Difference between revisions of "1989 AHSME Problems/Problem 2"
(Created page with "<math>\sqrt{\frac{1}{9}+\frac{1}{16}}=</math> (A) <math>\frac{1}{5}</math>\hskip.5in (B)<math>\frac{1}{4}</math>\hskip.5in (C)<math>\frac{2}{7}</math>\hskip.5in (D)<math>\...") |
m |
||
| (3 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| − | + | == Problem == | |
| − | + | <math> \sqrt{\frac{1}{9}+\frac{1}{16}}= </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 == | ||
| + | |||
| + | <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}} | ||
Latest revision as of 07:39, 22 October 2014
Problem
Solution
and hence the answer is 
.
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
