Salt flows into the tank at a rate of
(1/2 lb/gal) * (6 gal/min) = 3 lb/min
and flows out at a rate of
(Q(t)/60 lb/gal) * (6 gal/min) = 6Q(t) lb/min
The net rate of change of the amount of salt in the tank at time
is then governed by

Solve for
:


![\dfrac{\mathrm d}{\mathrm dt}[e^{6t}Q]=3e^{6t}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Be%5E%7B6t%7DQ%5D%3D3e%5E%7B6t%7D)


The tank starts with 10 lb of salt, so that Q(0) = 10. This gives us

so that the amount of salt in the tank at time
is given by

Answer:
I believe you multiply 18 and 6 so most likely your answer is 108 because 18 x 6 is 108
Answer:
There are three intersections on the graph, therefore there are three solutions to the system
Step-by-step explanation:
The points are A is (1,1) and B (3,3)
With that 2/2
in it simplified form it is = 1
So your slope is 2/2 or 1
Hope this helps
Answer:x=1, y=1
Step-by-step explanation:
2(1)-1=1
3(1)1-1=2