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Question Number 136827 Answers: 1 Comments: 4

Find the maximum and minimum values of: 1) y=5sinx 2cosx 2) y=sin^2 x+3cos^2 x

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum} \\ $$$$\mathrm{values}\:\mathrm{of}:\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}=\mathrm{5sinx}\:\mathrm{2cosx} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}=\mathrm{sin}^{\mathrm{2}} \mathrm{x}+\mathrm{3cos}^{\mathrm{2}} \mathrm{x} \\ $$

Question Number 136826 Answers: 0 Comments: 0

Question Number 136825 Answers: 0 Comments: 0

Question Number 136828 Answers: 0 Comments: 1

Question Number 136824 Answers: 0 Comments: 0

Question Number 136823 Answers: 0 Comments: 0

Question Number 136815 Answers: 2 Comments: 0

Question Number 136813 Answers: 1 Comments: 0

......advanced calculus(I).... if : Ω=∫_(π/4) ^( (π/2)) ((x.cos(2x).cos(x))/(sin^7 (x))) dx=((aπ)/(90)) then :: a=???

$$\:\:\:\:\: \\ $$$$\:\:\:......{advanced}\:\:\:\:{calculus}\left({I}\right).... \\ $$$$\:\:\:\:{if}\::\: \\ $$$$\:\:\Omega=\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{x}.{cos}\left(\mathrm{2}{x}\right).{cos}\left({x}\right)}{{sin}^{\mathrm{7}} \left({x}\right)}\:{dx}=\frac{{a}\pi}{\mathrm{90}} \\ $$$$\:\:\:{then}\:\:::\:\:{a}=??? \\ $$$$\:\: \\ $$$$\:\:\:\: \\ $$

Question Number 136806 Answers: 2 Comments: 0

Question Number 136805 Answers: 0 Comments: 0

Question Number 136803 Answers: 1 Comments: 0

.....Advanced ◂.............▶ Calculus..... 𝛗=∫_(−∞) ^( ∞) ((sin((1/x^3 )))/x)dx=......??? solution:: 𝛗=^((1/x) =t) 2∫_0 ^( ∞) ((sin(t^3 ))/(1/t)).(dt/t^2 ) ............. =2∫_0 ^( ∞) ((sin(t^3 ))/t)dt ......... =^(t^3 =y) (2/3)∫_0 ^( ∞) ((sin(y))/y^(1/3) ).(dy/y^(2/3) )=(2/3)∫_0 ^( ∞) ((sin(y))/y) =^(⟨∫_0 ^( ∞) ((sin(r))/r)dr =(π/2)⟩) (π/3) ........... ........𝛗=∫_(−∞) ^( ∞) ((sin((1/x))^3 )/x)dx=(π/3) ........✓✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.....\mathscr{A}{dvanced}\:\:\:\:\blacktriangleleft.............\blacktriangleright\:\:\:\mathscr{C}{alculus}..... \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{{x}}{dx}=......??? \\ $$$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\frac{\mathrm{1}}{{x}}\:={t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} }\:............. \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{{t}}{dt}\:......... \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{{t}^{\mathrm{3}} ={y}} {=}\:\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}^{\frac{\mathrm{1}}{\mathrm{3}}} }.\frac{{dy}}{{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}} \\ $$$$\:\:\:\:\overset{\langle\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({r}\right)}{{r}}{dr}\:=\frac{\pi}{\mathrm{2}}\rangle} {=}\:\frac{\pi}{\mathrm{3}}\:........... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:........\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} }{{x}}{dx}=\frac{\pi}{\mathrm{3}}\:\:........\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 136799 Answers: 0 Comments: 0

∫_0 ^1 Π_(n=1) ^∞ (1−q^n )dq

$$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{q}^{{n}} \right){dq} \\ $$

Question Number 136793 Answers: 1 Comments: 0

Question Number 136792 Answers: 1 Comments: 0

How do you solve the diophantine equation (1)3xy +2x +y = 12 ? (2) x^3 = 4y^2 +4y−3 ?

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{diophantine}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\right)\mathrm{3xy}\:+\mathrm{2x}\:+\mathrm{y}\:=\:\mathrm{12}\:? \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{3}} =\:\mathrm{4y}^{\mathrm{2}} +\mathrm{4y}−\mathrm{3}\:? \\ $$

Question Number 136783 Answers: 1 Comments: 2

a,b∈R (a+b)^n =Σ_(k=0) ^n ((n),(k) )a^k b^(n−k) demontration!

$${a},{b}\in{R} \\ $$$$\left({a}+{b}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{a}^{{k}} {b}^{{n}−{k}} \\ $$$${demontration}! \\ $$

Question Number 136781 Answers: 2 Comments: 0

What is the equation of the circle passing though (-2,-4) and (4,6) its center lies on the line 3x- 2y+20=0

$$ \\ $$What is the equation of the circle passing though (-2,-4) and (4,6) its center lies on the line 3x- 2y+20=0

Question Number 136765 Answers: 2 Comments: 0

∫_0 ^1 ((ln(x+(√(1−x^2 ))))/x)dx=(π^2 /(16))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$

Question Number 136762 Answers: 0 Comments: 0

let C(a,r)={z∈C, ∣z−a∣=r } Let u,v,w ∈C(a,r) such as u+v=2w Prove that ((u−a)/(v−a)) =1 , u=v=w It shows that the middle of a segment joining two points in a circle is not in that circle

$$\:{let}\:\:{C}\left({a},{r}\right)=\left\{{z}\in\mathbb{C},\:\:\mid{z}−{a}\mid={r}\:\right\} \\ $$$${Let}\:{u},{v},{w}\:\in{C}\left({a},{r}\right)\:{such}\:{as}\:\:\:{u}+{v}=\mathrm{2}{w} \\ $$$${Prove}\:{that}\:\:\frac{{u}−{a}}{{v}−{a}}\:=\mathrm{1}\:,\:{u}={v}={w} \\ $$$$ \\ $$$${It}\:{shows}\:{that}\:{the}\:{middle}\:{of}\:{a}\:{segment}\: \\ $$$${joining}\:{two}\:{points}\:{in}\:{a}\:{circle}\:\:{is}\:{not}\:{in}\:{that}\:{circle} \\ $$

Question Number 136761 Answers: 1 Comments: 1

Question Number 136760 Answers: 1 Comments: 0

Question Number 136758 Answers: 1 Comments: 0

1==> lim_(x→∞) ((1/π)+(1/π^2 )+(1/π^3 )+∙∙∙+(1/π^n ))=? 2=>lim_(x→∞) ((1+2^2 +3^2 +∙∙∙∙+n^2 )/(1−n^3 ))=?

$$\:\:\mathrm{1}==>\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\pi}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{\mathrm{1}}{\pi^{\mathrm{3}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{\pi^{{n}} }\right)=? \\ $$$$ \\ $$$$\:\mathrm{2}=>\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\centerdot\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} }{\mathrm{1}−{n}^{\mathrm{3}} }=? \\ $$

Question Number 136757 Answers: 1 Comments: 0

1+((1/2))^2 (1/(2.1!))+(((1.3)/2^2 ))^2 (1/(2^2 .2!))+(((1.3.5)/2^3 )).(1/(2^3 .3!))+....=(((Γ((1/4)))/((2π^3 )^(1/4) )))^2

$$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}+\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right).\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+....=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{2}\pi^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{4}} }\right)^{\mathrm{2}} \\ $$

Question Number 136750 Answers: 3 Comments: 0

.....advanced calculus..... prove that :: ... 𝛗=∫_0 ^( ∞) ((1−e^(−x^2 ) )/x^2 )dx=(√π)

$$\:\:\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:\:{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:...\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi}\:\:\:\: \\ $$

Question Number 136749 Answers: 2 Comments: 0

Given system equation { ((x^2 +3xy+y^2 +1=0)),((x^3 +y^3 −7=0)) :} has solution (x_1 ,y_1 ) &(x_2 ,y_2 ) for x,y∈R. Find the value of x_1 ^2 .y_2 +x_2 ^2 .y_1 .

$$\mathrm{Given}\:\mathrm{system}\:\mathrm{equation}\: \\ $$$$\:\begin{cases}{\mathrm{x}^{\mathrm{2}} +\mathrm{3xy}+\mathrm{y}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}}\\{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} −\mathrm{7}=\mathrm{0}}\end{cases}\:\mathrm{has}\:\mathrm{solution}\: \\ $$$$\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{1}} \right)\:\&\left(\mathrm{x}_{\mathrm{2}} ,\mathrm{y}_{\mathrm{2}} \right)\:\mathrm{for}\:\mathrm{x},\mathrm{y}\in\mathbb{R}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} .\mathrm{y}_{\mathrm{2}} \:+\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} .\mathrm{y}_{\mathrm{1}} .\: \\ $$

Question Number 136745 Answers: 0 Comments: 1

Question Number 136741 Answers: 1 Comments: 0

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