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VashaNatasha [74]
2 years ago
12

Which item is essential to know before sketching a navigation menu flowchart? template specifics, such as horizontal or vertical

menu layout whether or not the site will implement a search feature all of the pages in the site and the content each page will contain who will be using the site and the design theme selected for them
Computers and Technology
1 answer:
tekilochka [14]2 years ago
8 0

Answer:

Who will be using the site and the design theme selected for them.

Explanation:

The target user, will almost always define the way a website should be created as it should be optimised for your target audience. A search feature depends on the content of the site and weather it has lots of information or not. A horizontal or vertical menu also depends on the content, but it is not above the user, as it is meant to make it easier for them to navigate the site.

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Which perspective is usually used in process simulations?
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C. Omnipresent

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Which type of word processing programs enables us to include illustrations within the program?
bonufazy [111]

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C.  full featured

Explanation:

You need to make use of the full-featured version of the word processing programs to use the illustrations that are part of the program. And simple versions might not have that. By fully featured it means you need to have the license included or else you will be able to use it for a few days or a maximum of 3 months, and that is certainly not a good idea.

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Lucas put a lot of thought into the design for his company's new white paper. He made sure to include repeating design elements
Alenkasestr [34]
C. provide consistency
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2 years ago
Read 2 more answers
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the
zmey [24]

Answer:

part (a).

The program in cpp is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>0)

   {

       if(balance<payment)    

         { payment = balance;}

       else

       {  

           std::cout << balance <<"\t\t\t"<< payment << std::endl;

           balance = balance - payment;

       }

   }

   return 0;

}

part (b).

The modified program from part (a), is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   double charge=0.01;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   balance = balance +( balance*charge );

   std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>payment)

   {

           std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

           balance = balance +( balance*charge );

           balance = balance - payment;

       }

   if(balance<payment)    

         { payment = balance;}          

   std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

   return 0;

}

Explanation:

1. The variables to hold the loan balance and the monthly payment are declared as double.

2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.  

3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.

4. The computed values are displayed for each month till the loan balance becomes zero.

5. The output for both part (a) and part (b) are attached as images.

4 0
2 years ago
You are consulting for a trucking company that does a large amount ofbusiness shipping packages between New York and Boston. The
likoan [24]

Answer:

Answer explained with detail below

Explanation:

Consider the solution given by the greedy algorithm as a sequence of packages, here represented by indexes: 1, 2, 3, ... n. Each package i has a weight, w_i, and an assigned truck t_i. { t_i } is a non-decreasing sequence (as the k'th truck is sent out before anything is placed on the k+1'th truck). If t_n = m, that means our solution takes m trucks to send out the n packages.

If the greedy solution is non-optimal, then there exists another solution { t'_i }, with the same constraints, s.t. t'_n = m' < t_n = m.

Consider the optimal solution that matches the greedy solution as long as possible, so \for all i < k, t_i = t'_i, and t_k != t'_k.

t_k != t'_k => Either

1) t_k = 1 + t'_k

    i.e. the greedy solution switched trucks before the optimal solution.

    But the greedy solution only switches trucks when the current truck is full. Since t_i = t'_i i < k, the contents of the current truck after adding the k - 1'th package are identical for the greedy and the optimal solutions.

    So, if the greedy solution switched trucks, that means that the truck couldn't fit the k'th package, so the optimal solution must switch trucks as well.

    So this situation cannot arise.

  2) t'_k = 1 + t_k

     i.e. the optimal solution switches trucks before the greedy solution.

     Construct the sequence { t"_i } s.t.

       t"_i = t_i, i <= k

       t"_k = t'_i, i > k

     This is the same as the optimal solution, except package k has been moved from truck t'_k to truck (t'_k - 1). Truck t'_k cannot be overpacked, since it has one less packages than it did in the optimal solution, and truck (t'_k - 1)

     cannot be overpacked, since it has no more packages than it did in the greedy solution.

     So { t"_i } must be a valid solution. If k = n, then we may have decreased the number of trucks required, which is a contradiction of the optimality of { t'_i }. Otherwise, we did not increase the number of trucks, so we created an optimal solution that matches { t_i } longer than { t'_i } does, which is a contradiction of the definition of { t'_i }.

   So the greedy solution must be optimal.

6 0
2 years ago
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