July 11, 2020, 11:20:51 am

### AuthorTopic: VCE General & Further Maths Question Thread!  (Read 386944 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### DrDusk

• NSW MVP - 2019
• Posts: 511
• I exist in the fourth dimension
• Respect: +127
##### Re: VCE Further Maths Question Thread!
« Reply #2490 on: September 28, 2019, 04:58:20 pm »
+1
I agree, but the answer is a linear graph...
It should be linear because your graphing y vs 1/x. Think of it this way. A linear graph at it's most basic form has an equation y = m(x). Graphing y vs x means as x increases, y increases in a constant proportion to it. Now consider the equation y = k/x. If we graph y vs x, as x increases y decreases giving a hyperbola. Now if we graph y vs 1/x, isn't this equation the same as graphing y vs x for y = mx. Think about it, y = k/x is of the form y = k(1/x), meaning that 'm' the gradient is k and the 'x' is replaced by 1/x. You could have anything in the brackets for example y = k(e^x). In that case graphing y vs e^x will be a linear graph, as y changes by a constant proportion when comparing it to e^x.

So in the end basically in the equation y = mx, you can 'replace' x with anything, but you must graph y vs whatever that is in the brackets.
HSC/Prelim Physics tutor

#### TheEagle

• Forum Obsessive
• Posts: 290
• Respect: +4
##### Re: VCE Further Maths Question Thread!
« Reply #2491 on: September 28, 2019, 05:29:14 pm »
0
It should be linear because your graphing y vs 1/x. Think of it this way. A linear graph at it's most basic form has an equation y = m(x). Graphing y vs x means as x increases, y increases in a constant proportion to it. Now consider the equation y = k/x. If we graph y vs x, as x increases y decreases giving a hyperbola. Now if we graph y vs 1/x, isn't this equation the same as graphing y vs x for y = mx. Think about it, y = k/x is of the form y = k(1/x), meaning that 'm' the gradient is k and the 'x' is replaced by 1/x. You could have anything in the brackets for example y = k(e^x). In that case graphing y vs e^x will be a linear graph, as y changes by a constant proportion when comparing it to e^x.

So in the end basically in the equation y = mx, you can 'replace' x with anything, but you must graph y vs whatever that is in the brackets.

oh I think I understand now. So would it be y=60(1/time) right? but when I do it I get: 60(1/1) =60 but when I do time=2 I get 60(1/2)=30 which is incorrect

sorry I'm bothering you people

#### DrDusk

• NSW MVP - 2019
• Posts: 511
• I exist in the fourth dimension
• Respect: +127
##### Re: VCE Further Maths Question Thread!
« Reply #2492 on: September 28, 2019, 05:42:48 pm »
+1

oh I think I understand now. So would it be y=60(1/time) right? but when I do it I get: 60(1/1) =60 but when I do time=2 I get 60(1/2)=30 which is incorrect

sorry I'm bothering you people
The data is not incorrect. Remember in essence your graphing y vs 1/x. So your points generated by those data values that you said will be (1/1, 60),(1/2, 30). If you take time = 3, you end up with the point (1/3, 20). Now just to check it's correct we can do m = (y2-y1)/(x2-x1), which gives us:

$m = \dfrac{20 - 60}{\dfrac{1}{3} - 1} = 60$

Well look at that, the gradient is just the value of 'k' which makes sense because the equation is y = k(1/time) and were graphing y vs 1/time which is a linear graph of gradient k.

sorry I'm bothering you people
Not at all, were here to help! =)
« Last Edit: September 28, 2019, 05:50:06 pm by DrDusk »
HSC/Prelim Physics tutor

#### AngelWings

• Victorian Moderator
• Part of the furniture
• Posts: 1968
• "Angel wings, please guide me..."
• Respect: +1000
##### Re: VCE Further Maths Question Thread!
« Reply #2493 on: September 28, 2019, 05:44:09 pm »
+1

So for the Q1c)i) I have gotten k=60 by doing the following: 30=k/2 (chose the point 2,30 and solved for k)
However, for part ii, the equation is avg speed= 1/time hence its no longer 60/time?
Yup. Transformations tend to change the graph, equation and coordinates quite drastically, so you’re on the money there.

oh I think I understand now. So would it be y=60(1/time) right? but when I do it I get: 60(1/1) =60 but when I do time=2 I get 60(1/2)=30 which is incorrect

sorry I'm bothering you people
Really close! You’re missing a divided sign somewhere...

If you’re truly stuck...
Try to substitute the transformation applied on time (see Q1c)ii.) where (time) is in the original equation (the one with k). Then simplify this.

P.S. We’re more than happy to help; you’re not bothering us at all. We’d much rather see you understand the question than leave with you still quizzical.

Note: got beaten. Leaving this here anyway.
VCE: Psychology | English Language | LOTE | Mathematical Methods (CAS) | Further Mathematics | Chemistry
Uni: (Hons)
University Virtual Events and Open Days

#### yourfriendlyneighbourhoodghost

• Forum Obsessive
• Posts: 204
• sleep now and dream, study now and live your dream
• Respect: +33
##### Re: VCE Further Maths Question Thread!
« Reply #2494 on: September 29, 2019, 03:51:54 pm »
0
Hi,

The only subject I am worried about is math. I have done so many practice exams and have a really detailed summary book. It's weird because I answer all the textbook questions confidently and correct, but then on most of the vcaa practices I get them wrong or I don't understand in what what's to apply my knowledge. This is only for the data and finicancial math part.

Does any one have any notes or tips on how to get better?
2018: Studio Arts [37]
2019: English [38] Psychology [38] Vis Com [36] Software Development [40] Further Maths [35]
ATAR: 87.95 ❤️

2020-2023 Bachelor of Arts @ Unimelb

#### AngelWings

• Victorian Moderator
• Part of the furniture
• Posts: 1968
• "Angel wings, please guide me..."
• Respect: +1000
##### Re: VCE Further Maths Question Thread!
« Reply #2495 on: September 29, 2019, 08:45:15 pm »
+1
Hi,

The only subject I am worried about is math. I have done so many practice exams and have a really detailed summary book. It's weird because I answer all the textbook questions confidently and correct, but then on most of the vcaa practices I get them wrong or I don't understand in what what's to apply my knowledge. This is only for the data and finicancial math part.

Does any one have any notes or tips on how to get better?
What’s the reason you’re getting the VCAA questions wrong? Is it the wording, identifying what type of question it is, dumb mistakes or something else entirely? Perhaps if you could look into this, you’d know how to fix this issue.

If it’s the wording throwing you off, take your time reading the question properly to ensure you know what the question is actually asking. If it’s identification, perhaps have a look at how questions are worded in your textbooks and connect them to those that are in the VCAA exams. There should be some similarities in key words, as the textbook will be aligned to the study design/ exam. If it’s dumb mistakes, reflect on what sort of mistakes you’re making and make a log of them to avoid doing the same mistakes again.

Hopefully that helps.
VCE: Psychology | English Language | LOTE | Mathematical Methods (CAS) | Further Mathematics | Chemistry
Uni: (Hons)
University Virtual Events and Open Days

#### DrDusk

• NSW MVP - 2019
• Posts: 511
• I exist in the fourth dimension
• Respect: +127
##### Re: VCE Further Maths Question Thread!
« Reply #2496 on: September 29, 2019, 09:13:31 pm »
+1
Hi,

The only subject I am worried about is math. I have done so many practice exams and have a really detailed summary book. It's weird because I answer all the textbook questions confidently and correct, but then on most of the vcaa practices I get them wrong or I don't understand in what what's to apply my knowledge. This is only for the data and finicancial math part.

Does any one have any notes or tips on how to get better?
That's normal. Doing just textbook questions wont help for the final exam. What you need to be doing is past papers as this is what will build your ability to problem solve. Most textbooks only have very niche questions so you can learn the content, whereas the exam questions require you to apply this knowledge in many different formats that just aren't usually in textbooks. Most of the time textbook questions are easier compared to the actual final. So far I only know of one textbook here in nsw that has questions as hard as the actual paper.

You should only need to do a few textbook ones, then go straight to doing past papers.

All the best with your studies =)
HSC/Prelim Physics tutor

#### Sherlock.Holmes

• Trailblazer
• Posts: 41
• Respect: 0
##### Re: VCE Further Maths Question Thread!
« Reply #2497 on: October 12, 2019, 10:53:39 pm »
+1
Hey y'all
I have finished all further maths SACs this year and am now rank 1. Yes lots of pressure.
I have started doing practice exams but im not really doing extremely well. I'm making a lot of silly mistakes and not having enough time to properly finish the exams. I think my main weakness is the financial module, i make too mistakes in that, not just silly mistakes but actually not understanding what to do. Im not sure what i should do now, i was aiming for a 45+ study score but i feel like i wont get it anymore... :/

#### yourfriendlyneighbourhoodghost

• Forum Obsessive
• Posts: 204
• sleep now and dream, study now and live your dream
• Respect: +33
##### Re: VCE Further Maths Question Thread!
« Reply #2498 on: October 15, 2019, 08:22:37 am »
0
Hi,
Thank you for taking time out to help students, I love ATAR notes so so much.

Anyway, how does this work? I don't understand how they got the answer with this explanation. The question was, what is the maximum flow from source to sink? and it didnt have any cuts on the original graph

Thank you (:
« Last Edit: October 16, 2019, 08:05:43 am by yourfriendlyneighbourhoodghost »
2018: Studio Arts [37]
2019: English [38] Psychology [38] Vis Com [36] Software Development [40] Further Maths [35]
ATAR: 87.95 ❤️

2020-2023 Bachelor of Arts @ Unimelb

#### Hedlosa

• Fresh Poster
• Posts: 1
• Respect: 0
##### Re: VCE Further Maths Question Thread!
« Reply #2499 on: October 15, 2019, 12:44:25 pm »
0
Hi,
Thank you for taking time out to help students, I love ATAR notes so so much.

Anyway, how does this work? I don't understand how they got the answer with this explanation.

Thank you (:
sorry but are you able to state what question this is for what exam? Just so I can get some context into what the question is asking.

#### yourfriendlyneighbourhoodghost

• Forum Obsessive
• Posts: 204
• sleep now and dream, study now and live your dream
• Respect: +33
##### Re: VCE Further Maths Question Thread!
« Reply #2500 on: October 16, 2019, 08:06:47 am »
0
sorry but are you able to state what question this is for what exam? Just so I can get some context into what the question is asking.
it was what is the maximum flow from the source to the sink?
2018: Studio Arts [37]
2019: English [38] Psychology [38] Vis Com [36] Software Development [40] Further Maths [35]
ATAR: 87.95 ❤️

2020-2023 Bachelor of Arts @ Unimelb

#### AISHAB

• Trailblazer
• Posts: 42
• Respect: +4
##### Re: VCE Further Maths Question Thread!
« Reply #2501 on: October 17, 2019, 02:14:40 pm »
0
Hi,

How do you find the equation of a line, from a graph?

Hefferman 2011 Q8

The graph does not begin at 0

#### MEH0010

• Posts: 17
• Class of 2020
• Respect: +2
##### Re: VCE Further Maths Question Thread!
« Reply #2502 on: October 21, 2019, 12:30:04 pm »
+2
Hi,

How do you find the equation of a line, from a graph?

Hefferman 2011 Q8

The graph does not begin at 0

Hey,
For these type of questions, i would choose two points from the regression line, in this case (75,68) and (85,60) and then put these points on your CAS or find the equation by hand, finding the gradient, then substituting two points into the equation to find the y-intercept.
Another approach is by inspection, as the horizontal axis does not start at 0.
Hope this helps
2019 (VCE): Biology [37] & Further Maths [43]
2020 (VCE): Indonesian (SL), Maths Methods, English & HHD

#### AISHAB

• Trailblazer
• Posts: 42
• Respect: +4
##### Re: VCE Further Maths Question Thread!
« Reply #2503 on: October 22, 2019, 12:55:57 pm »
0
Hey,
For these type of questions, i would choose two points from the regression line, in this case (75,68) and (85,60) and then put these points on your CAS or find the equation by hand, finding the gradient, then substituting two points into the equation to find the y-intercept.
Another approach is by inspection, as the horizontal axis does not start at 0.
Hope this helps

Thanks!

What about when you are given a scatterplot with its points and an equation (2017 FURTHER EXAM 2 - QUESTION 3B)
Egg density = 191 + 31.3 x number of female moths

How do you draw the graph of this least squares regression line?

#### TheEagle

• Forum Obsessive
• Posts: 290
• Respect: +4
##### Re: VCE Further Maths Question Thread!
« Reply #2504 on: October 22, 2019, 07:09:05 pm »
+1
Hi,

How do you find the equation of a line, from a graph?

Hefferman 2011 Q8

The graph does not begin at 0

Hey!

If you look carefully, the point where the line crosses the y axis is not really the y intercept because the corresponding x value isn't zero. Instead it is (75, 68) and if you choose another point, i.e: (85,60), then if you find the gradient by doing (y2-y1) / (x2-x1), which would give -0.8. So at the moment our equation is Y=-0.8x+c      Now let's substitute one of our points that we had chosen before to find c --> 68= -0.8 *75 + c --> c= 128       hence our equation is Y=-0.8x + 128 --> Closest is OPTION E!

Another way to do this question faster (only works on exam 1 due to giving us the options):
substitute a pair of points that are clear into all options and see which gives the correct option
i.e: (subbing point (75, 68))
Option A) 68 = 49- 1.09 * 75 --> 68 is not equal to -32.75
Option B) 68= 68+1.09 *75 --> 68 is not equal to 49.75
Option C) 68=68-0.76*75 --> 68 is not equal to11
Option D) 68= 68+0.76 *75 --> 68 is not equal to 125
Option E) 68= 125 - 0.76 *75 --> 68 is equal to 68 (CORRECT)