Do you recall implicit differentiation? Have a look at the link below!

https://www.khanacademy.org/math/calculus/d...fferentiation-1

Aren't both x and y functions of s?

x=x(s), y=y(s)?

How could you differentiate sxy and only treat one as a function of s?

Sorry, I wasn't really sure what the question wanted actually. But I suppose v1n0d managed to solve the puzzle!

Edit: This is actually quite new to me to be frank. I have never seen composition of functions being expressed in this form so good stuff! Learnt something new today!

This post has been edited by studyboy: Jan 4 2014, 01:14 AM

Let me try to explain this,

1) For this, the arrangement of the vowels do not matter. As long as they are together, be it AEAI, AAEI, IEAA or etc, they count as 1 entity. Hence the are 8! ways of arranging M A T H E M A T I C S.

2) For the second question, let us think of it via a 2 stage process.

STAGE i) The band of vowels can be rearrange so we have 4!. However, as mentioned by v1n0d, given the 2 As we have to divide 4! by 2! so there are in total 4!/2! ways of arranging the A A E I.

STAGE ii) Now, from Question 1 we know that we have 8! ways of arranging M A T H E M A T I C S so that the vowels are together. However, there are 2 M's and 2 T's so in effect, we have 8!/2!2! ways of doing the arrangement.

Therefore, according to the multiplication principal, we will have (8!/2!2!) x (4!/2!) ways of arrangement.

Nevertheless, I have absolutely no idea how the words used in Question 1 implies the answer given.

Let us wait what there others have to say about this.

There is quite a bit going on here in this 3 minute video.

1) The rule of Sarrus is used to compute the determinant of a 3x3 square matrix.

http://en.wikipedia.org/wiki/Rule_of_Sarrus

2) An illustration of how a matrix with a zero determinant will lead to a system Ax = b having no unique solutions. The example used is one pertaining to a diagonal matrix

http://en.wikipedia.org/wiki/Diagonal_matrix

Using the the rule of Sarrus, the determinant for a diagonal matrix is just the product of the main diagonal since all the non diagonal entries are zero. If we have a system of Ax = b such that after row reductions, we end up with an augmented matrix with a diagonal matrix for A, then a case where |A| = 0 means one of the diagonal entries is zero. Hence, a unique solution cannot be obtained since the system cannot be solved.

In a nutshell |A| not zero iff A is invertible. If A is invertible, a unique solution can be found for the system Ax = b.

dv/dt = a (acceleration). Yes you are absolutely correct.

1 / tan (pi/2) = 0.