Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 12:41Z

∑_{k=1}^{n} k = n(n+1)/2

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

460K

∇ × E = −∂B/∂t

438K

4y² + 7y + 7 = 0

8ψ² + 7ψ + 0 = 0

186K

CH₄ + 2O₂ → CO₂ + 2H₂O

411K

∇²t = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

448K

\frac{2}{2} \sum_{i=1}^{n} i^{2}

22K

6β² + 2β + 2 = 0

det| 5 2 ; 3 6 | = 24

459K

∑_{k=1}^{n} k = n(n+1)/2

406K

2φ² + 2φ + 5 = 0

5α² + 7α + 6 = 0

\lim_{n\to 0} \frac{\sin n}{n} = 1