The figure above represents a network of one-way streets. The arrows indicate the direction of traffic flow and the numbers indicate the amount of traffic flow into or out of each of the four intersections during a certain hour. During that hour, what was the amount of traffic flow along the street from R to S if the total amount of traffic flow into P was 1, 200?
(Assume that none of the traffic originates or terminates in the network)
200
250
300
350
400
SP=1200-800=400
任意一个点流入=流出,S点流入=550+RS=S点流出=400+400
RS=800-550=250
[引用]-看题的时候要冷静,抓住关键点!!
图上都是单行道,箭头是行进方向,数字是进去每个点或者出每个点的车流量,某个小时里进p的总流量是1200,问R到S的流量
进P的有两个,一个是800一个是SP,加起来1200说明SP400,S相关的四个方向有进550出400出SP和进RS,RS+550=400+400,因此RS=250
1200-800+400(加回S流出的400)-550(减去流入的550)=250(PS)
P是1200, 流入P点的是800和S,流入S的是550和R,但是S还要流出400。
这是个纯加法题!看懂题目
题目绕而已,就是俩加法,400+400=550+x。x是250,
这种题我竟然算了6分钟我感觉我脑子进水了
我的天啊,这是加法,老师给的答案:800+ps=1200=》ps=400;rs+550=400+400=》rs=250
依然是老虎,简单明了直接计算,莫慌!!!!!
要被你笑死 真的 😄
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哪位大佬解释下,这题什么意思啊
into P = 1200
into S = out of S → 550+x=400+(1200-800)
1200-800+400=550+X
b