Sentence examples for by using substitutions from inspiring English sources

The optimal solution is obtained by using substitutions z = [ ∑ m ∑ k ∈ F r x km u km ] − 1 and y km = x km z, which basically transforms the fractional problem into a standard linear program as follows by using t as a dummy variable: max t (27) s. t γ r [ LT Ω k − 1 ∑ m y km u km − z ] ≥ t ∀ k ∈ F r, (28) ∑ k ∈ F r y km − z = 0 ∀ m (29) 0 ≤ y km ≤ z ∀ k, ∀ m, (30) ∑ m ∑ k ∈ F r y km u km = 1.

This is suitable for use in hand-held electronic devices and smartphones, and the clinical applicability is broadened by using substitutions for creatinine and Killip class.

(I'll note that while it is natural and effective to use calculus to solve the maximization problem in the induction step, it can be done without calculus by using substitution to put the function to be maximized in the form A – (f(x))^2, which is clearly maximized when f(x) = 0).

Here, by using substitution R2 = r, we obtain the desired result in Equation (6).

Moreover, the ATH35L secretion efficiency was reduced for NSP4b regardless of the increased total hydrophobicity by using substitution of amino acid residues in the h-domain with polyleucine (Fig. 2c; Table 4).

They are estimated by using substitution models and rate heterogeneity parameters, which correct for multiple substitutions at a site [ 51].

This sequence data matrix was partitioned by gene and codon position using substitution and site heterogeneity models determined in Modeltest (Additional file 7, see Phylogenetic Analysis section).

Keyboard substitutions use the pattern of a traditional American (QWERTY) keyboard pattern to use substitutions, generally by shifting the letters up, down, left or right by a certain number of places.

Arena helped by using three substitutions very effectively.

(8) By using successive substitutions in (8), the condition (M<1) implies that 1 is not an eigenvalue of the kernel (varphi t,s)).

Further, by using the substitutions t i = − u i, i ≠ 1, we obtain the following identity: ∫ R n − 1 K ( − u 1, …, − u i − 1, 1, − u i + 1, …, − u n ) ∏ j = 1, j ≠ i n | u j | a j d ˆ i u = k i ( a ) (2.2). for i = 1, 2, …, n, where we assume that the above integral converges.