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January 22, 2022, 06:54:42 am

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#### disney_princess

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##### Re: VCE Methods Question Thread!
« Reply #19275 on: November 21, 2021, 05:08:46 pm »
0
Hi! I'm new to this platform so if my post violates any platform rules pls let me know!

I was just hoping that if anyone has past commercial exams (Heffernan, Neap, MAV, Kilbaha, insight, tssm), including solutions, for methods and specialist maths and would be happy to share them with me? I know most school provide these resources to student, but my school is a (not so great) far north public school and teachers don't provide any of these extra resources.

Would be super super grateful for any of those commercial methods and spesh exams. thank you so much!

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19276 on: November 21, 2021, 05:44:45 pm »
+1
Hi! I'm new to this platform so if my post violates any platform rules pls let me know!

I was just hoping that if anyone has past commercial exams (Heffernan, Neap, MAV, Kilbaha, insight, tssm), including solutions, for methods and specialist maths and would be happy to share them with me? I know most school provide these resources to student, but my school is a (not so great) far north public school and teachers don't provide any of these extra resources.

Would be super super grateful for any of those commercial methods and spesh exams. thank you so much!

Hi disney_princess,

Although I am not a moderator, the rules of the atarnotes forum (which you can read here) state that:
ATAR Notes has zero tolerance for distribution of illegal materials, or materials which infringe on copyright, including textbooks, company exams, and similar. Equally, we have zero tolerance for requests of these materials. Any threads offering or requesting distribution of these materials will be locked/removed, and offenders will be receive warnings/bans as required. This includes linking to other sites where the material can be accessed, and definitely includes exchanges via personal message.
So although I cannot send you commercial exams, I can direct you to a few good resources that do not violate any rule:
- ATARNotes free methods exams
- TWM Publications free methods trial exams

I will edit this post if I come across any free practice exams that do not infringe on copyright.

#### trying.to.get.a.good.atar

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##### Re: VCE Methods Question Thread!
« Reply #19277 on: December 09, 2021, 02:52:14 pm »
0
Does anyone know how to solve this using simultaneous equations?

Two children had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an equal number. How many marbles did each child start with and how many did they finish with?

Thanks

#### Billuminati

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##### Re: VCE Methods Question Thread!
« Reply #19278 on: December 09, 2021, 05:33:29 pm »
+2
Does anyone know how to solve this using simultaneous equations?

Two children had 110 marbles between them. After one child had lost half her marbles and the other had lost 20 they had an equal number. How many marbles did each child start with and how many did they finish with?

Thanks

a + b = 110 (1)
a/2 = b-20 (2), rearrange this to a = 2b - 40
Sub a =  2b-40 into (1)
2b-40+b = 110

I think from here on you should be confident enough to solve for b, then sub the b value into (1) to find a. Note that these are the initial number of marbles they have. You just need to half for a and subtract 20 for b to find the final number of marbles
VCE 2016-2018

2017: Biology [38], Further Maths [44]

2018: Methods [37], French [38], Chem [40], English [44]

UMAT: 56/43/80, 57th percentile (LLLLOOOOOOOOOLLLLLLLL)

ATAR: 98.1

2019- : Bachelor of Biomedical Science at Monash (Scholars), minoring in Chemistry

GAMSAT September 2021: 65/67/86, 76 overall (98th percentile)

#### trying.to.get.a.good.atar

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##### Re: VCE Methods Question Thread!
« Reply #19279 on: December 11, 2021, 04:30:47 pm »
0
does anyone know how to solve this using simultaneous equations?

Linda thinks of a two-digit number. The sum of the digits is 8. If she reverses the digits, the new number is 36 greater than her original number. What was Linda’s original number?

i know that the answer will be 26 but i can't figure out how to find this answer using simultaneous equations

#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #19280 on: December 11, 2021, 05:17:58 pm »
+3
does anyone know how to solve this using simultaneous equations?

Linda thinks of a two-digit number. The sum of the digits is 8. If she reverses the digits, the new number is 36 greater than her original number. What was Linda’s original number?

i know that the answer will be 26 but i can't figure out how to find this answer using simultaneous equations

Linda thinks of a two-digit number. Let's say that the two-digit number is XY (to make it simple we'll call it z) which can be represented as Z =10X + Y. Where X represents the tens column and Y represents the units column.
The sume of the digits is 8
therefore, X + Y = 8
reversing the digits the new number is 36 greater than the original number therefore
reversing the digits can be represented as 10Y + X
(10Y + X) - (10X + Y) = 36

So our two equations are

1) X+ Y = 8
2) (10Y + X) - (10X + Y) = 36

2) can be further simplified down to 9Y - 9X = 36 and then Y - X = 4
so now 2) is Y - X = 4

1) - 2)

X + Y - (Y - X) = 8 - 4
2x  = 4
X = 2

now sub X = 2 back into X + Y = 8
Y = 6

Therefore our number is 26

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #19281 on: December 11, 2021, 05:21:45 pm »
+3
does anyone know how to solve this using simultaneous equations?

Linda thinks of a two-digit number. The sum of the digits is 8. If she reverses the digits, the new number is 36 greater than her original number. What was Linda’s original number?

i know that the answer will be 26 but i can't figure out how to find this answer using simultaneous equations

Lets assume that our 2 digit number will be in the form of $AB$

So we know the sum of the digits is 8, so $A+B=8$, and we also know the value of this particular 2 digit number will be $10A+B$ where $A$ is tens digit and $B$ is units digit.

When we reverse these digits the value of that number will be $10B+A$, as we have switched the positions of the tens and units digit.

To determine the difference expression of the original and reverse we will simply do $\left(10A+B\right)-\left(10B+A\right)\Rightarrow 9A-9B$

So in this question we have formed two simultaneous equations, that is...
$A+B=8$, and $9A-9B=36$

Hope this helps, if you have trouble solving these simultaneous equations don't hesitate to ask for help!

EDIT: Beaten by Sine by literally seconds lmao
« Last Edit: December 11, 2021, 05:23:47 pm by 1729 »

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19282 on: December 18, 2021, 12:53:52 am »
0
Hiii everyone!

Is there a rule to find the asymptotes of a graph (vertical and horizontal)?
Like, more complex ones:
2- 1/(2x-3)
Or some other hard ones?

Thank you so much!

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19283 on: December 18, 2021, 10:33:15 am »
+3
The fractional part ie. 1/(2x-3) will never be non-zero, so there is a horizontal asymptote at y=2. There isn't really a well-defined rule as I see it for horizontal asymptotes.

Usually, the "rule" for vertical asymptotes is just the values in the domain where the function is not defined (usually where a denominator of some sort is equal to zero). Here, it's just where 2x-3 = 0 ie. x=1.5
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HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
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#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19284 on: December 27, 2021, 12:53:15 pm »
0
Thanks that makes so much more sense!
I actually have this other question that I can't seem to understand.
Explain why the equation a(x − h)^2 + k = 0 has two solutions if and only if either a > 0 and k < 0 or if a < 0 and k > 0.

The solution is as follows:
2 solns if − k/a > 0
This occurs if k/a < 0
∴ a > 0 and k < 0, or a < 0 and k > 0

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #19285 on: December 27, 2021, 03:39:00 pm »
+5
You can have a look at the determinant, which is $4h^2a^2-4a(ah^2 + k) = -4ak$ (recall that $\Delta = b^2 - 4ac$ for a parabola of the form $ax^2+bx+c$).

For a parabola to have two solutions, this must be greater than zero ie. -4ak > 0, and the same solution follows.

Alternatively, observe that the parabola in the question has the same solutions as the parabola $(x-h)^2 + k/a$ -- if k/a is greater than zero, then there are no solutions since the leading coefficient is positive (it's just 1) and the vertex at x=h is above the x-axis. If k/a is zero, there is a single solution at x=h. This leaves k/a being less than zero, and the same solution follows.
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Subject Acceleration (2018)
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#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19286 on: December 28, 2021, 01:06:58 pm »
+1
Alternatively, observe that the parabola in the question has the same solutions as the parabola $(x-h)^2 + k/a$ -- if k/a is greater than zero, then there are no solutions since the leading coefficient is positive (it's just 1) and the vertex at x=h is above the x-axis. If k/a is zero, there is a single solution at x=h. This leaves k/a being less than zero, and the same solution follows.

I understand how you found the discriminant to be -4ak but I didn't understand the rest of the part.

#### Billuminati

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##### Re: VCE Methods Question Thread!
« Reply #19287 on: December 28, 2021, 01:31:05 pm »
+3
I understand how you found the discriminant to be -4ak but I didn't understand the rest of the part.

It might help to link this to transformations. k/a is the number of units the parabola is translated in the positive direction of the y-axis. When k/a = 0, the graph just touches the x-axis at its turning point ie (h,0). If you move that downwards ie when k/a is negative, the parabola will have 2 x-intercepts, but if you move it up ie when k/a is positive, the parabola won't have an x-intercept at all (no solutions)
VCE 2016-2018

2017: Biology [38], Further Maths [44]

2018: Methods [37], French [38], Chem [40], English [44]

UMAT: 56/43/80, 57th percentile (LLLLOOOOOOOOOLLLLLLLL)

ATAR: 98.1

2019- : Bachelor of Biomedical Science at Monash (Scholars), minoring in Chemistry

GAMSAT September 2021: 65/67/86, 76 overall (98th percentile)

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19288 on: December 28, 2021, 01:52:43 pm »
0
It might help to link this to transformations. k/a is the number of units the parabola is translated in the positive direction of the y-axis. When k/a = 0, the graph just touches the x-axis at its turning point ie (h,0). If you move that downwards ie when k/a is negative, the parabola will have 2 x-intercepts, but if you move it up ie when k/a is positive, the parabola won't have an x-intercept at all (no solutions)
YESS that is so helpful!!!! Thank you!

#### PizzaMaster

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##### Re: VCE Methods Question Thread!
« Reply #19289 on: December 30, 2021, 01:04:58 pm »
0
Hiii!
For the domain and range of hyperbolas, is it just that we basically eliminate the asymptotes and that's it?
I mean if there is a graph y = 2/(x+2)+ 3
The asymptotes are:
x= -2 and y=3
So the domain and range will be:
domain = R\{−2}
range = R\{3}
Or is there another way to find the domain and range?