The value of the limit \(\lim_{t \to 0} \frac{(5-t)^2 - 25}{t}\) is equal to:
The vectors \(\vec{a} = 4\mathbf{i} - 3\mathbf{j} - \mathbf{k}\) and \(\vec{b} = 3\mathbf{i} + 2\mathbf{j} + \lambda\mathbf{k}\) are perpendicular to each other. Then the value of \(\lambda\) is equal to:
Let \( f(x) = \begin{cases} x^2 - \alpha, & \text{if } x < 1 \\ \beta x - 3, & \text{if } x \geq 1 \end{cases} \). If \( f \) is continuous at \( x = 1 \), then the value of \( \alpha + \beta \) is:
Let \( f(x) = 2 - 7 \sin{\left( \frac{2x}{7} \right)} \). Then the maximum value of \( f(x) \) is:
If a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive directions of the x, y, and z-axis respectively, then \(\cos 2\alpha + \cos 2\beta + \cos 2\gamma\) equals:
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:
If \( 0 \leq \alpha \leq \frac{\pi}{2} \) and \(\sin \left(\alpha - \frac{\pi}{12}\right) = \frac{1}{2}\), then \(\alpha\) is equal to:
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:
If \( 0 \leq x \leq 5 \), then the greatest value of \( \alpha \) and the least value of \( \beta \) satisfying the inequalities \( \alpha \leq 3x + 5 \leq \beta \) are, respectively,
The value of the limit \(\lim_{x \to 0} \frac{(2 + \cos 3x) \sin^2 x}{x \tan(2x)}\) is equal to:
The foci of the ellipse \(\frac{x^2}{49} + \frac{y^2}{24} = 1\) are: