HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

[barbernator]

Re: HSC 2012 Marathon

bit of algebra practice:

rearrange to make the subject y, also, for the first one, put in the form (x-k)(y-h)=c

<a href="http://www.codecogs.com/eqnedit.php?latex=2xy-3x-y-2=0\\ and\\ y^2 - 8y @plus; 5x - x^2 - 15 = 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?2xy-3x-y-2=0\\ and\\ y^2 - 8y + 5x - x^2 - 15 = 0" title="2xy-3x-y-2=0\\ and\\ y^2 - 8y + 5x - x^2 - 15 = 0" /></a>

(rating: green curry)

 

[Timske]

Re: HSC 2012 Marathon

1. 2xy - 3x - y - 2 = 0
y(2x - 1) = 3x + 2
y = (3x+2)/(2x-1)
2xy - 3x - y - 2 = 0 -> (x - 1/2)(y - 3/2) = 1.75

2. y^2 - 8y + 5x - x^2 - 15 = 0
y^2 - 8y - 15 = x^2 - 5x
(y-4)^2 - 31 = x^2 - 5x
(y-4)^2 = x^2 - 5x + 31
y - 4 = sqrt(x^2 - 5x + 31)
:. y = sqrt(x^2 - 5x + 31) + 4

 

[Deliriously]

Re: HSC 2012 Marathon

The foot D of the altitude BD of a triangle ABC lies between A and C. Prove that:

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=TBP: ~CD-AD=\frac{(BC)^2-(AB^2)}{AC}\\ \\ let~AD=x,~DC=y,~DB=h\\ \\ by~pythagoras~theorum\\ \\ RHS=\frac{(BC)^2-(AB^2)}{AC}\\ \\ =\frac{(h^2@plus;y^2)-(h^2@plus;x^2)}{x@plus;y}\\ \\ =\frac{y^2-x^2}{x@plus;y}\\ \\ =y-x\\ \\ =CD-AD\\ \\ =LHS~as~required" target="_blank"><img src="http://latex.codecogs.com/gif.latex?TBP: ~CD-AD=\frac{(BC)^2-(AB^2)}{AC}\\ \\ let~AD=x,~DC=y,~DB=h\\ \\ by~pythagoras~theorum\\ \\ RHS=\frac{(BC)^2-(AB^2)}{AC}\\ \\ =\frac{(h^2+y^2)-(h^2+x^2)}{x+y}\\ \\ =\frac{y^2-x^2}{x+y}\\ \\ =y-x\\ \\ =CD-AD\\ \\ =LHS~as~required" title="TBP: ~CD-AD=\frac{(BC)^2-(AB^2)}{AC}\\ \\ let~AD=x,~DC=y,~DB=h\\ \\ by~pythagoras~theorum\\ \\ RHS=\frac{(BC)^2-(AB^2)}{AC}\\ \\ =\frac{(h^2+y^2)-(h^2+x^2)}{x+y}\\ \\ =\frac{y^2-x^2}{x+y}\\ \\ =y-x\\ \\ =CD-AD\\ \\ =LHS~as~required" /></a>
NEW QUESTION:
The letters of the word PERSEVERE are arranged in a row. How many different arrangements are possible.(easy)

for those who want to post solutions but dont know how to properly, http://www.codecogs.com/latex/eqneditor.php

lol whoops need to edit my solution

 

[Timske]

Re: HSC 2012 Marathon

^ I've used that before but i can't copy/paste it

<a href="http://www.codecogs.com/eqnedit.php?latex=x\\ @plus; \\ y" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x\\ + \\ y" title="x\\ + \\ y" /></a>

 

[barbernator]

Re: HSC 2012 Marathon

^ I've used that before but i can't copy/paste it

copy the code box from down the bottom and paste that

 

[bleakarcher]

Re: HSC 2012 Marathon

how come when i press space bar it doesnt actually show a space on the latex image?

 

[AAEldar]

Re: HSC 2012 Marathon

how come when i press space bar it doesnt actually show a space on the latex image?

To do a space in latex you can either do "\;" and it'll put a space (I think "\," does the same thing). Or you can use the \text{ } and put a space in the braces, OR you can use the $ $ tags and put a space there.

EDIT: Didn't know you could do what barbernator said!

 

[barbernator]

Re: HSC 2012 Marathon

how come when i press space bar it doesnt actually show a space on the latex image?

~ between words

 

[bleakarcher]

Re: HSC 2012 Marathon

my keyboard is fucked. it doesnt have that backward slash but it has the forward slash. FUUUUUUUUUUUUUUUUUUUUUUUUUUUUU

 

[bleakarcher]

Re: HSC 2012 Marathon

~ between words

thanks bruh

 

[dulip]

Re: HSC 2012 Marathon



new question:
If two of the roots of the equation are equal show that
[easy/medium]

 

[Carrotsticks]

Re: HSC 2012 Marathon

new question:
If two of the roots of the equation are equal show that
[easy/medium]

dulip, that question is more suited for an Extension 2 student under the topic "Polynomials". It is a very commonly asked question, and is essentially the 'Discriminant' of a cubic.

Here is one for the mean time:

By letting x = tan(u), evaluate the integral: (Medium/Hard)



Or:

By letting u = cos(x), show that: (Easy)

 

[nightweaver066]

Re: HSC 2012 Marathon

dulip, that question is more suited for an Extension 2 student under the topic "Polynomials". It is a very commonly asked question, and is essentially the 'Discriminant' of a cubic.

Here is one for the mean time:

By letting x = tan(u), evaluate the integral:

Hopefully i did it right in my head lol.

 

[Carrotsticks]

Re: HSC 2012 Marathon

Hopefully i did it right in my head lol.

Nope, but close.

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{2}~(\frac{x}{x^2@plus;1} ~ @plus; ~ \tan^{-1})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})" title="\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})" /></a>

 

[Carrotsticks]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{2}~(\frac{x}{x^2@plus;1} ~ @plus; ~ \tan^{-1})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})" title="\frac{1}{2}~(\frac{x}{x^2+1} ~ + ~ \tan^{-1})" /></a>

Correct!

I'll put up another question since you forgot to post one up yourself.

Find the condition such that the two lines:

ax+by+c=0 and dx+ey+f=0

Are:

a) Perpendicular.

b) Parallel.

 

[bleakarcher]

Re: HSC 2012 Marathon

Cmon Carrot.

ax+by+c=0 => y=[-ax-c]/b has gradient -a/b
dx+ey+f=0 => y=[-dx-f]/e has gradient -d/e
a) If the two lines are parallel then their gradient are equal,
i.e. -a/b=-d/e
Hence, bd=ae
b) If the two lines are parallel then the product of their gradient is -1,
i.e. (-a/b)(-d/e)=-1
Hence, ad=-eb

 

[RealiseNothing]

Re: HSC 2012 Marathon

Cmon Carrot.

ax+by+c=0 => y=[-ax-c]/b has gradient -a/b
dx+ey+f=0 => y=[-dx-f]/e has gradient -d/e
a) If the two lines are parallel then their gradient are equal,
i.e. -a/b=-d/e
Hence, bd=ae
b) If the two lines are parallel then the product of their gradient is -1,
i.e. (-a/b)(-d/e)=-1
Hence, ad=-eb

For the parallel one though, you also have to put in the condition that they aren't the same line.

 

[bleakarcher]

Re: HSC 2012 Marathon

For the parallel one though, you also have to put in the condition that they aren't the same line.

oh right completely forgot about that one. so you just add c/b =/= f/e
so bf=/=ec