Maths Methods 3/4 Help Thread 2011

[xZero]

guys... it doesnt matter, the only simplification you need to do is for fractions, if vcaa has nothing better to deduct your marks from besides finding brackets in the solution then that exam better be so damn easy that yr 9 can ace it.

/end argument, get back to exam revision :S

[golden]

1st Picture:
I was thinking about this question.

The answers say that for < 3 months, P = -450.

I don't get why this is seeing as P is the profit and that if the seller sells one battery, they get s-450 profit, but when they have to replace it 'for free' they effectively have: s - 450 - 450 = s - 900.

2nd Picture:
The answers treated it as if it were 6 months, I treated it as if it were 7 months because it said at the end of 6 months. Could I get a second opinion on this?

Edit: This is from NEAP 2008 Exam 2.

Refer to the pictures from page 37 - lol can someone please answer this question?

[nacho]

What is house percentage and how exactly is it calculated?
Also, how many 'picture cards' are there in a standard deck?
Are things like these usually explained/defied in the exams when they come up? (eg house percentage is ...)

[paulsterio]

What is house percentage and how exactly is it calculated?
Also, how many 'picture cards' are there in a standard deck?
Are things like these usually explained/defied in the exams when they come up? (eg house percentage is ...)

They're supposed to be defined in exams

[Andiio]

House percentage

[stephanton]

hey guys, i was just wondering, when doing antidifferenciation, when do you put the '+c' in the equation and when do you leave it off...
i am just a little confused... :s

[paulsterio]

Always put the +C unless it's a definite integral (one with bounds)

When it's asking for "an antiderivative" you can set c = 0, or you can just leave the c, both are fine, in my opinion, just leave the c, it's easier

[stephanton]

thankyou

makes sense.

[mickeymouse]

me needs help

insight 2011 exam 1

Question 8b) how did they get that calculation for steady state?

[ech_93]

Can't work out what I'm doing wrong for q22 in this exam: http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/cas/pastexams/2009/2009mmCAS2-w.pdf.... Answer is D

Can someone tell me how to do it?
Thanks!

[b^3]

First of all we are looking at the two graphs and both axis. We can look at it as a box from 0 to the intersection of height 3 take away the area between y=ln(x-1) and the x-axis for 2 to the intersection.
First find the intersection. ln(x-1)=3
then x=e³+1
And remember it is above the x-axis from 2 onwards, so that will be your teminal for the area between ln(x-1) and the x-axis.
(red+blue area take red area)

So we now have Area =



EDIT: Added image

[paulsterio]

b^3, you could have just done that as an area between two curves, with the top curve being y = 3 and the bottom curve being y=ln(x-1) - it's how I usually do it



(By CAS)

Now, to add on the area of the rectangle from x = 0 to x = 2

[b^3]

Yeh I was going to do it that way originally but it's kinda like doing the rectange bit twice (as in area of y=3 form 2 to e^3 +1 is a rectangle take the intergral). They are both quick anyway, and plus it's a MC so it doesn't really matter.

[Andiio]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

[Milkshake]

Found that, for this question, there are lots of ways of getting the right answer and that its also rather easy to get a wrong answer... I could do it a couple of ways but yeah, just interested.
So some help/explanations on the 'preferred'/'best' method to use would be appreciated

Given that cos(a) = -sin(b) where 0<b<pi/2, find a in terms of b for pi<a<3pi/2.

Thanks!

Since sin is positive in the first quad, then cos must be negative according to the equation. Sin and cos can only produce the same value at the same angle at pi/4 (hope you get what I mean )
So b must be pi/4
Which means -sin(pi/4) = negative square root 2/2
So a = 5pi/4
Therefore a = b + pi

Nvm, I get what I did wrong - only got the answer for one solution, it doesn't work for the other angles