Difference between revisions of "1991 AHSME Problems/Problem 3"

(Created page with "<math>(4^{-1}-3^{-1})^{-1}=</math> (A) <math>-12</math> (B) <math>-1</math> (C) <math>\frac{1}{12}</math> (D) <math>1</math> (E) <math>12</math>")
 
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(A) <math>-12</math>  (B) <math>-1</math>  (C) <math>\frac{1}{12}</math>  (D) <math>1</math>  (E) <math>12</math>
 
(A) <math>-12</math>  (B) <math>-1</math>  (C) <math>\frac{1}{12}</math>  (D) <math>1</math>  (E) <math>12</math>
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Revision as of 12:52, 5 July 2013

$(4^{-1}-3^{-1})^{-1}=$

(A) $-12$ (B) $-1$ (C) $\frac{1}{12}$ (D) $1$ (E) $12$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png