VCE Methods Question Thread!

[keltingmeith]

cheers i think c=4 based on your working

I make no claim on my ability to add numbers. ;P

[keltingmeith]

How would you do this question?

A rectangular area, 14m by 16m is to be divided into 4 equal garden beds.Between each section there is a x-metre wide path.

find a formula for the total area of the path?

Thanks 

There's a few ways to do this, the easiest would be to just consider the path and ignore the garden beds.

So, we have two rectangles, but they intersect at a point. For now, we'll pretend that the two rectangles are separate. This means that the area of the first is 16x, and the are of the second is 14x, giving us a total area of 30x. Now, remember that centre bit we ignored? Well, that's a square with side length x. Now, when we've added the two rectangles, we added that square twice, so now we'll subtract it once (kind of like in probability when you find the union of two sets). This means that the total area is given by the function:



However, you should also note that x cannot be bigger than either side, so an appropriate domain would be 0<x<14

[Springyboy]

You're right, its c=4, but regardless the rest is right!
Just a question though, why are you doing that question now, are you indulging in a headstart for next year or ?

My 1/2 exam is on tuesday, getting in some revision before that

[knightrider]

There's a few ways to do this, the easiest would be to just consider the path and ignore the garden beds.

So, we have two rectangles, but they intersect at a point. For now, we'll pretend that the two rectangles are separate. This means that the area of the first is 16x, and the are of the second is 14x, giving us a total area of 30x. Now, remember that centre bit we ignored? Well, that's a square with side length x. Now, when we've added the two rectangles, we added that square twice, so now we'll subtract it once (kind of like in probability when you find the union of two sets). This means that the total area is given by the function:






However, you should also note that x cannot be bigger than either side, so an appropriate domain would be 0<x<14

Thanks eulerfan 
if the rectangles are separate how is the area of the first 16x that's only one side of the separate rectangles i dont understand.

[keltingmeith]

Thanks eulerfan 

is this the formula for the total area of the path right?

Yes - this is the total area of the path.

[cosine]

My 1/2 exam is on tuesday, getting in some revision before that

Oh alright, good luck

[knightrider]

Yes - this is the total area of the path.

Thanks eulerfan 
if the rectangles are separate how is the area of the first 16x that's only one side of the separate rectangles i dont understand.

[knightrider]

i came across a question relating to trig that asked me to move a graph as quoted "now move it 60 degrees in a clockwise direction ".

How do you move trig graphs.

[keltingmeith]

i came across a question relating to trig that asked me to move a graph as quoted "now move it 60 degrees in a clockwise direction ".

How do you move trig graphs.

.... I mean, it's certainly possible to do this to a graph.

But like, it's entirely beyond the scope of methods.

(the answer to your other question is coming)

[Zues]

would we ever be asked to  "discuss transformations from its original" ? e.g. x^2 - > 2x^2 .. is this a exam question?

[cosine]

would we ever be asked to  "discuss transformations from its original" ? e.g. x^2 - > 2x^2 .. is this a exam question?

I have seen a few on practice exams, so possibly yes. Why though?

[Zues]

cause my book ask those questions.

Just a question, Do you want people from your school to do tuition with you? i know this may seem selfish but i could ask few others if they want tuition, but with this competition ranking thing im not too sure?

[Orb]

cause my book ask those questions.

Just a question, Do you want people from your school to do tuition with you? i know this may seem selfish but i could ask few others if they want tuition, but with this competition ranking thing im not too sure?

If you're only concerned about your ATAR, no.

But I would definitely suggest and recommend helping your school mates out. This is because friendships last far longer than just VCE. Sure, your 'friend' might be the only who squeezes you out of a few ranks and, at most, cost you a 1-2 study score. However, I find that your study score isn't the be all and end all, you'll enjoy school much more if you helped your friends out and developed longer lasting relationships because it also enhances your everyday life at school as well.

From the academics perspective, additionally after the SACs are done and dusted, your SAC score will be boosted if your cohort does well in general, so from that area you should be encouraging your friends to get tutoring and benefit itself. In the end, a lowered SAC ranking may not even cost you a study score and may benefit your study score instead!

Tl;dr get your friends to do tuition with you. The benefits far outweigh the harms

[IndefatigableLover]

If you're only concerned about your ATAR, no.

But I would definitely suggest and recommend helping your school mates out. This is because friendships last far longer than just VCE. Sure, your 'friend' might be the only who squeezes you out of a few ranks and, at most, cost you a 1-2 study score. However, I find that your study score isn't the be all and end all, you'll enjoy school much more if you helped your friends out and developed longer lasting relationships because it also enhances your everyday life at school as well.

From the academics perspective, additionally after the SACs are done and dusted, your SAC score will be boosted if your cohort does well in general, so from that area you should be encouraging your friends to get tutoring and benefit itself. In the end, a lowered SAC ranking may not even cost you a study score and may benefit your study score instead!

Tl;dr get your friends to do tuition with you. The benefits far outweigh the harms

^Definitely agree with this! If you want your SACs to scale up well then your cohort has to perform fairly well on the end of year exam so if doing tuition helps with them getting higher marks then that's both great for them and for you too.

As for doing tuition with friends at school on a social level, do keep in mind that it is primarily for studying rather than for having fun so doing tuition with friends can be great but know that the main goal is indeed to solidify your understandings and all

[cosine]

Regardless of scaling and all that jazz, you should help out your mates if they need tuition or are struggling with an area that you can assist in. You should do this from the heart, not for wanting your school to do good so your SACs get scaled lol, do good and good will come back to you. Just imagine you needed last minute help before the exam, and you ask someone but they refuse to help, because they want to do better than you... Not only would everyone dislike that competitive person, but no one helped you in need. Be humble, help others, and most importantly dont allow the competitive spirit get to you!