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July 27, 2025, 01:22:10 pm

Author Topic: VCE Methods Question Thread!  (Read 5580812 times)  Share 

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yawho

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Re: VCE Methods Question Thread!
« Reply #315 on: February 19, 2012, 11:09:30 pm »
0
define your mathematical definition of "part of"; please enlighten me.

and yes it does.
Which part of the article in the link suggests it does?

yawho

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Re: VCE Methods Question Thread!
« Reply #316 on: February 19, 2012, 11:13:02 pm »
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'I mean cut out as in the only way to get an even function out of p+(q-pr)/(x+r) is to remove (q-pr)/(x+r). If you want to be formal, I guess you can say I 'set' (q-pr)/(x+r)=0'

I did not comment further after that until you posted.

TrueTears

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Re: VCE Methods Question Thread!
« Reply #317 on: February 19, 2012, 11:14:06 pm »
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I used my definition of "part of" since you never gave me one, and I'm not saying it's the mathematical definition since I have never used that phrase before but nonetheless, I assumed you meant for something to be "part of" a function it needs to touch the function, an asymptote is a tangent to the function at infinity, if it's tangent, then it touches the function (albeit at infinity), hence it is "part of" the function. Again like I said, I want to avoid this in total because I don't want to start a controversial debate on what defines infinity.

If you disagree, which you will obviously, then please enlighten me with your mathematical definition of "part of", if not, my previous solution suffices.

You guys are making a meal of a simple question. Here is a more tangible explanation of TT's solution (without using controversial words like 'infinity', 'part of', 'cut' etc.):

The only hyperbola that's an even function is constant, so is constant. Now but clearly so
I thought you said not using infinity.
If you want be pedantic, pretend kamil only said

Quote
The only hyperbola that's an even function is constant

...which you seem to have a hard time interpreting too from your previous post.



When I said in my few previous posts before about the only way to make the hypebola even is analogous to what kamil just stated.

Quote
However only one part of any hyperbola is even, that is it's horizontal asymptote

and never did I state that the asymptote had to the "part of" the function

Quote
just happens p = horizontal asymptote
« Last Edit: February 19, 2012, 11:20:13 pm by TrueTears »
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

yawho

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Re: VCE Methods Question Thread!
« Reply #318 on: February 19, 2012, 11:43:07 pm »
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part of is not the term that I would use. I was quoting that to reply your posts. I have never seen the usage of such term in my study of mathematics. The closest mathematical terms to that are: an element of, a subset of. No element of (or no subset of) the asymptote is an element of (or a subset of) function f.

yawho

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Re: VCE Methods Question Thread!
« Reply #319 on: February 20, 2012, 12:01:44 am »
+1

Quote 'another way is to notice that since a hyperbola is not even in nature, we have to find some part of it that is, and a horizontal line is clearly even, hence f(x) = p (again the horizontal asymptote)'

Quote 'However only one part of any hyperbola is even, that is it's horizontal asymptote, so we pretty much "delete" off every part of the hyperbola and just leave it's horizontal asymptote intact, in other words p+(q-pr)/(x+r) just becomes p since we delete the (q-pr)/(x+r)
component, so f(x) = p.'

Quote 'However only one part of any hyperbola is even, that is it's horizontal asymptote' and never did I state that the asymptote had to the "part of" the function

One part of a human body is quite round, that is it's head. Is the head a part of a human body?

noname

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Re: VCE Methods Question Thread!
« Reply #320 on: February 20, 2012, 03:07:21 pm »
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I keep getting a close answer, but a wrong answer, for part e).
Apparently the answer is
629/14 square units

Phy124

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Re: VCE Methods Question Thread!
« Reply #321 on: February 20, 2012, 04:37:50 pm »
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I believe you would be look at the following:













I can't give you the exact answer without knowing the equations of the lines and the points where they meet. I'll be happy to give it a shot if you give me those :)
« Last Edit: February 20, 2012, 04:39:21 pm by Phy124 »
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noname

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Re: VCE Methods Question Thread!
« Reply #322 on: February 20, 2012, 04:58:20 pm »
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I believe you would be look at the following:













I can't give you the exact answer without knowing the equations of the lines and the points where they meet. I'll be happy to give it a shot if you give me those :)

A (7,2)
B (2,5)
C (8,15)
D (66/7,82/7)

Equation AB: y=-3x/5 + 31/5
Equation perpendicular bisector of AB: y=5x/3 -4
Equation AD: y=4x-26
Equation BC: y=5x/3 + 5/3

Phy124

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Re: VCE Methods Question Thread!
« Reply #323 on: February 20, 2012, 05:23:30 pm »
+1
Oh, my graph was right :P

Ok firstly we have to work out the line segment CD for the last part.

we know it goes through C and D so we can calculate the gradient from that:



The use y - y1 = m(x - x1)





Now to calculate the area, using the formula I showed in my previous post:

« Last Edit: February 20, 2012, 05:25:34 pm by Phy124 »
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noname

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Re: VCE Methods Question Thread!
« Reply #324 on: February 20, 2012, 05:31:53 pm »
+1
Oh, my graph was right :P

Ok firstly we have to work out the line segment CD for the last part.

we know it goes through C and D so we can calculate the gradient from that:



The use y - y1 = m(x - x1)





Now to calculate the area, using the formula I showed in my previous post:



Sorry to bother, but what exactly does that formula mean?
Like.. the cursive f and stuff? 

Phy124

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Re: VCE Methods Question Thread!
« Reply #325 on: February 20, 2012, 05:39:38 pm »
+1
It's an integral sign. Maybe you're not up to that yet? my bad :P

If you haven't learnt integration yet, you'll probably have to make a few triangles and rectangles to calculate the area. I'm going out now, if no one else has helped when I get back, I'll do that for you :)
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chocolatedaddy

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Re: VCE Methods Question Thread!
« Reply #326 on: February 20, 2012, 06:29:02 pm »
+2
(Image removed from quote.)

I keep getting a close answer, but a wrong answer, for part e).
Apparently the answer is
629/14 square units
To find the area of this quadrilateral. You will need to split it into a triangle and a trapezium.The triangle being DXA and the trapezium being BCDX. Work out the areas of both and add them together and you have the area.

noname

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Re: VCE Methods Question Thread!
« Reply #327 on: February 20, 2012, 07:05:01 pm »
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It's an integral sign. Maybe you're not up to that yet? my bad :P

If you haven't learnt integration yet, you'll probably have to make a few triangles and rectangles to calculate the area. I'm going out now, if no one else has helped when I get back, I'll do that for you :)

(Image removed from quote.)

I keep getting a close answer, but a wrong answer, for part e).
Apparently the answer is
629/14 square units
To find the area of this quadrilateral. You will need to split it into a triangle and a trapezium.The triangle being DXA and the trapezium being BCDX. Work out the areas of both and add them together and you have the area.

yeh trapezium and triangle works!
omg i tried doing that method for the past 2 days, but only succeeded today.
It turns out I probably keyed in the wrong values in the calculator... seeing as how there are so many terms in there -_-
THANKS ALL

ashoni

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Re: VCE Methods Question Thread!
« Reply #328 on: February 23, 2012, 06:12:46 pm »
+1
need help with quadratic question... D:

4a. Find the value of c such that y=x+c is a tangent to the parabola y=x^2 -x-12.
  b. Find the values of m for which the sraight line y=mx+6 is tangent to the parabola.

thanks guys!

Phy124

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Re: VCE Methods Question Thread!
« Reply #329 on: February 23, 2012, 06:53:44 pm »
+1
need help with quadratic question... D:

4a. Find the value of c such that y=x+c is a tangent to the parabola y=x^2 -x-12.
  b. Find the values of m for which the sraight line y=mx+6 is tangent to the parabola.

thanks guys!
a) The line will be at a tangent when the lines interact at a single point.



















b) Er... I wouldn't have thought any values of m would yield mx + 6 to be a tangent to the parabola x2 - x - 12, I'll investigate, maybe I've read something wrong.

edit: left out an "= 0"
« Last Edit: February 23, 2012, 07:16:24 pm by Phy124 »
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