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AlgebraQuestion and Answers: Page 295

Question Number 68835 Answers: 1 Comments: 4

reposting this because it′s not yet solved correctly. I criticize that most of you people do not test if your solutions fit the given equations in many cases x^2 +1−(√(x^3 +x))=6x please determine (1) how many real solutions (2) how many complex solutions we can expect (3) solve it (4) test your solutions

$$\mathrm{reposting}\:\mathrm{this}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{yet}\:\mathrm{solved} \\ $$$$\mathrm{correctly}.\:\mathrm{I}\:\mathrm{criticize}\:\mathrm{that}\:\mathrm{most}\:\mathrm{of}\:\mathrm{you}\:\mathrm{people} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{test}\:\mathrm{if}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equations}\:\mathrm{in}\:\mathrm{many}\:\mathrm{cases} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +\mathrm{1}−\sqrt{{x}^{\mathrm{3}} +{x}}=\mathrm{6}{x} \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{determine} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{how}\:\mathrm{many}\:\mathrm{complex}\:\mathrm{solutions} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{expect} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{solve}\:\mathrm{it} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{test}\:\mathrm{your}\:\mathrm{solutions} \\ $$

Question Number 68768 Answers: 1 Comments: 1

((2x−1))^(1/3) +((x−1))^(1/3) = 1

$$\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:=\:\mathrm{1} \\ $$

Question Number 68740 Answers: 1 Comments: 1

Question Number 68721 Answers: 1 Comments: 1

Question Number 68636 Answers: 2 Comments: 0

((−a^2 +a+1)/( a^2 +a+1))=((−b^2 +b+1)/( b^2 +b+1)) = (( 2a^2 −2ab+(b−a))/(−2a^2 +2ab+(b−a))) =((−2ab+(a+b)+2)/( 2ab+(a+b)+2)) Solve for a.

$$\frac{−{a}^{\mathrm{2}} +{a}+\mathrm{1}}{\:\:\:\:{a}^{\mathrm{2}} +{a}+\mathrm{1}}=\frac{−{b}^{\mathrm{2}} +{b}+\mathrm{1}}{\:\:\:\:{b}^{\mathrm{2}} +{b}+\mathrm{1}} \\ $$$$\:\:=\:\frac{\:\:\:\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ab}+\left({b}−{a}\right)}{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{ab}+\left({b}−{a}\right)} \\ $$$$\:\:=\frac{−\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}}{\:\:\:\:\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}} \\ $$$${Solve}\:{for}\:\boldsymbol{{a}}. \\ $$$$ \\ $$

Question Number 68624 Answers: 2 Comments: 0

Question Number 68549 Answers: 1 Comments: 3

y=(x^3 /(x^2 +1)) y^(−1) =...

$${y}=\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${y}^{−\mathrm{1}} =... \\ $$

Question Number 68433 Answers: 0 Comments: 0

hello i search som lectur about hypergeometric fonction2F_1 (a,b,c,x)=((Γ(c))/(Γ(a)Γ(b)))Σ_(n≥0) ((Γ(a+n)Γ(b+n))/(Γ(c+n)n!))x^n

$${hello} \\ $$$${i}\:{search}\:{som}\:{lectur}\:{about}\:{hypergeometric}\:{fonction}\mathrm{2}{F}_{\mathrm{1}} \left({a},{b},{c},{x}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({a}\right)\Gamma\left({b}\right)}\sum_{{n}\geqslant\mathrm{0}} \frac{\Gamma\left({a}+{n}\right)\Gamma\left({b}+{n}\right)}{\Gamma\left({c}+{n}\right){n}!}{x}^{{n}} \\ $$$$ \\ $$

Question Number 68422 Answers: 1 Comments: 0

Question Number 68414 Answers: 2 Comments: 1

Is it possible to find any value for a,b,c from below system of equetions? { ((sina+sinb=sinc)),((cosa+cosb=cosc)) :}

$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\mathrm{from}\:\mathrm{below}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equetions}? \\ $$$$\begin{cases}{\boldsymbol{\mathrm{sina}}+\boldsymbol{\mathrm{sinb}}=\boldsymbol{\mathrm{sinc}}}\\{\boldsymbol{\mathrm{cosa}}+\boldsymbol{\mathrm{cosb}}=\boldsymbol{\mathrm{cosc}}}\end{cases} \\ $$

Question Number 68336 Answers: 1 Comments: 1

A man gave $5,720.00 to be shared among his son and three daughters. If each of the daughter′s share is (3/4) of the son′s share, how much did the son receive?

$$\mathrm{A}\:\mathrm{man}\:\mathrm{gave}\:\$\mathrm{5},\mathrm{720}.\mathrm{00}\:\mathrm{to}\:\mathrm{be}\:\mathrm{shared}\:\mathrm{among} \\ $$$$\mathrm{his}\:\mathrm{son}\:\mathrm{and}\:\mathrm{three}\:\mathrm{daughters}.\:\mathrm{If}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{daughter}'\mathrm{s}\:\mathrm{share}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{son}'\mathrm{s}\:\mathrm{share}, \\ $$$$\mathrm{how}\:\mathrm{much}\:\mathrm{did}\:\mathrm{the}\:\mathrm{son}\:\mathrm{receive}? \\ $$

Question Number 68285 Answers: 1 Comments: 2

two students ngum ebon gave their ages as 124_4 and 33_x respectively.if both of them are of thesame ages .find in what base ebon gave her age

$${two}\:{students}\:{ngum}\:{ebon}\:{gave}\:{their}\:{ages}\:{as}\:\mathrm{124}_{\mathrm{4}} {and}\:\mathrm{33}_{{x}} {respectively}.{if}\:{both}\:{of}\:{them}\:{are}\:{of}\:{thesame}\:{ages}\:.{find}\:{in}\:{what}\:{base}\:{ebon}\:{gave}\:{her}\:{age} \\ $$

Question Number 68280 Answers: 1 Comments: 1

given that 432_n −413_n =11_(10) .find the value of n

$${given}\:{that}\:\mathrm{432}_{{n}} −\mathrm{413}_{{n}} =\mathrm{11}_{\mathrm{10}} .{find}\:{the}\:{value}\:{of}\:{n} \\ $$

Question Number 68234 Answers: 1 Comments: 5

Question Number 68207 Answers: 1 Comments: 1

solve for x∈C sin x=z (z=a+bi=re^(iθ) )

$${solve}\:{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{sin}\:{x}={z}\:\:\:\left({z}={a}+{bi}={re}^{{i}\theta} \right) \\ $$

Question Number 68161 Answers: 0 Comments: 7

In my textbook its written: In applying the nth−term test we can see that: Σ_(n=1) ^∞ (−1)^(n+1) diverges because lim_(n→∞) (−1)^(n+1) does not exist. But then why Σ_(n=1) ^∞ (−1)^(n+1) (1/n^2 ) , Σ_(n=1) ^∞ (−1)^(n+1) (1/(ln(n))) converges ?

$${In}\:{my}\:{textbook}\:{its}\:{written}: \\ $$$${In}\:{applying}\:{the}\:{nth}−{term}\:{test}\:{we}\: \\ $$$${can}\:{see}\:{that}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{diverges}\:{because}\: \\ $$$${lim}_{{n}\rightarrow\infty} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{does}\:{not}\:{exist}. \\ $$$${But}\:{then}\:{why}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:,\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{ln}\left({n}\right)} \\ $$$${converges}\:? \\ $$

Question Number 68122 Answers: 1 Comments: 1

Question Number 67997 Answers: 0 Comments: 0

5y^2 +2axy+b=0 ay^2 +2bx+5c=0 (5x+3a)y^2 +(4ax^2 )y−bx−5c=0 5y^2 −x(5x+2a)y−ax^3 −3b=0 Please solve simultaneously for x and y such that all four equations are obeyed.

$$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2}{axy}+{b}=\mathrm{0} \\ $$$${ay}^{\mathrm{2}} +\mathrm{2}{bx}+\mathrm{5}{c}=\mathrm{0} \\ $$$$\left(\mathrm{5}{x}+\mathrm{3}{a}\right){y}^{\mathrm{2}} +\left(\mathrm{4}{ax}^{\mathrm{2}} \right){y}−{bx}−\mathrm{5}{c}=\mathrm{0} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} −{x}\left(\mathrm{5}{x}+\mathrm{2}{a}\right){y}−{ax}^{\mathrm{3}} −\mathrm{3}{b}=\mathrm{0} \\ $$$${Please}\:{solve}\:{simultaneously} \\ $$$${for}\:{x}\:{and}\:{y}\:{such}\:{that}\:{all}\:{four} \\ $$$${equations}\:{are}\:{obeyed}. \\ $$

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.

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

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