Answer:
I'm sorry I don't have answer but.
it's not C!
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
Learn more about trigonometry here : brainly.com/question/7331447
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The answer is A because Alon could have rotated the figure and discovered that they are congruent
11. 5x^2-30=0
+30 to both sides
5x^2=30
÷5 both sides
x^2=6
square root both sides
x= +square root 6 ,- square root 6
12. 4x^2+10=26
-10 both sides
4x^2=16
÷4 both sides
x^2=4
square root both sides
x=+2,-2
13. a=72yrd^2
l=2w
w=?
a=l×w
72=2w×w
72=2w^2
÷2 both sides
36=w^2
square root both sides
w=+6,-6
a measurement can't be - so w=6
Plug into l=2w
l=2×6
l=12
w=6
Answer:
Step-by-step explanation:
Odd numbers backwards - 9, 7, 5, 3
1 = -2+9=7
2 = -1+7=6
3 = 0+5=5
4 = 1 +3=4