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--超趕 , 40 marks -- 各位知識朋友 , 入黎幫我手 ar !!! ( 數學F.1 ) (57)
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各位知識朋友 , 入黎幫我手 ar !!! 我有兩條數學問題唔識 , 幫手解答 ar( 數學F.1 )!!!!!!!!!!!!!!! (要步驟) ( 1 ) A special tea is produced by mixing tea A,B and C in the ratio of 4 : 5 : 1 . Given that the weight of tea B in the special tea is 125 g, find the weight of the special tea. ( 2 ) y : z = 5 : 3 , z : x = 7 : 10 更新: Sorry I forget type this : Question 2 is find x : y : z .
最佳解答:
( 1 ) Let the weight of the special tea be X g. weight of tea B / weight of the special tea = 125/x 125/x = 5/(4+5+1) x = 250 ∴The weight of the special tea is 250g. ( 2 ) 你條問題係唔係求x:y:z ? 如果係 : y : z = 5 : 3 , z : x = 7 : 10 x : z = 10 : 7 y : z = 5 : 3 x : y : z = 10×3 : 5×7 : 3×7 x : y : z = 30 : 35 : 21
其他解答:
1.設混合茶有x g. 125/x=5/(4+5+1) x=250 ∴混合茶有250g. 2. y:z=5:3,z:x=7:10,x:z=10:7 x:y:z=10×3:5×7:3×7 x:y:z=30:35:21|||||1. the weight of the special tea = 125 x (4+5+1)/5 =250g 2. y:z=5:3 z:x=7:10 y:z=35:21 Z:x=21:30 y:z:x= 35:21:30|||||(1) x/5=125 x=25 10x=250 the weight of the special tea is 250g. (2) more information.|||||1. 125g x (4+5+1) / 5 = 250 g 2. y/z = 5/3, z/x = 7/10 thus, y/z X z/x = (5x7)/(3x10) y/x = 7/6 By substitution, 7/z = 5/3 , so z= 35/3 so x : y : z = 6 : 7 : 35/3
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