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By considering relationship (2), the ordering cost (A) can be written as a linear function of (L), that is Aleft( L right) = x + yL (3)where (x = left( {1 - frac{1}{omega }} right)A_{0}) and (y = frac{{A_{0} }}{{omega L_{0} }}).
Think of it as an "L" so that the right goes to the top and the left to the bottom.
As a consequence, Φ ( L ) ∩ Ξ contains at least three points, precisely the nonempty intersections of Φ ( L left ) ∩ ( L left ′ − ( 2 k π, 0 ) ), Φ ( L right ) ∩ ( L left ′ − ( 2 k π, 0 ) ) and Φ ( L right ) ∩ ( L right ′ − ( 2 k π, 0 ) ). Suppose that j = m o ( T ) − 1 (in this case, j is even).
QUESTION FROM STEVE L: Am I right to take it that Mohseni lives in Dubai?
In particular Emmanuel Boateng tormented Tony Beltran all night down RSL's right flank.
And I am using a kind of an ultra device for L-series right now, for BB10.
The optimal ordering cost (Aleft( L right) = a + bln L)(for logarithmic case).
with equality if and only if K is homothetic to L. Note that the right-hand side of (14) is non-negative for x + r ( K, L ) L ⊆ K ( x ∈ R n ).
When (left| L right| = 2), it refers to a binary classification, and if (left| L right| > 2), it refers to a multi-class classification.
L-O-U, right?
i, k, l Right eye.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com