Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:33Z

11n² + 5n + 4 = 0

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

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

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

186K

det| 9 8 ; 3 10 | = 66

411K

4φ² + 6φ + 2 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

E = mc²

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

10ψ² + 5ψ + 0 = 0

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