So I heard you faggots will argue about anything

So I heard you faggots will argue about anything

Other urls found in this thread:

wolframalpha.com/input/?i=9-3÷⅓+1
twitter.com/AnonBabble

why argue? this is a shitty thread.

1

it's 8, fuck off nigger

This is shit. It's one.

1

-9

1

It's 1 faggots. BEDMAS.

1

you niggers if you use PEMDAS its 9

discalculics say 1
I say 19

4

9

9-3÷(1/3)+1 = 4

Heres how:

Now with bodmas we go left too right.

6÷(1/3)+1
6÷1 is 6
6÷3 is 2
So
(6/2)=3
3+1
4

Its just some simple nath guys, how in gods great name can you get this wrong?

It is 7, 3 / (1/3) = 1
9-1-1=7

If you actually use order of operations, you get 1. Multiplication/Addition comes before addition or subtraction so we can solve 3/(1/3). From that you get 9. And if we then substitute 9 back into the place of 3/(1/3) we get 9-9+1. Which is 1.

What the fuck

9-3÷(1/3)+1
9-3÷0.3+1
9-9+1
0+1
1

he's right you know

hey bud, just saying your wrong.

>bodmas
>pemdas
>jewdas
Fuck off it's spelt bidmas
Brackets, indicies, division, multiplication, addition subtraction.

arent your suppose to do addition before subtraction or am I just tripping?

>3 / (1/3) = 1
FFS

WOw the Long list if Fagots that cant do math..
Yes, this guy knows whats up . PEMDAS is the not just a suggestion , but a Rule ..

3/(1/3)=9 bud

Hey bud, just saying if you are going to make a dumb statement, at least show your work :)

Both subtraction and addition have the same priority. So you do which ever comes first left to right

WolfromAlpha says it's 1

you dont do (6/6)/(1/3), that like saying 1/(1/3).
You would have to use (6/1)/(1/3) for the original equation to be relevant

no

Since Sup Forums loves fractions lets try fractions inside of fractions.

Math is boring

Hate to break it to you, 6/6 is 1.

I was only using 6. Dont add extra numbers to try and make me look dumb. Cause youll only do it to yourself :)

Pay attention niggers, division goes first.

The correct answer is 0.9999...
Anyone who disagrees has nigger level math skills.

If I remember correctly, there are some cases where that order would actually matter, but not for this problem.

Also +1 for the answer equaling 1. It's astonishing how order of operations stumps so many people. Then again, it's hard to tell who's baiting and who actually can't do math.

Riddle me this. Why would you solve the 1/3 before the 3÷1/3?
Both are division.

1/3 = 0,333333333
3/0,333333333 = 9,0000009
9-9,0000009 = 0,00009
1+0,00000009 = 1,00000009
gg ez

What about this

0.9999
Also known as 1, you pretentious faggot

In this case it actually does matter. If you add first its -1. Sub(lile you should) first, 1

1/3 == 3^-1
is exponential

>there are some cases where that order would actually matter
This comes up in conditional convergence for infinite products and infinite series.

It's also important in non-Abelian group multiplication.

Oh man, I can't believe anyone saved these. I made a whole set of these a few years ago. I'll post them.

...

9-3*3+1
9-9+1
1

got it faggot?

...

...

0.9999... is also known as 1
0.9999 is not
nigger

...

retard... its 1.000000000000.....1

...

huh

well, shit

I actually thought the answer was 9, but it turns out the way it is written there are implied to be brackets around the 1/3

wolframalpha.com/input/?i=9-3÷⅓+1

the answer really is 1

...

>9-3/(1/3)+1
>9-9+1=1

...

Very tru there. But wouldnt it also be

3^-1/3

...

...

>inb4
>there are no ( ) so fuck off it's in order

9-3=6
6/1_3=2
2+1=3

There are no priorities so it's in order you niggers,

THERE IS NO GODDAMN PARENTHESIS THE DIVISION WONT GO FIRST NIGGER

This one here is actually true under the analytic continuation of the zeta function. However, the way it is represented here is facetious.

You're all mentally challenged if you think 3/(1/3) is 1. 9-9+1=1

...

>Brackets
Why is Captcha purpously trying to stop me from posting?

Autism intensifies

And here is an actual problem for any mathfags.

>
>
>
>
>
>
>
>
>
> WOw the Long list if Fagots that cant do math..
>
>Yes, this guy knows whats up . PEMDAS is the not just a suggestion , but a Rule ..
drink bleach

9-3/ONE THIRD+1

9-3/ 1_3 +1 =/= 9-3/ (1/3) + 1

You believe that there is an end to infinitely repeating 0s, and beyond that end, there sits a 1.
Excellent.

9, pemdas

Niggers pemdas isn't the solution for everything

I come up with...
From the creator of "the race card" comes...

The "pemdas card"

yes, there is a theoretical 1, not actual 1

U fuckin retard

lol

its 1.

Oh my god
9-3÷1/3+1
9-(3/1×3/1)+1
9-9+1
1

I'm a nigger. You are a fag.

THERE ARE PRIORITIES!!! YOU CANT JUST GO 1+2÷2 AND SAY ITS 3÷2 YOU DUMB CUNT PRIORITY × AND ÷ ALWAYSSSSSS

its an alternating series, either 1 or 0

>its an alternating series
It can be re-written as an alternating series.

>either 1 or 0
It's not. It's larger than 1.

>It's not. It's larger than 1.
Woops, meant to say smaller.

1

When dividing with fractions remember to keep change and flip

So 3 / 1/3 becomes 3 x 1/3
therefore, 9 - 9 + 1 = 1

But fuck this, i'm sick of seeing a bunch of fake ass smarty pants downplaying people who are doing just fine in life and have abandoned mathematics, for their own ego masturbation

Are you really that retarded that you have to write this cancerous pasta down instead of solving it in like 3 seconds in your head?

More specifically, 0

meant that when n is sufficiently large enough the limit is either 1 or 0, for a lower n it is indeed just lower than 1

goes to infinity

>meant that when n is sufficiently large enough the limit is either 1 or 0
This is still wrong.

No, idiot, 0>/\(3)

care to explain?

No. The factors tend to 1 fast enough that P(8) converges.

i/10

Do you want the answer, or do you just want me to say why it doesn't go to 1 or 0?

Im trying to help the less fortunate here. Im showing my work so maybe some of these retards might learn something.

every factor is larger than 1, multiplying an infinite amount of factors wich are all >1 goes to infinity

no, 3^-1/3 would be 1/cuberoot(3)
just like 3 ^-1/2 would be 1/sqrt(3)

You don't understand limits or infinite products.

tiny brain

how so, both those statements are correct

look, nomatter how large n becomes P will always get larger with every n. if you think you know better please give a valid explanation

niarb ynit

>if you think you know better please give a valid explanation
I do know better. Give me a moment to write it out.

Also, this is very well established mathematics.

9-3/1/3+1
9-1+1
8+1
9
Ez