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NEW QUESTION: 1
After reviewing the EVN configuration, a network administrator notices that a predefined EVN, which is known as "vnet global" was configured.
What is the purpose of this EVN?
A. It aggregates and carries all dot1q tagged traffic.
B. It refers to the global routing context and corresponds to the default RIB.
C. It safeguards the virtual network that is preconfigured to avoid mismatched routing instances
D. It defines the routing scope for each particular EVN edge interface.
Answer: B

NEW QUESTION: 2
Which are the two advantages of having Direct Internet Access? (Choose two.)
A. Secure WAN transport
B. Limits guest Wi-Fi at the branch
C. Leverage local internet path
D. Threat mitigation techniques
E. Protects remote employees
Answer: A,C
Explanation:
https://www.cisco.com/c/en/us/products/collateral/security/router-security/white-paper-c11-
736660.html

NEW QUESTION: 3
ZIEHEN UND LOSLASSEN
Sie haben einen Verfügbarkeitssatz mit dem Namen AS1, der drei virtuelle Maschinen mit den Namen VM1, VM2 und VM3 enthält.
Sie versuchen, VM1 neu zu konfigurieren, um eine größere Größe zu verwenden. Der Vorgang schlägt fehl und Sie erhalten eine Zuordnungsfehlermeldung.
Sie müssen sicherstellen, dass die Größenänderung erfolgreich ist.
Welche drei Aktionen sollten Sie nacheinander ausführen? Verschieben Sie zum Beantworten die entsprechenden Aktionen aus der Liste der Aktionen in den Antwortbereich und ordnen Sie sie in der richtigen Reihenfolge an.

Answer:
Explanation:

NEW QUESTION: 4
The network administrator is asked to configure 113 point-to-point links. Which IP addressing scheme defines the address range and subnet mask that meet the requirement and waste the fewest subnet and host addresses?
A. 10.10.0.0/23 subnetted with mask 255.255.255.252
B. 10.10.0.0/16 subnetted with mask 255.255.255.252
C. 10.10.1.0/24 subnetted with mask 255.255.255.252
D. 10.10.1.0/25 subnetted with mask 255.255.255.252
E. 10.10.0.0/18 subnetted with mask 255.255.255.252
Answer: A
Explanation:
We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7
bits (because 2