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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)
A drug has a decay rate of k = - ¼ ln(¾) / hr. How soon after an initial dose of 1600 mg will the drug reach its minimum therapeutic value of 900 mg in the body?
If we "break up" the root into the total of two pieces clearly we get different answers. Simplified radical form: We will simplify radicals shortly so we have to next
find k,is -2 a root of the equation 3x2
Hi, I''m looking for assistance/solutions to individual questions. I''ve already answered them but seek confirmation my answers are correct. I don''t want answers to a complete e
Utilizes the definition of the limit to prove the given limit. Solution Let M > 0 be any number and we'll have to choose a δ > 0 so that, 1/ x 2 > M
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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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