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:
405/14
Step-by-step explanation:
Answer:
C and D
Step-by-step explanation:
i=√-1
n/4=x.5
since 2 is the remainder
n=2x
then i^n=i^2x
when x is even, i^n =1
when x is odd, i^n=-1
Answer:
80/100 or 80%
Step-by-step explanation:
7/10 = 70/100
70/100 + 10/100 = 80/100