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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)
what is harmonic progression
Sketch the feasible region for the following set of constraints: 3y - 2x ≥ 0 y + 8x ≤ 53 y - 2x ≤ 2 x ≥ 3. Then find the maximum and minimum values of the objective
encoded with the matrix -3 -7 and 4 9. what lights up a soccer stadium? ecoded message: {-3 - 7} {3 2 } {3 6} {57 127} {52 127} {77 173} {23 51)
WHAT TWO SIX DIDGIT NUMBERS CAN YOU ADD 984,357
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x=ct,y=c/t d^2/dx^2
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A medical survey was conducted in order to establish the proportion of the population which was infected along with cancer. The results indicated that 40 percent of the population
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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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