Can someone please show me how to do this step by step please. I dont understand
ID: 3847805 • Letter: C
Question
Can someone please show me how to do this step by step please. I dont understand any of it
1.) Add a new virtual disk
a) Edit VMware player settings to add a new virtual hard disk.
b) Start or reboot the virtual machine and the system will recognize the new disk.
Using the terminal in Linux
2) Check device and partition information
a) Use “ls -l" command to view device files in the “/dev” directory. How many files are related to hard disk devices and partitions? Display them.
b) Use “df” and “du” command to check disk usage.
c) Use the following command to view partition information. Examine the information presented.
i) fdisk -l
ii) parted /dev/sda print
d) Take a screenshot in which the second hard disk is displayed ( Screenshot 6.2-1 )
3.) Try cfdisk to manage partitions
a) Delete all partitions on the disk, and recreate a primary partition and an extended partition with 2 logical partitions.
b)Take a screenshot that shows all new partitions(Screenshot 6.2-2).
c) Format and mount new partitions to the current file system.
Explanation / Answer
A tree is by definition a connected (hence non empty) graph GG while not cycles. Here may be a proof by induction of the quantity m m of edges that each such (finite) tree has m+1m+1 nodes. (Then in fact each tree with nn nodes has n1n1 edges). If m=0m=0 (no edges) one cannot have over eleven nodes (connectivity) and no less either (non-emptyness). within the case m>0m>0 select any edge (a,b)(a,b). If there have been a path from aa to pellet while not mistreatment that edge, then one would have a cycle in GG in conjunction with that edge. thus removing the sting (a,b)(a,b) disconnects aa from pellet within the remaining graph GG. conjointly each node is connected in GG either to aa or to pellet (every path in GG passing neither through aa nor pellet remains a path in GG). thus GG may be a disjoint union of 2 trees; say they need p,qp,q edges severally. Then m=p+q+1m=p+q+1, and by induction GG (hence GG) has (p+1)+(q+1)=p+q+2=m+1(p+1)+(q+1)=p+q+2=m+1 nodes.
A tree is by definition a connected (hence non empty) graph GG while not cycles. Here may be a proof by induction of the quantity m m of edges that each such (finite) tree has m+1m+1 nodes. (Then in fact each tree with nn nodes has n1n1 edges). If m=0m=0 (no edges) one cannot have over eleven nodes (connectivity) and no less either (non-emptyness). within the case m>0m>0 select any edge (a,b)(a,b). If there have been a path from aa to pellet while not mistreatment that edge, then one would have a cycle in GG in conjunction with that edge. thus removing the sting (a,b)(a,b) disconnects aa from pellet within the remaining graph GG. conjointly each node is connected in GG either to aa or to pellet (every path in GG passing neither through aa nor pellet remains a path in GG). thus GG may be a disjoint union of 2 trees; say they need p,qp,q edges severally. Then m=p+q+1m=p+q+1, and by induction GG (hence GG) has (p+1)+(q+1)=p+q+2=m+1(p+1)+(q+1)=p+q+2=m+1 nodes.
For any finite graph the total of the degrees of the nodes is double the quantity of edges. For a tree with nn nodes and n1n1 edges it follows that each one nodes cannot have degree22. in reality either there area unit a minimum of 2 nodes of degree one one (leaves) or one among degree zero 0; the latter solely happens for n=1n=1. within the proof on top of I needed to avoid mistreatment this truth en passant, as its direct proof is a couple of troublesome because the on top of proof itself.