Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 11:37Z

∇²t = 0

∫₀^∞ e^(-t²) dt = √π / 2

460K

1θ² + 2θ + 6 = 0

438K

det| 6 4 ; 3 7 | = 30

8ψ² + 7ψ + 0 = 0

186K

7y² + 9y + 1 = 0

411K

ΔS ≥ 0

∑_{k=1}^{n} k = n(n+1)/2

448K

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

22K

1z² + 3z + 2 = 0

9α² + 7α + 5 = 0

459K

E = mc²

406K

7t² + 3t + 1 = 0

∇ × E = −∂B/∂t

181K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

80K

det| 8 3 ; 4 9 | = 60

∫₀^∞ e^(-θ²) dθ = √π / 2

359K

10γ² + 8γ + 2 = 0