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Sub 505 Level|   Exponents|      
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Bunuel
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9^(-4)/27^(-3) = 02 Mar 2017, 08:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

5% (low)

Question Stats:

88% (00:36) correct 12% (00:53) wrong based on 110 sessions

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\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27
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C
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9^(-4)/27^(-3) = 02 Mar 2017, 08:41
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Bunuel
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27

\(\frac{9^{-4}}{27^{-3}}=9^{-4}/27^{-3} = \frac{1}{9^4} / \frac{1}{27^3}=\frac{27^3}{9^4}=\frac{3^9}{3^8}=3\)

The answer is C
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9^(-4)/27^(-3) = 02 Mar 2017, 08:49
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Inverting the signs.

Fraction becomes \(\frac{27^3}{9^4}\) = \(\frac{3^9}{3^8}\) = 3
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9^(-4)/27^(-3) = 02 Mar 2017, 09:13
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\(\frac{3^{-8}}{3^{-9}} = 3^{-8-(-9)} = 3^{1} = 3\)
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Abhishek009
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9^(-4)/27^(-3) = 02 Mar 2017, 12:40
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Bunuel
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27

\(\frac{9^{(-4)}}{27^{(-3)}}\)

\(= \frac{3^{(-8)}}{3^{(-9)}}\)

\(= 3^{( 9 - 8)}\)

\(= 3^1\)

Thus, answer must be (C) 3
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9^(-4)/27^(-3) = 02 Mar 2017, 14:34
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At a quick glance of the question you can easily eliminate A,D, & E because they are repeats from the question/too easily derived. That leaves B & C, inverting the fraction leaves exponents 9-8=1/base remains 3. 3^1. Answer=C.
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9^(-4)/27^(-3) = 02 Mar 2017, 14:38
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Bunuel
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27


it is obviously 3. the answer is C.

(3^-8)/(3^-9)=3^1=3
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9^(-4)/27^(-3) = 03 Mar 2017, 01:50
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Bunuel
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27

IMO C
the problem can be simplified as (27*27*27)/(9*9*9*9)= (3*3*3)/(9) = 3
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Bunuel
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