Search results

lol thxs

Yep harder 3u is defs the hardest, because it involves deeper understanding of the whole 3unit course and most of the questions are usally the hardest...in my own subjective opinion i say the 2nd hardest possible questions are from mechancis and yes...if it is in q8 it is hard...I believe the...

In general harder 3u is the hardest, however it is at its peak when they combine harder 3u with proving irrationality by contradiction as in the 2003 paper..that is among the hardest possible question...also sometimes they can make really hard series questions involving integrals where you need...

you should of worked out from iii) that: 2I_{1}=\frac{1}{2}, 4I_{3}=\frac{7}{12} now these are the first 2 odd terms of the sequence, if you look at what the question wants you to prove, you can then realise that: 2I_{1}=1-\frac{1}{2}, 4I_{3}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} then you...

If you find the first and second derivative you find x=-1 is a max turning point, and x=2 is a min turning point. If i sub x=2 into y, i get: y(2)=2(2)^{3}-3(2)^{2}-12(2)+k=k-20 so that means (2, k-20) is my mim turning point If i sub x=-1 into y, i get: y(-1)=2(-1)^{3}-3(-1)^{2}-12(-1)+k=k...

If we take 2 items from 4 distinct items then that occurs in 4x3=12 ways, which is just _{2}^{4}\textrm{P} now to pair these items off we divide by 2x2 as we are taking 2 items and we want to form pairs with these. therefore we get that p_{2}=3. The 2x2 represents that we are pairing off the 4...

the env's are arranged in a line 1-6, by pos 1 i am reffering to env 1, which is what we dont want A to be in

This is perfectly right, however in comb/perm questions you dont wanna be looking at every case, you wanna look at total cases and then minus number of cases in which our restriction is breached, but props to your soln

With my above reasoning i forgot to think of all cases: So no restrictions means there are 6!=720 ways.. Now we dont want A is position one, A can be in position 1 in 5!=120 ways Also, we dont want B in position two, B can be in position two in 5!=120 ways Now, the number of ways A and B are...

it's cause if it is A has 5 to choose from and B has 5 to choose from then they can coincide in one env, which is not allowed, only one letter is allowed in each env

well if above soln is wrong this is the only other explanation

yeh stupid perm questions...first look at the case where there are no restrictions: 6! next look at the case when A is in env 1 and B in env 2 (the only case we dont want), this occurs in 4! ways so total ways therefore where A not in env1 and B not in env 2 is: 6! - 4! = 696

i dont have time to answer them now, but Q11,14 have sigma notation, which means the sum of...so what you can do to make it easier for yourself is to expand the series by subbing in each value from the starting value on the bottom to the last value on the top, then it just becomes a normal...

you should have N equal to :(-2mb, a(2 + 4m^{2}+\frac{b}{a})) set x and y to from these new parametric eqns: x=-2mb, y=a(2 + 4m^{2}+\frac{b}{a}) now as m is fixed we dont need to sub for m, what we will do is sub for 2 as follows: \frac{-x}{mb}=2, y=a(\frac{-x}{mb} +...

for max distance we need to incorporate distance into this expression using: D = ut distance = speed(u) x time now we know F is the rate of fuel per hour in litres, therefore: F = \frac{dl}{dt} which reads fuel is rate of change of litres with respect to time. If we invert and integrate...

lol which school is that?

and we can use the change of base for anything, but we only know how to differentiate base e so thats why we change it to e in these questions

fine full solution is: \frac{d}{dx}(\log_{10}x)^{3}=\frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} using change of base rule. \frac{d}{dx}(\frac{\log_{e}x}{\log_{e}10})^{3} = 3(\frac{1}{x\log_{e}10})(\frac{\log_{e}x}{\log_{e}10})^{2} using power rule. Edit: full worked solution to...

i just saw the asnwer..to change my answer to the right answer you realise: \log_{e}10 = \frac{\log_{10}10}{\log_{10}e} this simplys to: \log_{e}10 = \frac{1}{\log_{10}e} and subbing this into my answer gives the desired result:

answer should be \frac{3}{xln10}(\log_{10}x)^{2}