Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 12:58Z
2ψ² + 5ψ + 4 = 0
N₂ + 3H₂ ⇌ 2NH₃
∑_{k=1}^{n} k = n(n+1)/2
276K
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
129K
\lim_{φ\to 0} \frac{\sin φ}{φ} = 1
det| 7 4 ; 4 8 | = 40
191K
∑_{k=1}^{n} k = n(n+1)/2
390K
∇²z = 0
N₂ + 3H₂ ⇌ 2NH₃
8t² + 2t + 3 = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
188K
\lim_{θ\to 0} \frac{\sin θ}{θ} = 1
F = ma
∫₀^∞ e^(-φ²) dφ = √π / 2