Las Vegas Raiders star defensive end Maxx Crosby has been named the AFC Defensive Player of the Week after a stellar performance in the Raiders’ 26-23 victory over the Baltimore Ravens. This marks the fifth time Crosby has earned this prestigious honor, highlighting his consistent dominance on the defensive line.
In the Week 2 matchup, Crosby was all over the field, playing every defensive snap for the Raiders. He recorded six total tackles, including an impressive four tackles for loss, two sacks, and a deflected pass. His relentless pressure on Ravens quarterback Lamar Jackson helped seal the victory for Las Vegas, as Crosby disrupted Baltimore’s offense from start to finish.
Crosby’s ability to perform at this high level comes as no surprise to fans and analysts alike. With this latest recognition, he has cemented his status as one of the premier defensive players in the NFL. Crosby currently leads the league in tackles for loss and continues to be the driving force behind the Raiders’ defense.
Looking ahead, the Raiders will return to Allegiant Stadium for their home opener against the Carolina Panthers, and all eyes will be on Crosby as he looks to continue his dominant season. With three sacks already through the first two games, Crosby is on pace for a career-best year.
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