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Incorrect, perhaps I should clarify. No girls seated next to each other means a formation (assume a circle) BGGBGBGBBGBBBBB is invalid as 2 girls are next to each other. ie. between any 2 girls, there must be at least 1 boy.

New question: How many ways are there to arrange 10 boys and 5 girls in a circle such that none of the girls are seated next to each other? EDIT: it means that between any 2 girls, there must be at least 1 boy

lolwat

70 WAM is piss easy. Most people i know doing com/adv maths have 80+ WAMs.

x=0 is not a solution (sub into original equation to see) and you have 5 solutions for degree 4 polynomial??

Correct. Algebraically, do this: |w|=10 and 0<=arg(w)<=pi/2 z=3+4i+w so w=z-3-4i So |z-3-4i|=|w|=10 with restrictions that 0<=arg(z-3-4i)<=pi/2, which is clearly the answer.

^Indeed you are correct. My bad. The above answer only really holds for b<=4.

Find the flaw in the reasoning below: \int \frac{1}{x} dx = x*\frac{1}{x} - \int -x*\frac{1}{x^2} dx = 1 + \int \frac{1}{x} dx\\ So hence 0=1 based on above Where the above uses integration by parts with u=1/x, du=-1/x^2, v=x, dv=1,

^correct

New question

I mean in between 0 and ln2 (not inclusive of ln2), d(e^x)/dx < d(2x)/dx which is why your logic still works. In general, the existence of changes in the relationship of derivatives of 2 functions can mean while they are still increasing functions, it is entirely possible for relationship...

The fact the relationship between the derivatives of the 2 functions don't change as x moves from 0 to ln2 is actually pretty vital in your argument.

GPA 7 = high distinction average

try another link... like the unsw maths website

^Obviously not looking hard enough Search "commerce/adv maths unsw" in google

Cos its a science course, as opposed to a commerce course

Similar to what you have tried to do but clearer. Just show the minimum of f(x) takes a positive value.

It is not immediate that this: "d/dx of e^x = d/dx of 2x when x=ln2. When x=ln2, 2>2ln2." justifies this: "Since both are increasing, .: e^x>2x in [0,ln2]"

Try this approach: Let f(x)=e^x-2x and show that f(x)>0 using simple calculus.

The result is correct, but this reasoning is rather incomplete.