Atank initially holds 200 gallons of fresh water (no salt). a brine solution containing 2 pounds of salt per gallon is poured in at a rate of 3 gallons per minute. the tank is kept well-mixed and drained at the rate of 3 gallons per minute. write and solve an initial value differential equation to model the amount of salt, a, in the tank at time, t. 2.

1.99 pounds per gallon of salt in t=2 in the tank.

Step-by-step explanation:

First we consider the matter balance equation that contemplates the input and output; the generation and consumption equal to the acomulation.  

In this case we have no Generation neither do Consumption so, if we consider Acomulation = A(t), the rate of change of A(t) in time is given by:

---(1)

where C is th concentration, with the initial value statement that A(t=0) = 0 because there is no salt in the time cero in the tank, only water.

Given the integral factor -> and multipying the entire (1) by it, we have:

Solving this integrals we obtain:

So given the initial value condition A(t=0)=0 we have:

,

and the solution is,

.

If we give the actual values we obtain then,

.

So in t= 2 we have .

hey seth! : )

i’m not 100% sure, but i think it’s b

hope this ya : p

miss u!

- nicole

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.