Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:59Z

11n² + 5n + 4 = 0

∇²β = 0

460K

∑_{k=1}^{n} k = n(n+1)/2

438K

∫₀^∞ e^(-x²) dx = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

186K

∇²ω = 0

411K

4φ² + 6φ + 2 = 0

det| 9 4 ; 1 10 | = 86

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

5z² + 8z + 3 = 0

459K

e^{i\pi} + 1 = 0

406K

∇²ω = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

181K

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

80K

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

ΔS ≥ 0

8β² + 5β + 0 = 0

det| 12 6 ; 6 13 | = 120