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April 03, 2026, 12:37:39 pm

Author Topic: VCE Methods Question Thread!  (Read 5998222 times)  Share 

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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #5430 on: July 21, 2014, 08:37:18 pm »
0
so
S1 = T^1So is first trip
S2 = T^2So is second trip
S3 = T^3So is third trip
S4 = T^4So is fourth trip

i dont get how S4= t^3So is fourth trip? am i missing something

Not quite - the INITIAL trip (S0) is the FIRST trip. So,

First trip: S0
Second trip: S1
Third Trip: S2
Fourth Trip: S3

Therefore, we raise it to the third power.

With these kinds of questions, it's pretty easy to be caught up in the the whole "fourth week, so I want S4". What you need to do, is figure out what was the INITIAL trip, and that's S0. From there, you count up to which time you're looking for, and that's the value of n you use - NOT whatever number they write in the question (unless they're the same).

LiquidPaperz

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Re: VCE Methods Question Thread!
« Reply #5431 on: July 21, 2014, 09:39:08 pm »
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so do i take it, if it says initial i treat it as the first week , day etc.

because i rememer doing questions on this and (cant find it now) few questions had initial and i was oblivious to it and still got same answer. Not sure if i remember it right though

knightrider

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Re: VCE Methods Question Thread!
« Reply #5432 on: July 21, 2014, 10:39:09 pm »
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How would you do this question

If 0° < a < 360°, then the equation sin (a) = cos (a)

How would you find how many solutions it has and also

If 0° ≤ a ≤ 180°, and sin (a) = cos (a), then a is equal to

Frozone

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Re: VCE Methods Question Thread!
« Reply #5433 on: July 21, 2014, 10:55:43 pm »
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could someone please show the working out for this:
 
calculate the exact area enclosed between the graph y= 2sin(2x), x=0, x=
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soNasty

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Re: VCE Methods Question Thread!
« Reply #5434 on: July 21, 2014, 10:59:12 pm »
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How would you do this question

If 0° < a < 360°, then the equation sin (a) = cos (a)

How would you find how many solutions it has and also

If 0° ≤ a ≤ 180°, and sin (a) = cos (a), then a is equal to

1. sin(a)=cos(a) becomes tan(a)=1 when u divide both sides by cos(a)
since the domain is [0,2pi] in radians, tan(a)=1 has two solutions. these are at a=pi/4 and 5pi/4

2. the other one has a domain of [0,pi] in radians, it is also tan(a)=1, the only solution that falls within this range is a=pi/4

soNasty

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Re: VCE Methods Question Thread!
« Reply #5435 on: July 21, 2014, 11:09:38 pm »
+1
could someone please show the working out for this:
 
calculate the exact area enclosed between the graph y= 2sin(2x), x=0, x=







   






« Last Edit: July 21, 2014, 11:14:05 pm by soNasty »

~V

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Re: VCE Methods Question Thread!
« Reply #5436 on: July 21, 2014, 11:23:29 pm »
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How do i find the value of "h" in linear approximation. Eg. 4.02^2, how is h=0.02?
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Frozone

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Re: VCE Methods Question Thread!
« Reply #5437 on: July 21, 2014, 11:37:42 pm »
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Thats exactly what i thought! But the answer says 4!
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soNasty

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Re: VCE Methods Question Thread!
« Reply #5438 on: July 21, 2014, 11:51:43 pm »
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How do i find the value of "h" in linear approximation. Eg. 4.02^2, how is h=0.02?

The function is f(x)=x^2
The closest  number to  4.02 is 4, the resultant difference is 0.02, which is the value of h

soNasty

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Re: VCE Methods Question Thread!
« Reply #5439 on: July 21, 2014, 11:52:23 pm »
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Thats exactly what i thought! But the answer says 4!

I put it on the calculator and got 0 as the answer! Maybe there's something missing

Phy124

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Re: VCE Methods Question Thread!
« Reply #5440 on: July 22, 2014, 12:00:01 am »
+3
could someone please show the working out for this:
 
calculate the exact area enclosed between the graph y= 2sin(2x), x=0, x=







   






It asks for the area enclosed so you have to split up the sections that are above and below the axis i.e.

Area enclosed above:



Area enclosed below:



Total Area = 2 + 2 = 4
« Last Edit: July 22, 2014, 12:01:41 am by Phy124 »
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~V

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Re: VCE Methods Question Thread!
« Reply #5441 on: July 22, 2014, 12:01:56 am »
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How does h=2 for ?
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LiquidPaperz

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Re: VCE Methods Question Thread!
« Reply #5442 on: July 22, 2014, 06:12:18 pm »
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why is the matrix written in a different order to the question? why cant you straight copy whats in the table into the matrix?

#J.Procrastinator

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Re: VCE Methods Question Thread!
« Reply #5443 on: July 22, 2014, 06:18:05 pm »
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How would I go about doing this question?

Consider the functions f: [0, infinity)  → R, f(x) = x-c and g: [-2, infinity)  → R, g(x)=x^2 + c, where c is a real number. How do we find all the possible values for c for which f composition g and g composition f exist?

Quick question, when we're given this for example,  f: [2, infinity)  → R , what does this mean exactly? And how would we read this?

Thanks! :D
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#J.Procrastinator

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Re: VCE Methods Question Thread!
« Reply #5444 on: July 22, 2014, 06:42:03 pm »
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why is the matrix written in a different order to the question? why cant you straight copy whats in the table into the matrix?

When you multiply matrices together, what you do is move across the top ROW of matrix A and down the first COLUMN of matrix B and multiply these numbers together and then add. If you just simply insert the numbers given into a matrix without arranging it, this will give you an incorrect total revenue.

The matrix provided in the solutions make sense as you will get :
(5 x 110) + (9 x 120) + (7 x 150) = $ 2680 (the total revenue that store A will get)

If you just write the values in the matrix as they appear, you will end up multiplying the amount of shirts at store A by the selling price of shirts, the amount of shirts at store B by the selling price of ties, and the amount of shirts at store C By the selling of shoes.

Hope this helps!
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