightarrow C = 2 \ln 50$.Thus,the equation is $2 \ln |P - 900| = t + 2 \ln 50$.To find the time $t$ when the population $P$ becomes $0$:$2 \ln |0 - 900| = t + 2 \ln 50$$2 \ln 900 = t + 2 \ln 50$$t = 2 \ln 900 - 2 \ln 50 = 2 \ln \left( \frac{900}{50} \right) = 2 \ln 18$.

The population $P=P(t)$ at time $t$ of a certain species follows the differential equation $\frac{dP}{dt}=0.5 P-450$. If $P(0)=850$,then the time at which the population becomes zero is

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