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

Revision as of 07:38, 22 October 2014

$(-1)^{5^{2}}+1^{2^{5}}=$

$\textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57$

$\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}$ 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

@@ Line 1: / Line 1: @@
-<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}</math> and hence the answer is D.
+== Problem ==
+<math> (-1)^{5^{2}}+1^{2^{5}}= </math>
+<math> \textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57 </math>
+== Solution ==
+<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=\sqrt{\frac{25}{9\cdot 16}}=\frac{5}{12}</math> and hence the answer is <math>\fbox{D}</math>.
+== See also ==
+{{AHSME box|year=1989|num-b=1|num-a=3}}
+[[Category: Introductory Algebra Problems]]
 {{MAA Notice}}