Band has 1-1x0+3/3 drummers

>band has 1-1x0+3/3 drummers

>band has 0 drummers

wtf are you talking about

It's 2, though you could see it as 4/3 if you did the math wrong. I don't know how you got 0.

Where's your parentheses nigga?

Nice bait.

not needed if you understand pemdas

it's zero retard

Name exactly 57 bands with 2 drummers

>band has -1/12 drummers

>1+1=0

It's 2
>1-1x0+3/3
No parenthesis or exponents
>1x0 = 0
>3/3 = 1
Now we have 1-0+1
>1-0=1
>1+1
>2

the fact that you even have to say this is really something

Amount of drummers equate to the solution of the Riemann Zeta function of 2

>1-1x0+3/3
>-1x0=0
>3/3=1
>1+1=2

>1+1=2

lmao nerd

isnt 1-0 = 0 though ?

>using x to signify multiplication
underage

I know what this number is and im scared of it

2+2=5

...

>band has sqrt(-1) members
Meme magic

1-1=0
0×0=0
0+3=3
3/3=1

How is that so hard?

he didn't explicitly state whether it's supposed to be x or * and he didn't have to because both of them work here.

1-1x0+3/3=0
1*x=x
x*0=0
1-(+3/3)=0
1-1=0

>1-1x0+3/3=0
what did he mean by this?

the point of these threads are to bait each other with bad math

good one

>band has -1*(1+sqrt(-π^2)+(sqrt(-π^2))^2)/2+(sqrt(-π^2))^3)/6+(sqrt(-π^2))^4)/24+(sqrt(-π^2))^5)/120+...) drummers

>1-(+3/3)=0

3/3 = 1
1-1=0

tards
1-1x0+3/3
1-1=0
=0x0=0
3/3=1
1+0=1
baka americans

>=5

>band has qtπ drummer

>band has sqrt(81)(sin2x+sin2y)/4.5 drummers

>band has 1x1 drummers
Only fa/tv/irgins will understand this reference

>ensemble has [ζ(3)]/|Aut(A_4)|*ϑ(π/2i) conductors

I'm sorry to hear that desu

Hi Terrence

>band has root 2 drummers