Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
Solve for S(t):
The left side is the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
Integrate both sides:
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
and so (b) the amount of sugar in the tank at time t is
As 
, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
Closed dot on 53, arrow going to the right
Answer: Each shirt costs $1.25, and each hat costs $2.5.
Step-by-step explanation:
Let x= Cost per T-shirt, y = Cost per Hat.
As per given,
3x+3y=11.25 (i)
4x+2y=10 (ii)
Multiply 4 to equation (i)
12x+12y=45 (iii)
12x+6y= 30 (iv)
Eliminate (iv) from (iii), we get
6y= 15
⇒ y = 2.5
Put value of y in (i)
3x+3(2.5)=11.25
⇒3x+7.5=11.25
⇒3x= 3.75
⇒x= 1.25
Hence, each shirt costs $1.25, and each hat costs $2.5.
Answer:
D)-3x+5
Step-by-step explanation:
Hope this helps ;)
Answer:
B) $7.50
Step-by-step explanation:
Interest (I) = Principal (P) x Interest Rate (r) x Time (t, in years)
I = 125(.03)(2)
I = 7.5
