Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:03Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
129K
F = ma
det| 9 4 ; 1 10 | = 86
191K
det| 3 7 ; 2 4 | = -2
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
1K
det| 4 8 ; 3 5 | = -4
153K
1x² + 5x + 7 = 0