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http://www.wolframalpha.com/input/?i=%28xy%29^2%2Bxy%3D2

Dude really sorry about this but the final answer is actually (-1,-1) and (1,1). Since if you sketch you have hyperboli with branches lying in opposite quadrant. Can someone explain why the equations gave incorrect solutions?

just one question. say you used the substitution x=sec^2(theta) would you end up in the denominator with abs[tan(theta)] or just tan(theta)? I stumbled across this thought the other day. 4U Cambridge just tells me tan(theta).

A tip for the future, with these questions (inequalities) you have a set of known statements and then from those you can derive the more non-obvious ones. [sqrt(p)-(1/sqrt(p))]^2>=0 where p>0 (p real) since the square of any real is always positive p+(1/p)-2>=0 p+(1/p)>=2

you're an expert lol, one problem with this question that may benefit the OP as well. When the line hits the cylinder how can you assume that it doesn't bend?

Let the height of the light source above the top of the cylinder after some time t from when the light source began moving be h. Let the radius of the shadow after this time t be r. By similar triangles it follows that: => r/4=(h+6)/h i.e. r=4+(24/h) Differentiating implicitly with respect to...

alright here is a picture I drew of the situation

one second mate, working on it for ya. why are the forums so inactive here lately?

So does largest possible domain refer to the set of all positive x or wut? lol, i dont even know

Re: HSC 2012 Marathon :) alright anyone gonna give the question a go?

I knew you couldn't do this but I never quite understood why. Can someone explain?

Beautiful :)

I'm going to assume that is the price after the discount. Let the original price of the ipod be x. 0.6*x=431 so x=431/0.6=$718.33

yep lol, sorry for not doing the entire question

To finish off: When x=y, =>x^2*x^2+x^2=2 i.e. x^4+x^2-2=0 (x^2+2)(x^2-1)=0 Hence, x=+/-1 So at the points where x=-1 or x=1 the gradient of the tangent to the curve is -1.

Re: HSC 2012 Marathon :) ^ a lot of you will probably recognise that one. If you do, please leave it for someone else lol.

Re: HSC 2012 Marathon :) A polynomials question.

Re: HSC 2012 Marathon :) oh right completely forgot about that one. so you just add c/b =/= f/e so bf=/=ec

Re: HSC 2012 Marathon :) Cmon Carrot. ax+by+c=0 => y=[-ax-c]/b has gradient -a/b dx+ey+f=0 => y=[-dx-f]/e has gradient -d/e a) If the two lines are parallel then their gradient are equal, i.e. -a/b=-d/e Hence, bd=ae b) If the two lines are parallel then the product of their gradient is -1...

In calculus, differentiation is an common operation used to determine the gradient of the tangent to a curve at any arbitary point which lie on it. If there is a function f(x) then the derivative of the function, also known as the gradient function, is denoted by f '(x) and now any values of x...