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VCE Specialist 3/4 Question Thread!
caffinatedloz:
--- Quote from: galaxysauce on December 27, 2021, 11:36:27 am ---P.S I've noticed that I have been struggling with proof questions and was wondering if anyone knew how to get better at them? Thanks!
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I know ArtyDreams has given you a headstart on the question but I thought I'd chuck in my two cents as well so that you can see how I would go about setting up a proof. Hope it's some help!
For proofs, my method was to write down all the things I know (the information in the question). For the question you've attached that would mean writing down AB=BC=CD=DA and AB=AX=BX.
Then you could also write down that as ABX is equilateral angles ABX, BAX and AXB are all 60 degrees. Because of this XBC and XAD are 30 degrees (due to the corners of a square being 90 degree angles).
As triangles AXD and BXC are isosceles (as AX=AB=DA and BX=AB=BC) angle ADX must be equal to AXD. Angle BXC must be equal to BCX. those angles are (180-XBC)/2. They are (180-30)/2 or 75 degrees each.
Then I would down what I have to show. This question doesn't ask you to prove anything, but if you wanted to prove that angle DXC is 150 degrees I would actually write that at the top of the page before I start my proof.
Then I would use the things from the "know" section to actually complete the proof.
My Proof LayoutSHOW: DXC = 150
LHS = DXC
= 360 - AXB - AXD - BXC
= 360 - 60 - 75 -75
= 150 = RHS as required
Best of luck!
galaxysauce:
Hey caffinatedloz!
Sorry if I'm being an airhead right now, but how is triangle AXD and BXC equilateral? I get that it would make sense if it was isosceles, but since it has a 30 degrees angle, it couldn't be an equilateral right?
Thanks!
caffinatedloz:
--- Quote from: galaxysauce on December 27, 2021, 12:32:20 pm ---Hey caffinatedloz!
Sorry if I'm being an airhead right now, but how is triangle AXD and BXC equilateral? I get that it would make sense if it was isosceles, but since it has a 30 degrees angle, it couldn't be an equilateral right?
Thanks!
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Sorry, yes you are right. I meant to say isosceles. I'll fix that now. Thank you!
^^^111^^^:
Hi AN,
I just have a quick question.
When we are taking reciprocals of a function, what happens to its vertical asymptote?
For example, if we have
We have a vertical asymptote at x = +sqrt(2) and x = - sqrt(2).
And if we were to take its reciprocal, it will then become
I know that the vertical asymptotes will now be the x-intercepts of the reciprocal, but my question is will there be a 'hole' at these x-values, since we are technically taking:
Thanks :)
mabajas76:
--- Quote from: ^^^111^^^ on December 29, 2021, 12:53:31 pm ---Hi AN,
I just have a quick question.
When we are taking reciprocals of a function, what happens to its vertical asymptote?
For example, if we have
We have a vertical asymptote at x = +sqrt(2) and x = - sqrt(2).
And if we were to take its reciprocal, it will then become
I know that the vertical asymptotes will now be the x-intercepts of the reciprocal, but my question is will there be a 'hole' at these x-values, since we are technically taking:
Thanks :)
--- End quote ---
The math symbols wern't working for me so i will refer to the 1/2x^2-4 as f(x).
So f(x)=1/0, that is how you get the asymtopes right? And then for the reciporcal f(x)^-1, the vertical asymtopes become the x intercepts as you know. Now, when you take the reciporcal of one side, you do the same to the other. so F(x)^-1=0/1=0. So there would not be holes because you can't find a value for which f(x)^-1=1/0, because the reciporcal f(x) has a value = to 1/0 and the new function is a reciporcal of this.
Hope that helped haha, sorry if it didn't.
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