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Simora [160]
3 years ago
15

You are configuring two PCs for your network, PC-A is given an IP address of 192.168.1.18 and PC-B is given an IP address of 192

.168.1.33. Both PCs receivces a subnet mask of 255.255.255.240A) what is network address of PC-A?B) what is network address of PC-B?C) will these PCs be able to communicate directly wiyh each other?D) what is the highest address that can be given to PC-B that allows it to be on the same network as PC-A ?
Computers and Technology
1 answer:
melisa1 [442]3 years ago
6 0

Answer:

Explanation:

a) 192.168.1.16

b) 192.168.1.32

c) No

d) 192.168.1.30

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Demonstrate the register addressing mode for the following instructions. Also what addressing mode belongs to these instructions
kotykmax [81]

Answer:

Demonstrate the register addressing mode for the following instructions. Also what addressing mode belongs to these instructions?

1. MOV CX, [BX+DI]

2. MOV AX, ARRAY[CX]

3. MOV BX, [CX+DI+6]

4 0
3 years ago
Interactive television with video-on-demand capabilities changes how people watch television and how consumers access the Intern
Alex787 [66]

Answer:

High learning

Explanation:

High Learning means that a product require significant customer education before customers understand how the product functions and that may make the product stay longer in the introduction stage whilst the customers are being familiarize with it. Examples are microwave ovens.

6 0
3 years ago
Finish and test the following two functions append and merge in the skeleton file:
avanturin [10]

Answer:

Explanation:

#include <iostream>

using namespace std;

int* append(int*,int,int*,int);

int* merge(int*,int,int*,int);

void print(int*,int);

int main()

{ int a[] = {11,33,55,77,99};

int b[] = {22,44,66,88};

print(a,5);

print(b,4);

int* c = append(a,5,b,4); // c points to the appended array=

print(c,9);

int* d = merge(a,5,b,4);

print(d,9);

}

void print(int* a, int n)

{ cout << "{" << a[0];

for (int i=1; i<n; i++)

cout << "," << a[i];

cout << "}\n";

}

int* append(int* a, int m, int* b, int n)

{

int * p= (int *)malloc(sizeof(int)*(m+n));

int i,index=0;

for(i=0;i<m;i++)

p[index++]=a[i];

for(i=0;i<n;i++)

p[index++]=b[i];

return p;

}

int* merge(int* a, int m, int* b, int n)

{

int i, j, k;

j = k = 0;

int *mergeRes = (int *)malloc(sizeof(int)*(m+n));

for (i = 0; i < m + n;) {

if (j < m && k < n) {

if (a[j] < b[k]) {

mergeRes[i] = a[j];

j++;

}

else {

mergeRes[i] = b[k];

k++;

}

i++;

}

// copying remaining elements from the b

else if (j == m) {

for (; i < m + n;) {

mergeRes[i] = b[k];

k++;

i++;

}

}

// copying remaining elements from the a

else {

for (; i < m + n;) {

mergeRes[i] = a[j];

j++;

i++;

}

}

}

return mergeRes;

}

4 0
3 years ago
Write a function that checks whether two words are anagrams. Two words are anagrams if they contain the same letters. For exampl
Masteriza [31]

Answer:

def isAnagram(s1, s2):

   list1=s1

   list2=s2

   sortedlist1 = sorted(list1)

   sortedlist2 = sorted(list2)

   if sortedlist1 == sortedlist2:

       print(list1+ " and "+list2+ " are anagram")

   else:

       print(list1+ " and "+list2+ " are not anagram")

Explanation:

Here is a call to the function isAnagram():

list1 =input("Enter String1 ")

list1 =input("Enter String2 ")

isAnagram(list1,list2)

Attached is the run and output for this program

4 0
3 years ago
How many possible password of length four to eight symbols can be formed using English alphabets both upper and lower case (A-Z
Fynjy0 [20]

Answer:

In a password, symbol/characters can be repeated. first calculate the total

symbols which can be used in a password.

So there are total 26(A-Z),26(a-z),10(0-9) and 2(_,$) symbols.

that is equal to 26+26+10+2=64.

Total number of password of length 4:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64=16777216

Total number of password of length 5:

here at each place can filled in total number of symbols i.e 64 way for each

place.Then total number of possible password is:

64*64*64*64*64=1073741824

Similarly,

Total number of password of length 6:

64*64*64*64*64*64=68719476736

Total number of password of length 7:

64*64*64*64*64*64*64=4398046511104

Total number of password of length 8:

64*64*64*64*64*64*64=281474976710656

Hence the total number of password possible is:285,942,833,217,536

7 0
3 years ago
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