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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 into function?
cos inver
4 friends have 235 marbles and want to share.How many marbles should each friend receive?
performs the mentioned operation and write the answers in standard form. ( -4 + 7 i ) + (5 -10 i ) Solution Actually there isn't much to do here other than add or subt
I have an assignment of set theory, please Explain Methods of set representation.
finding missing values from given triangle diagra m..
The null hypothesis It is the hypothesis being tested, the belief of a specific characteristic for illustration, US Bureau of Standards may walk to a sugar making company along
I have no clue how to do it
simplex methord
Q. How to divide two fractions?If you want to divide two fractions, You invert the second fraction (that means, turn it upside-down) and multiply (change the division to a
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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