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NEW QUESTION: 1
Sie haben die Datenbanken wie in der folgenden Tabelle gezeigt konfiguriert.

Welche beiden Datenbanken können die Funktion "Datenbank dehnen" verwenden? Jede richtige Antwort bietet eine vollständige Lösung.
HINWEIS; Jede richtige Auswahl ist einen Punkt wert.
A. DB4
B. DB1
C. DB2
D. DB3
Answer: B,D
Explanation:
References:
https://docs.microsoft.com/en-us/sql/sql-server/stretch-database/enable-stretch-database-for-a-database?
view=sql-server-2017

NEW QUESTION: 2

A. Option C
B. Option D
C. Option B
D. Option A
E. Option E
Answer: C
Explanation:
Explanation
The Information Technology Infrastructure Library (ITIL) is a commonly used service management practice within businesses. An organization can determine its level of maturity by verifying against one of the five maturity levels. The ITIL practice is helpful for determining maturity levels and establishing service level management practices throughout the business-led approach.
Alignment with business needs
Achievable service level agreements
Predictable and consistent processes
Service delivery efficiency
Measurable and improved processes
Establishes a common language

NEW QUESTION: 3
Case study.

Some notices from above configuration:
+The OSPF network type between R2&R3 is non broadcast.
Q1: Show ip ospf database
Q2: Show ip ospf interface serial 1/0
Q3: Show ip ospf
Q4: Show ip route
You have been asked to evaluate an OSPF network and to answer questions a customer has about its operation. Note: You are not allowed to use the show running-config command.

Although in this sim we are not allowed to use "show running-config" command but we post the configuration here so that you can understand more about the topology.
R1
interface Loopback0
ip address 1.1.1.1 255.255.255.255
no shut
interface Serial0/0
ip address 192.168.13.1 255.255.255.0
ip ospf network non-broadcast
no shut
router ospf 1
network 192.168.13.0 0.0.0.255 area 0
network 1.1.1.1 0.0.0.0 area 0
_______________________________________
R2
interface Loopback 0
ip address 2.2.2.2 255.255.255.255
no shut
interface S0/0
ip address 192.168.23.2 255.255.255.0
ip ospf network non-broadcast
no shut
router ospf 1
network 192.168.23.0 0.0.0.255 area 0
network 2.2.2.2 0.0.0.0 area 0
neighbor 192.168.23.3
_________________________________________________
R3
interface Loopback 0
ip address 3.3.3.3 255.255.255.255
no shut
interface fa0/0
ip address 192.168.34.3 255.255.255.0
no shut
interface S0/1
ip address 192.168.23.3 255.255.255.0
ip ospf network non-broadcast
no shut
interface S0/0
ip address 192.168.13.3 255.255.255.0
ip ospf network non-broadcast
no shut
router ospf 1
network 192.168.13.0 0.0.0.255 area 0
network 192.168.23.0 0.0.0.255 area 0
network 192.168.34.0 0.0.0.255 area 1
network 3.3.3.3 0.0.0.0 area 0
area 1 virtual-link 4.4.4.4
neighbor 192.168.23.2
_______________________________________________________________
R4
interface Loopback 0
ip address 4.4.4.4 255.255.255.255
interface FastEthernet0/0
ip address 192.168.34.4 255.255.255.0
interface Fa0/1
ip address 192.168.45.4 255.255.255.0
no shut
interface Fa1/0
ip address 192.168.46.4 255.255.255.0
no shut
router ospf 1
network 192.168.34.0 0.0.0.255 area 1
network 192.168.45.0 0.0.0.255 area 2
network 192.168.46.0 0.0.0.255 area 3
network 4.4.4.4 0.0.0.0 area 1
area 1 virtual-link 3.3.3.3
area 2 nssa
area 3 stub no-summary
_________________________________________________________________
R5
interface Loopback0
ip address 5.5.5.5 255.255.255.255
interface Loopback1
ip address 5.5.1.1 255.255.255.255
interface Loopback2
ip address 5.5.2.1 255.255.255.255
interface Loopback3
ip address 5.5.3.1 255.255.255.255
interface Loopback4
ip address 5.5.4.1 255.255.255.255
no shut
interface Fa0/0
ip address 192.168.45.5 255.255.255.0
no shut
router ospf 1
network 192.168.45.0 0.0.0.255 area 2
network 5.5.0.0 0.0.255.255 area 2
area 2 nssa
_______________________________________________
R6
interface Fa0/0
ip address 192.168.46.6 255.255.255.0
no shut
interface Loopback 0
ip address 6.6.6.6 255.255.255.255
no shut
router ospf 1
network 192.168.46.0 0.0.0.255 area 3
network 6.6.6.6 0.0.0.0 area 3
area 3 stub
Areas of Router 5 and 6 are not normal areas, inspect their routing tables and determine which statement is true?
A. Only R5′s loopback is present in R5′s Routing table
B. R5′s Loopback and R6′s Loopback are both present in R5′s Routing table
C. Only R5′s loopback is present in R6′s Routing table
D. R5′s Loopback and R6′s Loopback are both present in R6′s Routing table
E. Only R6′s loopback is present in R5′s Routing table
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Area 2 (of R5) is a Not-so-Stubby area (NSSA). You can check it by the "show ip ospf" command on R4 or R5 (in Area 2 section). For example, below is the output of "show ip ospf" command on R5:

In general, NSSA is same as normal area except that it can generate LSA Type 7 (redistribute from another domain) so we can see both Loopback interfaces of R5 & R6 in the routing table of R5.

Note: NSSA does not receive a default route by default so you will not see a default route on R5.
Area 3 (of R6) is a Totally-Stubby area so R6 only has one default route to outside world. You can check with the "show ip ospf" command on R4 and R6 (area 3 section):


Notice that on R4 you will get more detail (shows "stub area, no summary LSA") than on R6 (only shows
"stub area").
R6 is in a totally-stubby area so we will not see any R5's Loopback interfaces in R6 routing table:

Note: You can see a default (summary) route to the outside (O*IA 0.0.0.0/0 ...) Even though this exercise looks complicated, it can be solve with simple commands:
Q1: show ip ospf database
Q2: show ip ospf database int s0/1
Q3: Show ip ospf
Q4: show ip ospf and show ip route

NEW QUESTION: 4
What happens when you attempt to compile and run the following code?
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
using namespace std;
void myfunction(int i) {
cout << " " << i;
}
int main() {
int t[] = { 10, 5, 9, 6, 2, 4, 7, 8, 3, 1 };
set<int> s1(t, t+10);
vector<int> v1(s1.rbegin(), s1.rend());
swap(s1, v1);
for_each(v1.begin(), v1.end(), myfunction);
for_each(s1.begin(), s1.end(), myfunction);
return 0;
}
Program outputs:
A. compilation error
B. 10 9 8 7 6 5 4 3 2 1 10 9 8 7 6 5 4 3 2 1
C. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
D. 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10
Answer: A