字體：小 中 大 |
|
|
|
| 2017/07/03 23:34:48瀏覽59｜回應0｜推薦0 | |
| 標題: 2題英文數學~幫幫手! 發問: 1.some sweets are shared by a gorup of children. if each child gets 5 sweets,8sweets will be left.if each child wants to get 7 sweets,10 more sweets will be needed.how many sweets are there? 2.find two consecutive integers such yhat 3/5 of the larger number exceeds 1/4 of the smaller one by 5又2分1 更新: 我要搵人幫我計呀! 要列式!! 最佳解答:
此文章來自奇摩知識+如有不便請留言告知 1) let x be the no. of children 5x+8=7x-10 5x=7x-10-8 7x-5x=10+8 2x=18 x=9 because 5(9)+8 =45+8 =53 therefore , there are 53 sweets.// 2) 唔肯定... 驚教錯...其他解答: 1. Let x be the no. of sweets, x/ 5 + 8 = x/7 - 10 x/ 5 - x/ 7= 10-8 7x - 5x = 70 x = 35 Therefore, there are 34 sweets. 2. Let x be the smaller no. and x + 1 be the larger no. ( 3 / 5 )( x + 1 ) + ( x / 4 ) = 11/2 (3x + 3)/ 5 + x/4 = 5.5 (12x+12+5x)/20 = 5.5 7x =98 x = 14 Therefore, the smaller integer is 14 and the larger one is 15. 2007-09-02 19:13:14 補充： * Therefore, there are [35] sweets. 2007-09-02 19:21:02 補充： Sorry 我計錯左第一題-____-重做::Let x be the no of sweets,y be the no. of children{5y 8 = x ---(1){7y - 10 = x ---(2) 5y 8 = 7y - 10 2y = 18y = 9Putting y = 9 into (1)x = 5(9) 8x =53Therefore, there are 53 sweets. =] |
|
| ( 時事評論｜媒體出版 ) |
