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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)
Question: Consider a digraph D on 5 nodes, named x0, x1,.., x4, such that its adjacency matrix contains 1's in all the elements above the diagonal A[0,0], A[1,1], A[2,2],.., e
find the matrix of the linear transformations T:R2->R2 defined by T(x,y,z)=(x+2y,x-3z).
What are some of the interestingmodern developments in cruise control systems that contrast with comparatively basic old systems
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Can you help me with what goes into 54
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In the diagram points V,W,X,Y and Z are collinear, VZ=52, XZ= 20 AND WX=XY=YZ. Find the indicated length of WX, VW, WY, VX, WZ, and VY
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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