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Norma-Jean [14]
2 years ago
7

Find the 37th term of 27, 19, 11​

Mathematics
1 answer:
Akimi4 [234]2 years ago
6 0

Answer:

291

Step-by-step explanation:

37=11+(37-1)*8

37 is the term you want

11 is the first term from least to greatest

(37-1) is from the term you want minus 1

*8 because 8 is the difference between each term

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Think about the sets of rational numbers, integers, and whole numbers.
Rus_ich [418]

Answer:

A'

Step-by-step explanation:

i think it is, sorry if im wrong

7 0
2 years ago
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What could explain what happened when the time was equal to 120 minutes
tensa zangetsu [6.8K]

Answer:

Step-by-step explanation:

Alright, lets get started.

Lets split the given graph in three parts.

First part: 0 to 30 minutes

We can see between this period, 0 to 30 minutes, the distance from home keeps increasing. It means from 0 to 30 minutes, Eli is moving towards the library.

Second part: 30 to 120 minutes

between 30 to 120 minutes,the distance from home does not changes. It means during this period, Eli is at the library.

Third part : 120 minutes to 135 minutes

Between 120 to 135 minutes, the distance from home is decreasing.

It means Eli is returning home in that period.

It means, at 120minute, Eli started her bicycle , home from the library. Hence option (b)     :    Answer

Hope it will help :)

5 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
You want to change your cell phone service. The new company charges $40 per month for cell phone use. There is a one-time start-
faltersainse [42]
25+40*t=c
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4 0
2 years ago
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6. Use the image below to find the missing value.
Katarina [22]

Option C:

The value of m is 4.

Solution:

Given data:

AB = 4m – 15

BC = 5m –6

AC = 15

<u>To find the value of m:</u>

Using segment addition postulate,

AB + BC = AC

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Arrange the like terms together.

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Divide by 9 on both sides of the equation.

m = 4

The value of m is 4.

Option C is the correct answer.

3 0
2 years ago
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