The Soc for People of 'GRDCT2008' Mk IV Watch

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overclocked
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#6161
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#6161
my sister bought 'David Archuleta's' album today... :s
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Champagne Supernova
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#6162
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#6162
(Original post by ArchedEdge)
lol no-one told me there was a new thread!

I just assumed you all just went your own ways....

looks like it's grown a bit now though eh?
Fair enough....
:cry: I..*sniffles* I don't think I could do that :cry:

Well, a little. Grown in thread numbers..not in size
Malsy
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#6163
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#6163
i know that name but can't put a face to it. lol @ your brother.
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jaz_jaz
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#6164
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#6164
(Original post by ArchedEdge)
Well hello guys....lol :p:

I never realised you made a new thread all over again!
jeeesh is that who i think it is?!
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overclocked
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#6165
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#6165
is ghosh around?
i can't simplify this expression
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jaz_jaz
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#6166
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#6166
i like david archuleta :ninja:
crusssssssssssssh
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Malsy
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#6167
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#6167
he's.....cute
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ArchedEdge
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#6168
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#6168
(Original post by jaz_jaz)
jeeesh is that who i think it is?!
Some sexy beast by the name of Ed?
Why yes, I think it is :cool:

:ninja:
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clad in armour
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#6169
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#6169
(Original post by bob9001)
is ghosh around?
i can't simplify this expression
what expression
i bet its binomial expansion innit
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jaz_jaz
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#6170
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#6170
(Original post by ArchedEdge)
Some sexy beast by the name of Ed?
Why yes, I think it is :cool:

:ninja:
of course! hows it going shortie? :tongue:


(Original post by Malsi101)
he's.....cute
i didnt actually mean i had a crush on him, i meant the name of his song lol
but hes a cutie yeeees.
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ArchedEdge
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#6171
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#6171
(Original post by jaz_jaz)
of course! hows it going shortie? :tongue:
Oh great, nice to see some things don't change :p:

I've been great thanks! Got back from NZL on a trip for 3 weeks last Tuesday...was amazing!

But I've had so much work to do, and I've got a big maths test on Monday

You?
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laurenlodge
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#6172
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#6172
(Original post by ArchedEdge)
lol no-one told me there was a new thread!

I just assumed you all just went your own ways....

looks like it's grown a bit now though eh?
Well, well, well.
Look who it ain't.


:p:
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clad in armour
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#6173
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#6173
(Original post by lozz2601)
Well, well, well.
Look who it ain't.


:p:
hi lozz
how you been keeping
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laurenlodge
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#6174
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#6174
(Original post by clad in armour)
hi lozz
how you been keeping
Hiya!
I'm alrighty thanks, yourself?
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Unbounded
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#6175
Report Thread starter 9 years ago
#6175
bob:

Spoiler:
Show
general thing for fractions:

 \boxed{\displaystyle\frac{a+b+c}  {d} = \frac{a}{d} + \frac{b}{d} + \frac{c}{d}}

Spoiler:
Show
don't let this confuse you. the following thing is never right, unless d=0: (hence the 'not equals sign')
 \frac{d}{a+b+c} \not= \frac{d}{a} + \frac{d}{b} + \frac{d}{c}


we had
 \displaystyle\frac{x^4 + x - 2}{x^2} = \frac{x^4}{x^2} + \frac{x}{x^2} - \frac{2}{x^2}

can you use some indice rules to finish it now?
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jaz_jaz
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#6176
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#6176
(Original post by ArchedEdge)
Oh great, nice to see some things don't change :p:

I've been great thanks! Got back from NZL on a trip for 3 weeks last Tuesday...was amazing!

But I've had so much work to do, and I've got a big maths test on Monday

You?
eh you know, same same.
since starting a-levels, my appearance now makes me look permanently ill. - bags under eyes & horrible skin from stress. NOT GOOD.
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overclocked
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#6177
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#6177
(Original post by GHOSH-5)
bob:

Spoiler:
Show
general thing for fractions:

 \boxed{\displaystyle\frac{a+b+c}  {d} = \frac{a}{d} + \frac{b}{d} + \frac{c}{d}}

Spoiler:
Show
don't let this confuse you. the following thing is never right, unless d=0: (hence the 'not equals sign')
 \frac{d}{a+b+c} \not= \frac{d}{a} + \frac{d}{b} + \frac{d}{c}


we had
 \displaystyle\frac{x^4 + x - 2}{x^2} = \frac{x^4}{x^2} + \frac{x}{x^2} - \frac{2}{x^2}

can you use some indice rules to finish it now?

thank you so much

!
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natty_d
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#6178
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#6178
night night
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Unbounded
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#6179
Report Thread starter 9 years ago
#6179
sorry ive been absent from the thread over the past few days guys. been really busy.

how is everyone!

Spoiler:
Show
an interesting challenge for mathematicians: found my dad's old a-level maths stuff. found a question which can be solved with C2 level stuff, but is significantly difficult. see the spoiler if you are curious.
Spoiler:
Show
If  -1 < r < 1 and  S_n = \displaystyle\sum_{p=1}^n r^p , find:
a)  S_{\infty} in terms of r

b)  \displaystyle\sum_{n=1}^{\infty} nr^n in terms of n and r

c)  \displaystyle\sum_{q=1}^n S_q in terms of n and r

try writing out the first few terms of the sums for b and c, to see where it's going.

Just to say, this was on the apparently 'easiest' paper of 'Pure A-level Maths'. I'd say it's STEP II level maybe.
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laurenlodge
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#6180
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#6180
I have Haribo¬!!!
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