Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:57Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
129K
F = ma
\frac{8}{7} \sum_{i=1}^{n} i^{2}
∑_{k=1}^{n} k = n(n+1)/2
1K
det| 5 6 ; 2 6 | = 18
153K
1x² + 5x + 7 = 0
det| 3 3 ; 4 4 | = 0
473K
det| 9 4 ; 1 10 | = 86
188K