2012 Year 9 &10 Mathematics Marathon (4 Viewers)

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

I'm trying to explain it as well but I cannot even use LaTeX haha.

EDIT: Yay it worked!

 

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus:

Proof:
A hexagon has 6 corners. Using the formula , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is:

Proof:
If a polygon has 700 sides, how many diagonals are there?





= 243950 diagonals

 

[kazemagic]

Re: 2012 Year 9 &10 Mathematics Marathon

My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus:

Proof:
A hexagon has 6 corners. Using the formula , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is:

Proof:
If a polygon has 700 sides, how many diagonals are there?





= 243950 diagonals

Great explanation! And this one by ymcaec!

hard to explain. i'll try my best

If n = number of vertices (or sides)


n - 3 = a vertex can form diagonals with all other vertices except for the two next to it and itself
then times n (for each of the vertices)
divide by 2 because there are 2 ends for each diagonal (so the diagonals are counted twice)

 

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

Great explanation!

Thanks

 

[Demento1]

Re: 2012 Year 9 &10 Mathematics Marathon

My proofs and explanations are horrible but lets hope this is correct! Here it is:
Let's use a hexagon as an example and start from the bottom right corner. Draw 3 lines from the corner to show the diagonals you can get from that point, you will find that you get 3 diagonals. Lets move on to the next corner which is the middle right corner. You will find that you only get 2 diagonals. If you go to the last corner which is the top right corner, you will only get 1 diagonal. This proves that each diagonal connects to each corner except itself and its 2 neighbouring corners.

Let "n" be the number of corners (you can use sides as well though since the number of corners = the number of sides). Thus:

Proof:
A hexagon has 6 corners. Using the formula , we get 3 which means that there are 3 diagonals from a corner, which is true. However this formula is only for 1 corner

If we multiply the number of sides/corners by the formula we just found, we will get the number of diagonals. We will also need to divide it by 2 since diagonals only connect 2 corners once which halves the shape. Thus the formula is:

Proof:
If a polygon has 700 sides, how many diagonals are there?





= 243950 diagonals

I should type like this ^.

 

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

I should type like this ^.

Haha! It looks good doesn't it?

 

[iBibah]

Re: 2012 Year 9 &10 Mathematics Marathon

All spot on.

Simple when you think about it, each vertex can join to every other one bar itself and the two next to it hence (n-3). This can happen for every vertex, hence n(n-3). Then we must cancel double hence n(n-3)/2.

 

[Timske]

Re: 2012 Year 9 &10 Mathematics Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" title="\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" /></a>

 

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" title="\textup{If} ~A^x = 3 \\ \textup{Find}~ A^{10x - x} - 5" /></a>

I am going to give this a go!

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" title="\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3^{10}}{3}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{3^{10}}{3}- 5" title="\frac{3^{10}}{3}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex== (3)^{9} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?= (3)^{9} - 5" title="= (3)^{9} - 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\therefore A^{10-x}- 5 = 19678" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\therefore A^{10-x}- 5 = 19678" title="\therefore A^{10-x}- 5 = 19678" /></a>

 

[twinklegal19]

Re: 2012 Year 9 &10 Mathematics Marathon

^That looks right

 

[Ealdoon]

Re: 2012 Year 9 &10 Mathematics Marathon

^That looks right

Thanks

 

[Demento1]

Re: 2012 Year 9 &10 Mathematics Marathon

I am going to give this a go!

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" title="\frac{A^{10x}}{A^{x}}- 5 = \frac{(A^{x})^{10}}{A^{x}}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3^{10}}{3}- 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{3^{10}}{3}- 5" title="\frac{3^{10}}{3}- 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex== (3)^{9} - 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?= (3)^{9} - 5" title="= (3)^{9} - 5" /></a>

<a href="http://www.codecogs.com/eqnedit.php?latex=\therefore A^{10-x}- 5 = 19678" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\therefore A^{10-x}- 5 = 19678" title="\therefore A^{10-x}- 5 = 19678" /></a>

That's correct. This is my simplified format of solving it. Didn't even notice the Q until just then...





Edit: You can prove this is possible through basic logarithms.

 

[twinklegal19]

Re: 2012 Year 9 &10 Mathematics Marathon

Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

 

[Demento1]

Re: 2012 Year 9 &10 Mathematics Marathon

Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:



ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.



Therefore: P = $4616.73 (nearest cent)

 

[RealiseNothing]

Re: 2012 Year 9 &10 Mathematics Marathon

Let's have some financial maths/consumer arithmetic

Tom has been investing his money for 5 years. Initially his investment was earning interest at 9% per annum compounded monthly. From the start of the global financial crisis 18 months ago, his investment began to lose value at 15% per annum each month.

i. If Tom initially invested $P, what was the value of his portfolio immediately before the global financial crisis? Give your answer in unsimplified form in terms of P.
ii. Tom’s investment portfolio is now worth $10 000. What was the amount of his original investment?

I hate you for posting this question.

 

[twinklegal19]

Re: 2012 Year 9 &10 Mathematics Marathon

I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:



ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.



Therefore: P = $4616.73 (nearest cent)

You got the first part right but not the second, try again

(dw, I got it wrong the first time last year as well...)

I hate you for posting this question.

<3

 

[ymcaec]

Re: 2012 Year 9 &10 Mathematics Marathon

I think you may have discovered a slightly weaker area in my mathematics. But anyways, I can hopefully solve the Q that you throw at me.

i) Amount before the global financial crisis:



ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.



Therefore: P = $4616.73 (nearest cent)

for ii. i got 9163.05 instead

 

[twinklegal19]

Re: 2012 Year 9 &10 Mathematics Marathon

for ii. i got 9163.05 instead

correct! Show the lines of working out?

 

[Demento1]

Re: 2012 Year 9 &10 Mathematics Marathon

for ii. i got 9163.05 instead

That's correct.

 

[ymcaec]

Re: 2012 Year 9 &10 Mathematics Marathon

correct! Show the lines of working out?

Instead of "plus", it should be a "times"