Maths, Addmaths and Statistics: Post your doubts here!

[Minato112]

Yeah but you'll be getting 8sinycosy=-3. So can you solve that WITHOUT USING any double angle identities???

Im srry, divide everywhere by cos x.

U get, 8 tan x + 3 sec^2 x = 0

Use the identity : tan^2 x + 1 = sec^2 x and solve the problem.

Hope it helps.

 

[Mustafa Malik]

can someone help me wd question 17 of paper 1 2000 novemeber??? its in the topical also.... plz any1????

 

[Dashing Cancerian]

can someone help me wd question 17 of paper 1 2000 novemeber??? its in the topical also.... plz any1????

4024???if yes then plz post the question...

 

[marshall mathers]

Im srry, divide everywhere by cos x.

U get, 8 tan x + 3 sec^2 x = 0

Use the identity : tan^2 x + 1 = sec^2 x and solve the problem.

Hope it helps.

Mate but some solutions don't satisfy the equation....Thanks though

 

[leosco1995]

You need to specify which one is angle x and which one is angle y. Is x the angle for A or the angle for B??

lol. Thanks for the help buddy, but I asked that question and gave my Add Math examination 1.5 years ago.

 

[marshall mathers]

lol. Thanks for the help buddy, but I asked that question and gave my Add Math examination 1.5 years ago.

lol man didn't check the year!!!Doesn't matter cuz i'll thank you because you gave me the chance to revise this topic as my exams are on the 8 and 15 nov 2012

 

[Minato112]

Mate but some solutions don't satisfy the equation....Thanks though

Srry, didnt get U... If U want, I can solve it for U.

 

[marshall mathers]

Srry, didnt get U... If U want, I can solve it for U.

Nope actually i wrongly replaced one solution in the equation. Thanks man

 

[Haris Bin Zahid]

Calculate how many different 5 digit numbers can be formed from the 9 digits 1,2,3,4,5,6,7,8,9 used without repetition. In how many of these 5 digit numbers will the digit 8 and 9 be adjacent?
I need help in the second part of the question.
marshall mathers?

 

[Mustafa Malik]

Ok but its wd a diagram..........
but still i will try.
ABCD is a quadrilateral wd AB parallel to DC. AC and BD meet at X where CX= 8 cm and XA = 10 cm. (a) given that BD= 27 find BX.
(b) ratio of area of triangle BXC: area of triangle AXD,

sorry my scanner isn;t working well ans are 15 cm and 1:1

 

[Iffat]

Can sum1 plz explain 2 me how to do maths (4024) 2011 oct-nov p11 q24(d)(ii), n 2011 oct-nov p12 q27(b)(ii).
plz reply asap xam iz 2moro!!

 

[Minato112]

Can sum1 plz explain 2 me how to do maths (4024) 2011 oct-nov p11 q24(d)(ii), n 2011 oct-nov p12 q27(b)(ii).
plz reply asap xam iz 2moro!!

Plz post the questions.

 

[Iffat]

Plz post the questions.

the question is 4rm a diagram. if u can jus download the paper 4rm here only

 

[Minato112]

the question is 4rm a diagram. if u can jus download the paper 4rm here only

P12 - Qu. 24 (b) (ii) doesn't exist. The same applies for the other question. Plz be more precise.

 

[marshall mathers]

Calculate how many different 5 digit numbers can be formed from the 9 digits 1,2,3,4,5,6,7,8,9 used without repetition. In how many of these 5 digit numbers will the digit 8 and 9 be adjacent?
I need help in the second part of the question.
marshall mathers?

Well first thing you want to do is to put 8 and 9 adjacent. Now you are left with 3 numbers to arrange from 7. To arrange these 3 numbers it will be 7P3. Now it has not specified any order for the 8 and 9 as in whether the 8 is first or the 9 but the question has just said that they should be adjacent. So it can be 89 or 98. Therefore you have 2 ways in which you can arrange the 8 and 9 next to one another.

Since the 8 and 9 should be adjacent you count them as being only ONE element by enclosing both numbers inside a SINGLE square . Now taking the last sentence into consideration, you'll now have to permute or arrange these 4 elements. Why four? Simply because the 8 and 9 are now assumed to be one element. This square can be placed in 4 spaces(see below) . The 8 and 9 are in bold just to show you that they are ONE element. It can also be 98 but don't bother about this. So you can have the following possibilities:

89_ _ _
_89_ _
_ _89_
_ _ _89

So the this one element can be placed in 4 ways. So your answer will be 2x7P3x4
2- because you can arrange the 8 and 9 in 2 ways
7P3-because since the 8 and 9 are included in the 5 digit number, you are left with 3 more digits to be arranged from 9-2=7 digits
4-because you can place this 89 or 98 in 4 places as illustrated above.


A common mistake would be to multiply by 4! instead of 4 but you don't have to multiply by 4! as you are arranging only this single element and not all the digits since you've already arranged the other 3 in 7P3 ways.

 

[marshall mathers]

Calculate how many different 5 digit numbers can be formed from the 9 digits 1,2,3,4,5,6,7,8,9 used without repetition. In how many of these 5 digit numbers will the digit 8 and 9 be adjacent?
I need help in the second part of the question.
marshall mathers?

Also in some questions it can ask you in how many ways can these numbers be arranged such that there is exactly one digit in between the 8 and 9. This is slightly more demanding but same thing here. To proceed you enclose the 8_9 or 9_8 inside a SINGLE square and count it as one element. But here the permutation will be different. You can try it out and give me the answer. It's always good to look for other questions within a question!

 

[Haris Bin Zahid]

Also in some questions it can ask you in how many ways can these numbers be arranged such that there is exactly one digit in between the 8 and 9. This is slightly more demanding but same thing here. To proceed you enclose the 8_9 or 9_8 inside a SINGLE square and count it as one element. But here the permutation will be different. You can try it out and give me the answer. It's always good to look for other questions within a question!

Really appreciate your help. Thanks a lot.
So is this the right answer to your question: 7P3 x 3 x 2 ?

 

[marshall mathers]

Really appreciate your help. Thanks a lot.
So is this the right answer to your question: 7P3 x 3 x 2 ?

Well the 8_9 is one element. You can interchange the 8 and 9 i.e it can be 8_9 or 9_8. Now the blank between the 8 and 9 can be filled by only one number from seven so 7P1 and the remaining 2 can be arrange from 6 so it will be 6P2. Now there are 3 ways in which the 8_9 can be placed

8_9_ _
_8_9_
_ _8_9

So your answer will be 7P1 x 6P2 x 2 x 3 which is correct!

Well i made a slight mistake at first my bad!

 

[Arsallan]

After the initiative taken by Math_angel in the IGCSE section, I have made this topic for students of O'levels. You can post any question or any problem here and me along with other XPF users will try to help you as soon as possible. This topic is a feasible option as it won't bring a hassle of making new topics for every question and it will also allow the helpers to not miss any of your questions and have one common topic for answering your Mathematics queries.

So start posting your problems.

i forget the process of question how could i memorize it.

 

[Haris Bin Zahid]

Well the 8_9 is one element. You can interchange the 8 and 9 i.e it can be 8_9 or 9_8. Now the blank between the 8 and 9 can be filled by only one number from seven so 7P1 and the remaining 2 can be arrange from 6 so it will be 6P2. Now there are 3 ways in which the 8_9 can be placed

8_9_ _
_8_9_
_ _8_9

So your answer will be 7P1 x 6P2 x 2 x 3 which is correct!

Well i made a slight mistake at first my bad!

So does this mean I answered it correctly? It's the same as your final answer (7P1 x 6P2 = 7P3)!