Maths_0
2017/06/25 03:31
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標題:
Maths
發問:
Peter said the expression u = 40 + 4w + 300/w^2 can be expressed as Y = k/X where Y is a function of u, X is a function of w and k is a constant. Find Y, X and k. 更新: Correction : The expression should be u = 40 + 4w + 300/w.
u =40 + 4w + (300/w) u = 4 (10w + w2 + 75) / w u = 4 / [w / (w2 + 10w + 75)] Compare with Y = k/X : Y = u k = 4 X = w / (w2 + 10w + 75) There are many alternative answers, such as : Y = u k = 8 X = 2w / (w2 + 10w + 75) etc.
其他解答:
不明白本題的用意，由於沒有規定，很多 transformation 也成立，例如： u = 40 + 4w + 300/w u = (40w + 4w2 + 300)/w u = 1/[w/(40w + 4w2 + 300)] 那我可否令 Y = u, k = 1, X = w/(40w + 4w2 + 300)？ 明顯地可以，所以不確定你的題目要問什麼．．．
Maths
發問:
Peter said the expression u = 40 + 4w + 300/w^2 can be expressed as Y = k/X where Y is a function of u, X is a function of w and k is a constant. Find Y, X and k. 更新: Correction : The expression should be u = 40 + 4w + 300/w.
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最佳解答:u =40 + 4w + (300/w) u = 4 (10w + w2 + 75) / w u = 4 / [w / (w2 + 10w + 75)] Compare with Y = k/X : Y = u k = 4 X = w / (w2 + 10w + 75) There are many alternative answers, such as : Y = u k = 8 X = 2w / (w2 + 10w + 75) etc.
其他解答:
不明白本題的用意，由於沒有規定，很多 transformation 也成立，例如： u = 40 + 4w + 300/w u = (40w + 4w2 + 300)/w u = 1/[w/(40w + 4w2 + 300)] 那我可否令 Y = u, k = 1, X = w/(40w + 4w2 + 300)？ 明顯地可以，所以不確定你的題目要問什麼．．．
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