2012 Year 9 &10 Mathematics Marathon (1 Viewer)

kazemagic · Jul 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Here's another question. Solve for x and y:
$\left\{\begin{matrix} x^2-xy+y^2=19 \\ x^4+x^2y^2+y^4=741 \end{matrix}\right.$

russ3l · Jul 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

whats the point of posting up such difficult questions? everyone just searches the 'hardest question' they can find in hope to challenge someone with it. I mean comeon...we are doing 4unit math questions ffs TT.TT

kazemagic · Jul 18, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
whats the point of posting up such difficult questions? everyone just searches the 'hardest question' they can find in hope to challenge someone with it. I mean comeon...we are doing 4unit math questions ffs TT.TT

lol that triangle question can be done just using pythagoras, ymcaec just did it the way that looked complicated by using the cosine rule.
And that simultaneous question, i remember that you need to use completing square method and that's pretty much it I think

ymcaec · Jul 21, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

kazemagic said:
Here's another question. Solve for x and y:
$\left\{\begin{matrix} x^2-xy+y^2=19 \\ x^4+x^2y^2+y^4=741 \end{matrix}\right.$

$\left\{\begin{matrix}x^2-xy+y^2=19 ... (1)\\x^4+x^2y^2+y^4=741 ...(2) \end{matrix}\right\\\\$Rearrange $ (1)\\x^2+y^2=19+xy \\(x^2+y^2)^2=(19+xy)^2... (3)\\\\$Rearrange $(2)\\(x^2+y^2)^2=741+x^2y^2...(4)\\\\(3)=(4)\\(19+xy)^2=741+x^2y^2\\361+38xy+x^2y^2=741+x^2y^2\\38xy=380\\\therefore xy=10\\\therefore y=\frac{10}{x}\\\\$Substitute $y=\frac{10}{x}$ into $(1)\\x^2-10+\frac{100}{x^2}=19\\(x^2-25)(x^2-4)=0\\\therefore x=\pm 5$ or $\pm 2\\\\\therefore$ Solution: $(5,2), (-5,-2), (2,5), (-2,-5)$

saysesame · Jul 22, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

Any more delicious maths questions?

kazemagic · Jul 23, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

ymcaec said:
$\left\{\begin{matrix}x^2-xy+y^2=19 ... (1)\\x^4+x^2y^2+y^4=741 ...(2) \end{matrix}\right\\\\$Rearrange $ (1)\\x^2+y^2=19+xy \\(x^2+y^2)^2=(19+xy)^2... (3)\\\\$Rearrange $(2)\\(x^2+y^2)^2=741+x^2y^2...(4)\\\\(3)=(4)\\(19+xy)^2=741+x^2y^2\\361+38xy+x^2y^2=741+x^2y^2\\38xy=380\\\therefore xy=10\\\therefore y=\frac{10}{x}\\\\$Substitute $y=\frac{10}{x}$ into $(1)\\x^2-10+\frac{100}{x^2}=19\\(x^2-25)(x^2-4)=0\\\therefore x=\pm 5$ or $\pm 2\\\\\therefore$ Solution: $(5,2), (-5,-2), (2,5), (-2,-5)$

For your equation number 4, shouldn't it be 741-x^2y^2 instead?
I'll check it moar carefully 2morrow

EDIT:Nvm correct

zeebobDD · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

saysesame said:
Any more delicious maths questions?

heres one for ya, show that the some of the numbers 1+2+3...+99+100 = 5050

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

$$If $y=\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}$ then find the simplest expression for $y^2$

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

^ymcaec, please wait a few hours until no one else can answer haha

zeebobDD · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
$$If $y=\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}$ then find the simplest expression for $y^2$

y^2 = 1/ [x(1-x)]

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

zeebobDD said:
y^2 = 1/ [x(1-x)]

correct

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

another question (procrastinating from school maths hw lol)

zeebobDD · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

try my q, and are you sure thats 9&10:S

RealiseNothing · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

zeebobDD said:
try my q, and are you sure thats 9&10:S

It is.

bluecrisps · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
another question (procrastinating from school maths hw lol)

b) i) 2x+y=36 (width of material)
Therefore y=36-2x (rearrange the equation)

ii) Area= length x height
= x x y
Sub Equation from B)i)
= x(36-2x)
= 36x x 2x^2
iii) Axis of symmetry= -b/2a
= -36/2(-2)
= -36/-4
= 9
This is the maximum value of x as it is the axis of symmetry for a parabola

iv) A=36x x 2x^2
= 36(9) -2(9)^2
= 329-162
= 167 cm2

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

zeebobDD said:
try my q, and are you sure thats 9&10:S

yes

It might seem very HSC-like with the maximum/minimum stuff, but it's really just a simple application of quadratics in year 10

bluecrisps said:
b) i) 2x+y=36 (width of material)
Therefore y=36-2x (rearrange the equation)

ii) Area= length x height
= x x y
Sub Equation from B)i)
= x(36-2x)
= 36x x 2x^2
iii) Axis of symmetry= -b/2a
= -36/2(-2)
= -36/-4
= 9
This is the maximum value of x as it is the axis of symmetry for a parabola

iv) A=36x x 2x^2
= 36(9) -2(9)^2
= 329-162
= 167 cm2

Correct

however there's a calculation error in the last few steps,

$36\times9-2\times9^2=162$ not $167$ =)$$

russ3l · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

RealiseNothing said:
It is.

The q looks like part of a hsc paper to me...

twinklegal19 · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

russ3l said:
The q looks like part of a hsc paper to me...

no, it's actually cut and pasted from a past year 10 yearly exam

there's plenty more where that came from

(just PM me)

russ3l · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
no, it's actually cut and pasted from a past year 10 yearly exam

there's plenty more where that came from

(just PM me)

Oh alrighty

post up another question

bluecrisps · Jul 28, 2012

Re: 2012 Year 9 &10 Mathematics Marathon

twinklegal19 said:
yes

It might seem very HSC-like with the maximum/minimum stuff, but it's really just a simple application of quadratics in year 10

Correct

however there's a calculation error in the last few steps,

$36\times9-2\times9^2=162$ not $167$ =)$$

Oops, darned windows calc -_-

2012 Year 9 &10 Mathematics Marathon (1 Viewer)

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