Exact(1)
americanus to be equivalent (and thus synonymous) to L. fallax var.
Similar(59)
However, these two equations are equivalent and thus, do not offer a set of equations that can be solved simultaneously.
The noise reduction subsystems of the used hearing instruments were equivalent and thus satisfy the same requirements and design.
Interestingly, PEEP application prevented this spread even when tidal volume was equivalent and thus end-inspiratory pressure higher (Figure 11).
(ii) (iii) Let be an eventually positive solution of (3.2), the case where is an eventually negative solution to (3.3) is equivalent, and thus we omit it.
Since we have only equalities in the calculations above we conclude that (2.3) and (2.5) are equivalent and, thus, by Theorem 2.1, (a) is proved for the case (a1).
All FCD geometries are similar in that they induce an additional pressure drop as fluid travels between the reservoir and the completion base pipe so that the total pressure drop for any fluid flow path is equivalent and thus fluid conformance is maximized (Atkinson et al. 2004).
Then the two conditions '(operatorname{ess}limsup_{trightarrow a} E t)leq E_{0})' and '(limsup_{trightarrow a} E t)leq E_{0})' are equivalent, and, thus, the conclusions in Theorems 1.1-1.2 1.1-1.2(t)) defined by (27) still hold, in particular, withavE tE(t):=E[rho,mathbf{u} ](t)leq E_{infty}) for (t>T) in Theorem 1.1.
Interestingly, PEEP application prevented this spread even when tidal volume was equivalent and thus end-inspiratory pressure higher.
Note that since the process is reversible, reversions and convergences are equivalent, and thus, as a function of n, what we are computing is the n-step homoplasy probability, i.e. the probability that at a given site, two species separated by n > 0 substitutions along the tree will be found in the same character state.
In the case \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K=1$$\end{document} K = 1 these two values are equivalent, and thus, we have perfect agreement with the existing theory.
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