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well it could be 5/3 ln 3x +C (not 3/5) this is equivalent to 5/3 ln x +C, since ln 3x = ln x + ln3, and ln3 is just a constant

^ if you still haven't got the solution, like i said just use sum/product of roots you can find the other root + show multiplicity is 2 using what trebla said then you get an expression for A and B using sum of roots 1 and 2 at a time, then just show that 4B = A2 + 8

well the U has to be either to the right or the left of M, and there would be the same number of arrangements of each -> so half are to the right, half to the left. (I meant to include that in the solution)

if they ask you to calculate GPE of an object then they would specify that it's absolute GPE, in which case it's defined to be negative anyway. GPE doesn't necessarily mean absolute GPE and the OP was asking why it can be used as both a positive and negative term As affinity said, it just...

aww you sure you wouldn't just prefer hints so you can solve them yourself? solutions: 1. 2. 3. my solution to the third one is probably worded quite poorly

err I meant the same thing the other guy meant i was just pointing out to him that even though you're most likely never going to solve a problem like this in life, you will most likely be utilising the problem solving skills gained in maths

1) how many arrangements of uniform are there? what's the probability that the M is to the right of the U? 2) keep in mind that the other backpackers can go to the same hostel if they want, just not the one with the brother/sister 3) keep in mind each digit can only be used once, and the...

for the question, just use sum/product of roots :)

when will you use problem solving/reasoning skills in life?

yes it is correct. there's no reason it shouldn't be as it is just a simple binomial probability (after all of that...) 1/3 chance boy, 2/3 change girl, 5 prizes i see my error now. i assumed all combinations were equally likely (but they're not. need to keep this in mind in future and not...

wait so 80/243 is actually correct?

the trick is to include the prizes in with the permutation/combination

sorry, i meant combinations ^ but you should still be multiplying them (unless i misinterpreted your method?) you allow for a person to get more than one award using the method i posted

i think they should just remove this question from the exam

I'm convinced that my 112/429 method is the correct one :( :(

10/21?! surely not. that sounds much too high (the binomial one -overcounts- some combinations)

ahh the binomial one is not correct, because you can't multiply by 5C3 if one person got multiple awards edit:lol yes you can.. @hermand, i think you should be multiplying those probablilities (15c and 30 c pairs) not adding. and 15c2*30c3 should be added on the bottom as well apart from that...

woops, deleted it \\ no wait i looked at it again and i think it's 80/243 ? 5C3 (1/3)2(2/3)3 just from binomial probability

I just tried it and got 112/429. no idea if that's right though

when you add quantities, you take the number of decimal places as the quantity with the least decimal places, when multiplying you use the number of significant figures as the quantity with the least sf's. just use the rules you're given in chem as to how many dp's/sig figs to put your answer to