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nintendo switch 2 black friday - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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🔥 Popular Posts

A rectangular field has a length 4 meters more than twice its width. I...
Let width = \( w \). Then length = \( 2w + 4 \).
Area: \( w(2w + 4) = 104 \)
Expand: \( 2w^2 + 4w = 104 \)
Divide by 2: \( w^2 + 2w = 52 \)
Bring to standard form: \( w^2 + 2w - 52 = 0 \)
\( w = \frac{-2 \pm \sqrt{2^2 - 4(1)(-52)}}{2} = \frac{-2 \pm \sqrt{4...
\( \sqrt{212} = \sqrt{4 \times 53} = 2\sqrt{53} \approx 2 \times 7.28...
\( w = \frac{-2 + 14.56}{2} = \frac{12.56}{2} = 6.28 \)
Width ≈ 6.28 meters.

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Subtract from (4):
Now from (5): $ b = 2 - 3a $
Substitute into (6):
12a + 3(2 - 3a) = 10 \Rightarrow 12a + 6 - 9a = 10 \Rightarrow 3a = 4...
Then $ b = 2 - 3 \cdot \frac{4}{3} = 2 - 4 = -2 $
From (2): $ \frac{4}{3} - 2 + c = 3 \Rightarrow c = 3 + 2 - \frac{4}{3...
And $ d = 2 $
p(x) = \frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2
Thus, the remainder when $ p(x) $ is divided by $ x^4 - 1 $ is simply...
\boxed{\frac{4}{3}x^3 - 2x^2 + \frac{11}{3}x + 2}

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