Matematika

Pertanyaan

jika P=7per6, Q=1 1per2, dan R=3per4 ,maka tentukan nilai!
a. pq-qr
b.(p-q)²-(q-r)²

1 Jawaban

  • a. pq-qr
    =(7/6x11/2)-(11/2x3/4)
    =77/12-33/8
    =154/24-99/24
    =55/24

    b. (p-q)^2-(q-r)^2
    =(7/6-11/2)^2-(11/2-3/4)^2
    =(7/6-33/6)^2-(22/4-3/4)^2
    =-26/6^2-19/4^2
    =169/9-361/16
    =2704/144-3249/144
    =-545/1448

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