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i'm in today, probably till 4:00

my sister finished hsc when she was 15. Not because she was talented or anything, but because she skipped 3 grades when placed at one school. Long story short.

same happens to me, except to a lesser extent. In year 10, i would specifically skip hard questions so i wouldn't feel terrible that i couldn't solve them. If you are having trouble understanding something, it always helps to do your homework/study with someone who is experienced or can help you...

is this topic in 2 unit?

the amateur will now depart. off to do some light reading, then bed. cbb.

still working on my notes/study on chem mon and man.

amateur has joined the party. let's see how long this will last.

re: HSC Chemistry Marathon Archive doesn't the syllabus only specify to identify neutralization as an electron-transfer reaction?

what do you suggest as a good go-to drink for pulling an all-nighter? coffee/tea never works for me.

i feel sorry for everyone doing discovery...they literally have no resources to help them such as sample responses etc.

approached a person in the first group...the wording is a bit ambig..you could still approach a person in the first group if they are together.

it'a actually at a party. so..what are your thoughts now?

correct. i don't understand why the probability of choosing a women in the first group is 1/2. i thought that there was 21 people in total at the party, and there are a total of 9 people in group one. so she has a 9/21 chance of picking someone in group 1, since group 1 and group 2 are mixed...

for some reason i also don't get the answer to this simple probability question: There are two groups of people at a part and Minh is blind folded. In the first group there are 4 men, 3 women and 2 children. In the second there are 7 men and 5 women. Minh is spun and asked to select one person...

this is the actual working for the q: y = x^2 + 16 loge (5-x)^2 y' = 2x + 16 . 2 (loge 5-x) . -1/5-x where did u go wrong?

I also have another question: How would you differentiate: x^2 + 16 loge (5-x)^2 i got 2x (-32)/(5-x) but the answer is: 2x (-32loge (5-x)) / (x-5) why do they keep the loge?

thank you everyone!! This is the first time i have actually encountered something like this. many thanks :)

but where in the question does it say diff in respect to x???

so if we had 2y - 2x^2 and we were diff in respect to x it would still be -2x right? and thanks.

isen't the diff of (r^2 - x^2) = 2r - 2x why do you only take into account the -2x