Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:34Z

11n² + 5n + 4 = 0

∑_{k=1}^{n} k = n(n+1)/2

460K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

438K

∫₀^∞ e^(-x²) dx = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

186K

e^{i\pi} + 1 = 0

411K

4φ² + 6φ + 2 = 0

∑_{k=1}^{n} k = n(n+1)/2

448K

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

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

det| 5 2 ; 3 6 | = 24

459K

ω = 2πf

406K

∇²ω = 0

det| 9 4 ; 1 10 | = 86

181K

10γ² + 8γ + 7 = 0

80K

d/dr [r^2] = 2r^1

F = G m₁m₂ / r²

359K

ΔS ≥ 0