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
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
91K
d/dφ [φ^6] = 6φ^5
∑_{k=1}^{n} k = n(n+1)/2
310K
6n² + 3n + 0 = 0
det| 9 4 ; 1 10 | = 86
264K
∑_{k=1}^{n} k = n(n+1)/2
6K