Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:39Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
129K
∑_{k=1}^{n} k = n(n+1)/2
\lim_{r\to 0} \frac{\sin r}{r} = 1
∫₀^∞ e^(-ψ²) dψ = √π / 2
473K
det| 11 5 ; 0 12 | = 132
338K