0.900 percent = 0.90 percent = 0.9 percent = 9/10 percent 0.900 percent = 0.009 0.900 = 90 percent
1 percent of a percent is 1/10000 or 0.0001; this is the same as saying 1 percent percent. Example: 5 percent percent is 0.0005, so 5 percent percent of 100 is 0.0005 x 100 or 0.05.
Parsing the question: What percent of 10 percent of 50 is 25 percent of 100? What percent of 10 percent of 50 is 25? What percent of 5 is 25? Answer: 500%
60 percent as a percent is 60%
To find 30 percent of 20 percent of 150 percent, you first need to calculate 20 percent of 150 percent. To do this, you multiply 0.20 by 1.50, which equals 0.30. Next, you find 30 percent of 0.30 by multiplying 0.30 by 0.30, which equals 0.09. Therefore, 30 percent of 20 percent of 150 percent is 0.09.
.o1 uF
bulbasaur!i like charizard better!
#include<stdio.h> #include<conio.h> #include<string.h> #include<stdlib.h> #include<process.h> void main() { clrscr(); char *s1,*s2,*o1,*o2,temp1,temp2; printf("Enter first statement:"); gets(s1); printf("Enter second statement:"); gets(s2); if(s1[0]!=s2[0]) { printf("Sorry"); getch(); exit(0); } o1[0]=s1[0]; o1[1]='-'; o1[2]='>'; for(int i=3;s1[i]==s2[i];i++) o1[i]=s1[i]; temp1=i; temp2=i; o1[i++]='Z'; o1[i++]='\0'; o2[0]='Z'; o2[1]='-'; o2[2]='>'; int p=3; for(int j=temp1;j<strlen(s1);j++) { o2[p]=s1[j]; p++; } o2[p++]='/'; for(j=temp2;j<strlen(s2);j++) { o2[p]=s2[j]; p++; } o2[p++]='\0'; puts(o1); puts(o2); getch(); }
O1 refers to the complexity class of problems that can be solved in a constant amount of time regardless of the input size. It is considered the most efficient level of time complexity in algorithm analysis.
0.1 litre = 10 centilitres
O1 visa
Usually two oxegen atoms (O2) though can be one (O1).
Usually two oxegen atoms (O2) though can be one (O1).
they catch the ball. but there is to o1 on each team catches against the separate team
700,000,000,000,000.O1
12 teenagers can eat 12 pizzas in one-and-a-half days so they can eat 48 in 6 days I always had difficulty with these problems. It is computed as follows: If P1 people can eat O1 objects in T1 days then P2 people can eat O2 objects in T2 days P1T1/O1 = P2T2/O2 In the example P1 = 1.5 T1 = 1.5 O1 = 1.5 P2 = 12 T2 = 6 solve for O2: O2 = P2 O1 T2/ P1 T1 = 48
6 inches in one half foot.