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July 27, 2025, 02:48:19 am

Author Topic: Maths Methods 3/4 Help Thread 2011  (Read 126886 times)  Share 

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luffy

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Re: Maths Methods 3/4 Help Thread
« Reply #15 on: December 15, 2010, 04:21:03 pm »
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wow, well done thushan. You beat my study score by a Cone Question xD. Did you get a premier award? (if they have told you yet...)

thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #16 on: December 15, 2010, 10:43:59 pm »
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If only I could draw a diagram...

Picture triangle APB with AB = 7, AP = 4, and angle(BAP) = 5 degrees, as the difference between the two bearings is 80-75 = 5.

Draw a horizontal line 'l' from A, and extend BP to meet line 'l'. Let this intersection point be L.

Now all you need to do is find that angle between line 'l' and PB extended, as we are focusing on the bearing of P from B...the direction only. Extending BP does not change its direction.

You would have found (by cosine rule) that PB = 3.04... and by sine rule you can get sin(angle(APB)) = 0.20 => angle(APB) = 168.4, as this angle is obtuse.

an(APL) = 180 - an(APB) = 11.6

Hence angle between line 'l' and PB = 10 + an(APL) = 21.6 degrees

Hence the bearing is 90 - 21.6 = 68.4

True bearing = 068.4T

@luffy: Thanks! Premier's awards aren't announced until well into next year, and to get one for Mathematics, you need to do Specialist and Methods, and I have done only Methods so far; I'm doing Specialist next year.
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ruchika5

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Re: Maths Methods 3/4 Help Thread
« Reply #17 on: December 16, 2010, 02:25:30 pm »
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Hey guys, just wondering, is probability an important part of 3/4 methods? My school didn't go over ANY of it during year 11 so I was looking through 1/2 Essentials today. It seems okay but I don't understand the difference between independent and mutually exclusive probability? If anyone can explain the difference I'd appreciate it heaps!

thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #18 on: December 16, 2010, 02:37:36 pm »
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Independent events A and B are where one event does not AFFECT the other. So Pr (A I B) = Pr (A) which is the same as Pr (A int B) = Pr (A) Pr (B). Think about tossing a coin twice. Getting a tail the first time does not affect the probability of getting a tail the second time. A and B can happen simultaneously.

Mutually exclusive events are where A and B CANNOT simultaneously occur. If A happens, that means B cannot happen. If B happens, A CANNOT happen. I.e. Pr(A int B) = 0. Eg. Toss a coin once. A = 'head' and B = 'tail.' These are mutually exclusive events, as you can't get a head and a tail in the one coin toss!
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Re: Maths Methods 3/4 Help Thread
« Reply #19 on: December 16, 2010, 02:40:18 pm »
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Thanks for clearing that up  8-)

thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #20 on: December 16, 2010, 02:48:05 pm »
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Any time
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thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #21 on: February 05, 2011, 11:22:46 am »
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Just bumping this; keep asking your questions!
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pi

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Re: Maths Methods 3/4 Help Thread
« Reply #22 on: February 05, 2011, 02:10:37 pm »
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Is there an 'extended' chain rule for composite functions like f(g(h(x)))? Or do we have to use it twice?

kamil9876

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Re: Maths Methods 3/4 Help Thread
« Reply #23 on: February 05, 2011, 02:32:04 pm »
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That is the idea, use it twice in the general case and you get your 'extended' chain rule.

Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #24 on: February 05, 2011, 02:46:05 pm »
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That is the idea, use it twice in the general case and you get your 'extended' chain rule.



Sounds about right: dy/dx = dy/du . du/dv . dv/dx
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Re: Maths Methods 3/4 Help Thread
« Reply #25 on: February 05, 2011, 02:47:30 pm »
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That is the idea, use it twice in the general case and you get your 'extended' chain rule.




Sounds about right: dy/dx = dy/du . du/dv . dv/dx
Thats what I was thinking, just wasn't 100% sure! (now I am though)

Thanks thuhsan and kamil!

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Re: Maths Methods 3/4 Help Thread
« Reply #26 on: February 05, 2011, 05:49:58 pm »
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Thushan! the coefficient of x^3 in (3x^2- 5/x)^3? the answer is -135 is that correct?

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Re: Maths Methods 3/4 Help Thread
« Reply #27 on: February 05, 2011, 05:52:37 pm »
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hey Thusthan, noob question

Why is y greater or equal to 3x + 2 , not a function?
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thushan

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Re: Maths Methods 3/4 Help Thread
« Reply #28 on: February 05, 2011, 07:34:13 pm »
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hey Thusthan, noob question

Why is y greater or equal to 3x + 2 , not a function?

Water -

A function has to be either one-to-one or many-to-one.

The relation y>= 3x + 2 is neither; it is many-to-many. Take x = 4. y = 20 satisfies the relation and y = 1001 satisfies the relation. Also, when y = 20, x = -2 satisfies the relation. This shows that this relation is many-to-many and hence cannot be a function.

Thushan! the coefficient of x^3 in (3x^2- 5/x)^3? the answer is -135 is that correct?

Yup that's right
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Re: Maths Methods 3/4 Help Thread
« Reply #29 on: February 05, 2011, 08:10:34 pm »
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Hey thushan, I also have a simple question..
Correct me if i'm wrong (lol which i obviously am) but, the 5th term would be when k=4 ? Or is it k=7?
And after that, I would have a very large coefficient, due to the combinatoric i'd have to evaluate?
i.e. either 100.C.5 or 100.C.7 ... I'm confused here.
At one stage, I did arrive at an answer of
1/2[(7a^4)(b^49)] which was close to an answer shown in the options.

What confuses me here is that it's given 100.C.3 already and asking for the 5 term in the expansion? wouldn't it generally say k=0? Why is it k=3 in this scenario?
« Last Edit: February 05, 2011, 09:16:00 pm by nacho »
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