h) \( \left(3^{2}\right)^{7} \cdot\left(3^{3}\right)^{5}= \)

\( \begin{array}{l}\text { Simplify the expression. Assume that variables represent } \\ \text { non-zero real numbers. }\end{array} \) \( \begin{array}{ll}\frac{6 a^{5}\left(a^{-2}\right)^{6}}{\left(a^{-1}\right)^{-3}} & \frac{6 a^{5}\left(a^{-2}\right)^{6}}{\left(a^{-1}\right)^{-3}}=\square \\ \text { (Simplify your answer. Type in exponenti. } \\ \text { notation. Type positive exponents.) }\end{array} \)

Résoudre dans \( \mathbb{C}: \mathrm{z}^{7}=1 \)

Rewrite the following without an exponent \[ (-7)^{-2} \]

Three-fifths of Jesse Black's inventory was destroyed by fire. He sold the remaining part, which was slightly damaged, for two-thirds of its value and received \( \$ 3400 \). (a) What was the value of the destroyed part of the inventory? (b) What was the value of the inventory before the fire? (a) The value is \( \$ \square \). (Round to the nearest cent as needed.)

Simplify. \( [32-(15+5) \div 4]-[56 \div(8-1)]^{\prime} \) \( [32-(15+5) \div 4]-[56 \div(8-1)]=\square \)