Author Topic: Mathematics Question Thread (Read 1627328 times) Tweet Share

K9810 · « **Reply #2910 on:** October 21, 2017, 11:24:31 am »

Can you please help me with part iii) and iv)

RuiAce · « **Reply #2911 on:** October 21, 2017, 11:28:02 am »

Quote from: asd987 on October 21, 2017, 11:18:09 am

Hi can i get help with ii) please. thanks

Assuming that the point of intersection was (2,4)
$\text{For }0\le y \le 4\text{ the region is bounded by}\\ \boxed{x^2 = 6 - \frac{y}{2}}$
$\text{For }4\le y \le 12\text{ the region is bounded by}\\ \boxed{x^2 = y}$
$\text{So the required volume integral is}\\ V = \pi \int_0^4 \left(6-\frac{y}2\right)dy + \pi \int_4^{12} y\,dy$

_____ · « **Reply #2912 on:** October 21, 2017, 12:04:41 pm »

Is it OK to use a pencil for diagrams? Front of 2016 paper just says black pen.

RuiAce · « **Reply #2913 on:** October 21, 2017, 12:08:54 pm »

Quote from: K9810 on October 21, 2017, 11:24:31 am

Can you please help me with part iii) and iv)

Some comprehensive discussion on part iii can be found here.

Quote from: _____ on October 21, 2017, 12:04:41 pm

Is it OK to use a pencil for diagrams? Front of 2016 paper just says black pen.

You really should be using pen. From previous posts, the general advice is to draw the axes in pen, then if you want draw the curve itself in pencil before tracing it. At least, that's if you insist on using the pencil.

sidzeman · « **Reply #2914 on:** October 21, 2017, 12:09:26 pm »

Quote from: _____ on October 21, 2017, 12:04:41 pm

Is it OK to use a pencil for diagrams? Front of 2016 paper just says black pen.

^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?

Also, could someone help me with part iv of this question please, as well as part ii from b

RuiAce · « **Reply #2915 on:** October 21, 2017, 12:10:44 pm »

Quote from: sidzeman on October 21, 2017, 12:09:26 pm

^expanding on that, will we have marks deducted for sketching lets say number plane axis freehand rather than with a ruler?

The HSC has never given any instruction on this. So if you want to do this, you're gambling on chance (a nice/mean marker).

winstondarmawan · « **Reply #2916 on:** October 21, 2017, 12:16:06 pm »

Adding onto these technicality questions, what are the specifications for rulers
Do they have to be completely blank with just a brand name and measurements?

RuiAce · « **Reply #2917 on:** October 21, 2017, 12:17:28 pm »

Quote from: winstondarmawan on October 21, 2017, 12:16:06 pm

Adding onto these technicality questions, what are the specifications for rulers
Do they have to be completely blank with just a brand name and measurements?

Of course you can’t have a ruler with a bunch of formulas written over it.

They can’t do anything about your name or just the brand name though as far as I’m aware of.

bsdfjnlkasn · « **Reply #2918 on:** October 21, 2017, 12:22:27 pm »

Hey there!

I've got a bunch of questions from the 2001 but i'll stagger them as I post

Thanks so much!

These are the specifics for the questions I need help with:

Finance - ii and iii

Geometry - ii and iii

bsdfjnlkasn · « **Reply #2919 on:** October 21, 2017, 12:23:20 pm »

Hey there!

Here's the second round

Thanks so much!

These are the specifics for the questions I need help with:

Trapezoidal rule - legit don't think I applied it right but don't know how to correct it?
My working: 2(1.7 + 2(1.3))
Answer is 12 m/s ??

bsdfjnlkasn · « **Reply #2920 on:** October 21, 2017, 12:35:41 pm »

Hey there!

Here's the third round

Thanks so much!

These are the specifics for the questions I need help with:

Plane geometry: v) - I thought to find where BC intersects the x-axis and the x-coordinate would be the perpendicular distance but I don't know the coordinates of C...

Range: not sure how to find the range..

bsdfjnlkasn · « **Reply #2921 on:** October 21, 2017, 12:40:17 pm »

Hey there!

Here's the final round

Thanks so much!

Just need help with the first part, then I'll attempt the second

RuiAce · « **Reply #2922 on:** October 21, 2017, 02:03:07 pm »

Quote from: bsdfjnlkasn on October 21, 2017, 12:23:20 pm

Hey there!

Here's the second round

Thanks so much!

These are the specifics for the questions I need help with:

Trapezoidal rule - legit don't think I applied it right but don't know how to correct it?
My working: 2(1.7 + 2(1.3))
Answer is 12 m/s ??

1.3 and 1.7 are both "inbetweens" for the trapezoidal rule. It should be $ 2 \left[0 + 0 + 2(1.7+1.3) \right] $ which is equal to 12.

Basically, the "first" and the "last" were both 0.

Quote from: bsdfjnlkasn on October 21, 2017, 12:40:17 pm

Hey there!

Here's the final round

Thanks so much!

Just need help with the first part, then I'll attempt the second

This one is in the compilation. It's ridiculous.

RuiAce · « **Reply #2923 on:** October 21, 2017, 02:09:58 pm »

Quote from: bsdfjnlkasn on October 21, 2017, 12:35:41 pm

Hey there!

Here's the third round

Thanks so much!

These are the specifics for the questions I need help with:

Plane geometry: v) - I thought to find where BC intersects the x-axis and the x-coordinate would be the perpendicular distance but I don't know the coordinates of C...

Range: not sure how to find the range..

$\text{We know that }y=\sqrt{25-x^2}\text{ is just the upper}\\ \text{semicircle with radius }r=5.\\ \text{So it has range }0 \le y \le 5.$
$\text{So }y= 2\sqrt{25-x^2}\text{ has range }0\le y \le 10.$
This is basically because it's just a "stretch" of the semicircle up and down. (But there is no down, because this is the upper semicircle. So it's only stretched up.)
_______________________________
$\text{The coordinates of the point }C\text{ can be found}\\ \text{by recalling that }OABC\text{ must be a parallelogram.}\\ \text{Since opp. sides on a parallelogram are equal, we must then have }\boxed{OA = CB}$
$\text{We can simply use "common sense" now, with translations.}\\ \text{Since }A\text{ is the point }(-2,5)\text{ and }B\text{ is the point }(4,3)\\ \text{Because the point }O\text{ is }A\text{ translated by }(2,-5)\\ C\text{ is the point }B\text{ translated by }(2,-5)$
$\text{Hence }C\text{ is }(2 + 4, -5 + 3) = (6,-2)$

RuiAce · « **Reply #2924 on:** October 21, 2017, 02:17:29 pm »

Quote from: bsdfjnlkasn on October 21, 2017, 12:22:27 pm

Hey there!

I've got a bunch of questions from the 2001 but i'll stagger them as I post

Thanks so much!

These are the specifics for the questions I need help with:

Finance - ii and iii

Geometry - ii and iii

$\text{The similarity of }\triangle DBA\text{ and }\triangle ABC\text{ implies that}\\ \frac{DB}{AB}=\frac{BA}{BC}$
$\text{So subbing in, we have }\\ \begin{align*}\frac{x-1}{1}& = \frac1x\\ x(x-1)&=1\\ x^2-x-1&=0\end{align*}$
_______________________
$\text{Note that by the quadratic formula, the equation above}\\ \text{solves to give }x = \frac{1+\sqrt5}{2}$
$\text{Since }\triangle ABC\text{ is isosceles with }\angle BAC = \frac{3\pi}{5}\\ \text{we have }\angle ABC = \angle ACB = \frac\pi5$
$\text{The cosine rule in }\triangle ABC\text{ gives}\\ \begin{align*}\cos \angle ABC &= \frac{AB^2+BC^2-AC^2}{2\times AB\times BC}\\ \therefore \cos \frac\pi5&= \frac{1^2+x^2-1^2}{2(1)(x)}\\ &= \frac{x}{2}\\ &= \frac{1+\sqrt5}{4}\tag{using the above result}\end{align*}$

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1627328 times) Tweet Share

K9810

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

_____

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

sidzeman

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

winstondarmawan

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

bsdfjnlkasn

Re: Mathematics Question Thread

bsdfjnlkasn

Re: Mathematics Question Thread

bsdfjnlkasn

Re: Mathematics Question Thread

bsdfjnlkasn

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Recent Posts

User:
Password: