Answer
$$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})= \frac{40}{x^{3}y^{8}}$$
Work Step by Step
$$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})$$
Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$
Thus,
$$(-8x^{-5}y^{-3})(-5x^{2}y^{-5})$$ $$=(-8)(-5)(x^{-5+2}y^{-3+(-5)}$$ $$=40x^{-3}y^{-8}$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$40x^{-3}y^{-8} = \frac{40}{x^{3}y^{8}}$$
