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November 08, 2025, 02:01:13 pm

Author Topic: VCE Methods Question Thread!  (Read 5782845 times)  Share 

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BLACKCATT

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Re: VCE Methods Question Thread!
« Reply #6180 on: October 09, 2014, 10:04:05 pm »
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The vertical axis is x
the Horizontal axis is time t seconds.

How do we find the time between -15<x<15 ?
 
« Last Edit: October 09, 2014, 10:08:40 pm by BLACKCATT »

silverpixeli

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Re: VCE Methods Question Thread!
« Reply #6181 on: October 09, 2014, 11:23:02 pm »
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How do we find the time between -15<x<15 ?

find the t-value of points A and B, by letting x=15 and then -15 and solving for t in both cases, just finding the first solutions to the trig equation. The difference in these values will be the time, then it's the same time between C and D so you can double it to get the total time for the part of the graph sketched.

I'm assuming you already have the rule for the graph, if not you can figure it out by using a cos graph (max at t=0) with amplitude 30 and period 16 (from looking at the graph), and you should be able to figure out a rule from that information :D
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LiquidPaperz

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Re: VCE Methods Question Thread!
« Reply #6182 on: October 10, 2014, 12:53:09 am »
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Oh, wait, my bad.

Yeah, I don't see how their working actually makes sense. I'll look at it again when I'm not delirious on chemicals.

have you overcome this sir? haha :)

knightrider

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Re: VCE Methods Question Thread!
« Reply #6183 on: October 10, 2014, 04:23:18 pm »
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How would you do this question?

show an open cardboard box of dimensions x cm(width),x cm(length), y cm(height) and the lid  for the box of dimensions x cm , x cm and 2cm (height). The total amount of cardboard used for the box and the lid is 1400 cm ^2
Let V cm^3 be the volume of the box.


A) find y in terms of x
B)Find v in terms of x

helpmeplz

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Re: VCE Methods Question Thread!
« Reply #6184 on: October 10, 2014, 05:03:06 pm »
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I'm struggling with integration questions  :'( :'(

Given that ∫h(x) dx = 5, evaluate ∫2h(x)dx (upper limit of 4, lower limit of 0) and ∫(h(x) + 3)dx (also with an upper limit of 4 and lower of 0)

Thanks  :)

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #6185 on: October 10, 2014, 05:05:49 pm »
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have you overcome this sir? haha :)
Alright, sorry for the delay.

From a transition matrix, you can figure out two things:
1. How many things change from x to y,
2. The total change of things from x to y (or net change)

So, using the proportions from the transition matrix (as we did), we know that 200 people will change their vote.

However, if you calculate the actual second state matrix, you'll find there's actually only a net change of 180 people. That's due to the people that move into a party that has a lower amount of people than the initial state, or people moving out of a party that has a higher amount.

This is basically a matter of funny wording, and if you're asked to find a net change as they wanted, you're better off doing a full matrix calculation.

kinslayer

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Re: VCE Methods Question Thread!
« Reply #6186 on: October 10, 2014, 05:13:24 pm »
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I'm struggling with integration questions  :'( :'(

Given that ∫h(x) dx = 5, evaluate ∫2h(x)dx (upper limit of 4, lower limit of 0) and ∫(h(x) + 3)dx (also with an upper limit of 4 and lower of 0)

Thanks  :)

∫2h(x)dx = 2∫h(x)dx = 10

∫(h(x) + 3)dx = ∫h(x)dx + 3∫dx = ∫h(x)dx + 12 = 17

(assuming all integrals are definite from 0 to 4)

helpmeplz

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Re: VCE Methods Question Thread!
« Reply #6187 on: October 10, 2014, 05:19:26 pm »
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∫2h(x)dx = 2∫h(x)dx = 10

∫(h(x) + 3)dx = ∫h(x)dx + 3∫dx = ∫h(x)dx + 12 = 17

(assuming all integrals are definite from 0 to 4)

Can you please show how you got this? My brain is frizzled and I can't understand it at all............

kinslayer

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Re: VCE Methods Question Thread!
« Reply #6188 on: October 10, 2014, 05:33:39 pm »
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Can you please show how you got this? My brain is frizzled and I can't understand it at all............

We know that



Using integral rules,



and



http://www.mathsisfun.com/calculus/integration-rules.html
« Last Edit: October 10, 2014, 07:52:16 pm by kinslayer »

helpmeplz

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Re: VCE Methods Question Thread!
« Reply #6189 on: October 10, 2014, 05:47:06 pm »
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Thank you so so so much!!!!  :D

knightrider

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Re: VCE Methods Question Thread!
« Reply #6190 on: October 10, 2014, 07:08:55 pm »
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A piece of wire 60cm long is cut into 2 pieces which are bent in order to create a square and a circle. If the square has side length 'x' cm

c)Not using calculus, find the length of the two pieces of wire in order to minimize the sum of the two areas.

For this questions what does it mean by not using calculus

lzxnl

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Re: VCE Methods Question Thread!
« Reply #6191 on: October 10, 2014, 07:49:13 pm »
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It means no derivatives to find the minimum

You should get a quadratic function that you can rewrite in turning point form. Then you can read off the min value
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knightrider

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Re: VCE Methods Question Thread!
« Reply #6192 on: October 10, 2014, 07:58:29 pm »
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It means no derivatives to find the minimum

You should get a quadratic function that you can rewrite in turning point form. Then you can read off the min value

Thx Lzxnl

You can also graph the function and find the minimum right
And also when i do this i will get a x value how will i use this to  find the length of the two pieces of wire in order to minimize the sum of the two areas.

psyxwar

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Re: VCE Methods Question Thread!
« Reply #6193 on: October 10, 2014, 08:42:54 pm »
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Thx Lzxnl

You can also graph the function and find the minimum right
And also when i do this i will get a x value how will i use this to  find the length of the two pieces of wire in order to minimize the sum of the two areas.
well I mean you could, but if it's tech-free that seems to be defeating the point of the question.

The square is 4x (from the question), meaning the circle is 60-4x
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knightrider

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Re: VCE Methods Question Thread!
« Reply #6194 on: October 10, 2014, 09:27:37 pm »
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well I mean you could, but if it's tech-free that seems to be defeating the point of the question.

The square is 4x (from the question), meaning the circle is 60-4x

Thanks
What would would be the implied domain in terms of these questions.
Like what values of x would not be permitted into these equations