It is 10, and 15 I believe
![\bf f(x)=y=2x+sin(x) \\\\\\ inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x) \\\\\\ \textit{now, the "y" in the inverse, is really just g(x)} \\\\\\ \textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dy%3D2x%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0Ainverse%5Cimplies%20x%3D2y%2Bsin%28y%29%5Cleftarrow%20f%5E%7B-1%7D%28x%29%5Cleftarrow%20g%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20the%20%22y%22%20in%20the%20inverse%2C%20is%20really%20just%20g%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%20can%20write%20it%20as%20%7Dx%3D2g%28x%29%2Bsin%5Bg%28x%29%5D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
![\bf \textit{let's use implicit differentiation}\\\\ 1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor} \\\\\\ 1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Blet%27s%20use%20implicit%20differentiation%7D%5C%5C%5C%5C%0A1%3D2%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%2Bcos%5Bg%28x%29%5D%5Ccdot%20%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5Cimpliedby%20%5Ctextit%7Bcommon%20factor%7D%0A%5C%5C%5C%5C%5C%5C%0A1%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5B2%2Bcos%5Bg%28x%29%5D%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B%5B2%2Bcos%5Bg%28x%29%5D%5D%7D%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%3Dg%27%28x%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D)
now, if we just knew what g(2) is, we'd be golden, however, we dunno
BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)
for inverse expressions, the domain and range is the same as the original, just switched over
so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2
so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2
thus 2 = 2x+sin(x)
![\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2 \\\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}](https://tex.z-dn.net/?f=%5Cbf%202%3D2x%2Bsin%28x%29%5Cimplies%200%3D2x%2Bsin%28x%29-2%0A%5C%5C%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D%5Cimplies%20g%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5B2x%2Bsin%28x%29-2%5D%7D)
hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
Answer:
In this graph, the y-intercept of the line is -2, if the coordinates are needed, its (0,-2). The equation of the line is y= x-2 ( slope intercept form ).
Step-by-step explanation: I am not sure how to use the multiplication system below to write the equation, so sorry I couldn't help!
Because both of the polygons are congruent and are flipped so there for they are the same angle