Recent content by dragonxhell

Could someone help me with this question? Because I'm stuck and have no idea how to solve it & it's due tomorrow :( Let S be the following subset of the vector space P_3 of all real polynomials p of degree at most 3: S={p∈ P_3 p(1)=0, p' (1)=0} where p' is the derivative of p...

Well... that's the name of the math course I'm doing so yeah I think my p/f is wrong and my teacher way is right 1/s e^10s (s^2+4s+5) transform to and then you just look up the table =1/5 [(1/s -s+2/(s+2)^2+1 -2/(s+2)^2+1 ]e^-10s

Well I know the Laplace you are talking about, I was using it for question 1. For an example 3/(s+3)^3 Y(t)= 3/2 t^2 e^-3t I usually do the algebra way but this question is too hairy so that's why I'm not sure if my working out is correct so if you could work your way and compare to my...

Yes, I did. Sometime its hard to rearrange function for inverse lagrange. I managed to work out the solution for question 3. So could you or someone check my question 2 please!

I got 2 questions to ask! I have finished one but not sure if it's correct so I need to double check with someone :) http://imageshack.us/a/img708/1324/83u8.png Here is my worked solution, I took this picture with my S4 and I wrote is very neatly as I could! The reason I didn't type it all...

sorry I type it all up on my phone and when I submit it it's so messy... so you are saying my integral y^4/4 must equal to fz which is z So the final answer is F=x^4/4 -3x^2y^2/2 +y^4/4 +z^2/2

Can you please check this question for me because my answer is different to my friends and they say its wrong... Given the equation: G= (x^3 -3xy^2)I +(y^3-3x^2y)j +2k is conservative find the potential function for the fields that are conservative My answer f:grad f=F fx=x^3-3xy^2...

Is this right? Can somebody double check this for me? It seem right to me... (x−1)²/9 + y²/8 = 1 8(x−1)² + 9y² = 72 8x² − 16x + 8 + 9y² = 72 9(x²+y²) − x² − 16x + 8 = 72 9(x²+y²) = x² + 16x + 64 9(x²+y²) = (x + 8)² 9r² = (r cosθ + 8)²3r = r cosθ + 88 + r cosθ = 3r a = 8, b = 3 c)R...

o so the centre is (1,0) and the equation you show me I have already obtained But don't you move the 1 over? To just simply get x by it self? So it should be 4 and -3 (-4+1?)

so for x it's 4 and -4 and for y its 2sqrt2 and -2sqrt 2 right? i did a sketch of it, by any chance you know how to do this question? =( Mod note: The image was way too large. Please scale the image down to about 900 x 600 pixels and edit this post to include it again, or add it in a new...

Hello =] I'm having trouble with this question, can somebody please help me with it! I'll thanks/like your comment if help me =) ![Question][1] I know that for a ellipse the parametric is x=a sin t , b= b cos t t:0 to 2pi (?) for part a) I drew up the graph but not sure if it's...

Yea that what I understand as well when some people explain to me in the first place but when I explain to my friends they ask sin(0) is 0 so I don't know how to tell them. Further more explanations I was able to notice that sin(x) is by it self like a constant so yeah. The reason x^2 =0 cus of...

Well this is an assignment in uni & I guess the person asked is the one in my course as well for engineering. I got the answer and understand the problem Our lecture just teach us to test non-empty by sub in 0 to the equation so that exactly what the other person did as me so yeah. It end up...

OMG! thank you it's like crystal clear now! you are the life saver. THANK YOU LOTS! Have a nice day!

hm... Someone explain like that for me as well let y(1) and y(2) be the solution, if the set was a subspace then y(1)+y(2) be the solution but if we evaluate the left hand side Y(1)+y(2) the right hand side is 2*sin(x) <- that's the part where I do not understand why it's 2*sin(x) is it because...