| √3 | ||
a = tg30 = | ||
| 3 |
| √3 | ||
y = | x+b | |
| 3 |
| √3 | ||
2 = | *2+b | |
| 3 |
| √3 | ||
z tego wylicz b; podstaw do pierwszego wzoru a= | i wyliczone b | |
| 3 |
| −2 √3 | ||
b = | − 2
| |
| 3 |
| √3 | 2√3 | |||
y = | x + 2 − | |||
| 3 | 3 |
| √3 | ||
2 = | *(−2)+b | |
| 3 |
| 2√3 | 6+2√3 | |||
b = 2+ | = | |||
| 3 | 3 |
| √3 | 6+2√3 | |||
y = | x + | |||
| 2 | 3 |
| 6+2√3 | ||
Skąd sie wzięła ta 6? a dokładnie to | ![]() | |
| 3 |