Matematika

Pertanyaan

1)Bentuk sederhana dari
(3y+5)²-(y-7)² adalah
A*8y²+8y+74
B*8y²+16y+74
C*8y²+22y-24
D*8y²+44y-24

2)hasil dari(2³:½)²/6X 2½
A*256/15
B*144/15
C*16/15
D*15/16
3)jika P=1 ⅔, Q=¾,dan R =-⅛..maka nilai dari 6p+3q-2r adalah
A.14
B.12¹/¹²
C.11
D.9

2 Jawaban

  • 1. (3y+5)^2 -(y-7)^2
    = 9y^2+30y+25 -y^2+14y-49
    = 8y^2+44y-24 (D)

    2. (2^3/1/2)^2 / 6 × 1
    = 2^8 / 6
    = 128/3 pernyataan nomor 2 tidak jelas!!

    3. 6p+3q-2r
    = 6×2/3 + 3 × 3/4 - 2 × -1/8
    = 12/3 + 9/4 + 2/8
    = 13/2
  • (1)
    (3y+5)^2 - (y-7)^2
    = 9y^2 + 30y + 25 - (y^2 - 14y + 49)
    = 9y^2 + 30y + 25 - y^2 + 14y - 49
    = 8y^2 + 44y - 24 (D)
    (2)
    (2^3 :1/2)^2 /(6×2 1/2)
    (8 : 1/2)^2/(6 × 5/2)
    16^2/(30/2)
    256/15 (A)
    (3)
    6p + 3q - 2r
    = 6(1 2/3) + 3(3/4) - 2(-1/8)
    = 6(5/3) + 9/4 - (-2/8)
    = 30/3 + 9/4 + 2/8
    = 10 + 2 1/4 + 1/4
    = (10 + 2) + (1/4 + 1/4)
    = 12 + 2/4
    = 12 2/4
    = 12 1/2(....)

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