HSC 2015 Maths Marathon (archive) (3 Viewers)

[kawaiipotato]

Re: HSC 2015 2U Marathon

oh lol thanks. don't know why I only added it

 

[Flop21]

Re: HSC 2015 2U Marathon

What is the best way to find the focus, vertex, directrix etc. of a parabola (from equation)??

Is there more than one way I should know? E.g. putting in form (x-x0)^2 = 4a(y-y0). How do I know what form to put it in, I don't really understand it tbh.

 

[milkytea99]

Re: HSC 2015 2U Marathon

What is the best way to find the focus, vertex, directrix etc. of a parabola (from equation)??

Is there more than one way I should know? E.g. putting in form (x-x0)^2 = 4a(y-y0). How do I know what form to put it in, I don't really understand it tbh.

Draw a diagram and plot the points. Although it sounds stupid, but it works ESP for vertex which is not (0.0)

 

[Drsoccerball]

Re: HSC 2015 2U Marathon

What is the best way to find the focus, vertex, directrix etc. of a parabola (from equation)??

Is there more than one way I should know? E.g. putting in form (x-x0)^2 = 4a(y-y0). How do I know what form to put it in, I don't really understand it tbh.

The thing you must find is the vertex aka turning point. This is found by when both sides are =0 therefore in your equation it would be (x0,y0). From there what you Must do is draw the direction or at least think of the direction the parabola is facing in this case its similar to x^2 therefore the focus would be higher than the vertex and the directrix would be lower than the vertex. That's pretty much it...

 

[milkytea99]

Re: HSC 2015 2U Marathon

The thing you must find is the vertex aka turning point. This is found by when both sides are =0 therefore in your equation it would be (x0,y0). From there what you Must do is draw the direction or at least think of the direction the parabola is facing in this case its similar to x^2 therefore the focus would be higher than the vertex and the directrix would be lower than the vertex. That's pretty much it...

What I've said but reworded in a much better way lol. My communication skills are so bad haha

 

[Drsoccerball]

Re: HSC 2015 2U Marathon

What I've said but reworded in a much better way lol. My communication skills are so bad haha

I doubt it... Im getting crap in english

 

[Flop21]

Re: HSC 2015 2U Marathon

Thought this would be a good question for the 2015ers to do:


The zoom function in a software package multiplies the dimensions of an image by 1.2. In an image, the height of a building is 50 mm. After the zoom function is applied once, the height of the building in the image is 60 mm. After a second application, its height is 72 mm.

i) calculate the height of the building in the image after the zoom function has been applied 8 times.

ii) The height of the building in hte image is required to be more than 400mm. Starting from the original image, what is the least number of times the zoom function must be applied?

 

[999bottles]

Re: HSC 2015 2U Marathon

We need the 2U beast here: @Drsoccerball where you at?

 

[davidgoes4wce]

Re: HSC 2015 2U Marathon

Thought this would be a good question for the 2015ers to do:


The zoom function in a software package multiplies the dimensions of an image by 1.2. In an image, the height of a building is 50 mm. After the zoom function is applied once, the height of the building in the image is 60 mm. After a second application, its height is 72 mm.

i) calculate the height of the building in the image after the zoom function has been applied 8 times.

ii) The height of the building in hte image is required to be more than 400mm. Starting from the original image, what is the least number of times the zoom function must be applied?

part
(i) 50 x (1.2)^8 =214.99 mm

 

[davidgoes4wce]

Re: HSC 2015 2U Marathon

part (ii)

400=50 (1.2)^n
solving with a calculator
n=11.4

Therefore, the camera needs to be zoomed a minimum of 12 times for the image to be more than 400 mm.

 

[Flop21]

Re: HSC 2015 2U Marathon

Shouldn't this thread be for 2015ers to test each other, or for primarily 2015ers to answer?? I thought that was the whole point of these marathons.

Jumping in and helping those that ask for it is fine... but isn't the point of this for other HSC students to answer, then ask a question themselves, so we're all getting tested and learning??

 

[BlueGas]

Re: HSC 2015 2U Marathon

part (ii)

400=50 (1.2)^n
solving with a calculator
n=11.4

Therefore, the camera needs to be zoomed a minimum of 12 times for the image to be more than 400 mm.

It's 11."4", you don't you round down?

 

[Drsoccerball]

Re: HSC 2015 2U Marathon

It's 11."4", you don't you round down?

If you round down you wont get the height you want it would be less.

 

[Flop21]

Re: HSC 2015 2U Marathon

It's 11."4", you don't you round down?

Note that n = the nth image. Not the number of times the zoom function has been applied.

You get n=13, the 13th image, but the number of times the function has been applied is one less than that, 12.

 

[leehuan]

Re: HSC 2015 2U Marathon

Shouldn't this thread be for 2015ers to test each other, or for primarily 2015ers to answer?? I thought that was the whole point of these marathons.

Jumping in and helping those that ask for it is fine... but isn't the point of this for other HSC students to answer, then ask a question themselves, so we're all getting tested and learning??

We lack 2015ers on here.
_________________________________
NEXT QUESTION:



A royal flush in poker is when the cards are specifically ranked 10, J, Q, K and A, whilst being of the same suit.
_________________________________
If that question is too hard for a 2U student...

 

[braintic]

Re: HSC 2015 2U Marathon

Shouldn't this thread be for 2015ers to test each other, or for primarily 2015ers to answer?? I thought that was the whole point of these marathons.

Jumping in and helping those that ask for it is fine... but isn't the point of this for other HSC students to answer, then ask a question themselves, so we're all getting tested and learning??

The person you are referring to IS a 2015er.

 

[Flop21]

Re: HSC 2015 2U Marathon

The person you are referring to IS a 2015er.

lol what, who am I referring to?

EDIT: It was a general statement I made after seeing this thread turn into 'post a question you can't do and someone post HSC will answer' (which is great, but you get what I'm saying).

And forgive me for mistaking a 2009 engineering graduate for a non-2015er.

 

[Drsoccerball]

Re: HSC 2015 2U Marathon

We lack 2015ers on here.
_________________________________
NEXT QUESTION:



A royal flush in poker is when the cards are specifically ranked 10, J, Q, K and A, whilst being of the same suit.
_________________________________
If that question is too hard for a 2U student...

Lets focus on one suit


Therefore multiply by 4 since there is four suits:

 

[flavurr]

Re: HSC 2015 2U Marathon

Lets focus on one suit


Therefore multiply by 4 since there is four suits:

hey dr wouldn't it be 20/52 * 4/51 * 3/50 * 2/49 *1/48 as once you pick the first card, there is only a possibility of 4/51 to pick the 2nd of that same suit then 3/50 to pick the 3rd of that same suit ect

 

[flavurr]

Re: HSC 2015 2U Marathon

We lack 2015ers on here.
_________________________________

If that question is too hard for a 2U student...

dy/dx=m= 1/2squareroot(x-4)

Simultaneously solve y=mx and y=squareroot(x-4) subbing in m=1/2squareroot(x-4)

x/2squareroot(x-4)=squareroot(x-4)

x= 2(x-4)

x=8