Your strategy seems fine to me. I'd recommend using different terminology for clarity's sake though. Since a, b, and c are the sides of the triangle,...
It seems okay to me, but your instructor may want you to justify that the absolute value operator "distributes" over a limit like that. I imagine you...
The math they did in their last step is absolutely correct, but they screwed it all up by plugging in the wrong value. The answer that the height is...
Well, one side of the equality is trivial. You automatically know that S <= X2, because X2 is the entire area of the square. Proving that the area of...
This logic doesn't work at all. If you were trying to prove that |x| is not differentiable (i.e. the first derivative is undefined) at the point x =...
Well, your images are sideways and a bit hard to read, but they appear to all be the same. Was that intentional? In any case, here's what I think the...
That's alright. At least now you know what areas to study and brush up on. If you're finding your class notes, textbook, and/or any handouts provided...
I'm assuming that the N and Q in your image are meant to be \mathbb{N} (the natural numbers) and \mathbb{Q} (the rational numbers), respectively. If...
Ehm... I think you may have uploaded the wrong image. What you uploaded is a table with properties of various bolts being compared. This contains no...
I suspect you're correct that there is a typo in part (a), but it doesn't actually change the answer. Interestingly enough, imposing the condition x...
Well, I'd begin by noting that your expression for the denominator is not correct. If that were true, the denominators would be {x, x4, x9, x16, ...}...
Erm... sorry, but I don't think I understand what you mean. If you mean just interchanging the symbols, then there's absolutely nothing stopping you...
You're very very close to the correct answer. We're told that the original graph is y = x2, so the new graph is going to be some variant of that. The...