Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:39Z

11n² + 5n + 4 = 0

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

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

186K

CH₄ + 2O₂ → CO₂ + 2H₂O

411K

4φ² + 6φ + 2 = 0

C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

448K

11φ² + 4φ + 0 = 0

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

459K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

406K

∇²ω = 0

8ψ² + 7ψ + 0 = 0

181K

d/dα [α^3] = 3α^2

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

ΔS ≥ 0