Which expression is equivalent to (20a^2 b^0 c^-4)(3a^15 b^12 c^0) for all values of a,b, and c where the expression is defined?
By using the property of exponents that $a^0=1$ for all $a\neq 0$, we know that $b^0=1$ and $c^0=1$ for all values of $b$ and $c$. So we can remove these factors from our expression, which leaves us with $(20a^2)(3a^{15})$.
To multiply these terms, we first multiply the coefficient $20$ by $3$ to get $20\cdot 3=60$. Then, we multiply $a^2$ by $a^{15}$ using the power of a power rule, which tells us that $a^m\cdot a^n= a^{m+n}$ for all $m,n$. Applying this rule, we have $a^2\cdot a^{15}=a^{2+15}=a^{17}$.
So the expression $(20a^2 b^0 c^{-4})(3a^{15} b^{12} c^0)$ simplifies to $60a^{17}$ for all values of $a$, $b$, and $c$ where the expression is defined.