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http://danielslake.virtualave.net/jrexams.html i have a few bio trials uploaded

let u=sinx du/dx=cosx therefore dx=du/cosx sub it in integral of cosx(sinx)^2 = integral of u^2 du = (u^3)/3 + C = (sinx)^3 / 3 + C

ok, the papers are all nicely tabulated and in pdf format now. enjoy! http://danielslake.virtualave.net/jrexams.html

the solutions are included. hmm ill see if i can convert these papers to pdf's. they seem to be easier to manage

the solutions are included this is the only open book assessment we have, and that means its <b>especially</b> hard :P

lets say we have 1M NaOH solution it will be neutralised with 1M HCl and 1M CH3COOH, even though one is a weak acid and one is a strong acid. this is because as the H+ ions in CH3COOH are neutralised by OH- in NaOH, more H+ ions are ionised into solution (Le Chatelier's principle)...

if the cation is more reactive, then it should displace a less reactive one.... i think have a look at the back of your periodic table. the 'standard potentials' bit the higher up the cation, the more reactive it is

sorry, i put the wrong URL here it is :P http://danielslake.virtualave.net/public/2Umathpapers.zip

http://danielslake.virtualave.net/2Umathpapers.zip (778kB)

1) <pre> let y =2secx-tanx =(2/cosx)-(sinx/cosx) =(2-sinx)/cosx dy/dx=[(cosx)(-cosx)-(2-sinx)(-sinx)]/(cosx)^2 =2sinx-1 = 0 (for stationary points) therefore x = pi/6 test nature (blahdiblah do this yourelf, you will find it is a local minimum) x=pi/6...