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Home » Muk’s Greatest Weakness On Display — The Devastating Secret Revealed Now! - Baxtercollege

Muk’s Greatest Weakness On Display — The Devastating Secret Revealed Now! - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Find intersection with \(y = 2x + 1\):

📌 2x + 1 = -\frac{1}{2}x + \frac{11}{2}

📌 Multiply both sides by 2:

📌 4x + 2 = -x + 11

📌 \Rightarrow 5x = 9 \Rightarrow x = \frac{9}{5}

📌 y = 2 \cdot \frac{9}{5} + 1 = \frac{18}{5} + \frac{5}{5} = \frac{23}{5}

📌 Thus, the closest point is \(\boxed{\left( \frac{9}{5}, \frac{23}{5} \right)}\).

📌 Compute the maximum value of \((\sec x + \csc x)^2\) for \(0 < x < \frac{\pi}{2}\).

🔥 Popular Posts

x^2\left( \frac{1}{100} - \frac{1}{64} \right) = 0
x^2 \left( \frac{64 - 100}{6400} \right) = 0 \Rightarrow x^2(-36/6400)...
So \( x = 0 \) is the only solution — but this implies the circle is t...
At \( x = 0 \): \( y^2 = 64 \), so \( y = \pm8 \).
Check in ellipse: \( 0 + \frac{64}{64} = 1 \) — yes.
Now check if other intersections exist:
The circle has radius 8, ellipse semi-minor axis is 8 — so the circle...
But does it intersect in more points? No — since the circle lies entir...
Actually, the circle is inscribed: at \( x=0 \), both have \( y = \pm8...
Try solving numerically:

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