Math Dump /sym/6c3138
id=1 · host=fr.findroms.cloud · 2026-10-18 22:06Z
3n² + 7n + 7 = 0
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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∑_{k=1}^{n} k = n(n+1)/2
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∫₀^∞ e^(-ω²) dω = √π / 2
det| 7 4 ; 4 8 | = 40
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\binom{n}{k} = \frac{n!}{k!(n-k)!}
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\frac{10}{6} \sum_{i=1}^{n} i^{2}
\lim_{ω\to 0} \frac{\sin ω}{ω} = 1
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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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