Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:57Z

11n² + 5n + 4 = 0

e^{i\pi} + 1 = 0

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

186K

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

411K

4φ² + 6φ + 2 = 0

∑_{k=1}^{n} k = n(n+1)/2

448K

ω = 2πf

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

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

459K

6x² + 6x + 1 = 0

406K

det| 9 4 ; 1 10 | = 86

181K

e^{i\pi} + 1 = 0

80K

∑_{k=1}^{n} k = n(n+1)/2

359K