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Answer:
yes
Step-by-step explanation:
using the rule of exponents
[tex]\frac{a^{m} }{a^{n} }[/tex] = [tex]\frac{1}{a^{(n-m)} }[/tex] , then
[tex]\frac{9^2}{9^{7} }[/tex] = [tex]\frac{1}{9^{(7-2)} }[/tex] = [tex]\frac{1}{9^{5} }[/tex]