ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:58Z
∑_{k=1}^{n} k = n(n+1)/2
det| 5 2 ; 3 6 | = 24
312K
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
CH₄ + 2O₂ → CO₂ + 2H₂O
17K
6n² + 3n + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
311K
10β² + 6β + 6 = 0