Mathematics: Post your doubts here!

[rz123]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

post your question or visit http://www.examsolutions.co.uk , mymaths.co.uk

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

URGENT HELP NEEDED FOR No. 9 and 10

Thank You in anticipation =)

Q9) Just select the desired formula and plug in the values:




Q10)
I already solved this question before in the same thread:
viewtopic.php?f=26&t=10916&start=10

However, i did this again because i figured it out late. lol!

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

HEy!! Can u plz help me out , i dont know a shit about Permutation and Combination...

i really need help

Here:
http://www.youtube.com/watch?v=2MSiQ3efT3c
http://www.youtube.com/watch?v=oQpKtm5T ... ure=relmfu

also look at the related videos on the right side.

 

[student92]

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please answer this question
describe the main features and uses of a box plot 2 marks
thankyou in advance

 

[bamteck]

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HELP =(

 

[bamteck]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

HELP =(

 

[IgoforA]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

can someone help me with no 7 part iii) which is the least value?? o/n 2007 p3

 

[rz123]

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please someone help me solving Q.7 P4 M/J 2006. i and ii part , M/J 2008 M1 Q.5 and Q.7 i part, please mention steps clearly..thanks a lot :| :| :|

http://www.xtremepapers.com/CIE/Interna ... 8_qp_4.pdf

 

[Scafalon40]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

if h is a function where h(x)=4x+1 and g[h(x)]=4x+2 find the function g(x)

through quick deduction you can probably figure out that g(x)=x+1

but what is the proper process of solving this?

in need a step by step guide for this
thank you
btw an answer today will be appreciated

 

[mrroland]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Hi scafolon40,

I can't think of a step by step solution to this problem because there is too many variation to gx.
It can have a varying coefficient of x terms with different powers.
It can also be a combination of different x terms.
Hence, I feel that this question is testing deduction skill.

My two cents' worth.

 

[Scafalon40]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

that makes sense
i didn't think of it that way
thankyou

 

[rz123]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

where is sir destined007 ? o.0(waiting for ur help bro)

 

[ambitious1]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Hey,
I'd like some notes about: 1- one-one functions; 2- domain; 3- range.
I get them right at times, but it always confuses me. I'll be of great help if you give me an explanaiton with examples
Thanks.

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

HELP =(

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

can someone help me with no 7 part iii) which is the least value?? o/n 2007 p3

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

please someone help me solving Q.7 P4 M/J 2006. i and ii part , M/J 2008 M1 Q.5 and Q.7 i part, please mention steps clearly..thanks a lot :| :| :|

http://www.xtremepapers.com/CIE/Interna ... 8_qp_4.pdf

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

if h is a function where h(x)=4x+1 and g[h(x)]=4x+2 find the function g(x)

through quick deduction you can probably figure out that g(x)=x+1

but what is the proper process of solving this?

in need a step by step guide for this
thank you
btw an answer today will be appreciated

Try to change gh(x) function to get h(x)

gh(x)=4x+2
gh(x)=4x+1+1 where 4x+1 is h(x)
gh(x)=h(x)+1
g(x)=x+1

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

Hey,
I'd like some notes about: 1- one-one functions; 2- domain; 3- range.
I get them right at times, but it always confuses me. I'll be of great help if you give me an explanaiton with examples
Thanks.

here:
viewtopic.php?f=26&t=10873

 

[1357913579]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

NO. 4 OF THIS PAPER AND TELL ME DID YOU ANSWER THIS QUESTION BY YOURE SELF FIRST TIME OR DID SOMEONE HELP YOU IT WAS HARD FOR ME COULNT ANSWER ANYTHING OF THIS QUESTION

 

[destined007]

Re: Maths help available here!!! Stuck somewhere?? Ask here!

NO. 4 OF THIS PAPER AND TELL ME DID YOU ANSWER THIS QUESTION BY YOURE SELF FIRST TIME OR DID SOMEONE HELP YOU IT WAS HARD FOR ME COULNT ANSWER ANYTHING OF THIS QUESTION

This question is simple.
asinbx+c means that you have to multiply every value of sinx by a and then add c to it. like sin90 is 1 so (1xa)+c. But when you multiple sin0 which is zero like (0xa)+c you will get c. Use this principal to calculate a and c in the question.
so sin0=0
a(sin0)+c=3
a(0)+c=3
c=3

sin90=1
a(sin90)+3=9
a(1)+3=9
a=6

now to calculate b, use period=2pi/coefficient of x
so pi=2pi/b
b=2