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  1. The chanllenges of supply chain in HK today that are- how to maintain JIT (just-in-time) inventory...

  2. ...其中一個(x²-1)的次方是正另一個是負" Ans. : 用 quotient rule, chain rule 照樣代入 . 你的 first step 做對了啦 !

    分類:科學及數學 > 數學 2016年09月21日

  3. 每間舖都一樣價錢, 但未必有同一款手錶, 因每間舖位置都有少少出入.

  4. dy/dx = d[sin(cos(tan(3x +1)))]/d[cos(tan(3x+1))] times d[cos(tan(3x+1))]/d[tan(3x +1)] times d[tan(3x+1)]/d(3x+1) times d(3x +1)/dx = cos(cos(tan(3x+1)) times -sin(tan(3x+1)) times sec^2(3x+1) times 3 = -3[cos(cos(tan(3x+1)))][sin(tan(3x+1))][sec^2(3x+1)].

    分類:科學及數學 > 數學 2008年10月28日

  5. ...like this d/dx[sin(1/(x^2+1)^6)] let f=[sin(1/(x^2+1)^6)] v=(1/(x^2+1)^6) Chain rule equation 應該係咁樣 df/dx=df/dv .dv/dx 但係我有一個咁o既step =df/dv...

    分類:科學及數學 > 數學 2006年12月09日

  6. a) y = (sin(π/2+2x))^3 dy/dx= 3 (sin(π/2+2x))^2 * cos(π/2+2x) * 2 dy/dx= 6 (sin(π/2+2x))^2 * cos(π/2+2x) b) y= √(sin (x^2)) dy/dx = (1/2) (sin (x^2)) ^ (-1/2) * cos (x^2) * 2x dy/dx = x * cos (x^2) * (sin (x^2)) ^ (-1/2)

    分類:科學及數學 > 數學 2009年01月23日

  7. y = sin^5 x + cos^5 x dy/dx = 5 sin^4 x (cos x) + 5 cos^4 x (- sin x) = 5 sin^4 x cos x - 5 cos^4 x sin x = 5 sin x cos x [sin^3 x - cos^3 x]

  8. ...of this increased demand at particular times of the year, on the supply chain ? A: While the demand can not be easily predicted, the special...

  9. ...enterprise resources planning is the essential system for supply chain management.

  10. My suggestion is OAK colour, I like this colour than black. In Spring & Summer I think use light colour bag is easy to matching.