ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:05Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
∫₀^∞ e^(-x²) dx = √π / 2
125K
d/dφ [φ^6] = 6φ^5
det| 11 9 ; 0 12 | = 132
17K
det| 9 5 ; 1 10 | = 85
310K
6n² + 3n + 0 = 0
10β² + 6β + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
39K
F = ma