consider a concrete mixing duct shown in Figure 1:
a). show that w^2 - 2Rw + R^2 over 4 = 0
b). Find the value of w when R = 2m.
圖片參考：http://imgcld.yimg.com/8/n/HA07162721/o/701101170145113873370810.jpg
唔識計!?20110117
2011-01-18 7:22 am
回答 (2)
2011-01-18 7:37 am
✔ 最佳答案
a)R² = (R - w)² + (R/2)²R² = R² - 2Rw + w² + R²/4w² - 2Rw + R²/4 = 0 ........(*)b)Sub 2m = R to (*) ,w² - 2(2m)w + (2m)²/4 = 0w² - 4mw + m² = 0w = [4m ± √(16m² - 4m²)] / 2w = [4m ± √(12m²))/2w = 2m ± √3 mw = (2 ± √3) mSince w < R ,w = (2 - √3) m
2011-01-18 7:43 am
lks...唔該你唔好胡亂刪除問題吧!
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