Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 14:35Z

3γ² + 8γ + 4 = 0

E = mc²

460K

∇ × E = −∂B/∂t

438K

PV = nRT

∇²z = 0

186K

∫₀^∞ e^(-x²) dx = √π / 2

411K

det| 1 9 ; 5 2 | = -43

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

∫₀^∞ e^(-θ²) dθ = √π / 2

22K

ω = 2πf

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

459K

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

406K

det| 7 4 ; 4 8 | = 40

181K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

80K

PV = nRT

359K

F = G m₁m₂ / r²

455K