Telekommunikatsiya fakulteti
Munosabatlar kompozitsiyasi
4 Metodichka 2011
- Bu sahifa navigatsiya:
- kopаytmаsi yoki kompozi ts iyasi
- 1.6.15. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(2,γ),(1,α),(1,β)} 1.6.1.
- 1.6.16. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(1,γ),(3,α),(1,β)} 1.6.2.
- 1.6.17. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(1,γ),(1,α),(3,β)} 1.6.3.
- 1.6.18. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(3,γ),(2,α),(2,β)} 1.6.4.
- 1.6.19. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(2,γ),(3,α),(2,β)} 1.6.5.
- 1.6.20. R 1 ={(a,3),(a,2),(a,1)}, R 2 ={(2,γ),(2,α),(3,β)} 1.6.6.
- 1.6.21. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(1,α),(1,β)} 1.6.7.
- 1.6.22. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(1,α),(1,γ)} 1.6.8.
- 1.6.23. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(1,α),(1,β)} 1.6.9.
- 1.6.24. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(2,α),(2,β)} 1.6.10.
- 1.6.25. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(2,α),(2,γ)} 1.6.11.
- 1.6.26. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(2,β),(2,γ),(3,α)} 1.6.12.
- 1.6.27. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(3,α),(2,γ)} 1.6.13.
- 1.6.28. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(1,β),(3,α),(3,γ)} 1.6.14.
- 1.6.29. R 1 ={(b,3),(b,2),(b,1)}, R 2 ={(3,β),(3,γ),(2,β)} 0-topshiriqning ishlanishi.
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