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Question Number 28501    Answers: 2   Comments: 1

Question Number 28504    Answers: 0   Comments: 0

Find the direction cosines of two lines which are connected by relation l+m+n=0 mn−2nl−2lm=0 my solution l=−(m+n) mn+2n(m+n)+2(n+m)n=0 2m^2 +5mn+2n^2 =0 (2m+n)(m+2n)=0 m=−2n, or m=−(1/2)n case 1: m=−2n l=−(m+n)=−n l^2 +m^2 +n^2 =1 6n^2 =1⇒n=±(1/(√6)) (l,m,n)=±(−(1/(√6)),−(2/(√6)),(1/(√6))) case 2:m=−(1/2)n l=−(1/2)n l^2 +m^2 +n^2 =1 (n^2 /4)+(n^2 /4)+n^2 =1⇒n=±(2/(√6)) (l,m,n)=±(−(1/(√6)),−(1/(√6)),(2/(√6))) so i get a total of 4 solution. Books answer is 2 lines ((1/(√6)),(1/(√6)),−(2/(√6))) and ((1/(√6)),−(2/(√6)),(1/(√6))) please help.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{cosines}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{lines}\:\mathrm{which}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{relation} \\ $$$${l}+{m}+{n}=\mathrm{0} \\ $$$${mn}−\mathrm{2}{nl}−\mathrm{2}{lm}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{solution}} \\ $$$${l}=−\left({m}+{n}\right) \\ $$$${mn}+\mathrm{2}{n}\left({m}+{n}\right)+\mathrm{2}\left({n}+{m}\right){n}=\mathrm{0} \\ $$$$\mathrm{2}{m}^{\mathrm{2}} +\mathrm{5}{mn}+\mathrm{2}{n}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{m}+{n}\right)\left({m}+\mathrm{2}{n}\right)=\mathrm{0} \\ $$$${m}=−\mathrm{2}{n},\:{or}\:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${case}\:\mathrm{1}:\:{m}=−\mathrm{2}{n} \\ $$$${l}=−\left({m}+{n}\right)=−{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{6}{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$${case}\:\mathrm{2}:{m}=−\left(\mathrm{1}/\mathrm{2}\right){n} \\ $$$${l}=−\frac{\mathrm{1}}{\mathrm{2}}{n} \\ $$$${l}^{\mathrm{2}} +{m}^{\mathrm{2}} +{n}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+\frac{{n}^{\mathrm{2}} }{\mathrm{4}}+{n}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\pm\frac{\mathrm{2}}{\sqrt{\mathrm{6}}} \\ $$$$\left({l},{m},{n}\right)=\pm\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{so}\:\mathrm{i}\:\mathrm{get}\:\mathrm{a}\:\mathrm{total}\:\mathrm{of}\:\mathrm{4}\:\mathrm{solution}. \\ $$$$\mathrm{Books}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}\:\mathrm{lines} \\ $$$$\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}}\right)\:\mathrm{and}\:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{6}}},−\frac{\mathrm{2}}{\sqrt{\mathrm{6}}},\frac{\mathrm{1}}{\sqrt{\mathrm{6}}}\right) \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$

Question Number 28490    Answers: 2   Comments: 0

lim_(x → a) (((cos(x) − cos(a))/(x − a))) Don′t use L′ho^^ spital rule.

$$\underset{\mathrm{x}\:\rightarrow\:\mathrm{a}} {\mathrm{lim}}\:\left(\frac{\mathrm{cos}\left(\mathrm{x}\right)\:−\:\mathrm{cos}\left(\mathrm{a}\right)}{\mathrm{x}\:−\:\mathrm{a}}\right) \\ $$$$ \\ $$$$\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{h}\bar {\mathrm{o}spital}\:\mathrm{rule}. \\ $$

Question Number 28483    Answers: 0   Comments: 0

Question Number 28481    Answers: 0   Comments: 1

Question Number 28480    Answers: 1   Comments: 1

Question Number 28479    Answers: 1   Comments: 0

Question Number 28475    Answers: 0   Comments: 0

Question Number 28468    Answers: 0   Comments: 0

Prove that for all values of u, the point [a cosh(u) , b sinh(u)] lies on the hyperbola whose equation is. (x^2 /a^2 ) − (y^2 /b^2 ) = 1 And that the tangent at that point is : (x/a) cosh(u) − (y/b) sinh(u)

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{u},\:\mathrm{the}\:\mathrm{point}\:\left[\mathrm{a}\:\mathrm{cosh}\left(\mathrm{u}\right)\:,\:\:\mathrm{b}\:\mathrm{sinh}\left(\mathrm{u}\right)\right]\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\mathrm{hyperbola}\:\mathrm{whose}\:\mathrm{equation}\:\mathrm{is}.\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$\mathrm{And}\:\mathrm{that}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{that}\:\mathrm{point}\:\mathrm{is}\::\:\:\:\frac{\mathrm{x}}{\mathrm{a}}\:\mathrm{cosh}\left(\mathrm{u}\right)\:\:−\:\:\frac{\mathrm{y}}{\mathrm{b}}\:\mathrm{sinh}\left(\mathrm{u}\right) \\ $$

Question Number 28467    Answers: 2   Comments: 1

Question Number 28464    Answers: 1   Comments: 0

Question Number 28461    Answers: 0   Comments: 2

Question Number 28456    Answers: 1   Comments: 1

Question Number 28448    Answers: 1   Comments: 0

find ∫∫_D ((xy)/(1+x^2 +y^2 ))dxdy with D= {(x,y)∈R^2 / x^2 +y^2 ≥1 } .

$${find}\:\int\int_{{D}} \:\:\:\:\frac{{xy}}{\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{dxdy}\:{with} \\ $$$${D}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{1}\:\:\right\}\:\:. \\ $$

Question Number 28449    Answers: 2   Comments: 0

Question Number 28446    Answers: 0   Comments: 0

find ∫∫_A (x+y) e^(−x) e^(−y) dxdy with A= {(x,y)∈R^2 /x≥0 ,y≥0 , x+y ≤1 } .

$${find}\:\int\int_{{A}} \left({x}+{y}\right)\:{e}^{−{x}} \:{e}^{−{y}} \:{dxdy}\:{with} \\ $$$${A}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:\:/{x}\geqslant\mathrm{0}\:,{y}\geqslant\mathrm{0}\:,\:{x}+{y}\:\leqslant\mathrm{1}\:\right\}\:\:. \\ $$

Question Number 28445    Answers: 0   Comments: 0

find ∫∫_(x≤x^2 +y^2 ≤1) ((dxdy)/((1+x^2 +y^2 ))) .

$${find}\:\int\int_{{x}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\mathrm{1}} \:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}\:\:. \\ $$

Question Number 28444    Answers: 0   Comments: 0

let give 1<a<b and I= ∫_0 ^π ∫_a ^b (du/(x−cosu)) dt find the value of ∫_0 ^π ln(((b−cost)/(a−cost)))dt .

$${let}\:{give}\:\:\mathrm{1}<{a}<{b}\:\:{and}\:{I}=\:\int_{\mathrm{0}} ^{\pi} \:\int_{{a}} ^{{b}} \:\:\frac{{du}}{{x}−{cosu}}\:{dt}\:\:{find}\:{the} \\ $$$${value}\:{of}\:\int_{\mathrm{0}} ^{\pi} \:\:{ln}\left(\frac{{b}−{cost}}{{a}−{cost}}\right){dt}\:. \\ $$

Question Number 28442    Answers: 0   Comments: 0

let give B(x,y)= ∫_0 ^1 u^(x−1) (1−u)^(y−1) du and (beta function) and Γ(x) =∫_0 ^∞ u^(x−1) e^(−u) du (x>0)(gamma function of euler) 1) prove that Γ(x)= 2∫_0 ^∞ u^(2x−1) e^(−u^2 ) du . 2) prove that B(x,y) = ((Γ(x).Γ(y))/(Γ(x+y))) .

$${let}\:{give}\:{B}\left({x},{y}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{x}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{y}−\mathrm{1}} {du}\:\:{and}\:\left({beta}\:{function}\right) \\ $$$${and}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{u}^{{x}−\mathrm{1}} \:{e}^{−{u}} \:{du}\:\:\:\:\:\left({x}>\mathrm{0}\right)\left({gamma}\:{function}\:{of}\:{euler}\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:\:\:\Gamma\left({x}\right)=\:\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:{u}^{\mathrm{2}{x}−\mathrm{1}} \:{e}^{−{u}^{\mathrm{2}} \:} {du}\:. \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\:{B}\left({x},{y}\right)\:=\:\frac{\Gamma\left({x}\right).\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)}\:. \\ $$$$ \\ $$

Question Number 28688    Answers: 0   Comments: 0

Question Number 28439    Answers: 0   Comments: 0

find ∫ ((1+tanx)/(1+sin^2 x))dx

$${find}\:\:\int\:\:\frac{\mathrm{1}+{tanx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$

Question Number 28438    Answers: 0   Comments: 0

find I= ∫_0 ^∞ (dt/((1+t^4 )^n )) with n integr and n≠0.

$${find}\:\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)^{{n}} }\:\:{with}\:{n}\:{integr}\:{and}\:{n}\neq\mathrm{0}. \\ $$

Question Number 28437    Answers: 0   Comments: 0

let give f(x)= sin ((π/x)) find f^((n)) .

$${let}\:{give}\:{f}\left({x}\right)=\:{sin}\:\left(\frac{\pi}{{x}}\right)\:\:\:{find}\:{f}^{\left({n}\right)} . \\ $$

Question Number 28436    Answers: 0   Comments: 0

study the nature of sequence (u_n ) / u_0 =1 and u_(n+1) = (1/(u_(n ) + e^(−n) )) .

$${study}\:{the}\:{nature}\:{of}\:{sequence}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{0}} =\mathrm{1}\:{and} \\ $$$${u}_{{n}+\mathrm{1}} =\:\:\frac{\mathrm{1}}{{u}_{{n}\:} +\:{e}^{−{n}} }\:. \\ $$

Question Number 28435    Answers: 0   Comments: 0

find Q(x) / nx^(n+1) −(n+1)x^n +1 =(x−1)^2 Q(x).

$${find}\:{Q}\left({x}\right)\:/\:{nx}^{{n}+\mathrm{1}} −\left({n}+\mathrm{1}\right){x}^{{n}} +\mathrm{1}\:=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}\left({x}\right). \\ $$

Question Number 28434    Answers: 0   Comments: 0

let give the polynomial p(x)=(x+1)^n −(x−1)^n with n from N^∗ 1) give the factorisation of p(x) inside C[x] 2) prove that Π_(k=0) ^(n−1) cotan(((kπ)/(2p+1)))=(1/(√(2p+1)))

$${let}\:{give}\:{the}\:{polynomial}\:{p}\left({x}\right)=\left({x}+\mathrm{1}\right)^{{n}} −\left({x}−\mathrm{1}\right)^{{n}} {with}\:{n} \\ $$$${from}\:{N}^{\ast} \\ $$$$\left.\mathrm{1}\right)\:{give}\:{the}\:{factorisation}\:{of}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} {cotan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}{p}+\mathrm{1}}} \\ $$

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