ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:58Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
det| 6 4 ; 4 7 | = 26
17K
det| 1 1 ; 5 2 | = -3
310K
6n² + 3n + 0 = 0
det| 4 8 ; 0 5 | = 20
264K
10β² + 6β + 6 = 0
CH₄ + 2O₂ → CO₂ + 2H₂O
68K
det| 9 4 ; 1 10 | = 86
39K
∑_{k=1}^{n} k = n(n+1)/2
364K