A 300 m long straight path on a hill rises from 150 m to 200 m above the sea-level . Find the gradient of the path . 圖片參考：http://hk.geocities.com/namsm4e/p0118.jpg 如圖所示，Path 的水平長度為 295.8m 所以gradient = 295.8 : 50
from above , we get f(x) = (x + 3) Q1(x) +10 f(x) = (x - 3) Q2(x) - 8 ...
...r(x)(x+2) - 10 put x = -2, we have -8 - 2a + b = -10 Upon solving above two equations, we have a = -5, b = -12 Therefore, f(x) = x^3...
...1/(1 + exp(x)) ~ 1/(1 + 1 + x) = 1/(x + 2) Now, as powers of x above the second to be neglected (x + 2)(2 - x) = 4 - x^2 ~ 4 Then...
... the exam. Students obtained grade 'C' or above : 200 · (144°+72°+18°)/360° = 130 .·. There are 130 students...
The above one has got something wrong. 圖片參考：http://i238.photobucket.com/...
The above claim is wrong already! The correct identity is abc[a....M.! The only thing I can say is, the above Yahoo knowledge http://hk.knowledge.yahoo.com/question/question?qid=7011060701220...