Answer:
Virtual storage configuration and management refers to the process of configuring and managing the storage resources of a virtualized environment. This typically involves creating and configuring virtual disks, assigning storage space to virtual machines, and managing the allocation of storage resources among different virtual machines.
Virtual machine management, on the other hand, refers to the process of managing the virtual machines in a virtualized environment. This typically involves creating and configuring virtual machines, installing and configuring operating systems and applications on the virtual machines, and managing the allocation of computing resources such as CPU, memory, and network bandwidth among different virtual machines.
An example of virtual storage configuration and management would be creating a virtual disk with a specified size and assigning it to a virtual machine to use as its primary storage. An example of virtual machine management would be creating a new virtual machine with a specified operating system and allocated resources, and then installing and configuring applications on it.
Both virtual storage configuration and management and virtual machine management are important tasks in a virtualized environment, as they help to ensure that the virtual machines have access to the necessary storage and computing resources to function properly. These tasks are typically performed by system administrators or other IT professionals who have expertise in virtualization and IT infrastructure management.
Electric switches are very important because all the equipment we are using is useless without a switch.
<span>END POINT : Snaps to the closest endpoint or corner of a geometric object. MID POINT : Snaps to the midpoint of a geometric object. CENTER : Snaps to the center of an arc, circle, ellipse, or elliptical arc.</span>
Answer:
True: In binary search algorithm, we follow the below steps sequentially:
Input: A sorted array B[1,2,...n] of n items and one item x to be searched.
Output: The index of x in B if exists in B, 0 otherwise.
- low=1
- high=n
- while( low < high )
- { mid=low + (high-low)/2
- if( B[mid]==x)
- {
- return(mid) //returns mid as the index of x
- }
- else
- {
- if( B[mid] < x) //takes only right half of the array
- {
- low=mid+1
- }
- else // takes only the left half of the array
- {
- high=mid-1
- }
- }
- }
- return( 0 )
Explanation:
For each iteration the line number 11 or line number 15 will be executed.
Both lines, cut the array size to half of it and takes as the input for next iteration.