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Question Number 119472    Answers: 1   Comments: 0

Question Number 122799    Answers: 1   Comments: 0

Question Number 122798    Answers: 2   Comments: 1

Question Number 119464    Answers: 2   Comments: 0

Expand as far as the term in x^3 (1) (3−2x−x^2 )^9 (2) (1−x+x^2 )^8

$$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$

Question Number 119463    Answers: 1   Comments: 2

one cube of side length 3cm is painted by three colour say (red blue and black) opposte face same colour Now cube is cut into 27 small cube i)How many cube has three face coloured ii)How many cube two face coloured iii)How many cube has no colour in its face

$${one}\:{cube}\:{of}\:{side}\:{length}\:\mathrm{3}{cm}\:{is}\:{painted}\:{by}\:{three}\:{colour} \\ $$$${say}\:\left({red}\:{blue}\:{and}\:{black}\right)\:{opposte}\:{face}\:{same}\:{colour} \\ $$$${Now}\:{cube}\:{is}\:{cut}\:{into}\:\mathrm{27}\:{small}\:{cube} \\ $$$$\left.{i}\right){How}\:{many}\:{cube}\:{has}\:{three}\:{face}\:{coloured} \\ $$$$\left.{ii}\right){How}\:{many}\:{cube}\:{two}\:{face}\:{coloured} \\ $$$$\left.{iii}\right){How}\:{many}\:{cube}\:\:{has}\:{no}\:{colour}\:{in}\:{its}\:{face} \\ $$

Question Number 119462    Answers: 2   Comments: 0

... advanced calculus... evaluate :: Ω=∫_0 ^( ∞) ((tan^(−1) (x))/(e^(2πx) −1))dx =? m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119452    Answers: 1   Comments: 1

Question Number 119451    Answers: 1   Comments: 0

Question Number 119446    Answers: 0   Comments: 0

calculste ∫_0 ^∞ ((ln(2+x^2 ))/(1+x^3 ))dx

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

Question Number 119445    Answers: 0   Comments: 0

...nice calculus... prove that:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(n^3 (((2n)),(n) ))) =^(???) ζ(3) m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119442    Answers: 0   Comments: 0

... advanced calculus... prove that : Σ_(n=1) ^∞ (1/(n^2 (((2n)),(n) ))) =^(???) ((ζ(2))/3) solution:: Σ_(n=1) ^∞ (1/(n^2 ∗(((2n)!)/((n!)^2 ))))=Σ_(n=1) ^∞ ((n!∗n!)/(n^2 ∗(2n)!)) =Σ_(n=1) ^∞ ((Γ(n)Γ(n+1))/(nΓ(2n+1)))=Σ_(n=1) ^∞ ((β(n,n+1))/n) =Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n−1) (1−x)^n =∫_0 ^( 1) (1/x)Σ(((x−x^2 )^n )/n)dx =−∫_0 ^( 1) ((ln(1−x+x^2 ))/x)dx =−∫_0 ^( 1) ((ln(1+x^3 )−ln(1+x))/x)dx =∫_0 ^( 1) ((ln(1+x))/x)dx −∫_0 ^( 1) ((ln(1+x^3 ))/x)dx =−li_2 (−1) −∫_0 ^( 1) ((Σ_(n=1) (((−1)^(n+1) x^(3n) )/n))/x) dx =(π^2 /(12))+Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^( 1) x^(3n−1) dx =(π^2 /(12)) +(1/3)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(12))−(π^2 /(36)) =(π^2 /(18)) =((ζ(2))/3) ✓✓ m.n.july.1970..

$$\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${solution}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \ast\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!\ast{n}!}{{n}^{\mathrm{2}} \ast\left(\mathrm{2}{n}\right)!}\: \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\Gamma\left({n}\right)\Gamma\left({n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left({n},{n}+\mathrm{1}\right)}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}}\Sigma\frac{\left({x}−{x}^{\mathrm{2}} \right)^{{n}} }{{n}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}{dx} \\ $$$$\:=−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\underset{{n}=\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}} }{{n}}}{{x}}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{1}} {dx}\:\:\:\:\: \\ $$$$ \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}.. \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 119440    Answers: 1   Comments: 1

Determine the integer which can be written: N=xyz^(−) ^7 =zyx^(−) ^(11)

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{which}\:\mathrm{can}\: \\ $$$$\mathrm{be}\:\mathrm{written}:\:\mathrm{N}=\overline {\mathrm{xyz}}\:^{\mathrm{7}} =\overline {\mathrm{zyx}}\:^{\mathrm{11}} \\ $$

Question Number 119425    Answers: 1   Comments: 0

decompose F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) and calculate ∫_(√2) ^(+∞) F(x)dx

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

Question Number 119421    Answers: 2   Comments: 1

find the value of Π_(i=1) ^(999) cos (ia) ; where a = ((2π)/(1999))

$${find}\:{the}\:{value}\:{of}\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{999}} {\prod}}\:\mathrm{cos}\:\left({ia}\right)\:;\:{where}\:{a}\:=\:\frac{\mathrm{2}\pi}{\mathrm{1999}} \\ $$

Question Number 119408    Answers: 1   Comments: 1

If α,β,γ are the three roots of the 3x^3 +12x^2 −77x+11=0, find the value of (α−1)(β−1)(γ−1).

$$\mathrm{If}\:\alpha,\beta,\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{three}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{3}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} −\mathrm{77}{x}+\mathrm{11}=\mathrm{0}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\alpha−\mathrm{1}\right)\left(\beta−\mathrm{1}\right)\left(\gamma−\mathrm{1}\right). \\ $$

Question Number 119401    Answers: 2   Comments: 0

let d be an application d:R^2 →R_+ d(x,y)=ln(1+((∣x−y∣)/(1+∣x−y∣))) shown that d is a distance on R^2 please help ★especially on triangular inequality

$$\boldsymbol{{let}}\:\boldsymbol{{d}}\:\boldsymbol{{be}}\:\boldsymbol{{an}}\:\boldsymbol{{application}} \\ $$$$\boldsymbol{{d}}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}_{+} \\ $$$$\boldsymbol{{d}}\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)=\boldsymbol{{ln}}\left(\mathrm{1}+\frac{\mid\boldsymbol{{x}}−\boldsymbol{{y}}\mid}{\mathrm{1}+\mid\boldsymbol{{x}}−\boldsymbol{{y}}\mid}\right) \\ $$$$\boldsymbol{{shown}}\:\boldsymbol{{that}}\:\boldsymbol{{d}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{distance}} \\ $$$$\boldsymbol{{on}}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{please}}\:\boldsymbol{{help}}\: \\ $$$$\:\bigstar\boldsymbol{{especially}}\:\boldsymbol{{on}}\:\boldsymbol{{triangular}} \\ $$$$\boldsymbol{{inequality}} \\ $$

Question Number 119397    Answers: 2   Comments: 1

Find the number of possible arrangements of the letters in the word PENCILS if (a) ′E′ is next to ′I′ (b) E comes before I (c) there are three letters between E and I

$${Find}\:{the}\:{number}\:{of}\:{possible}\:{arrangements} \\ $$$${of}\:{the}\:{letters}\:{in}\:{the}\:{word}\:{PENCILS}\:{if} \\ $$$$\left({a}\right)\:'{E}'\:{is}\:{next}\:{to}\:'{I}' \\ $$$$\left({b}\right)\:{E}\:{comes}\:{before}\:{I} \\ $$$$\left({c}\right)\:{there}\:{are}\:{three}\:{letters}\:{between}\:{E}\:{and}\:{I} \\ $$

Question Number 119396    Answers: 3   Comments: 1

If the roots of the equation 24x^4 −52x^3 +18x^2 +13x−6=0 are α , −α , β and (1/β). Find the value of α and β.

$${If}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\mathrm{24}{x}^{\mathrm{4}} −\mathrm{52}{x}^{\mathrm{3}} +\mathrm{18}{x}^{\mathrm{2}} +\mathrm{13}{x}−\mathrm{6}=\mathrm{0}\:{are}\: \\ $$$$\alpha\:,\:−\alpha\:,\:\beta\:{and}\:\frac{\mathrm{1}}{\beta}.\:{Find}\:{the}\:{value}\:{of}\: \\ $$$$\alpha\:{and}\:\beta. \\ $$

Question Number 119395    Answers: 1   Comments: 0

Question Number 119391    Answers: 1   Comments: 0

Given that f(x−3) = x^2 − 12x + 41 find an explicit expression for f(x) please I need the procedure

$$\:\mathrm{Given}\:\mathrm{that}\:{f}\left({x}−\mathrm{3}\right)\:=\:{x}^{\mathrm{2}} \:−\:\mathrm{12}{x}\:+\:\mathrm{41} \\ $$$$\:\mathrm{find}\:\mathrm{an}\:\mathrm{explicit}\:\mathrm{expression}\:\mathrm{for}\:{f}\left({x}\right) \\ $$$$ \\ $$$$\:{please}\:{I}\:\:{need}\:{the}\:{procedure} \\ $$

Question Number 119390    Answers: 2   Comments: 0

we have 15 different mathematics books, 10 different physics books and 12 different chemistry books. we should choose 6 books such that they contain all three kinds of books. in how many ways can we do this?

$${we}\:{have}\:\mathrm{15}\:{different}\:{mathematics} \\ $$$${books},\:\mathrm{10}\:{different}\:{physics}\:{books}\:{and} \\ $$$$\mathrm{12}\:{different}\:{chemistry}\:{books}.\:{we}\:{should} \\ $$$${choose}\:\mathrm{6}\:{books}\:{such}\:{that}\:{they}\:{contain} \\ $$$${all}\:{three}\:{kinds}\:{of}\:{books}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{we}\:{do}\:{this}? \\ $$

Question Number 119386    Answers: 2   Comments: 0

The length of a rectangle is decreased by 20%, and the width is increased by x%, but the area remains the same. Find the value of x.

$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{decreased}\:\mathrm{by}\:\mathrm{20\%}, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{width}\:\mathrm{is}\:\mathrm{increased}\:\mathrm{by}\:{x\%}, \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{area}\:\mathrm{remains}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}. \\ $$

Question Number 119376    Answers: 3   Comments: 2

Solve for x in the equation below ax^2 +bx + c = 0.

$${Solve}\:{for}\:\boldsymbol{{x}}\:{in}\:{the}\:{equation}\:{below} \\ $$$${ax}^{\mathrm{2}} \:+{bx}\:+\:{c}\:=\:\mathrm{0}. \\ $$

Question Number 119366    Answers: 3   Comments: 0

lim_(x→−1) (((√(1+(√(x+5))))−(√3))/(x+1)) =?

$$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{x}+\mathrm{5}}}−\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\:=? \\ $$

Question Number 119373    Answers: 4   Comments: 1

Find all sum of all x interval [ 0, 2π ] such that 3cot^2 x + 8 cot x + 3 = 0

$${Find}\:{all}\:{sum}\:{of}\:{all}\:{x}\:{interval} \\ $$$$\left[\:\mathrm{0},\:\mathrm{2}\pi\:\right]\:{such}\:{that}\:\mathrm{3cot}\:^{\mathrm{2}} {x}\:+\:\mathrm{8}\:\mathrm{cot}\:{x}\:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$

Question Number 119372    Answers: 2   Comments: 0

Determine minimum value of ((sec^4 α)/(tan^2 β)) + ((sec^4 β)/(tan^2 α)) , over all α,β ≠ ((kπ)/2) and k∈Z

$${Determine}\:{minimum}\:{value}\:{of}\: \\ $$$$\:\frac{\mathrm{sec}\:^{\mathrm{4}} \alpha}{\mathrm{tan}\:^{\mathrm{2}} \beta}\:+\:\frac{\mathrm{sec}\:^{\mathrm{4}} \beta}{\mathrm{tan}\:^{\mathrm{2}} \alpha}\:,\:{over}\:{all}\:\alpha,\beta\:\neq\:\frac{{k}\pi}{\mathrm{2}} \\ $$$${and}\:{k}\in\mathbb{Z} \\ $$

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