Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:52Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

iħ ∂ψ/∂t = Ĥψ

438K

∫₀^∞ e^(-x²) dx = √π / 2

5z² + 8z + 3 = 0

186K

N₂ + 3H₂ ⇌ 2NH₃

411K

4φ² + 6φ + 2 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

det| 3 7 ; 2 4 | = -2

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

det| 9 8 ; 3 10 | = 66

459K

∑_{k=1}^{n} k = n(n+1)/2

406K

det| 2 8 ; 7 3 | = -50

181K

det| 7 4 ; 4 8 | = 40

80K

det| 8 9 ; 4 9 | = 36

359K