Answer
$\frac{b^{7}}{121}$
Work Step by Step
We are given the expression $\frac{11^{-9}b^{3}}{11^{-7}b^{-4}}$.
To simplify, we can use the quotient rule, which holds that $\frac{a^{m}}{a^{n}}=a^{m-n}$ (where a is a nonzero real number, and m and n are integers).
$(11^{-9-(-7)})\times(b^{3-(-4)})=11^{-2}\times b^{7}$
To simplify this into only positive exponents, we know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer).
$\frac{b^{7}}{11^{2}}=\frac{b^{7}}{11\times11}=\frac{b^{7}}{121}$