September 21, 2020, 01:27:58 pm

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#### Azila2004

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##### Re: VCE Methods Question Thread!
« Reply #18630 on: July 16, 2020, 02:21:14 pm »
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Hey is it possible you have misread the question? Because if you sub in 2 as z it would be -4 and 20

I just checked now and you're right :O
Yeah, I write my 2s very similarly to Zs, whoops!
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#### chemistrykind

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##### Re: VCE Methods Question Thread!
« Reply #18631 on: July 16, 2020, 02:32:31 pm »
+2
Could you please what's the permutation method as that is the way that my teacher + textbook would like me to do it

Sure! So for this question, we look at permutations rather than combinations (because we specifically need an even number in the units slot). Because we have 5 numbers and we want to know how many arrangements there are of a two-digit even number, we create a formula for each slot (10s with 5 choices and 1s with 3 choices).

10s = 5!/(5-1)! = 5
1s = 3!/(3-1)! = 3

In total, this is 5*3 = 15 permutations, including all number repetitions. However, we need to exclude those (44, 66, 88 ), so we obtain 15-3=12, which is the same answer from earlier
2018, VCE: Literature
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#### rozmaaate

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##### Re: VCE Methods Question Thread!
« Reply #18632 on: July 18, 2020, 06:12:29 pm »
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The angle of depression of a boat from a cliff 60m high is 10°. How far ,to the nearest metre ,is the boat from the base of the cliff?

Sorry for the basic question I'm not great at math, correct answer is 340m apparently btw

#### Owlbird83

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##### Re: VCE Methods Question Thread!
« Reply #18633 on: July 18, 2020, 06:52:09 pm »
+5
The angle of depression of a boat from a cliff 60m high is 10°. How far ,to the nearest metre ,is the boat from the base of the cliff?

Sorry for the basic question I'm not great at math, correct answer is 340m apparently btw

Always draw the triangle first. Then add the numbers and angles you know. (angle of 'depression' means that from the horizontal it's 10degrees, so you'd use 80degrees as the angle in the triangle)
Think about SOH CAH TOA, and pick the formula for the info you have.
In this diagram we have the length adjacent (60m) and we want the opposite length. Using this, pick out the formula with o and a, which is the one with tan, and put the numbers into the formula, and solve for 'o' (or x as I've called it here).

I hope this helps
« Last Edit: July 18, 2020, 06:55:48 pm by Owlbird83 »
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#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18634 on: July 19, 2020, 12:33:06 pm »
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I've been doing some past exams, and I've come across "linear approximation" questions a few times. Is it in the current Methods Study Design Still?

Thanks,
James
VCE Class of 2021
2020: Maths Methods [?]
2021: English [?] Biology [?] Chemistry [?] Physics [?] Specialist Maths [?]

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18635 on: July 19, 2020, 12:45:19 pm »
+4
I've been doing some past exams, and I've come across "linear approximation" questions a few times. Is it in the current Methods Study Design Still?

Thanks,
James

Oh man, haven't heard those words since I did VCE. Linear approximation is well and truly gone (note: it's still technically in Specialist as Euler's method, so maybe not "well and truly")
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#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18636 on: July 19, 2020, 12:59:14 pm »
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Oh man, haven't heard those words since I did VCE. Linear approximation is well and truly gone (note: it's still technically in Specialist as Euler's method, so maybe not "well and truly")

Ok thanks! I was wondering since I haven't ever heard the type of questions before
VCE Class of 2021
2020: Maths Methods [?]
2021: English [?] Biology [?] Chemistry [?] Physics [?] Specialist Maths [?]

#### Oynx

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##### Re: VCE Methods Question Thread!
« Reply #18637 on: July 20, 2020, 02:42:40 pm »
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Hey need some help with a question:

If cosx=−cos(π6) and π2 < x < π find the value of x.

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #18638 on: July 20, 2020, 03:03:41 pm »
+4
Hey need some help with a question:

If cosx=−cos(π6) and π2 < x < π find the value of x.

Hey there Oynx!

Did you mean $\cos(x) = -\cos(\pi \cdot 6)$ or $\cos(x) = -\cos\left(\frac{\pi}{6}\right)$

I'm guessing the second one, for one thing, cos(u) = cos(-u); for another thing, cos(u) = -cos(u + pi)

So $-\cos\left(\frac{\pi}{6}\right) = -\cos\left(\frac{-\pi}{6}\right) = \cos\left(\pi - \frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)$ Assuming I didn't make any silly mistakes.

(and then it's pretty obvious x is 5pi/6) both properties used there should be fairly obvious from just, like, looking at the unit circle.

Hope this makes sense, if it doesn't don't hesitate to ask!

EDIT: Just in case if anybody is confused.
The reason behind it is because of the different quadrants, because if you did 180 - 60 = x for the second question you would get 120, which doesn't satisfy 180 < x < 270.

Start by thinking about what -cos(60) is on the unit circle; cos(60) is the horizontal distance at 60 degrees, so "negative" of that is reflecting 60 degrees across the middle which gives you cos(180 - 60) = cos(120), which is the first one but cos is horizontal distance, and there are two points on the other side of the unit circle with that. Because if you change the vertical distance only, the horizontal distance is the same and the point with the same horizontal distance but different vertical distance you can get by reflecting it over the other axis, the x axis. Reflecting over the x axis you can think of as negating the thing on the inside, so cos(-120), and then add 360 because negative numbers are inconvenient and you get 240.
« Last Edit: July 20, 2020, 07:19:23 pm by 1729 »

#### MoonChild1234

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##### Re: VCE Methods Question Thread!
« Reply #18639 on: July 22, 2020, 09:31:36 am »
0
is x^2 - (integral from 0 to 2 of x^3+4) considered an integral expression, or do both components need to be integrals?

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #18640 on: July 22, 2020, 09:58:27 am »
+4
is x^2 - (integral from 0 to 2 of x^3+4) considered an integral expression, or do both components need to be integrals?
$\int _0^2x^3+4dx$ Is an integral expression.

However, x^2 by itself is not an integral expression because it has no integrals in it. An integral expression is an expression with an integral in it, so yes that entire thing is an integral expression.
« Last Edit: July 22, 2020, 10:12:28 am by 1729 »

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18641 on: July 26, 2020, 12:54:10 pm »
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Hey guys,
I gave this as my answer to a revision question and it is incorrect. I cannot see what I did that is an error. Do you guys know what Im doing wrong?
Many thanks,
Corey

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18642 on: July 26, 2020, 12:57:00 pm »
+1
Last attachment was too low res

#### 1729

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##### Re: VCE Methods Question Thread!
« Reply #18643 on: July 26, 2020, 01:04:47 pm »
+3
Last attachment was too low res
Hey Corey King. Welcome to the forums!

You messed your answer up with the third step. You are actually meant to expand the minus sign.

So it should be :

12y-15 -(2y+1) = 6y. In this case you can look it as 12y-15 -1(2y+1)=6y, try and expand the -1. So you should get
12y-15-2y+1=6y

You should be able to solve from there, and get the answer to be 3.5.

If you do not understand don't hesitate to ask!