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August 01, 2025, 01:21:01 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2555150 times)  Share 

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Stick

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Re: Specialist 3/4 Question Thread!
« Reply #1275 on: February 11, 2013, 09:10:16 pm »
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I actually saw that, but I thought it was wrong and didn't do it. :P Thanks for the help! :)
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Homer

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Re: Specialist 3/4 Question Thread!
« Reply #1276 on: February 13, 2013, 06:47:22 pm »
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How would I convert 3 degrees per hour to degrees per second?
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Re: Specialist 3/4 Question Thread!
« Reply #1277 on: February 13, 2013, 06:55:11 pm »
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How would I convert 3 degrees per hour to degrees per second?
divide by 3600
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pi

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Re: Specialist 3/4 Question Thread!
« Reply #1278 on: February 13, 2013, 08:50:31 pm »
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How would I convert 3 degrees per hour to degrees per second?

Is this a serious spesh question..?

For future reference, use the units:
deg/1hour = deg/60mins = deg/(60*60)s = deg/3600s

Hence, multiply by 1/3600 or divide by 3600 to make the conversion.

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Re: Specialist 3/4 Question Thread!
« Reply #1279 on: February 14, 2013, 10:56:58 am »
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If we're given a vector proof with a shape that already has vectors defined in the question, are we allowed to define new ones? :-\ I don't know if I make sense, but here's an example: Consider the quadrilateral OABC where OA=a and OB=b. Prove... etc. Can I then make OC=c then use it in my proof or will I lose marks as I'm supposed to use a and b?

Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #1280 on: February 14, 2013, 12:31:53 pm »
+1
If we're given a vector proof with a shape that already has vectors defined in the question, are we allowed to define new ones? :-\ I don't know if I make sense, but here's an example: Consider the quadrilateral OABC where OA=a and OB=b. Prove... etc. Can I then make OC=c then use it in my proof or will I lose marks as I'm supposed to use a and b?

Its okay to let OC=c but you need to write c in term of a and b so that in the end you can sub it back into your proof to get the answer  (as your proof will contain c while question requires the proof in term of a and b only)
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jono88

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Re: Specialist 3/4 Question Thread!
« Reply #1281 on: February 14, 2013, 11:32:36 pm »
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Find all real values of r for which ri is a solution of the equation z^4-2z^3+11z^2-18z+18=0
(z^2+ri)(z^2-ri)
=z^2+r^2
therefore (z^2+r^2)(az^2+bz+c)=the equation in question
so far ive solved a=1
b=-2
what do i to find c?

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Re: Specialist 3/4 Question Thread!
« Reply #1282 on: February 15, 2013, 01:08:26 am »
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You should be subbing z = ri into the polynomial equation. You then get real and imaginary parts, which both need to equal zero. You need to find the values of r for which both of them are zero simultaneously.
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Jaswinder

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Re: Specialist 3/4 Question Thread!
« Reply #1283 on: February 15, 2013, 10:18:14 pm »
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How would I anti differentiate x^2/(x^2+4)

polar

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Re: Specialist 3/4 Question Thread!
« Reply #1284 on: February 15, 2013, 10:20:29 pm »
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How would I anti differentiate x^2/(x^2+4)



hence,

the rest of the question
« Last Edit: February 15, 2013, 10:22:59 pm by polar »

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Re: Specialist 3/4 Question Thread!
« Reply #1285 on: February 15, 2013, 10:30:51 pm »
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Woah you are one smart cookie polar

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Re: Specialist 3/4 Question Thread!
« Reply #1286 on: February 16, 2013, 10:17:36 am »
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Chapter 2E, question 18:
Coplanar points A, B, C, D, and E have position vectors a, b, c, d, and e respectively to a relative origin O. A is the midpoint of OB and E divides AC in the ratio 1:2. If e = d, show that OCDB is a parallelogram.

I've drawn a diagram, but it hasn't really helped. I know I need to show that OC and BD are equal (or CD and OB).

The Worked solutions say the following:


However, between the second and third lines of the working for OC, they change DC to -b, which is using the fact that it is a parallelogram (what we're trying to prove) to show that it is a parallelogram, isn't it?
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Re: Specialist 3/4 Question Thread!
« Reply #1287 on: February 16, 2013, 11:36:59 am »
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^^ Had the same problem with that question :L

Just stuck on part c for this one.


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Re: Specialist 3/4 Question Thread!
« Reply #1288 on: February 16, 2013, 12:00:11 pm »
+2
^^ Had the same problem with that question :L

Just stuck on part c for this one.


picture AB with MN intersecting it - lets call the point of intersection X. X cuts AB and also MN so:

firstly, there must exist a such that



similarly, there must exist a such that



now, equate these two expressions (I'll use matrices because they're easier to do on latex, use i-j-k notation here if you want)



now, write out simultaneous equations and solve them. this gives
hence, substitute into any expression for . for example

« Last Edit: February 16, 2013, 12:04:36 pm by polar »

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Re: Specialist 3/4 Question Thread!
« Reply #1289 on: February 16, 2013, 12:22:39 pm »
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Nice one, thanks.