A large quantity of waste material contains metal. At the successive passes through a recovery process the mass of the metal recovered is:
32kg at the first pass,
24kg at the second pass;
18kg at the third pass;
13,5kg at the fourth pass, and so on, to form a geometric sequence,
1. How much metal would be recovered at the eighth pass?
(Round off the answer to TWO decimal places)
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The amount of metal recovered at the eight pass is approximately 4.27 kg

Step-by-step explanation:

The mass of metal recovered formed the following geometric sequence;

32 kg, 24 kg, 18 kg, 13.5 kg

∴ The first term, a = 32 kg

The common ratio in the geometric sequence is found as follows;

r = a·r/a = 24 kg/(32 kg) = 0.75

r = a·r²/a·r = 18 kg/(24 kg) = 0.75

r = a·r³/a·r² = 13.5 kg/(18 kg) = 0.75

The mass of the metal recovered in the eight pass will be the 8th term of the geometric series given as follows;

nth term = a·rⁿ⁻¹

a = 32, r = 0.75, n = 8

∴ 8th term = 32·(0.75)⁸⁻¹ = 32·0.75⁷ = 2187/512 ≈ 4.27

The amount of metal recovered at the eight pass ≈ 4.27 kg.

side xy = 10

sin(x) = 4/5

cos(x) = 3/5

tan(x) = 4/3

step-by-step explanation:

the answer to this question is in the pic