Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:55Z

11n² + 5n + 4 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

460K

PV = nRT

438K

∫₀^∞ e^(-x²) dx = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

186K

det| 9 4 ; 1 10 | = 86

411K

4φ² + 6φ + 2 = 0

8θ² + 8θ + 6 = 0

448K

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

22K

∫₀^∞ e^(-φ²) dφ = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

459K

\lim_{ω\to 0} \frac{\sin ω}{ω} = 1

406K

10t² + 9t + 5 = 0

181K