Probably why u've been getting things wrong...lol... The last math class i have ever taken was in High school and I AP'd out of taking math ever again...That was in 2002...so the fact that I remember PEMDAS from middle school math should be quite impressive. Regardless..if your goal is to find x...you have to rearrange the value to equal x and then plug in your values...
The correct answer is 288. It's a flawed question. Sure, you can rearrange the original problem algebraically and then determine that x = 0. But you can't plug in the values for all the variables and properly do the arithmetic on the right side of the problem to prove both sides are equal after determining x to be 0. If I was asked this in school, I'd probably write a paragraph.
Not quite. And btw. I'm a mathematian fockassing on descrete numbers not b****es that hide from me 0 to the 0 power is an argument Yep
It's an interesting problem and more tricky than the "48 or 2" or whatever it was. Yes, when solving problems, one only puts in values at the very end. And then one gets x = zy - y*y. If you then put in z=0, y=0 you get x=0. But that is not correct because in the previous step you multiplied both sides by y. You can multiply both sides of an equation by any number and get the same equation as long as it is NOT 0.
You cannot divide by zero. Anything divided by zero is undetermined. This equation is unsolvable with algebra .
There are a couple of people in this thread who know what they are talking about. The rest of you want to stay far away from math problem. I gave an example before, it's no different from the original prob: Given y/x = 2, and x = 0, what is y ?
Ok so I asked a couple professors over the weekend and they finally got back to me, something about Labor Day weekend....eh? Anyway, both answers were basically in this unfolding..... That is not possible since something undefined "times zero" remains undefined.
