Answer: The measure of the complement of angle x=90-x=90-30=60º
Step-by-step explanation:
Answer:
Step-by-step explanation:
hello :
3b^4−6b^2+27b = 3b(b^3-2b+9)
First you simplify both sides of the equation to make v+-41/9=-3/4.
Then add -41/9 to both sides to get the answer which is 137/36
v=137/36
Answer:
5/6 and 6/7
Step-by-step explanation:
In the top of the fraction its going up by one each time (1,2,3,4 so the 5 and 6 would be the top half of the fraction), and in the bottom half its also going up by one but it started at 2 so its one more than the top half (2,3,4,5, so the 6 and 7 would be the bottom half of the fraction)
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.
