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Question Number 67992    Answers: 1   Comments: 0

(1) z=a+bi (2) z=re^(iθ) express the values of (a) real (z^z ) [real part] (b) imag (z^z ) [imaginary part] (c) abs (z^z ) [absolute value] (d) arg (z^z ) [argument = angle]

$$\left(\mathrm{1}\right)\:{z}={a}+{b}\mathrm{i} \\ $$$$\left(\mathrm{2}\right)\:{z}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{express}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{real}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{real}\:\mathrm{part}\right] \\ $$$$\left(\mathrm{b}\right)\:\mathrm{imag}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{imaginary}\:\mathrm{part}\right] \\ $$$$\left(\mathrm{c}\right)\:\mathrm{abs}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{absolute}\:\mathrm{value}\right] \\ $$$$\left(\mathrm{d}\right)\:\mathrm{arg}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{argument}\:=\:\mathrm{angle}\right] \\ $$

Question Number 67881    Answers: 1   Comments: 0

find all x,y ∈R such that (x+yi)^(2019) =x−yi

$${find}\:{all}\:{x},{y}\:\in{R}\:{such}\:{that} \\ $$$$\left({x}+{yi}\right)^{\mathrm{2019}} ={x}−{yi} \\ $$

Question Number 67826    Answers: 2   Comments: 4

Question Number 67819    Answers: 0   Comments: 1

y=x^5 +ax^4 +bx^3 +cx^2 +dx+e If we let x=t+h can we find h in terms of a,b,c,d,e such that y=(t+R)(t^2 +pt+q)(t^2 +s) this means two roots are of opposite sign, of course its possible by shifting the curve along x, but can we find the shift h ?

$${y}={x}^{\mathrm{5}} +{ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e} \\ $$$${If}\:{we}\:{let}\:{x}={t}+{h} \\ $$$${can}\:{we}\:{find}\:{h}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d},{e} \\ $$$${such}\:{that} \\ $$$${y}=\left({t}+{R}\right)\left({t}^{\mathrm{2}} +{pt}+{q}\right)\left({t}^{\mathrm{2}} +{s}\right) \\ $$$${this}\:{means}\:{two}\:{roots}\:{are}\:{of} \\ $$$${opposite}\:{sign},\:{of}\:{course}\:{its} \\ $$$${possible}\:{by}\:{shifting}\:{the}\:{curve} \\ $$$${along}\:{x},\:{but}\:{can}\:{we}\:{find}\:{the} \\ $$$${shift}\:\boldsymbol{{h}}\:? \\ $$

Question Number 67743    Answers: 0   Comments: 0

^

$$\:^{} \\ $$

Question Number 67711    Answers: 0   Comments: 0

Find the value of (1/(cos^2 (10°))) + (1/(sin^2 (20°))) + (1/(sin^2 (40°)))

$${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{10}°\right)}\:+\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{20}°\right)}\:+\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{40}°\right)}\: \\ $$

Question Number 67574    Answers: 0   Comments: 1

Solve x^2 +1<−5

$$\mathrm{Solve}\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}<−\mathrm{5} \\ $$

Question Number 67501    Answers: 2   Comments: 2

Show that 1n^3 + 2n + 3n^2 is divisible by 2 and 3 for all positive integers n.

$$\mathrm{Show}\:\mathrm{that}\:\:\mathrm{1n}^{\mathrm{3}} \:+\:\mathrm{2n}\:+\:\mathrm{3n}^{\mathrm{2}} \:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}. \\ $$

Question Number 67492    Answers: 0   Comments: 1

please check my comment to qu. 67471 I′ve been confusing myself...

$$\mathrm{please}\:\mathrm{check}\:\mathrm{my}\:\mathrm{comment}\:\mathrm{to}\:\mathrm{qu}.\:\mathrm{67471} \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{confusing}\:\mathrm{myself}... \\ $$

Question Number 67413    Answers: 0   Comments: 0

Question Number 67371    Answers: 1   Comments: 0

Question Number 67333    Answers: 0   Comments: 0

evaluate Σ_(n=0) ^(+∞) (1/((1+8n)^2 ))

$${evaluate}\:\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{8}{n}\right)^{\mathrm{2}} } \\ $$

Question Number 67337    Answers: 1   Comments: 0

m(p+1)=a ((q(p+1))/(r+1))=b & ((p(p+1))/(q+1))=c and ((r(p+1))/(m+1))=d find either of p,q,r,m in terms of a,b,c,d.

$${m}\left({p}+\mathrm{1}\right)={a} \\ $$$$\frac{{q}\left({p}+\mathrm{1}\right)}{{r}+\mathrm{1}}={b}\:\:\&\:\:\frac{{p}\left({p}+\mathrm{1}\right)}{{q}+\mathrm{1}}={c} \\ $$$${and}\:\:\frac{{r}\left({p}+\mathrm{1}\right)}{{m}+\mathrm{1}}={d} \\ $$$${find}\:{either}\:{of}\:{p},{q},{r},{m}\:{in}\:{terms}\:{of} \\ $$$${a},{b},{c},{d}. \\ $$

Question Number 67244    Answers: 0   Comments: 7

Which of the series converge and which diverge? Check by the limit comparison test. 1) Σ_(n=2) ^∞ ((1+n ln(n))/(n^2 +5)) 2) Σ_(n=1) ^∞ ((ln(n))/n^(3/2) ) 3) Σ_(n=3) ^∞ (1/(ln(lnn))) 4) Σ_(n=1) ^∞ (1/(n (n)^(1/n) )) ??

$${Which}\:{of}\:{the}\:{series}\:{converge}\:{and}\: \\ $$$${which}\:{diverge}?\:{Check}\:{by}\:{the}\:{limit} \\ $$$${comparison}\:{test}. \\ $$$$\left.\mathrm{1}\right)\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}+{n}\:{ln}\left({n}\right)}{{n}^{\mathrm{2}} +\mathrm{5}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{ln}\left({n}\right)}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\left.\mathrm{3}\right)\:\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{ln}\left({lnn}\right)} \\ $$$$\left.\mathrm{4}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\:\left({n}\right)^{\frac{\mathrm{1}}{{n}}} }\:\: \\ $$$$?? \\ $$

Question Number 67208    Answers: 1   Comments: 6

Find the times in a day when the hour′s, minute′s and second′s hand of a clock occupy the same angular position. [old question reposted]

$${Find}\:{the}\:{times}\:{in}\:{a}\:{day}\:{when} \\ $$$${the}\:{hour}'{s},\:{minute}'{s}\:{and}\:{second}'{s} \\ $$$${hand}\:{of}\:{a}\:{clock}\:{occupy}\:{the}\:{same} \\ $$$${angular}\:{position}. \\ $$$$\left[{old}\:{question}\:{reposted}\right] \\ $$

Question Number 67167    Answers: 4   Comments: 2

solve for real x and y:[a,b∈R] a. { ((x^3 +1=y^3 )),((x^2 +1=y^2 )) :} b. { ((x^3 +x^2 +1=y^3 )),((x^2 +x+1=y^2 )) :} c. { ((x^3 +y^2 =9xy)),((x^2 +y^3 =8xy)) :} d. { ((ax+by=2ab)),((x^2 +y^2 =4abxy)) :}

$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{real}}\:\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{y}}:\left[\mathrm{a},\mathrm{b}\in\mathrm{R}\right] \\ $$$$\boldsymbol{\mathrm{a}}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{1}=\boldsymbol{\mathrm{y}}^{\mathrm{3}} }\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}=\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\end{cases}\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{\mathrm{b}}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{1}=\boldsymbol{\mathrm{y}}^{\mathrm{3}} }\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{1}=\boldsymbol{\mathrm{y}}^{\mathrm{2}} }\end{cases} \\ $$$$\boldsymbol{\mathrm{c}}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{9}\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\mathrm{8}\boldsymbol{\mathrm{xy}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{d}}.\begin{cases}{\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{by}}=\mathrm{2}\boldsymbol{\mathrm{ab}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}\boldsymbol{\mathrm{abxy}}}\end{cases} \\ $$

Question Number 67136    Answers: 2   Comments: 0

factorize 2x^3 −1

$${factorize}\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1} \\ $$

Question Number 67025    Answers: 0   Comments: 3

Question Number 66995    Answers: 0   Comments: 4

Question Number 66969    Answers: 1   Comments: 1

Question Number 66863    Answers: 0   Comments: 0

x^3 (b−x)^2 +aex(x−b)+e^2 =0 solve for x.

$$\:\:\:{x}^{\mathrm{3}} \left({b}−{x}\right)^{\mathrm{2}} +{aex}\left({x}−{b}\right)+{e}^{\mathrm{2}} =\mathrm{0}\: \\ $$$${solve}\:{for}\:{x}. \\ $$

Question Number 66857    Answers: 0   Comments: 3

Question Number 66855    Answers: 2   Comments: 0

simplify ((p^2 +2pq+q^2 )/(p^3 −pq^2 +p^2 q−q^3 ))

$${simplify} \\ $$$$\frac{{p}^{\mathrm{2}} +\mathrm{2}{pq}+{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} −{pq}^{\mathrm{2}} +{p}^{\mathrm{2}} {q}−{q}^{\mathrm{3}} } \\ $$

Question Number 66833    Answers: 0   Comments: 0

Somewhere , there are scorpio, snake and mouse. We ascertain that : Every morning , each snake eats a mouse . Every afternoon ,each scorpio kills a snake. And every night , each mouse eats a scorpio. Two weeks passed and we find that there was remaining only one animal . 1) If that animal is a snake , how many snake were there two week ago? 2)If that animal is a scorpio , how many scorpio were there two weeks ago? 3)If that animal is a mouse ,how many mouse were there two weeks ago? 4)If there are remaining one animal of each breed , How many were they for each breed.

$${Somewhere}\:,\:{there}\:{are}\:{scorpio},\:{snake}\:{and}\:{mouse}. \\ $$$${We}\:{ascertain}\:{that}\:: \\ $$$${Every}\:{morning}\:,\:{each}\:{snake}\:{eats}\:{a}\:{mouse}\:. \\ $$$${Every}\:{afternoon}\:,{each}\:{scorpio}\:\:{kills}\:{a}\:{snake}. \\ $$$${And}\:{every}\:{night}\:,\:{each}\:{mouse}\:{eats}\:{a}\:{scorpio}. \\ $$$${Two}\:{weeks}\:{passed}\:{and}\:{we}\:{find}\:{that}\:{there}\:{was}\:{remaining}\:{only}\:{one}\:{animal}\:. \\ $$$$\left.\mathrm{1}\right)\:{If}\:{that}\:{animal}\:{is}\:{a}\:{snake}\:,\:{how}\:{many}\:{snake}\:{were}\:{there}\:{two}\:{week}\:{ago}? \\ $$$$\left.\mathrm{2}\right){If}\:{that}\:{animal}\:{is}\:{a}\:{scorpio}\:,\:{how}\:{many}\:{scorpio}\:{were}\:{there}\:{two}\:{weeks}\:{ago}? \\ $$$$\left.\mathrm{3}\right){If}\:{that}\:{animal}\:{is}\:{a}\:{mouse}\:,{how}\:{many}\:{mouse}\:{were}\:{there}\:\:{two}\:{weeks}\:{ago}? \\ $$$$\left.\mathrm{4}\right){If}\:{there}\:{are}\:{remaining}\:{one}\:{animal}\:{of}\:{each}\:{breed}\:,\:{How}\:{many}\:{were}\:{they}\:{for}\:{each}\:{breed}. \\ $$

Question Number 66827    Answers: 0   Comments: 3

Question Number 66769    Answers: 1   Comments: 2

simplify ((x+4)/(x−4))−((5x+20)/(x^2 −16))

$${simplify} \\ $$$$\frac{{x}+\mathrm{4}}{{x}−\mathrm{4}}−\frac{\mathrm{5}{x}+\mathrm{20}}{{x}^{\mathrm{2}} −\mathrm{16}} \\ $$

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