Question

Tolong jwb nomor 2 dan 3 aja ya pake cara. Terima kasih​

Answer 1

Jawab:

Nomor 2 :

S∞ = 8/17

a) a = 2, r = (2/p) / 2 = 1/p

S∞ = a. (1 - r)

8/17 = 2. (1 - r)

(1 - r) = 8/17 ÷ 2

(1 - r) = 4/17

1 - 1/p = 4/17

1/p = 1 - 4/17

1/p = 17/17 - 4/17

1/p = 3/17

p =17/3

b) a = 3, r = (6/p) / 3 = 2/p

S∞ = a. (1 - r)

8/17 = 3. (1 - r)

(1 - r) = 8/17 ÷ 3

(1 - r) = 8/51

1 - 2/p = 8/51

2/p = 1 - 8/51

2/p = 51/51 - 8/51

2/p = 43/51

p = 2 / (43/51)

p = 2 × 51 ÷ 43

p = 102/43

Nomor 3 :

Option a :

Bola jatuh = 20 meter.

Lintasan Pantulan ke-1 = 20 × 4/5 = 16 meter.

Lintasan Turun ke-1 = 16 meter.

Lintasan Pantulan ke-2 = 16 × 4/5 = 12,8 meter.

Lintasan Turun ke-2 = 12,8 meter.

Lintasan Pantulan ke-3 = 12,8 × 4/5 = 10,24 meter.

Lintasan Turun ke-3 = 10,24 meter.

Lintasan Pantulan ke-4 = 10,24 × 4/5 = 8,192 meter.

Panjang Lintasan Bola sampai Pantulan ke-4

= 20 + 16 + 16 + 12,8 + 12,8 + 10,24 + 10,24 + 8,192

= 106,272 meter

Option b :

Panjang lintasan = ketinggian bola jatuh + (2 × deret takhingga)

Ketinggian bola jatuh = 20 m

a = pantuklan ke-1 = 4/5 × 20 = 16

r = 4/5

S∞ = a / (1-r) = 16 / (1 - 4/5) = 16 ÷ 1/5 = 80 meter

Panjang lintasan = 20 m + 80 m = 100 meter