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
125K
d/dφ [φ^6] = 6φ^5
det| 7 4 ; 4 8 | = 40
310K
6n² + 3n + 0 = 0
det| 5 6 ; 4 6 | = 6
311K
10β² + 6β + 6 = 0
4β² + 3β + 0 = 0
F = ma
d/dα [α^1] = 1α^0
det| 10 8 ; 4 11 | = 78