Recent content by johnnyamerica

Solved: x=((384x/(y+384))*y+384(384x/(y+384)))/384 I solved for y1 and y2. I can't edit the title of the original post but this is solved.

Any value of y will satisfy that X is indeed equal to X? Oh, thank you. So, like if (e^(iy) + e^(-iy))/2=x, and y were 5, let's say, then x=x? Next you'll tell me that x-x=0. I'd require proof, however.

I don't know what a tautology is or what it is about this equation that makes it impossible to find y, but for any x other than 960 for which there can only be one value of y, there are two values of y - between -144 and 576. I'm guessing making y the subject of these functions is beyond the...

They're a mess because they're the result of weeks of work. Sorry to be blunt but ALL of the equations in my first post are correct on both sides of the equation. They're given just as different possible starting locations in case someone dislikes one of them...

Lol, yea, that tends to happen. I found four redundancies in the original equation. I've tested it repeatedly and for every y between -144 and 576 it's correct. x=((sqrt(768^2-(((-1280)(-y))/(-y-384))^2)))((y+384)/(384)) ---- v= 384x/(y+384) v=...

((sqrt(768^2-(((-1280)(-y))/(-y-384))^2)))((y+384)/(y+384-y)) = x If you swap x and y and type y=((sqrt(768^2-(((-1280)(-x))/(-x-384))^2)))((x+384)/(x+384-x)) into even the simplest graphing calculator you can see the curve. I'm trying to solve y=? Or x in the calculator-friendly...

If x= ((384x/(y+384))*y+384(384x/(y+384)))/384 y= ? If y^2=x, y must either be y1= sqrt(x), y2= -sqrt(x) If ((384x/(y+384))*y+384(384x/(y+384)))/384 = x = 0 Then y1=576 and y2=-144 It's a curve. There are two values of y for any x except at the cusp of the curve. How can I...

I need to solve for y. Without already knowing y. All these equations depend on already knowing what y is. If x=2y. y=x/2. If x=(e^(iy) + e^(-iy))/2, y= either -ilog(x-sqrt(x^2-1)) or -ilog(x+sqrt(x^2-1)) If x= ((384x/(y+384))*y+384(384x/(y+384)))/384 y= ?

If x=y, x-x=0. If x=(e^(it) + e^(-it))/2, x-x=0. If cos(x)=5, x-x=0. If x=0, x-x=0.

I get that. I just want to solve this to equal y, so that if I know what x is I can solve for y. If I know that y=216, I know that x=960. But what if I knew that x=961, let's say? I'd have zero clue what y=, and I'd have to guess. I tried to find x=y, and by guestimation the closet I...

x-x=0, you say? I wasn't aware. Any value of x or y can result in a different value of x or y. I can find any x if I know y, but how can I find any y without already knowing y?

Nascent's wrong, it isn't 0=0. It's ( y(384x/(y+384)) + 384(384x/(y+384)) )/384 =x So how do I find y=...?

Affirmative. Every equation I've written as x= gives an identical answer for a given y. If y=216 x=960. 960 = (((384*960)/(216+384))*216+384((384*960)/(216+384)))/384 960 = (1280((sqrt(768^2-(((-1280)(-216))/(-216-384))^2))))/(-sqrt(768^2-((768^2-(((-1280)(-216))/(-216-384))^2)))+1280)...

Homework Statement These are different way's I've phrased the question: x = ((384x/(y+384))*y+384(384x/(y+384)))/384 x = (1280((sqrt(768^2-(((-1280)(-y))/(-y-384))^2))))/(-sqrt(768^2-((768^2-(((-1280)(-y))/(-y-384))^2)))+1280) x =...

Homework Statement The actual numbers aren't completely relevant. I made a graphic. http://i168.photobucket.com/albums/u193/kamikazehighland/calculus.png I'm actually not in calculus or engineering. I'm actually writing a program, but I've done enough research to know how to...