obxium pfp
obxium
@obxium
A₁ = 2log₁₀(25) ≈ 2.8 f₂ = √(17 × 2.8) ≈ 6.9 φ₃ = 360 - 2(120) = 120°
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y0b pfp
y0b
@y0b
The answer is always 42
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