Complete the do-while loop to output 0 to the value of countLimit using printVal. Assume the user will only input a positive number. For example, if countLimit is 5 the expected output will be

Question not complete

Complete the do-while loop to output every number form 0 to countLimit using printVal. Assume the user will only input a positive number. For example, if countLimit is 5 the expected output will be 0 1 2 3 4 5

import java.util.Scanner;

public class CountToLimit {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

int countLimit = 0;

int printVal = 0;

// Get user input

countLimit = scnr.nextInt();

printVal = 0;

do {

System.out.print(printVal + " ");

printVal = printVal + 1;

} while ( /* Your solution goes here */ );

System.out.println("");

return;

}

}

Replace while ( /* Your solution goes here */ );

With

while ( printVal <= countLimit );

This will check if the value of variable printVal is still within print range of variable countLimit.

The full program becomes

import java.util.Scanner;

public class CountToLimit {

public static void main (String [] args) {

Scanner scnr = new Scanner(System.in);

int countLimit = 0;

int printVal = 0;

// Get user input

countLimit = scnr.nextInt();

printVal = 0;

do {

System.out.print(printVal + " ");

printVal = printVal + 1;

} while ( printVal <= countLimit );

System.out.println("");

return;

}

}

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

$75.00

step-by-step explanation:

i = prt

i = 1,5000 * 0.025 * 2

i = 75