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2 years ago

10

Need help with this

1 answer:

kvasek [131]2 years ago

6 0

Answer:

The first answer.

Step-by-step explanation:

First of all, the circle on -3 did not have filling so that means it is not equal to z. Then we look on the circle on 6. It does have a filling which means it is equal to and greater than z. Now we look in between -3 and 6. Is there any white spots? If no, then we don’t have to use an OR. Therefore, the first answer is the only plausible answer.

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The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i

Anit [1.1K]

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:

$a_n = ar^{n-1}$

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

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

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

4 0

2 years ago

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Viefleur [7K]

Answer:

This is only true with a cube

Step-by-step explanation:

Only a cube has 6 equal faces of equal size.

I can't really put a diagram here so i hope that explanation helped!

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