InterviewSolution
Saved Bookmarks
| 1. |
Find the value of `root(3)(16.51)` approximately. |
|
Answer» Let P = `root(3)(16.51)` log P = log `(16.51)^(1//3)` `=(1)/(3)log16.51` `=(1)/(3) (1.2178)=0.4059` log P = 0.4059 P = antilog (0.4059) `therefore` P = 2.546. |
|