Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:34Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

N₂ + 3H₂ ⇌ 2NH₃

438K

∫₀^∞ e^(-x²) dx = √π / 2

det| 4 8 ; 3 5 | = -4

186K

∇²t = 0

411K

4φ² + 6φ + 2 = 0

det| 7 4 ; 4 8 | = 40

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

\lim_{t\to 0} \frac{\sin t}{t} = 1

459K

∑_{k=1}^{n} k = n(n+1)/2

406K

∫₀^∞ e^(-x²) dx = √π / 2

181K

det| 5 9 ; 7 6 | = -33

80K