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yeah i guess so..

damn theres too much too remember :( Thanks though.

Yeah hehe i agree with Rivet. If i couldnt do the question i originally posted..id be stupid to think that i could do 4 unit maths. But yeah..

hahahah woops i missed something out.. :| i read your post and i thought "what the hell is he talking about" and yeah i meant to write Integrate sqrt(16 - x2)/x2 Sorry for wasting both your time rivet and wholovesmaths

Integrate (16 - x2)/x2 Thanks in advance!

We still have volume, mechanics and harder 3 unit to go aswell... but it doesnt seem like were gonna be starting volumes this term :| were still on integration..

The second part: (2/3)int(0,1) nxn-1(1-x)sqrt(1-x)dx = (2n/3)int(0,1) {xn-1sqrt(1-x)}(1-x)dx = (2n/3)int(0,1) [ {xn-1sqrt(1-x)} - x{xn-1sqrt(1-x)}]dx = (2n/3)int(0,1) [ {xn-1sqrt(1-x)} - {xnsqrt(1-x)}]dx ? Is that what you meant?

I dunno if you want the working out, so ill put that within a "spoiler tag". Final Answer: dy/dx = 1/(1-x2)(1/2) EDIT: Erm... i just realised i never simplified it..sorry its cuz entering on the computer is so crappy..its confusing to look back at stuff cuz of all the sup/sub tags and etc.

Hehe cool. I like your sig and your little quote thing "Nucking Futs". That movie was cool... its got me saying "Shizit" now around people whom i cant/shouldnt normally swear infront of. Started out as a joke though.. now its a habbit..

What you said is pretty much what i said: Also the electric field working against the electron is the same as saying that work is being done by the electrons by moving against the electric field, except the perspective is reversed. So does that mean everything that i said in my original...

W = ∆KE = KEf - KEi So if a certain voltage Vo is applied and the motion of the photoelectron is completely opposed, KEf = 0. Meaning, W = - KEi. Is it safe to say that in this case, W is negative as the photoelectron is doing work travelling against the electric field set up. Such that W...

Woah... i really dont understand that..hehehe. Thanks though ill look at it and try to figure it out..

Could somebody please help me with this question? Ive tried rewriting it as (1+x)n-1 * (1+x) for the denominator and alot of other variants... Cant get the answer :(! If In = 0∫t 1/(1+x2)n dx) where n >= 1, show that: 2nIn+1 = (2n-1)In + t/(1+t2)n. Thanks in advance.

Q1. An insulated lithium surface is irradiated with light of requency 7 x 10^14 Hz and photoelectrons are emitted until it acquires a positive potential of 0.7v. Calculate the work function of lithium in eV. Take h= 6.6 x 10^-34 and e=1.6 x 10^-19" I don't get the "until it acquires a...

fuck yeah..im such a dickhead...it does equal zero...

Question 8, 33d..thanks

Ok fitzpatrick has confused the hell out of me. WIth partial fractions with a repeated denominator, e.g. (3x2 + 10x -5)/(x+1)2(x-2) it seperated them out as the following, a/(x+1)2 + b/(x+1) + c/(x-2), but the answer doesnt have b/(x+1)... So im confused..

Yeah 4 unit maths really does help out with 3 unit... Its funny though... in year 11 i was doing 3 unit..and my marks for 2 unit were 100% for the first two tests...where my 3 unit marks were in the 80s.. now that i picked 4 unit my 3 unit marks went up to 100% for the first two tests... And...

i had glandular fever last year.. and i was stuffed.. i had a huge lump on my neck... i didnt feel like eating anything at all..and i slept SO much.. the doctor even told me i could take as much time off school i wanted... and shed give a note etc.. but i didnt want to miss school and that was...

:( I wish it wasnt fake.. i get 98.65