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AlgebraQuestion and Answers: Page 62

Question Number 197922 Answers: 0 Comments: 1

Question Number 197895 Answers: 1 Comments: 0

Solve the equation: x^4 − x^3 − 4x^2 + 3x + 2 = 0

$${Solve}\:{the}\:{equation}: \\ $$$${x}^{\mathrm{4}} \:−\:{x}^{\mathrm{3}} \:−\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$

Question Number 197880 Answers: 1 Comments: 0

find the sum of infinite series (1/2^1 )∙(1/3^2 ) + (1/2^2 )∙(1/3^4 )(1^2 +2^2 +3^2 ) + (1/2^3 )∙(1/3^6 )(1^2 +2^2 +3^2 +...+7^2 )+ (1/2^4 )∙(1/3^8 )(1^2 +2^2 +3^2 +...+15^2 )+........

$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{7}^{\mathrm{2}} \right)+ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{8}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{15}^{\mathrm{2}} \right)+........ \\ $$

Question Number 197876 Answers: 2 Comments: 1

Question Number 197860 Answers: 3 Comments: 0

((2^(17) +2^(16) +2^(15) +…+1)/(2^8 +2^7 +2^6 +…+1)) = ?

$$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$

Question Number 197843 Answers: 0 Comments: 2

Exercice 2

$$\boldsymbol{\mathrm{Exercice}}\:\:\mathrm{2} \\ $$

Question Number 197838 Answers: 2 Comments: 0

Question Number 197794 Answers: 1 Comments: 0

if x = log tan((π/4)+(y/2)), prove that y = −ilog tan(((ix)/2) + (π/4)) here i = (√(−1))

$$\:\:\:\:\mathrm{if}\:\mathrm{x}\:\:\:=\:\:\:\mathrm{log}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{y}}{\mathrm{2}}\right),\:\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\mathrm{y}\:\:\:\:=\:\:\:−{i}\mathrm{log}\:\mathrm{tan}\left(\frac{{ix}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\right)\:\:\:\:\:\mathrm{here}\:{i}\:\:=\:\sqrt{−\mathrm{1}} \\ $$

Question Number 197784 Answers: 1 Comments: 0

((x−2+3((x−3))^(1/3) (1+((x−3))^(1/3) )))^(1/3) + ((x+5+6((x−3))^(1/3) (1+2((x−3))^(1/3) )))^(1/3) = 5

$$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$

Question Number 197708 Answers: 1 Comments: 0

x^a =x^(a+4) where a∈Z solve for x by showing steps

$$ \\ $$$${x}^{{a}} ={x}^{{a}+\mathrm{4}} \:\:\mathrm{where}\:{a}\in\mathbb{Z} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{by}\:\mathrm{showing}\:\mathrm{steps} \\ $$

Question Number 197706 Answers: 1 Comments: 0

Question Number 197623 Answers: 1 Comments: 1

x,y∈N 162 ∙ x^2 = y^3 min(x+y)=?

$$\mathrm{x},\mathrm{y}\in\mathbb{N} \\ $$$$\mathrm{162}\:\centerdot\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{min}\left(\mathrm{x}+\mathrm{y}\right)=? \\ $$

Question Number 197618 Answers: 1 Comments: 0

Question Number 197609 Answers: 2 Comments: 0

((x+3)/(2022)) + ((x+2)/(2023)) + ((x+1)/(2024)) + (x/(2025)) = −4

$$\:\:\:\frac{{x}+\mathrm{3}}{\mathrm{2022}}\:+\:\frac{{x}+\mathrm{2}}{\mathrm{2023}}\:+\:\frac{{x}+\mathrm{1}}{\mathrm{2024}}\:+\:\frac{{x}}{\mathrm{2025}}\:=\:−\mathrm{4} \\ $$

Question Number 197567 Answers: 1 Comments: 0

Question Number 197530 Answers: 2 Comments: 0

f(x)−(x^2 /2)f′′(x)=0 f(x)=?

$$\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{f}''\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$

Question Number 197476 Answers: 1 Comments: 7

find the sum (1/(x+1))+(2/(x^2 +1))+(4/(x^4 +1))+.........+(2^n /(x^2^n +1)) = ??

$$ \\ $$$$\:\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{x}^{\mathrm{4}} +\mathrm{1}}+.........+\frac{\mathrm{2}^{{n}} }{{x}^{\mathrm{2}^{{n}} } +\mathrm{1}}\:\:=\:?? \\ $$

Question Number 197464 Answers: 1 Comments: 0

Question Number 197455 Answers: 1 Comments: 0

f(x)=((2x+6)/2)+6−x g(x)=(√(x^(99) +x^(98) +x^(97) +.....+x)) f(g(1))+f(g(2))+.........+f(g(100))=?

$${f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{6}}{\mathrm{2}}+\mathrm{6}−{x} \\ $$$${g}\left({x}\right)=\sqrt{{x}^{\mathrm{99}} +{x}^{\mathrm{98}} +{x}^{\mathrm{97}} +.....+{x}} \\ $$$${f}\left({g}\left(\mathrm{1}\right)\right)+{f}\left({g}\left(\mathrm{2}\right)\right)+.........+{f}\left({g}\left(\mathrm{100}\right)\right)=? \\ $$

Question Number 197422 Answers: 2 Comments: 0

Question Number 197419 Answers: 1 Comments: 0

Question Number 197409 Answers: 1 Comments: 4

Question Number 197393 Answers: 1 Comments: 0

Question Number 197360 Answers: 1 Comments: 0

Find: ∫_0 ^( ∞) sin^2 ( (√x) ) e^(−x) dx = ?

$$\mathrm{Find}: \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \:\left(\:\sqrt{\mathrm{x}}\:\right)\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:? \\ $$

Question Number 197312 Answers: 1 Comments: 0

(((log_2 20)^2 −(log_2 5)^2 )/(log_2 10))=?

$$\frac{\left({log}_{\mathrm{2}} \mathrm{20}\right)^{\mathrm{2}} −\left({log}_{\mathrm{2}} \mathrm{5}\right)^{\mathrm{2}} }{{log}_{\mathrm{2}} \mathrm{10}}=? \\ $$

Question Number 197275 Answers: 1 Comments: 0

how do i prove this, help please. ∣((x^2 −2x−3)/(x^2 +2x+4))∣≤(5/4),∣x∣≤2

$$ \\ $$$$\:{how}\:{do}\:{i}\:{prove}\:{this},\:{help}\:{please}. \\ $$$$\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\mid\leqslant\frac{\mathrm{5}}{\mathrm{4}},\mid{x}\mid\leqslant\mathrm{2} \\ $$$$ \\ $$$$ \\ $$

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