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VLD [36.1K]
3 years ago
6

Which of the following is a geometric sequence? Help pleaseee!

Mathematics
2 answers:
ioda3 years ago
7 0

Answer: B

Step-by-step explanation:

Division of components are consistent  - the same

Natalija [7]3 years ago
3 0

Answer:

B. -3, 3, -3, 3...

Step-by-step explanation:

There's two types of sequences, arithmetic and geometric.

Arithmetic equations are<u> sequences that increase or decrease by adding or subtracting the previous number.</u>

For example, take a look at the following sequence:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20...

Here, the numbers are increasing by +2. [adding]

So, this the sequence is arithmetic, since its <u>adding</u>.

Geometric sequences are <u>sequences that increase or decrease by multiplying or dividing the previous number.</u>

For example, take a look at the following sequence:

2, 4, 16, 32, 64, 128, 256, 512...

Here, the numbers are icnreasing by x2. [multiplying]

So, the sequence is geometric since its <u>multiplying</u>.

Based on this information, the correct answer is "B. -3, 3, -3, 3..." since its being multiplyed by -1.

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It is 3. Sorry if it's wrong.

Step-by-step explanation:

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3 years ago
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How do you write 6/72 in simplest form
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1/12
5 0
3 years ago
5+4g+8=1 (g=-3;g =-1; g =2)
makkiz [27]
When g equals- 3 answer is 1

4 0
3 years ago
What is the value of x?
Fittoniya [83]
Let the 2 equal lengths be represented by 2p.  Then you can write the equation of ratios

  p       2p
---- = -------
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8 0
3 years ago
Suppose the names of the months of the year are placed in a hat and a name is drawn at random. (a) list the sample space for thi
ollegr [7]

(a)

The sample space is a set whose elements are all the possible outcomes for the experiment. Since we will extract one of the months of the years, the sample space is the set composed by all the 12 months:

\Omega = \{ \text{January},\ \text{February},\ldots,\text{December}\}

(b)

An event is a subset of the sample space. Events are often defined by their properties. In this example, the event E is the subset of the sample space defined as

E = \{ x \in \Omega: x \text{ starts with the letter J}\}

So, we have

E = \{\text{January},\ \text{June},\ \text{July}\}

(c)

If all outcomes have equal probability, then the probability of an event is the ratio bewteen its cardinality, and the cardinality of the whole sample space:

P(E) = \dfrac{n(E)}{n(\Omega)} = \dfrac{3}{12} = \dfrac{1}{4}

In words, since there are three months beginning with J out of 12 months, we have a probability of 3 over 12 to pick a month starting with J, which simplifies to 1 over 4.

3 0
3 years ago
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