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Hey guys Just wanted to clarify something: often in exams they will have a structure chart with parameters being received by a module and returned by it. For example the module algorithm2 receives the parameters a and b and returns c. Would it be ok to write algorithm 2 as BEGIN...

Yes, you have to write something that does the same thing, in a similar way. In terms of arrays BTW the course specs use, (), but it shouldn't matter.

isn't that usually a big A? Thanks though I knew I was missing some information in this formula Edit: nm, I checked it and it is amplitude. Thank you for your help.

Hi guys I just wanted to clarify, in the formula v^2 = n^2(a^2 - x^2) (for SMH). Is a acceleration, or just an unfortunate procedural usage? Thanks

Shoot, your right ... I hope I don't lose marks for writing the extra line x= sinh(7srqt(3)/26) :/

It depends on the algorithm, if its an integer operation most computers (and languages) will round down. An Int() function will also round down, round() will round up. In this case both give the same answer

Except that's what sinx equalled, not what x equalled. It was sinx=7sqrt(3)/26 or x=sinh( 7 sqrt(3)/26 ) Did anyone manage to get it simpler than that?

It depends if you look at speed or velocity. Its speed could be taken as -3, positive acceleration then indicates that the speed is increasing (towards positive infinity). This is how I read the question and it is a valid interpretation. However, you can also take it a speed = |velocity|...

either 2 or three depending if there was a mark for expansion or integration (or both), the last mark will be for the correct answer.

Sadly it's a very ambiguous question, looks like the majority is going with D. Both ways could be argued pretty fairly.

Yes, but I think you can also leave it as a decimal since you can have .123 of a month? (.123 is approximately 4 days)

You'd be correct ... if speed was a vector, speed is not a vector. The current speed of the particle is negative, if it has positive acceleration than speed is increasing. Velocity on the other hand would be decreasing. (Nb. force, acceleration and velocity are vectors, speed is not. Vectors...

Yep, that's right.

I missed that, it that case you'd have to integrate dP/dt, get the equation for P and then sub that back in to dP/dt ... what where the marks in v for then, I assumed at least one was for integrating P?

Hey guys So SDD's on the same day as MX1, I would have assumed they had a pretty similar cohort. So is that the case, who else is doing MX1 as well as SDD?

Don't think that's right, it doesn't take into account the other player's chance of winning before he wins. I think the right answer might be the limiting sum of (1/36)^2 subbed in as a power of 1/36

wolfram says 3... not sure I got that right (right method, but I swear my answer was a multiple of 2)

I couldn't eliminate P, so I didn't get an answer (and pens down before I had time to make one up). How did you get 40?

with b) iv. I got the derivative of N and set it equal to dP/dt (which is probably 2/3 marks), but I couldn't work out how to eliminate P.

why did you shade only the top right excluding? The region is anything outside the circle in all sectors. I think you needed to draw the whole circle I just double checked on wolfram, and that is the case http://www.wolframalpha.com/input/?i=%28x-2%29^2+%2B+%28y+-3%29^2+%3E%3D+4