PLEASE HELP ASAP i don’t have much time.

Answer:
[tex]x=-3, d=2, e=3[/tex]
Step-by-step explanation:
[tex]\frac{9a^3b^5}{-3ab^2} =\frac{9}{-3} \cdot\frac{a^3}{a} \cdot \frac{b^5}{b^2}[/tex]
Recall how to divide exponents. For instance, [tex]a^3/a^1=a^{3-1}=a^2[/tex].
[tex]=-3a^2b^3[/tex]
Thus, [tex]x=-3, d=2, e=3[/tex]