Recent content by hahutzy

I still don't understand why I would need to use cosine law for this, as claimed by my teacher...

So are you saying [itex]|\vec{AB} + \vec{BC}|[/tex] = [itex]|\vec{AC}|[/tex]? If so, why does my teacher insist that I use the cosine law?

I tried. What I'm confused about is the interpretation of [itex]|\vec{AB} + \vec{BC}|[/tex] I mean, mathematically, I believe I can do this: [itex]|\vec{AB} + \vec{BC}|[/tex] = [itex]|\vec{AC}|[/tex] Because [itex]\vec{AB} + \vec{BC}[/tex] = [itex]\vec{AC}[/tex] In which case, [itex]|\vec{AB}...

Homework Statement Given: A (-3,7) B (5,22) C (8,18) are 3 points in 2D space. Find [itex]|\vec{AB} + \vec{BC}|[/tex] Homework Equations ||\vec{v} - \vec{w}||^2 = ||\vec{v}||^2 + ||\vec{w}||^2 - 2 ||\vec{v}|| \cdot ||\vec{w}|| \cos \theta The Attempt at a Solution Isn't...

I understand that, however, what I can't wrap my head around is doing the same thing if the term is being raised to the power of something. Namely, ((y/x)+3)N where N is integer > 1 Let's say I had N=2, ((y/x)+3)2 Expanding gives y2/x2 + 6(y/x) + 9 y2/x2 + 6(y/x)(x/x) + 9(x2/x2) [y2...

Thanks! But just making sure I'm getting this. Multiply both sides cx = |y/x-1| / (|y+3x|^5 ) cx*(|y+3x|^5 ) = |y/x-1| Should be Multiply both sides c(1/|x|) = |y/x-1| / (|y+3x|^5 ) c*(|y+3x|^5 ) = |y/x-1|*|x| c*(|y+3x|^5 ) = |y-x| Right? Also, is there any easy way to remember that...

Hey guys, I've been stuck on this problem for a good hour... I have no idea how to finish it up. Homework Statement Solve: dy/dx = 4y-3x / (2x-y) Use homogeneous equations method. Homework Equations Answer : |y - x| = c|y + 3x|5 (also y = -3x)The Attempt at a Solution dy/dx = 4y-3x / (2x-y)...

Thank you!

Homework Statement So the problem goes like this: dy/dx = ( x^2 + xy + y^2 ) / x^2 a) Show that it is a homogeneous equation. b) Let v = y/x and express the eqn in x and v c) Solve for y Homework Equations (Included) The Attempt at a Solution a) dy/dx = ( x^2 + xy + y^2 ) / x^2 * [(1/xy)...