1986 AJHSME Problems/Problem 20: Difference between revisions
5849206328x (talk | contribs) mNo edit summary |
|||
| Line 8: | Line 8: | ||
<cmath> \frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}</cmath> | <cmath> \frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}</cmath> | ||
Which is closest to <math>3\rightarrow\boxed{\text{D}}</math>. | |||
(The original expression is approximately equal to <math>3.44921198</math>.) | (The original expression is approximately equal to <math>3.44921198</math>.) | ||
Revision as of 19:59, 9 October 2012
Problem
The value of the expression
is closest to
Solution
Which is closest to
.
(The original expression is approximately equal to
.)
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||