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svlad2 [7]
3 years ago
5

#Java What is the solution? ​

Computers and Technology
1 answer:
inysia [295]3 years ago
6 0

Answer:

I have answered the first part in another question of yours, but the second part is in the provided screenshot!

Explanation:

Steps to solve this problem:

Go through all the numbers from 1 -> n1, every time you go through each of these numbers, go through all the numbers from 1 -> n2. Multiply these numbers and add them to an output string. Print each line of output.

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2 Write<br>a program to and area of<br>4<br>walls. A = 2h (l+b)​
Zanzabum

Answer:

ADGH AVFH VVBLA FDHLAVRGLBHSRGLHFSgh HGK g gH G avoshhbv so bhsf vhbfsb gfsb gb gh g hhow gw

Explanation:

eehhfhsghOhrrbfghedhchggfgsbbfbhdhfhehyfnfnhhsh

3 0
3 years ago
WILL GIVE A BRAINLIEST!!! PLS HELP!!!
DIA [1.3K]

Answer:

never gonna give you uppp

Explanation:

criiiiii

7 0
3 years ago
What refrigerant has been approved for new household refrigerators and freezers
Alexxx [7]
Bsbsnsnsnsn jajwjemoeododhdhdw whajsjshtzzvsssnxbba. 3
8 0
4 years ago
What is the range of the well-known ports used by tcp and udp?
melamori03 [73]
/etc/services on my CentOS machine goes up to 49000. Reserved ports are less than 1024.

Well known and reserved ports are two different things. Well known ports only need a registration with IANA. Reserved ports, on *nixes, requires root for a deamon to bind to the port.
8 0
3 years ago
Consider a company that needs to sort an array of structures of type Customer by balance, with the largest balance first. Here i
8090 [49]

Answer:

#include <iostream>

#include <string>

#include <cstring>

#include <cstdlib>

using namespace std;

struct Person

{

string name;

int age;

};

int main()

{

struct Person data[10];

struct Person *pData[10],*temp;

string names[] = {"a","b","c","g","z","l","p","q","r","w"};

int num[] = {4,6,34,8,13,90,33,22,18,23};

for(int i=0;i<9;i++)

{

data[i].name = names[i];

data[i].age = num[i];

pData[i] = &data[i];

}

for(int i=0;i<9;i++)

{

for(int j=i+1;j<9;j++)

{

if(pData[i]->name.compare(pData[j]->name)>0)

{

temp = pData[i];

pData[i] = pData[j];

pData[j] = temp;

}

}

}

for(int i=0;i<9;i++)

{

cout<<pData[i]->name<<" "<<pData[i]->age<<endl;

}

}

Explanation:

The line #include <iostream> initializes the program.

This program created an array of only 10 structures, Customer data. Also defined an auxiliary array Customer *pData.

The program uses these parameters to sorts the array of pointers so that when you go through pData in increasing order of index k, the entries pData[k] point to Customer objects in decreasing order by balance, meaning that pdata[0] now points to the customer with the highest balance, and pData[9] points to the customer with the smallest balance.

4 0
3 years ago
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