ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-17 23:17Z
\lim_{β\to 0} \frac{\sin β}{β} = 1
AgNO₃ + NaCl → AgCl↓ + NaNO₃
211K
∫₀^∞ e^(-α²) dα = √π / 2
∑_{k=1}^{n} k = n(n+1)/2
311K
7y² + 6y + 1 = 0
det| 7 4 ; 4 8 | = 40
177K
E = mc²
\lim_{φ\to 0} \frac{\sin φ}{φ} = 1
45K
\lim_{α\to 0} \frac{\sin α}{α} = 1
207K
∇²y = 0
AgNO₃ + NaCl → AgCl↓ + NaNO₃
6r² + 8r + 4 = 0