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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)
identify 4 sketch the quadric surfaces
1+2+3+.....+n=1/2n(n+1)
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how does sin of x equal negative 1/3
We know that one has to deal with numbers in day-to-day life irrespective of his inclination and field of work. Also one cannot refute the fact
Calculate the value of the following limits. Solution From the graph of this function illustrated below, We can illustrate that both of the one-sided limits suffer
Mike sells on the average 15 newspapers per week (Monday – Friday). Find the probability that 2.1 In a given week he will sell all the newspapers [7] 2.2 In a given day he will sel
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I don''t know how to do the next step like if I had 73 divided by 9 wouldn''t 7 go into nine 1 time then you have to do something else but that is the part I don''t understand
matrix
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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