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
Y=-2/5-1 i hope this helps!
Easy there on the caps, partner.
The nine, using the distributive property, distributes itself across the brackets, and so 9 times t is 9t and 9 times -1 is -9.
Therefore, 9(t-1) = 9t -9
Answer: $4.20 for 7
This problem can be solved by dividing the pay by the number of items.
$6.30 / 9 = $0.70
$5.50 / 5 = $1.10
$4.20 / 7 = $0.60
$0.80 / 1 = $0.80
This shows that for the third option, $4.20 for 7, the rate per item is the lowest.
