Author Topic: Additional Math Help HERE ONLY...!  (Read 75556 times)

Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #255 on: June 02, 2010, 05:46:40 pm »
yeah, if u get more than one value it basically tells u that the particle passes the origin at more than one point, i.e. it moves first in the positive direction and then in the negative direction, and then u have to find the distance travelled in the positive and the negative direction, and theres also one more way to find if the particle goes in the positive direction, if u find the maximum displacement and it comes to a value not equal to ur final displacement, i.e. in this q if the maximum displacement is not -25m then u get to knw that the particle has travelled in the positive direction also, for example in this u get the maximum displacement at t=5, i.e. x=7.5m, thus u get to knw that the particle has also travelled in the positive direction and u can take the movement in the positive direction into account

hahaha, okay! thanks ;D

x - q = px / y

(x - q) / x = (px / y) * (1 / x)

(x - q) / x = p / y

1 - q / x = p / y

y = p / (1 - q / x)

would it works ?

lol sry umm ..checkk teh ms..  :-\
You said I must eat so many lemons cause I am so bitter.

Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #256 on: June 02, 2010, 05:52:34 pm »
No idea lol, but I don't presume those sort of Qs would appear in this year's paper, somewhat beyond our scope ...
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Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #257 on: June 02, 2010, 05:56:28 pm »
No idea lol, but I don't presume those sort of Qs would appear in this year's paper, somewhat beyond our scope ...

LOL? are you sureeeee.. ? hahaha thanks anyways! ;) ;)
You said I must eat so many lemons cause I am so bitter.

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #258 on: June 02, 2010, 06:01:46 pm »
px = xy - qy

qy  = xy-px

qy = x(y-p)

x = (qy)/(y-p)

(reciprocal of both sides)

1/x = (y-p)/(qy)

1/x = y/qy - p/qy

1/x = 1/q - (p/q)(1/y)

1/x = (1/y)(-p/q) + 1/q

  y  =  xm + c 

y = 1/x

x = 1/y

m = (-p/q)

c = (1/q)
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Offline cooldude

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Re: Additional Math Help HERE ONLY...!
« Reply #259 on: June 02, 2010, 06:04:27 pm »
hahaha, okay! thanks ;D

lol sry umm ..checkk teh ms..  :-\

the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p

now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it

now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
 
and blah blah blah


Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #260 on: June 02, 2010, 06:10:50 pm »
the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p

now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it

now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
 
and blah blah blah


:o :o :o ... which table did you get those values from?  :o :o :o lol
You said I must eat so many lemons cause I am so bitter.

Offline J.Darren

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Re: Additional Math Help HERE ONLY...!
« Reply #261 on: June 02, 2010, 06:13:18 pm »
:o :o :o ... which table did you get those values from?  :o :o :o lol

The marking scheme thingie ...
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Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #262 on: June 02, 2010, 06:13:29 pm »
Cooldude went backwards, check my post , ive gone forwards. the same thing....
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Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #263 on: June 02, 2010, 06:15:05 pm »
the first one is impossible, the one above the table, ill show u the explanation (ill go backwards)-->
1/y=-9/px+1/p
reciprocate--> y=-px/9+p
write as single fraction--> y=(-px+9p)/9
cross multiply--> 9y=9p-px
rearrange--> 9y+px=9p

now the first one in the table-->
y=qy/x+p
now just multiply evrything by x ull get it

now the 2nd one-->
y/x=y/q-(p/q)
multiply evrything by q--> qy/x=y-p
cross multiply--> qy=xy-px
rearrange ull get it
 
and blah blah blah
The marking scheme thingie ...
Cooldude went backwards, check my post , ive gone forwards. the same thing....

OH. okay, THANKS GUYS! thaaaank you so much :P
You said I must eat so many lemons cause I am so bitter.

Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #264 on: June 03, 2010, 06:09:33 pm »
lol.. hi again :)
i need help with 8ii) D:
thanks :) :)!
You said I must eat so many lemons cause I am so bitter.

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #265 on: June 03, 2010, 06:51:54 pm »
INCOMPLETE Q...post the paper link and ms..
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #266 on: June 03, 2010, 07:37:39 pm »
answer to the venn diagram question

Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #267 on: June 03, 2010, 08:45:25 pm »
lol.. hi again :)
i need help with 8ii) D:
thanks :) :)!
Check the attachment. Hope it helps =]

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #268 on: June 03, 2010, 10:03:03 pm »
it says C intersection d=165000

Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #269 on: June 03, 2010, 10:12:36 pm »
it says C intersection d=165000
nope, it says people who don't have both a comp and a dishwasher so that would be everything outside of (C intersection D) i think...
ok, let me go check the mark scheme and i'll get back to you!