bleakarcher
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Re: 2012 HSC MX1 Marathon
wait i got 6pi^2 when i did it again
wait i got 6pi^2 when i did it again
HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)
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Carrotsticks
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largarithmic
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Carrotsticks
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Re: 2012 HSC MX1 Marathon
2. If a girl is behind each boy, that basically means the girls must be on the outside ring and boys on the inside. Arrange boys 4! Arrange girls 5! Answer same as (1)
3. C(10,5) to pick 5 people to go to first table. This obviously forces the other 5 to go to the other table. Multiply by 4! to arrange one table, and 4! to arrange the other. Use the factorial fraction notation for C(n,k) to arrive at the required answer.
Apologies in advance if I made a silly error. Very tired now.
1. Fix and arrange 5 boys in a circle, so 4!. Insert 5 girls into each gap and arrange them, so 5!. Answer: 4!5!Actually I needed help with this probability question:
\ one\ ring,\ with\ no\ two\ girls\ next\ to\ one\ another;\\ b)\ two\ cocentric\ rings,\ with\ a\ girl\ behind\ each\ boy?\\ ii)\ Show\ that\ 10\ persons\ may\ be\ seated\ at\ two\ round\ tables,\ 5\ persons\ at\ each,\ in\ \frac{10!}{5^2}\ ways.)
2. If a girl is behind each boy, that basically means the girls must be on the outside ring and boys on the inside. Arrange boys 4! Arrange girls 5! Answer same as (1)
3. C(10,5) to pick 5 people to go to first table. This obviously forces the other 5 to go to the other table. Multiply by 4! to arrange one table, and 4! to arrange the other. Use the factorial fraction notation for C(n,k) to arrive at the required answer.
Apologies in advance if I made a silly error. Very tired now.
Skeptyks
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Re: 2012 HSC MX1 Marathon
Thankyou! Managed to work out 1) and 2) during the time but for 3), I had no idea how to approach the question.
Re: 2012 HSC MX1 Marathon
Sought of question, and if you don't knOw it then
Your teacher is bad lol. I went through a chord of
Contact question involvIng the hyperbola with my
Class the other day and if you didnt know this method
That would of turned Into possibly the worst quadratic
Equation you could face. But yeh all you 3/4 unit kids
Viewing this forum look at how carrot did this sought of
Question and make sure you understand how to apply it,
If not ask your teacher or the forum
This should be the only method you use for thisNooblet, might I suggest something to you. This is a very useful trick that will save you a lot of time during exams.
You wanted to find the midpoint of PQ. To do this, you combined the equations for the parabola and the line simultaneously. Afterwards, you used the quadratic formula, added the two solutions, then divided by two.
However if you didn't notice, the two x co-ordinates of the points of intersections are the roots of the quadratic that you acquired by combining the two equations.
What you can recognise is that the midpoint of the x coordinates is the 'average' of the x coordinates.
The 'average' as taught in Year 7 (or earlier) can be found by adding the two numbers, then dividing them by 2.
Instead of actually finding both the points of intersections, then adding them, you could have used the 'Sum of Roots' method on that quadratic, which would have acquired the 'Adding of the two numbers' as I mentioned above.
Now, all you need to do is divide by two!
Allow me to demonstrate... Maths is better shown than said:
 gives us $ -2at $, which is exactly what you had.$)
Sought of question, and if you don't knOw it then
Your teacher is bad lol. I went through a chord of
Contact question involvIng the hyperbola with my
Class the other day and if you didnt know this method
That would of turned Into possibly the worst quadratic
Equation you could face. But yeh all you 3/4 unit kids
Viewing this forum look at how carrot did this sought of
Question and make sure you understand how to apply it,
If not ask your teacher or the forum
Carrotsticks
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Re: 2012 HSC MX1 Marathon
Also, a bit of a 3D problem. Not necessarily asked in HSC, but good way to force you to visualise:
The first 'octant' in the XYZ plane refers to where X, Y and Z are all positive. Show that a sphere tangential to the XYZ axes simultaneously has centre (r,r,r), where r is the radius of the sphere.
Sorry to be grammar nazi, but this really bothered me lol. iPhone spellcheck?
Also, a bit of a 3D problem. Not necessarily asked in HSC, but good way to force you to visualise:
The first 'octant' in the XYZ plane refers to where X, Y and Z are all positive. Show that a sphere tangential to the XYZ axes simultaneously has centre (r,r,r), where r is the radius of the sphere.
Skeptyks
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Re: 2012 HSC MX1 Marathon
Actually, I was wondering, why did you multiply with the second 4!? 10C5 determines the combinations for the first table, 4! for the second table. Obviously your calculations are correct as 10!/(5!5!) x 4!4! = 10!/5^2 but I don't understand how it works.
bleakarcher
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deswa1
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bleakarcher
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Re: 2012 HSC MX1 Marathon
I can not believe it...Carrotsticks and largarithmic, turns out you need sleep
I can not believe it...Carrotsticks and largarithmic, turns out you need sleep
Timske
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Re: 2012 HSC MX1 Marathon
How would u make x^2 the subject?New question:
The region inside the circle <a href="http://www.codecogs.com/eqnedit.php?latex=(x-3)^2@plus;y^2=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x-3)^2+y^2=1" title="(x-3)^2+y^2=1" /></a> is rotated about the y-axis.
deswa1
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Re: 2012 HSC MX1 Marathon
<a href="http://www.codecogs.com/eqnedit.php?latex=(x-3)^2@plus;y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" title="(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=(x-3)^2@plus;y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" title="(x-3)^2+y^2=1 \\ (x-3)^2=1-y^2 \\ (x-3)=\pm \sqrt(1-y^2) \\ x=3\pm \sqrt(1-y^2)" /></a>