Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:19Z

11n² + 5n + 4 = 0

ΔS ≥ 0

460K

∫₀^∞ e^(-z²) dz = √π / 2

438K

∫₀^∞ e^(-x²) dx = √π / 2

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

186K

∑_{k=1}^{n} k = n(n+1)/2

411K

4φ² + 6φ + 2 = 0

∂γ/∂θ = 2γ

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

det| 9 4 ; 1 10 | = 86

406K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

181K

9α² + 7α + 5 = 0

80K

10γ² + 8γ + 7 = 0

359K

∑_{k=1}^{n} k = n(n+1)/2

455K