Answer:
\franc{\left(2ab^{^3}\right)^{^{-1}}}{20\cdot \left(2^{^{-1}}\cdot \:\:ba^5\right)^{^2}}\cdot \left(\frac{4ab^{^{-1}}}{5b}\right)^{^{-2}}
Explanation:
B. 2x5x5x5x5
4.9m questions
6.4m answers