Author Topic: Additional Math Help HERE ONLY...!  (Read 75515 times)

Offline shan2391

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Re: Additional Math Help HERE ONLY...!
« Reply #165 on: June 04, 2009, 11:11:17 pm »
guys help Q.2 in this paper(attached)

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #166 on: June 04, 2009, 11:34:49 pm »
guys help Q.2 in this paper(attached)

c=3 since (0,0) nusves to (0,3) b=1 since there is no scaling in the x direction and c=2 since tanpi/4=1 but 5-3=2 so there is a scaing factor 2 in the y direction
« Last Edit: June 04, 2009, 11:37:41 pm by astarmathsandphysics »

Offline divineobsidian

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Re: Additional Math Help HERE ONLY...!
« Reply #167 on: June 05, 2009, 12:07:20 am »
Quote

yess =)

so the eqn will be joined together, since y is the common subject
becomes x2-5x+18=mx+2
then bring everything over
x2-5x-mx+18-2=0
x2-x(5+m)+16=0
so here the components of the quad eqn is fulfilled =)
so ax2+bx+c=0
a=1 b=(5+m) c=16

since tangent never touches the curve,
we use b2-4(a)(c) < 0
so substitute....
(5+m)2-4(1)(16)<0
25+10m+m2-64<0
m2+10m-29<0

solve quadratically, im sure u'll get ure answer  :D

k thanks arc sorry for long reply i was sleeping in asia

edit:yeah By the way vanib i did OR on the P1 but i think i kinda messed up
« Last Edit: June 05, 2009, 12:20:26 am by divineobsidian »

Offline Sweet_03

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Re: Additional Math Help HERE ONLY...!
« Reply #168 on: June 05, 2009, 04:42:54 am »
I have a Q on vectors

If You Have a vector AB
and You need to find th unit vector
is this the right formula ?

1 / |AB| x AB

and to find out the magnitude ..
SQUARE ROOT of a2 + b2

Is this Right ?

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #169 on: June 05, 2009, 05:19:24 am »
yes correct Sweet_03
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Offline Sweet_03

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Re: Additional Math Help HERE ONLY...!
« Reply #170 on: June 05, 2009, 05:33:36 am »
aha ! Finally i got it :-DD
Thnnnx aLot

and Good Luck everyone

Offline vanibharutham

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Re: Additional Math Help HERE ONLY...!
« Reply #171 on: June 05, 2009, 06:00:31 am »
i have a feeling realtive velocity might shwo up :( im scared  :'(
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Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #172 on: June 05, 2009, 07:04:37 am »
guys help Q.2 in this paper(attached)

c=3 since (0,0) nusves to (0,3) b=1 since there is no scaling in the x direction and c=2 since tanpi/4=1 but 5-3=2 so there is a scaing factor 2 in the y direction
how do u find the value for a and b?"??
plzz explain clearly
« Last Edit: June 05, 2009, 07:11:39 am by Adi...\m/ »
divine intervention!

Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #173 on: June 05, 2009, 07:12:08 am »
I meant a=2

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #174 on: September 25, 2009, 10:38:29 am »
y is b=1....didnt get "there is no scaling in the x-direction" ... any other method
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nid404

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Re: Additional Math Help HERE ONLY...!
« Reply #175 on: September 25, 2009, 10:43:27 am »
To a cyclist travelling due south on a straight horizontal road at 7 m/s, the wind appears to be
blowing from the north-east. Given that the wind has a constant speed of 12 m/s, find the direction
from which the wind is blowing.

Have a look aadi.....if you get the other part lemme know

Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #176 on: September 25, 2009, 10:45:18 am »
shudnt 135 come between 7 and 12?
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nid404

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Re: Additional Math Help HERE ONLY...!
« Reply #177 on: September 25, 2009, 10:49:45 am »
shudnt 135 come between 7 and 12?


cos inverse of 7/12 =/ 135...equal to abt 55


Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #178 on: September 25, 2009, 11:20:17 am »
but 135 shud come between 7 and 12 righ?
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nid404

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Re: Additional Math Help HERE ONLY...!
« Reply #179 on: September 25, 2009, 11:24:22 am »
but 135 shud come between 7 and 12 righ?

not possible....how r u saying that??