Tugas 02 : Konversi Bilangan Ke Biner

Nama : pandu wiberson

Nim : 1500018021

Kelas : E

Berikut soal bserta jawabanya :

1. 01000101₂= …………. ₁₀

Jawab:

(0 x 2 ^n-1) + (1 x 2 ^n-2) + (0 x 2 ^n-3) + (0 x 2 ^n-4) + (0 x 2 ^n-5) + (1 x 2 ^n-6) + (0 x 2 ^n-7) + (1 x 2 ^n-8)

(0 x 2 ⁷) + (1 x 2 ⁶) + (0 x 2 ⁵) + (0 x 2 ⁴) + (0 x 2 ³) + (1 x 2 ²) + (0 x 2 ¹) + (1 x 2 ⁰)

= 0 + 64 + 0 + 0 + 0 + 4 + 0 + 2

= 70 ₁₀

2. 01110000₂= …………. ₁₀

Jawab:

(0 x 2 ^n-1) + (1 x 2 ^n-2) + (1 x 2 ^n-3) + (1 x 2 ^n-4) + (0 x 2 ^n-5) + (0 x 2 ^n-6) + (0 x 2 ^n-7) + (0 x 2 ^n-8)

(0 x 2 ⁷) + (1 x 2 ⁶) + (1 x 2 ⁵) + (1 x 2 ⁴) + (0 x 2 ³) + (0 x 2 ²) + (0 x 2 ¹) + (0 x 2 ⁰)

= 0 + 64 + 32 + 16 + 0 + 0 + 0 + 0

= 112₁₀

3. 01010101₂= …………. ₁₀

Jawab:

(0 x 2 ^n-1) + (1 x 2 ^n-2) + (0 x 2 ^n-3) + (1 x 2 ^n-4) + (0 x 2 ^n-5) + (1 x 2 ^n-6) + (0 x 2 ^n-7) + (1 x 2 ^n-8)

(0 x 2 ⁷) + (1 x 2 ⁶) + (0 x 2 ⁵) + (1 x 2 ⁴) + (0 x 2 ³) + (1 x 2 ²) + (0 x 2 ¹) + (1 x 2 ⁰)

= 0 + 64 + 0 + 16 + 0 + 4 + 0 + 2

= 86₁₀

4. 01100101₂= …………. ₁₀

Jawab:

(0 x 2 ^n-1) + (1 x 2 ^n-2) + (1 x 2 ^n-3) + (0 x 2 ^n-4) + (0 x 2 ^n-5) + (1 x 2 ^n-6) + (0 x 2 ^n-7) + (1 x 2 ^n-8)

(0 x 2 ⁷) + (1 x 2 ⁶) + (1 x 2 ⁵) + (0 x 2 ⁴) + (0 x 2 ³) + (1 x 2 ²) + (0 x 2 ¹) + (1 x 2 ⁰)

= 0 + 64 + 32 + 0 + 0 + 4 + 0 + 2

= 100₁₀

5. 1000110 ₂= …………. ₁₀

Jawab:

(1 x 2 ^n-1) + (0 x 2 ^n-2) + (0 x 2 ^n-3) + (0 x 2 ^n-4) + (1 x 2 ^n-5) + (1 x 2 ^n-6) + (0 x 2 ^n-7)

(1 x 2 ⁶) + (0 x 2 ⁵) + (0 x 2 ⁴) + (0 x 2 ³) + (1 x 2 ²) + (1 x 2 ¹) + (0 x 2 ⁰)

= 64 + 0 + 0 + 0 + 4 + 2 + 0

= 70₁₀

6. 345 ₈= …………. ₁₀

Jawab:

(3 x 8^n-1) + (4 x 8^n-2) + (5 x 8^n-3)

(3 x 8 ²) + (4 x 8 ¹) + (5 x 8 ⁰)

= 192 + 32 + 40

= 264

7. 12AB₁₆= …………. ₁₀

Jawab:

(1 x 16^n-1) + (2 x 16^n-2) + (10 x 16^n-3) + (11 x 16 ^n-4)

(1 x 16 ³) + (2 x 16 ²) + (10 x 16 ¹) + (11 x 16 ⁰)

= 4096 + 512 + 160 + 76

= 4944

3zoan