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Lunna [17]
3 years ago
11

True or false: Every sequence is either arithmetic or geometric. If this is true, explain. If false,

Mathematics
1 answer:
tiny-mole [99]3 years ago
5 0

In an arithmetic sequence, the difference between consecutive terms is constant. In formulas, there exists a number r such that

a_{n+1}-a_n=r\quad \forall n\geq 1

In an geometric sequence, the ratio between consecutive terms is constant. In formulas, there exists a number r such that

\dfrac{a_{n+1}}{a_n}=r\quad \forall n\geq 1

So, there exists infinite sequences that are not arithmetic nor geometric. Simply choose a sequence where neither the difference nor the ratio between consecutive terms is constant.

For example, any sequence starting with

1, 15, -3,\ldots

Won't be arithmetic nor geometric. It's not arithmetic (no matter how you continue it, indefinitely), because the difference between the first two numbers is 14, and between the second and the third is -18, and thus it's not constant. It's not geometric either, because the ratio between the first two numbers is 15, and between the second and the third is -1/5, and thus it's not constant.

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How to do this problem​
Mumz [18]

Answer:

y = 3/2x -14

Step-by-step explanation:

do the inverse of -2/3 what is 3/2

and now sole the equation of y+8 = 3/2(x-4)

distribute 3/2 ---> y+8 = 3/2x - 8

subtract 8 from both sided getting you y=3/2x-14

7 0
3 years ago
F (x)=5x+1 9(x)=-2x+15
german

Answer:

x = 0

Step-by-step explanation:

(x)=5x+19(x)=-2x+15

We move all terms to the left:

(x)-(5x+19(x))=0

We add all the numbers together, and all the variables

x-(+24x)=0

We get rid of parentheses

x-24x=0

We add all the numbers together, and all the variables

-23x=0

x=0/-23

x=0

4 0
3 years ago
Please helppppp me What’s the answer?
taurus [48]

Answer:

She shaded above the line when she should have shaded below the line.

Step-by-step explanation:

hope this helps and is right. p.s. i really need brainliest :)

if this is wrong then I am incredibly sorry!

5 0
2 years ago
 An athlete runs an equal distance 2 days a week. The other 5 days of the​ week, he runs a total of 11 miles. Write an equation
katovenus [111]
Ok since the athlete an equal distance of 2 days a week and he runs a total of 11 miles in 5 days you will need to think back so before he had 11 miles done next weak he had a total of 19 miles so you will start at 11 then count up to 19.

Use a number line to show how many days it took for the athlete to get 19 miles.

Then once you do the number line count the lumps and that’s how many days it took for the athlete to get 19 miles.

Hope this helped.
3 0
3 years ago
Please help very urgent
Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 1

Explanation:

The geometric mean can be represented by \sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}.

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

a_{n} = a_{1} • r^{n-1}

Where a_{n} is the nth term, a_{1} is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, a_{1} will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

\sqrt[5]{\frac{a_{6}}{a_{1}}} =

\sqrt[5]{\frac{1}{32}} =

\frac{1}{2}.

Therefore: r = \frac{1}{2}.

Using all the information we have, we can find the explicit rule:

a_{n} = a_{1} • r^{n-1}

  • a_{1} = 32
  • r = \frac{1}{2}

a_{n} = a_{1} • r^{n-1} →

\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

a_{1} = 32 • (\frac{1}{2})^{1-1}

a_{1} = 32 • (\frac{1}{2})^{0}

a_{1} = 32 • 1

a_{1} = 32

a_{6} = 32 • (\frac{1}{2})^{6-1}

a_{6} = 32 • (\frac{1}{2})^{5}

a_{6} = 32 • \frac{1}{32}

a_{6} = 1

3 0
3 years ago
Read 2 more answers
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