2012 Year 9 &10 Mathematics Marathon (1 Viewer)

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

iBibah said:
You use y-y1=m(x-x1) (a formula for working out the equation of a line given a point and gradient) and your given Q(-1,0).........

so y1=0
x1=-1

I think you can do the rest

So given that $m=tan\theta$

$y=tan\theta(x+1)$

$y=xtan\theta + tan\theta$

I am still not certain about my answer because it simply doesn't sound legit to me but I'm happy to learn (correct me because it doesn't sound right).

Demento1 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Fawun said:
Similar working out

I will never be able to understand how to use latex

It is quite easy. I learnt through trial and error and people would generally laugh at me (still do) about my beginner LaTeX skills. Try this for for LaTeXing and copy paste your text from the site onto your post with $and$ surrounding it.

http://www.codecogs.com/latex/eqneditor.php

kat- said:
78?

That's what I got.

twinklegal19 · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
So given that $m=tan\theta$

$y=tan\theta(x+1)$

$y=xtan\theta + tan\theta$

I am still not certain about my answer because it simply doesn't sound legit to me but I'm happy to learn (correct me because it doesn't sound right).

lol replace $\theta$ with $\beta$ in line with the angles in the question

my example was just in general

iBibah · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
So given that $m=tan\theta$

$y=tan\theta(x+1)$

$y=xtan\theta + tan\theta$

I am still not certain about my answer because it simply doesn't sound legit to me but I'm happy to learn (correct me because it doesn't sound right).

The question didn't specify but iirc you were supposed to leave it as m, so y=mx+m (so ur right

)

kazemagic · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento smashing all the questions like a baws lol
Here's another flavour to add to the ice cream.
Simplify the following:

enoilgam · Aug 9, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kat- said:
78?

EDIT: my working out
x + 2y = 318 (1)
x = 240 - y (2)

(240 - y) + 2y = 318
y = 318 - 240
= 78

therefore amount of $2 coins is 78.

Where's the latex button? Or do i need to download it separately

Looks good to me.

russ3l · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

find the centre of this circle and the radius in its most simplified form in $x^2 - 2x + y^2 +4y -3 = 0$

twinklegal19 · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
find the centre of this circle and the radius in its most simplified form in $x^2 - 2x + y^2 +4y -3 = 0$

lol that sort of question literally appears in almost every single exam paper.

this one is definitely worth mastering young year 10s

Demento1 · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Demento smashing all the questions like a baws lol
Here's another flavour to add to the ice cream.
Simplify the following:

Sorry that I didn't get to your Q earlier but here it is now:

$\frac{6^n+3^n}{2^{n+1}+2}$

$\frac{3^n(2^n+1)}{2(2^n+1)}$

$=\frac{3^n}{2}$

Demento1 · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
find the centre of this circle and the radius in its most simplified form in $x^2 - 2x + y^2 +4y -3 = 0$

The working out on LaTeX for the equation of the circle was slightly tiresome but worth it:

$x^2 - 2x + y^2 +4y -3 = 0$

$x^2 - 2x + y^2 +4y = 3$

$(x^2-2x-(1)^2)+(y^2+4y+(2)^2)=3+1+4$

$(x-1)^2+(y+2)^2 = 8$

Therefore, centre = (1, -2) and radius = $\sqrt{8}=2\sqrt{2}$ units.

russ3l · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Demento1 said:
The working out on LaTeX for the equation of the circle was slightly tiresome but worth it:

$x^2 - 2x + y^2 +4y -3 = 0$

$x^2 - 2x + y^2 +4y = 3$

$(x^2-2x-(1)^2)+(y^2+4y+(2)^2)=3+1+4$

$(x-1)^2+(y+2)^2 = 8$

Therefore, centre = (1, -2) and radius = $\sqrt{8}=2\sqrt{2}$ units.

correct!

Demento1 · Aug 10, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
correct!

I'll go and post an interest algebraic question I found for tomorrow. I'm sure you'll be able to do it if you put some thought into it.

maths_genius · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Prove the quadratic equation from the general form of the parabola, f(x) = ax^2 + bx + c Hint: You have to use the "completing the square" technique

kev-kun · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

$\\ax^2+bx+c=0 \\* ax^2+bx=-c \\* x^2+\frac{b}{a}x=\frac{-c}{a}\\*x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\\*(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\*=\frac{-4ac+b}{4a^2}\\*x+\frac{b}{2a}=\pm \sqrt{\frac{-4ac+b^2}{4a^2}}\\*=\pm\frac{\sqrt b^2-4ac}{2a}\\*\therefore f(x)=\frac{-b\pm\sqrt(b^2-4ac)}{2a}$
Phew took a while XD

Demento1 · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
$\\ax^2+bx+c=0 \\* ax^2+bx=-c \\* x^2+\frac{b}{a}x=\frac{-c}{a}\\*x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\\*(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\*=\frac{-4ac+b}{4a^2}\\*x+\frac{b}{2a}=\pm \sqrt{\frac{-4ac+b^2}{4a^2}}\\*=\pm\frac{\sqrt b^2-4ac}{2a}\\*\therefore f(x)=\frac{-b\pm\sqrt(b^2-4ac)}{2a}$
Phew took a while XD

Beat me to it. So quick. ^

kazemagic · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Here's another one lol

Fawun · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
$\\ax^2+bx+c=0 \\* ax^2+bx=-c \\* x^2+\frac{b}{a}x=\frac{-c}{a}\\*x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\\*(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\*=\frac{-4ac+b}{4a^2}\\*x+\frac{b}{2a}=\pm \sqrt{\frac{-4ac+b^2}{4a^2}}\\*=\pm\frac{\sqrt b^2-4ac}{2a}\\*\therefore f(x)=\frac{-b\pm\sqrt(b^2-4ac)}{2a}$
Phew took a while XD

tutor me babe

kazemagic said:
Here's another one lol

what the hell

If this was an exam question, this will be one of those questions that you will sit thinking about for like 5 minutes before you say "fuck it" and move on to the next question.

iBibah · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kev-kun said:
$\\ax^2+bx+c=0 \\* ax^2+bx=-c \\* x^2+\frac{b}{a}x=\frac{-c}{a}\\*x^2+\frac{b}{a}x+(\frac{b}{2a})^2=\frac{-c}{a}+(\frac{b}{2a})^2\\*(x+\frac{b}{2a})^2=\frac{-c}{a}+\frac{b^2}{4a^2}\\*=\frac{-4ac+b}{4a^2}\\*x+\frac{b}{2a}=\pm \sqrt{\frac{-4ac+b^2}{4a^2}}\\*=\pm\frac{\sqrt b^2-4ac}{2a}\\*\therefore f(x)=\frac{-b\pm\sqrt(b^2-4ac)}{2a}$
Phew took a while XD

well done, but the last line is x= not f(x)= .

kev-kun · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

maths_genius said:
Prove the quadratic equation from the general form of the parabola, f(x) = ax^2 + bx + c Hint: You have to use the "completing the square" technique

iBibah said:
well done, but the last line is x= not f(x)= .

Well tried to stay to the question, guess I was wrong ==''

BTW just curious, are we allowed to submit questions from our textbooks?

maths_genius · Aug 12, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Find all solutions to the intersection of xy=4 and y=sqrt(16-(x^2))

2012 Year 9 &10 Mathematics Marathon (1 Viewer)

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