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3 years ago

What is the inverse of the following conditional statement?

Mathematics

1 answer:

anygoal [31]3 years ago

6 0

"If we are the state champions then we won the game" or C is your answer

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Assume that in the absence of immigration and emigration, the growth of a country's population P(t) satisfies dP/dt = kP for som

Morgarella [4.7K]

Answer:

A) Differential equation for population growth in case of individual immigration is:

$\frac{dP}{dt}=kP+r$

B) Differential equation for population growth in case of individual emigration is:

$\frac{dP}{dt}=kP-r$

Step-by-step explanation:

Population growth rate in the absence of immigration and emigration is given as:

$\frac{dP}{dt}=kP--(1)$

A) When individuals are allowed to immigrate:

Let r be the constant rate of individual immigration given that r >0.

Differential equation for population growth in this case is:

$\frac{dP}{dt}=kP+r$

B) In case of individual emigration:

Let r be the constant rate of individual emigration given that r >0.

Differential equation for population growth in this case is:

$\frac{dP}{dt}=kP-r$

8 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$