Difference between revisions of "1989 AHSME Problems/Problem 2"

Latest revision as of 07:39, 22 October 2014

$\sqrt{\frac{1}{9}+\frac{1}{16}}=$

$\textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12}$

$\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}$ and hence the answer is $\fbox{D}$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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 == Problem ==
-<math> (-1)^{5^{2}}+1^{2^{5}}= </math>
+<math> \sqrt{\frac{1}{9}+\frac{1}{16}}= </math>
-<math> \textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57 </math>
+<math> \textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12} </math>
 == Solution ==