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8_murik_8 [283]
3 years ago
9

8) The higher the drag coefficient, the_____ the car will go a)Siower b)Faster

Computers and Technology
2 answers:
Murrr4er [49]3 years ago
6 0

the answer is * slower *

yanalaym [24]3 years ago
5 0
The correct answer is a)Slower
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I need help thanks please!
maw [93]

Answer:

a) to persuade

she is doing this to propose an idea to the boss of the company

6 0
3 years ago
Read 2 more answers
Case 2-2 Jack has a computer at home that he uses to access the Internet, store and edit personal photos, and create and edit do
bekas [8.4K]

Answer:

Check button under error checking

Explanation:

Under the Tools tab there are two options:

Error checking and Optimize and defragment drive option.

clicking the check button with administrative permission under error checking option will examine the hard drive for errors.

3 0
3 years ago
We are building a word processor and we would like to implement a "reflow" functionality that full-justifies text.Given a maximu
adoni [48]

Answer:

The code is given below in Java with appropriate comments

Explanation:

Reflow is an function which helps you to justify your functionality

You need to input a string array and the max width and you will get the

output in a list(string).

public List<String> Reflow(String[] input, int width) {

   List<String> result = new ArrayList<String>();

 

   if(input==null || input.length==0){

       return result;

   }

 

 

   int count=0;

   int last=0;

   ArrayList<String> list = new ArrayList<String>();

   for(int i=0; i<input.length; i++){

       count = count + input[i].length();

 

       if(count+i-last>width){

           int wl = count-input[i].length();

           int sl = maxWidth-wl;

           int each = 1;

           int el = 0;

 

           if(i-last-1>0){

               each = sl/(i-last-1);

               el = sl%(i-last-1);

           }

 

           StringBuilder sb = new StringBuilder();

 

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

               sb.append(input[k]);

 

               int ce = 0;

               while(ce<each){

                   sb.append(" ");

                   ce++;

               }

 

               if(el>0){

                   sb.append(" ");

                   el--;

               }

           }

 

           sb.append(input[i-1]);//last input in the line

           //if only one word in this line, need to fill left with space

           while(sb.length()<width){

               sb.append(" ");

           }

 

           result.add(sb.toString());

 

           last = i;

           count=input[i].length();

       }

   }

 

   int lastLen = 0;

   StringBuilder sb = new StringBuilder();

 

   for(int i=last; i<input.length-1; i++){

       count = count+input[i].length();

       sb.append(input[i]+" ");

   }

 

   sb.append(input[input.length-1]);

   int d=0;

   while(sb.length()<maxWidth){

       sb.append(" ");

   }

   result.add(sb.toString());

 

   return result;

}

6 0
3 years ago
Find the gear ratio of the gear in the picture. <br><br> 25 points+brainliest
ankoles [38]

The gear ratios would look like A:B=1:9 and C:D=1:32 if you are just needing the gear teeth ratio. Not a lot of information to go off of from the question though

8 0
3 years ago
Consider an array inarr containing atleast two non-zero unique positive integers. Identify and print, outnum, the number of uniq
Delvig [45]

Answer:

Program.java  

import java.util.Scanner;  

public class Program {  

   public static boolean isPalindrome(String str){

       int start = 0, end = str.length()-1;

       while (start < end){

           if(str.charAt(start) != str.charAt(end)){

               return false;

           }

           start++;

           end--;

       }

       return true;

   }  

   public static int countPalindromePairs(String[] inarr){

       int count = 0;

       for(int i=0; i<inarr.length; i++){

           for(int j=i+1; j<inarr.length; j++){

               StringBuilder sb = new StringBuilder();

               sb.append(inarr[i]).append(inarr[j]);

               if(isPalindrome(sb.toString())){

                   count++;

               }

           }

       }

       return count == 0 ? -1 : count;

   }

   public static void main(String[] args) {

       Scanner sc = new Scanner(System.in);

       String line = sc.next();

       String[] inarr = line.split(",");

       int count = countPalindromePairs(inarr);

       System.out.println("RESULT: "+count);

   }

}

Explanation:

OUTPUT:

3 0
3 years ago
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