Answer:
Step-by-step explanation:
[tex](-4a^2)(2a^{-3})^{-4}\qquad\text{use}\ (xy)^n=x^ny^n\ \text{and}\ (x^n)^m=x^{nm}\\\\=(-4a^2)\bigg(2^{-4}a^{(-3)(-4)}\bigg)\qquad\text{use}\ 4=2^2\\\\=(-2^2a^2)(2^{-4}a^{12})\\\\=-(2^2)(2^{-4})(a^2a^{12})\qquad\text{use}\ x^nx^m=x^{n+m}\\\\=-2^{2+(-4)}a^{2+12}\\\\=-2^{-2}a^{14}\qquad\text{use}\ x^{-n}=\dfrac{1}{x^n}\\\\=-\dfrac{1}{2^2}a^{14}\\\\=-\dfrac{a^{14}}{4}[/tex]