某人要到相距2.4千米的地方办事,要求在18分内到达,以知步行每分钟90米...

发布网友 发布时间：2024-10-04 09:10

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热心网友 时间：2024-10-05 15:26

跑步x km 步行(2400-x) km => x/210 +(2400-x)/90 <=18
=>630*x/210 +630*(2400-x)/90 <=630*18
=>3x +7*(2400-x) <=11340
=> -4x +16800<=11340 => -4x<= -5460 => x>=1365 即 1365<=x<=2400
=> 1365/210<=x/210<=2400/210 => 6.5<= x/210(跑步时间) <=11又 3/7
ans 6.5分与11又 3/7分之间 含两端

热心网友 时间：2024-10-05 15:25

设时间为X，210X=2400，X≈11.43

热心网友 时间：2024-10-05 15:18

解：跑步x km 步行(2400-x) km => x/210 +(2400-x)/90 <=18
=>630*x/210 +630*(2400-x)/90 <=630*18
=>3x +7*(2400-x) <=11340
=> -4x +16800<=11340 => -4x<= -5460 => x>=1365 即 1365<=x<=2400
=> 1365/210<=x/210<=2400/210 => 6.5<= x/210(跑步时间) <=11又 3/7
ans 6.5分与11又 3/7分之间 含两端

热心网友 时间：2024-10-05 15:20

跑步x km 步行(2400-x) km => x/210 +(2400-x)/90 <=18
=>630*x/210 +630*(2400-x)/90 <=630*18
=>3x +7*(2400-x) <=11340
=> -4x +16800<=11340 => -4x<= -5460 => x>=1365 即 1365<=x<=2400
=> 1365/210<=x/210<=2400/210 => 6.5<= x/210(跑步时间) <=11又 3/7
ans 6.5分与11又 3/7分之间 含两端

热心网友 时间：2024-10-05 15:23

要11.42857142857142857......