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Re: HSC 2015 3U Marathon You need to draw a diagram to show that h^2\cot^2{40^{\circ}}+20^2=h^2\cot^2{30^{\circ}} , where h is the height of the building. After re-arranging and simplifying you should get h=\frac{20}{\sqrt{\cot^2{30^{\circ}}-\cot^2{40^{\circ}}}}

Note that \sin{3x}=3\sin{x}-4\sin^3{x} , so that you will have to solve the trig. equation 4\sin{x}-4\sin^3{x}=\cos{x} after substituition. This will turn out be \cos{x}(2\sin{2x}-1)=0 (after re-arranging and simplifying), making it easy to find your general solutions.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread c is a constant and t is a parameter, you've mixed the two up. Instead, after eliminating your parameter t, you should have the Cartesian equation xy=c^2 . For parts (ii) and (iii), you are on the right track but try to see how...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread When q=\frac{3}{2} , sub. it into Q to find that the coordinates of Q are Q(4,3) (which is the same as P, hence you disregard this). The solution q=-\frac{2}{3} will give the coordinates of Q to be Q(-\frac{16}{9}...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread For (i) you are correct, a=\frac{4}{3} . For (ii), leave Q as Q(\frac{8}{3}q, \frac{4}{3}q^2) and use point-gradient formula to find the equation of the chord. For (iii), since PQ is a focal chord, it passes through the focus...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread For part (i), you simply substitute in the values and show that the LHS and RHS are equal. For part (ii), find the Cartesian equation of the chord, using the points P(-2, \frac{2}{3}) and (1,1) you should find that the...

Re: HSC 2015 3U Marathon (A) 48^{\circ} ? Never mind, got beaten above.

Might be hard to get some insight here, since there seems to be < 90 students who do this course per year (and the fact that it's also a 3rd year ECON course), making it less popular than other subjects running within the Economics faculty.

Re: HSC 2015 3U Marathon Consider \frac{\text{d}}{\text{d}x}(1+x)^{50} by expanding out (1+x)^{50} and differentiating it w.r.t. x term by term. Substitute x = 1 and the result should appear.

I know what you mean, but with the given information, I am not able to see anything else. Don't know whether it is to scale or not but it's the closest thing I can think of. As I said, I could be completely wrong.

I could be completely wrong, but I simply considered the latus rectum of the parabola (which I assumed was 10). The latus rectum of a parabola is the line segment through the focus perpendicular to the major axis (which has both endpoints on the curve). The length of this latus rectum is given...

Re: MX2 2015 Integration Marathon You're right, I had an extra u factor.

Not sure about this, but is it F(0, \frac{5}{2}) ?

Re: MX2 2015 Integration Marathon ^ Use the substitution x=u^6 to simplify your integral to 6\int u^2-u+1-\dfrac{1}{1+u} \text{ d}u .

Re: HSC 2015 3U Marathon Construct a circle with diameter AB and let C be a point lying on the circle. Let P and Q lie on the minor arcs AC and BC respectively. Construct the chords AP, PC, CQ and BQ. Construct another chord PB so that \angle{APQ} = 90^{\circ} . Now, since BPCQ is a cyclic...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread Remember, focal length is always positive, so it should be a=\dfrac{2}{3} , since your parabola is of the form x^2=-4ay (concave down). Parametric equation of the parabola will given by x=2at, y=-at^2 , where a is a positive...

Hey man, getting this message whilst trying to PM you: 'bleakarcher has exceeded their stored private messages quota and cannot accept further messages until they clear some space.' Let me know when I can send messages to you once again :)

Re: UNSW Chit Chat Thread 2015. MATH1251 - Exactly the same assessment structure as 1151, only difference is lecturers and material you learn. ACTL1101 - Sherris is a solid lecturer, teaches the content quite well but sometimes he may become boring to listen too. He is very well organised and...

Yes, the inequality states that the graph is strictly positive for all real x Well 'touching' the x-axis refers to the graph of the quadratic being tangent to the x-axis at one point, which means the graph will consist of two equal real roots and so the value of the quadratic will equal 0 at...

Entire graph is above the x-axis, since the questions states the quadratic must be positive for all real x.