Answer
$\dfrac{t^{11}}{27m^2}$
Work Step by Step
RECALL:
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$
(iii) $a^0=1, a\ne0$
Use the rules above to have:
$=3^{2-5}m^{5-7}t^{6-(-5)}
\\=3^{-3}m^{-2}t^{6+5}
\\=\dfrac{1}{3^3} \cdot \dfrac{1}{m^2} \cdot t^{11}
\\=\dfrac{t^{11}}{27m^2}$
