Answer:
Yes
Step-by-step explanation:
A rhombus is a parallelogram with 4 congruent sides.
Answer:
hang on give me a sec to edit this
it is.
Answer:
Population of mosquitoes in the area at any time t is:
![P(t) =504,943.26 -104,943.26e^{0.693t}](https://tex.z-dn.net/?f=P%28t%29%20%3D504%2C943.26%20%20-104%2C943.26e%5E%7B0.693t%7D)
Step-by-step explanation:
assume population at any time t = P(t)
population increases at a rate proportional to the current population:
⇒dP/dt ∝ P
----(1)
where k is constant rate at which population is doubled
solving (1)
![ln|P(t)|=kt +C\\P(t)= e^{kt+C}\\P(t)=Ce^{kt}](https://tex.z-dn.net/?f=ln%7CP%28t%29%7C%3Dkt%20%2BC%5C%5CP%28t%29%3D%20e%5E%7Bkt%2BC%7D%5C%5CP%28t%29%3DCe%5E%7Bkt%7D)
---- (2)
initial population = 400,000
population is doubled every week
⇒P(1)=2P(0)
Using (2)
![P_{o}e^{k(1)} = 2P_{o} e^{k(0)}\\](https://tex.z-dn.net/?f=P_%7Bo%7De%5E%7Bk%281%29%7D%20%3D%202P_%7Bo%7D%20e%5E%7Bk%280%29%7D%5C%5C)
![e^{k} =2\\k=ln|2|\\](https://tex.z-dn.net/?f=e%5E%7Bk%7D%20%3D2%5C%5Ck%3Dln%7C2%7C%5C%5C)
In presence of predators amount is decreased by 50,000 per day
Then amount decreased per week = 350,000
In this case (1) becomes
---(3)
solving (3) by calculating integrating factor
![I.F=e^{\int-k dt}](https://tex.z-dn.net/?f=I.F%3De%5E%7B%5Cint-k%20dt%7D)
Multiplying I.F with all terms of (3)
![e^{-kt}\frac{dP}{dt} - ke^{-kt}P =-350,000 e^{-kt}\\\frac{d}{dt}(e^{-kt}P) = -350,000 e^{-kt}](https://tex.z-dn.net/?f=e%5E%7B-kt%7D%5Cfrac%7BdP%7D%7Bdt%7D%20-%20ke%5E%7B-kt%7DP%20%3D-350%2C000%20e%5E%7B-kt%7D%5C%5C%5Cfrac%7Bd%7D%7Bdt%7D%28e%5E%7B-kt%7DP%29%20%3D%20%20-350%2C000%20e%5E%7B-kt%7D)
Integrating w.r.to t
![e^{-kt}P(t)= \frac{350,000e^{-kt}}{k} +C](https://tex.z-dn.net/?f=e%5E%7B-kt%7DP%28t%29%3D%20%5Cfrac%7B350%2C000e%5E%7B-kt%7D%7D%7Bk%7D%20%2BC)
![P(t) =\frac{350,000}{k} +Ce^{kt}\\](https://tex.z-dn.net/?f=P%28t%29%20%3D%5Cfrac%7B350%2C000%7D%7Bk%7D%20%2BCe%5E%7Bkt%7D%5C%5C)
![k=ln|2| =0.693](https://tex.z-dn.net/?f=k%3Dln%7C2%7C%20%3D0.693)
![P(t) =504,943.26 + Ce^{0.693t}\\](https://tex.z-dn.net/?f=P%28t%29%20%3D504%2C943.26%20%2B%20Ce%5E%7B0.693t%7D%5C%5C)
at t=0
![P(0) =504,943.26 + Ce^{0.693(0)}](https://tex.z-dn.net/?f=P%280%29%20%3D504%2C943.26%20%2B%20Ce%5E%7B0.693%280%29%7D)
![400,000 =504,943.26 + C](https://tex.z-dn.net/?f=400%2C000%20%3D504%2C943.26%20%2B%20C)
![C = -104,943.26](https://tex.z-dn.net/?f=C%20%3D%20-104%2C943.26%20)
So, population of mosquitoes in the area at any time t is
![P(t) =504,943.26 -104,943.26e^{0.693t}](https://tex.z-dn.net/?f=P%28t%29%20%3D504%2C943.26%20%20-104%2C943.26e%5E%7B0.693t%7D)