Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:03Z

11n² + 5n + 4 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

460K

∂γ/∂θ = 2γ

438K

∫₀^∞ e^(-x²) dx = √π / 2

∂y/∂n = 5y

186K

12y² + 5y + 7 = 0

411K

4φ² + 6φ + 2 = 0

d/dt [t^3] = 3t^2

448K

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

22K

∑_{k=1}^{n} k = n(n+1)/2

459K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

406K

∇²ω = 0

181K

det| 9 4 ; 1 10 | = 86

80K

N₂ + 3H₂ ⇌ 2NH₃

359K