2012 Year 9 &10 Mathematics Marathon (5 Viewers)
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Ealdoon
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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:}{2})
Proof:
If a polygon has 700 sides, how many diagonals are there?
}{2})
= 243950 diagonals
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
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
Re: 2012 Year 9 &10 Mathematics Marathon
Great explanation! And this one by ymcaec!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
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
Last edited:
Ealdoon
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Re: 2012 Year 9 &10 Mathematics Marathon
I should type like this ^.
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
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
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Re: 2012 Year 9 &10 Mathematics Marathon
Haha! It looks good doesn't it?
I should type like this ^.
iBibah
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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.
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
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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>
<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
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Re: 2012 Year 9 &10 Mathematics Marathon
<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>
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>
Re: 2012 Year 9 &10 Mathematics Marathon
^That looks right
^That looks right
Ealdoon
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Last edited:
Re: 2012 Year 9 &10 Mathematics Marathon
Edit: You can prove this is possible through basic logarithms.
That's correct. This is my simplified format of solving it. Didn't even notice the Q until just then...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>
Edit: You can prove this is possible through basic logarithms.
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?
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?
Re: 2012 Year 9 &10 Mathematics Marathon
i) Amount before the global financial crisis:
^{42} )
ii) I cannot bother typing all the working out, although the working out soon ended up as this with me.
^{42} + \left ( \frac{79}{80} \right )^{18}} )
Therefore: P = $4616.73 (nearest cent)
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.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) 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)
Last edited:
RealiseNothing
what is that?It is Cowpea
Re: 2012 Year 9 &10 Mathematics Marathon
I hate you for posting this question.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?
Re: 2012 Year 9 &10 Mathematics Marathon
(dw, I got it wrong the first time last year as well...)
You got the first part right but not the second, try againI 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)
(dw, I got it wrong the first time last year as well...)
<3
Re: 2012 Year 9 &10 Mathematics Marathon
for ii. i got 9163.05 insteadI 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 = $4616.73 (nearest cent)
Re: 2012 Year 9 &10 Mathematics Marathon
^{42} \times \left ( \frac{79}{80} \right )^{18}} \\ = 9163.05$ (2dp)$)
Instead of "plus", it should be a "times"