****Leaving Certificate: Higher Level Maths Discussion****

booblefoop wrote: »

I found a list of all the theorems we need to know about! I can't post the URL because I'm a 'new user' but if anyone needs it feel free to PM me

Just take out the http://www. part and post it here I'm sure the mods won't mind since it's helping us :pac:

Here's the link: http://gofree.indigo.ie/~hallinan/ProjectMaths/LC_Geometry_Strand2/Leaving%20Cert%20Geometry_summary_theorems.htm

Do we really need to know all of that :pac:

Aspiring wrote: »

Here's what it was like on the DEB mock for anybody that doesn't have it. Blanked out the answers so you can use it to practice on.

are you able to show the answer? I need to see if the one im learning is right

Nimrod 7 wrote: »

Is that not theorem 13?

**** it is...oh god Im screwed..so I only got 9 marks in mocks for number 13...Sigh

We only need to be able to prove theorems 11, 12 and 13 ya?

Aspiring wrote: »

Here's the Project Maths powerpoint on theorem 12.

http://www.projectmaths.ie/students/cd-strand1and2/docs/theorem12.ppsx

They do it differently but everything in Concise Maths does seem to make sense, what part of it do you think we are missing?

It looks like we're just missing that part in the middle before they state that the lines cut of equal segments on the other line, it looks like they state that the sum of all the sections equals the length, which I think I can manage to write if it comes up :pac:

We could also be asked on some of the Junior Cert theorems, unlikely but theorems 4,6,9,14 and 19 are still there.

FatRat wrote: »

There better be another easy stats 75 mark question. After the f*ck up on Friday with no financial maths and hardly any algebra, Im almost thinking they'all do the same with paper 2 and not go by the same layout they've had for the past few years. A 75 mark question mostly made up of probability instead of stats would be sick.

Those z score things get you a handy few marks. What else would the put in a 75m probability? Bernoulli trials? Venn diagrams? :pac:

Hi,I would really appreciate some help here...
I was trying out the Paper One exam paper and I've run into a bit of difficulty with Q9b which ye had to do in the paper on Friday!

I can get out the answer most of ye are getting ...k=-0.0339 but there's one step I would've done differently which gives me a different answer and I'd really appreciate if someone could point out what's wrong with the step I've done...

So...
Q: After minutes the temperature of the water is 88 degrees C.
Find the value of K, correct to three significant figures.
I'm using 77 as my A value from part a and y is equalled to 65... by taking water temp from constant room temp
y=Ae^kt
Here it goes:

65=77e^kt ... but I'm going to use t as 5 as it stated so in Q
65=77e^5k

Here's where I need help... to get your answer ye obviously took the 77 down and divided it under the 65 but why can't I just multiply the 77 by 'e' and THEN turn it into log to the base 209.307 of 65 is equalled to 5k and work it out from there... it gives a different answer?

So basically my question is why do you bring the 77 down and divide it under the 66 rather than multiplying 77 by the e and proceed by turn it into a log to the base etc....?


I'd really appreciate some clarification sorry for the long post..

I can't post the url because im a new user but if you search 'oconailldcs' on youtube he has videos of the theorem 11,12,13 proofs. it really really helped me and its really easy to follow

yoyojc wrote: »

Hi,I would really appreciate some help here...
I was trying out the Paper One exam paper and I've run into a bit of difficulty with Q9b which ye had to do in the paper on Friday!

I can get out the answer most of ye are getting ...k=-0.0339 but there's one step I would've done differently which gives me a different answer and I'd really appreciate if someone could point out what's wrong with the step I've done...

So...
Q: After minutes the temperature of the water is 88 degrees C.
Find the value of K, correct to three significant figures.
I'm using 77 as my A value from part a and y is equalled to 65... by taking water temp from constant room temp
y=Ae^kt
Here it goes:

65=77e^kt ... but I'm going to use t as 5 as it stated so in Q
65=77e^5k

Here's where I need help... to get your answer ye obviously took the 77 down and divided it under the 65 but why can't I just multiply the 77 by 'e' and THEN turn it into log to the base 209.307 of 65 is equalled to 5k and work it out from there... it gives a different answer?

So basically my question is why do you bring the 77 down and divide it under the 66 rather than multiplying 77 by the e and proceed by turn it into a log to the base etc....?


I'd really appreciate some clarification sorry for the long post..

Not sure what you mean exactly, but did you multiply the 65 by e as well?

yoyojc wrote: »

Hi,I would really appreciate some help here...
I was trying out the Paper One exam paper and I've run into a bit of difficulty with Q9b which ye had to do in the paper on Friday!

I can get out the answer most of ye are getting ...k=-0.0339 but there's one step I would've done differently which gives me a different answer and I'd really appreciate if someone could point out what's wrong with the step I've done...

So...
Q: After minutes the temperature of the water is 88 degrees C.
Find the value of K, correct to three significant figures.
I'm using 77 as my A value from part a and y is equalled to 65... by taking water temp from constant room temp
y=Ae^kt
Here it goes:

65=77e^kt ... but I'm going to use t as 5 as it stated so in Q
65=77e^5k

Here's where I need help... to get your answer ye obviously took the 77 down and divided it under the 65 but why can't I just multiply the 77 by 'e' and THEN turn it into log to the base 209.307 of 65 is equalled to 5k and work it out from there... it gives a different answer?

So basically my question is why do you bring the 77 down and divide it under the 66 rather than multiplying 77 by the e and proceed by turn it into a log to the base etc....?


I'd really appreciate some clarification sorry for the long post..

If I'm understanding you correctly, you're wondering why you can't multiply the 77 by e? Because the e is raised to a power. You can't touch the e until you deal with the exponent (BEMDAS). That's why you divide across by 77 and then get the natural log (ln or log to the base e) of both sides.

Chonaic17 wrote: »

We could also be asked on some of the Junior Cert theorems, unlikely but theorems 4,6,9,14 and 19 are still there.

Syllabus says we only have to be able to prove Theorems 11, 12, 13.

Calvin Leong wrote: »

For probability, what is the difference between the two buttons P and C on the calculator? And when is one used as opposed to the other? ie 7P3 or 7C3

P is for Permutations, C is for combinations.

A permutation is an arrangement of objects in a specific order, while a combination is a selection of objects in any order.

7P3 is for when you have 7 objects and want to find out how many ways you can arrange any 3 of them. E.g. how many arrangements of three letters can you make from A, B, C, D, E, F, G?

7C3 is for when you have 7 objects and want to find out how many different selections of 3 objects you can make. E.g. how many different selections of 3 letters can you make from A, B, C, D, E, F, G?

7P3 = 210, 7C3 = 35.
The reason there are so many more permutations is because order is taken into account. In the first instance, the permutation ABC is distinct from CBA so they are counted separately. However, in the second instance ABC and CBA are the same combination of letters, so they are not counted twice.

I hope that makes sense.

Considering we got the exact same proof by induction question on paper one that was on the 2014 sample, what do people think the chances of something similar happening in paper 2 are? Just coincidence? Or maybe something more?