| √5−1 | ||
Mając dane cos(48+482)= | ||
| 4 |
| √5 − 1 | ||
cos(72o) = sin(18o) = | ||
| 4 |
| 1 | ||
cos(48o) = cos(30o + 18o) = | ( √30 + 6√5 + 1 − √5) | |
| 8 |
| √5−1 | ||
4t3−3t− | =0 | |
| 4 |
| 3 | √5−1 | |||
t3− | t− | =0 | ||
| 4 | 16 |
https://matematykaszkolna.pl/forum/394655.html
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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| Twój nick | |