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2012 Year 9 &10 Mathematics Marathon (1 Viewer)
- Thread starter TerenceM
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Re: 2012 Year 9 &10 Mathematics Marathon
CorrectSorry that I didn't get to your Q earlier but here it is now:
iBibah
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Re: 2012 Year 9 &10 Mathematics Marathon
f(x)=*insert quadratic function*
then x=*insert quadratic equation you just proved*
Your not trying to find another way of expressing it, but rather a general formula to find the roots, hence x=.
Just trying to help you understand what your doing a little more
And yes, you can post questions from texts.
haha its ok, its just that if
f(x)=*insert quadratic function*
then x=*insert quadratic equation you just proved*
Your not trying to find another way of expressing it, but rather a general formula to find the roots, hence x=.
Just trying to help you understand what your doing a little more
And yes, you can post questions from texts.
Re: 2012 Year 9 &10 Mathematics Marathon
First we know BE = ED (diagonals of parallelogram bisect each other)
Then using Pythagoras:
\\\\{AB}^2+{BD}^2={AD}^2\\{AB}^2+{(2ED)}^2={AD}^2...(2)\\\\$Substitute (1) into (2)$\\{AE}^2-{ED}^2+4{ED}^2={AD}^2\\{AD}^2={AE}^2+3{ED}^2)
PythagorasHere's another one lol
First we know BE = ED (diagonals of parallelogram bisect each other)
Then using Pythagoras:
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Re: 2012 Year 9 &10 Mathematics Marathon
Have fun with this guys
Use a suitable substitution to solve
^2-9\left ( x^2+\frac{1}{x^2} \right )+20=0)
Have fun with this guys
Use a suitable substitution to solve
kev-kun
Member
Re: 2012 Year 9 &10 Mathematics Marathon
^2-9(x^2+\frac{1}{x^2})+20=0 \\Let\ y=(x^2+\frac{1}{x^2}) \\y^2-9y+20=0 \\(y-5)(y-4)=0 \\Sub\ y=(x^2+\frac{1}{x^2}) \\(x^2+\frac{1}{x^2}-5)(x^2+\frac{1}{x^2}-4)=0 \\That's\ all\ I \ got\ ==" )
Have fun with this guys
Use a suitable substitution to solve
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Re: 2012 Year 9 &10 Mathematics Marathon
i.e. y=5,4 and from there you can replace y with what you originally substituted
finish solving your quadratic equation^2-9(x^2+\frac{1}{x^2})+20=0 \\Let\y=(x^2+\frac{1}{x^2}) \\y^2-9y+20=0 \\(y-5)(y-4)=0 \\Sub\ y=(x^2+\frac{1}{x^2}) \\(x^2+\frac{1}{x^2}-5)(x^2+\frac{1}{x^2}-4)=0 \\That's\ all\ I \ got\ ==" )
i.e. y=5,4 and from there you can replace y with what you originally substitutedRe: 2012 Year 9 &10 Mathematics Marathon
@Twinkle happy birthday m8 *insert twinkle twinkle little stars song here*
Yep good job lolPythagoras
First we know BE = ED (diagonals of parallelogram bisect each other)
Then using Pythagoras:
@Twinkle happy birthday m8 *insert twinkle twinkle little stars song here*
Carrotsticks
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Re: 2012 Year 9 &10 Mathematics Marathon
A quadratic within a quadratic.finish solving your quadratic equation
kev-kun
Member
Re: 2012 Year 9 &10 Mathematics Marathon
This isnt really a syllabus problem, but more of a logic problem - nonetheless though I think the skills needed are pretty important in maths. Some people may recognise it from somewhere, but I wont say where...
A person has a five litre jug and a three litre jug and they need to fill the five litre jug with exactly four litres of water. The person cannot guess the amount and they have no measurement aids other then the two jugs. Assuming they have an unlimited supply of water, how would they complete this task?
This isnt really a syllabus problem, but more of a logic problem - nonetheless though I think the skills needed are pretty important in maths. Some people may recognise it from somewhere, but I wont say where...
A person has a five litre jug and a three litre jug and they need to fill the five litre jug with exactly four litres of water. The person cannot guess the amount and they have no measurement aids other then the two jugs. Assuming they have an unlimited supply of water, how would they complete this task?
Re: 2012 Year 9 &10 Mathematics Marathon
To do it, you fill the 5 litre jug completely, which means you have 5 litres, then you pour the contents into the 3 litre jug until it is completely full. Which means that you now have 2 litres left in the 5 litre jug. You do it twice which means you will ahve 4 litres.
Re: 2012 Year 9 &10 Mathematics Marathon
Well since the question stated that you can have unlimited amount of water, after pouring the contents into the three litre jug Now you have 2 litres left in the 5 litre jug. You then pour the 2 litre jug in the other jug, then empty the three litre jug and start again to get another 2 which gives you four.
Well since the question stated that you can have unlimited amount of water, after pouring the contents into the three litre jug Now you have 2 litres left in the 5 litre jug. You then pour the 2 litre jug in the other jug, then empty the three litre jug and start again to get another 2 which gives you four.
nightweaver066
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Re: 2012 Year 9 &10 Mathematics Marathon
@enoilgam, i've been playing a game exactly like this on my phone and it's so much fun haha
i think you only have one 5L jug and one 3L jug. are you taking it to be 3 jugs?
@enoilgam, i've been playing a game exactly like this on my phone and it's so much fun haha
Re: 2012 Year 9 &10 Mathematics Marathon
There are only two jugs fawun. You are close, but the end is where you are going wrong.i think you only have one 5L jug and one 3L jug. are you taking it to be 3 jugs?
@enoilgam, i've been playing a game exactly like this on my phone and it's so much fun haha
Re: 2012 Year 9 &10 Mathematics Marathon
y=3 litre jug
Fill up y, pour into x so x has 3 litre. Fill up y, then pour it to x so x has 5 litre and y has 1 litre. Pour x into sink, so nothing remains in it. Pour the 1 litre from y to x. Fill up y then pour it to x, then u get der 4 litres in 5 litre jug
Let x= 5 litre jug
y=3 litre jug
Fill up y, pour into x so x has 3 litre. Fill up y, then pour it to x so x has 5 litre and y has 1 litre. Pour x into sink, so nothing remains in it. Pour the 1 litre from y to x. Fill up y then pour it to x, then u get der 4 litres in 5 litre jug
Re: 2012 Year 9 &10 Mathematics Marathon
Nice - that wasnt the method I had in mind but it works nonetheless.
Re: 2012 Year 9 &10 Mathematics Marathon
The only information that is provided is their capacity. There is another way to do it, can anyone think of it.