VCE General & Further Maths Question Thread!

[Math_lord]

Yes, but the whole question pretty much revolves around the "clearly doesn't seem to occur" bit, so that's not areally a valid reason? 

plezz.... The question is wrong because when preparing the task, iam pretty sure the teacher was meant to change the mean to something else

[@#035;3]

Need help with Q9 MODULE 2 EXAM 1 2009.
Cheers.

[MightyBeh]

Need help with Q9 MODULE 2 EXAM 1 2009.
Cheers.

This one took a few steps, and there's probably a few ways to go about it. Here's how I did it though:

Firstly, the only part of the diagram we currently have enough to work with is the triangle at the bottom.


We have the three side lengths, so we can only work out the area (Heron's rule) or any of the angles (cosine rule). Given that the question has nothing to do with finding the area, cosine rule is the obvious choice here. For simplicity, I'm finding the angle QMT. (Note that TQM is a right angle, I just forgot to add that to my diagram )

For simplicity's sake, let the angle QMT = theta



Next we need to find the length TQ (the distance between the pole and the wall) because the pole will fall along this line and (conveniently? ) create a triangle for us to work with.


(again note that the angle MQT is 90 degrees, I just didn't mark it (whoops))

Finally, it's just some pythagoras'


Hope that's clear, it was a kind of round-about question.

[BakedDwarf]

I managed to get the right answer for the attached question below. However, my working out was overly complicated and, although i got the right answer, i believe there is a much simpler way of working it out.

Can someone please attempt the question and post their working out?

[MightyBeh]

I managed to get the right answer for the attached question below. However, my working out was overly complicated and, although i got the right answer, i believe there is a much simpler way of working it out.

Can someone please attempt the question and post their working out?

What's the answer? I think I got it but my method feels too simple 

[BakedDwarf]

What's the answer? I think I got it but my method feels too simple 

Post your answer and i'll say if it's right or wrong. This has been the most challenging question i've seen in further maths so far, so it's good practice.

[StupidProdigy]

Post your answer and i'll say if it's right or wrong. This has been the most challenging question i've seen in further maths so far, so it's good practice.

1/sqrt(2pi)?

[BakedDwarf]

1/sqrt(2pi)?

Nope. But remember the question asked for three decimal places, so your answer would be 0.399, but it's still not right.

[TheMereCat]

Nope. But remember the question asked for three decimal places, so your answer would be 0.399, but it's still not right.

Wouldn't it just be finding the area of the triangle: using herons formula. Which comes to 0.43301

Then since area ratio is 1:2. Divide area of the triangle by two: 0.21650
To find area of circle


Then since pi*r^2 = area of the circle

Divide 0.21650 by pi = 0.06891 and then sqrt of 0.06891 = 0.263

Isn't that it?

[MightyBeh]

I got 0.214m, yay or nay?

[BakedDwarf]

Wouldn't it just be finding the area of the triangle: using herons formula. Which comes to 0.43301

Then since area ratio is 1:2. Divide area of the triangle by two: 0.21650
To find area of circle


Then since pi*r^2 = area of the circle

Divide 0.21650 by pi = 0.06891 and then sqrt of 0.06891 = 0.263

Isn't that it?

I see where you're coming from, and that's how i originally thought of it, but that's not it.

The scale factor is for the area of the SHADED region of the triangle to the area of the REFLECTIVE material. Not for the area of the WHOLE triangle to the area of the REFLECTIVE material.

You got mixed up here.

[BakedDwarf]

I got 0.214m, yay or nay?

Yay

Is this how you worked it out?

A(shaded region) = [0.5 x sin(60)] - [pi x r^2]
A(reflective material) = [0.25 x sin(60)] - [0.5 x pi x r^2]
pi x r^2 = [0.25 x sin(60)] - [0.5 x pi x r^2]
r = 0.214 m

This question was in fact a VCAA question. Only 4% got full marks.

[TheMereCat]

Yay

Is this how you worked it out?

A(shaded region) = [0.5 x sin(60)] - [pi x r^2]
A(reflective material) = [0.25 x sin(60)] - [0.5 x pi x r^2]
pi x r^2 = [0.25 x sin(60)] - [0.5 x pi x r^2]
r = 0.214 m

This question was in fact a VCAA question. Only 4% got full marks.

Oh right I see, I'm on my phone so I didn't really read the question properly lol. Anyways good question to post. What year is this from?

[MightyBeh]

Yay

Is this how you worked it out?

A(shaded region) = 0.5 x sin(60) - pi x r^2
A(reflective material) = 0.25 x sin(60) - 0.5 x pi x r^2
pi x r^2 = 0.25 x sin(60) - 0.5 x pi x r^2
r = 0.214 m

This question was in fact a VCAA question. Only 4% for full marks.

Nope! Working's attached

Edit: Although the method was very similar

[BakedDwarf]

Oh right I see, I'm on my phone so I didn't really read the question properly lol. Anyways good question to post. What year is this from?

2004 exam 2

Nope! Working's attached

Edit: Although the method was very similar

Oh, i see. So you just found the area of the triangle differently.