how damn hard is kilbaha 2010 Exam 1?Tech Free

[/0]

your explanations are brilliant Martoman, would you be kind enough to explain how to solve with matrices as well

When the determinant is 0 you have to use something along the lines of Martoman's reasoning to determine whether there are no solutions or infinitely many solutions.

[m@tty]

matricies? I would find p with the determinate = 0 so (-3)(-5) -4p = 0 if there is no solution so p = 15/4

for q I'd then apply my logic.

Ah but can be infinite or no solutions

No unique solution.

[TyErd]

okay thnx for the help but im now stumped on question 5

[Duck]

Just did kilbaha 09 exam 1 and have been scarred for life. No way am I attempting the 10 exam.

[the.watchman]

okay thnx for the help but im now stumped on question 5

When finding the inverse, I made the equation: by multiplying everything by
Then you can use the quadratic formula to get in terms of x

[Daniel15]

Just did kilbaha 09 exam 1 and have been scarred for life. No way am I attempting the 10 exam.

Haha, the Kilbaha exams were really incredibly hard when I attempted them, back when I was in high school

[stonecold]

okay thnx for the help but im now stumped on question 5

When finding the inverse, I made the equation: by multiplying everything by
Then you can use the quadratic formula to get in terms of x

This and question 11 were the only hard things IMO.  Despite still managing to make my regulatory careless errors, all the other questions were fairly straight forward.

[the.watchman]

okay thnx for the help but im now stumped on question 5

When finding the inverse, I made the equation: by multiplying everything by
Then you can use the quadratic formula to get in terms of x

This and question 11 were the only hard things IMO.  Despite still managing to make my regulatory careless errors, all the other questions were fairly straight forward.

Lol, Q11 was pretty annoying, I got sick of writing everything out...

[thushan]

Just did Kilbaha 2010 Exam 1...
I liked that paper; it was quite challenging

[thushan]

okay thnx for the help but im now stumped on question 5

As the.watchman said, multiply through by e^(2y)
Solve the equation like a quadratic, and prove that one of the solutions is invalid
--> x - sqrt(x^2+4)<0 for all x

[dopplereffect]

I liked how they put a pure maths probability question in there, it wasn't particularly hard but the fractions were ridiculous.

Question 5 was a real bitch, especially using quadratic formula and realising that x-(x^2+4)^1/2 wasn't valid for x is an element of R.

But hopefully by doing these ultra hard exams the hard questions in the VCAA exam are a cake walk.

[LachyMc]

you mean a "walk in the park" or a "piece of cake"

[kyzoo]

But hopefully by doing these ultra hard exams the hard questions in the VCAA exam are a cake walk.

Then you have the problem of overconfidence =/

[dopplereffect]

you mean a "walk in the park" or a "piece of cake"

Yeah, my cliches suck.

But hopefully by doing these ultra hard exams the hard questions in the VCAA exam are a cake walk.

Then you have the problem of overconfidence =/

More like confident enough to be able to do all the hard questions without getting too stressed out.