The value of \[ \left(\frac{10i}{(2-i)(3-i)}\right)^{2024} \] is equal to:
The foci of the ellipse \(\frac{x^2}{49} + \frac{y^2}{24} = 1\) are:
The value of the limit \(\lim_{t \to 0} \frac{(5-t)^2 - 25}{t}\) is equal to:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
For a hyperbola, the vertices are at \( (6, 0) \) and \( (-6, 0) \). If the foci are at \( (2\sqrt{10}, 0) \) and \( -2\sqrt{10}, 0) \), then the equation of the hyperbola is:
If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
For \(1 \leq x<\infty\), let \(f(x) = \sin^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{x}\right)\). Then \(f'(x) =\)
\[ \int_0^{\frac{\pi}{4}} (\tan^3 x + \tan^5 x) \, dx \]
If \( a = \frac{1 + \tan \theta + \sec \theta}{2 \sec \theta} \) and \( b = \frac{\sin \theta}{1 - \sec \theta + \tan \theta} \), then \( \frac{a}{b} \) is equal to:
The integral \(\int e^x \sqrt{e^x} \, dx\) equals:
The value of the limit \(\lim_{x \to 0} \frac{(2 + \cos 3x) \sin^2 x}{x \tan(2x)}\) is equal to:
If \(\sec \theta + \tan \theta = 2 + \sqrt{3}\), then \(\sec \theta - \tan \theta\) is:
An assignment of probabilities for outcomes of the sample space \( S = \{1, 2, 3, 4, 5, 6\} \) is as follows:
\[ \begin{array}{c|c c c c c c} 1 & 2 & 3 & 4 & 5 & 6 \\ \hline k & 3k & 5k & 7k & 9k & 11k \end{array} \]
If this assignment is valid, then the value of \( k \) is: