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Re: MX2 Integration Marathon , I was about to do partial fractions lol...

Re: HSC 2014 2U Marathon Yes, that's correct :D

Re: HSC 2014 2U Marathon Question:

Re: HSC 2014 4U Marathon - Advanced Level Holy shit, if you don't SR this year, I would be really surprised...

Re: HSC 2014 2U Marathon Question:

Re: MX2 Integration Marathon

Oh I see

This question has appeared so many times...

Re: HSC 2013 3U Marathon Thread

Let's call the side length of the cube 'x' We are given: dV/dt = 23 & we seek to find: dA/dt Generally what I do is express the Surface Area(A), Volume(V) in terms of 'x' So A=6x^2 and V=x^3 To find dA/dt we need to somehow incorporate dV/dt via chain rule. The way we can do that is by using the...

Re: HSC 2014 2U Marathon bump, was going to generalise the question but that's a bit too harsh

Re: MX2 Integration Marathon -> I forgot I was integrating in terms of d(tanx) :p

Yeah, it was a question in a past paper that I did. They're starting to link influences/strategies to achieve/assist/affect something very specific, yet dense enough to be able write a case study on and is within the boundaries of the rubric. Another example: How can limitations in financial...

Re: MX2 Integration Marathon Nice Question!

Re: HSC 2014 4U Marathon - Advanced Level α+β+γ=180 -> cot(α+β)=-cot(γ) [cot(α)cot(β)-1]/[cot(α)+cot(β)] = -cot(γ) cot(γ)/[cot(α)+cot(β)] = cot^2(γ)/[cot(α)cot(β)-1] = [-sin(α)sin(β)cos^2(γ)]/[sin^2(γ)cos(α+β)] = [sin(α)sin(β)cos(γ)]/sin^2(γ) But, cos(γ) = [a^2+b^2-c^2]/2ab = [c^2(k-1)]/2ab...

Re: HSC 2014 4U Marathon sec(x)+tan(x)=A 1/(1-sin(x)) = A sin(x) = 1 - 1/A cos^2(x) = 1 - (1-(1/A))^2 = (2A-1)/A^2 cos(x) = ±[sqrt(2A-1)]/A 1/(1 - cos(x)) = A/(A±sqrt(2A-1)) Therefore, cosec(x) + cot(x) = A/(A±sqrt(2A-1)) where A>= 1/2

Re: HSC 2014 2U Marathon (i) 1/2ln(17) (ii) Using trapezoidal rule with 5 function values; 1/2ln(17)≈1/2(4/17 + 1 + 7/5) ln(17)≈224/85 ----------------------------------------------------------------------------------------- Question: At the beginning of each year Danny invests in his...

Also, is 0 < theta < 90? or 270 < theta < 360? Depending on which of these conditions are specified will determine whether tan(theta/2) is positive or negative.

Let t=tan(x/2), (1-t^2)/(1+t^2) =1/5, cross multiply, it comes out in 1 line.

Took me a while to figure this out; so find the equation of the displacement of both particles. You should get x(1)=-(1/k)[gt+(1/k)(g+ku)(e^(-kt)-1)] and x(2)=(1/k)[gt+(1/k)(g-ku)(e^(-kt)-1)]. These two particles will intersect when x(1)+x(2)=d. Make t the subject and it comes out :D