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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)
The probability that a leap year will have 53 sunday is ? and how please explain it ? (a)1/7 (b) 2/7 (c) 5/7 (d)6/7 Sol) A leap year has 366 days, therefore 52 weeks i.e
Minima, Maxima and points of inflexion a) Test for relative maximum Consider the given function of x whose graph is presented by the figure given below
Diffrent type of rectillinar figure..
Trace the curve (x/a)^3/2+(y/b)^2/3=1
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For a population with a mean of μ=70 and a standard deviation of o=20, how much error, on average, would you expect between the sample mean (M) and the population mean for each of
1+2i=a+ib so find a and b
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is this free for LIFE that means forever never ever going to pay
1. Solve the right triangle. B = 135 c = 3.72 A ≈ ____° (round to the nearest tenth as needed) 2. Solve the right triangle, where a =4 and b =10 The length of
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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