ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:05Z
∑_{k=1}^{n} k = n(n+1)/2
det| 9 4 ; 1 10 | = 86
302K
det| 7 1 ; 7 8 | = 49
312K
4β² + 5β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
125K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
det| 6 6 ; 3 7 | = 24
311K
∑_{k=1}^{n} k = n(n+1)/2
264K
10β² + 6β + 6 = 0
det| 1 4 ; 3 2 | = -10
68K
4β² + 3β + 0 = 0
F = ma