How many bowlingballs are there, Sup Forums?
How many bowlingballs are there, Sup Forums?
Other urls found in this thread:
m.wolframalpha.com
twitter.com
there's at least 10
a decent amount
More than 2.
A whole damn stack of em
165
Impossible to tell but quite a lot
45+36+28+21+15+10+6+3+1
165
9x9^9 / 3
A plethora
81
In view, obscuring what could be something what is not bowling balls behind and inside.
lots
165 is the correct answers. The easiest way to calculate it is to use the binomial coefficient representation of the tetrahedral numbers. A tetrahedral number with a base edge length of n is calculated as (n+2 choose 3). So we have (9 choose 3)=165.
Shit I fucked up. It would be (11 choose 3) = 165.
Far too many to get in my ass.... but I'd try.
At least 2.
( x = x +1 ) ^2 * 9
285?
361 niggers
...
9 width
so
9*9+8*8+7*7+6*6+5*5+4*4+3*3+2*2+1= 285
the correct answer is 165. Easiest way to get the answer without being a fucking mathematician is to find the number of balls in the bottom row. After that just subtract how many balls are in the longest row from the stack below it then add them all together. Example: Bottom row has 45 balls,subtract 9(amount of balls in the longest row) which leaves you with 36,the amount of balls in the second to bottom row.
this
That would be for a square,ya dingus
Thats what I got aswell. I read engineering math recently but not sure if i fucked up kek
262
well i presume it's a square pyramid, it could be a triangle, but i forgot how to calculate that.
n^2 + n
81. i win
i counted all the balls
no joke
1- 1
2- 3
3- 6
4- 10
5- 15
6- 21
7- 28
8- 36
9- 45
total is 165
165
honest question, how do you calculate this.
pascals triangle, or the equation for a triangular based pyramid
0. Those aren't bowling balls, they're alien eggs. We're all gonna die...
square pyramidal number for a square and tetrahedral number for triangular. you can easily find the answer on google with just knowing how many stacks of balls there are.
>pascals triangle
thanks
Refer to
about three fiddy
On the assumption it is individual layers form the ascending triangular numbers ranging from T1 to T9, if we take the sum of these numbers from 1 to 9 using the following summation formula (which can be proved by induction):
n(n+1)(n+2)/6
substitute n = 9
9(10)(11)/6
This gives 165.