Answer:
12x+9=1
Step-by-step explanation:
9 more than comes after
12 times a number
12x
9 more than
12x+9
equals 1
12x+9=1
If g(x) is the inverse of f(x), then by definition of inverse function,
f(g(x)) = g(f(x)) = x
Given that f(x) = 5x - 4, we have
f(g(x)) = 5 g(x) - 4 = x
Solve for g(x) :
5 g(x) - 4 = x
5 g(x) = x + 4
g(x) = (x + 4)/5
To prove:
![$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7B1-%5Ccos%20x%7D-%5Cfrac%7B%5Ccos%20x%7D%7B1%2B%5Ccos%20x%7D%3D2%20%5Ccot%20%5E%7B2%7D%20x%2B1)
Solution:
![$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}](https://tex.z-dn.net/?f=%24LHS%20%3D%20%5Cfrac%7B1%7D%7B1-%5Ccos%20x%7D-%5Cfrac%7B%5Ccos%20x%7D%7B1%2B%5Ccos%20x%7D)
Multiply first term by
and second term by
.
![$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}](https://tex.z-dn.net/?f=%24%3D%20%5Cfrac%7B1%281%2B%5Ccos%20x%29%7D%7B%281-%5Ccos%20x%29%281%2B%5Ccos%20x%29%7D-%5Cfrac%7B%5Ccos%20x%281-%5Ccos%20x%29%7D%7B%281%2B%5Ccos%20x%29%281-%5Ccos%20x%29%7D)
Using the identity: ![(a-b)(a+b)=(a^2-b^2)](https://tex.z-dn.net/?f=%28a-b%29%28a%2Bb%29%3D%28a%5E2-b%5E2%29)
![$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}](https://tex.z-dn.net/?f=%24%3D%20%5Cfrac%7B1%2B%5Ccos%20x%7D%7B%281%5E2-%5Ccos%5E2%20x%29%7D-%5Cfrac%7B%5Ccos%20x-%5Ccos%5E2%20x%7D%7B%281%5E2-%5Ccos%5E2%20x%29%7D)
Denominators are same, you can subtract the fractions.
![$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}](https://tex.z-dn.net/?f=%24%3D%20%5Cfrac%7B1%2B%5Ccos%20x-%5Ccos%20x%2B%5Ccos%5E2%20x%7D%7B%281%5E2-%5Ccos%5E2%20x%29%7D)
Using the identity: ![1-\cos ^{2}(x)=\sin ^{2}(x)](https://tex.z-dn.net/?f=1-%5Ccos%20%5E%7B2%7D%28x%29%3D%5Csin%20%5E%7B2%7D%28x%29)
![$= \frac{1+\cos^2 x}{\sin^2x}](https://tex.z-dn.net/?f=%24%3D%20%5Cfrac%7B1%2B%5Ccos%5E2%20x%7D%7B%5Csin%5E2x%7D)
Using the identity: ![1=\cos ^{2}(x)+\sin ^{2}(x)](https://tex.z-dn.net/?f=1%3D%5Ccos%20%5E%7B2%7D%28x%29%2B%5Csin%20%5E%7B2%7D%28x%29)
![$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%5Ccos%20%5E%7B2%7Dx%2B%5Ccos%20%5E%7B2%7Dx%2B%5Csin%20%5E%7B2%7Dx%7D%7B%5Csin%20%5E%7B2%7Dx%7D)
------------ (1)
![RHS=2 \cot ^{2} x+1](https://tex.z-dn.net/?f=RHS%3D2%20%5Ccot%20%5E%7B2%7D%20x%2B1)
Using the identity: ![\cot (x)=\frac{\cos (x)}{\sin (x)}](https://tex.z-dn.net/?f=%5Ccot%20%28x%29%3D%5Cfrac%7B%5Ccos%20%28x%29%7D%7B%5Csin%20%28x%29%7D)
![$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}](https://tex.z-dn.net/?f=%24%3D1%2B2%5Cleft%28%5Cfrac%7B%5Ccos%20x%7D%7B%5Csin%20x%7D%5Cright%29%5E%7B2%7D)
![$=1+2\frac{\cos^{2} x}{\sin^{2} x}](https://tex.z-dn.net/?f=%24%3D1%2B2%5Cfrac%7B%5Ccos%5E%7B2%7D%20x%7D%7B%5Csin%5E%7B2%7D%20x%7D)
------------ (2)
Equation (1) = Equation (2)
LHS = RHS
![$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7B1-%5Ccos%20x%7D-%5Cfrac%7B%5Ccos%20x%7D%7B1%2B%5Ccos%20x%7D%3D2%20%5Ccot%20%5E%7B2%7D%20x%2B1)
Hence proved.
Ask Siri and she will probably gone to the answer
Answer:
6, I believe?
Step-by-step explanation:
300/6=50
750/6=125, and
900/6=150, so I think it's 6 bracelets, if i'm wrong you can call me dumb.