Author Topic: Additional Math Help HERE ONLY...!  (Read 64690 times)

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #15 on: May 29, 2009, 07:58:54 am »
dont see those questions coming for additional maths :S

have no idea how to solve them
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Offline Padapop

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Re: Additional Math Help HERE ONLY...!
« Reply #16 on: May 29, 2009, 12:17:14 pm »
:D

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #17 on: May 29, 2009, 02:03:53 pm »
Closing speed is68-8 equals 60 so t equals d/v=2.5

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #18 on: May 29, 2009, 07:15:00 pm »
Q. A train 150m long is running with a speed of 68mph. In whate time will it pass a man who is running at 8mph in the same direction which the train is going.

Q. A and B undertake to do a peice of work for $600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C
     they finish it in 3 days. Find the share of each??
hope u help..

man a does $100 of work a day so in 3 days earns $300
man b does $75 dollars of work a day so earns $225

Mean c earns $600 -$225-$300=$75

ans is 300:225:75 or 4:3:1

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #19 on: May 29, 2009, 08:22:24 pm »
Closing speed is68-8 equals 60 so t equals d/v=2.5

how can u be sure that they start from the same position?

isnt this relative velocity?

Surely if The train starts off at lets say (10i + 20j) and has a velocity of 68m/s

Its position would be (10i + 20j) + 68*t


Isnt the whole thing dependent on position?

i dont know?  ??? could you explain
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #20 on: May 29, 2009, 09:06:13 pm »
They start 150 apart and mud in same direction. The one behind catches up at a seed of 60 so t=150/60

Offline lil^$tar

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Re: Additional Math Help HERE ONLY...!
« Reply #21 on: May 29, 2009, 10:41:47 pm »
can any1 help me with relative velocity...i dont understand it
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Offline ramezamgad

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Re: Additional Math Help HERE ONLY...!
« Reply #22 on: May 30, 2009, 01:21:36 am »
it simply means the velocity of 1 particle relative to another 1
there are only 2 probabilities
both are moving in the same direction so their velocities are subtracted
they are moving towards each other so their velocities are added

waiting for ur reply lil^$tar

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #23 on: May 30, 2009, 04:36:59 am »
The relative velocity is the rate at which the distance is increasing or decreasing

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #24 on: May 30, 2009, 11:09:05 am »
hey got it...tks astar..
 :)

@astarmathandphysics
do u have the timetable for igcse november 2009 final..

if so can u plz upload it here.,.
or send it to eddie_adi619@hotmail.com
thanks in advance
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Offline sweetsh

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Re: Additional Math Help HERE ONLY...!
« Reply #25 on: May 30, 2009, 11:11:38 am »
It is not published, too early, when iit's published  it appears on CIE's website.

Offline Anonymous

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Re: Additional Math Help HERE ONLY...!
« Reply #26 on: May 30, 2009, 11:15:45 am »
hey, does anyone have the Oct/Nov 08 Examiner's report?

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #27 on: May 30, 2009, 11:17:12 am »
but the cie website says..it has been uploaded in the cie direct which requires login

i got hold of the provisional one..but no final??
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Offline lil^$tar

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Re: Additional Math Help HERE ONLY...!
« Reply #28 on: May 30, 2009, 05:26:12 pm »
it simply means the velocity of 1 particle relative to another 1
there are only 2 probabilities
both are moving in the same direction so their velocities are subtracted
they are moving towards each other so their velocities are added

waiting for ur reply lil^$tar
ok thanx can u giv me a question on it so dat i cud see how gud i understand it
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Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #29 on: May 30, 2009, 07:57:41 pm »
Hi,

i was doing the following question:

The function f(x) = [ (ex + 1) / 4 ] for the domain x ? 0


i) Obtain an expression for f-1
ii) State the domain and range for f-1

The first part was easy...
i got the answer, and when i checked it with the mark scheme i got the right answer, which was:
ln(4x-1)

The second part of my answer didn't match the mark scheme, and i dont seem to understand the why.

The domain, according to me should be:

x > 1/4

The range should be:

f-1(x) > -3.2188

However, the mark scheme says that the answers are:

DOMAIN: x > 1/2
RANGE: f-1(x) > 0

could some one please explain my mistake?
« Last Edit: May 30, 2009, 08:19:00 pm by vanibharutham »
A genius is 1% intelligence, 99% effort.