a.
This is the geometric sequence: a1 = 1.5 m = 150
cm, r = 0.74
The formula now is an = a1 x r ^ n-1
Just substitute a1 = 150 cm and r = 0.74
an = 150 x (0.74) ^ n – 1
b.
The second part is:
A6 = 150 x 0.74 ^ 6-1
= 150 x 0.74 ^5
= 150 x 0.221906624
= 33.28509936 cm
Answer:
Possible lengths of the silk fibre 'x' are 100, 75, 60, 50 and 30 yards respectively.
Step-by-step explanation:
Let the length of the silk fiber = x yards.
Let the number of silk fibres = a.
It is given that the silk fibers between 3 to 10 yards are combined to form a silk thread of 300 yards.
So, we get the equation,
, where 
i.e.
, 
So, the possible values of 'a' are 3, 4, 5, 6 an 10.
Thus, the possible lengths of the silk fibre 'x' are 100, 75, 60, 50 and 30 yards respectively.
<h3>I'll teach you how to solve 125 1/3</h3>
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125 1/3
Convert mixed numbers to improper fractions:
125 1/3=
125*3+1/3=
376/3
Your Answer Is 376/3
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