Recent content by powerless

AB = AO + OB AB = -OA + OB AB = -a + b CD = CO + OD CD = -OC + OD CD = -(3a-b) + (-a + 2b) CD = -3a + b -a + 2b CD = -4a + 3b midpoint of AB = (-a + b)/2 midpoint of CD = (-4a + 3b)/2 Is that correct?

Let a and b be non-zero parallel vectorsand let the points A,B,C and D be given by OA=a, OB=b, OC= 3a-b,OD = -a+2b a) express the vectors AB,AC,AD,BC,BD and CD in terms of a and b. b) Find, in terms of a and b, vectors from the origin to the midpoints of the line segments AB and CD...

The limit is 1. Am I right?

btw, did i use the correct expression in a, in the numerator, when x approaching 2 from the negative side?

Are we allowed to cancel out? We haven't studied L'Hôpital's rule but I don't know what to do with that one..

I started learning limits and I have some difficulties qith this question; Find the limits or say why they do not exist. a)lim_{x\rightarrow2}\frac{\left|x^2 + x -6\right|}{x-2} b) lim_{x\rightarrow0}\frac{\sqrt{4-x}-2}{\sqrt{x+1}-1} 3. The Attempt at a Solution a)we have an...

Thank you guys for the links! Hi snipez! I'm starting to understand these. And yes I like the hird link as well. Here's the 2nd part of my question: Simplify cos(2tan^{-1}(x)) y = 2tan^{-1}(x) => cos(y) cos 2\theta = cos^2 \theta - sin^2 \theta = 1-2sin^2\theta cos (2...

Where can I learn about the unit circle(from the very basics)? some links would be helpful! Because my textbooks haven't elaborated much on the basics of the unit circle, i don't know where to learn it from.

Thanks for your message! So can I use √1² + 3²? (using pythagoras to find the hypothenuse) I'm not sure what I should do next.

Evaluate sin(tan^{-1}(\frac{1}{3})) 3. The Attempt at a Solution Let y = tan^{-1}(\frac{1}{3}) <=> tan(y)=1/3 y = ? sin(tan^{-1}(\frac{1}{3})) = sin y I think for solution to exist tan^{-1}(\frac{1}{3}) must be in the range of sin, (that is \left\frac{-\pi}{2}...

Thank you Tiny Tim! I finally learned how to do it! :smile: Wow there are so many solutions to this equation!

Hi Tiny Tim! OK I do it again; \theta = \frac{4\pi}{6} . \frac{1}{3} + \frac{n\pi}{3} \theta = \frac{2\pi}{9}+\frac{n \pi}{3} The solutions must be within [0, 2π] If n = 0 \theta = \frac{2\pi}{9} If n = 1 \theta = \frac{2\pi}{9}+\frac{\pi}{3} = \frac{5 \pi}{9} If n...

please help! It's very difficult for me because i don't have any worked examples here to assist me. I will attempt this one: tan(3 \theta + \frac{\pi}{6}) = \frac{-1}{\sqrt{3}} (3 \theta + \frac{\pi}{6}) = \frac{5\pi}{6} 3 \theta = \frac{5\pi}{6}-\frac{\pi}{6} \theta =...

0.99999... Yes, now I have to show that 0.9999999 is rational. I want to see the surprises you were talking about! i don't see any recurring blocks, so i think i should just take 9 because it is 9 that is repeating. 10x-x= 9.9999999... - 0.9999999... 9x= 9 x= \frac{9}{9} = 1 ? Does...

Do you mean; 10^5x-x= 37124.37124... - 0.3712437124... 100000x-x= 37124 99999x= 37124 x= \frac{37124}{99999} \in Q Am I right or what?