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:
11
Step-by-step explanation:
5l=15.25
1l=3.05
11l=33.55
It depends if it is negative that's being timesed by 2 more negatives because then that would make it negative but if it's positive times a negatives times a negative then it would be positive and no matter what a positive times any positives and no negatives that will be positive. Hope this helped! :)
Answer:
it's a terminating decimal
Answer:
x = -10; y = 9
Step-by-step explanation:
(1) 7x + 5y = -25
(2) 5x + 3y = -23 Multiply (1) by 3
(3) 21x + 15y = -75 Multiply (2) by 5
(4) 25x + 15y = -115 Subtract (3) from (4)
4x = -40 Divide each side by 4
(5) x = -10 Substitute (5) into (2)
5(-10) + 3y = -23
-50 + 3y = -23 Add 50 to each side
3y = 27 Divide each side by 3
y = 9
x = -10; y = 9
Check:
(1) 7(-10) + 5×9 = -25
-70 + 45 = -25
-25 = -25
(2) 5(-10) + 3×9 = -23
50 + 27 = -23
-23 = -23
