which four inequalities can be used to find the solution to this absolute value inequality 3<|x+2|<6

\[\begin{array}{l}

\frac{{106}}{{204}}\,\,\& \,\,\frac{{107}}{{206}}\\

106.206 = 106.\left( {204 + 2} \right) = 106.204 + 212\\

107.204 = \left( {106 + 1} \right).204 = 106.204 + 204\\

 \Rightarrow 106.204 + 212 > 106.204 + 204\\

 \Rightarrow 106.206 > 107.204\\

 \Rightarrow \frac{{106}}{{204}}\, > \,\frac{{107}}{{206}}

\end{array}\]

whatever you can spend with $11.53 or under.

step-by-step explanation:

20 - 8.47 = 11.53. take away the money he used for lunch from the starting amount of 20.