Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 07:47Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
326K
∫₀^∞ e^(-t²) dt = √π / 2
198K
7r² + 8r + 6 = 0
det| 9 4 ; 1 10 | = 86
155K
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
366K
det| 10 9 ; 3 11 | = 83
2ψ² + 6ψ + 0 = 0
∫₀^∞ e^(-φ²) dφ = √π / 2
359K
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
263K
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
233K