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HSC 2013 Maths Marathon (archive) (1 Viewer)

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Carrotsticks

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Re: HSC 2013 2U Marathon

I can't get this question? How do you do this.
It's out of the 2U course, so no need to worry.

That's gay
No, it's not 'gay'.

The point of the question is for you to practise deducing some expression from another given expression. Accepting the required result and playing with it until you get a known identity is technically valid for that question, but defeats the purpose of it.

Sorry if I'm being a bit snappy, but people saying "Oh.. that's gay" really rustles my jimmies.
 

leesh95

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Re: HSC 2013 2U Marathon

Not quite, the angles are in AP but the actual terms may not necessarily be in AP.

Hint: Consider the identity sin(x) = cos(90-x)
Can you please explain a bit more? I can't understand.
 

Carrotsticks

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Re: HSC 2013 2U Marathon

Can you please explain a bit more? I can't understand.
Just because we have A + B + C + D + ..., doesn't always mean we have an AP.

The series was not an AP. You can easily check using a calculator that T_2 - T_1 is not equal to T_3 - T_2.
 

Menomaths

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Re: HSC 2013 2U Marathon


Show working/thought process
 

mahmoudali

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Re: HSC 2013 2U Marathon

How did you know you had to add 9?
because you look at 6y and you need y^2 +6y +c to be a perfect square and only 9 makes it so
so add 9 to both sides since genral equation of a circle is (x-h)^2 +(y-k)^2 = r
 

mahmoudali

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Re: HSC 2013 2U Marathon

because you look at 6y and you need y^2 +6y +c to be a perfect square and only 9 makes it so
so add 9 to both sides since genral equation of a circle is (x-h)^2 +(y-k)^2 = r
idk what im talking about lol this was the pattern i saw in solutions
 

Menomaths

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Re: HSC 2013 2U Marathon

I'm unable to do this integral can you please help me and post your solution right now

-----------





bounded by pi/4 and 0



 
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RealiseNothing

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Re: HSC 2013 2U Marathon

Sy posted this up I think ages ago, but it's still probably the best 2U integral I've seen:

 
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