Question

PLEASE HELP ASAP i don’t have much time.

Answer 1

Answer:

$x=-3, d=2, e=3$

Step-by-step explanation:

$\frac{9a^3b^5}{-3ab^2} =\frac{9}{-3} \cdot\frac{a^3}{a} \cdot \frac{b^5}{b^2}$

Recall how to divide exponents. For instance, $a^3/a^1=a^{3-1}=a^2$.

$=-3a^2b^3$

Thus, $x=-3, d=2, e=3$