1. **Stating the problem:** We are given a weighted graph with edges and their weights: $A-B=4$, $A-C=3$, $B-C=2$, $B-D=5$, $C-D=7$, $C-E=8$, $D-E=6$. We want to find the total weight of the Minimum Spanning Tree (MST) starting from vertex $A$. 2. **Formula and rules:** The MST is a subset of edges connecting all vertices with the minimum total edge weight and no cycles. We can use Prim's algorithm starting from $A$. 3. **Step-by-step MST construction using Prim's algorithm:** - Start at $A$. The edges from $A$ are $A-B=4$ and $A-C=3$. Choose the smallest edge: $A-C=3$. - Now vertices in MST: $A, C$. Edges from these to outside vertices: $A-B=4$, $C-B=2$, $C-D=7$, $C-E=8$. Choose smallest: $C-B=2$. - MST vertices: $A, C, B$. Edges to outside: $B-D=5$, $C-D=7$, $C-E=8$. Choose smallest: $B-D=5$. - MST vertices: $A, B, C, D$. Edges to outside: $C-E=8$, $D-E=6$. Choose smallest: $D-E=6$. - MST vertices: $A, B, C, D, E$. All vertices included. 4. **Calculate total MST weight:** $$3 + 2 + 5 + 6 = 16$$ 5. **Answer:** The total weight of the MST starting from $A$ is **16**.