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djverab [1.8K]
3 years ago
6

Solve for w ........................

Mathematics
1 answer:
Kobotan [32]3 years ago
4 0

Answer:

w = -2 + 2z(2x-1)/y

Step-by-step explanation:

(2x-1)/y = (w+2)/2z

2z(2x-1)/y = w+2

-2 + 2z(2x-1)/y = w

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Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

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a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}

the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

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