Math Dump /sym/6c3138

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

11n² + 5n + 4 = 0

ω = 2πf

460K

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

438K

∫₀^∞ e^(-x²) dx = √π / 2

λ = h/p

186K

∇²β = 0

411K

4φ² + 6φ + 2 = 0

N₂ + 3H₂ ⇌ 2NH₃

448K

∑_{k=1}^{n} k = n(n+1)/2

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

\oint_C \vec{F}\cdot d\vec{r} = 0

459K

d/dφ [φ^1] = 1φ^0

406K

∇²ω = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

181K

11φ² + 4φ + 0 = 0

80K

d/dr [r^2] = 2r^1

det| 9 4 ; 1 10 | = 86

359K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

455K