Search results

In my humble opinion the difficulty is similiar to that of 4U maths. The thing is that in PHYS1121/31 at UNSW one of the lecturers (Joe Wolfe) makes you think and he does encourage it too in his lectures and sets exams accordingly. But the problem is that people are trained to be 'formula...

It depends on the person. All my friends got a CR or D and found it really hard. I got a HD and found it fun/interesting... But I barely passed a subject that was apparently very easy (friends got D's and HDs). So go figure.

I don't think it's 14. I know at least 2 specific cases [BGBBGBBGBBGBBGB and BBGBGBGBGBGBGBG] with 48 and 6!*7! number of ways each... EDIT: Oh 14! ... that's the total number of ways without restrictions.

Have you got any tutees yet?

If all 5 girls are together then no of ways that could happen is 1*10!*5! 4 girls ... 1*11!*4! 3 girls ... 1*12*3! 2 girls ... 1*13!*2! total number of ways = 1*14! So that's how I got that... But then again I only do General maths so what would I know... EDIT: The above may be wrong since...

I had another answer up there but I thought it was wrong (in the order of 10^10). EDIT: Yeah I think my other answer was correct: 1*14! - (10!*5!+11!*4!+12!*3!+13!*2!) = 7.05*10^10 (3 sf) ... maybe not though.

In that case, it would be 1*6!*2!*7! = 7257600 I think...

New question: x^2-(p+iq)x+3i=0 (p, q are real) has roots \alpha and \beta. The sum of the square of the roots is 8. (i) Write down expressions for the sum of roots and product of roots. (ii) Hence find the possible values of p and q.

Re: Is there any use bringing a laptop to uni? They probably failed.

Consider the girls seated next to each other. This is done in 5! ways. This leaves 15-5+1=11 people units. So No of ways girls sitting next to each other = 1*10!*5! So No of ways girls not sitting next to each other = 1*14!-1*10!*5!=8.67*10^10 (3 sf).

Or you can see it because the equation would be undefined (involves 1/x).

(i) If P(x)=3x^4-11x^3+14x^2-11x+3 show that P(x)=x^2(3(x+\frac{1}{x})^2-11(x+\frac{1}{x})+8) (ii) Hence solve P(x)=0 over C.

Probablility is fun.

*Looks at timetable* I have 3 * 4hr of classes straight and 1 * 5hr of classes straight. ... so go figure.

There's an explanation in Coroneos.

What textbook is that? I would write it up. But my solution uses a circle geometry theorem. I don't think the solution should since the textbook I got it from was written in 1960, way before circle geometry was introduced in 1981. So I'm sure that my method is bad.

What about for more than 3 function values?

The courses are very similiar I believe. I think the main difference is the order in which they do stuff. Eg, they do probability in S1 and complex numbers in S2. Whereas for 1141 it's the other way around. And they go into a bit more detail I think.

How do u do that?

Well, technically it is. But still it's boring.