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
Plot a point at (0,-1)
Then from that point, go down 3 points, and to the right one point. Repeat that until you hit the bottom of the graph or run out of room. From the first point (0,-1), go up 3 and to the left 1 until you run out of room on the graph. Should be a straight line
Answer:
Step-by-step explanation:
the shape is a rhombus the are is 24
Everything that is not 4/9
