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Home » a^4 r^6 = 20^4 \Rightarrow (ar^1.5)^4 = 20^4 \Rightarrow ar^1.5 = 20 - jntua results

a^4 r^6 = 20^4 \Rightarrow (ar^1.5)^4 = 20^4 \Rightarrow ar^1.5 = 20 - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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So \( z^3 = \omega \) or \( z^3 = \omega^2 \), where \( \omega = e^{2\...
We now solve \( z^3 = \omega \) and \( z^3 = \omega^2 \), each giving...
Let’s find the roots of \( z^3 = \omega \). Write \( \omega = e^{2\pi...
z = \omega^{1/3} \cdot e^{2\pi i k / 3} = e^{2\pi i/9} \cdot e^{2\pi i...
Similarly, for \( z^3 = \omega^2 = e^{-2\pi i/3} \):
z = e^{-2\pi i/9} \cdot e^{2\pi i k / 3} = e^{-2\pi i (1 - 3k)/9} = e^...
So all roots have arguments:
From first set: \( \theta = \frac{2\pi(3k+1)}{9} \), \( k=0,1,2 \) → \...
From second set: \( \theta = \frac{2\pi(3k+2)}{9} \), \( k=0,1,2 \) →...
List the real parts: \( \cos\theta \) for each:

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Dividiendo por 9, obtenemos \( x = 20 \).
Así, los ángulos son \( 2x = 40^\circ \), \( 3x = 60^\circ \) y \( 4x...
#### 40, 60, 80
Una ecuación cuadrática \( x^2 - 5x + 6
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