Atank initially contains 40 ounces of salt mixed in 100 gallons of water. a solution containing 4 oz of salt per gallon is then pumped into the tank at a rate of 5 gal/min. the stirred mixture flows out of the tank at the same rate. how much salt is in the tank after 20 minutes ?

The amount of salt that is in the tank after 20 minutes is;

A(20) = 267.56 Oz

Initial amount of salt in tank; A(0) = 40 Ounces

A solution containing 4 oz salt per gallon is pumped into the tank at a rate of 5 gal/min.

This means rate in oz/min = 4/1 × 5/1 = 20 oz/min

A'(t) = 20 - [(A(t)/(100)) × 5/1]

A'(t) = 20 - A(t)/20

Rearranging gives;

A'(t) + A(t)/20 = 20

Multiplying through by the integrating factor gives;

A'(t) + (A(t)/20) = 20

A(t) = 400  + C

Divide through by to get;

A(t) = 400 + C

At initial condition of A(0) = 40, we have;

40 = 400 + C

C = -360

Thus; at time (t), the amount of salt left is given by;

A(t) = 400 - 360

A(20) = 400 - 360

A(20) = 400 - 132.44

A(20) = 267.56 Oz

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  21.3

step-by-step explanation:

multiplying the given equation by c/sin(25°), we get

  c = 9/sin(25°) ≈ 21.2958 ≈ 21.3

we assume "c" is the length of ab, the side opposite angle c in triangle abc.

hey seth! : )

i’m not 100% sure, but i think it’s b

hope this ya : p

miss u!

- nicole