1. **Problem Statement:** Determine if the given graph with vertices A, B, C, D, F and edges as described is Eulerian. If yes, find the Euler circuit; if no, explain why. 2. **Eulerian Graph Criteria:** A graph is Eulerian if it is connected and every vertex has an even degree (an even number of edges). 3. **Calculate the degree of each vertex:** - Vertex A connects to F, B, C → degree 3 (odd) - Vertex F connects to A, B, D → degree 3 (odd) - Vertex B connects to A, F, C, D → degree 4 (even) - Vertex C connects to A, B, D → degree 3 (odd) - Vertex D connects to F, B, C → degree 3 (odd) 4. **Check degrees:** Vertices A, F, C, and D have odd degrees (3), only B has even degree (4). 5. **Conclusion:** Since more than two vertices have odd degree, the graph is **not Eulerian**. An Euler circuit requires all vertices to have even degree. 6. **Additional note:** A graph with exactly two vertices of odd degree has an Euler path but not an Euler circuit. Here, there are four odd-degree vertices, so neither an Euler path nor circuit exists. **Final answer:** The graph is not Eulerian because it has four vertices with odd degree, which violates the Euler circuit condition.