Search results

Re: 2015 permutation X2 marathon This is the approach I was looking for, although I wasn't sure if suggesting recursion made it too easy.

Re: 2015 permutation X2 marathon In how many ways can a coin be tossed n times such that no two consecutive tosses are heads?

Re: HSC 2015 4U Marathon - Advanced Level Are people avoiding this due to length/irrelevance/perceived difficulty? If so I can assure you it's all bark and no bite, and a decent test of simple integration and series, whilst also showing a cool result.

Re: HSC 2015 4U Marathon - Advanced Level No, I wrote the question.

Re: HSC 2015 4U Marathon - Advanced Level Edit: The first (b)(iii) should be labeled (b)(ii) and the coefficient in front of the integral expression for a_n should be \frac{1}{\pi}.

Yes, those expressions are correct. To check that they are unit vectors (ie normalised), compute the norm of the new vectors. If it is 1, then they are normalised.

Why do you have parameters? The cross product should just give you a vector. Also you shouldn't need reduced row echelon form or augmented matrices, it's a determinant calculation for the cross product: Given two vectors A=(a1,a2,a3) and B=(b1,b2,b3) in R^3...

The answer to this question has nothing to do with the properties of P, multiplication of diagonal matrices is commutative.

10 vectors with components all 0 except for one which is 5, so 5e1, 5e2, etc where ei are the standard basis vectors in R^10. You can then see that there is no 11th vector that satisfies the given conditions.

Re: HSC 2014 4U Marathon I am sure there is a more efficient method, but here is my attempt: Edit: both the plus signs in the square roots for the final expression should be minuses.

The second derivative doesn't fail. A point of inflection occurs when there is a change of concavity. To test this, you substitute a point on either side of the point that makes the second derivative zero. Substituting the points -1 and 1 into the second derivative of y=x^4 gives no sign change...

c) Using V=25-10r, substitute 12.5 (half of the velocity of projection) for V 12.5=25-10r 10r=25-12.5=12.5 r=1.25 a) Remember, acceleration is the derivative of velocity with respect to time so: v=49 -49e^-0.5t dv/dt=24.5e^-0.5t

Try (1-0.01)^13

No that is an x, he expanded the (3+x) bracket out, multiplying (1-x)^15 by each term of (3+x)

Re: HSC 2014 4U Marathon So to get 1225 for part (i) would you do: 50!/(48!x2!) ?

Re: HSC 2014 4U Marathon i) 48! ii) 47!/48!=1/48

Re: HSC 2014 4U Marathon - Advanced Level Cos(5theta) = cos(5x2pi/5) = cos(2pi) = 1 So substituting back into the RHS of the original equation we get 16cos^5(theta) -20cos^3(theta) + 5cos(theta) = 1 Then subtract the 1

Re: HSC 2014 4U Marathon - Advanced Level Expand [cos(theta) + isin(theta)]^5 using De Moivre's Theorem for the LHS and Binomial Theorem on the RHS. Equate the real parts