Answer: (c)
Step-by-step explanation:
Given
![f(x)=\dfrac{1}{x-3}\\\\g(x)=\sqrt{x+5}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7B1%7D%7Bx-3%7D%5C%5C%5C%5Cg%28x%29%3D%5Csqrt%7Bx%2B5%7D)
Here, ![\sqrt{x+5}\ \text{is always greater than equal to 0}\\\Rightarrow x+5\geq 0\\\Rightarrow x\geq -5\quad \ldots(i)](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B5%7D%5C%20%5Ctext%7Bis%20always%20greater%20than%20equal%20to%200%7D%5C%5C%5CRightarrow%20x%2B5%5Cgeq%200%5C%5C%5CRightarrow%20x%5Cgeq%20-5%5Cquad%20%5Cldots%28i%29)
To get
, replace
in
by ![g(x)\ \text{i.e. by}\ \sqrt{x+5}](https://tex.z-dn.net/?f=g%28x%29%5C%20%5Ctext%7Bi.e.%20by%7D%5C%20%5Csqrt%7Bx%2B5%7D)
![\Rightarrow f\left(g(x)\right)=\dfrac{1}{\sqrt{x+5}-3}\\\\\text{Denominator must not be equal to 0}\\\\\therefore \sqrt{x+5}-3\neq0\\\Rightarrow \sqrt{x+5}\neq 3\\\Rightarrow x+5\neq 9\\\Rightarrow x\neq 4\quad \ldots(ii)](https://tex.z-dn.net/?f=%5CRightarrow%20f%5Cleft%28g%28x%29%5Cright%29%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%2B5%7D-3%7D%5C%5C%5C%5C%5Ctext%7BDenominator%20must%20not%20be%20equal%20to%200%7D%5C%5C%5C%5C%5Ctherefore%20%5Csqrt%7Bx%2B5%7D-3%5Cneq0%5C%5C%5CRightarrow%20%5Csqrt%7Bx%2B5%7D%5Cneq%203%5C%5C%5CRightarrow%20x%2B5%5Cneq%209%5C%5C%5CRightarrow%20x%5Cneq%204%5Cquad%20%5Cldots%28ii%29)
Using
and
it can be concluded that the domain of
is all real numbers except 0.
Therefore, its domain is given by
![x\in [-5,4)\cup (4,\infty)](https://tex.z-dn.net/?f=x%5Cin%20%5B-5%2C4%29%5Ccup%20%284%2C%5Cinfty%29)
Option (c) is correct.
I think B. Would be the answer
Answer is 0.31. Hope this helps.
Answer:
It is an octagon if and only if it has eight sides.
Step-by-step explanation: