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yeeyong Female
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  #61 Old 13-04-2009 Default Re: Stpm 2009

i'm so sorry that i didn't realise i typed 2 instead of 1....
is the question 1 i don know how....
and to my knowledge, if not mistaken the question 2, that essay writing, all r writing bout facts rite??? correct me if i'm wrong..

how to interpret the graph and the table? i have read questions that contain few graphs and table in one question... how to start? how to write?
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sugarspice Female
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  #62 Old 14-04-2009 Default Re: Stpm 2009

Yes for essay part 1 we have to INTERGRATE the stimuli given besides identifying major trends.

Sometimes it's really tough as we can't see the link between the stimuli, and so we can't intergrate much...

the essay in 2nd part is always the argumentative type.

Last edited by sugarspice; 14-04-2009 at 05:40 AM.
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lawteoh Male
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  #63 Old 15-04-2009 Default Re: Stpm 2009

Anyway, if you guys free, try out this question. Maths question. I ought to be able to do it, but somehow got stuck.. actually, just helping my friend out only.. lolz.

log(base2)y = 3 + log(base2)x + log(base2)(x+1)

It says to write X in terms of Y...

I got until here.

y/ [x(x+1)] = 8

Guess there must be another special method to do it..
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tsar Male
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  #64 Old 15-04-2009 Default Re: Stpm 2009

Very interesting question lol. I'm no maths expert, but may I suggest

x = (-8 +- (64+32y)^1/2)/16

Last edited by tsar; 15-04-2009 at 01:00 PM.
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lawteoh Male
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  #65 Old 15-04-2009 Default Re: Stpm 2009

Quote:
Originally Posted by tsar View Post
Very interesting question lol. I'm no maths expert, but may I suggest

x = (-8 +- (64+32y)^1/2)/16
lolz.. u actually square root it? can wan meh? haha.. i nv see this in my stpm book also last time.. mayb the ques is type wrongly la..
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uoykcuf
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  #66 Old 15-04-2009 Default Re: Stpm 2009

use (-b+-(sqrt(b^2-4ac)))/2a after got y/ [x(x+1)] = 8
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lawteoh Male
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  #67 Old 15-04-2009 Default Re: Stpm 2009

Quote:
Originally Posted by uoykcuf View Post
use (-b+-(sqrt(b^2-4ac)))/2a after got y/ [x(x+1)] = 8
If use ur method,

8 x^2 +8x -y= 0

and after the formula , u will get x = [-8 +- sqrt (64+32y)] / 16 .. got the y in the square root.. can solve meh? hmm..
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tsar Male
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  #68 Old 15-04-2009 Default Re: Stpm 2009

Lol his method is my method.

I thought the requirement of the question was to put x in terms of y? If you want to go for a solution you'll need 2 equations for 2 unknowns.
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uoykcuf
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  #69 Old 15-04-2009 Default Re: Stpm 2009

Quote:
Originally Posted by lawteoh View Post
If use ur method,

8 x^2 +8x -y= 0

and after the formula , u will get x = [-8 +- sqrt (64+32y)] / 16 .. got the y in the square root.. can solve meh? hmm..
ya can
if you don't want use tat formula, you can use completing the square which will give you same thing.
y/ [x(x+1)] = 8
y=8[x(x+1)]
y=8((x+1/2)^2-1/4)
.
.

Quote:
Originally Posted by tsar View Post
Lol his method is my method.

I thought the requirement of the question was to put x in terms of y? If you want to go for a solution you'll need 2 equations for 2 unknowns.
i also just know that

Last edited by uoykcuf; 15-04-2009 at 03:29 PM. Reason: Automerged Doublepost
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markwongsk Male
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  #70 Old 15-04-2009 Default Re: Stpm 2009

log(base 2)y = 3 + log(base 2)x + log(Base 2)(x+1)
log(base 2)y = log(base 2)[(8x)(x+1)]
y/8 = x(x+1)
y/8 = x^2 + x
y/8 = x^2 + x + 1/4 - 1/4
y/8 = (x+1/2)^2 - 1/4
(y/8 + 1/4)^1/2 = x + 1/2 [y>0 since it's a logarithm]
therefore

X = (y/8 + 1/4)^1/2 - 1/2
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