ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:13Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
91K
d/dφ [φ^6] = 6φ^5
CH₄ + 2O₂ → CO₂ + 2H₂O
17K
det| 7 4 ; 4 8 | = 40
310K
6n² + 3n + 0 = 0
10β² + 6β + 6 = 0