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.
Simplified it's 576x^8y^7
We know that
scale factor=1 in/7.5 ft
scale factor=measurements on the blueprint/measurements <span> in the actual
</span>measurements on the blueprint=[measurements in the actual*scale factor]
so
for 18 ft
measurements on the blueprint=[18 ft*(1 in/7.5 ft)-----> 2.4 in
for 16 ft
measurements on the blueprint=[16 ft*(1 in/7.5 ft)-----> 2.1 in
the dimensions on the blueprint are
2.4 in x 2.1 in
Answer:
-3x-15
Step-by-step explanation:
you use the distributive property. So it would be 3 times -x and 3 times -5 so you get -3x-15.
From 2000 to 2011, there are 11 years. From the given function, we can calculate the profit for a small business at the end of 2011 by substituting 11 to the x of the given equation.
y = (30,000)(1.06)(11)
y = 349,800
Thus, the profit at the end of 2011 is approximately $349,800.
