Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:00Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

CH₄ + 2O₂ → CO₂ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

∇²r = 0

186K

N₂ + 3H₂ ⇌ 2NH₃

411K

4φ² + 6φ + 2 = 0

CH₄ + 2O₂ → CO₂ + 2H₂O

448K

det| 7 4 ; 4 8 | = 40

22K

det| 5 6 ; 2 6 | = 18

459K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

406K

∑_{k=1}^{n} k = n(n+1)/2

181K

E = mc²

80K