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9th grade+ math help?


Picc84

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I was arguing with my teacher today, about something more complicated than it needed to be, but i wanted to know about the more complicated version, and she did not want to tell me, so now i need to ask someone on hear! please help me...

Ok, we were working with patterns... and this was the pattern...

2,5,11,23,47,95,192,386

The Math for the pattern the easy way is...

n2+1

Also there is another pattern...

n3 (when "3") doubles each time...

How would i write that in a mathamatical formula? such as n2+1?

If you do not understand, i will try and explain more, if you need it, but please help, thank you!

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n3 (when "3") doubles each time...

Would be written:

n(3^n)

When you say "doubles each time" that implies an exponential function (3 to the power of n, in your example).

If n=1 then you get 1(3^1)=1(3)=3

If n=2 then you get 2(3^2)=2(9)=18

If n=3 then you get 3(3^3)=3(27)=

Hmmm... I don't know, I can't count that high. :P

Is that what you were looking for?

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n3 (when "3") doubles each time...

Would be written:

n(3^n)

When you say "doubles each time" that implies an exponential function (3 to the power of n, in your example).

If n=1 then you get 1(3^1)=1(3)=3

If n=2 then you get 2(3^2)=2(9)=18

If n=3 then you get 3(3^3)=3(27)=

Hmmm... I don't know, I can't count that high. :P

Is that what you were looking for?

well you cannot use the powers of 3, because you need to use odd numbers such as "6,12,18,ect..." to keep the pattern going...

and 3(27)=81 i belive...

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well i do not want to use the -1 method... thats why i was wondering about this...

hm... lemme see, hears a better discription if its evan better? lol

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

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Ah, yes. Now I understand what you want.

Yes, trickman is on to it. I would just use * instead of x for multiplication to be less confusing. Or, just leave it out.

n+3(n-1)

EDIT: trickman, you made a small error... remember, that any number raised to the power of 0 = 1.

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well i do not want to use the -1 method... thats why i was wondering about this...

hm... lemme see, hears a better discription if its evan better? lol

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

Well I think the only way to do this is the n-1 method...

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Like i said, i do not want to use the (-1) method, thats the easier way, i wanted to use the harder way like this...

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

But your right about that way! lol

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n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

In your first post you said that the 3 should "double" each time, but going from 6 to 9 is not doubling.

So then, you want this pattern?

n+3, where n=1

n+6, where n=2

n+9, where n=3

n+12, where n=4

n+15, where n=5

If so, your formula would be:

n+3n

or simply

4n

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  • 1 year later...

You think that's hard.

Try graphing absolute inequality equations. :D

I'm doing that in algebra II right now.

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Ooh, those are my favorite. :-)

 

The Doctor: There was a goblin, or a trickster, or a warrior... A nameless, terrible thing, soaked in the blood of a billion galaxies. The most feared being in all the cosmos. And nothing could stop it, or hold it, or reason with it. One day it would just drop out of the sky and tear down your world.
Amy: But how did it end up in there?
The Doctor: You know fairy tales. A good wizard tricked it.
River Song: I hate good wizards in fairy tales; they always turn out to be him.

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