Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
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Answer:
0.9905,0.9837,0.8066
Step-by-step explanation:
Given that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped).
Each shaft is independent of the other and probability for non conforming is the same for each trial.
Hence X the among these that are nonconforming and can be reworked. is Bin with n =200 and p = 0.10
a) 
b) 
c) 
Answer:
y=-5/2x+3
Step-by-step explanation:
Assume uniformly accelerated motion
Vf = Vo - at => a= [Vo - Vf] / t
Vf = 0, t = 20 s => a = Vo / 20
x = Vot - at^2 / 2
x = Vot - (Vo / 20) * t^2 /2
x = 1775 feet * [1 mile / 5280 feet] = 0.336 mile
t = 20 s
=> 0.336 = 20Vo - Vo(20)^2 /(40) = 20Vo - 10 Vo = 10 Vo
=> Vo = 0.336 /10 = 0.0336 mile /s
Vo = 0.0336 mile/s * 3600 s/h = 120.96 mile / h
Therefore, the car was way over the speed limit.
