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Question Number 107999    Answers: 3   Comments: 5

Question Number 108000    Answers: 1   Comments: 0

If x,y>0 log x+log y=2, What is the minimum value of (1/x)+(1/y)

$$\mathrm{If}\:\mathrm{x},\mathrm{y}>\mathrm{0}\: \\ $$$$\mathrm{log}\:\mathrm{x}+\mathrm{log}\:\mathrm{y}=\mathrm{2}, \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}} \\ $$

Question Number 107995    Answers: 3   Comments: 0

Solve the differential equation; x′′(t)+2x′(t)+x(t)=1+t (using the method of variation of parameters)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}; \\ $$$$\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{2x}'\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{1}+\mathrm{t} \\ $$$$\left(\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{variation}\:\mathrm{of}\:\mathrm{parameters}\right) \\ $$

Question Number 107993    Answers: 3   Comments: 0

Question Number 107978    Answers: 1   Comments: 0

on the interval of [0,π] solve tan^2 x+3secx= −3

$${on}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\pi\right]\:{solve} \\ $$$$\mathrm{tan}^{\mathrm{2}} {x}+\mathrm{3}{secx}=\:−\mathrm{3} \\ $$$$ \\ $$

Question Number 107977    Answers: 2   Comments: 0

On the interval of [0,2π] solve sin 6x +sin 2x=0

$${On}\:{the}\:{interval}\:{of}\:\left[\mathrm{0},\mathrm{2}\pi\right]\:{solve} \\ $$$$\mathrm{sin}\:\mathrm{6}{x}\:+\mathrm{sin}\:\mathrm{2}{x}=\mathrm{0} \\ $$$$ \\ $$

Question Number 107976    Answers: 1   Comments: 0

Rewrite cos6xcos 4x as a sum or difference

$${Rewrite}\:\mathrm{cos6}{x}\mathrm{cos}\:\mathrm{4}{x}\:{as}\:{a}\:{sum}\:{or} \\ $$$${difference} \\ $$

Question Number 107975    Answers: 3   Comments: 1

((♣JS♥)/(•≡•)) (1) lim_(x→0) ((sin (tan x)−tan (sin x))/(x−sin x )) (2)lim_(x→∞) x^2 (√((1−cos ((2/x)))(√((1−cos ((2/x)))(√((1−cos ((2/x)))(√(...)))))))) ?

$$\:\:\:\:\:\:\frac{\clubsuit{JS}\heartsuit}{\bullet\equiv\bullet} \\ $$$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\:\left(\mathrm{sin}\:{x}\right)}{{x}−\mathrm{sin}\:{x}\:} \\ $$$$\:\left(\mathrm{2}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{2}}{{x}}\right)\right)\sqrt{...}}}}\:?\: \\ $$$$ \\ $$

Question Number 107974    Answers: 1   Comments: 0

Σ_(k=1) ^n (√(1+(1/k^2 )+(1/((1+k)^2 ))))

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{k}\right)^{\mathrm{2}} }} \\ $$

Question Number 107871    Answers: 1   Comments: 1

((✓JS✓)/♥) lim_(x→0) (√((x tan x)/(sin 2x−cos 2x +1))) ?

$$\:\:\:\:\:\:\:\frac{\checkmark\mathcal{JS}\checkmark}{\heartsuit} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{{x}\:\mathrm{tan}\:{x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{1}}}\:?\: \\ $$

Question Number 107965    Answers: 2   Comments: 0

((⊚BeMath⊚)/) ∫ x (√(x/(2a−x))) dx ?

$$\:\:\:\:\frac{\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc}{} \\ $$$$\int\:{x}\:\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}\:{dx}\:?\: \\ $$

Question Number 107947    Answers: 1   Comments: 0

Question Number 107946    Answers: 0   Comments: 0

Let a sequence {a_n } satisfies a_n = { ((2, n=1)),((2ln(a_(n−1) )+(1/a_(n−1) ) , n≥2)) :} Prove that a_n ≥1+(1/n) for all n∈N.

$$\mathrm{Let}\:\mathrm{a}\:\mathrm{sequence}\:\left\{{a}_{\mathrm{n}} \right\}\:\mathrm{satisfies} \\ $$$${a}_{\mathrm{n}} =\begin{cases}{\mathrm{2},\:\mathrm{n}=\mathrm{1}}\\{\mathrm{2ln}\left({a}_{\mathrm{n}−\mathrm{1}} \right)+\frac{\mathrm{1}}{{a}_{\mathrm{n}−\mathrm{1}} }\:,\:\mathrm{n}\geqslant\mathrm{2}}\end{cases} \\ $$$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$${a}_{\mathrm{n}} \geqslant\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\in\mathbb{N}. \\ $$

Question Number 107945    Answers: 3   Comments: 0

((○BeMath○)/(∧⌣∧)) { ((x^4 +(1/x^4 ) = 23)),((x^3 −(1/x^3 ) = ?)) :}

$$\:\:\:\:\:\:\frac{\circ\mathbb{B}{e}\mathbb{M}{ath}\circ}{\wedge\smile\wedge} \\ $$$$\:\:\:\begin{cases}{{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\mathrm{23}}\\{{x}^{\mathrm{3}} −\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:?}\end{cases} \\ $$

Question Number 107941    Answers: 0   Comments: 1

Question Number 107932    Answers: 1   Comments: 3

((BeMath)/•) Given { ((tan (x−y)=(3/4))),((tan x = 2 )) :} find tan y ?

$$\:\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet} \\ $$$${Given}\:\begin{cases}{\mathrm{tan}\:\left({x}−{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{tan}\:{x}\:=\:\mathrm{2}\:}\end{cases} \\ $$$${find}\:\:\mathrm{tan}\:{y}\:? \\ $$

Question Number 107930    Answers: 3   Comments: 0

((BeMath)/(•∩•)) lim_(x→0) (sin x)^(1/(ln (√x))) ?

$$\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet\cap\bullet} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\sqrt{{x}}}} \:? \\ $$

Question Number 107929    Answers: 0   Comments: 0

(1/(1+2^2 +3^3 ))+(1/(2^2 +3^3 +4^4 ))+(1/(3^3 +4^4 +5^5 ))+....

$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{4}} +\mathrm{5}^{\mathrm{5}} }+.... \\ $$

Question Number 107928    Answers: 0   Comments: 0

Question Number 107927    Answers: 0   Comments: 0

Question Number 107926    Answers: 2   Comments: 0

Question Number 107925    Answers: 2   Comments: 0

Question Number 107924    Answers: 1   Comments: 0

Question Number 107923    Answers: 3   Comments: 0

((⊚BeMath⊚)/) ∫_0 ^1 x^(9/2) (1−x)^(5/2) dx ?

$$\:\:\:\:\:\frac{\circledcirc\mathbb{B}{e}\mathcal{M}{ath}\circledcirc}{} \\ $$$$\:\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{x}^{\frac{\mathrm{9}}{\mathrm{2}}} \:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:{dx}\:? \\ $$

Question Number 107922    Answers: 1   Comments: 0

if: y=(x/(x^2 +1)) then find (d(√y)/d(√x)) ?

$${if}:\:{y}=\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{then}\:{find}\:\frac{{d}\sqrt{{y}}}{{d}\sqrt{{x}}}\:? \\ $$

Question Number 107913    Answers: 0   Comments: 0

X_0 = ((X_1 .F_(y1) + X_2 .F_(y2) + .....)/(F_(y2) + F_(y2 ) + .....)) Y_0 = ((Y_1 .F_(x1) + Y_2 .F_(x2) + ......)/(F_(x1) + F_(x2) + ......)) X_0 = ((x_1 .m_1 + x_2 .m_2 + x_3 .m_3 + ....)/(m_1 + m_2 + m_3 + ......)) Y_0 = ((y_1 .m_1 + y_2 .m_2 + y_3 .m_3 + ....)/(m_1 + m_2 + m_3 + .....)) X_z = ((l_1 .x_1 + l_2 .x_2 + l_3 .x_3 + .......)/(l_1 + l_2 + l_3 + ......)) Y_z = ((l_1 .y_1 + l_2 .y_2 + l_3 .y_3 + ...... )/(l_1 + l_2 + l_3 + ......))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{X}_{\mathrm{0}} \:=\:\frac{\mathrm{X}_{\mathrm{1}} .\mathrm{F}_{\mathrm{y1}} \:+\:\mathrm{X}_{\mathrm{2}} .\mathrm{F}_{\mathrm{y2}} \:+\:.....}{\mathrm{F}_{\mathrm{y2}} \:+\:\mathrm{F}_{\mathrm{y2}\:} +\:.....} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Y}_{\mathrm{0}} \:=\:\frac{\mathrm{Y}_{\mathrm{1}} .\mathrm{F}_{\mathrm{x1}} \:+\:\mathrm{Y}_{\mathrm{2}} .\mathrm{F}_{\mathrm{x2}} \:+\:......}{\mathrm{F}_{\mathrm{x1}} \:+\:\mathrm{F}_{\mathrm{x2}} +\:......} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{X}_{\mathrm{0}} \:=\:\frac{\mathrm{x}_{\mathrm{1}} .\mathrm{m}_{\mathrm{1}} \:+\:\mathrm{x}_{\mathrm{2}} .\mathrm{m}_{\mathrm{2}} \:+\:\mathrm{x}_{\mathrm{3}} .\mathrm{m}_{\mathrm{3}} \:+\:....}{\mathrm{m}_{\mathrm{1}} \:+\:\mathrm{m}_{\mathrm{2}} \:+\:\mathrm{m}_{\mathrm{3}} \:+\:......} \\ $$$$\:\:\:\:\:\:\:\mathrm{Y}_{\mathrm{0}} \:=\:\frac{\mathrm{y}_{\mathrm{1}} .\mathrm{m}_{\mathrm{1}} \:+\:\mathrm{y}_{\mathrm{2}} .\mathrm{m}_{\mathrm{2}} \:+\:\mathrm{y}_{\mathrm{3}} .\mathrm{m}_{\mathrm{3}} \:+\:....}{\mathrm{m}_{\mathrm{1}} \:+\:\mathrm{m}_{\mathrm{2}} \:+\:\mathrm{m}_{\mathrm{3}} \:+\:.....} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{X}_{\mathrm{z}} \:=\:\frac{{l}_{\mathrm{1}} .\mathrm{x}_{\mathrm{1}} \:+\:{l}_{\mathrm{2}} \:.\mathrm{x}_{\mathrm{2}} \:+\:{l}_{\mathrm{3}} .\mathrm{x}_{\mathrm{3}} \:+\:.......}{{l}_{\mathrm{1}} \:+\:{l}_{\mathrm{2}} \:+\:{l}_{\mathrm{3}} \:+\:......} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Y}_{\mathrm{z}} \:=\:\frac{{l}_{\mathrm{1}} .{y}_{\mathrm{1}} \:+\:{l}_{\mathrm{2}} .{y}_{\mathrm{2}} \:+\:{l}_{\mathrm{3}} .{y}_{\mathrm{3}} \:+\:......\:}{{l}_{\mathrm{1}} \:+\:{l}_{\mathrm{2}} \:+\:{l}_{\mathrm{3}} \:+\:......} \\ $$$$ \\ $$

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