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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)
Let g be a function from the set G = {1,2,3,...34,35,36). Let f be a function from the set F = {1,2,3,...34,35,36}. Set G and F contain 36 identical elements (a - z and 0 - 9).
The two sides of a triangle are 17cm and 28cm long, and the length of the median drawn to the third side is equal to 19.5 cm. What is the distance from an endpoint of the median to
I need marketing management sample assignment as a guide
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(a) An unordered pair fm; ng with 1 ≤ m ≠ n ≤ 6 is called a duad. List the 15 duads. (b) There are 15 ways to partition {1, ......, 6 } into 3 duads, such as { {1; 2}, {3, 4},
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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