Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-17 22:41Z

N₂ + 3H₂ ⇌ 2NH₃

∑_{k=1}^{n} k = n(n+1)/2

246K

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

118K

9ψ² + 4ψ + 2 = 0

iħ ∂ψ/∂t = Ĥψ

43K

7r² + 8r + 5 = 0

261K

d/dβ [β^6] = 6β^5

∑_{k=1}^{n} k = n(n+1)/2

424K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

76K

det| 6 1 ; 4 7 | = 38

det| 9 4 ; 1 10 | = 86

41K

∑_{k=1}^{n} k = n(n+1)/2

298K

\frac{6}{5} \sum_{i=1}^{n} i^{2}