Difference between revisions of "1971 AHSME Problems/Problem 7"
Logical231 (talk | contribs) |
Logical231 (talk | contribs) |
||
| Line 1: | Line 1: | ||
| − | <math>Let\ x\ equal\ 2^{-2k} | + | <math>Let\ x\ equal\ 2^{-2k}\ \\* |
| − | \ | + | From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule.\ |
| − | From\ this\ we\ get \frac{x}{2}-(\frac{-x}{\frac{-1}{2}})+x\ by\ using\ power\ rule\ | + | \\*Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\ |
| − | Now\ we\ can\ see\ this\ simplies\ to\ \frac{-x}{2}\ | + | \\*Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{-x}{2} is\ equal\ to\ -2^{-(2k+1)}\ |
| − | Looking\ at\ \frac{x}{2} we\ can\ clearly\ see\ that\ \frac{-x}{2} is\ equal\ to\ -2^{-(2k+1)}\ | + | \\*Thus\ our\ answer\ is\ c</math> |
| − | Thus\ our\ answer\ is\ c</math> | ||
Revision as of 20:52, 29 April 2020
