Answer:
a) 
b) 
Step-by-step explanation:
By definition, we have that the change rate of salt in the tank is
, where
is the rate of salt entering and
is the rate of salt going outside.
Then we have,
, and

So we obtain.
, then
, and using the integrating factor
, therefore
, we get
, after integrating both sides
, therefore
, to find
we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions
, so 

Finally we can write an expression for the amount of salt in the tank at any time t, it is 
b) The tank will overflow due Rin>Rout, at a rate of
, due we have 500 L to overflow
, so we can evualuate the expression of a)
, is the salt concentration when the tank overflows
Answer:
sin70°
Step-by-step explanation:
Given sin110° where 110° is an obtuse angle in the second quadrant.
The related acute angle in the first quadrant is 180° - 110° = 70°
Thus
sin110° = sin70°
I believe it would be the second option....