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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)
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Kyra receives a 5% commission on every car she sells. She received a $1,325 commission on the last car she sold. What was the cost of the car? Use the proportion part/whole =
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48 more than the quotientvof a number and 64
Proof of: ∫ f(x) + g(x) dx = ∫ f(x) dx + ∫g(x) dx It is also a very easy proof. Assume that F(x) is an anti-derivative of f(x) and that G(x) is an anti-derivative of
in a given figure a,b,c and d are points on a circle such that ABC =40 and DAB= 60 find the measure of DBA
Lines EF and GH are graphed on this coordinate plane. Which point is the intersection of lines EF and GH?
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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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