k(k + 5)
It is because 3 is a factor of 9. So if 9 is a factor of some number X, then X = k*9 for the factor pair 9 and k But X = k*9 = k*(3*3) = (k*3)*3 so that X is k*3 times 3 ie it is a multiple of 3.
k = 5
7(k + 5)(k - 4)
Consider a number k. Let m be the square root of k : since k is not a square number, m is not an integer and so m cannot be a factor of k.Suppose p is a factor of k and suppose p< m.Then k has another factor q such that k = p*q and since p < m, then q> m.Thus, for every factor smaller than the square root, there is another factor that is larger than the square root. That is, the factors come is distinct pairs and so there is an even number of factors.
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The duration of The K Factor is 300.0 seconds.
The K Factor was created on 2010-02-06.
The K Factor ended on 2010-03-20.
YES.If k is even then 2 is a factor of k.If 5 is also a factor of k then 2 x 5 = 10 is another factor of k.NOTE : The product of two or more prime factors of a number produces another factor.
It is a factor in all numbers of the form 310*k where k is an integer.
k(k + 5)
What is a sprinkler's "K-factor The K-factor is the nozzles's / sprinkler's constant at a given volume flow rate and is generally calculated with a formula: K = Q / √ p K = sprinkler's / nozzle's constantQ = volume flow rate (l/min)p = pressure at the sprinkler / nozzle (bar) examples: K-factor 57 = 171 l/min / √ 10 barK-factor 80 = 240 l/min / √ 9 barK-factor 115 = 345 l/min / √ 9 bar Generally the thread (DN) is associated with a K-factor. 3/8″ DN 10 K-factor 57½″ DN 15 K-factor 80½″ DN 15 K-factor 115¾″ DN 20 K-factor 160 == ==
K-factor is a weighting of the harmonic load currents according to their effects on transformer heating, as derived from ANSI/IEEE C57.110. A K-factor of 1.0 indicates a linear load (no harmonics). The higher the K-factor, the greater the harmonic heating effects.