Recent content by hmm?

Alright this kinda clarifies my question, so -5 is like the min? Which would grant -5<x-2<5--since epsilon can never be E<0 - the absolute value is necessary? Sorry if I come off a bit slow, but it's just that I'm trying to justify every step so I completely understand the concept.

Hello, I'm having trouble with a step;I was wondering if someone could shed some light on me. Thanks. Prove that lim x->-2 (x^2-1)=3 0<|f(x)-L|<epsilon whenever 0<|x-a|<delta 0<|(x^2-1)-3|<epsilon whenever 0<|x-(-2)<delta = |x^2-4|<epsilon =...

If you could provide me a demonstration of let's say \sin(-\frac{7\pi}{6}) = -\sin(\frac{7\pi}{6}) I would be most greatful, Data. As you can see, this is what excites the most trouble in my understanding--when \theta \geq 180 . My apologies if I come off tenacious, but not understanding this...

I feel the true satisfaction in obtaining immortality is not the fact of the immortal life, but that human-kind has surpassed the threshold of being controlled by natural course, arguably entropy. As for the contingency of outside forces acting upon, and eliminating us without our consent is...

From what I gather, I think your explanation and the books is similiar. So as far as I can understand Cos(-60)=1/2 and Sin(-30)=-1/2 which satisfies the Sin(-x)=-Sinx and Cos(-x)=Cosx--this is not hard for me to comprehend, but I was thinking, what about Cos(-120)? Does this not equal -1/2, or...

Hello, I'm curious if anyone can shed some light on my seemingly opaque brain as to why Sine is an odd function and Cosine is an even function?

You're absolutely right--all those -1 from omitted thetas and xs from test questions still haven't sunk in :/...haha.

In addition to what these people said: you should try to remember identities that will serve as shortcuts when proving identities. like: Sin^2 + Cos^2=1 <= this may be the most important one Sin^2= 1 - Cos^2 Cos^2= 1 - Sin^2

Ahh, thanks! I can't believe I couldn't think of this, haha. (logx)^2-logx=0 logx(logx-1)=0 logx=0 , log(x-1)=0 x=1, 10 A quick two questions: is there any other way to solve this, other than the y^2-y=0, and still maintain solutions 1, 10? For (logx)^n n>1, how would you treat...

y=logx x=10^(logx)^1/2 This is where I'm stuck, because I feel it would be illegal to assume 10^logx^1/2...shouldn't it be (logx)^1/2. I don't know how to simplify after this point.

Hello, I'm having trouble recovering all the solutions for this equation; here is the work I've managed to produce thus far. logx=\sqrt{logx} (logx)^2=logx 10^logx^2=x <--I'm not sure of this move? x^2=x x^2-x=0 x(x-1)=0 x= 1, 0 I know 0 cannot work, and 10...

the funny thing about this problem is I answered it completely different, which was also wrong; I guess it was spur of the moment, and I was enduring a brain fart of epic proportion.

well, since I know the sin of that triangle is sqrt(36-x^2)/6 Hypotenuse=6 opp=sqrt(36-x^2) using pathagorean theorem x^2 + y^2=6^2 y^2= 36-x^2 y=/sqrt(36-x^2)...

for some reason, I feel it would be cos^-1(x/6)...

Hello, \arcsin( \frac{\sqrt{36-x^2}}{6})= \arccos(?) I was wondering how one would solve this equation; it was given as an extra credit problem on a test I took today. Being the impulsive person I am, I can't wait until monday to receive the solution.