ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:01Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
det| 9 4 ; 1 10 | = 86
264K
10β² + 6β + 6 = 0
F = ma
det| 11 5 ; 0 12 | = 132
6K
det| 10 8 ; 4 11 | = 78