what is the answer to this maths question?

[Poverty]

Who can solve this equation?

 

[Danoff]

Solve it for what? X? is C12 C^12? or is C12 a variable (or constant)... completely different from C14? On the face of it it looks like either C14, x , or u1 is zero... but I'm not sure I'm reading the equation correctly.

 

[Greycap]

I tried, but the system came tumbling down as I didn't know what is the correct thing to have on the left side of the equation mark...

- R -

 

[Danoff]

D - not enough information.

Done.

 

[Poverty]

I see, I took this of another forum and no one could even get close to working it out so I wanted to see what my reliable GTP had to say.

 

[Sharky.]

Looks like differentiation.

 

[SS69]

it depends on what you're trying to solve for. There is enough information, it would just have to be solved in terms of the variables given.

 

[Danoff]

it depends on what you're trying to solve for. There is enough information, it would just have to be solved in terms of the variables given.

If C12, C14 and U1 are non-zero constants, and you're solving for x, the answer is zero.

But not knowing what to solve for, or what the heck C12 and C14 mean (whether it's an exponent, or they're constants, or functions of x) means there isn't enough information.

Because you said this:

it depends on what you're trying to solve for.

We know that there is not enough information given. But even if I said we're solving for x, would you know the answer? If you say yes, you'd be wrong.


(edit: For starter's there's the whole "which x?" question. I mean, are we solving for the X, or the x? Is the little x multiplication? Are they the same variable? We don't know.)

 

[SS69]

If C12, C14 and U1 are non-zero constants, and you're solving for x, the answer is zero.

But not knowing what to solve for, or what the heck C12 and C14 mean (whether it's an exponent, or they're constants, or functions of x) means there isn't enough information.

Because you said this:


We know that there is not enough information given. But even if I said we're solving for x, would you know the answer? If you say yes, you'd be wrong.


(edit: For starter's there's the whole "which x?" question. I mean, are we solving for the X, or the x? Is the little x multiplication? Are they the same variable? We don't know.)

You can solve for X, you just wouldnt get a whole number, there would still be left over equation. This is assuming c12 and c14 are (c)(12) and (c)(14) and u1 is (u)(1). The equation would have to be solved in those terms. And assuming the small x is for multiplication. Personally I think everyone is over thinking this and what ive said is the correct assumtions. And if I am corrent than this is a really simple equation to solve. But then again it's been a while since I've dealt with anything other than geometry and simple algebra.

 

[Danoff]

You can solve for X, you just wouldnt get a whole number, there would still be left over equation.

Not without more information, like the information you assume below... which is my point.

This is assuming c12 and c14 are (c)(12) and (c)(14) and u1 is (u)(1). The equation would have to be solved in those terms. And assuming the small x is for multiplication.

Why would the small x be multiplication if the equation means (c)(12) instead of c12?

Why does it not read

d/dx(c x 12)=c x 14 x u x 1

...in which case no x variable appears in the equation. However, if we assume that u1 means (u)(1), then why write the 1 at all? It doesn't make any sense.

Seriously, there is more to this problem (or it's really sloppy... or both).

 

[Tornado]

However, if we assume that u1 means (u)(1), then why write the 1 at all? It doesn't make any sense.

Perhaps based on the font that the "1" is actually an "L"?

 

[PLmatt91]

You add the above, divide by what ever you get, then multiply my finger up that equations bottom.

kthxbye!

*I was never good in math*

 

[tait]

3

 

[opendriver]

The answer is 42

 

[Evolution.]

I was waiting for someone to say that .

 

[Kylehnat]

This is un-solvable gibberish. Is this an actual problem? (If so, what the heck is it from?). Or is this supposed to be one of those "joke" problems, which works out to x = bugger off.

Side note: I once used Fick's Law of Diffusion to insult someone. I think it boiled down to "U = DoUcHe", or something like that. Quite clever, if I do say so myself .

 

[dougiemeats]

Does c12 and c14 have anything to do with carbon-12 and carbon-14?

My chemistry is weak, so I don't know how you derive one from the other involving the "u1" as the d/dx indicates.

 

[Evolution.]

Ouch! Nice one Danoff .

 

[KieranMurphy]

Weird looking equation. C12 and C14 make me think it's got something to do with radioactove dating or something. But the uppercase D on the left make it look like an equation put together by someone who didn't really know what he was doing (assuming it's meant to be a differential).

Is the X on the right meant to be a variable called X or a multiplication sign - it's seperated from the other items as if for some reason (e.g. it's a multiplication symbol). I suppose the obvious question is what to solve for? Uppercase X or lowercase X? Are they even separate unknowns, the same unknown (badly designed equation) or is one of them not even an unknown but a multiplication sign?

As it is, I'm pretty confident that no-one can "solve" that equation without a lot more information.

KM.

 

[socomplayer2]

I would love to help but .... I suck balls at math.

 

[Kylehnat]

Weird looking equation. C12 and C14 make me think it's got something to do with radioactove dating or something.

Unlikely. Radioactivity follows an exponential function, not a simple derivative. Even if "X" were defined as some e^rt function, c12 and c14 would need to be switched around to even remotely resemble radioactive decay. Then there's this whole "ul" bit... If "c14 x ul" is actually a cross product, we could start dipping into the realm of quantum physics . I vote we not.

Poverty, what is the context around this? I'm curious. It's obvious that someone didn't just pull this out of their ass. (If they did, they deserve to be smacked for wasting people's time).

 

[OZZYGT]

I bet you can't even see your feet.

So....do you want to answer the question or you just want to keep making a fool of yourself? Let's not try to win an award here please.

-----------------------

As for the math question, I could be awfully wrong but I don't think you have to solve for a variable here, since we have too many and apparently the small "x" is a multiplication sign because it's smaller than the other X.
Also, if Toronado was right, then we have yet another variable to solve for which is "L" although it looks like a 1 for me.

I did some research and it looks like (c12), (c14) and u14 are numbers values related to chemistry and radiology...and I'm not touching that one with a ten foot pole.