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Mathematics: Post your doubts here!

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Hi all, I would please like to know the solution of question 4 in this paper. I just reached that point where I should leave the question or I will cause myself problem. Thanks in advance :)
cosec2@ = sec@ + cot @
so,
1/sin2@ =(1/cos@) + (cos@/sin@)
1/sin2@ = (sin@ + cos^2@)/(cos@sin@)
1/sin2@ = [sin@ +(1- sin^2@)]/ (cos@sin@)
1/sin2@ = (sin@+1- sin^2@)/(cos@sin@)
i am multiplying right hand side with (2/2) so,

1/sin2@ = 2(sin@+1-sin^2@)/2sin@cos@
1/sin2@ = 2(sin@+1-sin^2@) / sin2@
sin2@ gets cancelled out therefore we are left with:
1 = 2(sin@+1-sin^2@)
1 = 2sin@ +2 - 2sin^2@
2sin^2@ - 2sin@ -1 = 0
now u can find the angle (y)
 
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Best of luck for 2mrw 2 awl of u !!! THNX 4 helping out at this eleventh hour!!!
:D
So yes i m signing out 4 2day!!!
Pray 4 me!

Especially regards to
Esme
iKhaled
PhyZac
knowitall10
Rutzaba
LMGD33
Alice123
SararaIH
VampBhums
..... n all others! :) (Excuse me if i missed any1s name here!!!)

Thank you so much for remembering me among all these amazing people and tagging me. Feels good :)
Insha'Allah the paper will be good and easy for everyone.
Pray for all of us sis :)
 
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@ anyone?

Q10i
da/dt = KV
da/dt = k(4/3πr^3)
da/dt = 4/3πkr^3

A=4πr^2 V = 4/3πr^3

da/dr = 8πr dv/dr = 4πr^3

dr/dt = dr/da x da/dt
dr/dt = (1/8πr) x 4/3πkr^3 substitute r and dr/dt to find K then u will end up with the given equation

Posted by iKhaled previously
 
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Posted by PhyZac previously:

Q110ii)
okay first, move value to make two integration
(let ! be the integration sign)
dr/dt = 0.08r^2
1/r^2 dr = 0.08 dt
! 1/r^2 dr = !0.08 dt
! r^-2 dr = 0.08t +c
-1/r = 0.08t + c
sub values given in question
-1/5 = 0.08(0) + c
c = -1/5

-1/r = 0.08t - 1/5
r = -1 / (0.08 - 1/5)
r = -5 /0.4t - 1


Now for part iii
i am not sure
but what i think is the answer must be positive.
to do this the value shudnt be smaller than 0, (try calculating)
and even not more than 2.5
i got 2.5 by
0.4t-1 =0
0.4t =1
t= 2.5
 
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