>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