How to make 93 out of these numbers? 1 9 9 2 (in that order)

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I need to make 1 9 9 2(in that order) equal 93. I can use any operation that doesnt require another number(e.g factorials but not cube roots). IDK what to do
 
I need to make 1 9 9 2(in that order) equal 93. I can use any operation that doesnt require another number(e.g factorials but not cube roots). IDK what to do

Are you allowed to do things like combining digits into a single number, or inserting decimal points or vincula (for repeating decimals)? Or is it only actual "operations"?

This is, of course, just a puzzle, with no standard way to "solve" it. "What to do" is just to play with ideas.
 
I need to make 1 9 9 2(in that order) equal 93. I can use any operation that doesnt require another number(e.g factorials but not cube roots). IDK what to do
Finally i see one of these. 19 + 92 = 93
 
Since we're giving answers, here's mine:
\(\displaystyle .\overline{1}\times 9+92 = 93\)
 
\(\displaystyle \Gamma{(n)} \ = \ (n - 1)! \ \ \ \ \ \ \ Each \ \ n \ \ is \ \ a \ \ positive \ \ integer. \ \ \)

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\(\displaystyle -1 \ + \ \Gamma{(\sqrt{9})} \ + \ 92 \ = \ 93\)

\(\displaystyle \Gamma{(-1 \ + \ \sqrt{9})} \ + \ 92 \ = \ 93\)

\(\displaystyle \Gamma({\Gamma{(1*\sqrt{9})})} \ + \ 92 \ = \ 93\)
 
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