The answer is 1.
n = 0
[tex] (\frac{9 ^{2(0)-1} }{(3 ^{0+1} ) ^{0} } ) ( \frac{(81 ^{0-1}) ^{0+1} }{(27 ^{0+2} ) ^{0-1} } [/tex]
[tex] (\frac{9 ^{-1} }{3 ^{0} } ) ( \frac{(81 ^{-1}) ^{1} }{(27 ^{0+2}) ^{-1} } )[/tex]
[tex] (\frac{9 ^{-1} }{1} ) ( \frac{81 ^{-1} }{27 ^{-2} } )[/tex]
[tex]( \frac{1}{9}) ( \frac{27 ^{2} }{81} ) [/tex]
= 27²
(9) (81)
= 729
729
= 1
FINAL ANSWER: 1
Remember that if the exponent of a number(base) is negative, it becomes a denominator, leave its factor 1. And if the denominator has a negative exponent, it becomes a numerator, leave its factor 1.
