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HSC 2013 MX2 Marathon (archive) (20 Viewers)

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SpiralFlex

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Re: HSC 2013 4U Marathon

:O Back to complex numbers are we?



 
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hamstar

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Re: HSC 2013 4U Marathon

In hsc maths, can we use matrices/determinants, even though it is not in the course, to find the area of a polygon if they haven't specified "hence".
 

SpiralFlex

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Re: HSC 2013 4U Marathon

In hsc maths, can we use matrices/determinants, even though it is not in the course, to find the area of a polygon if they haven't specified "hence".
Yes

But some teachers may not like it. Or even know the method. There was a conics question I remember that it would be more faster to use it. I think it's better practice to use the hence.
 
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hamstar

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Re: HSC 2013 4U Marathon

^ Thanks a lot. Yeh matrices/determinants are really helpful and fast in finding the area, especially in conics. In the 2003 hsc conics question, 4 marks for the area of the triangle, using only 2-3 lines of working.
 

seanieg89

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Re: HSC 2013 4U Marathon

Do you mean yes as in it is mathematically valid or yes as in it will be awarded marks according to the marking guidelines set forth by the board? I am extremely skeptical about the latter.
 

SpiralFlex

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Re: HSC 2013 4U Marathon

Do you mean yes as in it is mathematically valid or yes as in it will be awarded marks according to the marking guidelines set forth by the board? I am extremely skeptical about the latter.
I was informed that if you use a method that is not within the bounds of the syllabus they will still mark it. Could I be wrong?
 

bangladesh

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Re: HSC 2013 4U Marathon

Would someone please do question 9?
I really struggle to understand it and have no idea as to how to approach it.


Screen shot 2013-10-20 at 2.39.49 PM.png



EDIT: Sorry, i don't know how to use the maths tool on this forum so i just screenshoted the questions.
 
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hayabusaboston

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Re: HSC 2013 4U Marathon

Would someone please do question 9?
I really struggle to understand it and have no idea as to how to approach it.


View attachment 28831



EDIT: Sorry, i don't know how to use the maths tool on this forum so i just screenshoted the questions.

Let cos^-1(x)=a, then cosa=x, draw up a triangle and make an angle a, and u can see that cosa=x/1, so hypotenuse is 1, this means that the remaining side is sqrt(1-x^2) according to pythagoras' theorem. Then, since u let cos^-1=a, just replace it in the original thing, so u get tana=? and u have ur triangle drawn up, so its easy to see the answer is C
 

hayabusaboston

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Re: HSC 2013 4U Marathon

can someone explain the perms/combs 2008 q7a
i) U want 3 balls of the same colour. FIRST THING U GOTTA DO, is pick a colour which u want to "replicate" three times. So 3c1, three colours. You want to have three balls out of n balls, so times that by nC3.

You divide this by the total number of possibilities, which is choosing three balls out of 3n balls, THIS IS BECAUSE you have n balls of each colour, and 3 colours, so 3nC3.



ii) Here you want to choose 1 ball of each colour, (all different colours) so you choose 1 from n, ie nC1, and you do this for each colour, so nC1^3, and again, divide by total possibilities, which is 3nC3.


iii) Here, you want to choose 1 colour, and have 2 balls of that colour, then choose another colour FROM THE TWO THAT REMAIN, to have 1 ball of one of those colours. So 3C1 for one colour, and u choose 2 balls from n balls, so nC2. To choose the remaining ball, you can choose from 2 colours, so pick one as per 2C1, and you choose 1 ball of that colour from n balls, so nC1. Times 3C1.nC2.2C1.nC1 and divide again by total possibilities, 3nC3 for the answer.


iv) Put the results in a ratio against one another, and use the standard transformation or idk what u call it, Nck=n!/(n-k)!k!, for each part of the ratio. If n is large, n, n-1, n-2 etc, are all around the same value n, so u can write n.n-1.n-2 as n^3 etc. Simplify and u have result.
 

obliviousninja

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do u have the solutions as well, coz I have what u have. but the answers are wierd.
i) (n-1)/(3n-1) x (n-2)/(3n-2)
 

bangladesh

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Re: HSC 2013 4U Marathon

Let cos^-1(x)=a, then cosa=x, draw up a triangle and make an angle a, and u can see that cosa=x/1, so hypotenuse is 1, this means that the remaining side is sqrt(1-x^2) according to pythagoras' theorem. Then, since u let cos^-1=a, just replace it in the original thing, so u get tana=? and u have ur triangle drawn up, so its easy to see the answer is C
ahhh yeah... i never even thought of drawing the triangle.

Thanks a lot.
 

VBN2470

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Re: HSC 2013 4U Marathon

Can someone please explain why the limits are as such in the following volumes question? Thanks.

Volume Question.PNG
 

hayabusaboston

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Re: HSC 2013 4U Marathon

Can someone please explain why the limits are as such in the following volumes question? Thanks.

View attachment 28833
At height H is the max volume of water u can have. So u get rsintheta as this is describing the amount of the radius of the cone contained in the height of the water, thus enabling u to find volume in terms of r.

HOLY SHIT WAT A WEIRDASS EXPLANATION


lol ehh idk how to explain differently. Realisenothing will.
 
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VBN2470

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Re: HSC 2013 4U Marathon

At height H is the max volume of water u can have. So u get rsintheta as this is describing the amount of the radius of the cone contained in the height of the water, thus enabling u to find volume in terms of r.

HOLY SHIT WAT A WEIRDASS EXPLANATION


lol ehh idk how to explain differently. Realisenothing will.
Yeah... the fact that the bowl is tilted is playing with my mind...
 

hayabusaboston

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Re: HSC 2013 4U Marathon

Yeah... the fact that the bowl is tilted is playing with my mind...
U know why its pi(r^2-h^2) though right? just confused about limits?

Idk how good of an explanation mine is lol, essentially its "what proportion of r is contained in the max height of the water".
 

VBN2470

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Re: HSC 2013 4U Marathon

U know why its pi(r^2-h^2) though right? just confused about limits?

Idk how good of an explanation mine is lol, essentially its "what proportion of r is contained in the max height of the water".
Yeah I understand the area of the slice etc. it's just the limits that's confusing me...
 
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