The solution set of the inequation $\frac{7+5x}{4} \geq \frac{x}{3} - 10$, $x \in \mathbb{R}$ is :

Let $P = \begin{bmatrix} 5 & 2 \\ 7 & 4 \end{bmatrix}$, $Q = \begin{bmatrix} 2 & 5 \\ 3 & 8 \end{bmatrix}$, $R = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$, then the matrix S such that QS - RP = 0 will be :

If $-\frac{1}{3x - 5} \leq 0$, then :

If a square matrix B satisfies $B^2 = I - B$ and $B^n = 5I - 8B$, then the value of n is :

If the matrix $A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix}$ satisfies the equation $A^T A = I_3$, then $x^2 + y^2 + z^2$ is :

Which of the following are true for the given graph.

(A) Break even point is at x = 10

(B) Break even point is at x = 8

(C) Break even point is at x = 6

(D) Fixed cost is 20

(E) Fixed cost is 0

Choose the correct answer from the options :

If $A=\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$, then $A^{1999}$ is equal to