Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:18Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
129K
F = ma
∑_{k=1}^{n} k = n(n+1)/2
191K
∫₀^∞ e^(-x²) dx = √π / 2
1K
∑_{k=1}^{n} k = n(n+1)/2
153K
det| 5 2 ; 3 6 | = 24
473K