VCE General & Further Maths Question Thread!

[MightyBeh]

You may have calculated that 2011 is twice the original (hence 200%), but this is an INCREASE by only 100%, since the original of 2005 is already 100% of the 2005.

Damn yeah I was about to ask for working out 

[n.a]

You may have calculated that 2011 is twice the original (hence 200%), but this is an INCREASE by only 100%, since the original of 2005 is already 100% of the 2005.

Yeah, that's pretty much what I did. Thanks!

[bedigursimran]

Hey guys. 2013 VCAA Exam 1: Can you please help me with question 8 Trig and question 7 and 8 in graphs and relations. Thanks so much!

[TheMereCat]

Hey guys. 2013 VCAA Exam 1: Can you please help me with question 8 Trig and question 7 and 8 in graphs and relations. Thanks so much!

For Question 8, Geo and Trig, basically you'd have to draw a diagram to get the answer, *was sort of lazy drawing it out right now, so here's the solution from itute. Basically you find the angle of the larger triangle as 51 degrees, and then since it lies on a 180 degree line you take away 51 from 180 to get the bearing of Black tower from white tower.


For Question 7 Graphs and Relations, all you really have to do is find the gradient of the straight line, as it represent the value for k in the rule. so the rule is H = k/d^2, to find k just find the gradient of the straight line which you could use any two points for. You should get the gradient as 58.4. Hence the rule is H=58.4/d^2.

[bedigursimran]

For Question 8, Geo and Trig, basically you'd have to draw a diagram to get the answer, *was sort of lazy drawing it out right now, so here's the solution from itute. Basically you find the angle of the larger triangle as 51 degrees, and then since it lies on a 180 degree line you take away 51 from 180 to get the bearing of Black tower from white tower.


For Question 7 Graphs and Relations, all you really have to do is find the gradient of the straight line, as it represent the value for k in the rule. so the rule is H = k/d^2, to find k just find the gradient of the straight line which you could use any two points for. You should get the gradient as 58.4. Hence the rule is H=58.4/d^2.

Why can't I substitute in numbers for x and y to get k? 25.7 = k/(0.44^2). The answer I got was C; H = 4.98/d^2. I substituted in the x coordinate H = 4.98/(0.44^2) and the answer is 25.7, the y value in the question. What am I doing wrong? And can you help me with question 8? Thanks you so much!!

[Losingmotivation]

Hey everyone, I'm aware that the SS you get is quite dependent upon both your sacs and your exam, but considering you have an A average in your sacs (most likely going to scale up from the level of difficulty??), what kind of percentage should you be aiming for in the exam to have a chance of obtaining a SS of over 40?

Thanks!

[MightyBeh]

Why can't I substitute in numbers for x and y to get k? 25.7 = k/(0.44^2). The answer I got was C; H = 4.98/d^2. I substituted in the x coordinate H = 4.98/(0.44^2) and the answer is 25.7, the y value in the question. What am I doing wrong? And can you help me with question 8? Thanks you so much!!

I've explained this one before! I think Latex is being weird though so if the worded explanation isn't enough I'll re-do the working out.

Although I'd also like to add since doing that the first time I've run into this rule which I think was mentioned in the itute and VCAA solutions:
(x/a) + (y/b) = 1,
where a and b are the x and y intercepts, respectively. (pretty sure it's relevant )

Hey everyone, I'm aware that the SS you get is quite dependent upon both your sacs and your exam, but considering you have an A average in your sacs (most likely going to scale up from the level of difficulty??), what kind of percentage should you be aiming for in the exam to have a chance of obtaining a SS of over 40?

Thanks!

Definitely above 90%. No more than a couple of marks each exam, it gets pretty competitive above 40

EDIT; Fixed the link

[Losingmotivation]

o.O okayyy.. will try!

thanks

[bedigursimran]

I've explained this one before! I think Latex is being weird though so if the worded explanation isn't enough I'll re-do the working out.

Although I'd also like to add since doing that the first time I've run into this rule which I think was mentioned in the itute and VCAA solutions:
(x/a) + (y/b) = 1,
where a and b are the x and y intercepts, respectively. (pretty sure it's relevant )

That link won't work. Can you copy the worked solution or tell me how to find it? And how would I use that formula? Both the x and y ints are 0, so 0/a + 0/b != 1.

[MightyBeh]

That link won't work. Can you copy the worked solution or tell me how to find it? And how would I use that formula? Both the x and y ints are 0, so 0/a + 0/b != 1.

Fixed the link, whoops. I don't understand how the x and y intercepts are 0, but I also didn't do the question using that rule. Just thought I'd put it out there

[@#035;3]

What should I do over the next two days to prepare for the exam.
I've ran out of exam 1's..

[TheMereCat]

What should I do over the next two days to prepare for the exam.
I've ran out of exam 1's..

Maybe redo VCAA exams you made some mistakes on or perfect your bound reference.

[TheMereCat]

Why can't I substitute in numbers for x and y to get k? 25.7 = k/(0.44^2). The answer I got was C; H = 4.98/d^2. I substituted in the x coordinate H = 4.98/(0.44^2) and the answer is 25.7, the y value in the question. What am I doing wrong? And can you help me with question 8? Thanks you so much!!

You can't substitute the coordinates to find k here, because it's plotted as H against 1/d^2, if you had the original graph, which would be H against D, you would be able to use it to find the coefficient. Since you have the transformed graph, you'll need to find the K using the gradient of the straight line.

[j0yce]

Hi
Could someone please explain Q.3a (matrices module) from last years exam 2?
I've read the examiners report but still don't understand it!
I thought element 0.4 (row 2, column 3) in matrix T  would have some role in the answer??

[TheMereCat]

Hi
Could someone please explain Q.3a (matrices module) from last years exam 2?
I've read the examiners report but still don't understand it!
I thought element 0.4 (row 2, column 3) in matrix T  would have some role in the answer??

I'll try explain this as best as i can, in the first transition matrix, you see that 10% of B changes to C before C withdraws, That's clear right? You can see from the matrix.

Next, in the second matrix, which depicts what would happen after Mr Choi withdraws, 10% would split equally, 5% will go to A (10% to 15%) and 5% to B (80% to 85%), you can see how the 10% would split equally between A and B, right? since 5% of a total 10% goes to B, 5/10 * 100 = 50%