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May 20, 2022, 08:12:44 am

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ArtyDreams

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9855 on: December 27, 2021, 11:50:06 am »
+3
Hi.. I'm back again haha.
Could someone please help me with this question? 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!
(Image removed from quote.)

Hi!

For this question, the main point is that ABX is an equilateral triangle, therefore, all angles and sides are equal. The angles in an equilateral triangle are 60º, so you use angle properties to deduce the sizes of the other angles. (For example, angle XAD is 30º, since XAB is 60º, and they add up to a right angle).

Hope this helps, let me know if you need some more guidance!

caffinatedloz

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9856 on: December 27, 2021, 12:20:10 pm »
+2
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!
(Image removed from quote.)

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 Layout
SHOW: DXC = 150

LHS = DXC
= 360 - AXB - AXD - BXC
= 360 - 60 - 75 -75
= 150 = RHS                as required

Best of luck!
« Last Edit: December 27, 2021, 01:39:08 pm by caffinatedloz »

galaxysauce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9857 on: December 27, 2021, 12:32:20 pm »
0
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

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9858 on: December 27, 2021, 01:38:18 pm »
0
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!

Sorry, yes you are right. I meant to say isosceles. I'll fix that now. Thank you!

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9859 on: December 29, 2021, 12:53:31 pm »
0
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

$y =\frac{1}{2x^2-4}$

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

$y =2x^2 -4$

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:
$y = \frac{1}{\frac{1}{2x^2-4}}$

Thanks

mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9860 on: December 29, 2021, 01:43:04 pm »
+2
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

$y =\frac{1}{2x^2-4}$

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

$y =2x^2 -4$

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:
$y = \frac{1}{\frac{1}{2x^2-4}}$

Thanks
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.
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9861 on: December 29, 2021, 03:45:09 pm »
0
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.

Ohh ok,

so just to confirm I'm understanding this, since we are taking the reciprocal, it would be to both sides, so that means it's 1/y = 1/f(x) or y = f(x). So there won't be holes, as this would be 0/1 (0 is numerator and 1 is denominator), instead of 1/ (1/0) (for example).

Thank you!

mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9862 on: December 29, 2021, 06:03:14 pm »
0
Well when you have f(x), that isn't an asymtope right? It is only when x= + or - sqrt of 2 that it becomes an asymptope, because + or - sqrt of 2 is equal to 1/0, so with 1/f(x), the y value for any x will become 1/y. So thats why at sqrt 2 it goes from 1/0(undefined) to 0/1.

Now what ur saying is if 1/(1/0) has an asymtope, I got what ur saying because its like saying 1/undefined which doesn't make sense. or 1/f(x), but f(x) has values which are undefined. Well the thing is undefined can be exprezsed as 1/0, that is its definition in numbers, a division by 0. And 1/(1/0) will actually give it a value. So no there aren't any values that 1/f(x) would still have asymtopes because even though the f(x) can still give a 1/0, it isn't ready to be evaluatated because its divided by 1.
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

galaxysauce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9863 on: January 02, 2022, 12:07:23 pm »
0
Hey everyone!

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9864 on: January 02, 2022, 01:54:21 pm »
+1
Recall that $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}||||\vec{v}||}.$

Choose some arbitrary vector $\vec{v} = (x_1, y_1, z_1)$. Then, we can say that it makes an angle of $\alpha$ with the unit vector $(1, 0, 0)$.

Then,
$\cos \alpha = \frac{x_1}{\sqrt{x_1^2+y_1^2+z_1^2}} \implies \cos^2 \alpha = \frac{x_1^2}{x_1^2 + y_1^2+z_1^2}.$

The rest of the proof follows by doing the same thing on the other coordinate axes then summing the results up.
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galaxysauce

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9865 on: January 07, 2022, 09:47:47 pm »
0
Hi.. could someone please help me with this question? It's a pretty simple one, but I'm stuck on drawing the "other" triangle, and use it to find the other "BC".

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9866 on: January 07, 2022, 10:07:36 pm »
+1
Remember that sine is positive in both the first and second quadrant, so the other value can be found by using the identity $\sin x = \sin (\pi - x)$. (Search up the ambiguous case if you want more information, or query again)

BC can be found in much the same way as you did the first time (using the cosine rule or otherwise).
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HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
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beep boop

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9867 on: January 28, 2022, 11:33:21 am »
0
I've done all the parts except b and d.

b. I tried to find the cube roots.

d. I'm confused as to use polar formula of z1z2 or use another method completely.
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mabajas76

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9868 on: January 28, 2022, 06:58:39 pm »
+2
So for b try this:
z=rcis(a)
z^3=z1
so r^3cis(3a)=to the z1
r^3=200^1/2, then cube root both. Then get 3a=-3pi/4 and solve for each, then the most important thing which is adding/subtracting the period to get all soloutions in the Argument form.
"Don't give up, and don't put too much effort into things that don't matter"-Albert Einstein, probably.

SpopyMoose

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9869 on: February 06, 2022, 08:30:04 pm »
0
I'm not sure if someone else has already asked this but is the sacs in 3/4 (both spesh and methods but I'm primarily concerned about spesh) all application questions and run 4-6 periods long? If so what do they look like and if anyone can send any example sacs that would be greatly appreciated!