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Question Number 146960 Answers: 2 Comments: 0

Question Number 146959 Answers: 1 Comments: 0

∫ln(1+(√(x^2 +2x+4)))dx

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

Question Number 147764 Answers: 1 Comments: 0

Question Number 146943 Answers: 3 Comments: 0

b_(n+2) ∙ b_(n+3) ∙ b_(n+4) = 3^(3n+3) geometric series b_8 = ?

$${b}_{\boldsymbol{{n}}+\mathrm{2}} \:\centerdot\:{b}_{\boldsymbol{{n}}+\mathrm{3}} \:\centerdot\:{b}_{\boldsymbol{{n}}+\mathrm{4}} \:=\:\mathrm{3}^{\mathrm{3}\boldsymbol{{n}}+\mathrm{3}} \\ $$$${geometric}\:{series}\:\:\boldsymbol{{b}}_{\mathrm{8}} \:=\:? \\ $$

Question Number 146938 Answers: 0 Comments: 0

Question Number 146936 Answers: 3 Comments: 0

{ ((x^2 +2y^2 +xy=37)),((y^2 +2x^2 +2xy=26)) :} ⇒ x^2 +y^2 =?

$$\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} +{xy}=\mathrm{37}}\\{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{26}}\end{cases}\:\Rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =? \\ $$

Question Number 146933 Answers: 0 Comments: 0

.....# advanced calculu#...... I := ∫_(−∞) ^( +∞) sin(cosh(x).cos(sinhx))dx=? ....solution .... I:=(1/2) ∫_(−∞) ^( +∞) {sin (cosh(x)+sinh(x))+sin(cosh(x)−sinh (x))} :=_(sinh(x)=((e^( x) −e^( −x) )/2)) ^(cosh(x)=((e^( x) +e^( −x) )/2)) (1/2) ∫_(−∞) ^( +∞) {sin(e^x )+sin (e^( −x) )}dx := (1/2) ∫_(−∞) ^( ∞) sin(e^( x) )dx +[(1/2)∫_(−∞) ^( +∞) sin(e^( x) )dx :: x=^(sub) −x] := ∫_(−∞) ^( +∞) sin(e^( x) ) dx =^(e^( x) =y) ∫_0 ^( ∞) ((sin(t))/t) dt ...... I:= (π/2) ..... ...m.n.1970...

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....#\:{advanced}\:\:{calculu}#...... \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{−\infty} ^{\:+\infty} {sin}\left({cosh}\left({x}\right).{cos}\left({sinhx}\right)\right){dx}=? \\ $$$$\:\:\:\:\:....{solution}\:.... \\ $$$$\:\:\:\:\:\:\:\mathrm{I}:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{\:+\infty} \left\{{sin}\:\left({cosh}\left({x}\right)+{sinh}\left({x}\right)\right)+{sin}\left({cosh}\left({x}\right)−{sinh}\:\left({x}\right)\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\::\underset{{sinh}\left({x}\right)=\frac{{e}^{\:{x}} −{e}^{\:−{x}} }{\mathrm{2}}} {\overset{{cosh}\left({x}\right)=\frac{{e}^{\:{x}} +{e}^{\:−{x}} }{\mathrm{2}}} {=}}\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{\:+\infty} \left\{{sin}\left({e}^{{x}} \right)+{sin}\:\left({e}^{\:−{x}} \right)\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\::=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{\:\infty} {sin}\left({e}^{\:{x}} \right){dx}\:+\left[\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\:+\infty} {sin}\left({e}^{\:{x}} \right){dx}\:::\:\:{x}\overset{{sub}} {=}−{x}\right] \\ $$$$\:\:\:\:\:\:\:\::=\:\int_{−\infty} ^{\:+\infty} {sin}\left({e}^{\:{x}} \:\right)\:{dx}\:\overset{{e}^{\:{x}} ={y}} {=}\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}\right)}{{t}}\:{dt}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\:\mathrm{I}:=\:\frac{\pi}{\mathrm{2}}\:..... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$\: \\ $$$$ \\ $$

Question Number 146929 Answers: 1 Comments: 5

Question Number 146927 Answers: 1 Comments: 0

Question Number 146926 Answers: 1 Comments: 0

Question Number 146924 Answers: 1 Comments: 0

Montrer que Σ_(k=0) ^(2n−1) cos^(2n) (𝛉+((k𝛑)/(2n)))= ((nC_(2n) ^n )/2^(2n−1) )

$$\boldsymbol{\mathrm{Montrer}}\:\boldsymbol{\mathrm{que}} \\ $$$$\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} {\sum}}\boldsymbol{{cos}}^{\mathrm{2}\boldsymbol{{n}}} \left(\boldsymbol{\theta}+\frac{\boldsymbol{{k}\pi}}{\mathrm{2}\boldsymbol{{n}}}\right)=\:\frac{\boldsymbol{{nC}}_{\mathrm{2}\boldsymbol{{n}}} ^{\boldsymbol{{n}}} }{\mathrm{2}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{1}} } \\ $$

Question Number 146923 Answers: 1 Comments: 0

if arg (((i - z)/i)) = (π/4) find RemZ + ImZ = ?

$${if}\:\:\:\:{arg}\:\left(\frac{{i}\:-\:{z}}{{i}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$$${find}\:\:\:\:{RemZ}\:+\:{ImZ}\:=\:? \\ $$

Question Number 146914 Answers: 2 Comments: 0

∣x^2 −3x−4∣ = ∣x−4∣ ⇒ x=?

$$\mid{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\mid\:=\:\mid{x}−\mathrm{4}\mid\:\Rightarrow\:{x}=? \\ $$

Question Number 146913 Answers: 1 Comments: 0

ax=by=cz=(2/3) and ab+bc+ac=36abc find x+y+z=?

$${ax}={by}={cz}=\frac{\mathrm{2}}{\mathrm{3}}\:{and}\:{ab}+{bc}+{ac}=\mathrm{36}{abc} \\ $$$${find}\:\:{x}+{y}+{z}=? \\ $$

Question Number 146906 Answers: 0 Comments: 0

Question Number 147078 Answers: 0 Comments: 1

Question Number 146904 Answers: 0 Comments: 0

F = ((Gm_1 m_2 )/r^2 ) Find the value of r with the following equation.

$$\boldsymbol{\mathrm{F}}\:=\:\frac{\boldsymbol{\mathrm{Gm}}_{\mathrm{1}} \boldsymbol{\mathrm{m}}_{\mathrm{2}} }{\boldsymbol{\mathrm{r}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{r}}\:\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}}\: \\ $$$$\boldsymbol{\mathrm{equation}}. \\ $$

Question Number 146902 Answers: 1 Comments: 0

let α and β roots of z^2 +3z+5=0 simlify U_n = Σ_(k=0) ^n (α^k +β^k ) and V_n =Σ_(k=0) ^n ((1/α^k )+(1/β^k ))

$$\mathrm{let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{roots}\:\mathrm{of}\:\:\mathrm{z}^{\mathrm{2}} +\mathrm{3z}+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{simlify}\:\mathrm{U}_{\mathrm{n}} =\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\alpha^{\mathrm{k}} \:+\beta^{\mathrm{k}} \right) \\ $$$$\mathrm{and}\:\mathrm{V}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\frac{\mathrm{1}}{\alpha^{\mathrm{k}} }+\frac{\mathrm{1}}{\beta^{\mathrm{k}} }\right) \\ $$

Question Number 146901 Answers: 1 Comments: 0

g(x)=cos(2arcsinx) calculate (dg/dx) and (d^2 g/dx^2 ) 2)find ∫_(−(1/2)) ^(1/2) g(x)dx

$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{cos}\left(\mathrm{2arcsinx}\right)\:\: \\ $$$$\mathrm{calculate}\:\frac{\mathrm{dg}}{\mathrm{dx}}\:\mathrm{and}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{g}}{\mathrm{dx}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\mathrm{find}\:\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{g}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 146899 Answers: 1 Comments: 0

f(x)=sin^5 x calculate f^((5)) ((π/2))

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{\mathrm{5}} \mathrm{x}\:\:\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{5}\right)} \left(\frac{\pi}{\mathrm{2}}\right) \\ $$

Question Number 146898 Answers: 2 Comments: 0

calculate ∫_0 ^∞ ((cosx)/((x^2 +1)(x^2 +2)(x^2 +3)))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{cosx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)}\mathrm{dx} \\ $$

Question Number 146896 Answers: 2 Comments: 0

(5/( (6)^(1/8) + 1)) ∙ (1/( (6)^(1/4) + 1)) ∙ (1/( (√6) + 1)) + 1 = ?

$$\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:=\:? \\ $$

Question Number 146895 Answers: 1 Comments: 0

(1/(2 + log_3 (25))) + (1/(2 + log_5 (9))) = ?

$$\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{3}} \left(\mathrm{25}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{5}} \left(\mathrm{9}\right)}\:=\:? \\ $$

Question Number 146886 Answers: 1 Comments: 1

Question Number 146878 Answers: 0 Comments: 0

in ΔABC if sin^2 A sin B sin C+ cos Bcos C=1 then the triangle is

$${in}\:\Delta{ABC}\:{if}\:\mathrm{sin}\:^{\mathrm{2}} {A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}+ \\ $$$$\mathrm{cos}\:{B}\mathrm{cos}\:{C}=\mathrm{1}\:{then}\:{the}\:{triangle}\:{is} \\ $$

Question Number 146877 Answers: 0 Comments: 0

if the sides a,b,c of a triangle ABC are in A.P. and if sin A =(sin B +sin C)cos α sin B =(sin C+sin A)cos β sin C =(sin A +sin B)cos γ then find the value of tan^2 (α/2)+tan^2 (γ/2)

$${if}\:{the}\:{sides}\:{a},{b},{c}\:{of}\:{a}\:{triangle}\:{ABC} \\ $$$${are}\:{in}\:{A}.{P}.\:{and}\:{if}\: \\ $$$$\mathrm{sin}\:{A}\:=\left(\mathrm{sin}\:{B}\:+\mathrm{sin}\:{C}\right)\mathrm{cos}\:\alpha \\ $$$$\mathrm{sin}\:{B}\:=\left(\mathrm{sin}\:{C}+\mathrm{sin}\:{A}\right)\mathrm{cos}\:\beta \\ $$$$\mathrm{sin}\:{C}\:=\left(\mathrm{sin}\:{A}\:+\mathrm{sin}\:{B}\right)\mathrm{cos}\:\gamma \\ $$$${then}\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\mathrm{tan}\:^{\mathrm{2}} \frac{\alpha}{\mathrm{2}}+\mathrm{tan}\:^{\mathrm{2}} \frac{\gamma}{\mathrm{2}} \\ $$

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