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JulsSmile [24]
3 years ago
9

For each obstacle, select the best solution. failing to find other members for a club: failing a class: failing to get the credi

ts necessary to graduate from high school: not having enough money to pay for culinary school: not getting a job or promotion:
Computers and Technology
2 answers:
WARRIOR [948]3 years ago
5 0

1. Hang around posters, or send a massive email to anyone to apply

2. Study more. Write down some notes on paper for the parts you dont get

3. Ask for extra credits, or re-do some of the works and keep reminding the teacher to check your work.

4. Try to work your hardest and get a scholorship, or try to get jobs in order to get money

5. Try other Jobs, or keep trying your best in every way to get that next promotion.

bazaltina [42]3 years ago
4 0

Answer:

Explanation:

1. Hang around posters, or send a massive email to anyone to apply

2. Study more. Write down some notes on paper for the parts you dont get

3. Ask for extra credits, or re-do some of the works and keep reminding the teacher to check your work.

4. Try to work your hardest and get a scholorship, or try to get jobs in order to get money

5. Try other Jobs, or keep trying your best in every way to get that next promotion.

Hope this helped!

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3 0
1 year ago
KDS Company has 3 departments: 1. Dept X 2. Dept Y 3. Dept Z. Design a program that will read as input each department's sales f
AVprozaik [17]

Answer:

In C++:

#include<iostream>  

using namespace std;  

int main() {  

  int depts[3][4];

  char dptcod [3] ={'X','Y','Z'};

  for (int rrw = 0; rrw < 3; rrw++) {  

      for (int cll = 0; cll < 4; cll++) {  

          cout<<"Dept "<<dptcod[rrw]<<" Q"<<(cll+1)<<": ";

          cin>>depts[rrw][cll];  }  }  

      for (int rrw = 0; rrw < 3; rrw++) {  

          int total = 0;

          for (int cll = 0; cll < 4; cll++) {  

              total+=depts[rrw][cll]; }

              cout<<"Dept "<<dptcod[rrw]<<" Total Sales: "<<total<<endl;}  

  return 0; }

Explanation:

This declares the 3 by 4 array which represents the sales of the 3 departments

  int depts[3][4];

This declares a character array which represents the character code of each array

  char dptcod [3] ={'X','Y','Z'};

This iterates through the row of the 2d array

  for (int rrw = 0; rrw < 3; rrw++) {  

This iterates through the column

      for (int cll = 0; cll < 4; cll++) {

This prompts the user for input

          cout<<"Dept "<<dptcod[rrw]<<" Q"<<(cll+1)<<": ";

This gets the input

          cin>>depts[rrw][cll];  }  }  

This iterates through the row of the array

      for (int rrw = 0; rrw < 3; rrw++) {

This initializes total sum to 0

          int total = 0;

This iterates through the column

          for (int cll = 0; cll < 4; cll++) {

This calculates the total sales of each department

              total+=depts[rrw][cll]; }

This prints the total sales of each department

              cout<<"Dept "<<dptcod[rrw]<<" Total Sales: "<<total<<endl;}  

I hope this helps a little bit

6 0
3 years ago
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