This is the famous fibonacci sequence, where each term in the sequence is the sum of the previous two. Fn=Fn-1 + Fn-2 F0 = 1 and F1 = 1 are the initial values to begin the sequence. F2 = F1 + F0 = 1 + 1 = 2 F3 = F2 + F1 = 2 + 1 = 3 and so on
It looks like there is no end date, so that means that 2 years of interest generate 1000: F3 * 1.182 - F3 = 1000 ; F3 = 2548.42, Then calculate the present value from F3:F3 = 2548.42 = P * (1.18)3 --> P = 1551.05
1 Get your number 2 hit the catalog key (2nd key + custom key) 3 press up to go to the bottom of the list 4 scroll til you see the ">frac" command 5 hit enter If you plan on using the fraction conversion feature a lot, you can make it a "custom" command, which works sort of like a hot key. Do the above steps, skipping step 5. Instead, hit F3 (which corresponds to the "CUSTM" command on the screen) with ">frac" selected. Then select which F key you want for it to occupy (F1-F5). Whenever you need to convert a fraction (or any of the commands) you can hit the "Custom" key and it will have 5 boxes on the bottom of the screen that you can use the F keys to choose an action. Possibly wrong answers follow. These may be instructions for other models of calculator. Type in your fraction then hit MATH > ENTER > ENTER EXAMPLE: .8 > MATH > ENTER > ENTER 4/5 alternate solution, since I don't know how the above one works. Type in your fraction then hit MATH > MISC/f5 > MORE> Frac/f1>ENTER EXAMPLE: .8 > MATH > MISC/f5 > MORE> Frac/f1>ENTER 4/5
find the constant difference for a hyperbola with foci f1 (5,0) and f2(5,0) and the point on the hyperbola (1,0).
10x10=10010+10+10+10+10+10+10+10+10+10=10010 times 10 is 100.
<html> <head> <title>asp</title> </head> <% dim f1,f2,f3,c f1=0 f2=1 response.Write(f1 &"<br>") response.write(f2 &"<br>") c=1 while(c<=10) f3=f1+f2 response.write(f3 &"<br>") c=c+1 f1=f2 f2=f3 wend %> </html>
Following logic for Fibonocci int f1=0; intf2=1; int i=0; while(i
#include<stdio.h> #include<conio.h> void main() { int i,n,f1=0,f2=1,f3; clrscr(); for (i-0;i<10;i++) { f3=f1+f2; f1=f2; f2=f3; printf("%d \n",f3); } getch(); } output 1 2 3 5 8 13 21 34 55 89
esc, f1, f2, f3, f4, f5, f6, f7, f8, f9 :)
f1 f2 f3 f4 f5
try this out...#include#includefloat valcal(float);int main(){float a,b,f,x,f1,f2,f3,i;a=2;b=3;int flag=0;f1=valcal(a);f2=valcal(b);printf("\n\n%f %f",f1,f2);for(i=0;i
#include<stdio.h> #include<conio.h> int main(void) { int n,f1=0,f2=1,f3=0,i=1; printf("Enter the Number :\n\t\t"); scanf("%d",&n); printf("%d %d",f1,f2); while(i<=n-3) { f3=f1+f2; printf("%d",f3); f1=f2; f2=f3;i++; } getch(); }
class lift { public static void main(String args[]) { int f1,f2,f3; f1=1; f2=2; f3=3; ch=3; switch(ch) { case 1: f1=1; System.out.println(f1); break; case 2: f2=2; System.out.println(f2); break; case 3: f3=3; System.out.println(f3); break; default: System.out.println("remaining floors"); } } }
f1 f2 f3 f4 f5
You sum up the frequencies upto and including the current band. So if you have data bands b1, b2, b3 and so on with frequencies f1, f2, f3 etc then the cumulative frequency for b1 = f1 b2 = f1+f2 b3 = f1+f2+f3 and so on.
The F1, F2, and F3 designation on a dog breed designates the breeding. A Pure Bred Poodle and a Pure Bred Maltese makes a F1 MaltiPoo. When a MaltiPoo and MaltiPoo are bred, it gets a F2 designation.
#include<conio.h> #include<stdio.h> main () { int f1=0,f2=1,f3=1,i,j,n; clrscr(); printf("enter n value"): scanf("%d",&n); if(n<=3) { for(i=0;i<=n;i++) { f1=f2; f2=f3; printf("%dfibonic value is\n",f3); f3=f1+f2; } else { printf("re ente n value"); } getch(); }