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Question Number 176581 Answers: 1 Comments: 0

Given { ((sin a+sin b=((√2)/2))),((cos a+cos b=((√6)/2))) :} for a,b real numbers. Evaluate sin (a+b). (A)((√3)/2) (D) −((√3)/2) (B) (2/( (√3))) (E)−(2/( (√3))) (C) ((√3)/4)

$$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{a}+\mathrm{sin}\:\mathrm{b}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{a}+\mathrm{cos}\:\mathrm{b}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Evaluate} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{a}+\mathrm{b}\right). \\ $$$$\:\left(\mathrm{A}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\left(\mathrm{E}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\left(\mathrm{C}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$

Question Number 176580 Answers: 1 Comments: 0

4^(x^2 −2x+2) −2^(x^2 −2x+3) +2=2^(x^2 −2x+2) x=?

$$\mathrm{4}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} −\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} +\mathrm{2}=\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} \\ $$$${x}=? \\ $$

Question Number 176571 Answers: 1 Comments: 0

lim_(x→1) ((lnx)/(1+lnx−1))=?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{lnx}}{\mathrm{1}+{lnx}−\mathrm{1}}=? \\ $$

Question Number 176570 Answers: 2 Comments: 0

(1) ∫^(π/2) _(π/3) ((1+sinx)/(cosx)) dx=?

$$\left(\mathrm{1}\right)\:\:\underset{\frac{\pi}{\mathrm{3}}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{sinx}}{{cosx}}\:{dx}=? \\ $$

Question Number 176566 Answers: 0 Comments: 0

−−−− calculate: Φ = Σ_(n=0) ^( ∞) (( 1)/((2n+1 ).e^( 4n+2) )) = ? where ” e ” is euler number. ≺ solution ≻ Φ = Σ_(n=0) ^∞ (1/e^( 4n+2) ) ∫_0 ^( 1) x^( 2n) dx = (1/e^( 2) ) ∫_0 ^( 1) Σ_(n=0) ^∞ ((( x^2 )/e^( 4) ) )^( n) dx = (1/e^( 2) ) ∫_0 ^( 1) (( 1)/(1− ((x/e^( 2) ) )^( 2) )) dx=(1/(2e^( 2) )) ∫_0 ^( 1) (1/(1−(x/e^( 2) ))) +(1/(1+(x/e^( 2) )))dx = (1/2) ln ( ((1+(1/e^( 2) ))/(1−(1/e^( 2) ))) ) = tanh^( −1) ((( 1)/e^( 2) ) ) ∴ Φ = coth^( −1) ( e^( 2) ) ■ m.n

$$−−−− \\ $$$$\:\:{calculate}:\:\:\:\:\Phi\:=\:\underset{{n}=\mathrm{0}} {\overset{\:\infty} {\sum}}\:\frac{\:\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\:\right).{e}^{\:\mathrm{4}{n}+\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:{where}\:\:''\:\:{e}\:\:''\:\:{is}\:\:{euler}\:{number}. \\ $$$$\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:\:\Phi\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{e}^{\:\mathrm{4}{n}+\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\:\:{x}^{\mathrm{2}} }{{e}^{\:\mathrm{4}} }\:\right)^{\:{n}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}}{\mathrm{1}−\:\left(\frac{{x}}{{e}^{\:\mathrm{2}} }\:\right)^{\:\mathrm{2}} }\:{dx}=\frac{\mathrm{1}}{\mathrm{2}{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−\frac{{x}}{{e}^{\:\mathrm{2}} }}\:+\frac{\mathrm{1}}{\mathrm{1}+\frac{{x}}{{e}^{\:\mathrm{2}} }}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:{ln}\:\left(\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }}\:\:\right)\:\:\:\:=\:{tanh}^{\:−\mathrm{1}} \left(\frac{\:\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\Phi\:=\:{coth}^{\:−\mathrm{1}} \left(\:{e}^{\:\mathrm{2}} \right)\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\: \\ $$$$\:\:\:\: \\ $$

Question Number 176565 Answers: 0 Comments: 0

Question Number 176564 Answers: 0 Comments: 0

Question Number 176563 Answers: 1 Comments: 0

is there an iOS version for this app? please

$$\mathrm{is}\:\mathrm{there}\:\mathrm{an}\:\mathrm{iOS}\:\mathrm{version}\:\mathrm{for}\:\mathrm{this}\:\mathrm{app}?\:\mathrm{please} \\ $$

Question Number 176555 Answers: 1 Comments: 2

ABCD−convex quadrilateral M∈Int(ABCD) , F−area , s−semiperimetr a , b , c−sides. Prove that: ((MA^4 )/b) + ((MB^4 )/c^4 ) + ((MC^4 )/d) + ((MD^4 )/a) ≥ ((2F^2 )/s)

$$\mathrm{ABCD}−\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{M}\in\mathrm{Int}\left(\mathrm{ABCD}\right)\:,\:\mathrm{F}−\mathrm{area}\:,\:\mathrm{s}−\mathrm{semiperimetr} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}−\mathrm{sides}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{MA}^{\mathrm{4}} }{\mathrm{b}}\:+\:\frac{\mathrm{MB}^{\mathrm{4}} }{\mathrm{c}^{\mathrm{4}} }\:+\:\frac{\mathrm{MC}^{\mathrm{4}} }{\mathrm{d}}\:+\:\frac{\mathrm{MD}^{\mathrm{4}} }{\mathrm{a}}\:\geqslant\:\frac{\mathrm{2F}^{\mathrm{2}} }{\mathrm{s}} \\ $$

Question Number 176554 Answers: 2 Comments: 0

x + (1/x) =ϕ (Golden ratio) then x^( 2000) +(( 1)/x^( 2000) ) = ?

$$ \\ $$$$\:\:\:\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\varphi\:\:\left({Golden}\:{ratio}\right) \\ $$$$\:\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\:\mathrm{2000}} \:+\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2000}} }\:=\:? \\ $$$$ \\ $$

Question Number 176576 Answers: 0 Comments: 0

Question Number 176549 Answers: 1 Comments: 0

donner la forme trigonometrique de 1/4(cosΠ/9+isinΠ/9)

$${donner}\:{la}\:{forme}\:{trigonometrique}\:{de}\:\mathrm{1}/\mathrm{4}\left({cos}\Pi/\mathrm{9}+{isin}\Pi/\mathrm{9}\right) \\ $$

Question Number 176547 Answers: 0 Comments: 0

Question Number 176546 Answers: 0 Comments: 0

Question Number 176542 Answers: 1 Comments: 0

Ω = ∫_0 ^( 1) (( x.tanh^( −1) (x))/((1+x)^( 2) ))dx= (1/(24)) (π^( 2) −6)

$$ \\ $$$$\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.{tanh}^{\:−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}\right)^{\:\mathrm{2}} }{dx}=\:\frac{\mathrm{1}}{\mathrm{24}}\:\left(\pi^{\:\mathrm{2}} −\mathrm{6}\right) \\ $$

Question Number 176540 Answers: 1 Comments: 0

Question Number 176531 Answers: 1 Comments: 4

Dans la figure ci−joint AB∣∣ A^′ B^′ AA^′ =BB^′ =CC^′ Determiner le rapport ((aire Δ(A^′ B^′ C^′ ))/(aire Δ(ABC)))=?

$${Dans}\:{la}\:{figure}\:{ci}−{joint} \\ $$$${AB}\mid\mid\:{A}^{'} {B}^{'} \:\:\:{AA}^{'} ={BB}^{'} ={CC}^{'} \\ $$$${Determiner}\:{le}\:{rapport} \\ $$$$\:\:\:\:\:\frac{{aire}\:\Delta\left({A}^{'} {B}^{'} {C}^{'} \right)}{{aire}\:\Delta\left({ABC}\right)}=? \\ $$

Question Number 176514 Answers: 1 Comments: 0

△ABC, sinA=cosB=tanC find the value of cos^3 A+cos^2 A−cosA.

$$\bigtriangleup{ABC},\:\mathrm{sin}{A}=\mathrm{cos}{B}=\mathrm{tan}{C} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}^{\mathrm{3}} {A}+\mathrm{cos}^{\mathrm{2}} {A}−\mathrm{cos}{A}. \\ $$

Question Number 176513 Answers: 1 Comments: 0

z∈C, ((z−3i)/(z+i))∈R^− , ((z−3)/(z+1))∈I find z.

$${z}\in\mathbb{C},\:\frac{{z}−\mathrm{3i}}{{z}+\mathrm{i}}\in\mathbb{R}^{−} \:,\:\:\frac{{z}−\mathrm{3}}{{z}+\mathrm{1}}\in\mathbb{I} \\ $$$$\mathrm{find}\:{z}. \\ $$

Question Number 176501 Answers: 1 Comments: 1

find the range of x+y such that (x−2)^2 + (y−4)^2 = 49

$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\:\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{49} \\ $$

Question Number 176490 Answers: 2 Comments: 0

solve ( x,y ∈ R ) { (( tan(x ) + tan (y )=2)),(( tan(2x ) + tan( 2y ) = 2)) :} −−−−−−−−

$$ \\ $$$$\:\:\:\:\:{solve}\:\:\:\left(\:{x},{y}\:\in\:\mathbb{R}\:\right) \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\begin{cases}{\:\mathrm{tan}\left({x}\:\right)\:+\:\mathrm{tan}\:\left({y}\:\right)=\mathrm{2}}\\{\:\mathrm{tan}\left(\mathrm{2}{x}\:\right)\:+\:\mathrm{tan}\left(\:\mathrm{2}{y}\:\right)\:=\:\mathrm{2}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−−− \\ $$

Question Number 176486 Answers: 1 Comments: 0

lim_(x→0) [ ((tan x−x)/x^5 ) −(1/(3x^2 )) ]=?

$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:−\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }\:\right]=? \\ $$

Question Number 176485 Answers: 2 Comments: 1

Question Number 176482 Answers: 0 Comments: 0

L(E) est l′algebre des endomorphisme continus d′un espace de Banach E, muni de la norme d′application lineaire ; GL(E) est le sous ensemble des elements inversibles de L(E) a)Montrer que GL(E) est ouvert dans L(E) b)montrer que l′application ∅:GL(E)→GL(E) u→u^(−1) =∅(u) est continu dans GL(E) c)montrer que ∅ est differentiable dans GL(E) dt calculer d∅

$${L}\left({E}\right)\:{est}\:{l}'{algebre}\:{des}\:{endomorphisme} \\ $$$${continus}\:{d}'{un}\:{espace}\:{de}\:{Banach}\:{E}, \\ $$$${muni}\:{de}\:{la}\:{norme}\:{d}'{application}\:{lineaire} \\ $$$$;\:{GL}\left({E}\right)\:{est}\:{le}\:{sous}\:{ensemble}\:{des} \\ $$$${elements}\:{inversibles}\:{de}\:{L}\left({E}\right) \\ $$$$\left.{a}\right){Montrer}\:{que}\:{GL}\left({E}\right)\:{est}\:{ouvert}\:{dans} \\ $$$${L}\left({E}\right) \\ $$$$\left.{b}\right){montrer}\:{que}\:{l}'{application}\: \\ $$$$\emptyset:{GL}\left({E}\right)\rightarrow{GL}\left({E}\right)\: \\ $$$$\:\:\:\:\:\:\:{u}\rightarrow{u}^{−\mathrm{1}} =\emptyset\left({u}\right)\:{est}\:{continu}\:{dans}\:{GL}\left({E}\right) \\ $$$$\left.{c}\right){montrer}\:{que}\:\emptyset\:{est}\:{differentiable} \\ $$$${dans}\:{GL}\left({E}\right)\:{dt}\:{calculer}\:{d}\emptyset \\ $$

Question Number 176472 Answers: 1 Comments: 0

solve for x,y,z with x+y+z=x^2 +y^2 +z^2 =x^3 +y^3 +z^3 =5

$${solve}\:{for}\:{x},{y},{z}\:{with} \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{5} \\ $$

Question Number 176468 Answers: 1 Comments: 1

((sin (2x+18°))/(sin (2x+12°))) =(√((sin 36°)/(sin 48°))) tan 2x = (√(tan M)) .(√(tan N)) 0°<M,N<90° ⇒M+N=?°

$$\:\:\frac{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{18}°\right)}{\mathrm{sin}\:\left(\mathrm{2}{x}+\mathrm{12}°\right)}\:=\sqrt{\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{sin}\:\mathrm{48}°}}\: \\ $$$$\:\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\sqrt{\mathrm{tan}\:{M}}\:.\sqrt{\mathrm{tan}\:{N}} \\ $$$$\:\:\mathrm{0}°<{M},{N}<\mathrm{90}°\:\Rightarrow{M}+{N}=?° \\ $$

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