kakpoinyya 26 tolong no 2 sampe 5
Matematika
siskavalen
Pertanyaan
kakpoinyya 26 tolong no 2 sampe 5
1 Jawaban
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1. Jawaban arsetpopeye
2) a = (1 , 1) => panjang vektor a = |a| = √(1^2 + 1^2) = √(1 + 1) = √2
Sudut yang dibentuk a dan sumbu x positif :
Tan ¢ = y/x = 1/1 = 1 = Tan 45°
¢ = 45° = π/4
3) a = (-3 , -2), b = (-5 , 4), c = (9 , -16)
A. c = 3a - 2b = (-9, -6) - (-10, 8) = (1, -14) ..... salah
B. c = -3a + 2b = (9, 6) + (-10, 8) = (-1, 14) ..... salah
C. c = 2a - 3b = (-6, -4) - (-15, 12) = (9, -16) .... BENAR
4) PQ = 1/7 PR => PR = 7 PQ
SR = 1/2 QR
QR = QP + PR = -PQ + 7PQ = 6PQ
PS = PR + RS
= PR - SR
= 7PQ - 1/2 QR
= 7PQ - 1/2 (6PQ)
= 7PQ - 3PQ
= 4PQ
Jadi PR : PS = 7 PQ : 4 PQ = 7 : 4
5) buat gambar segienam ABCDEF
CD = CB + BA + AD
= CB + BA + 2(BC)
= b - c + a - b + 2(c - b)
= -c + a + 2c - 2b
= a - 2b + c