XXXVIII is '38' That is x = 10 x = 10 x = 10 V = 5 I = 1 I = 1 I = 1 Add 10 + 10 + 10 + 5 + 1 + 1 + 1 = 38
It depends with your speed but it can take you 10^18 light years to count from 1 to 10 sextillion.
50 + 10 +(10-1) = 69
The I before the X means 10 minus 1. This converect to 19 in Arabic numerals.
Locate 6,1Input "Your Name Please ",Nam$LOCATE 8,1FOR Count=1 to 10Print TAB(5) Count " > ";Nam$Next Count
you cna you can only count backwards form 10-1
1-wahed 2-ethnan 3-thalatha 4-araba'a 5-khamsa 6-setta 7-saba3a 8-thamania 9-tesa'a 10-ashra
XXXVIII is '38' That is x = 10 x = 10 x = 10 V = 5 I = 1 I = 1 I = 1 Add 10 + 10 + 10 + 5 + 1 + 1 + 1 = 38
It depends with your speed but it can take you 10^18 light years to count from 1 to 10 sextillion.
int i; for (i=1; i<=10; i++) printf ("%d %d\n", i, i*i);
11
50 + 10 +(10-1) = 69
The I before the X means 10 minus 1. This converect to 19 in Arabic numerals.
Locate 6,1Input "Your Name Please ",Nam$LOCATE 8,1FOR Count=1 to 10Print TAB(5) Count " > ";Nam$Next Count
2016 (1000 + 1000 + 10 + 5 + 1)
public class DoWhileLoopDemo {public static void main(String[] args) {int count = 1;System.out.println("Printing Numbers from 1 to 10");do{System.out.println(count++);}while( count
it is 10. First, you need to count how many 1's there are. 1,2,3,4,5,6,7,8,9,10! There are 10.