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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)
Unit circle: The unit circle is one of the most valuable tools to come out in trig. Unluckily, most people don't study it as well. Below is the unit circle with just the first
difference between PERT and CPM
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6987+746-212*7665
Variation of Parameters Notice there the differential equation, y′′ + q (t) y′ + r (t) y = g (t) Suppose that y 1 (t) and y 2 (t) are a fundamental set of solutions for
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A HOSPITAL CURRENTLY ORDERS SALINE AT THE BEGINNING OF EACH MONTH. THIS MONTH, THEY HAD 178 BAGS OF SALINE IN STOCK AND ORDERED 1,277 BAGS. DEMAND FOR SALINE IS NORMALLY DISTRIBUTE
Ask if tanA+sinA=m and m^2-n^2=4 rute mn show that tanA-sinA=n
How can I solve the in-equations? Assist me.
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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