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Question Number 97303    Answers: 0   Comments: 0

Given x_1 +x_2 +x_3 = 0 , y_1 + y_2 +y_3 = 0 and x_1 y_1 + x_2 y_2 + x_3 y_3 = 0 . The value of (x_1 ^2 /(x_1 ^2 +x_2 ^2 +x_3 ^2 )) + (y_1 ^2 /(y_1 ^2 +y_2 ^2 +y_3 ^2 )) = ?

$$\boldsymbol{\mathrm{G}}\mathrm{iven}\:\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{0}\:,\:\mathrm{y}_{\mathrm{1}} \:+\:\mathrm{y}_{\mathrm{2}} +\mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} +\:\mathrm{x}_{\mathrm{2}} \mathrm{y}_{\mathrm{2}} \:+\:\mathrm{x}_{\mathrm{3}} \mathrm{y}_{\mathrm{3}} \:=\:\mathrm{0}\:.\:\mathrm{The}\:\mathrm{value} \\ $$$$\mathrm{of}\:\frac{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} ^{\mathrm{2}} }\:+\:\frac{\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{y}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{y}_{\mathrm{3}} ^{\mathrm{2}} }\:=\:?\: \\ $$

Question Number 97283    Answers: 0   Comments: 0

Question Number 97275    Answers: 1   Comments: 14

Trial version with additional colors is now available.

$$\mathrm{Trial}\:\mathrm{version}\:\mathrm{with}\:\mathrm{additional} \\ $$$$\mathrm{colors}\:\mathrm{is}\:\mathrm{now}\:\mathrm{available}. \\ $$

Question Number 97272    Answers: 0   Comments: 0

Question Number 97271    Answers: 1   Comments: 2

hello every one why do planets of the solar system revolve around the sun in an eliptical not circular orbit

$${hello}\:{every}\:{one} \\ $$$${why}\:{do}\:{planets}\:{of}\:{the}\:{solar}\:{system} \\ $$$${revolve}\:{around}\:{the}\:{sun}\:{in}\:{an}\:{eliptical} \\ $$$${not}\:{circular}\:{orbit} \\ $$

Question Number 97270    Answers: 1   Comments: 3

solve for all real value of x,y and z giving answer the form (x,y,z) { ((x(x+y)+z(x−y)= 4)),((y(y+z)+x(y−z) = −4)),((z(z+x)+y(z−x) = 5)) :}

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{real}}\:\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{z}}\:\boldsymbol{\mathrm{giving}} \\ $$$$\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{form}}\:\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)\:\begin{cases}{\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{z}}\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}\right)=\:\mathrm{4}}\\{\boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}\right)+\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{z}}\right)\:=\:−\mathrm{4}}\\{\boldsymbol{\mathrm{z}}\left(\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{z}}−\boldsymbol{\mathrm{x}}\right)\:=\:\mathrm{5}}\end{cases} \\ $$

Question Number 97250    Answers: 1   Comments: 0

Question Number 97239    Answers: 0   Comments: 3

∫ ((sec^3 x dx)/(√(tan x))) ?

$$\int\:\frac{\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}}{\sqrt{\mathrm{tan}\:{x}}}\:?\: \\ $$

Question Number 97238    Answers: 0   Comments: 1

Question Number 97236    Answers: 3   Comments: 3

Question Number 97235    Answers: 1   Comments: 0

∫_0 ^1 ln(x) ln(1−x) dx ?

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\mathrm{ln}\left(\mathrm{x}\right)\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:\mathrm{dx}\:?\: \\ $$

Question Number 97231    Answers: 0   Comments: 0

developp at fourier serie f(x) =(1/(cos^2 x −3cosx +2))

$$\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:−\mathrm{3cosx}\:+\mathrm{2}} \\ $$

Question Number 97230    Answers: 2   Comments: 0

1) developp at fourier serie f(x)=ln(sinx) 2) developp at fourier serie g(x)=ln(cosx +sinx) 3)developp at fourier seri e h(x) =ln(cosx +2sinx)

$$\left.\mathrm{1}\right)\:\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{sinx}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{cosx}\:+\mathrm{sinx}\right) \\ $$$$\left.\mathrm{3}\right)\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{seri}\:\mathrm{e}\:\mathrm{h}\left(\mathrm{x}\right)\:=\mathrm{ln}\left(\mathrm{cosx}\:+\mathrm{2sinx}\right) \\ $$

Question Number 97227    Answers: 2   Comments: 1

Question Number 97226    Answers: 0   Comments: 0

let a_n the sequence wich verify a_n +a_(n+1) =(((−1)^n )/n^2 ) calculate Σ_(n=1) ^∞ a_n x^n

$$\mathrm{let}\:\:\mathrm{a}_{\mathrm{n}} \:\mathrm{the}\:\mathrm{sequence}\:\mathrm{wich}\:\mathrm{verify}\:\mathrm{a}_{\mathrm{n}} \:+\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}} \\ $$

Question Number 97218    Answers: 2   Comments: 0

∫_(−2) ^3 ∣x−2∣ ⌊ (x/2) ⌋ sgn (x−1) dx

$$\underset{−\mathrm{2}} {\overset{\mathrm{3}} {\int}}\:\mid{x}−\mathrm{2}\mid\:\lfloor\:\frac{{x}}{\mathrm{2}}\:\rfloor\:\mathrm{sgn}\:\left({x}−\mathrm{1}\right)\:{dx}\: \\ $$

Question Number 97206    Answers: 2   Comments: 0

Question Number 97200    Answers: 0   Comments: 8

App was updated about a week back to use new backend server for forum as we were facing problems on old server. Please update app to the latest version 2.079 or above. Thanks for ur cooperation. Note: Do not uninstall/reinstall as android removes app data on uninstalls. Backup all inapp saved equation to sdcard before uninstall.

$$\mathrm{App}\:\mathrm{was}\:\mathrm{updated}\:\mathrm{about}\:\:\mathrm{a}\:\mathrm{week} \\ $$$$\mathrm{back}\:\mathrm{to}\:\mathrm{use}\:\mathrm{new}\:\mathrm{backend}\:\mathrm{server} \\ $$$$\mathrm{for}\:\mathrm{forum}\:\mathrm{as}\:\mathrm{we}\:\mathrm{were}\:\mathrm{facing}\:\mathrm{problems}\:\mathrm{on}\:\mathrm{old} \\ $$$$\mathrm{server}. \\ $$$$\mathrm{Please}\:\mathrm{update}\:\mathrm{app}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{latest}\:\mathrm{version}\:\mathrm{2}.\mathrm{079}\:\mathrm{or}\:\mathrm{above}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{ur}\:\mathrm{cooperation}. \\ $$$$\mathrm{Note}:\:\mathrm{Do}\:\mathrm{not}\:\mathrm{uninstall}/\mathrm{reinstall} \\ $$$$\mathrm{as}\:\mathrm{android}\:\mathrm{removes}\:\mathrm{app}\:\mathrm{data}\:\mathrm{on} \\ $$$$\mathrm{uninstalls}.\:\mathrm{Backup}\:\mathrm{all}\:\mathrm{inapp} \\ $$$$\mathrm{saved}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{sdcard}\:\mathrm{before} \\ $$$$\mathrm{uninstall}. \\ $$

Question Number 97199    Answers: 1   Comments: 0

lim_(x→0) ((sin x−tan x)/((((1+x^2 ))^(1/(3 )) −1)((√(1+sin x))−1))) = ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}}{\left(\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\right)\left(\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}−\mathrm{1}\right)}\:=\:? \\ $$

Question Number 97192    Answers: 1   Comments: 4

Question Number 97189    Answers: 0   Comments: 0

Question Number 97182    Answers: 0   Comments: 0

Question Number 97181    Answers: 0   Comments: 0

Question Number 97174    Answers: 0   Comments: 0

Exercise_(−) ABC is a triangle. AB=AC=2 and BC=2(√2). I is midle of [BC]. J is a point such as AJ^(→) =(2/3)AI^(→) . J is the center of gravity of the triangle. 1)a) we define the set(T) of ∀ point M of plane: AM^2 +BM^2 +CM^2 =8. Show that BM^2 +CM^2 =2IM^2 +4 and AM^2 +2IM^2 =3JM^2 +(4/3) b)deduct that AM^2 +BM^2 +CM^2 =3JM^2 +((16)/3) c)Deduct the nature of (T)

$$\underset{−} {{Exercise}} \\ $$$${ABC}\:{is}\:{a}\:{triangle}.\:{AB}={AC}=\mathrm{2}\:{and} \\ $$$${BC}=\mathrm{2}\sqrt{\mathrm{2}}.\:{I}\:{is}\:{midle}\:{of}\:\left[{BC}\right]. \\ $$$${J}\:{is}\:{a}\:{point}\:{such}\:{as}\:\overset{\rightarrow} {{AJ}}=\frac{\mathrm{2}}{\mathrm{3}}\overset{\rightarrow} {{AI}}.\:{J}\:{is} \\ $$$${the}\:{center}\:{of}\:{gravity}\:{of}\:{the}\:{triangle}. \\ $$$$\left.\mathrm{1}\left.\right){a}\right)\:{we}\:{define}\:{the}\:{set}\left({T}\right)\:{of}\:\forall\:{point}\:{M} \\ $$$${of}\:{plane}: \\ $$$${AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{8}. \\ $$$${Show}\:{that}\:{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{2}{IM}^{\mathrm{2}} +\mathrm{4} \\ $$$${and}\:{AM}^{\mathrm{2}} +\mathrm{2}{IM}^{\mathrm{2}} =\mathrm{3}{JM}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\left.{b}\right){deduct}\:{that}\: \\ $$$${AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} +{CM}^{\mathrm{2}} =\mathrm{3}{JM}^{\mathrm{2}} +\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\left.{c}\right){Deduct}\:{the}\:{nature}\:{of}\:\left({T}\right) \\ $$

Question Number 97173    Answers: 1   Comments: 0

Question Number 97157    Answers: 3   Comments: 1

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