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8-n

Barbados \n . Botswana \n . Bulgaria \n . Cameroon \n . Colombia \n . Ethopia \n . Hondurus \n . Kiribati \n . Malaysia \n . Mongolia \n . Pakistan \n . Paraguay \n . Portugal \n . Slovakia \n .

n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed

n ,n ,n,n,,n ,,n,n

(n*n)+n

'n'

N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N N - 5*N = 4*N

n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42

n n n n n n n n n n n n n n n o o o o o o o o o o o o o t t t t t t t t t t t h h h h h h h h h h h h h h i i i i i i i i i i n n n n n n n n n n g g g g g g gg g g g g g g g gg gg g

n to n

n nn n n n n n

Assuming you mean the first n counting numbers then: let S{n} be the sum; then: S{n} = 1 + 2 + ... + (n-1) + n As addition is commutative, the sum can be reversed to give: S{n} = n + (n-1) + ... + 2 + 1 Now add the two versions together (term by term), giving: S{n} + S{n} = (1 + n) + (2 + (n-1)) + ... + ((n-1) + 2) + (n + 1) → 2S{n} = (n+1) + (n+1) + ... + (n+1) + (n+1) As there were originally n terms, this is (n+1) added n times, giving: 2S{n} = n(n+1) → S{n} = ½n(n+1) The sum of the first n counting numbers is ½n(n+1).

the code for the box is: # include <iostream> using namespace std; void main () { cout << "*********\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*\t*\n"; cout << "*********\n"; the code for the oval is: # include <iostream> using namespace std; void main () { cout << " ***\n"; cout <<" * *\n"; cout <<"* *\n"; cout <<"* *\n"; cout <<"* *\n"; cout <<"* *\n"; cout <<"* *\n"; cout <<" * *\n"; cout << " ***\n"; the code for the arrow is: # include <iostream> using namespace std; void main () { cout << " * \n"; cout <<" ***\n"; cout <<"*****\n"; cout <<" * \n"; cout <<" * \n"; cout <<" * \n"; cout <<" * \n"; cout <<" * \n"; the code for the diamond: # include <iostream> using namespace std; void main () { cout << " *\n "; cout <<" * *\n"; cout<<" * *\n"; cout<<" * *\n"; cout<<"* *\n"; cout<<" * *\n"; cout<<" * *\n"; cout <<" * *\n"; cout << " *\n "; }

n^3 (n cubed)

Successor of n=n+1 Predecessor of n=n-1 Sum=[n+1]+[n-1] Plus 1 plus minus 1= [1]+[-1]=0 [n][n] =2n Hope this helps

n n n n

It is n/2.

2 N cubed (3)

If n=60 then n+n= n, Next n.

n2 + n = n(n + 1)

n cubed = n times n times n

No. n is not a factor of n + 5.

n = 5

jazz has been around for a billion years

Suppose the number is N then 4N = 3*abs(35-N) where abs(35-N) is the difference between 35 and N If N > 35 then 4*N = 3*(N-35) = 3*N - 105 So N = -105 If N < 35 then 4*N = 3*(35-N) = 105 - 3*N So 7*N = 105 ie N = 15 So N = -105 or 15