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Question Number 210926    Answers: 0   Comments: 1

valeur de : tan^2 ((π/7))+tan^2 (((2π)/7))+tan^2 (((3π)/7)) = ???

$$\mathrm{valeur}\:\mathrm{de}\::\: \\ $$$$\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=\:??? \\ $$

Question Number 210922    Answers: 0   Comments: 0

Question Number 210920    Answers: 0   Comments: 0

Question Number 210919    Answers: 1   Comments: 0

Question Number 210918    Answers: 1   Comments: 0

Question Number 210917    Answers: 1   Comments: 0

Find the area intersected by three circles of radius 1, centered at the origin, at (1, 0) and (1, 1) respectively.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{intersected}\:\mathrm{by}\:\mathrm{three} \\ $$$$\mathrm{circles}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1},\:\mathrm{centered}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{origin},\:\mathrm{at}\:\left(\mathrm{1},\:\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{1},\:\mathrm{1}\right)\:\mathrm{respectively}. \\ $$

Question Number 210912    Answers: 2   Comments: 0

A uniform stick AB of length L and mass M is balanced horizontally on a knife edge 10.0cm from A when an object of mass 400g is suspended at A. When the knife edge is moved 5cm further, the object has to be moved to a point 9.00cm from A for the stick to balance. Represent the balance system with a suitable diagram, determine the mass and Length of the stick

A uniform stick AB of length L and mass M is balanced horizontally on a knife edge 10.0cm from A when an object of mass 400g is suspended at A. When the knife edge is moved 5cm further, the object has to be moved to a point 9.00cm from A for the stick to balance. Represent the balance system with a suitable diagram, determine the mass and Length of the stick

Question Number 210906    Answers: 1   Comments: 0

((19x − x^2 )/(x + 1)) ∙ (x + ((19 − x)/(x + 1))) = 78 find: x = ?

$$\frac{\mathrm{19x}\:−\:\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:+\:\mathrm{1}}\:\centerdot\:\left(\mathrm{x}\:+\:\frac{\mathrm{19}\:−\:\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:=\:\mathrm{78} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210896    Answers: 0   Comments: 0

Σ_(n=1) ^∞ (H_n ^− /((2n+1)^q ))=??

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\overset{−} {{H}}_{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{q}} }=?? \\ $$

Question Number 210895    Answers: 4   Comments: 0

Question Number 210875    Answers: 4   Comments: 0

Question Number 210874    Answers: 0   Comments: 0

Question Number 210873    Answers: 0   Comments: 0

Question Number 210871    Answers: 0   Comments: 1

7(x−2)((√(10 + x)) − ((7−x))^(1/3) )−15x + 6 = 0 x = ?

$$\mathrm{7}\left(\mathrm{x}−\mathrm{2}\right)\left(\sqrt{\mathrm{10}\:+\:\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{x}}\right)−\mathrm{15x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210870    Answers: 0   Comments: 1

8x^2 − 13x + 11 = (2/x) + (1 + (3/x)) ((3x^2 −2))^(1/3) x = ?

$$\mathrm{8x}^{\mathrm{2}} \:−\:\mathrm{13x}\:+\:\mathrm{11}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{x}}\right)\:\sqrt[{\mathrm{3}}]{\mathrm{3x}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210868    Answers: 1   Comments: 5

let p ,q be reals such that p>q>0 define the sequence {x_n } where x_1 = p+q and x_n = x_1 −((pq)/x_(n−1) ) for n≥2 for all n then x_n = ??

$$\mathrm{let}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{be}\:\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}>\mathrm{q}>\mathrm{0}\:\mathrm{define} \\ $$$$\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{x}_{\mathrm{n}} \right\}\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{p}+\mathrm{q}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{x}_{\mathrm{1}} −\frac{\mathrm{pq}}{\mathrm{x}_{\mathrm{n}−\mathrm{1}} }\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\:\mathrm{then}\:\mathrm{x}_{\mathrm{n}} \:=\:?? \\ $$

Question Number 210858    Answers: 2   Comments: 0

Solve The Equation: (7x+1)(9x+1)(21x+1)(63x+1)= ((160)/(189))

$$\mathrm{Solve}\:\mathrm{The}\:\mathrm{Equation}: \\ $$$$\left(\mathrm{7x}+\mathrm{1}\right)\left(\mathrm{9x}+\mathrm{1}\right)\left(\mathrm{21x}+\mathrm{1}\right)\left(\mathrm{63x}+\mathrm{1}\right)=\:\frac{\mathrm{160}}{\mathrm{189}} \\ $$

Question Number 210855    Answers: 3   Comments: 0

Question Number 210843    Answers: 1   Comments: 0

Question Number 210842    Answers: 1   Comments: 0

Question Number 210851    Answers: 1   Comments: 1

Question Number 210840    Answers: 1   Comments: 0

Prove that if x, y are rational numbers satisfying the equation x^5 + y^5 = 2(x^2)(y^2) then 1 - xy is the square of rational number

$$ \\ $$Prove that if x, y are rational numbers satisfying the equation x^5 + y^5 = 2(x^2)(y^2) then 1 - xy is the square of rational number

Question Number 210824    Answers: 0   Comments: 0

Question Number 210820    Answers: 2   Comments: 0

prove ∫_0 ^∞ ((arctan (√(x^2 +2)))/((x^2 +1)(√(x^2 +2)))) dx=(π^2 /(12))

$$\mathrm{prove} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{arctan}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Question Number 210839    Answers: 0   Comments: 1

show that ∫_0 ^x e^(xt) e^(−t^2 ) dt=e^(x^2 /4) ∫^x _0 e^(−(t^2 /4))

$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\int_{\mathrm{0}} ^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{xt}}} \boldsymbol{\mathrm{e}}^{−\mathrm{t}^{\mathrm{2}} } \boldsymbol{\mathrm{dt}}=\boldsymbol{\mathrm{e}}^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{4}}} \underset{\mathrm{0}} {\int}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{e}}^{−\frac{\boldsymbol{\mathrm{t}}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$ \\ $$

Question Number 210836    Answers: 0   Comments: 0

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