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Question Number 86343    Answers: 0   Comments: 2

Dear mr w. i want discuss for equation find minimum and maximum value of xy +2 with constraint x^2 +y^(2 ) = 6. my way (short cut) ⇒ x^2 = y^2 = 3 { ((max = ((√3))^2 +2 = 5)),((min = −((√3))^2 +1 = −1)) :} it correct?

$$\mathrm{Dear}\:\mathrm{mr}\:\mathrm{w}.\:\mathrm{i}\:\mathrm{want}\: \\ $$$$\mathrm{discuss}\:\mathrm{for}\:\mathrm{equation} \\ $$$$\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{xy}\:+\mathrm{2}\:\mathrm{with}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}\:} =\:\mathrm{6}. \\ $$$$\mathrm{my}\:\mathrm{way}\:\left(\mathrm{short}\:\mathrm{cut}\right) \\ $$$$\Rightarrow\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\begin{cases}{\mathrm{max}\:=\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}\:=\:\mathrm{5}}\\{\mathrm{min}\:=\:−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:+\mathrm{1}\:=\:−\mathrm{1}}\end{cases} \\ $$$$\mathrm{it}\:\mathrm{correct}? \\ $$

Question Number 86341    Answers: 0   Comments: 0

Can anyone pls check question no. 76808

$${Can}\:{anyone}\:{pls}\:{check}\: \\ $$$${question}\:{no}.\:\mathrm{76808} \\ $$

Question Number 86339    Answers: 1   Comments: 0

4^x +6^x =9^x

$$\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} \\ $$$$ \\ $$$$ \\ $$

Question Number 86337    Answers: 0   Comments: 1

[4

$$\left[\mathrm{4}\right. \\ $$

Question Number 86328    Answers: 1   Comments: 0

∫((x^2 −1)/((x^2 +1)(√(1+x^4 )))) dx

$$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:{dx} \\ $$

Question Number 86326    Answers: 0   Comments: 0

Let f a continue function acknowleding α as a fix point on [0,1].F a function such as (dF/dx)=f(x) ∀ n , u_(n+1) =((F(u_n )−F(α))/(u_n −α)) Prove that lim_(n→∞) u_n =α

$${Let}\:{f}\:{a}\:{continue}\:{function}\:{acknowleding} \\ $$$$\alpha\:{as}\:{a}\:{fix}\:{point}\:{on}\:\left[\mathrm{0},\mathrm{1}\right].{F}\:\:{a}\:{function}\:{such}\:{as}\:\frac{{dF}}{{dx}}={f}\left({x}\right) \\ $$$$\forall\:{n}\:,\:\:{u}_{{n}+\mathrm{1}} =\frac{{F}\left({u}_{{n}} \right)−{F}\left(\alpha\right)}{{u}_{{n}} −\alpha}\: \\ $$$${Prove}\:{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} \:=\alpha \\ $$

Question Number 86324    Answers: 0   Comments: 2

∫_( 0) ^( (π/4)) ((1 )/((√(sin x)) + (√(cos x)))) dx

$$\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{1}\:}{\sqrt{\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}}\:\:\:+\:\:\sqrt{\boldsymbol{\mathrm{cos}}\:\boldsymbol{\mathrm{x}}}}\:\boldsymbol{\mathrm{dx}} \\ $$

Question Number 86313    Answers: 2   Comments: 2

∫_0 ^(π/2) (dx/(√(1+tan x)))

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}} \\ $$

Question Number 86307    Answers: 1   Comments: 0

Solve the following equation: (d^2 y/dx^2 ) −2 (dy/dx) +y = x e^x sin x.

$$\:\:\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\mathrm{2}\:\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:+\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\mathrm{x}}. \\ $$

Question Number 86302    Answers: 1   Comments: 1

∫ (dx/((x^2 +1)(√(x^2 +4)))) =?

$$\int\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}\:=? \\ $$

Question Number 86298    Answers: 0   Comments: 5

let x^x^x^⋰ =2 x^2 =2 x=±(√2) then let x^x^x^⋰ =4 x^4 =4 x=±(4)^(1/4) =±(√2) so we had prove 2=4 right?

$${let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\pm\sqrt{\mathrm{2}} \\ $$$${then}\:{let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{4} \\ $$$${x}^{\mathrm{4}} =\mathrm{4} \\ $$$${x}=\pm\sqrt[{\mathrm{4}}]{\mathrm{4}}=\pm\sqrt{\mathrm{2}} \\ $$$${so}\:{we}\:{had}\:{prove}\:\mathrm{2}=\mathrm{4}\:{right}? \\ $$

Question Number 86295    Answers: 0   Comments: 0

Question Number 86294    Answers: 1   Comments: 0

y = 2x + (y′)^2 −4y′

$$\mathrm{y}\:=\:\mathrm{2x}\:+\:\left(\mathrm{y}'\right)^{\mathrm{2}} −\mathrm{4y}' \\ $$

Question Number 86374    Answers: 0   Comments: 1

calculate bycomplex method ∫_1 ^(+∞) (dx/(1+x^2 ))

$${calculate}\:{bycomplex}\:{method}\:\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Question Number 86282    Answers: 0   Comments: 7

Question Number 86288    Answers: 1   Comments: 0

∫ (dx/((1+x^4 )^(1/4) ))

$$\int\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{4}} } \\ $$

Question Number 86269    Answers: 0   Comments: 1

∫((sec^2 (x))/((tan(x)−1)^4 (tan(x)−2))) dx

$$\int\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\left({tan}\left({x}\right)−\mathrm{1}\right)^{\mathrm{4}} \left({tan}\left({x}\right)−\mathrm{2}\right)}\:{dx} \\ $$

Question Number 86254    Answers: 1   Comments: 0

∫ (√(x(√(x(√(x(√(x.......)))))))) dx

$$\int\:\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}.......}}}}\:\:\:\mathrm{dx} \\ $$

Question Number 86247    Answers: 0   Comments: 3

∫_0 ^8 ∫_0 ^x^(2/3) (√(x^2 +y^2 +1)) dy dx

$$\int_{\mathrm{0}} ^{\mathrm{8}} \int_{\mathrm{0}} ^{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} } \:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}}\:{dy}\:{dx} \\ $$

Question Number 86246    Answers: 1   Comments: 0

∫((x^2 −1)/(x^2 +1)) (1/(√(1 + x^4 ))) dx = ?

$$\int\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:{dx}\:=\:? \\ $$

Question Number 86244    Answers: 0   Comments: 0

Question Number 86242    Answers: 1   Comments: 0

Find y=CF+PI in following differential equation: (d^2 y/dx^2 )+3(dy/dx)+2y= e^(2x) sinx .

$$\:\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{C}}\mathrm{F}+\boldsymbol{\mathrm{P}}\mathrm{I}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{following}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }+\mathrm{3}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\mathrm{2}\boldsymbol{\mathrm{y}}=\:\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{sinx}}\:. \\ $$$$\: \\ $$$$ \\ $$

Question Number 86240    Answers: 1   Comments: 0

use the Chinese Remainder theorem to find x such that x ≡ 2(mod 3) 2x ≡ 3(mod 5) 3x≡ 4( mod 7)

$$\mathrm{use}\:\mathrm{the}\:\mathrm{Chinese}\:\mathrm{Remainder}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{find} \\ $$$$\:\:{x}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{x}\:\equiv\:\mathrm{2}\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\mathrm{2}{x}\:\equiv\:\mathrm{3}\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\mathrm{3}{x}\equiv\:\mathrm{4}\left(\:\mathrm{mod}\:\mathrm{7}\right) \\ $$

Question Number 86231    Answers: 2   Comments: 1

Question Number 86230    Answers: 1   Comments: 0

is (−1)^(m/n) =((((−1 )^m ))^(1/n) ) or =(((−1))^(1/n) )^m or both of them are fault and why ?

$${is}\:\:\left(−\mathrm{1}\right)^{\frac{{m}}{{n}}} \:=\left(\sqrt[{{n}}]{\left(−\mathrm{1}\:\right)^{{m}} }\right)\:{or}\:=\left(\sqrt[{{n}}]{−\mathrm{1}}\right)^{{m}} \\ $$$${or}\:{both}\:{of}\:{them}\:{are}\:{fault}\:{and}\:{why}\:? \\ $$

Question Number 86258    Answers: 1   Comments: 3

calculate I=∫_1 ^(+∞) ((x^2 −1)/(x^4 −x^2 +1))dx

$${calculate}\:{I}=\int_{\mathrm{1}} ^{+\infty} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$

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