Recent content by Mysteek

Thank you. Question has been solved =] You were a great help. Edit: Crap, it won't let me edit the title or first post to say that it's been solved. :(

Darn.. I feel sooo lost but I'm starting to have an idea of what is happening I'll start here again: (cos kx2, kx1*(kx2)3) is the end point, because from here I cannot isolate k to prove the definition of linearity. So to show a counter example, can I just substitute in any number other than 1...

I see. So a few notes to keep in mind: Domain -> Columns of original function Codomain -> Columns of mapped (if I'm even using the right terminology) So would ii be: g(x1, x2) = (cos x2, x1x23) Domain: R2 Codomain: R2 And g(k(x1, x2)) = g(kx1, kx2) = (cos kx2, kx1x23) (Am I supposed...

Ah thank you. I'm still reaally confused and it's making me so frustrated. I've been working at the same question for 5 hours.. but at least I can give you guys a better picture of where I'm at now. So you're saying that I should be able to show f(kx) = kf(x), if it is linear. I've learned...

Crap, I'm really sorry, I completely messed up part ii. ii was supposed to say g(x1, x2) = (cos x2, x1x2^3) It's hard to see, but what the right side says is cosine of xsubscript2, xsubscript1xsubscript 2, to the exponent 3. Thanks for your example! Based on that, I'm guessing that I can do...

Homework Statement For the following mappings, state the domain and the codomain. Determine whether the mapping is linear by using the definition of linearity: either prove it is linear or give a counterexample to show why it cannot be linear. i.) f(x1,x2,x3)=(2x2, x1−x3) ii.) g(x1, x2) =...

Sorry for double posting - for some reason my browser isn't letting me click edit on my last post :( When finding the direction of the force applied, could I use sine law? Ex: 6.77N is the magnitude of the change of momentum, thus it would represent the hypotenuse of a right angle triangle...

22.4 N would be the magnitude of the change of momentum, if I were to divide this by time I would get 22.4 / 3.31s = 6.77 N Oh, right! That makes perfect sense all the sudden, a bit more reading had me discover that momentum is in Newton seconds, so if I divided by time, seconds would...

Ah, momentum, brings back memories. So then: Px(initial) = mass of object * initial velocity = 1.2kg(14.4m/s) = 17.28 Px(final) = mass of object * x component final velocity = 1.2kg(23.88m/s) = 28.656 Py(final) = mass of object...

First post =] Requesting guidance with this particular problem: Homework Statement A 1.20 kg object is moving in the x direction at 14.4 m/s. Just 3.31 s later, it is moving at 28.8 m/s at 34.0 degrees to the x axis. a.) What is the magnitude of the force applied during this time...