Answer:
y≥-4 or (-4,∞)
Step-by-step explanation:
<span>B. {-5,-4,-3…. }
............</span>
Answer:
D
Step-by-step explanation:
they will sue for the rest of the money, because they are entitled to it if you are the cause of an accident.
Let
= amount of salt (in pounds) in the tank at time
(in minutes). Then
.
Salt flows in at a rate

and flows out at a rate

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.
Then the net rate of salt flow is given by the differential equation

which I'll solve with the integrating factor method.



Integrate both sides. By the fundamental theorem of calculus,





After 1 hour = 60 minutes, the tank will contain

pounds of salt.
Answer:
30
Step-by-step explanation:
If CUB is 78 and that’s the full angle just to 48 minus 78 and you get your answer.