Sentence examples for double contributing from inspiring English sources

Exact(1)

The midfielder bounced back to inspire Bayern to a league and cup double, contributing to his haul of eight Bundesliga titles, seven DFB-Pokals and the 2013 Champions League title with the club.

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Musial's Game 1 triple and Game 2 double contributed to the Cardinals' two-games-to-none series victory.

The Sea Kings earned their second straight win in the Beach Pit Classic with seven hits and no errors Tuesday at CdM. Sam Younesi (one for two, two RBIs, double), Nolan McCarthy (two for three, three RBIs), Evan Larsen (RBI) and Brett Olson (double) contributed to help CdM (5-12) rally from a 5-1 deficit.

While Hingis could go on to concentrate on her quarterfinal singles victory over Monica Seles, Gambill said that playing mixed doubles contributed to his third-round singles loss to Thomas Johansson.

Many studies used double tasks contributing greatly to the field of postural control [1], [2].

Previously, Hirschorn headed programming at VH1, where over a span of four years he doubled ratings, contributing to a more than tripling in revenue for the network.

He notched a near triple-double in contributing 19 points, 13 rebounds and nine blocks in the semifinals against the Rebels.

Similarly, isolated double bonds contribute no colour to a hydrocarbon, but, in a molecule containing a chain of alternating single and double bonds, intense coloration results from conjugation.

Poulter opened with three straight birdies before four bogeys and a double bogey contributed to his defeat in the three-man group format.

Seven Sharapova double faults contributed to her demise in the decider and Cibulkova showed no nerves as she matched her French Open victory over the Russian of five years ago.

If we expand M 2 = ( C k 1 + C k 2 + … + C k m ) 2, the double products contribute only even (~0) entries: M 2 = C 2 k 1 + C 2 k 2 + … + C 2 k m mod 2. If N = 2 n, squaring matrix M n-times gives M N = C N k 1 + C N k 2 + … + C N k m mod 2. Because C N k 1 = ( C N ) k 1, and C N = I (the unit matrix), M N = m·I mod 2. This proves the following.

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