What do you need help on.
Answer:
option 3
Step-by-step explanation:
not sure if right
Answer: 36.9
Step-by-step explanation:
For this question you have to find out whether you are using Sin, Cos or Tan.
To work out which one you are using you can use SOH CAH TOA to help you.
SOH- The 'S' is for Sin, the 'O' and the 'H' is for Opposite over Hypotenuse.
CAH- The 'C' is for Cos and the 'A' and the 'H' is for Adjacent over Hypotenuse.
TOA- The 'T' is for Tan, the 'O' and the 'A is for Opposite over Adjacent.
As you have the Opposite (the side opposite the angle we're looking for) and the Adjacent (the side next to the angle were looking for) we would be using Tan.
So first you would have to write:
Tanθ=3/4
Then because we want Theta (θ) on it's own we would have to do the inverse to 'undo' the equation. Tan-1 is the inverse of Tan.
θ=Tan-1 3/4
To find the answer you would have to put it in your calculator.
θ=36.9
Hope this helps :)
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.
Answer:
the answer is C. 20/27 cubic cm.
Step-by-step explanation:
V = LWH
length 5(1/3 cm)
width 2(1/3 cm)
height 2(1/3 cm)
V = (5/3 cm) (2/3 cm) (2/3 cm)
V = 20/27 cubic cm.
