If you want any help with maths, ask me here

pinkbear

Hi,
I know the leaving cert maths courses pretty well. If anyone wants help with any question, any topic, any level, just post here and I'll answer back as soon as I can.

LLAMAMILK

I have a test on patterns and sequences on Wednesday as part of my report. I'm in Ordinary so apart from Sn and Tn what do I need to know ?

pinkbear

Sequence: A sequence is a list of numbers. E.g. 1,2,3,4,5…
Series: A series is the sum of the terms of a sequence. E.g. 1+2+3+4+5+….

To Find if a Sequence is Linear/Arithmetic, Quadratic, Exponential:
Calculate 1st differences. If they're the same, it's linear/arithmetic.
If they're not calculate 2nd differences. If they're the same, it's quadratic.
If they're not, calculate the ratio between the terms. If they're the same, it's geometric.
If they're not, it could be none of the above, but that's highly unlikely so it's more likely you made a mistake.

Linear / arithmetic: Tn=a+(n-1)d, Sn = n/2 [ 2a + (n – 1)d ] where a is the first term, d is the (1st) difference, and n is the number you sub in.

Quadratic: Tn=an^2 + bn + c, where a=half the 2nd difference
You have to use simultaneous equations to work out b and c. (Example at the bottom)

Geometric: Sn= a(1-r^n) / (1-r), where a is the first term, r is the ratio between one term and the next, n is the number of terms in the series.

These formulas are all in the tables (I think pg 22), except Tn for the quadratic.

E.g. Identify the sequence 1, 5, 13, 25, 41… as linear, quadratic, exponential or none of these. Find Tn and Sn of the series.

Solution:
1st Differences: 4, 8, 12, 16 ….⇒ Not the same ⇒ Not linear
2nd Differences: 4, 4, 4 ….⇒ The same ⇒ Quadratic

Tn = an^2 + bn + c, where a = half the 2nd difference = ½ of 4 = 2
⇒Tn = 2n^2 + bn + c

T1 = 2(1)^2 + b(1) + c = 4 ⇒ 2+ b + c = 1 ⇒ b + c = -1
T2 = 2(2)^2 + b(2) + c = 5 ⇒ 8 + 2b + c = 5 ⇒ 2b + c = -3

Applying simultaneous equations: 2b + c = -3
-b - c = 1
b = -2

Subbing b=-2 into b + c = -1: -2+c = -1 ⇒ c = 1 ⇒Tn = 2n^2 - 2n + 1

Test this: T5=2(52)-2(5)+1 = 2(25) -10+ 1 = 41 ✓

Here's one for you to try:
Q. 13,15,19,25,33 are the first 5 numbers in a pattern.
(i) Follow the pattern to write the next 3 numbers.
(ii) Show that the pattern is quadratic.
(iii) Un=an^2+bn+c. Find the values of a, b, and c.

Oathkeeper

Hi I'm doing higher level maths and I was just wondering how to get good at higher maths in general I usually just barely pass maths so I'm in need of help

pinkbear

Hi Oathkeeper,
There is not just one way to get good at higher maths, but a few points are:
- I would recommend grinds and grinds schools. These can teach you much more efficiently.... you'd probably learn as much in one hours grinds / grinds school as you would in 3 hours on your own. A good grinds teacher should be able to cover almost a full topic in a class.... it might be rushed, but if you get an overview of a full topic, and see 10-12 exam questions done, you'll be in such a better position. (I know they are expensive, so it's not an option for many people, but I think they have a good pay back in terms of extra points).
- You have to start doing the exam questions early. Many teachers, unfortunately, just go by the books and don't look much at exam papers. It's only once you start doing lots and lots of exam papers that you start to see patterns emerging and you can see really quickly what they are looking for in each question.
- When you're doing your homework, if you can't do every question, really focus on the ones you can't get. The ones you got right you can forget about, but do whatever you have to do to get the other ones right. Post here and I'll help you, or send whatsapp messages to your class looking for assistance. If you are going into your leaving cert having got every homework question fully right, you are in a much stronger position.
- Studyclix is great. The paid accounts are even better as you have access to so much material. If you were revising a topic - say trigonometry - rather than wading through your book, I would recommend quickly looking at OL JC questions first to make sure you can do those, then moving onto to HL JC, and OL LC. Looking at those questions shouldn't take too long and will be a great revision of the basics, then should you move onto HL LC. Try to do a question, then click on the Marking Scheme, and so on.
- I read before that the part of the brain that deals with maths doesn't fully develop until about age 26. (I'm over 26). Certainly this was true for me. While I could manage the questions in school, and always did quite well, I didn't really understand them fully. As I got older, I was studying maths in university, but also I could understand concepts with a clarity that I didn't have when I was in school, and I think a lot of it was just due to age. So if anyone is considering studying engineering / maths / anything with a lot of maths in it, I wouldn't let the fact that you find maths difficult now put you off. It will definitely become easier in time.

Hope some of that is of use to you.
PinkBear

EireLemon

How would you go about doing part iii) & iv)? Getting rid of the 5dx in part iii) and 6 in part iv) is causing me issues.

josha1

jinx257

I know how to do calculus, how to apply the formulas etc. But I dont understand it, like where does it come from, whats behind it etc?

pinkbear

jinx257 - Answer on attached pdf.

EireLemon

josha1 wrote: »

Why do you use the differentiation formula instead of the integration formula?

josha1

EireLemon wrote: »

Why do you use the differentiation formula instead of the integration formula?

That is the integration formula. It should be in the log tables under inverse trigonometric formulae for integration.

They're both the same thing really I'm not too sure which one is in the log tables. If you want to use your formula you keep the constant outside the integration sign. The constant with negate the 1/a in front of you formula, so it ends up the same as my formula.

thetalker

Hi if you have any ideas there's a math Q I was wondering how to do.

In a private sweep 100 tickets are sold. A man buys one ticket. A prize is awarded on each draw of a ticket. Assuming replacement of a ticket after each draw, find the least number of prizes which must be awarded so that the probability of the man winning at least one prize is at least 1/4.

I checked the back and the answer is 29 but I have no idea how to approach the Q or how that was found.

pinkbear

I'm getting 28 instead of 29, and don't have time to check it fully, but I'm confident that my method is okay, so you might be able to spot my error (or maybe the back of the book is wrong). Let me know if you spot it!

---

In any particular draw the chance of him winning is 1/100. So for the probability to be 1/4 or greater....

(Chances he wins in first) + (Chances he loses first & wins in 2nd) + (chances he loses first two & wins in 3rd)+ ...... >1/4

1/100 + (99/100)(1/100) + (99/100)(99/100)(1/100) + ..... > 1/4

This is a geometric series that we don't know how many terms are in.

Sn = a (1-r^n)
1 - r
where a = 1/100, r = 99/100

Sn = (1/100) ( 1 - (99/100)^n)

---

1 - 99/100

= (1/100) ( 1 - (99/100)^n) = 1 - (99/100)^n

---

1/100

1 - (99/100)^n > 1/4

1 - (99/100)^n > 1/4
1 - 0.25 > (99/100)^n
0.75 > (99/100)^n
ln 0.75 > ln (0.99)^n

ln 0.75 > n

---

ln 0.99

28.62 > n
n = 28

pinkbear

Basically you're looking for what power of .99 (his chances of not winning) would give a value just under 0.25 but not greater than it.
.99^30=0.7513
.99^29=0.7471

thetalker

Wow thanks for the quick reply, I would never have imagined to do that tbh.

Im a bit confused as to why you're adding the probabilties though. I thought you would only do that for mutually exclusive events. I can see your finding all the possibilities that will add to 1/4 but why is what I don't understand.

I think you pretty much nailed the method though, I think you forgot to flip the inequality sign once you divided by a negative number hence n is actually greater than 28.62 not smaller.

thetalker

Would it not be a geometric sequence where n is the amount of prizes given out/ amount of times game is played and r is 99/100
So Tn=ar^n-1, where a is just 1
Like you posted that would mean were trying to find what power aka n is needed to give
99/100^n-1= 0.75 and then use ln on both sides and then approximate n-1, so that gives 29.62

Or is there a flaw to this method

Paulieniceguy

pinkbear wrote: »

Hi,
I know the leaving cert maths courses pretty well. If anyone wants help with any question, any topic, any level, just post here and I'll answer back as soon as I can.

Fair play to you for offering your help, I will be sure to let my niece know as she is finding higher maths quite difficult

pinkbear

thetalker:
Ah yes, ln .99 is negative, so I was dividing by a negative number but didn't know it! That's where I went wrong.

I'm adding the probabilities because if this line:
(Chances he wins in first) + (Chances he loses first & wins in 2nd) + (chances he loses first two & wins in 3rd)+ ...... >1/4

Winning and losing are mutually exclusive, so you add the probabilities.

Regarding your last post, where you talked about the Tn=an^n-1 method, there is a flaw with this method (I think). You are not looking for which turn has a greater than 1 in 4 chance of winning, you are looking for the sum of them, as you will only get to that turn if you have gone through all the others.

(p.s. I'm not a genius at maths by any means, just more familiar with the course than many! So I do get things wrong.)

thetalker

Ah the thing is I read that line but was confused, you see if you take the probability of winning initially, why do you add the portability of him winning and losing and etc

It feels like if someone was rolling a die to get a 6 and went

1/6 + 1/6(5/6) + 1/6(5/6)^2 etc. to calculate the odds of rolling another 6, it just confuses me because I would understand if it was a series so you had it in a way which was

1/6, 1/6(5/6), 1/6(5/6)^2 etc where we can calculate the odds of getting the the next success or loss based on the sequence.

Tbh we only recently went over this chapter so maybe I'm misunderstanding the theory itself. But I would've thought winning and losing were directly related to each other.

"You are not looking for which turn has a greater than 1 in 4 chance of winning, you are looking for the sum of them, as you will only get to that turn if you have gone through all the others."
But doesn't the term formula do that? since you multiply probabilities to get the next not add them. Looking for the sum feels confusing as like I pointed out I don't think they are exclusive. You have to do the previous one to get to the next after all.

Either way you've been a great help as I was going after the question in the wrong direction!

pinkbear

Ok, hopefully this will clarify it for you, but otherwise maybe ask your teacher.

Where you said ".... where we can calculate the odds of getting the the next success or loss based on the sequence"

the thing to remember is that in ANY go, the chances of success are 1/100. The chances of success are not changing. What is changing is your chances of getting that far in the series (Note... series, not sequence, but I digress).

E.g. if there were loads prizes, sooner or later you're extremely likely to win.

The only way you can be in the 10th draw is if you lost every one of the first 9.

So the total probability of the series is
Chances you win the first OR the second OR the third OR the fourth etc.

But the chances of you winning the second draw means that you also had to lose the first.
And the chances of you winning third draw means that you also had to lose the first and the second.
etc.

thetalker

Hmm I think I'll read over this a few more times to get it, it feels like my brain will eventually click and I'll see why.
I can't really ask for help of the math teacher as these aren't hw Q's and they're busy most days.

Since you already helped I saw a new Q today that I wasn't sure how to do, maybe you can help

"In the equation below, calculate the value of x + y

3^2x - 2^2y = 17 where x and y are natural numbers"

I was trying to use the log rules but couldn't figure out which one. I then thought I could drop the base or something but that doesn't seem possible. Do you have any idea?

pinkbear

I did it this way, as I can't figure out a way to do it neatly with logs either.

Notice that 3^2x - 2^2y is actually a difference of squares.

(3^x)^2 - (2^y)^2 = 17
(3^x - 2^y)(3^x + 2^y) = 17

As 17 is a prime number =>
3^x - 2^y = 1 and 3^x + 2^y = 17

Trying the first few numbers, and we know that x and y can't possibly be very big, we see:
3^2 - 2^3 = 1 and 3^2 + 2^8 = 17

Or by simultaneous equations: 2.3^x = 18
3^x = 9
x = 2

If you were stuck, you could also graph it on desmos or geogebra, and look through the graph until you see a natural number value of x and y.

Keem 'em coming!

Ponguin

In financial maths regarding loans, pensions etc when do you use present value and when do you use future?

pinkbear

Ponguin wrote: »

In financial maths regarding loans, pensions etc when do you use present value and when do you use future?

P means present / before / principal
F means future / after

Use F=P(1+i)^t when you want to find the future value of money, e.g. how much an investment will amount to at some point in the future. E.g. If I invest 10,000 now how much will it be worth in 5 years time? If I put in 400/month in a pension, how much does it amount to?

Use P = F / (1+i)^t when you want to find the present value of a future investment or amount. E.g. how much do I have to invest NOW so that I have 10,000 in 5 years time? If I want to get 1000/month from a pension fund what value has to be in the fund on my retirement date?

Very often the two are combined in a question, e.g. in a pension question. Someone is investing every month into a pension, then withdrawing every month from it. Use the F=.... for the first part, and P=.... for the second part.

thetalker

Ah once again that would never have crossed my mind because I was so focused on the log rules. Guess you really do have to analyze the Q. Anyway since you said keep it coming here's one my teacher couldn't do :P We thought substitution might be what was needed but it didn't seem to work.

"Let x and y have opposite signs and satisy the following equations;

x + y + xy = -2
And
x^2 + y^2 +(xy)^2 = 64

Find the numerical value of x+y"

Oh and here's another like that which also stumped us. I have a feeling we need to apply the same line of thinking as the other though.

x and y are positive integers, if
sqrt x + sqrt Y = sqrt 333, find x+y

These are probably the last Q's I'll be sending here for a while

pinkbear

Hi thetalker,

I haven't had much time, but I haven't been able to solve either of those questions. I can get the first one by popping it into desmos or geogebra or something like that - x=-1.162, y=5.62. But I'm afraid I can't figure out how you could get it mathematically.

PinkBear

DaleyBlind

Question about percentiles.
My maths book is saying two different things. For example, 20 students in a class. If I got the 5th best score on a test what is my percentile. In the example the the book, they say you count the amount that did worse than you, in this case 15. However later on in the problems there are questions on percentiles and the solutions say you count your position, in the case 16th worst.

So in the first case the answer is 1500/20 = 75% and the second is 1600/20 = 80%. Clearly, at least on is wrong. But I'm thinking that the correct thing to do is (15.5/20) x 100, because when calculating percentiles if you get the answer 15, you get the average of 15 and 16, and if you get 15.5 you move it up to 16, so the correct thing to do is put 15.5 into the formula. The inconsistency in the book has slightly confused me and I was just wondering what the correct procedure is. Sorry if anything I said is unclear because I went on for a bit.

thetalker

Thats fine, youve been a great help anyway with the solutions to the other Q's!

pinkbear

DaleyBlind wrote: »

Question about percentiles.
My maths book is saying two different things. For example, 20 students in a class. If I got the 5th best score on a test what is my percentile. In the example the the book, they say you count the amount that did worse than you, in this case 15. However later on in the problems there are questions on percentiles and the solutions say you count your position, in the case 16th worst.

So in the first case the answer is 1500/20 = 75% and the second is 1600/20 = 80%. Clearly, at least on is wrong. But I'm thinking that the correct thing to do is (15.5/20) x 100, because when calculating percentiles if you get the answer 15, you get the average of 15 and 16, and if you get 15.5 you move it up to 16, so the correct thing to do is put 15.5 into the formula. The inconsistency in the book has slightly confused me and I was just wondering what the correct procedure is. Sorry if anything I said is unclear because I went on for a bit.

Hi,

The confusion arises because there isn't an agreed on definition, and the books are inconsistent. E.g. Consise maths doesn't give formula, and active maths says that 75th percentile is upper quartile, but then goes on to say the formula to use to get percentile is:
Percentile = 100x/(total number of values)
Where x is the number of students who had a smaller score than you.
By upper quartile you get 15.5/20 =77.5
By formula you get: (100)(15)/(20) = 75

Either is fine, and unlikely to be asked, as syllabus says that you need to understand about how percentiles relate to relative standing of results only.

DaleyBlind

pinkbear wrote: »

Hi,

The confusion arises because there isn't an agreed on definition, and the books are inconsistent. E.g. Consise maths doesn't give formula, and active maths says that 75th percentile is upper quartile, but then goes on to say the formula to use to get percentile is:
Percentile = 100x/(total number of values)
Where x is the number of students who had a smaller score than you.
By upper quartile you get 15.5/20 =77.5
By formula you get: (100)(15)/(20) = 75

Either is fine, and unlikely to be asked, as syllabus says that you need to understand about how percentiles relate to relative standing of results only.

Thanks a lot! Can't say I'm huge fan of certain parts of statistics, as it doesn't follow basic logic as most of the other chapters in LC maths do. But you definitely cleared that one up for me.

Account Number

Can someone help me with part (ii) prove logbase5 24 = (3a + b) over (1-a)
when a = logbase10 2 and b = log base10 3?? I have up to logbase10 24 over logbase10 5 and am lost cos the change of base doesn't work?? Sorry about the way I laid it out in the post,have no clue where half the buttons are on a laptop keyboard.
https://scontent-fra3-1.xx.fbcdn.net/v/t34.0-0/p280x280/16468846_733399326814859_351788346_n.jpg?oh=143de0052eab1bb6fcc19e4c73f2f89b&oe=589523DD