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NEW QUESTION: 1
An advantage of native compiling over cross compiling is that:
A. Build scripts can detect details of the target, and automatically configure the build to match.
B. It allows greater parallelism when building code by utilizing many processors.
C. It can enable the final code to be smaller, and execute more quickly.
D. The compiler is able to produce error and warning messages in a range of languages.
Answer: A

NEW QUESTION: 2
JS Industriesは、最初のリモートオフィスを追加してビジネスを拡大しました。リモートオフィスルーター（R3）は以前に構成されており、すべての企業サブネットはR3から到達可能です。 JS Industriesは、EIGRPスタブルーティング機能とともにルートサマライズを使用して、ネットワークの安定性を高め、メモリ使用量と帯域幅使用率をR3に削減することに関心があります。別のネットワーク専門家がこのソリューションの実装を任されました。ただし、R3から離れたリモートネットワークデバイスとのEIGRPスタブルーティング接続を設定するプロセスで失われました。
現在、EIGRPはネットワーク内のすべてのルーターR2、R3、およびR4で構成されています。タスクは、リモートオフィスのルーターR3との接続障害の原因を特定して解決することです。問題が解決したら、リモートオフィスルータR3のみにルート要約を設定して、タスクを完了する必要があります。
R2からR3 LANインターフェイスへのpingが成功し、R3 IPルーティングテーブルに含まれるサブネットが2つの10.0.0.0のみである場合、障害を修正しました。




Answer:
Explanation:
See the explanation part for details.
Explanation
First we have to figure out why R3 and R4 can not communicate with each other. Use the show running-config command on router R3.

[eigrp10.jpg]
Notice that R3 is configured as a stub receive-only router. The receive-only keyword will restrict the router from sharing any of its routes with any other router in that EIGRP autonomous system. This keyword will also prevent any type of route from being sent. Therefore we will remove this command and replace it with the eigrp stub command:
R3# configure terminal R3(config)# router eigrp 123 R3(config-router)# no eigrp stub receive-only R3(config-router)# eigrp stub R3(config-router)# end Now R3 will send updates containing its connected and summary routes to other routers. Notice that the eigrp stub command equals to the eigrp stub connected summary because the connected and summary options are enabled by default.
Next we will configure router R3 so that it has only 2 subnets of 10.0.0.0 network. Use the show ip route command on R3 to view its routing table:
[ee1%255B5%255D.jpg]

Because we want the routing table of R3 only have 2 subnets so we have to summary sub-networks at the interface which is connected with R3, the s0/0 interface of R4.
There is one interesting thing about the output of the show ip route shown above: the 10.2.3.0/24, which is a directly connected network of R3. We can't get rid of it in the routing table no matter what technique we use to summary the networks. Therefore, to make the routing table of R3 has only 2 subnets we have to summary other subnets into one subnet.
In the output if we don't see the summary line (like 10.0.0.0/8 is a summary...) then we should use the command ip summary-address eigrp 123 10.2.0.0 255.255.0.0 so that all the ping can work well.
In conclusion, we will use the ip summary-address eigrp 123 10.2.0.0 255.255.0.0 at the interface s0/0 of R4 to summary.
R4> enable R4# conf t
R4(config)# interface s0/0 R4(config-if)# ip summary-address eigrp 123 10.2.0.0 255.255.0.0 Now we jump back to R3 and use the show ip route command to verify the effect, the output is shown below:
[ee2%255B5%255D.jpg]

Note: Please notice that the IP addresses and the subnet masks in your real exam might be different so you might use different ones to solve this question.
Just for your information, notice that if you use another network than 10.0.0.0/8 to summary, for example, if you use the command ip summary-address eigrp 123 10.2.0.0 255.255.0.0 you will leave a /16 network in the output of the show ip route command.
[ee3%255B5%255D.jpg]

But in your real exam, if you don't see the line "10.0.0.0/8 is a summary, Null0" then you can summarize using the network 10.2.0.0/16. This summarization is better because all the pings can work well.
Finally don't forget to use the copy run start command on routers R3 and R4 to save the configurations.
R3(config-if)# end
R3# copy run start
R4(config-if)# end
R4# copy run start
If the "copy run start" command doesn't work then use "write memory".

NEW QUESTION: 3
Which two Connector modes are available for use in EasyETL? (Choose two.)
A. Iterator
B. Lookup
C. Delete
D. AddOnly
E. Call/Reply
Answer: A,D

NEW QUESTION: 4
Examine the structure of the PROGRAMS table:

Which two SQL statements would execute successfully? (Choose two.)
A. SELECT NVL(TO_CHAR(MONTHS_BETWEEN(start-date,end_date)),'Ongoing')FROM programs;
B. SELECT NVL(ADD_MONTHS(END_DATE,1)SYSDATE)FROM programs;
C. SELECT TO_DATE(NVL(SYSDATE-END_DATE,SYSDATE))FROM programs;
D. SELECT NVL(MONTHS_BETWEEN(start_date,end_date),'Ongoing')FROM programs;
Answer: A,C