 Another point in that direction is to look at the mechanical equilibrium condition (2) which is used as the basic equation for workspace and tension ... If this is the case let us define m6 as the number of valid cables sextuplets. Therefore, it seems to us that the preference of M3 with respect to M6 is not intuitive. Moreover, the technique proposed by Brewka seems, in some case, not capturing the intuitive meaning of preferences. Example 7. M6 is estimated to be 100 million years old; M7 is more than twice that age. At a distance of ~1,600 light-years, ... same size – though it appears twice as large in our skies (owing to the fact that it is half as far from us as M6). (ii) If M is a KS-submodule of V, then the displayed equation above, applied to the elements of So, tells us that S" preserves the subspace M6 of V. It is clear that M and M� have the same dimension, since � is invertible. These 30 points of difference between 96% of SOCA-DSEM and 67% of M6 allows us to conclude that SOCA-DSEM is balanced in agility-rigor, while M9 can be classified as more rigorous than agile (i.e. a non balanced development methodology)�... Let us note that if i2 is offered in M11 after executing i1/o1 then only o1 can be produced. ... However, we have M11 conf□ M6 because all sequences concerned by M11 that appear in M6 (in fact only the sequence i1/o1 ,i2/o1) are�... This identity is called the interchange rule . ... It is worthwhile to reconsider equation ( M6.69 ) using an argument due to Van Mieghem ( 1973 ) . ... surfaces , and volumes Let us M6.6 The general form of the budget equation 103. Let us consider, on the multilattice M6", the commutative conjunctor &, defined, for each ar, ye M6*, as T if a = T and y = T a & y = ( L if a € {l, b} or y € { L, b} Ol. Otherwise It easy to see that & is left-continuous. However , control of the M6 toll to relieve pressures in the area ? it remains the fact that there has been so much Mr ... does not provide for us to and the local authorities to evaluate safety impacts . direct the concessionaire . 0 0 0 The right-hand side of this inequality is finite with probability one and, therefore, � (M6)�dt < co. ... Let us show that T T –2 M / A#dt < |Mer (-/ st) • (7.95) Indeed, according to the Karhunen expansion (see�...