Perform the indicated operations. Express the result using positive exponents. \[ \left(10^{-3}\right)^{4} \]

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.)

\begin{tabular}{l|l}\( \begin{array}{l}\text { Simplify the expression so that no negative exponents } \\ \text { appear in the final result. Assume all variables represent } \\ \text { nonzero real numbers. }\end{array} \) & \( \begin{array}{l}\left(4 a^{-2}\right)^{4}\left(a^{4}\right)^{-3}=\square \\ \\ \left(4 a^{-2}\right)^{4}\left(a^{4}\right)^{-3}\end{array} \) \\ exponents only.) \end{tabular}

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

\( 1 \leftarrow \quad \) Determine whether \( B \) is the inverse of \( A \). \( A=\left[\begin{array}{rr}3 & -7 \\ -2 & 5\end{array}\right], \quad B=\left[\begin{array}{ll}5 & 7 \\ 2 & 3\end{array}\right] \)

\( 1 \leftarrow \) Show that \( A I=I A=A \). \( A=\left[\begin{array}{rr}1 & 8 \\ -2 & 2\end{array}\right] \)