Reaction Log /sym/649cc0
id=1 · host=co.gridbatch.net · 2026-10-18 23:04Z
6n² + 4n + 1 = 0
∑_{k=1}^{n} k = n(n+1)/2
278K
ΔS ≥ 0
\frac{8}{9} \sum_{i=1}^{n} i^{2}
∫₀^∞ e^(-ψ²) dψ = √π / 2
250K
d/dω [ω^10] = 10ω^9
∑_{k=1}^{n} k = n(n+1)/2
352K
∇²β = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
d/dr [r^4] = 4r^3