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AllQuestion and Answers: Page 26

Question Number 214753 Answers: 0 Comments: 1

Question Number 214750 Answers: 1 Comments: 0

Question Number 214742 Answers: 2 Comments: 2

Question Number 214731 Answers: 1 Comments: 7

TINKU TARA Why are you don′t solve the plot bug? Whenever I type in x^2 and plot from drawing, It got an error.

$$\boldsymbol{\mathrm{TINKU}}\:\boldsymbol{\mathrm{TARA}} \\ $$$$\mathrm{Why}\:\mathrm{are}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{bug}? \\ $$$$\mathrm{Whenever}\:\mathrm{I}\:\mathrm{type}\:\mathrm{in}\:{x}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{plot}\:\mathrm{from}\:\mathrm{drawing},\:\mathrm{It}\:\mathrm{got}\:\mathrm{an}\:\mathrm{error}. \\ $$

Question Number 214729 Answers: 1 Comments: 0

Emergency ((x(x+1))/2)=((2x−3)/3)+1

$$\mathrm{Emergency} \\ $$$$\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{3}}+\mathrm{1} \\ $$

Question Number 214726 Answers: 0 Comments: 0

for the function z = xtan^(−1) ((y/x))+ysin^(−1) ((x/y))+2 then the value of x(∂z/∂x)+y(∂z/∂y)=?

$$\mathrm{for}\:\mathrm{the}\:\mathrm{function}\:{z}\:=\:{x}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)+{y}\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{{y}}\right)+\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\frac{\partial{z}}{\partial{x}}+{y}\frac{\partial{z}}{\partial{y}}=?\: \\ $$

Question Number 214713 Answers: 2 Comments: 0

Find matrix B if given AB=BA= (((0 0)),((0 0)) ) where A= (((5 3)),((5 3)) ) and B ≠ (((0 0)),((0 0)) )

$$\:\:\mathrm{Find}\:\mathrm{matrix}\:\mathrm{B}\:\mathrm{if}\:\mathrm{given}\:\mathrm{AB}=\mathrm{BA}=\begin{pmatrix}{\mathrm{0}\:\:\mathrm{0}}\\{\mathrm{0}\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\mathrm{where}\:\mathrm{A}=\:\begin{pmatrix}{\mathrm{5}\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\mathrm{3}}\end{pmatrix}\:\mathrm{and}\:\mathrm{B}\:\neq\:\begin{pmatrix}{\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$ \\ $$

Question Number 214712 Answers: 2 Comments: 0

$$\:\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$

Question Number 214722 Answers: 1 Comments: 1

Question Number 214719 Answers: 4 Comments: 0

Question Number 214701 Answers: 0 Comments: 2

Hey tinku tara, is the plot bug fixing?

$$\mathrm{Hey}\:\mathrm{tinku}\:\mathrm{tara}, \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{plot}\:\mathrm{bug}\:\mathrm{fixing}? \\ $$

Question Number 214696 Answers: 1 Comments: 1

Question Number 214683 Answers: 1 Comments: 1

Question Number 214678 Answers: 1 Comments: 0

solve partial differantial equation x((∂f(x,y))/∂x)+y((∂f(x,y))/∂y)=f(x,y)ln(x^2 +y^2 ) ((∂^2 f(x,y))/∂x^2 )+((∂^2 f(x,y))/∂y^2 )=0

$$\mathrm{solve} \\ $$$$\mathrm{partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$${x}\frac{\partial{f}\left({x},{y}\right)}{\partial{x}}+{y}\frac{\partial{f}\left({x},{y}\right)}{\partial{y}}={f}\left({x},{y}\right)\mathrm{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {f}\left({x},{y}\right)}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$

Question Number 214675 Answers: 1 Comments: 0

Question Number 214667 Answers: 0 Comments: 0

let′s define linear differantial operator D as D=z∙((d )/dz)(z∙((d )/dz))+z(1−((α/z))^2 ) when Df(z)={z∙((d )/dz)(z∙((d )/dz))+z(1−((α/z))^2 )}f(z)=0 f(z)=?

$$\mathrm{let}'\mathrm{s}\:\mathrm{define}\:\mathrm{linear}\:\mathrm{differantial}\:\mathrm{operator}\:\mathcal{D} \\ $$$$\mathrm{as}\:\mathcal{D}={z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right) \\ $$$$\mathrm{when} \\ $$$$\mathcal{D}{f}\left({z}\right)=\left\{{z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\left({z}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\right)+{z}\left(\mathrm{1}−\left(\frac{\alpha}{{z}}\right)^{\mathrm{2}} \right)\right\}{f}\left({z}\right)=\mathrm{0} \\ $$$${f}\left({z}\right)=? \\ $$

Question Number 214658 Answers: 2 Comments: 0

1) The function H is defined by H(x) =3cosh(x/3)+sinh(x/3). Find the value of λ for which H(lnλ^3 )=4 2) Prove that ∫_2 ^4 ((6x+1)/((2x−3)(3x−2)))dx= ln 10. 3)Show that ((sinθ + sin2θ)/(1+ cosθ+ cos2θ))≡ tanθ 4) If z=cosθ+ i sinθ, Show that z+(1/z)=2cosθ and that z^n +(1/z^n )=2cos nθ hence or otherwise show that 32cos^6 θ= cos 6θ +6cos 4θ + 15cos 2θ+ 10. Mr Hans

$$\left.\mathrm{1}\right)\:\mathrm{The}\:\mathrm{function}\:\mathrm{H}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{H}\left(\mathrm{x}\right)\:=\mathrm{3cosh}\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{sinh}\frac{\mathrm{x}}{\mathrm{3}}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\lambda\:\mathrm{for}\:\mathrm{which}\:\mathrm{H}\left(\mathrm{ln}\lambda^{\mathrm{3}} \right)=\mathrm{4} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{2}} ^{\mathrm{4}} \:\frac{\mathrm{6x}+\mathrm{1}}{\left(\mathrm{2x}−\mathrm{3}\right)\left(\mathrm{3x}−\mathrm{2}\right)}\mathrm{dx}=\:\mathrm{ln}\:\mathrm{10}. \\ $$$$\left.\mathrm{3}\right)\mathrm{Show}\:\mathrm{that}\:\frac{\mathrm{sin}\theta\:+\:\mathrm{sin2}\theta}{\mathrm{1}+\:\mathrm{cos}\theta+\:\mathrm{cos2}\theta}\equiv\:\mathrm{tan}\theta \\ $$$$\left.\mathrm{4}\right)\:\mathrm{If}\:\mathrm{z}=\mathrm{cos}\theta+\:\mathrm{i}\:\mathrm{sin}\theta,\:\mathrm{Show}\:\mathrm{that}\:\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}}=\mathrm{2cos}\theta\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{z}^{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}} }=\mathrm{2cos}\:\mathrm{n}\theta\:\mathrm{hence}\:\mathrm{or}\:\mathrm{otherwise}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{32cos}^{\mathrm{6}} \theta=\:\mathrm{cos}\:\mathrm{6}\theta\:+\mathrm{6cos}\:\mathrm{4}\theta\:+\:\mathrm{15cos}\:\mathrm{2}\theta+\:\mathrm{10}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Mr}\:\mathrm{Hans} \\ $$

Question Number 214656 Answers: 1 Comments: 1