Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:56Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

det| 7 4 ; 4 8 | = 40

438K

∫₀^∞ e^(-x²) dx = √π / 2

11φ² + 4φ + 0 = 0

186K

N₂ + 3H₂ ⇌ 2NH₃

411K

4φ² + 6φ + 2 = 0

∑_{k=1}^{n} k = n(n+1)/2

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

∂γ/∂θ = 2γ

459K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

406K

ΔS ≥ 0

181K

∂ψ/∂β = 3ψ

80K

F = G m₁m₂ / r²

359K