车场做26000辆车,实际比计划多做1300辆,实际做了计划的百分之几?

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设小车为x辆,中型车为y辆,
x+y=50
5x+8y=340
解方程组
x=?, y=?

因只有4个空,所以恰有4空连着就只有4联排了.则根据位置不同,有1-4 2-5.13-16,共计13种

乙的速度是16÷2/5=40千米
甲行完全程的时间是180÷40-2/5=4.5-0.4=4.1小时

23－12＝11
﹙89＋46﹚÷﹙1＋2﹚＝45
89－45＝44
44÷11＝4天

黑—明
雨来得及时的特点
春雨来得及时
早晨起来,那潮湿的泥土上一定满是红色的花瓣,成都的大街小巷必将是一片万紫千红的景象.

At peak hours,to take the train from the rear yard line to turn back legislation to line break.Start time because of work 7 minutes,start in the compact (3 minute interval) cases must be made to take up 4 lines,will be made to Unit 4 in the first half for the start line,the second half 3 shares issued to the line for vehicle access (As the final 6 minutes of time to work,only three lines can not be made to access 3 minutes interval in a row to reach the train).
In the off-peak hours,the train line to take this break legislation.7 shares issued to the line can be carried out next train operations.
In the next train,the train originating from,to start again from the arrival terminal,at least 45 minutes 10 seconds (downlink direction) / 44 minutes 15 seconds (uplink direction),only reached the next train (Tianjin Station) ,The peak hours up to be able to start 15; as early summit and to meet the needs of start,Tianjin night stand at least 10 more parking.
At departure time,since the peak hours then made 15 pairs of trains (because of the train station stop in the middle),end-to-fat line of departure and arrival capacity is tight,so the train departure point in time there is no law that can not be established standardized plans to run .

大汽车x辆；小汽车9-x辆
6x+5(9-x)=44
6x+45-5x=44
x=-1?
大汽车有6个轮子,小汽车有4个轮子吧?
6x+4(9-x)=44
2x=8
x=4
9-x=5
大汽车4辆；小汽车5辆

应交停车费 5+（7-2）x1.5=5+5x1.5=12.5元

答第一个问题:第一车间调24人到第二车间,第一车间40人,第二车间80人,刚好一半!

解题思路：（1）与围墙垂直的边长=与围墙平行的一边长-（m-n）；
（2）护栏的长度=2×与围墙垂直的边长+与围墙平行的一边长；
（3）把m、n的值代入（2）中的代数式进行求值即可．

（1）依题意得 （2m+3n）-（m-n）=m+4n；（2）护栏的长度=2（m+4n）+（2m+3n）=4m+11n；答：护栏的长度是：4m+11n．（3）由（2）知，护栏的长度是4m+11n．则依题意得（4×30+11×10）×80=18 400（元）．答：若m=30...

点评：
本题考点： 列代数式；代数式求值．

考点点评： 本题考查了列代数式和代数式求值．解题时要数形结合，该护栏的长度是由三条边组成的．

恰有紧靠在一起的4股道无车的概率是
(12-4+1)/C(12,8)=9/495=1/55
解释：
“恰有紧靠在一起的4股道无车”包括
第i至i+3股道无车(i=1 to 9) 其余8股道有车
共12-4+1=9种情况

上山速度比1.25:1=5:4
甲到山顶,乙走到距离山顶 1-4/5＝1/5
甲下山速度：乙上山速度 120％x1.25:1=3:2
两点距离占总长 1/5x2/(3+2)=2/25
50x2/25=4km

一元一次方程 ：设自行车X辆,三轮车（42-X）辆
2X+3（42-X）=100
2X+126-3X=100
-X=-26
X=26
42-26=16
答：自行车26辆,三轮车16辆
假设法：假设全是三轮车
42*3=126（个）
126-100=26（个）
26除以（3-2）=26（辆）
42-26=16（辆）
答：自行车26辆,三轮车16辆
二元一次方程组：设自行车X辆,三轮车Y辆
X+Y=42
2X+3Y=100
方程组解得：X=26,Y=16
答：自行车26辆,三轮车16辆

只会游泳的有120-65=55人.
分析：会骑自行车的65人中包含两样都会的30人,而只会游泳的是就是不会骑自行车的（因为每人至少会一样）.

时速28.8千米一小时=8米/秒
105÷15=7米/秒
8-7=1米/秒=3.6千米/小时
因为是从身后开来,所以速度就是火车与人的速度差,

就是鸡兔同笼啊