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Mathematics: Post your doubts here!

S

sessionmay2013

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Hey fellas, if any of you could help me a little with this super simple question, it would be much appreciated! (y)
When it comes to partial fractions in paper 3, I know how to express a rational fraction as partial fractions when the denominator (bottom part of fraction) is
(ax + b)(cx + d)(ex + f),
OR
(ax + b)(cx + d)²

BUT I wouldn't know how to approach a problem where you have to express a fraction with the denominator
(ax + b)(x²+c²)
as partial fractions.
I'm sure it just a simple step I haven't come across to before, so if anyone could explain this to me it would be GREATLY appreciated! Thanks beforehand! Cheers! :)
 
syed1995

syed1995

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sessionmay2013 said:
Hey fellas, if any of you could help me a little with this super simple question, it would be much appreciated! (y)
When it comes to partial fractions in paper 3, I know how to express a rational fraction as partial fractions when the denominator (bottom part of fraction) is
(ax + b)(cx + d)(ex + f),
OR
(ax + b)(cx + d)²

BUT I wouldn't know how to approach a problem where you have to express a fraction with the denominator
(ax + b)(x²+c²)
as partial fractions.
I'm sure it just a simple step I haven't come across to before, so if anyone could explain this to me it would be GREATLY appreciated! Thanks beforehand! Cheers! :)
Click to expand...

the (ax+b)(x^2+c^2) will become (Ax+B)/x^2+c^2 + C/(ax+b) I guess.. I would like you to confirm this from someone else as well before you start using it :) Link me a question and I will be able to help better.. I don't do well in the question with "in terms of"
 
Namehere

Namehere

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Can someone explain how to do Q6 first part?
 

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ZaqZainab

ZaqZainab

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Gehad Mohamed said:
Hello can someone please explain Q 6 (iii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_62.pdf
Click to expand...
ZaqZainab said:
okay so method one which in my attachment is 3
is when Not all next to each other lets take an example
3 Blue pens and 4 green
Find the possible arrangements when not ALL green next to each other
we can have it like this
|||||||
but if its method 2 which is in my attachment 4
you cant have it like this you are suppose to have it
||||||| or any other way in which a green is not next to another green
BUT sometime both the methods work when you have
for example 3 blue and 2 green
just because the number is 2 that's why
Click to expand...
we have a total of 12 tress lets take tress other than hibiscus
so we have 8 tress
8! diffrenent arrangemnts
coming to the hibiscus tree 4 of them have 9 different places
so
9P4
Multiply them
 
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