1 | ||
y = − | x + 3 | |
2 |
1 | ||
l:⊥k ⇔ al*{− | ) = −1 ⇔ al=2 | |
2 |
5 | 1 | ||
=2*(− | )+b | ||
6 | 3 |
5 | 2 | 5 | 4 | 9 | 3 | |||||||
b = | + | = | + | = | = | |||||||
6 | 3 | 6 | 6 | 6 | 2 |
3 | ||
y = 2x+ | ||
2 |
1 | 5 | |||
to p⊥k więc p: −2(x+ | )+(y− | )=0 | ||
3 | 6 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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