Matematika

Pertanyaan

1 per x²y × 2x per 3y² × 5 per 3x × (4y)²

1 Jawaban

  • 1/x²y × 2x/3y² × 5/3x × (4y)²
    = 1/x²y × 2x/3y² × 5/3x × 16y²
    = (1×(2/3)×(5/3)×16) . x^(-2+1-1) . y^(1-2+2)
    = 160/9 . x^-2 . y^1
    = 160y/9x²

    ^ tanda pangkat

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