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Andreas93 [3]
3 years ago
14

Sum of 18/7 and 13/7 is *​

Computers and Technology
2 answers:
NARA [144]3 years ago
7 0
The sum would be 31/7, or 4.428571428571429 in decimal form.
Dima020 [189]3 years ago
4 0

Answer:

\frac{31}{7}

Explanation:

\frac{18}{7} + \frac{13}{7}

\frac{18 + 13}{7}

\frac{31}{7}

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Construct sequence using a specified number of integers within a range. The sequence must be strictly increasing at first and th
natulia [17]

Answer:

Explanation:

The following code is written in Java. It first creates a function that takes in the three parameters to create the sequence, if the sequence is valid it returns the created sequence, otherwise it prints -1 and returns an empty array. The second function called compareSequences takes in to sequences as parameters and compares them, printing out the winner. A test case has been provided in the main method and the output can be seen in the image below.

import java.util.Arrays;

class Brainly {

   public static void main(String[] args) {

       int[] seq1 = createSequence(5, 4, 2);

       int[] seq2 = createSequence(5, 3, 3);

       System.out.println("Winner: " + Arrays.toString(compareSequences(seq1, seq2)));

   }

   public static int[] createSequence(int num, int upperEnd, int lowerEnd) {

       int[] myArray = new int[num];

       int middle = (int) Math.floor(num/2.0);

       myArray[0] = upperEnd;

       myArray[num-1] = lowerEnd;

       int currentValue = upperEnd;

       for (int i = 1; i < num-1; i++) {

           if (i <= middle) {

               myArray[i] = currentValue + 1;

               currentValue += 1;

           } else {

               myArray[i] = currentValue - 1;

               currentValue += 1;

           }

       }

       System.out.println(Arrays.toString(myArray));

       if (myArray[num-2] < lowerEnd) {

           System.out.println("-1");

           int[] empty = {};

           return empty;

       } else {

           return myArray;

       }

   }

   public static int[] compareSequences(int[] seq1, int[] seq2) {

       int lowestLength;

       if (seq1.length > seq2.length) {

           lowestLength = seq2.length;

       } else {

           lowestLength = seq1.length;

       }

       for (int i = 0; i < lowestLength; i++) {

           if (seq1[i] > seq2[i]) {

               return seq1;

           } else if (seq1[i] < seq2[i]) {

               return seq2;

           }

       }

       if (seq1.length > seq2.length) {

           return seq1;

       } else {

           return seq2;

       }

   }

}

7 0
3 years ago
Write a program to find all integer solutions to the equation 4x + 3y -9z = 5 for values of x, y, and z between 0 to 100.
rjkz [21]

Answer:

Following are the code to the given question:

#include <iostream>//header file

using namespace std;

int main()//main method

{

  int c= 0;//defining integer variable to count number of solutions

  int x,y,z;//defining integer variable which is used in the loop

  for (x = 0; x <= 100; x++)//defining for loop to x value

     for (y = 0; y <= 100; y++)//defining for loop to y value

        for (z = 0; z <= 100; z++)//defining for loop to z value

           if (4 * x + 3 * y - 9 * z == 5)//use if to check given condition

           {

              c++;//increment count value

              cout << "(" << x << "," << y << "," << z << ")";//print solutions  

              cout << (c % 11? " " : "\n");//use for add file solution in a row

           }

  cout << "\n\nThere are " << c << " solution(s)\n";//print solution couts

  return 0;

}

Output:

Please find the attached file.

Explanation:

In the above-given code four integer variable "x,y,z, and c" is declared, in which "x,y, and z" is used in the for loop with the if conditional statement that checks the given condition and prints the solution and the "c" variable is used to counts the number of solution, and at the last, it uses the print method to print its values.  

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Answer:

is there some oppsense you can choose from

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3 years ago
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Home is the answer. Hope this helps. :)
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