<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).
</span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).
</span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.
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<span>I hope this helps! </span>
The value of y = 40.
5x = 40
x = 8
180 = 40 + 40 + (2y + 20)
40 = y
Answer: A) is the right answer. 
Step-by-step explanation:
Given product : 
To multiply above expression, first we need to combine like terms and then we need to use the law of exponents.
![2.3\times3\times10^{-3}\times10^8\\=(2.3\times3)\times(10^{-3}\times10^8)\\=6.9\times10^{-3+8}......[\text{by law of exponents }a^n\cdot\ a^m=a^{m+n}]\\=6.9\times10^{5}](https://tex.z-dn.net/?f=2.3%5Ctimes3%5Ctimes10%5E%7B-3%7D%5Ctimes10%5E8%5C%5C%3D%282.3%5Ctimes3%29%5Ctimes%2810%5E%7B-3%7D%5Ctimes10%5E8%29%5C%5C%3D6.9%5Ctimes10%5E%7B-3%2B8%7D......%5B%5Ctext%7Bby%20law%20of%20exponents%20%7Da%5En%5Ccdot%5C%20a%5Em%3Da%5E%7Bm%2Bn%7D%5D%5C%5C%3D6.9%5Ctimes10%5E%7B5%7D)
Thus, the answer is .
If your looking how to plot this on a graph idk but I know its something like 15.00x>200
0.15y<200
I think... I'm not sure... but it would make sense lol :)
2|3x + 5| = -10 . Divide both sides by 2:
|3x + 5| = -5 ===>3x+5=-5 & -3x-5=-5 In both cases x=0
