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4 years ago

How do you determine if a sequence is geometric

Mathematics

2 answers:

alexira [117]4 years ago

6 0

Answer:

If a sequence is geometric, it will repeat itself.

Examples:

1, 2, 3, 4.....

3, 9, 27, 81.....

-1, 3, -9, 27.....

3, 6, 9, 12.....

An example of a sequence that is not geometrical would be:

2, 7, 3, 18, 26, 472....

Lubov Fominskaja [6]4 years ago

5 0

(I told you in the definition)

A geometric sequence is a sequence in which each term of the sequence is obtained by multiplying/dividing by a common value, called the common ratio, to the preceding term. Given a sequence, we can determine whether the sequence is arithmetic, geometric or neither by comparing the terms of the sequence.

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These dot plots show the heights (in feet) from a sample of two different

sweet-ann [11.9K]

Answer:a

Step-by-step explanation:

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3 years ago

Jenny multiplies the square root of her favorite positive integer by $\sqrt{2}$. Her product is an integer. a) Name three number

Dmitry [639]

Answer:

Part a) $2,50,18$

Part b) When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

Step-by-step explanation:

Let

x-------> the favorite positive integer

Part a)

1) For $x=2$

$\sqrt{2}*\sqrt{2}=\sqrt{4}=2$ -----> the product is an integer

The number $x=2$ could be Jenny favorite positive integer

2) For $x=50$

$\sqrt{50}*\sqrt{2}=\sqrt{100}=10$ -----> the product is an integer

The number $x=50$ could be Jenny favorite positive integer

3) For $x=18$

$\sqrt{18}*\sqrt{2}=\sqrt{36}=6$ -----> the product is an integer

The number $x=18$ could be Jenny favorite positive integer

Part B)

1) For $x=2$

$\sqrt{2}/\sqrt{2}=\sqrt{1}=1$ -----> the result is an integer

2) For $x=50$

$\sqrt{50}/\sqrt{2}=\sqrt{25}=5$ -----> the result is an integer

3) For $x=18$

$\sqrt{18}/\sqrt{2}=\sqrt{9}=3$ -----> the result is an integer

Therefore

When Jenny divides the square root of her favorite positive integer by $\sqrt{2}$ , she gets an integer

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3 years ago

An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c

Solnce55 [7]

The answer is 2 only

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3 years ago

Roger served 5/8 pound of cracker wich was 2/3 of the entire box what was the weight of the crackers originally in the box

horrorfan [7]

Answer:
original weight of the box = 15 / 16 pounds

Explanation:
Assume that the original amount in the box is x.
We are given that:
5/8 pounds represent 2/3 of the total amount (x).
This can be translated into the following equation:
(2/3) x = 5 / 8

Now, we will solve for x as follows:
(2/3) x = 5 / 8
Multiply both sides by 24 to get rid of the denominators as follows:
(2/3) x * 24 = (5 / 8) * 24
16 x = 15
Divide both sides by 16 to isolate the x as follows:
x = 15 / 16 pounds

Hope this helps :)

8 0

4 years ago

Whats the answer to this question

guapka [62]

Answer:

a=20

Step-by-step explanation:

7+13=20

20-7=13

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3 years ago

How do you determine if a sequence is geometric​

How do you determine if a sequence is geometric