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maths _30
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[IMG]http://i293.photobucket.com/albums/mm67/zaza520/0011-18.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0012-17.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0013-20.jpg[/IMG] 更新: 要詳解
(9) The curve is opening downward, so coeff of x2 is negative. Also constant term = 4 So ans = C (16) 10x-y = x x - y = log x y = x - log x (20) ∠DEB = 38 since △BDE is isos. ∠DBA = 76 (Ext. ∠ of triangle) ∠DCF = ∠DBA = 76 (Angle in the same segment) ∠CFD = 180 - 76 - 38 = 66 (21) ∠BCD = 116 (Opp. ∠s, cyclic quad.) ∠CBD = ∠CDB = (180 - 116)/2 = 32 (Base ∠s, isos. triangle) ∠ABD = 90 (Angle in semi-circle) ∠ABC = 90 + 32 = 122 ∠ADB = ∠ABC = 122 (Ext. ∠s, cyclic quad.)
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maths _30
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[IMG]http://i293.photobucket.com/albums/mm67/zaza520/0011-18.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0012-17.jpg[/IMG] [IMG]http://i293.photobucket.com/albums/mm67/zaza520/0013-20.jpg[/IMG] 更新: 要詳解
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最佳解答:(9) The curve is opening downward, so coeff of x2 is negative. Also constant term = 4 So ans = C (16) 10x-y = x x - y = log x y = x - log x (20) ∠DEB = 38 since △BDE is isos. ∠DBA = 76 (Ext. ∠ of triangle) ∠DCF = ∠DBA = 76 (Angle in the same segment) ∠CFD = 180 - 76 - 38 = 66 (21) ∠BCD = 116 (Opp. ∠s, cyclic quad.) ∠CBD = ∠CDB = (180 - 116)/2 = 32 (Base ∠s, isos. triangle) ∠ABD = 90 (Angle in semi-circle) ∠ABC = 90 + 32 = 122 ∠ADB = ∠ABC = 122 (Ext. ∠s, cyclic quad.)
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