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Question Number 206160 Answers: 2 Comments: 1

If x and y are real numbers then is it possible that secθ = ((2xy)/(x^2 + y^2 )) ?

$$\mathrm{If}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{then}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{possible}\:\mathrm{that}\:\mathrm{sec}\theta\:=\:\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }\:? \\ $$

Question Number 206157 Answers: 2 Comments: 3

Question Number 206156 Answers: 1 Comments: 4

Question Number 206155 Answers: 4 Comments: 0

x = 2+(2/(2+(2/(2+(2/(2+(2/(2+(2/x))))))))) find x.

$$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{x}}}}}} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{x}.\: \\ $$

Question Number 206150 Answers: 1 Comments: 0

Calcul ∣x−α∣ = ????? • x=−1 ; α ∈ [−1; −(3/4)] • x=2 ; α ∈ [0 ; 3] help please

$$\mathrm{Calcul}\:\:\:\mid\mathrm{x}−\alpha\mid\:=\:????? \\ $$$$\:\:\bullet\:\:\:\:\:\:\mathrm{x}=−\mathrm{1}\:\:;\:\:\:\alpha\:\in\:\left[−\mathrm{1};\:−\frac{\mathrm{3}}{\mathrm{4}}\right] \\ $$$$\:\:\bullet\:\:\:\:\mathrm{x}=\mathrm{2}\:\:;\:\:\:\:\alpha\:\in\:\left[\mathrm{0}\:;\:\mathrm{3}\right]\:\:\:\:\:\:\:\mathrm{help}\:\mathrm{please}\:\: \\ $$

Question Number 206151 Answers: 1 Comments: 0

Question Number 206146 Answers: 0 Comments: 0

Question Number 206142 Answers: 1 Comments: 0

let (d^2 y/dx^2 )+p(x)(dy/dx)+q(x)y=0 , x∈R where p(x) and q(x) are continuous function if y_1 = sinx−2cosx and y_2 = 2sinx +cosx are L.I (linearly independent) solution then ∣4p(0)+2q(1)∣ = ?

$$\:\:\:\:\mathrm{let}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{p}\left({x}\right)\frac{{dy}}{{dx}}+{q}\left({x}\right){y}=\mathrm{0}\:,\:{x}\in\mathbb{R}\:\mathrm{where}\: \\ $$$$\:\:\:\:{p}\left({x}\right)\:\mathrm{and}\:{q}\left({x}\right)\:\mathrm{are}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{if} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} =\:\mathrm{sin}{x}−\mathrm{2cos}{x}\:{and}\:{y}_{\mathrm{2}} \:=\:\mathrm{2sin}{x}\:+\mathrm{cos}{x} \\ $$$$\:\:\:\:\mathrm{are}\:{L}.{I}\:\left(\mathrm{linearly}\:\mathrm{independent}\right)\:\mathrm{solution} \\ $$$$\:\:\:\:\:\mathrm{then}\:\:\mid\mathrm{4}{p}\left(\mathrm{0}\right)+\mathrm{2}{q}\left(\mathrm{1}\right)\mid\:=\:?\:\:\: \\ $$

Question Number 206136 Answers: 1 Comments: 0

Question Number 206129 Answers: 1 Comments: 0

Question Number 206111 Answers: 3 Comments: 1

Question Number 206108 Answers: 3 Comments: 0

If x = (√(((√5) + 1)/( (√5) − 1))) then x^(12) = ?

$$\mathrm{If}\:{x}\:=\:\sqrt{\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}}\:\mathrm{then}\:{x}^{\mathrm{12}} \:=\:? \\ $$

Question Number 206107 Answers: 0 Comments: 4

can you guy explian why we need a eighgen value eighgen vector and mean?? such as Av=𝛌v

$$\mathrm{can}\:\mathrm{you}\:\mathrm{guy}\:\mathrm{explian}\:\mathrm{why} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{a}\:\mathrm{eighgen}\:\mathrm{value}\:\mathrm{eighgen}\:\mathrm{vector} \\ $$$$\mathrm{and}\:\mathrm{mean}?? \\ $$$$\mathrm{such}\:\mathrm{as}\:\mathrm{A}\boldsymbol{\mathrm{v}}=\boldsymbol{\lambda\mathrm{v}} \\ $$$$ \\ $$

Question Number 206104 Answers: 2 Comments: 0

⇃

$$\:\:\:\:\downharpoonleft\underline{\:} \\ $$

Question Number 206096 Answers: 1 Comments: 0

∫(1/(x^3 (√(x^2 −1)))) .dx

$$\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:.{dx} \\ $$

Question Number 206095 Answers: 2 Comments: 0

Question Number 206093 Answers: 2 Comments: 0

Question Number 206082 Answers: 2 Comments: 0

If cos𝛂 = sin𝛂 + (1/( (√3))) Find sin2𝛂 = ?

$$\mathrm{If}\:\:\:\mathrm{cos}\boldsymbol{\alpha}\:=\:\mathrm{sin}\boldsymbol{\alpha}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{sin2}\boldsymbol{\alpha}\:=\:? \\ $$

Question Number 206079 Answers: 3 Comments: 0

Question Number 206078 Answers: 0 Comments: 1

Question Number 206074 Answers: 0 Comments: 1

(x_1 − x_2 + x_3 )^2 = x_2 (x_4 + x_5 − x_2 ) (x_2 − x_3 + x_4 )^2 = x_3 (x_5 + x_1 − x_3 ) (x_3 − x_4 + x_5 )^2 = x_4 (x_1 + x_2 − x_4 ) (x_4 − x_5 + x_1 )^2 = x_5 (x_2 + x_3 − x_5 ) (x_5 − x_1 + x_2 )^2 = x_1 (x_3 + x_4 − x_1 ) Find ((2x_1 + x_2 + x_3 )/(3x_4 − x_5 )) .

$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{2}} \left({x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \:−\:{x}_{\mathrm{2}} \right) \\ $$$$\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{3}} \left({x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right) \\ $$$$\left({x}_{\mathrm{3}} \:−\:{x}_{\mathrm{4}} \:+\:{x}_{\mathrm{5}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{4}} \left({x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}} \right) \\ $$$$\left({x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} \:+\:{x}_{\mathrm{1}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{5}} \left({x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} \:−\:{x}_{\mathrm{5}} \right) \\ $$$$\left({x}_{\mathrm{5}} \:−\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:=\:{x}_{\mathrm{1}} \left({x}_{\mathrm{3}} \:+\:{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{1}} \right) \\ $$$$\mathrm{Find}\:\frac{\mathrm{2}{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} \:+\:{x}_{\mathrm{3}} }{\mathrm{3}{x}_{\mathrm{4}} \:−\:{x}_{\mathrm{5}} }\:. \\ $$

Question Number 206072 Answers: 0 Comments: 0

∫_0 ^π arctan(((ln(sin(x)))/x))dx=...?

$$\int_{\mathrm{0}} ^{\pi} {arctan}\left(\frac{{ln}\left({sin}\left({x}\right)\right)}{{x}}\right){dx}=...? \\ $$

Question Number 206069 Answers: 1 Comments: 0

lim_(x→0) ((−x^3 +x)/(sin x))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−{x}^{\mathrm{3}} +{x}}{\mathrm{sin}\:{x}} \\ $$

Question Number 206066 Answers: 2 Comments: 0

Find: (3/4) ∙ (8/9) ∙ ((15)/(16)) ∙ ... ∙ ((120)/(121)) = ?

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{8}}{\mathrm{9}}\:\centerdot\:\frac{\mathrm{15}}{\mathrm{16}}\:\centerdot\:...\:\centerdot\:\frac{\mathrm{120}}{\mathrm{121}}\:=\:? \\ $$

Question Number 206064 Answers: 3 Comments: 1

Question Number 206063 Answers: 2 Comments: 0

If (x + (√(1 + x^2 )))(y + (√(1 + y^2 ))) = 1 then find (x + y)^2 .

$$\mathrm{If}\:\left({x}\:+\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\right)\left({y}\:+\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{2}} }\right)\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{find}\:\left({x}\:+\:{y}\right)^{\mathrm{2}} . \\ $$

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