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Y=θ[SIN(INθ)+COS(INθ)],THEN FIND dy÷dθ.
Solution) Y=θ[SIN(INθ)+COS(INθ)]applying u.v rulethen dy÷dθ={[ SIN(INθ)+COS(INθ) ] dθ÷dθ }+ {θ[ d÷dθ{SIN(INθ)+COS(INθ) ] } => SIN(INθ)+COS(INθ) + θ{ (COS(INθ)÷ θ) - (SIN(INθ)÷θ) } θ is canceled and sin(ln θ ) is also canceled then u will get => 2COS(INθ)
find a quadratic polynomial whose zeroes are 2 and -6.verify the relationship between the coefficients and zeroes of the polynomial
A small square is located inside a bigger square. The length of the small square is 3 in. The length of the large square is 7m. What is the area of the big square if you take out t
2 5 - 6 7 4 6 8 -10 8- 6 1 4
Find out if the following sets of functions are linearly dependent or independent. (a) f ( x ) = 9 cos ( 2 x ) g ( x ) = 2 cos2 ( x ) - 2 sin 2 ( x ) (b) f
Equation s(in Tth second)=u+at-a/2 seems to be dimensionally incorrect.why?
A reliability system consists of 15 independent components. The probability that a component works is 0.9 for each. The system works if at least 11 of the components work. Fi
log2(x^2)=(log2(x))2
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y=Θ[sin(lnΘ)+cos(lnΘ)] dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule) dy/dΘ=2cosΘ.
y=Θ[sin(lnΘ)+cos(lnΘ)]
dy/dΘ=[sin(lnΘ)+cos(lnΘ)] + Θ[cos(lnΘ)-sin(lnΘ)]*1/Θ ---->(Use Multiplication rule)
dy/dΘ=2cosΘ.
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