Apond contains 2760 l of pure water and an uknown amount of an undesirable chemical. water contaninig 0.01 kg of this chemical per liter flows into the pond at a rate of 4 l/h. the mixture flows out at the same rate, so the amount of water in the pond remains constant. assume that the chemical is uniformly distributed throughout the pond.
let q(t) be the amount of chemical (in kg) in the pond at time t hours.
(a) write a differential equation for the amount of chemical in the pond? at any time time (enter q for q(t))
(b) how much chemical will be in the pond after a long time?

Let be the unknown amount of the chemical originally in the pond, so .

a. The incoming water introduces the chemical at a rate of

and the mixture flows out at a rate of

so that the net rate of change (in kg/hr) of the chemical in the pond is given by the differential equation,

b. The ODE is linear; multiplying both sides by gives

Condense the left side into the derivative of a product:

Integrate both sides to get

and solve for to get

The pond starts with kg of the chemical, so when we have

so that the amount of chemical in the water at time is

As , the exponential term will converge to 0, leaving a fixed amount of 276 kg of the chemical in the pond.

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.

its 60

4 x 12 = 48

3 x 4 = 12

48 + 12 = 60

yeeeet