VCE General & Further Maths Question Thread!

[LiquidPaperz]

with q6, idk what they did, i got C personally

with q4 i got B, i dont know where the 4 x 6 came from? the question didnt make much sense to me there was something missing i think?
- i didnt do 1/2 lol

[Sunshine98]

Matrices multiple-choice question.  Half an hour later...  still can't work it out.

If A =  |  1  4  1 |        and X =  |  p |        and AX = |  b  |         then,
          |  2  -3 0 |                      | q  |                        | 2a |
           | 1  0  0  |                      |  r |                        |  a  |

A.  r = b - a
B.  r = a - b
C.  r = -1
D.  r = 1
E.  r = a + b

The way I would go about the question is by using the simultaneous equation's techniques in matrices. Since u haven't actually said which is the answer I'm not sure if the way I did it is correct.
Firstly , my answer was option a, so if that's not the answer disregard my explanation.
So what I did is I acted as if my unknown were in matrix x (which it is ) , so the question itself does A*X=AX , to find X I did the inverse of A multiplied by the product matrix AX, so A ^-1*AX , this would give u the value of X .
Input in calculator and answer is the third value in the 3*1 matrix , which for me was b-1a which is equivalent to b-a ,meaning that the answer is a
 I hope that helps.
As I said , if my answer was incorrect plz disregard it , and tell me so we can figure out how to actually do the question.

[Sunshine98]

with q6, idk what they did, i got C personally

with q4 i got B, i dont know where the 4 x 6 came from? the question didnt make much sense to me there was something missing i think?
- i didnt do 1/2 lol

Wow , q6 is baffling , I  got the same answer , I don't understand .
Which trial is it from ?
Also ,it reminds me of a VCAA question which was similar, however, the answer in the multiple choice were intervals so instead of having the exact value it said 'between 10 and 15' so I don't think they'll ask such a question, now.
But I am going to ask my teacher, u should too, its better safe than sorry.

[benobrien]

I dont think VCAA will ask for such specificity.

[LiquidPaperz]

Wow , q6 is baffling , I  got the same answer , I don't understand .
Which trial is it from ?
Also ,it reminds me of a VCAA question which was similar, however, the answer in the multiple choice were intervals so instead of having the exact value it said 'between 10 and 15' so I don't think they'll ask such a question, now.
But I am going to ask my teacher, u should too, its better safe than sorry.

tell me how it goes and what she/he says.

it was a mav one 2008/2009 off the top of my head

[myanacondadont]

with q6, idk what they did, i got C personally

with q4 i got B, i dont know where the 4 x 6 came from? the question didnt make much sense to me there was something missing i think?
- i didnt do 1/2 lol

If you still are confused about Q4 I can help. Q6 looks a bit wonky and yeah, answers for that are pretty self-explanatory.

Let C = running total of perimeter or some shiznit like that.

Since we know the radius and the circumference of a circle: We get

Since the circumference in this question is 1/4 of a circle then So that covers the outside perimeter.

Now for the inside. We know on each side there is 12cm of wire. Bringing the total Now all is left is the inside line.

Using Pythagoras: 

Now adding the previous C:   

Thought I might aswell do it completely worked out for whatever reason. Your question was directed at the 6x4, so that comes from the perimeter on each side of the shape. It's actually 12x2 in my working but they left it as 6x4.


Edit: Does anyone have any 2014 further maths trial exams? My school doesn't give us any aside from like 2008-2012 and all of which I've completed. If anyone can throw a dog a bone I'd greatly appreciate it.

[engton1796]

The way I would go about the question is by using the simultaneous equation's techniques in matrices. Since u haven't actually said which is the answer I'm not sure if the way I did it is correct.
Firstly , my answer was option a, so if that's not the answer disregard my explanation.
So what I did is I acted as if my unknown were in matrix x (which it is ) , so the question itself does A*X=AX , to find X I did the inverse of A multiplied by the product matrix AX, so A ^-1*AX , this would give u the value of X .
Input in calculator and answer is the third value in the 3*1 matrix , which for me was b-1a which is equivalent to b-a ,meaning that the answer is a
 I hope that helps.
As I said , if my answer was incorrect plz disregard it , and tell me so we can figure out how to actually do the question.

Thanks for the reply.  Your answer was indeed correct as I found out today, so your explanation would be on the money. 
Just one more problem I had. 
When reading your explanation, you mentioned multiplying the inverse of A by the product matrix of AX.  I went ahead and multiplied the two, substituting "b" with a random 1, "a" with 2 and "2a" with 4.  I got 2, 0 and -1 for X.  How did you figure out the answer was A instead of A or D?  For A, b-a / 2-1 = 1  and then for D, a+b / -1+2 = 1, so I got the same answer.

[Sunshine98]

Thanks for the reply.  Your answer was indeed correct as I found out today, so your explanation would be on the money. 
Just one more problem I had. 
When reading your explanation, you mentioned multiplying the inverse of A by the product matrix of AX.  I went ahead and multiplied the two, substituting "b" with a random 1, "a" with 2 and "2a" with 4.  I got 2, 0 and -1 for X.  How did you figure out the answer was A instead of A or D?  For A, b-a / 2-1 = 1  and then for D, a+b / -1+2 = 1, so I got the same answer.

There is no need to  substitute numbers , in fact you simply place the matrix calculations  into ur CAS and it will give you the expression , which is option a .There is no need to over complicate the question (I should take my own advice some times, as I too have a tendency to do such  )
anyways I hope that helps 

[LiquidPaperz]

why is this a straight line? (the cross section)

[AngelWings]

why is this a straight line? (the cross section)

I would assume that this is because your intervals are almost completely equal themselves and the height is constantly increasing by 2 metres, so your cross section would be almost perfectly linear (almost, as it still deviates slightly from the bumps here and there).

However, I would require the assistance of others to clarify this, before saying that's correct. (i.e. Somebody please check this. I am not saying this is completely right, so don't take my word for it until it's said to be right by somebody else as well. )

[Billion]

How do you differentiate between options A and C?
The answer is A, but I chose C as the balance of the loan reaches near 0 assuming that it has been paid off?

Thanks

[Billion]

why is this a straight line? (the cross section)

Technically, the line shouldn't be a 'straight' line, as the gaps in between the 2m intervals are not the same in length.
However the assessor's report did say they accepted answers with little variance from a straight line. So assuming you didn't have substantial changes in height you were awarded full marks.

[LiquidPaperz]

ah yes thought so, i thought there wold be some slight rigid movements.

however i dont get the concept why it is a straight line at all?

[LiquidPaperz]

why have they said it has reached 60m? it in within the 50m contour line

[Billion]

why have they said it has reached 60m? it in within the 50m contour line

I presume that the information before the question would have stated Z to be 60m high.
If not, technically speaking the line could read 59.99m and not reach the 60m mark in which the interval of 60m would not be shown on the map.