|KL|+a | ||
PBCLK= | *h | |
2 |
2 | ||
|KL|= | a | |
3 |
5a | ||
PBCLK= | *h | |
6 |
1 | a√3 | |||
(3a)2+( | * | )2=h2⇔ | ||
3 | 2 |
109 | ||
h2= | a2 | |
12 |
a√327 | ||
h= | ||
6 |
5a | |
*h=5√327 | |
6 |
5a | a√327 | ||
* | =5√327⇔ | ||
6 | 6 |
2 | ||
mam pytanie skąd wiemy, że |KL|= | a ![]() | |
3 |
2 | ||
|A'O|= | |A'D| | |
3 |
|A'O| | |A'D| | ||
= | ⇔ | ||
|KL| | a |
| |A'D| | 2 | |||||||||
= | ⇔|KL|= | a | |||||||||
|KL| | a | 3 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
Kliknij po więcej przykładów | |
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Twój nick | |