A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. Required:a. How much sugar is in the tank at the begining?b. Find the amount of sugar after t minutes. c. As t becomes large, what value is y(t) approaching ?

(a) Let denote the amount of sugar in the tank at time . The tank starts with only pure water, so .

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(A(t)/1080 kg/L) * (7 L/min) = 7A(t)/1080 kg/min

so that the net rate of change of is governed by the ODE,

or

Multiply both sides by the integrating factor to condense the left side into the derivative of a product:

Integrate both sides:

Solve for :

Given that , we find

so that the amount of sugar at any time is

(c) As , the exponential term converges to 0 and we're left with

or 75.6 kg of sugar.

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

step-by-step explanation:

why? because i noticed a pattern. 60 120. those numbers without the zero, make the 6 times tables leaving your answer to be b since 6 times 4 is 24 and adding a zero making it 240.