Maths, Addmaths and Statistics: Post your doubts here!

[:: A^]

Can someone help me out with this question from O/N/2011/11

12^1/2
______
3^3/2

ASAP!!

 

[MustafaMotani]

WoW sab ki probs solved hai..

 

[AlphaBeta]

WoW sab ki probs solved hai..

reply above post

 

[MustafaMotani]

only English plz i dont understand

In shearing area of figure does not change, while in stretching it changes.... and the change in area in stretching is k times that stretch factor

 

[kitecrystal]

Can someone help me out with this question from O/N/2011/11

12^1/2
______
3^3/2

ASAP!!

use the concept x^1\2=square root of x.......​

answer is 3\2​

 

[AlphaBeta]

In shearing area of figure does not change, while in stretching it changes.... and the change in area in stretching is k times that stretch factor

there was some perpperpendicular stuff my teacher taught i addition to areas difference

 

[moni]

use the concept x^1\2=square root of x.......​

answer is 3\2 ​

can u please solve?

 

[AlphaBeta]

In shearing area of figure does not change, while in stretching it changes.... and the change in area in stretching is k times that stretch factor

explain da answer for part 23 )b)

 

[:: A^]

use the concept x^1\2=square root of x.......​

answer is 3\2 ​

can you please give step by step instructions pls.

 

[AlphaBeta]

can you please give step by step instructions pls.

dont bother only 1 mark nd it migth not come this time ;notes for indices

 

[MustafaMotani]

reply above post

temperatur changed from 16 to -4 as we go up the cliff
so for every 1m rise change in temperature is 20/3000 = 1/150 C
for 1800m rise change will be 1800 x 1/150 12 C so new temperature = 16-12=4C

ii) when temperature is 0 C it means temp fall by 16 C
so X x 1/150 = 16
X = 0150 x 16 = 2400 m

iii) it is sort of finding the equation ...
cordinate = (3000,-4) (0,16)

eq will be 150y = 2400 - x
and so 150T = 2400 - x

 

[MustafaMotani]

there was some perpperpendicular stuff my teacher taught i addition to areas difference

There maybe, I am not aware of that..

 

[AlphaBeta]

temperatur changed from 16 to -4 as we go up the cliff
so for every 1m rise change in temperature is 20/3000 = 1/150 C
for 1800m rise change will be 1800 x 1/150 12 C so new temperature = 16-12=4C

ii) when temperature is 0 C it means temp fall by 16 C
so X x 1/150 = 16
X = 0150 x 16 = 2400 m

iii) it is sort of finding the equation ...
cordinate = (3000,-4) (0,16)

eq will be 150y = 2400 - x
and so temperature = 2400 - x

thanks

 

[:: A^]

dont bother only 1 mark nd it migth not come this time ;notes for indices

dude?? i dont care whther it comes for this time or not, i just want my doubt cleared... and 1 mark?? Sometimes a mark can make the difference...U dont wanna answer? then simply leave it... Dont go off-topic....

 

[MustafaMotani]

thanks

in last equtaion its not Temperature = 2400 - x
its "150Temperature = 2400 - x "

welcum ..

 

[MustafaMotani]

dude?? i dont care whther it comes for this time or not, i just want my doubt cleared... and 1 mark?? Sometimes a mark can make the difference...U dont wanna answer? then simply leave it... Dont go off-topic....

dont get angry... its simple inidices rule.. do you know indices rules.. ?

 

[MustafaMotani]

dude equation is 16 -X/150

its the same!
waha jo 150T hai us 150 ko next side per divide karo u ll get ur equation

 

[AlphaBeta]

its the same!
waha jo 150T hai us 150 ko next side per divide karo u ll get ur equation

i know i didnt see ur latest postnd about shear nd stretch can u clarify plz

 

[MustafaMotani]

moni
@::A^
(12)^0.5 ko yun b likh saktay hai (2 x 2 x 3) ^0.5 thus 2 ka ek pair hai to take 2 out ull get 2(3)^0.5
3^3/2 = 27^0.5 = (3 x 3 x 3)^0.5 yaha b 3 ka ek pair ban raha hai to take 3 out it will become 3(3)^0.5

now 3^0.5 are cancelled leaving 2/3

tumhe ye tab he samajh aega if you know how to take out numbers from square roots

 

[AlphaBeta]

anyone explain sin cos for paper 1