Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:01Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

∇ × E = −∂B/∂t

438K

∫₀^∞ e^(-x²) dx = √π / 2

det| 11 5 ; 0 12 | = 132

186K

det| 2 3 ; 5 3 | = -9

411K

4φ² + 6φ + 2 = 0

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

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

6r² + 3r + 1 = 0

459K

8ψ² + 7ψ + 0 = 0

406K

∑_{k=1}^{n} k = n(n+1)/2

181K

∑_{k=1}^{n} k = n(n+1)/2

80K