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July 29, 2021, 12:49:25 pm

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#### eloisegrace

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##### Re: VCE Methods Question Thread!
« Reply #18825 on: September 30, 2020, 05:39:53 pm »
0
Hello,

I have a question from VCAA Exam 2 2011 MCQ10, I am not certain if it is relevant to the current study design as I have never seen this question type but have seen it in multiple old VCAA exams.

Maria is a very enthusiastic young tennis player who plays a match every day. If she wins on one particular day, the probability that the wins the next day is 0.8. If she loses one day, the probability that she loses the next day is 0.6.
The long term probability that Maria wins a match is:

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2020 - mathematical methods [42] | further mathematics [45]
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#### ArtyDreams

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##### Re: VCE Methods Question Thread!
« Reply #18826 on: September 30, 2020, 06:13:36 pm »
+4
Hello,

I have a question from VCAA Exam 2 2011 MCQ10, I am not certain if it is relevant to the current study design as I have never seen this question type but have seen it in multiple old VCAA exams.

Maria is a very enthusiastic young tennis player who plays a match every day. If she wins on one particular day, the probability that the wins the next day is 0.8. If she loses one day, the probability that she loses the next day is 0.6.
The long term probability that Maria wins a match is:

Spoiler

Hi Eloise! Long term probability is not part of the new study design Pretty sure you work it out with matricies or something, but you don't need to worry about it

#### eloisegrace

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##### Re: VCE Methods Question Thread!
« Reply #18827 on: September 30, 2020, 07:42:50 pm »
+1
Hi Eloise! Long term probability is not part of the new study design Pretty sure you work it out with matricies or something, but you don't need to worry about it
Thank you ArtyDreams Nice to know that I haven't just forgotten probability haha
2020 - mathematical methods [42] | further mathematics [45]
2021 - english language | chemistry | french | physical education
my vce journey !

#### FlammaZ

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##### Re: VCE Methods Question Thread!
« Reply #18828 on: October 01, 2020, 06:23:16 pm »
0
Hello everyone, could someone clear up part C)

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18829 on: October 02, 2020, 12:03:25 am »
+3
Hello everyone, could someone clear up part C)

Cross-sectional area = the area if you cut into the shape. For example, the cross-sectional area of a cube will be a square, of a sphere will be a circle, of a cylinder will be a circle if you cut it short-ways, or a rectangle if you cut it long-ways. To go in reverse is easy for the cylinder if you cut it short-ways, you can find the volume by multiplying the circular cross-section by the height of the cylinder (in fact, that's what the formula of the volume of a cylinder is - V=pi*r^2*h)

With that in mind, how much papier mache can you make with a 0.5 m * 0.5 m * 30 cm box? And how much of that can you put into the volume of that ramp?

#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18830 on: October 03, 2020, 04:42:13 pm »
+7
Hey Larry, welcome to the Atar Notes forums!

Wow this question looks wayyyy beyond the scope of the Maths Methods course, but I will attempt to answer it.

For this question you will need to use the results from both parts a) and b), and do quite a bit of algebraic manipulation for it to simplify, as well as being familiar with the trigonometric formulae

The full working is attached in the image below.

Hope this helps!
James
VCE Class of 2021
2020: Maths Methods [49 ~ 50.15]
2021: English [?] Chemistry [?] Physics [?] Specialist Maths [?] UMEP Maths [?]

Maths Methods Tutoring

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18831 on: October 04, 2020, 04:10:53 pm »
0
Hey guys,
I don't understand why the book states (x-a) and (y-0) in the middle of this (https://gyazo.com/dc410472548eba8a407eb79dafac38bf) screenshot.
When I go through the algebraic process I get this (attached file)
Where is my mistake?
I get -b=-b in the end, which is true, but I'm clearly not getting what they're doing.
Many thanks,
Corey

#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18832 on: October 04, 2020, 05:51:01 pm »
+8
Hey Corey!

Great to see that you're not taking formulas in the book for granted, and deriving it on your own instead.

Your first step in finding the gradient is correct using the typical gradient formula. For the next step you have to apply the point gradient form.

Recall that if you know a point (x1, y1) and the gradient, you can write a linear equation in the y - y1 = m (x - x1) form. Now you can sub in the gradient and the point (a, 0)

y - y1 = m (x - x1)
y - 0 = - b/a (x - a)
y = - b/a (x - a)

Your mistake was trying to sub in both of your points, and in the process got rid of the x and y. I'll leave you to explore the rest of this formula. Let us know how you go and come back on this thread if you have any other questions!

Hope this helps,
James
VCE Class of 2021
2020: Maths Methods [49 ~ 50.15]
2021: English [?] Chemistry [?] Physics [?] Specialist Maths [?] UMEP Maths [?]

Maths Methods Tutoring

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18833 on: October 05, 2020, 10:52:35 am »
0
Hey Corey!

Great to see that you're not taking formulas in the book for granted, and deriving it on your own instead.

Your first step in finding the gradient is correct using the typical gradient formula. For the next step you have to apply the point gradient form.

Recall that if you know a point (x1, y1) and the gradient, you can write a linear equation in the y - y1 = m (x - x1) form. Now you can sub in the gradient and the point (a, 0)

y - y1 = m (x - x1)
y - 0 = - b/a (x - a)
y = - b/a (x - a)

Your mistake was trying to sub in both of your points, and in the process got rid of the x and y. I'll leave you to explore the rest of this formula. Let us know how you go and come back on this thread if you have any other questions!

Hope this helps,
James

Ah I see. I kept substituting either both x1 y2 or y1 and x2 instead for some reason.
I think the change of labelling mixed me up as well. Changing the y2 to just y and x2 to just x. I can see why they do that though
Thank you for the help James!
Corey

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18834 on: October 05, 2020, 11:07:59 am »
0
Hey guys,
I got indeterminate as an answer for the gradient of a straight line graph.
I'm just wondering what I put as the gradient in the point-gradient formula now? Do I just write zero, and treat it as such?
Many thanks,
Corey

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18835 on: October 05, 2020, 11:08:58 am »
0
Hey guys,
I got Undefined* as an answer for the gradient of a straight line graph.
I'm just wondering what I put as the gradient in the point-gradient formula now? Do I just write zero, and treat it as such?
Many thanks,
Corey

#### p0kem0n21

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##### Re: VCE Methods Question Thread!
« Reply #18836 on: October 05, 2020, 02:16:17 pm »
+3

If the gradient for a straight line is undefined, then that should mean that it is a vertical line of the formula x=a, where a is any real number (or basically the x-coordinate for your two points). An undefined gradient is not the same as when your gradient is equal to zero. If the gradient of the line was zero, then it would be a horizontal line instead of equation y=b (where b is the y-coordinate of your points).

#### svnflower

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##### Re: VCE Methods Question Thread!
« Reply #18837 on: October 05, 2020, 02:20:53 pm »
0
Hello,

How do we graph y=tan2x given a y=tanx graph?

I don't quite understand how to graph trig functions with power 2 e.g. cos2x and sin2x

#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #18838 on: October 05, 2020, 02:37:52 pm »
+8
Hey there!

General tips for graphing $(f(x))^2$ given $f(x)$:
- Start by drawing $|f(x)|$ (reflecting all negative parts of $f(x)$ around the x-axis, since $\sqrt{ (f(x))^2} = |f(x)|$
- $(f(x))^2$ will be below $|f(x)|$ where $|f(x)| < 1$, and $(f(x))^2$ will be above $|f(x)|$ where $|f(x)| > 1$
- $(f(x))^2$ will always be greater than 0 for all x in the domain of $|f(x)|$
- Turn all cusps formed by $|f(x)|$ into smooth turning points (like on a parabola)
- Keep all vertical asymptotes
- Square all horizontal asymptotes (move them accordingly)
- Indicate clearly the slope of $(f(x))^2$ - $\frac{d}{dx} (f(x))^2 = 2f'(x)f(x)$.

Can't think of anything else atm, but hope this is enough to get you started
« Last Edit: October 05, 2020, 02:40:00 pm by fun_jirachi »
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#### svnflower

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##### Re: VCE Methods Question Thread!
« Reply #18839 on: October 07, 2020, 09:15:43 am »
0
Hey there!

General tips for graphing $(f(x))^2$ given $f(x)$:
- Start by drawing $|f(x)|$ (reflecting all negative parts of $f(x)$ around the x-axis, since $\sqrt{ (f(x))^2} = |f(x)|$
- $(f(x))^2$ will be below $|f(x)|$ where $|f(x)| < 1$, and $(f(x))^2$ will be above $|f(x)|$ where $|f(x)| > 1$
- $(f(x))^2$ will always be greater than 0 for all x in the domain of $|f(x)|$
- Turn all cusps formed by $|f(x)|$ into smooth turning points (like on a parabola)
- Keep all vertical asymptotes
- Square all horizontal asymptotes (move them accordingly)
- Indicate clearly the slope of $(f(x))^2$ - $\frac{d}{dx} (f(x))^2 = 2f'(x)f(x)$.

Can't think of anything else atm, but hope this is enough to get you started

Thankyou so much, helps a lot!!

Also, what does it mean for a function to be defined? Is it the values where the curve is above the x-axis?