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stealing notes...thats just insane

"There are two groups of people *at a part*..." *at a park? *apart? I'm assuming its "There are 2 groups of people APART..."

(i) Prob of choosing FIRST group is 1/2 Prob of choosing a woman in the first group is 3/9 Prob (first group, woman chosen) = 1/2 x 3/9 = 1/6 (ii) Prob of choosing SECOND group = 1/2 Prob of choosing a lady in second group = 5/12 Prob (second group, woman chosen) = 1/2 x 5/12 = 5/24 Total...

keep in mind, i do extra steps so u can see the light \alpha= x^2 + 16\left ( ln(5-x)\right)^2 \\ \\ \frac{d\alpha}{dx} = 2x + (2)(16)ln(5-x)\cdot \frac{d(ln(5-x))}{dx} \\ \\ \frac{d\alpha}{dx} = 2x + (32ln(5-x))\cdot \frac{-1}{5-x} FOR THE BOLDED PART - can u show me ur working out?

\alpha = (r^2 - x^2)^{\frac{1}{2}} \\ \\ \therefore \frac{d\alpha}{dx} = \frac{1}{2}(r^2 - x^2)^{\frac{1}{2}-1}\left ( \frac{d(-x^2)}{dx} \right ) \\ \\ \therefore \frac{d\alpha}{dx} = \frac{1}{2}(r^2 - x^2)^{\frac{-1}{2}} (-2x) \\ \\ \therefore \frac{d\alpha}{dx} = -x(r^2 - x^2)^{\frac{-1}{2}}...

a confirmation is all i need! (eg. i get it, me understand, etc) even if i waste 2 hours of life writing up a solution to help a person, if he/she says "yes" i'll be happy

at night i feel weird... (its when i take a break from studying)

how do i know 'lovepink' FULLY understands how i did the question?

bit messy, making it look nice will waste time A = 98x - \frac{21x^2}{2} \\ \frac{dA}{dx} = 98 - 21x \\ $At $\frac{dA}{dx}=0,$ stationary points exist \\ \\98 - 21x = 0, x= $\frac{14}{3} \\ \\ $Test for max or min \\ $\frac{d^2A}{dx^2} = -21 \\ $Since $\frac{d^2A}{dx^2} <0, $then for all...

i've wasted my time...

"(c) Show that A is a maximum when half the wire fencing has been placed parallel to the brick wall" (i've helped a human with this question on facebook, and that human sent me a photo of the question) EDIT: damnnn whoever gave u that, editted the question

l = 2 + (r^2 - x^2)^{\frac{1}{2}}- (1-2x + r^2)^{\frac{1}{2}} \\ \\ \frac{dl}{dx} = -x(r^2 - x^2)^{\frac{-1}{2}} + (1-2x + r^2)^{\frac{-1}{2}}\\\\ \\ \frac{dl}{dx} = \frac{-x}{\sqrt{r^2 - x^2}} + \frac{1}{\sqrt{1-2x+r^2}} \\\\\\ \frac{dl}{dx} = \frac{-x\sqrt{1-2x+r^2}+\sqrt{r^2 - x^2}}{\sqrt{r^2...

same lol

To remember this formula, like did, i remember PDHPE, STD = sexually transmitted disease S x T = D

1 sec, scanner is having a go at me

do u want me to scan my solution? (it'll be easy to see) but u must promise me not to just skim through it and go "yes i see, i get the idea" then forget about it

haah i used these when i studied prelim chem, the setting is really beautiful

D = S x T Distance = Speed x Time T = 10 min = 10 x 60 = 600 seconds Distance = Change in distance from first boat to second boat I thought general math don't learn trigonometry Say something, so i can decide the future

Re: HSC 2013 3U Marathon Thread ooo thats a nice approach!

Re: HSC 2013 4U Marathon a^{m-1} + \dots +1 = a^{m-1} + \dots + 1 \therefore \underbrace{a^m + a^{m-1}+ \dots +1}_{m+1 terms} > \underbrace{a^{m-1} + \dots +1}_{ mterms} \therefore $The average of the LHS $> RHS \\ \\ \therefore \frac{a^m + a^{m-1} + \dots +1}{m+1} > \frac{a^{m-1}+\dots +1}{m}