I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Neither it has 12 You c u n t
c + n
12/n
For n=number 12xn, (12)(n), (12)n, 12(n), or 12n
12 numbers on a clock(face)
12 Numbers on a Clock Face
The answer depends on if you can choose the same kind of donuts more then once. Or in other words, is repetition permitted. If you can only choose the same kind of donuts only once, it is a 21 choose 12 problem: C(n,k) = n! / (k! (n - k)!) C(21, 12) = 21! / (12! (21 - 12)!) = 21! / (12! (9)!) = 293,930 If you can choose the same kind of donuts more then once, it is a combination with repetition problem. P(n+k-1,k) = (n+k-1)! / (k! (n-1)!) or put it into C(n,k) with n+k-1 as 21 + 12 - 1 = 32 and k as 12 so C(21+12-1,12) = C(32, 12) = 32! / (12! (32 - 12)!) = 32! / (12! (20)!) = 225,792,840
12 Noon on a clock face. 12 Numbers on a clock face.
Vitamin B-12 contain Co, C, H, O and N.
I think that you're doing the same thing that I did for IB. Here's what I think you're asking for : 0, 2, 6, 12, 20, 30, 42... c₁ = 0 c₂ = 2 c₃ = 6 (c₂ + 4 = 2 + 4 = 6) c₄ = 12 (c₃ + 6 = 2 + 4 + 6 = 12) c₅ = 20 (c₄ + 8 = 2 + 4 + 6 + 8 = 20) c₆ = 2 + 4 + 6 + 8 + 10 etc... dn = (n/2) <2c₁ + (n-1) 2> dn = (n/2) <2 (2) + (n-1) 2> dn = (n/2) (4 + 2n - 2) dn = (n/2) (2 + 2n) dn = (2n/2) + (2n²/2) dn = n + n²
Neither it has 12 You c u n t
c
N-14 has more protons. It contains one more proton than C.
HERE IS IT!!!!!#include#includevoid main(){int a[12],b[12],c[12],i,j=0,k=0;clrscr();printf("Enter an integer\n");for(i=0;i
0.3333
A-N-C-I-E-N-T-E-G-Y-P-T {12 Letters}