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NEW QUESTION: 1
네트워크에 contoso.com이라는 Active Directory 도메인이 있습니다. 도메인에는 Server1과 Server2라는 두 서버가 있습니다. Server1은 Windows Server 2012 R2를 실행합니다. Server2는 Windows Server 2008 R2 서비스 팩 1 (SP1)을 실행하며 DHCP 서버 서버 역할이 설치되어 있습니다.
Server1의 DHCP 콘솔을 사용하여 Server2의 DHCP를 관리해야 합니다.
먼저 무엇을 해야 합니까?
A. Server1의 서버 관리자에서 기능을 설치하십시오.
B. Server2의 Internet Explorer에서 Windows Management Framework 3.0을 다운로드하여 설치하십시오.
C. Server2의 Windows PowerShell에서 PSRemoting 사용을 실행합니다.
D. Server2의 고급 보안이 설정된 Windows 방화벽에서 인바운드 규칙을 만듭니다.
Answer: A
Explanation:
설명
Enable-PSRemoting cmdlet은 WS-Management 기술을 사용하여 전송 된 Windows PowerShell 원격 명령을 받도록 컴퓨터를 구성합니다. Windows Server 2012 R2에서는 Windows PowerShell 원격이 기본적으로 사용됩니다. Enable-PSRemoting을 사용하여 지원되는 다른 버전의 Windows에서 Windows PowerShell 원격을 사용하도록 설정하고 Windows Server 2012에서 원격을 사용하지 않도록 설정 한 경우 원격을 다시 사용하도록 설정할 수 있습니다. 명령을 수신 할 각 컴퓨터에서 이 명령을 한 번만 실행하면 됩니다. 명령 만 보내는 컴퓨터에서는 실행할 필요가 없습니다. 구성은 리스너를 활성화하므로 필요한 곳에서만 실행하는 것이 좋습니다.
참고 : (B 아님) 서버 관리자를 사용하여 Windows Server 2008 및 Windows Server 2008 R2를 실행하는 원격 서버를 관리 할 수 있지만 이러한 이전 운영 체제를 완전히 관리하려면 다음 업데이트가 필요합니다.
NEW QUESTION: 2
Two cars are driving towards one another. The first car is traveling at a speed of 120 Km/h, which is 28% faster than the second car. If the distance between the cars is 855 Km, how long will it takes the cars to meet (in hours)?
A. 4.
B. 4.5.
C. 3.5.
D. 3.
E. 2.5.
Answer: A
Explanation:
Explanation/Reference:
Explanation:
The speed of the second car is X, (X + 0.28X = 120) X = 93.75 Km/h.
In order to find the time it will take the cars to meet, you should divide the total distance by the sum of the car's speeds: (855 / (120+93.75) = 4). Therefore the answer is D, four hours.
NEW QUESTION: 3
A developer is writing an application with three java Persistence API entities: order, customer, and Address. There is a many-to-one relationship between order and customer, and a one to-many relationship between customer and Address.
Which two Criteria queries will return the orders of all customers who have an address whose value specified by the String parameter postalcode? (Choose two)
A. String postalCode = . . .
Criteria Builder cb = . . .
CriteriaQuery<order> cq = cb.createQuery (Order.class);
Root <order> order = cq.from(order.class);
Join <order, Customer> customer = order.join(Order_.customer);
Root <Order> order = cq.from (Order.class);
Join <customer, Address> address = customer join (Order_.customer)
cq.where (cb.equal (address.get(Address_.postalCode), postalCode));
cq.select (order). Distinct (true);
// query execution code here
.. .
B. String postalCode = ...
CriteriaBuilder cb = ...
Root<order> order = cq.from (Order . class ) ,
Join<order, Address> address = order . join (Order_. customer).join (Customer_.addresses); cq.where <cb.equal (address.get(Address_.postalCode) , postalCode) ) ; cq.selec:(order).distinct (true);
// query execution code here
.. .
C. String postalCode = . . .
Criteria Builder cb = . . .
Root <Order> order = cq.from (Order.class);
order.join (order_. customer).join(Customer_.addresses);
cq.where (cb.equal (address.get(Address_.postalCode), postalCode));
cq.select (order). Distinct (true);
// query execution code here
D. String postalCode = ...
CriteriaBuilder cb = ...
Root<order> order = cq.from (Order . class) ,
Join<order, Address> address = order.join(Customer_.addresses);
cq.where(ct>.equal(address.get(Address_.postalCode), postalCode));
cq-select(order).distinct(true);
// query execution code here
.. .
Answer: A,B
Explanation:
Explanation/Reference:
A: Join Order and Customer and join Customer and Address. Works fine.
Not B: Chained joined not set up correctly.
C: Join Order and Address through first joining Order and Customer.
Not D: Cannot Join Order Address it just one single join. Need to chain the join.
Note: Querying Relationships Using Joins
For queries that navigate to related entity classes, the query must define a join to the related entity by calling one of the From.join methods on the query root object or another join object. The join methods are similar to the JOIN keyword in JPQL.
The target of the join uses the Metamodel class of type EntityType<T> to specify the persistent field or property of the joined entity.
The join methods return an object of type Join<X, Y>, where X is the source entity and Y is the target of the join. In the following code snippet, Pet is the source entity, Owner is the target, and Pet_ is a statically generated metamodel class:
CriteriaQuery<Pet> cq = cb.createQuery(Pet.class);
Root<Pet> pet = cq.from(Pet.class);
Join<Pet, Owner> owner = pet.join(Pet_.owners);
Joins can be chained together to navigate to related entities of the target entity without having to create a Join<X, Y> instance for each join:
CriteriaQuery<Pet> cq = cb.createQuery(Pet.class);
Root<Pet> pet = cq.from(Pet.class);
Join<Owner, Address> address = cq.join(Pet_.owners).join(Owner_.addresses); Reference: Using the Criteria API and Metamodel API to Create Basic Typesafe Queries