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Home » Multiply the number of ways to choose the red and blue marbles: - Coaching Toolbox

Multiply the number of ways to choose the red and blue marbles: - Coaching Toolbox

📅 May 6, 2026 👤 scraface

May 06, 2026

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\[ x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
pour \( a = 1 \), \( b = -6 \), \( c = 8 \) :
\[ x = rac{6 \pm \sqrt{36 - 32}}{2} \]
\[ x = rac{6 \pm 2}{2} \]
Les racines sont \( x = 4 \) et \( x = 2 \).
Les nouvelles racines sont \( 2 imes 4 = 8 \) et \( 2 imes 2 = 4 \).
La nouvelle Ã©quation quadratique avec les racines 8 et 4 est :
\[ (x - 8)(x - 4) = 0 \]
\[ x^2 - 12x + 32 = 0 \]
Ainsi, \( a = 1 \), \( b = -12 \), \( c = 32 \).

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Thus $ n^2 = \frac{25 \cdot 9}{400} = \frac{225}{400} = \frac{9}{16} \...
If $ n = \frac{3}{4} $, $ m = -\frac{16}{9} \cdot \frac{3}{4} = -\frac...
Check $ \vec{OG} \cdot (\vec{OA} + \vec{OB}) = 0 $: Both cases satisfy...
\boxed{\left( -\dfrac{4}{3}, \dfrac{3}{4} \right)}
Question: The maximum imaginary part of a root of $ z^5 + z^3 + z + 1...
Solution: Factor the polynomial:
z^5 + z^3 + z + 1 = z^3(z^2 + 1) + 1(z + 1).
Group terms:
Roots: $ z^3 = -1 \implies z = e^{i\pi/3 + 2k\pi/3} $, and $ z^2 = -1...
The roots with maximum imaginary part are $ e^{i\pi/3} $ and $ e^{

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