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Question Number 81364    Answers: 0   Comments: 2

log_2 (log_3 (log_2 (2/x))−1) < 1 has solution (1/a^(26) )<x<b. find a ?

$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{x}}\right)−\mathrm{1}\right)\:<\:\mathrm{1}\: \\ $$$$\mathrm{has}\:\mathrm{solution}\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{26}} }<\mathrm{x}<\mathrm{b}.\:\mathrm{find}\:\mathrm{a}\:? \\ $$

Question Number 81354    Answers: 0   Comments: 2

given y = sin ((π/3)−2x)+2cos ((π/(12))+x)+1 where x ∈(0,2π) has maximum and minimum value is p and q. find p^2 −q^2 ? (A) −18 (B) −16 (C) ((63)/4) (D) 16

$$\mathrm{given}\:\mathrm{y}\:=\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\mathrm{2x}\right)+\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{12}}+\mathrm{x}\right)+\mathrm{1} \\ $$$$\mathrm{where}\:\mathrm{x}\:\in\left(\mathrm{0},\mathrm{2}\pi\right)\:\mathrm{has}\:\mathrm{maximum}\:\mathrm{and} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{is}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}. \\ $$$$\mathrm{find}\:\mathrm{p}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} \:? \\ $$$$\left(\mathrm{A}\right)\:−\mathrm{18}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:−\mathrm{16} \\ $$$$\left(\mathrm{C}\right)\:\frac{\mathrm{63}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{16} \\ $$

Question Number 81340    Answers: 1   Comments: 4

Quation posted Times ago i can′t find it .after Somme Try i got close Fofme ∫_0 ^π sin(x^2 )dx=(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4),−(π^4 /4))? sin(x)=Σ_(k≥0) (((−1)^k x^(2k+1) )/((2k+1)!)) ⇒sin(x^2 )=Σ_(k≥0) (((−1)^k x^(4k+2) )/((2k+1)!)) ∫_0 ^π sin(x^2 )dx=Σ_(k≥0) (((−1)^k )/((2k+1)!))∫_0 ^π x^(4k+2) dx =Σ_(k≥0) (((−1)^k π^(4k+3) )/((2k+1)!(4k+3))) =(π^3 /3)Σ_(k≥0) ((3(−1)^k π^(4k) )/((2k+1)!(4k+3))) (2k+1)!=k!.Π_(j=1) ^(k+1) (k+j)=k!.2^k .3....(2k+1) =(π^3 /3).Σ_(k≥0) ((3(−1)^k π^(4k) )/(k!.2^k .3....(2k+1)(4k+3))) =(π^3 /3)Σ_(k≥0) ((2^k .3)/(3...(2k+1).(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) ((2^k .3.....(4k−1))/(3.....(2k+1).(7.....(4k−1)(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) (((1/4^k ).(3)...(4k−1))/((1/2^k ).(3)...(2k+1).(1/4^k )(7)...(4k+3))).(−(π^4 /4))^k .(1/(k!)) =(π^3 /3)Σ_(k≥0) ((((3/4))....((3/4)+k−1))/(((3/2))....((3/2)+k−1).((7/4))....((7/4)+k−1))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4);−(π^4 /4))

$${Quation}\:\:{posted}\:{Times}\:{ago} \\ $$$${i}\:{can}'{t}\:{find}\:{it}\:.{after}\:{Somme}\:{Try}\:{i}\:{got}\:{close}\:{Fofme} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}},−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)? \\ $$$${sin}\left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{sin}\left({x}^{\mathrm{2}} \right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{4}{k}+\mathrm{2}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\pi} {x}^{\mathrm{4}{k}+\mathrm{2}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}+\mathrm{3}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)!={k}!.\underset{{j}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\prod}}\left({k}+{j}\right)={k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{{k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}}{\mathrm{3}...\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}.....\left(\mathrm{4}{k}−\mathrm{1}\right)}{\mathrm{3}.....\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{7}.....\left(\mathrm{4}{k}−\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)\right.}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\frac{\mathrm{1}}{\mathrm{4}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{4}{k}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{2}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{2}{k}+\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{4}^{{k}} }\left(\mathrm{7}\right)...\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)....\left(\frac{\mathrm{3}}{\mathrm{4}}+{k}−\mathrm{1}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)....\left(\frac{\mathrm{3}}{\mathrm{2}}+{k}−\mathrm{1}\right).\left(\frac{\mathrm{7}}{\mathrm{4}}\right)....\left(\frac{\mathrm{7}}{\mathrm{4}}+{k}−\mathrm{1}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}};−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right) \\ $$$$ \\ $$

Question Number 81336    Answers: 1   Comments: 7

decompose F(x)=(1/((x−1)^3 (x+3)^7 )) and detrmine ∫ F(x)dx 2) calculate ∫_2 ^(+∞) F(x)dx

$${decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{3}\right)^{\mathrm{7}} } \\ $$$${and}\:{detrmine}\:\:\int\:{F}\left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:{F}\left({x}\right){dx} \\ $$

Question Number 81331    Answers: 0   Comments: 10

Question Number 81322    Answers: 1   Comments: 1

Question Number 81320    Answers: 0   Comments: 0

Question Number 81319    Answers: 1   Comments: 0

Question Number 81318    Answers: 0   Comments: 1

Question Number 81313    Answers: 1   Comments: 5

∫(((x−2)^3 )/((x+2)^5 )) dx

$$\int\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{5}} }\:\mathrm{dx} \\ $$

Question Number 81311    Answers: 0   Comments: 3

Question Number 81308    Answers: 1   Comments: 2

Question Number 81303    Answers: 1   Comments: 1

Question Number 81301    Answers: 0   Comments: 2

Question Number 81295    Answers: 1   Comments: 2

what is vector unit orthogonal to (1,2,−2) and parallel to yz−plane?

$${what}\:{is}\:{vector}\:{unit}\:{orthogonal} \\ $$$${to}\:\left(\mathrm{1},\mathrm{2},−\mathrm{2}\right)\:{and}\:{parallel}\:{to}\: \\ $$$${yz}−{plane}? \\ $$

Question Number 81290    Answers: 0   Comments: 3

∫_(−1) ^1 arctan(x)arctan((x/(√(1−x^2 ))))arctan(((1+x)/(1−x)))dx

$$\int_{−\mathrm{1}} ^{\mathrm{1}} {arctan}\left({x}\right){arctan}\left(\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right){arctan}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right){dx} \\ $$

Question Number 81289    Answers: 0   Comments: 5

discussthesymmetryofthefollowingcurves squareofx+squareofy=1

$${discussthesymmetryofthefollowingcurves} \\ $$$${squareofx}+{squareofy}=\mathrm{1} \\ $$

Question Number 81286    Answers: 1   Comments: 0

what is tan ((3π)/(11)) + 4sin ((2π)/(11)) ?

$${what}\:{is}\: \\ $$$$\mathrm{tan}\:\frac{\mathrm{3}\pi}{\mathrm{11}}\:+\:\mathrm{4sin}\:\frac{\mathrm{2}\pi}{\mathrm{11}}\:? \\ $$

Question Number 81279    Answers: 0   Comments: 2

∫ ((√(2x+1))/(3x)) dx = ?

$$\int\:\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}\:{dx}\:=\:? \\ $$

Question Number 81275    Answers: 0   Comments: 0

Question Number 81274    Answers: 0   Comments: 0

Question Number 81267    Answers: 0   Comments: 5

p,is a point,inside ,onside ,outside of equilateral triangle.find side of triangle if distance of :p from vertices of triangle be equail to: 5,7,11. (study each conditions separately). find side of ABC and p_1 p_2 p_3 in a special case that: { ((Ap_1 =11)),((Bp_2 =5)),((Cp_3 =7)) :}

$$\boldsymbol{\mathrm{p}},\mathrm{is}\:\mathrm{a}\:\mathrm{point},\boldsymbol{\mathrm{inside}}\:,\boldsymbol{\mathrm{onside}}\:,\boldsymbol{\mathrm{outside}}\:\mathrm{of} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}.\mathrm{find}\:\mathrm{side}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{if}\:\mathrm{distance}\:\mathrm{of}\::\boldsymbol{\mathrm{p}}\:\mathrm{from}\:\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{be}\:\mathrm{equail}\:\mathrm{to}:\:\mathrm{5},\mathrm{7},\mathrm{11}. \\ $$$$\left(\mathrm{study}\:\mathrm{each}\:\mathrm{conditions}\:\mathrm{separately}\right). \\ $$$$\mathrm{find}\:\mathrm{side}\:\mathrm{of}\:\mathrm{ABC}\:\mathrm{and}\:\mathrm{p}_{\mathrm{1}} \mathrm{p}_{\mathrm{2}} \mathrm{p}_{\mathrm{3}} \:\mathrm{in}\:\mathrm{a}\: \\ $$$$\mathrm{special}\:\mathrm{case}\:\mathrm{that}:\begin{cases}{\mathrm{Ap}_{\mathrm{1}} =\mathrm{11}}\\{\mathrm{Bp}_{\mathrm{2}} =\mathrm{5}}\\{\mathrm{Cp}_{\mathrm{3}} =\mathrm{7}}\end{cases} \\ $$

Question Number 81242    Answers: 1   Comments: 3

Question Number 81226    Answers: 1   Comments: 0

(d/dx)(x!) and (d/dx)(1+2+3+...+x)

$$\frac{{d}}{{dx}}\left({x}!\right)\:{and}\:\frac{{d}}{{dx}}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+{x}\right) \\ $$

Question Number 81221    Answers: 0   Comments: 1

Question Number 81219    Answers: 1   Comments: 4

given a probability function f(x)= (1/3), 1≤x≤4 and f(x)=0 in other x. find the value of σ^(2 ) ?

$${given}\:{a}\:{probability}\: \\ $$$${function}\: \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{1}\leqslant{x}\leqslant\mathrm{4}\:{and}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${in}\:{other}\:{x}.\:{find}\:{the}\:{value}\: \\ $$$${of}\:\sigma^{\mathrm{2}\:} \:? \\ $$

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