Kunci Jawaban Matematika Kelas 9 Halaman 58-59 Soal Nomor 1-13 tentang Pangkat Nol, Pangkat Negatif

Pembahasan: ((xⁿ⁻⊃1; yⁿ)⊃3;)/(x⊃2;ⁿ y⁶⁺ⁿ)
= (x⊃3;ⁿ⁻⊃3; y⊃3;ⁿ)/(x⊃2;ⁿ y⁶⁺ⁿ)
= x⊃3;ⁿ⁻⊃3;⁻⊃2;ⁿ y⊃3;ⁿ⁻⁶⁻ⁿ
= xⁿ⁻⊃3; y⊃2;ⁿ⁻⁶
maka,
xⁿ⁻⊃3; y⊃2;ⁿ⁻⁶ = xᵃyᵇ
a = n-3 dan b = 2n-6
Nilai b/a =
b/a = (2n-6)/(n-3)
b/a = (2(n-3))/(n-3)
b/a = 2

Soal nomor 4

a. y⊃3; × (3y)⊃2;
= y⊃3; × 3⊃2; × y⊃2;
= 9 × y⊃3;⁺⊃2;
= 9 y⁵

b. b 2y⁵ × b⊃3; 6 y⊃2;
= b⊃1;⸍⊃2; . 2 y⁵ × b⊃3; . 6 y⊃2;
= b⊃1;⸍⊃2;⁺⊃3;. 12 . y⁵⁺⊃2;
= b⁷⸍⊃2; . 12 . y⁷
= b⁷⸍⊃2; 12 y⁷

c. (tn⊃3;)⁴ × 4t⊃3;
= t⁴ n⊃1;⊃2; × 4t⊃3;
= 4 n⊃1;⊃2; t⁴⁺⊃3;
= 4 n⊃1;⊃2; t⁷

d. (2x⊃3;) × 3(x⊃2;y⊃2;)⊃3; × 5y⁴
= 2 x⊃3; . 3 ×⁶ y⁶ . 5 y⁴
= 2 . 3 . 5 . x⊃3; . x⁶ . y⁶ . y⁴
= 30 x⊃3;⁺⁶ y⁶⁺⁴
= 30 x⁹ y⊃1;⁰

Soal nomor 5

a. 0,00000056 = 5,6 × 10⁻⁷

b. 2.500.000 = 2,5 × 10⁶

c. 0,98 = 9,8 × 10⁻⊃1;

d. 10.000.000.000.000 = 1 × 10⊃1;⊃3;

Baca juga: Kunci Jawaban Buku Bahasa Indonesia Kelas 9 Halaman 11 tentang Mengidentifikasi Teks

Soal nomor 6

a. a. 12 × 2⊃3;
= 12 × 8
= 96
= 9,6 × 10

b. 7,27 × 10⊃2; – 0,5 × 10⊃3;
= (7,27 × 100) – (0,5 × 1000)
= 727 – 500
= 227
= 2,27 × 10⊃2;

c. (8,32 × 10⁴) : (4 × 10⁻⁶)
= (8,32 : 4) × 10⁴⁻⁽⁻⁶⁾
= 2,08 × 10⊃1;⁰