It is not except when n = 1.
1
We know that the complete graph has n(n-1)/2 edges and we want to find out n such that n(n-1)/2 greater or equal to 500. Thus n(n-1) greater or equal to 1000. Taking n=33, we have, n(n-1)=33(33-1)=1056>1000. Therefore required smallest integer is n=33.
Pressure can only be expressed as n/m2 (not n.m) 1 n/m2 = 1 pascal
Difficult to tell without brackets, but (n-1)*(n+1) = n^2 - 1 [in other words, n squared minus 1] is a useful algebraic identity.
Let n be any number and n/n = 1 and n1/n1 = n1-1 which is n0 that must equal 1
It is not except when n = 1.
If: 2n = 1 Then: n = 1/2 or 0.5
52 = 4 + (n - 1)³ → (n - 1)³ = 48 → n - 1 = ³√48 → n = 1 + ³√48 ≈ 4.634
As stated in the question, it equals 1.
Any value with a 'zero' exponent is equaL TO '1'. A^(0) = 1 proof Let a^(0) =. a^(n - n) = a^(n) / a^(n) Cancel down by a^(n) hence it equals '1'.
Any monomial in the format: axn has a derivative equal to: nax(n - 1) In this case, "a" is equal to 1 and "n" is equal to 2. So the derivative of x2 is equal to 2x.
1 N is equal to 1/1000 kN. 23400 is equal to 23400/1000=23.4 kN.
2n5 = 1 000 000 n5 = 500 000 n = 5√500 000) n = (500 000)1/5 n = 13.8
No.
For n not equal to -1, it is 1/(n+1)*xn+1 while for n = -1, it is ln(|x|), the logarithm to base e.
1