Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:35Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
F = ma
\frac{8}{7} \sum_{i=1}^{n} i^{2}
det| 5 6 ; 2 6 | = 18
153K
1x² + 5x + 7 = 0
det| 3 6 ; 0 4 | = 12
473K
det| 9 4 ; 1 10 | = 86
338K
∇²ψ = 0
∑_{k=1}^{n} k = n(n+1)/2
188K