ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:03Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-n²) dn = √π / 2
302K
CH₄ + 2O₂ → CO₂ + 2H₂O
312K
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
∑_{k=1}^{n} k = n(n+1)/2
17K
∑_{k=1}^{n} k = n(n+1)/2
311K
det| 9 4 ; 1 10 | = 86
364K