VCE Methods Question Thread!

[hobbitle]

...width of w dots.

a.)  which rectangular number has a width of 6?

I've pointed out the two key pieces of information in the question, if it helps.
The width is represented by 'w'.
You are given a width - 6.
w = width = 6.

Make sure you understand that the variable (represented by x, or w, or y, or whatever) ACTUALLY REPRESENTS SOMETHING.  It's not simply a letter.  Think of it as actually a numerical value of some kind.  It might represent the width of a rectangle, or the height of a person in a rocketship, or the speed of a ball through the air. 

I'm just trying to help by figuring out what made you put the 6 'in front of' the w instead of replacing the w with 6....

[Only Cheating Yourself]

I've pointed out the two key pieces of information in the question, if it helps.
The width is represented by 'w'.
You are given a width - 6.
w = width = 6.

Make sure you understand that the variable (represented by x, or w, or y, or whatever) ACTUALLY REPRESENTS SOMETHING.  It's not simply a letter.  Think of it as actually a numerical value of some kind.  It might represent the width of a rectangle, or the height of a person in a rocketship, or the speed of a ball through the air. 

I'm just trying to help by figuring out what made you put the 6 'in front of' the w instead of replacing the w with 6....

I don't know how to break an equation, so because i've learnt some formulas etc i think it has to relate to that etc…

[hobbitle]

break an equation

I don't know what you mean by this.

You started a thread about thinking like a mathematician.
Yes, sometimes it helps to know the chapter you are studying and have it guide you to find your answer.
But if you want to think like a mathematician, just deal with the problem at hand.
The last couple of questions you have asked, you have overcomplicated your working because you are trying to predict what the textbook is asking you.
Real life problems don't come with a textbook.
I'm not being rude - you are trying very hard, it's clear - I'm actually trying to make your life easier.
Have an awareness of the chapter you are studying, but tackle each new question with an open mind.  Forget the last 10 questions you did.  This one might be different.
That way when you get thrown curveballs in exams, you'll be able to deal with them.
Sort the information, write it down, draw diagrams.  Stop dwelling on what chapter in the textbook you are in.  Just treat each question as an individual.  You'll be good.

[M_BONG]

area = - int^(pi/4)_(0) (sinx - cosx) dx + int^(pi)_(pi/4) (sinx - cosx) dx
= - [-cosx - sinx]^(pi/4)_(0) + [-cosx - sinx]^(pi)_(pi/4)
= - (-sqrt(2) + 1) + (1 + sqrt(2))
= sqrt(2) - 1 + 1 + sqrt(2)
= 2sqrt(2) units^2

Thanks heaps! Found my mistake.

Also, when I work answers out, my calculator seems to spit out answers with "In" in it.
Eg. In(4) which can be approximated to an actual value.
What does In actually mean? Is it inverse of? Is it part of the Methods course; if so, what topic is it covered in?

[Conic]

Thanks heaps! Found my mistake.

Also, when I work answers out, my calculator seems to spit out answers with "In" in it.
Eg. In(4) which can be approximated to an actual value.
What does In actually mean? Is it inverse of? Is it part of the Methods course; if so, what topic is it covered in?

ln is the natural logarithm (base e), so .

[M_BONG]

Can someone help me with spotting my misconception/mistake here?






divide both sides by log e



Actual answer: k = In (4)


EDIT: Made working clearer

[clıppy]

Can someone help me with spotting my misconception/mistake here?








Actual answer: k = In (4)

Just remember that

[clıppy]

Can someone help me with spotting my misconception/mistake here?








Actual answer: k = In (4)

I think you tried dividing the loge(e) to the other side [in the 2nd step] and thought the loge would cancel out. Dividing loge(e) wouldn't be the problem if it was simplified down to 1 but you forgot about the log law and mistakenly left it as e.

[Only Cheating Yourself]

The perimeter of a rectangle is 20cm and its area is 24cm^2.  Calculate the length and width of the rectangle

I have no clue what to do, this is really getting frustrating!  I actually don't know where to start!!

[Nato]

Can someone help me with spotting my misconception/mistake here?






divide both sides by log e



Actual answer: k = In (4)


EDIT: Made working clearer

so from
we get
and now we can divide both sides by -1
this results in
now, following one of the log laws, we can 'move' the -1 on top and it will act as an an exponent
you may have seen this is textbooks as
we get,
now this will simplify to 
and since
that means

[Orb]

The perimeter of a rectangle is 20cm and its area is 24cm^2.  Calculate the length and width of the rectangle

I have no clue what to do, this is really getting frustrating!  I actually don't know where to start!!

Okay so let the length of the rectangle be A, let the width be B (not an irregular rectangle )

Therefore, 2A+2B = 20
A+B = 10

Area is AxB = 24

straight from A+B = 10 it means that both numbers (as they are positive) are 1 digit.

Easily, you can tell that A is 6 (length) and B is 4 (width).

I'll try to find a mathematical way but logically it seems pretty simple to derive.

[hobbitle]

The perimeter of a rectangle is 20cm and its area is 24cm^2.  Calculate the length and width of the rectangle

Hamo above is correct but if they want you to find it mathematically..... hm, have you done simultaneous equations/substitution yet?  If not, ignore the following....

First, you acknowledge that both the perimeter of a rectangle and the area of a rectangle are both functions of length (L) and width (W).
So, let L be the length of the rectangle, and W be the width.  You get two functions:

1)   L x W = 24 (using the information provided about the area)
2)   2L + 2W = 20 ---> divide by 2 ---> L + W = 10 (using the information provided about the perimeter)

If you take 2), and rearrange it, it becomes L = 10 - W

Then you take this result and 'plug it in' to equation 1).

W(10 - W) = 24

expand

10W - W^2 = 24

rearrange to get a quadratic

W^2 - 10W + 24 = 0

which factorises to, by inspection

(W - 6)(W - 4) = 0

therefore W = 6 or W = 4.

If W = 6, and we know that L + W = 10, then we know that L must be 4.
If W = 4, and we know that L + W = 10, then we know that L must be 6.
Length and width are essentially interchangeable so you know that the length is 6cm and the width is 4cm.

[Orb]

Hamo above is correct but if they want you to find it mathematically..... hm, have you done simultaneous equations/substitution yet?  If not, ignore the following....

First, you acknowledge that both the perimeter of a rectangle and the area of a rectangle are both functions of length (L) and width (W).
So, let L be the length of the rectangle, and W be the width.  You get two functions:

1)   L x W = 24 (using the information provided about the area)
2)   2L + 2W = 20 ---> divide by 2 ---> L + W = 10 (using the information provided about the perimeter)

If you take 2), and rearrange it, it becomes L = 10 - W

Then you take this result and 'plug it in' to equation 1).

W(10 - W) = 24

expand

10W - W^2 = 24

rearrange to get a quadratic

W^2 - 10W + 24 = 0

which factorises to, by inspection

(W - 6)(W - 4) = 0

therefore W = 6 or W = 4.

If W = 6, and we know that L + W = 10, then we know that L must be 4.
If W = 4, and we know that L + W = 10, then we know that L must be 6.
Length and width are essentially interchangeable so you know that the length is 6cm and the width is 4cm.

Yeah, nice thinking.

It never crossed my mind, I think the answer just came straight away haha

[hobbitle]

Yeah, nice thinking.

It never crossed my mind, I think the answer just came straight away haha

Yes of course it can be done without working... but I assume they wanted you to extract the quadratic!

[Only Cheating Yourself]

Thanks guys, but I'm still really confused on how to structure a worded response! Where does the 2 come from?  why is it 2A+2B=20?

And yes i have done simultaneous equations.