По теореме Пифагора (ну почти):
|OA| = √((–4 – 0)² + (2 – 0)² + (4 – 0)²) = √(16 + 4 + 16) = √36 = 6.
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По теореме Пифагора (ну почти):
|OA| = √((–4 – 0)² + (2 – 0)² + (4 – 0)²) = √(16 + 4 + 16) = √36 = 6.