HSC 2015 MX2 Marathon (archive) (7 Viewers)

[braintic]

Re: HSC 2015 4U Marathon

close but you have to select them before arranging. Solution:
Lock people a and b on one side and then arrange them 1x2! but you still have to fill up the other spots... therfore 2!x(14C2)x2!x(12C2)x2!..... take 2 factorial out and then type it into your calculator you get 1362160800 and since there is 8 side you multiply by 8 to get 1.09x10^10 with the assumption that one can differenciate between sides

You are not answering the question you asked. Problem is, I haven't yet figured out what question you are answering.

Ekman's solution seems to be the right one to me.

Why would you even need to consider pairs for the remaining 14 people? There is absolutely no restriction on the placement of the other 14 (except that they can't sit in the 2 spots already occupied).

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

because theyre on different sides

 

[braintic]

Re: HSC 2015 4U Marathon

because theyre on different sides

Sorry. As I said, there is no restriction on the placement of the other 14.
Randomly placing the other 14 automatically deals with every outcome for each side.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

if you take it that way its considered as if the rest is a line. By the way how do you reply where it shows the previous message

 

[braintic]

Re: HSC 2015 4U Marathon

Just realised that your answer is exactly the same - you must have typed it in wrongly.

14C2 times 2! = 14P2 = 14 times 13

12C2 times 2! = 12P2 = 12 times 11

...

etc.

So your answer is 2 times (14 times 13 times 12 times 11 times .....) = 2 times 14!

But definitely not a nice way to think about it.

 

[InteGrand]

Re: HSC 2015 4U Marathon

if you take it that way its considered as if the rest is a line. By the way how do you reply where it shows the previous message

Click "Reply With Quote".

 

[braintic]

Re: HSC 2015 4U Marathon

how do you reply where it shows the previous message

"Reply With Quote"

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

oh silly mistake i took out 2 and its multiply therefore2^8(14C2)(12C2).... thanks again

 

[braintic]

Re: HSC 2015 4U Marathon

oh silly mistake i took out 2 and its multiply therefore2^8(14C2)(12C2).... thanks again

Just to explain the concept further:

You could have filled the remaining 14 spots using any grouping you wanted, and the answer would have been the same.

So, say you wanted to first fill 3 particular spots, then fill another 6 particular spots, then fill the last 5 spots.

Your calculation would have been:
(14C3 times 3!) times (11C6 times 6!) times (5C5 times 5!) = the same thing

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

Just to explain the concept further:

You could have filled the remaining 14 spots using any grouping you wanted, and the answer would have been the same.

So, say you wanted to first fill 3 particular spots, then fill another 6 particular spots, then fill the last 5 spots.

Your calculation would have been:
(14C3 times 3!) times (11C6 times 6!) times (5C5 times 5!) = the same thing

its a different answer though

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

its a different answer though[/QUOTEmy bad again ahahahah x2!

 

[braintic]

Re: HSC 2015 4U Marathon

Yes, I was only counting the ways of filling the other 14 positions.

And .... when you hit "Reply With Quote", make sure you don't edit the part that says [QUOTE = blah blah] or the part that says [/QUOTE].
You deleted the final "]", which ruined the quote.
If you edit your comment now and type in the final "]", you should then see the quote once you post again.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

First time making an inequality question prove that for x>0 that 1< ((x^2 +2x^(3/2) +x)^1/2 )/x

 

[Ekman]

Re: HSC 2015 4U Marathon

First time making an inequality question prove that for x>0 that 1< ((x^2 +2x^(3/2) +x)^1/2 )/x

1< ((x^2 +2x^(3/2) +x)^1/2 )/x
1 < (x^2 + 2x^(3/2) + x) / x^2 (squared both sides)
1 < 1 + (2x^(1/2)) / x + 1/x
(2x^(1/2)) / x + 1/x is >0 since x>0
Thus 1 + something that is >0 it is then >1
Therefore 1< ((x^2 +2x^(3/2) +x)^1/2 )/x

 

[Ekman]

Re: HSC 2015 4U Marathon

First time making an inequality question prove that for x>0 that 1< ((x^2 +2x^(3/2) +x)^1/2 )/x

OR:
1< ((x^2 +2x^(3/2) +x)^1/2 )/x
x^2 < x^2 + 2x^(3/2) + x
0 < 2x^(3/2) + x
x^2 < 4x^3
0 < x^2 ( 4x -1)
x> 0 x> 1/4 (Since there is an overlap, there is no need for the x>1/4)
Thus 1< ((x^2 +2x^(3/2) +x)^1/2 )/x only works if x>0 as proven and stated by the question.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

OR:
1< ((x^2 +2x^(3/2) +x)^1/2 )/x
x^2 < x^2 + 2x^(3/2) + x
0 < 2x^(3/2) + x
x^2 < 4x^3
0 < x^2 ( 4x -1)
x> 0 x> 1/4 (Since there is an overlap, there is no need for the x>1/4)
Thus 1< ((x^2 +2x^(3/2) +x)^1/2 )/x only works if x>0 as proven and stated by the question.

with the inequality questions in extension two i dont think you can start from what is given notsure

 

[RealiseNothing]

Re: HSC 2015 4U Marathon

with the inequality questions in extension two i dont think you can start from what is given notsure

If all your steps are biconditional (work both ways essentially) then it's all gee.

 

[Drsoccerball]

Re: HSC 2015 4U Marathon

If all your steps are biconditional (work both ways essentially) then it's all gee.

Sounds good. Anyways next question:
The lines x=2 x= -2 y=3 are asymptotes to the curve with the equation
y=(ax^2)/((x)^2 +bc+c)
By using the above information show that:
y= (3x^2)/((x)^2 -4)

 

[braintic]

Re: HSC 2015 4U Marathon

Sounds good. Anyways next question:
The lines x=2 x= -2 y=3 are asymptotes to the curve with the equation
y=(ax^2)/((x)^2 +bc+c)
By using the above information show that:
y= (3x^2)/((x)^2 -4)

Correction: bx, not bc

 

[Kaido]

Re: HSC 2015 4U Marathon

x^2+bx+x? or x^2+bx+cx?

Solve simultaneously by subbing 2,-2,3 and setting =0 ? pre-guess