Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-19 00:57Z

d/dy [y^8] = 8y^7

∑_{k=1}^{n} k = n(n+1)/2

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

264K

\lim_{x\to 0} \frac{\sin x}{x} = 1

6r² + 3r + 1 = 0

297K

d/dt [t^3] = 3t^2

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d/dn [n^3] = 3n^2

iħ ∂ψ/∂t = Ĥψ

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iħ ∂ψ/∂t = Ĥψ

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6t² + 4t + 0 = 0

∂y/∂γ = 8y

377K

\lim_{β\to 0} \frac{\sin β}{β} = 1

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∇ × E = −∂B/∂t

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∫₀^∞ e^(-t²) dt = √π / 2

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8ψ² + 7ψ + 0 = 0

109K