"g(t)= \\dfrac{1}{5t+10}"
let`s call g(t) as y:
"y= \\dfrac{1}{5t+10}" then "( \\dfrac{1}{y})\t-10=5t"
"t = \\dfrac{1}{5y}-2"
"g(t)^{-1} = \\dfrac{1}{5t}-2" is inverse.
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