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Home » Thus, the total number of arrangements is \(\boxed241920\). - Abbey Badges

Thus, the total number of arrangements is \(\boxed241920\). - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Solution: The diameter of the semicircular drum is 12 meters, so the radius is half the diameter. Thus, $ r = \frac{12}{2} = 6 $ meters. The radius is $ \boxed{6} $ meters.Question: Find $ x $ such that the vectors $ \begin{pmatrix} x \\ 3 \end{pmatrix} $ and $ \begin{pmatrix} 2 \\ -x \end{pmatrix} $ are orthogonal.

📌 Solution: Two vectors are orthogonal if their dot product equals zero. Compute the dot product: $ x \cdot 2 + 3 \cdot (-x) = 2x - 3x = -x $. Set this equal to zero: $ -x = 0 $. Solving gives $ x = 0 $. \boxed{0}

📌 Question: Determine the minimum value of $ (\tan x + \cot x)^2 $ for $ x $ in the domain where $ \tan x $ and $ \cot x $ are defined.

📌 Solution: Expand $ (\tan x + \cot x)^2 = \tan^2 x + 2 \tan x \cot x + \cot^2 x $. Simplify using $ \tan x \cot x = 1 $: $ \tan^2 x + 2 + \cot^2 x $. Use identity $ \tan^2 x + \cot^2 x = (\tan x - \cot x)^2 + 2 $, so expression becomes $ (\tan x - \cot x)^2 + 4 $. The minimum occurs when $ \tan x = \cot x $, i.e., $ x = \frac{\pi}{4} + k\frac{\pi}{2} $, giving $ 0 + 4 = 4 $. \boxed{4}

📌 Question: Compute $ (\cos \theta + i \sin \theta)^5 $ using De Moivre’s theorem.

📌 Solution: By De Moivre’s theorem, $ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) $. Applying $ n = 5 $, the result is $ \cos(5\theta) + i \sin(5\theta) $. \boxed{\cos(5\theta) + i \sin(5\theta)}

📌 Question: Find the range of $ f(x) = \frac{2\sin x + 1}{\sin x - 3} $ for $ x $ where the denominator is nonzero.

📌 Solution: Let $ y = \frac{2\sin x + 1}{\sin x - 3} $. Solve for $ \sin x $: $ y(\sin x - 3) = 2\sin x + 1 \Rightarrow y\sin x - 3y = 2\sin x + 1 \Rightarrow \sin x(y - 2) = 3y + 1 $. Thus, $ \sin x = \frac{3y + 1}{y - 2} $. Since $ |\sin x| \leq 1 $, solve $ \left| \frac{3y + 1}{y - 2} \right| \leq 1 $. This leads to $ -1 \leq \frac{3y + 1}{y - 2} \leq 1 $. Sol

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ho = rac{ed}{1 + e\cos\phi} $ identifies conic type.
Rewrite:
ho = rac{4}{2 + \cos\phi} = rac{2}{1 + rac{1}{2}\cos\phi}.
This matches $
ho = rac{2}{1 + e\cos\phi} $ with $ e = rac{1}{2} $, which is **elli...
Thus, the surface is an **ellipsoid-centered ellipse (in axial cross-s...
Standard form confirms it's an **ellipsoid** (quasi-spherical distorti...
Thus, the conic section (in the context of conic classification and ph...
But the exact conic type defined by $
ho = rac{ed}{1 + e\cos\phi} $ with $ e = 1/2 $ is **elliptical**.

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