September 2018 1 20 Report
Rozwiąż równania:
2sin3x=-pierw. z 2
3xtg(2x+π)=-pierw z 3
1+tg^2[(π-x)/2]={1+tg[(π-x)/2]}^2
More Questions From This User See All

Recommend Questions



Life Enjoy

" Life is not a problem to be solved but a reality to be experienced! "

Get in touch

Social

© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.