Recent content by jawad1

Oh I am so sorry. I forgot to post it. I need to show that if f(a)=0 for some a \in \mathbb{R}, the either f is constant or f(x)=(x-a)g(x), where g is a polynomial function of degree n-1

ok I see. I do the same thing then for question 2 right ?

For the first part, I understood: we will get f(0)=c_0=0 which is a constant However show that f(x)=xg(x) doesn't seem so obvious to me

is that correct ?

Since 0 is a root of f, the polynomial x divides the polynomial f. Therefore, there exists g such that xg(x)=f(x). However, if we write p the degree of g, degree of f= p+1 (since xg(x)=f(x)). Therefore if f equals zero, g equals zero. If not p=n-1

Let [ tex ]f(x)=\sum_{i=0}^n c_i x^i[ / tex ] be an arbitrary polynomial function of degree n Show that if f(0)=0 then either f is constant or f(x)=xg(x), where g is a polynomial function of degree n-1 I don't know how to start. Please help Thank you in advance

d(Total)= (5/4)(u/(u+v))d I do not have any values for u,v and d so..

I am sorry if it isn't so clear. How do we use LaTeX on this forum ?

Let d_0: distance traveled during the first flight from Aville toward the train near Bville. d_0=ut t:some time d': distance each train travels on the first flight d'=vt We have d_0+d'=d d_0/d'=(ut)/(vt) <=> d'/d_0=v/u <=> d'=(v/u)d_0 d_0+(v/u)d_0=d <=> d_0=(u/(u+v))d...

Just want to verify question e: E=hf=6.626*10^-34*1=7*10^-34 Question f(i): 1/2mv^2=hf Which means that: m=2hf/v^2 Therefore: m=(2*7*10^-34)/(1.0*10^-3)=1.4*10^-30 g

I still don't understand how to proceed..

Ok. Is the sketch similar to the one I used in order to answer question c)(i) If so, how to I use it for question d) ?

for f) the formula is 1/2mv^2=hf-[work function]

I do not have a sketch of the setup. It is only given as a text

Homework Statement Consider two trains moving in opposite directions on the same track. The trains start simultaneously from two towns, Aville and Bville, separated by a distance d. Each train travels toward each other with constant speed v. A bee is initially located in front of the train...