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ICSE 10th board maths paper 2013 

P_{1} V_{1} / T_{1 }= P_{2} V_{2} / T_{2} (for an ideal gas)
T_{1} = 27 + 273 = 300 K and T_{2} = 327 + 273 = 600 K. Substituting gives
V_{2} = (P_{1} / P_{2}) (T_{2} / T_{1}) V_{1 }= (1 / 2) (600 / 300) V_{1 }= V_{1}.
Since V_{2 }= V_{1}, volume remains same, i.e., 4 liters.
bsl 1 day ago

ISC Specimen 2011 Mathematics 

Take √x = t,
Then dt/dx = 1/(2√x )
So, dt/dx = 1/2t
Therefore, dx = 2tdt
Replacing √x with t and dx with 2tdt, we get:
∫2tsintdt
Now we can integrate by parts and we get:
2t∫sintdt  ∫[{2d(t)/dt}{∫sintdt}]dt + c
= 2t(cost)  ∫(2(cost))dt
= 2tcost + 2sint
= 2√xcos√x + 2sin√x
vivek_venkat_subramaniam 2 days ago

Mathematics 2012 board paper 

(i) ABCD IS A CYCLIC QUAD
so DCB+ DAB = 180 (sum of opp angles of cyclic quadrilateral is supplementary)
ang DAB = 50
(ii) join DB
angle ADB = 90 (angle in a semicircle is 90 )
in triangle ADB 50 + 90+ angDBA = 180
hence angle DBA = 40
scorewellmarks_2014 2 days ago



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