We are given with two equations to find the values of two variables, hence the problem can be solved.
Adding the two equations:
x + y = 12<u>x - y = 10
</u>2x = 22
<u />x =11
y = 1
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
#SPJ9
C(x) = 1200(40) + 100
or
= mx+c
where m = 1200, x = 40, y = 100
hope it helped,
click thx if it did.
Hi there!
»»————- ★ ————-««
I believe your answer is:
4
»»————- ★ ————-««
Here’s why:
⸻⸻⸻⸻
- I am assuming that the fraction is supposed to be the exponent.
⸻⸻⸻⸻
![64^{\frac{1}{3}}\\--------\\\rightarrow \text{Recall the exponent rule: } a^{\frac{m}{n}}=(\sqrt[n]{a})^m\\\\\\\rightarrow \sqrt[3]{64}\\\\\rightarrow \boxed{4}](https://tex.z-dn.net/?f=64%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C--------%5C%5C%5Crightarrow%20%5Ctext%7BRecall%20the%20exponent%20rule%3A%20%7D%20a%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%3D%28%5Csqrt%5Bn%5D%7Ba%7D%29%5Em%5C%5C%5C%5C%5C%5C%5Crightarrow%20%5Csqrt%5B3%5D%7B64%7D%5C%5C%5C%5C%5Crightarrow%20%20%5Cboxed%7B4%7D)
⸻⸻⸻⸻
»»————- ★ ————-««
Hope this helps you. I apologize if it’s incorrect.
If something is a function, every x-value, only has one (1) y-value. If the line x = 0 intersects the graph in two points, there are to values for y (instead of one). Therefore the graph is not a function of x.
Your answer is A.
