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NEW QUESTION: 1
How does EIGRP derive the metric for manual summary routes?
A. It uses the best metric vectors of all component routes in the topology table.
B. It uses the worst composite metric of any component route in the topology table.
C. It uses the worst metric vectors of all component routes in the topology table.
D. It uses the best composite metric of any component route in the topology table.
Answer: D
Explanation:
For example if your router has a routing table like this: D 192.168.8.0/24 [90/2632528] via 192.168.0.1, 00:00:12, Serial0/0 D 192.168.9.0/24 [90/2323456] via 192.168.0.1, 00:00:12, Serial0/0
D 192.168.10.0/24 [90/2195456] via 192.168.0.1, 00:00:12, Serial0/0 D 192.168.11.0/24 [90/2323456] via 192.168.0.1, 00:00:12, Serial0/0 Now suppose you want to manually summarize all the routes above, you can use this
command (on the router that advertised these routes to our router):
Router(config-if)#ip summary-address eigrp 1 192.168.8.0 255.255.248.0
After that the routing table of your router will look like this:
D 192.168.8.0/21 [90/2195456] via 192.168.0.1, 00:01:42, Serial0/0
And we can see the manual summary route takes the smallest metric of the specific routes.

NEW QUESTION: 2
You are working for a company that designs mobile applications. They maintain a server where player records are assigned to their different games. The tracking system is new and in development.
The application uses Entity Framework to connect to an Azure Database. The database holds a Player table and Game table.
When adding a player, the code should insert a new player record, and add a relationship between an existing game record and the new player record.
The application will call CreatePlayerWithGame with the correct gameIdand the playerId to start the process. (Line numbers are included for reference only.)

For each of the following statements, select Yes if the statement is true. Otherwise, select No.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation:
Many-to-many relationships without an entity class to represent the join table are not yet supported. However, you can represent a many-to-many relationship by including an entity class for the join table and mapping two separate one-to-many relationships.
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<PostTag>()
.HasKey(t => new { t.PostId, t.TagId });
modelBuilder.Entity<PostTag>()
.HasOne(pt => pt.Post)
.WithMany(p => p.PostTags)
.HasForeignKey(pt => pt.PostId);
modelBuilder.Entity<PostTag>()
.HasOne(pt => pt.Tag)
.WithMany(t => t.PostTags)
.HasForeignKey(pt => pt.TagId);
}
}

NEW QUESTION: 3
You are configuring a switch so that it accepts traffic from a maximum of two dynamic MAC addresses. Drag and drop the required configuration commands on the left into the correct sequence on the right. (Not all commands are used.)

Answer:
Explanation:

NEW QUESTION: 4
You are starting to understand the needed package dependencies for chaincode. What are the two required packages? (Select two.)
A. Peer Protobuf
B. Shim
C. Init
D. Start
Answer: A,B