n2
(n - 4)(n - 12)
If you mean: n2+20n+100 then yes it is because (n+10)(n+10) when factored
169
t(n) = n2
n=40 n2+n+41=1681 which is not a prime.
n x n = n2
The hybridization of N i n N2 is sp.
P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
.04m2-n2=(.2m+n)(.2m-n)
There are many sorting algorithms with worst case of complexity O(n2). These algorithms have different average and best cases. They are:Best caseAverage caseWorst case1) Quick sortO(n*log n)O(n*log n)O(n2)2) Insertion sortO(n)O(n2)O(n2)3) Bubble sortO(n)O(n2)O(n2)4) Selection sortO(n2)O(n2)O(n2)
n2
n2 = 20 + n so n2 - n - 20 = 0 that is (n - 5)(n + 4) = 0 so that n - 5 = 0 or n + 4 = 0 n = 5 or n = -4
n2-1 and n2-4 are trivial cases because of n2-m2=(n-m)(n+m). So the only prime of the form n2-1 is 3 and of the form n2-4 is 5.
If you mean: 6/n times 5/n-1 = 1/3 Then: 30/n2-n = 1/3 Multiplying both sides by n2-n: 30 = n2-n/3 Multiplying both sides by 3: 90 = n2-n Subtracting 90 from both sides: 0 = n2-n-90 or n2-n-90 = 0 Solving the above quadratic equation: n = -9 or n =10 If n is of a material value its more likely to be 10 Note that n2 means n squared
n2 + 9n + 18 = (n + 6)(n + 3)
n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - n + 1)