Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:55Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

∫₀^∞ e^(-t²) dt = √π / 2

438K

∫₀^∞ e^(-x²) dx = √π / 2

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

186K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

411K

4φ² + 6φ + 2 = 0

6r² + 3r + 1 = 0

448K

det| 9 4 ; 1 10 | = 86

22K

PV = nRT

459K

∑_{k=1}^{n} k = n(n+1)/2

406K

9α² + 7α + 5 = 0

181K