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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 radiograph is made of an object with a width of 3 mm using an x-ray tube with a 2 mm focal spot at a source-to-film distance of 100 cm. The object being imaged is 15 cm from the
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Question: Use Gauss elimination method to solve the following system of equations. -y +3z=4 2x-y-2z= 2 2x-2y+z =6 4x-y-7z= 0
Two angles are supplementary. The evaluate of one is 30 more than twice the measure of the other. Determine the measure of the larger angle. a. 130° b. 20° c. 50° d. 70
sin30+cos30=
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what is 1/5 + 1/8 equals?
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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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