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This is one of possible solutions:
66
73.333 % repeating For future reference you can figure this out yourself by using the following equation (22 / 30) * 100 = Percentage
22% of 22= 22% * 22= 0.22 * 22= 4.84
All of the following are: -31, -30, -29, -28, -27, -26, -25, -24, -23, -22, -21, -20, -19, and -18
Each digit is repeated the number of times equal to its value. 1 22 333 4444 55555 666666 7777777 88888888 999999999
22 x 11 x 101 = 4444
the product of 22 by 2 is 4444
1. ................is an alternative of if statement. 2. which of the following is an entry controlled loop? (a) for (b) switch (c) while (d) both b and c 3. which of the following is an entry controlled loop? (A) for (B) while (c)do while (d)both a and b (e)none of these 4.consider the given code fragment: const int n=25; int i=0,s=0; while(i { s+=i; i++; } cout<<s; write equivalent for loop for this. 5.write a programme to display the following output. 1 22 333 4444 55555 666666 6.write a programme to find the sum of digits of an integer number using (a) entry controlled loop (b) exit controlled loop
22^333^444^55555^666666^7777777^88888888^999999999 repeat this step 4096 times
#include <stdio.h> int main (void) { puts ("4 1 44 11 444 111 4444 1111 000000000 444444 3333 22 1"); return 0; }
The sum of 4,44,444... for 22 positions would 3.25611 E+22 (22 zeroes). In the thousands position would be a 0.
This is one of possible solutions:
int i = 4;while (i>0) { for (int n=0; n<i; ++n) { printf ("%d", i); } printf (" "); } printf ("\n");
Any number that is not 1, 2, 11 and 22
66
it is n+11 = y - 11