???
Kaja : (1+cosx)(1/sinx−ctgx)=sinx
21 maj 20:42
irena_1:
P=sinx
| | 1 | | 1 | | cosx | |
L=(1+cosx)( |
| −ctgx)=(1+cosx)( |
| − |
| )= |
| | sinx | | sinx | | sinx | |
| | 1−cosx | | (1+cosx)(1−cosx) | | 1−cos2x | |
=(1+cosx)* |
| = |
| = |
| = |
| | sinx | | sinx | | sinx | |
L=P
21 maj 20:47
siwy: a=log2 i b=log3 wyznacz log192 w zależności od a i b
21 maj 20:51
irena_1:
log192=log(26*3)=log26+log3=6log2+log3=6a+b
21 maj 21:00