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AlgebraQuestion and Answers: Page 241

Question Number 120324 Answers: 1 Comments: 0

we are in C. (E): z^3 +(4−5i)z^2 +(8−20i)z−40i=0 1) Show that (E) has one imaginary pure root 2) solve (E)

$$\mathrm{we}\:\mathrm{are}\:\mathrm{in}\:\mathbb{C}. \\ $$$$\left(\mathrm{E}\right):\:\mathrm{z}^{\mathrm{3}} +\left(\mathrm{4}−\mathrm{5i}\right)\mathrm{z}^{\mathrm{2}} +\left(\mathrm{8}−\mathrm{20i}\right)\mathrm{z}−\mathrm{40i}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{E}\right)\:\mathrm{has}\:\mathrm{one}\:\mathrm{imaginary}\:\mathrm{pure}\:\mathrm{root} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{solve}\:\left(\mathrm{E}\right) \\ $$

Question Number 120240 Answers: 1 Comments: 0

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Question Number 120229 Answers: 2 Comments: 0

Question Number 120228 Answers: 0 Comments: 0

Please solve using special function like LambertW function or any other function (if possibe) ((8/7))^x +17^x =25x

$${Please}\:{solve}\:{using}\:{special}\:{function}\:{like}\:{LambertW} \\ $$$${function}\:{or}\:{any}\:{other}\:{function}\:\left({if}\:{possibe}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$

Question Number 120185 Answers: 1 Comments: 0

solve for x, y∈Z (√x)+(√y)=(√(2020))

$${solve}\:{for}\:{x},\:{y}\in{Z} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2020}} \\ $$

Question Number 120154 Answers: 1 Comments: 0

solve using LambertW function ((8/7))^x +17=25x

$${solve}\:{using}\:{LambertW}\:{function} \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}=\mathrm{25}{x} \\ $$

Question Number 120152 Answers: 0 Comments: 0

Determinate and construct the set of points M which have as affix z in each case: 1) arg(i−z)=0[π] 2) arg(z+1−i)=(π/6)[2π]

$$\mathrm{Determinate}\:\mathrm{and}\:\mathrm{construct}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points}\:\mathrm{M} \\ $$$$\mathrm{which}\:\mathrm{have}\:\mathrm{as}\:\mathrm{affix}\:\mathrm{z}\:\:\mathrm{in}\:\mathrm{each}\:\mathrm{case}: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{arg}\left(\mathrm{i}−\mathrm{z}\right)=\mathrm{0}\left[\pi\right] \\ $$$$\left.\mathrm{2}\right)\:\mathrm{arg}\left(\mathrm{z}+\mathrm{1}−\mathrm{i}\right)=\frac{\pi}{\mathrm{6}}\left[\mathrm{2}\pi\right] \\ $$$$ \\ $$

Question Number 120151 Answers: 0 Comments: 0

Represent in complex plane the set of points M which have as affix z such that ∣z∣=2 and arg(z+1)=(π/4)[π]

$$\mathrm{Represent}\:\mathrm{in}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{points} \\ $$$$\mathrm{M}\:\mathrm{which}\:\mathrm{have}\:\mathrm{as}\:\mathrm{affix}\:\mathrm{z}\:\mathrm{such}\:\mathrm{that}\:\mid\mathrm{z}\mid=\mathrm{2}\:\mathrm{and} \\ $$$$\mathrm{arg}\left(\mathrm{z}+\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}\left[\pi\right] \\ $$

Question Number 120110 Answers: 2 Comments: 0

{ ((((4x^2 )/(4x^2 +1)) = y)),((((4y^2 )/(4y^2 +1)) = z )),((((4z^2 )/(4z^2 +1)) = x)) :} where x,y,z ≠ 0

$$\begin{cases}{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:=\:{y}}\\{\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}}\:=\:{z}\:}\\{\frac{\mathrm{4}{z}^{\mathrm{2}} }{\mathrm{4}{z}^{\mathrm{2}} +\mathrm{1}}\:=\:{x}}\end{cases} \\ $$$${where}\:{x},{y},{z}\:\neq\:\mathrm{0}\: \\ $$

Question Number 120108 Answers: 0 Comments: 0

(x^4 +ax^2 +a^2 )^2 +x^6 =1 solve for: x,a∈R

$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{6}} =\mathrm{1} \\ $$$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}:\:\:\mathrm{x},\mathrm{a}\in\boldsymbol{\mathrm{R}} \\ $$

Question Number 120036 Answers: 1 Comments: 0

Question Number 120035 Answers: 1 Comments: 0

Question Number 120006 Answers: 0 Comments: 0

Question Number 119997 Answers: 4 Comments: 0

solve for x,a∈R. (√(x^2 +ax+a^2 ))+(√(x^2 −ax+a^2 ))=1

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\mathrm{a}\in\boldsymbol{\mathrm{R}}. \\ $$$$\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }+\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{ax}}+\boldsymbol{{a}}^{\mathrm{2}} }=\mathrm{1} \\ $$

Question Number 119996 Answers: 1 Comments: 0

{ ((x^3 +y^2 =a)),((x^2 +y^3 =b)) :} [solve for:x,y,a≠b∈R]

$$\begin{cases}{\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{a}}}\\{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{3}} =\boldsymbol{{b}}}\end{cases}\:\:\:\left[\boldsymbol{{solve}}\:\boldsymbol{{for}}:\mathrm{x},\mathrm{y},\mathrm{a}\neq\mathrm{b}\in\boldsymbol{\mathrm{R}}\right] \\ $$

Question Number 119979 Answers: 3 Comments: 1

Question Number 119939 Answers: 3 Comments: 0

Question Number 119921 Answers: 2 Comments: 0

solving the following system of equations { ((((3x−y)/(x−3y))=x^2 )),((((3y−z)/(y−3z))=y^2 )),((((3z−x)/(z−3x))=z^2 )) :}

$${solving}\:{the}\:{following}\:{system}\:{of}\:{equations} \\ $$$$\begin{cases}{\frac{\mathrm{3}{x}−{y}}{{x}−\mathrm{3}{y}}={x}^{\mathrm{2}} }\\{\frac{\mathrm{3}{y}−{z}}{{y}−\mathrm{3}{z}}={y}^{\mathrm{2}} }\\{\frac{\mathrm{3}{z}−{x}}{{z}−\mathrm{3}{x}}={z}^{\mathrm{2}} }\end{cases} \\ $$

Question Number 119894 Answers: 2 Comments: 0

Question Number 119849 Answers: 1 Comments: 0

Find all pair(x,y) of real numbers that are the solutions to the system { ((x^4 +2x^3 −y=−(1/4)+(√3))),((y^4 +2y^3 −x=−(1/4)−(√3))) :}

$${Find}\:{all}\:{pair}\left({x},{y}\right)\:{of}\:{real}\:{numbers} \\ $$$${that}\:{are}\:{the}\:{solutions}\:{to}\:{the}\:{system} \\ $$$$\begin{cases}{{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −{y}=−\frac{\mathrm{1}}{\mathrm{4}}+\sqrt{\mathrm{3}}}\\{{y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} −{x}=−\frac{\mathrm{1}}{\mathrm{4}}−\sqrt{\mathrm{3}}}\end{cases} \\ $$

Question Number 119848 Answers: 1 Comments: 0

Solve in real numbers the equation (x)^(1/(3 )) + ((x−1))^(1/(3 )) + ((x+1))^(1/(3 )) = 0

$${Solve}\:{in}\:{real}\:{numbers}\:{the}\:{equation} \\ $$$$\sqrt[{\mathrm{3}\:}]{{x}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}−\mathrm{1}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}+\mathrm{1}}\:=\:\mathrm{0} \\ $$

Question Number 119832 Answers: 0 Comments: 0

evaluate: I = ∫_0 ^( 1) (((x+1)/x))^(x!) dx

$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}!} {dx} \\ $$$$ \\ $$$$ \\ $$

Question Number 119831 Answers: 0 Comments: 1

evaluate: I = ∫_1 ^( ∞) ((1/x))^x dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{evaluate}:\:\:\:{I}\:\:=\:\int_{\mathrm{1}} ^{\:\infty} \left(\frac{\mathrm{1}}{{x}}\right)^{{x}} {dx} \\ $$$$\: \\ $$$$\: \\ $$

Question Number 119807 Answers: 1 Comments: 0

Let f be a real-valued function defined on the inte- rval [−1, 1]. If the area of the equilateral triangle with (0, 0) and (x, f(x)) as two vertices is (√3)/4, then f(x) is equal to (A) (√(1−x^2 )) (B) (√(1+x^2 )) (C) −(√(1−x^2 )) (D) −(√(1+x^2 ))

$$\mathrm{Let}\:{f}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}-\mathrm{valued}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{on}\:\mathrm{the}\:\mathrm{inte}- \\ $$$$\mathrm{rval}\:\left[−\mathrm{1},\:\mathrm{1}\right].\:\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{with} \\ $$$$\left(\mathrm{0},\:\mathrm{0}\right)\:\mathrm{and}\:\left(\mathrm{x},\:{f}\left(\mathrm{x}\right)\right)\:\mathrm{as}\:\mathrm{two}\:\mathrm{vertices}\:\mathrm{is}\:\sqrt{\mathrm{3}}/\mathrm{4},\:\mathrm{then}\:{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 119800 Answers: 0 Comments: 2

Examples of functions such that f(x+y)=f(x)+f(y) for all x,y∈R

$$\mathrm{Examples}\:\mathrm{of}\:\mathrm{functions}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left(\mathrm{x}+\mathrm{y}\right)={f}\left(\mathrm{x}\right)+{f}\left(\mathrm{y}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\mathrm{y}\in\mathbb{R} \\ $$

Question Number 119801 Answers: 0 Comments: 0

If f:R→R is a function such that f(0)=1 and f(x+f(y))= f(x)+y for all x, y∈R, then (A) 1 is a period of f (B) f(n)=1 for all integers n (C) f(n)=n for all integers n (D) f(−1)=0

$$\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a}\:\mathrm{function}\:\mathrm{such}\:\mathrm{that}\:{f}\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:{f}\left(\mathrm{x}+{f}\left(\mathrm{y}\right)\right)= \\ $$$${f}\left(\mathrm{x}\right)+\mathrm{y}\:\mathrm{for}\:\mathrm{all}\:\mathrm{x},\:\mathrm{y}\in\mathbb{R},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f} \\ $$$$\left(\mathrm{B}\right)\:{f}\left({n}\right)=\mathrm{1}\:\mathrm{for}\:\mathrm{all}\:\mathrm{integers}\:{n} \\ $$$$\left(\mathrm{C}\right)\:{f}\left({n}\right)={n}\:\mathrm{for}\:\mathrm{all}\:\mathrm{integers}\:{n} \\ $$$$\left(\mathrm{D}\right)\:{f}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$

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