Find the number of triangles in the given figure.
Step 1: Analyze the figure.
The given figure consists of a series of intersecting lines, forming multiple triangles. By counting the individual and combined triangles formed by these lines, we can conclude the total number.
Step 2: Conclusion.
The total number of triangles formed is 48. Hence, the correct answer is 3. 48.
A regular dodecagon (12-sided regular polygon) is inscribed in a circle of radius \( r \) cm as shown in the figure. The side of the dodecagon is \( d \) cm. All the triangles (numbered 1 to 12 in the figure) are used to form squares of side \( r \) cm, and each numbered triangle is used only once to form a square. The number of squares that can be formed and the number of triangles required to form each square, respectively, are:
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?
In C language, mat[i][j] is equivalent to: (where mat[i][j] is a two-dimensional array)
Suppose a minimum spanning tree is to be generated for a graph whose edge weights are given below. Identify the graph which represents a valid minimum spanning tree?
\[\begin{array}{|c|c|}\hline \text{Edges through Vertex points} & \text{Weight of the corresponding Edge} \\ \hline (1,2) & 11 \\ \hline (3,6) & 14 \\ \hline (4,6) & 21 \\ \hline (2,6) & 24 \\ \hline (1,4) & 31 \\ \hline (3,5) & 36 \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match LIST-I with LIST-II
Choose the correct answer from the options given below:
Consider the following set of processes, assumed to have arrived at time 0 in the order P1, P2, P3, P4, and P5, with the given length of the CPU burst (in milliseconds) and their priority:
\[\begin{array}{|c|c|c|}\hline \text{Process} & \text{Burst Time (ms)} & \text{Priority} \\ \hline \text{P1} & 10 & 3 \\ \hline \text{P2} & 1 & 1 \\ \hline \text{P3} & 4 & 4 \\ \hline \text{P4} & 1 & 2 \\ \hline \text{P5} & 5 & 5 \\ \hline \end{array}\]
Using priority scheduling (where priority 1 denotes the highest priority and priority 5 denotes the lowest priority), find the average waiting time.