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April 13, 2026, 09:31:22 am

Author Topic: VCE Methods Question Thread!  (Read 6015746 times)  Share 

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Cogglesnatch Cuttlefish

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Re: VCE Methods Question Thread!
« Reply #9165 on: March 09, 2015, 04:22:22 pm »
+1
Can someone help with:



Thank you in advance!

Take out a factor of 2e^2x from the numerator and 5e^x from the denominator
Then the numerator and denominator should simplify to -2e^x/5

Im sorry If I couldn't explain it properly; cant be bothered using the tex thingy and it would be too messy without it   :P
Science at uom

knightrider

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Re: VCE Methods Question Thread!
« Reply #9166 on: March 09, 2015, 04:50:06 pm »
0
How would you do this question?

A sand timer consists of two cones joined at the apex. Each cone has height h, radius r and an angle at the apex of 60°.

Express the radius of the top cone in terms of its height. Give your answer in exact form.

Floatzel98

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Re: VCE Methods Question Thread!
« Reply #9167 on: March 09, 2015, 06:16:27 pm »
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I need help with d) ii). Are we supposed to look at the graphs or equate and solve the two equations? I'm kind of lost with it.
Anyone?
2016 - 2019: Bachelor of Science Advanced - Research (Honours) [Physics and Mathematics] @ Monash University

Eiffel

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Re: VCE Methods Question Thread!
« Reply #9168 on: March 09, 2015, 06:30:21 pm »
0
this is a property of even functions: f(-x)=f(x)
it means that: the y value of an x value is equal to the y value of -x
for y=(x-h)^2+k to be symmetrical, h can only be 0 because if it is shifted left or right, f(-x)=f(x) is not true
Even functions mean that f(x) = f (-x). Therefore, they are concerned with reflection in the y axis.

Think about it, what is the only value of x will always be true for x = -x... 0!

1 =/= -1
2 =/= -2
etc...

But 0 = -0
0 = 0

Therefore, if your function is to be even, then x can have no horizontal translation.

X - h can only be equal to -(x-h) if the value of h is 0.

Hope that made sense.

Edit: Oh and K can be any real number because symmetry around the y-axis is not affected by vertical translations.


sorry, i still dont understand.

i do understand the property

Cosec

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Re: VCE Methods Question Thread!
« Reply #9169 on: March 09, 2015, 06:40:56 pm »
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Anyone?

What question in particular? Or the whole thing?

Floatzel98

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Re: VCE Methods Question Thread!
« Reply #9170 on: March 09, 2015, 07:24:41 pm »
0
What question in particular? Or the whole thing?
I need help with d)ii). The very last part of question 6
2016 - 2019: Bachelor of Science Advanced - Research (Honours) [Physics and Mathematics] @ Monash University

knightrider

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Re: VCE Methods Question Thread!
« Reply #9171 on: March 09, 2015, 07:25:22 pm »
0
How would you do this question?

A sand timer consists of two cones joined at the apex. Each cone has height h, radius r and an angle at the apex of 60°.

Express the radius of the top cone in terms of its height. Give your answer in exact form.

Question?can someone answer this?

kinslayer

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Re: VCE Methods Question Thread!
« Reply #9172 on: March 09, 2015, 07:28:15 pm »
+1
sorry, i still dont understand.

i do understand the property

f(x) = (x-h)^2 + k

Assume f(x) is even. Then:

f(x) = f(-x)

(x-h)^2 + k = (-x - h)^2 + k

(x-h)^2 = (-(x+h))^2

(x-h)^2 = (x+h)^2

x^2 - 2h + h^2 = x^2 + 2h + h^2

-2h = 2h

-4h = 0

h = 0


kinslayer

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Re: VCE Methods Question Thread!
« Reply #9173 on: March 09, 2015, 07:31:10 pm »
0
Question?can someone answer this?

Draw a diagram. You are looking for a right-angled triangle with opposite and adjacent sides h and r, respectively. (The apex angle will give you the angle, you will need to use symmetry). Then trigonometry tells us that tan(pi/3) = h/r.

cosine

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Re: VCE Methods Question Thread!
« Reply #9174 on: March 09, 2015, 08:34:24 pm »
0
Draw a diagram. You are looking for a right-angled triangle with opposite and adjacent sides h and r, respectively. (The apex angle will give you the angle, you will need to use symmetry). Then trigonometry tells us that tan(pi/3) = h/r.
Can someone give me a quick run down on how to solve equations using matrices? Apparently its on my sac tomorrow, i didnt know :(
2016-2019: Bachelor of Biomedicine
2015: VCE (ATAR: 94.85)

dankfrank420

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Re: VCE Methods Question Thread!
« Reply #9175 on: March 09, 2015, 08:36:33 pm »
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Can someone give me a quick run down on how to solve equations using matrices? Apparently its on my sac tomorrow, i didnt know :(

Transformations? Or just equations?

cosine

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Re: VCE Methods Question Thread!
« Reply #9176 on: March 09, 2015, 08:47:23 pm »
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Transformations? Or just equations?

Equations man, thanks :3
2016-2019: Bachelor of Biomedicine
2015: VCE (ATAR: 94.85)

knightrider

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Re: VCE Methods Question Thread!
« Reply #9177 on: March 09, 2015, 09:38:09 pm »
0
How would you do 0.06 divided 12 by hand?

dankfrank420

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Re: VCE Methods Question Thread!
« Reply #9178 on: March 09, 2015, 10:01:42 pm »
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Equations man, thanks :3

It's pretty basic, but inverse/matrix multiplication can be tricky at the start.

Try Khan Academy?

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #9179 on: March 09, 2015, 10:04:34 pm »
+3
Equations man, thanks :3

Consider the following transformations:

-Dilation from x-axis by 1/3
-Reflection in the y-axis
-Translate left 5 units
-Translate up 2 units.

So, consider the mappings
x to x'
y to y'

Well, x and y will both have a relationship with this transformed variables. So first, dilation from the x-axis:

y'=y/3

Next, reflection in the y-axis:

x'=-x

Now, the translations:

x'=-x-5
y'=y/3+2

Now, solving for x and y, we get:

x=-x'-5
y=3y'-6

We can put this into the following matrix equation:



Basically, just write out the transformations, change into a matrix equation (you should be able to go from a set of linear equations into a matrix equation for methods - becomes even more important for Markov Chains later on)

EDIT: Wait, just realised how wrong I read that LOL. Ah well, still here if people need help with matrix transformations. :P dankfrank420's on the ball with this one.

How would you do 0.06 divided 12 by hand?

It's 0.06/12.
Done. Exact form. (of course, normally we take out the decimals, so I'd multiply top and bottom by 100)