QUOTE(come come ji @ Jul 24 2008, 09:30 AM) [snapback]2990289[/snapback]
koi tou riddle solve kare 
app he kar dain me wait kairng
Full Version: Riddle # 251----->solved
app he kar dain me wait kairng
QUOTE(Cupid Princess @ Jul 23 2008, 01:41 PM) [snapback]2988585[/snapback]
aab life ka kia karein gi?
yah to riddle pir hi dawa ho gaya
mein lahroon case?
haan plz life aab solve akr dein na buhat din ho gey ais riddle mein dimag martay
sure i can do it ..par u sab ko mana pary ga u r not able for all kinda riddle
fir i can jee
me k kalaaf caseyee din din bi me ko deikna tha ab
QUOTE(*Life* @ Jul 24 2008, 12:38 PM) [snapback]2990724[/snapback]
yee riddle hi hai 
ab kis ko nahi ata tu asa hee boly gey na sab
ab kis ko nahi ata tu asa hee boly gey na sab
hum nahin maantay issay riddle
QUOTE(Kashif @ Jul 24 2008, 01:28 PM) [snapback]2990792[/snapback]
hum nahin maantay issay riddle
right
This solution is based on the observation that two sinewaves of equal amplitude but differing phase have a maximum absolute sum where they intersect; this is shown by the fact that their slopes are equal in terms of |dy/dx| at the intersection, and as you move away, the decreasing function has a steeper slope than the increasing one. Any time that happens, you have a maximum sum.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
QUOTE(Silent-Viewer @ Jul 29 2008, 12:14 PM) [snapback]2999383[/snapback]
This solution is based on the observation that two sinewaves of equal amplitude but differing phase have a maximum absolute sum where they intersect; this is shown by the fact that their slopes are equal in terms of |dy/dx| at the intersection, and as you move away, the decreasing function has a steeper slope than the increasing one. Any time that happens, you have a maximum sum.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
link bhi de daitey na,,,,,, jahan se copy paste mara hey
QUOTE(MJB @ Jul 29 2008, 03:27 PM) [snapback]2999406[/snapback]
link bhi de daitey na,,,,,, jahan se copy paste mara hey
copy hee sahi batha tu
abi tak tu kisi nay copy bi nahi mara tha
QUOTE(Silent-Viewer @ Jul 29 2008, 03:14 PM) [snapback]2999383[/snapback]
This solution is based on the observation that two sinewaves of equal amplitude but differing phase have a maximum absolute sum where they intersect; this is shown by the fact that their slopes are equal in terms of |dy/dx| at the intersection, and as you move away, the decreasing function has a steeper slope than the increasing one. Any time that happens, you have a maximum sum.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
So we can say in this case that each amplitude = 0.725A. Then half the phase difference Δθ/2 = arccos(0.725) = 0.759762 rad; Δθ = 1.5195 rad = 87.06 deg = 0.2418λ.
al leaset aik nay tu try ki na 
QUOTE(*Life* @ Jul 30 2008, 07:25 AM) [snapback]3000345[/snapback]
al leaset aik nay tu try ki na chalo acha hay SV nay copy/paste kar liya
waisay agli baar Physics/Chemistry walay riddle post kiyay to aik bhi nahin milay ga copy/paste karnay wala
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