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DJ's are good at math right?
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| JohnSmith |
i have an interesting problem, i am trying to solve it.
observe the following set of numbers:
PHP:
there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..
i will not rest until i find this formula. |
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| skywarp |
| And WHAT exactly does this have to do with playing records ?! |
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| DJ LIQUID |
| hey...im really good at math....it must say something bout my djing skillz :o :D :D :D |
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| JohnSmith |
| quote: | Originally posted by skywarp
And WHAT exactly does this have to do with playing records ?! |
has nothing to do with playing records.
but, like i said, DJs are good at math right? |
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| Special_K |
| im not good at math..... |
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| Dmatrox |
im pretty good at math, but im better at biology :D
Yeah, what does math have to do with playing records? |
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| patticus |
wtf? are those derivatives of some sort??
being asian, i used to be good at math, but then my overall capacity for the arts (in other words, stupidity) came to the forefront..
i got 78 x 150 wrong today. i said like 340 or something stupid:eek: |
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| sym |
Well what looks like to me is the following...
Find the distance between the first x and the first y. This number we will call z.
Than take the next x value you have and need the y for. Find the distance betweent his 2nd x and the first x you used to find z.
Than subtract that number from z and add the result to x2 to get your y2.
At least thats how I saw it... You'd have to be more specific with what you want... You can AIM me @ SyM2oo1 |
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| IronDragon |
| quote: | Originally posted by JohnSmith
i have an interesting problem, i am trying to solve it.
observe the following set of numbers:
PHP:
there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..
i will not rest until i find this formula. |
::giggles like a bimbo:: math is hard.
But really, I am only so-so at best in math. I assume you'll find that everyone else in here will range from friggin' genius to friggin' idiot. Maybe you'll get lucky and some'll know what the hell you're trying to do.
Suerte,
Dave |
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| uberclkgtr |
ah, this looks kinda fun. :rolleyes:
i can get it to work for all except the 1st set. here's how i set it up and solved it:
y - x = a + b
where a = the set number (ie 1, 2, 3, 4 or 5)
and b we have to figure out
if you make table of y and b values for a given a such as the following:
code:
for a = 5
y b
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10 4
11 3
12 2
13 1
14 0
for a = 4
y b
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8 3
9 2
10 1
11 0
you can see that b can be expressed as
b = 3a - y - 1
so we have the two equations
x - y = a + b
b = 3a - y -1
which we can solve by substitution to get
y = (4a + x - 1) / 2
this only works if a > 1. :whip:
i suppose you could use an iterative approach, where you specify y1, and based on the value of a and y1, you get y2. but i've done enough of your homework for tonight. :D
i'm better at math than spinning records... ;) |
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| hapamoto |
| haha.. this.. im asian too but that doesn't mean im good at math because of it.. im just good in math because I am.. anyways if i wanted to be doing math problems i'd be taking math classes.. sorry, i could probably help you but seems like other people already figured it out.. yay less work for me :D |
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| Xquisite |
| heh i got a messy equation. I dont think you want to see it. It has a ton of mods in it. |
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