Recent content by m00npirate

x^2 = y^2 \Rightarrow x = \pm y isn't if and only if?

Yea that was a typo, I've figured out what went wrong though. -1 just isn't in the range of the square root function. It has no solution.

If I square both sides from the beginning I still get x = +/- 1, neither of which solve the equation.

I have been trying for the past hour or so to figure out the solution to this equation: \sqrt{x} + 1 = 0 but to no avail! Here's what i tried... \sqrt{x} = -1 \sqrt{x} = i^4 x = \pm i^8 x = \pm 1 But this is most obviously wrong. What did I do wrong and what is the real...

a time like in the future? i don't understand the question.

No, and my apologies if my input is therefore irrelevant! I was just saying that that is normal and I don't think any technique other than repetition/relearning will solve it ( i really don't know anything about the technique you mentioned though.)

Unfortunately that is pretty much just how memory works. Though after learning something 2 or 3 times it tends to stick longer. Use it or lose it, basically.

well if you can prove it then you are all set =P

compute it using determinants and you should see why

Adopting a frame moving at c would cause dividing-by-zero type absurdities. The first of which being that space would contract 100%. How can you be moving at c if there's nowhere to go? xD

b is an arbitrary real number as well, he's asking why you reverse the sign of b and not a, for which i do not have an answer. my best guess is that its just more useful to define it that way.

If there's no variable then the function doesnt...vary. ie no matter what x is, the function is just constant.

On the "left" of 1, the function is defined as F(x)=x^2 +1, then F'(1)=_____ On the "right" of 1, the function is defined as F(x)=2x, then F'(1)=_____ If these two values exist and are equal, then the function is differentiable at 1.

For a function to be differentiable at a point, its left sided and right sided derivatives must be equal. Usually the function is the same on both sides of a point, but for this function it has two different pieces.

I was taking awareness to be binary. Either I am aware or I am not.