[x²y²/a³b³]n plz answer me faster i will mark u branliest​

Question

[x²y²/a³b³]n

plz answer me faster i will mark u branliest​

in progress 0
Eliza 4 weeks 2021-11-03T20:17:17+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-03T20:18:22+00:00

    Answer:

    Hi ,

    p = a²b³

    q = a³b

    HCF ( p,q ) = a²b

     [ ∵Product of the smallest power of each

         common prime factors in the numbers ]

    LCM ( p , q ) = a³b³

    [ ∵ Product of the greatest power of each 

      prime factors , in the numbers ]

    Now ,

    HCF ( p , q ) × LCM ( p , q ) = a²b × a³b³ 

    = a∧5b∧4 ——–( 1 )

    [∵ a∧m × b∧n = a∧m+n ]

    pq = a²b³ × a³b

    = a∧5 b∧4 —————( 2 ) 

    from ( 1 ) and ( 2 ) , we conclude 

    HCF ( p , q ) × LCM ( p ,q ) = pq

    Step-by-step explanation:

    hope it helps

    mark me brainiest

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )