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Home » $ a^3 = \frac79 \cdot 6241160 \cdot 25600 = \frac4930394096000 $, - Dyverse

$ a^3 = \frac79 \cdot 6241160 \cdot 25600 = \frac4930394096000 $, - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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Solve first system:
From \(a + 2b = 4\), \(a = 4 - 2b\). Plug into \(3a - b = 5\):
3(4 - 2b) - b = 12 - 6b - b = 12 - 7b = 5 \Rightarrow 7b = 7 \Rightarr...
Second system: \(c + 2d = 1\), \(3c - d = -2\)
From \(c = 1 - 2d\), plug:
3(1 - 2d) - d = 3 - 6d - d = 3 - 7d = -2 \Rightarrow 7d = 5 \Rightarro...
\mathbf{M} = egin{bmatrix} 2 & 1 \ -rac{3}{7} & rac{5}{7} \end{bmat...
oxed{egin{bmatrix} 2 & 1 \ -\dfrac{3}{7} & \dfrac{5}{7} \end{bmatrix...
Question: Find the maximum value of \((\sin x + \cos x)^2 + ( an x + \...
Solution: First, expand:

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