Please help me with this question.

Answer: C. [tex]\frac{x-3}{3x+1}[/tex]
Step-by-step explanation:
[tex]\frac{x+2}{x^2+5x+6}[/tex] ÷ [tex]\frac{3x+1}{x^2-9}[/tex] =
[tex]\frac{x+2}{x^2+5x+6} * \frac{x^2-9}{3x+1} = \frac{x+2}{(x+3)(x+2)} * \frac{(x-3)(x+3)}{3x+1} =[/tex]
multiply and cancel out same terms / simplify
first x+3 and x+3 cancel out
then x+2/x+2 cancels out to make 1
[tex]\frac{x+2}{(x+2)} * \frac{(x-3)}{3x+1} = \frac{x-3}{3x+1}[/tex]