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AlgebraQuestion and Answers: Page 29

Question Number 211103 Answers: 0 Comments: 0

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Question Number 210934 Answers: 2 Comments: 0

If ((x^2 + ax − 18)/(x^2 + 7x + 2b)) = ((x − c)/(x + 5)) Find a + b + c = ?

$$\mathrm{If}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{ax}\:−\:\mathrm{18}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{7x}\:+\:\mathrm{2b}}\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{c}}{\mathrm{x}\:+\:\mathrm{5}} \\ $$$$\mathrm{Find}\:\:\:\boldsymbol{\mathrm{a}}\:+\:\boldsymbol{\mathrm{b}}\:+\:\boldsymbol{\mathrm{c}}\:=\:? \\ $$

Question Number 210922 Answers: 0 Comments: 0

Question Number 210919 Answers: 1 Comments: 0

Question Number 210906 Answers: 1 Comments: 0

((19x − x^2 )/(x + 1)) ∙ (x + ((19 − x)/(x + 1))) = 78 find: x = ?

$$\frac{\mathrm{19x}\:−\:\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:+\:\mathrm{1}}\:\centerdot\:\left(\mathrm{x}\:+\:\frac{\mathrm{19}\:−\:\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}\right)\:=\:\mathrm{78} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210895 Answers: 4 Comments: 0

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Question Number 210871 Answers: 0 Comments: 1

7(x−2)((√(10 + x)) − ((7−x))^(1/3) )−15x + 6 = 0 x = ?

$$\mathrm{7}\left(\mathrm{x}−\mathrm{2}\right)\left(\sqrt{\mathrm{10}\:+\:\mathrm{x}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{7}−\mathrm{x}}\right)−\mathrm{15x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210870 Answers: 0 Comments: 1

8x^2 − 13x + 11 = (2/x) + (1 + (3/x)) ((3x^2 −2))^(1/3) x = ?

$$\mathrm{8x}^{\mathrm{2}} \:−\:\mathrm{13x}\:+\:\mathrm{11}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{x}}\right)\:\sqrt[{\mathrm{3}}]{\mathrm{3x}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 210868 Answers: 1 Comments: 5

let p ,q be reals such that p>q>0 define the sequence {x_n } where x_1 = p+q and x_n = x_1 −((pq)/x_(n−1) ) for n≥2 for all n then x_n = ??

$$\mathrm{let}\:\mathrm{p}\:,\mathrm{q}\:\mathrm{be}\:\mathrm{reals}\:\mathrm{such}\:\mathrm{that}\:\mathrm{p}>\mathrm{q}>\mathrm{0}\:\mathrm{define} \\ $$$$\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{x}_{\mathrm{n}} \right\}\:\mathrm{where}\:\mathrm{x}_{\mathrm{1}} =\:\mathrm{p}+\mathrm{q}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} \:=\:\mathrm{x}_{\mathrm{1}} −\frac{\mathrm{pq}}{\mathrm{x}_{\mathrm{n}−\mathrm{1}} }\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{for}\:\mathrm{all}\:\mathrm{n}\:\mathrm{then}\:\mathrm{x}_{\mathrm{n}} \:=\:?? \\ $$

Question Number 210858 Answers: 2 Comments: 0

Solve The Equation: (7x+1)(9x+1)(21x+1)(63x+1)= ((160)/(189))

$$\mathrm{Solve}\:\mathrm{The}\:\mathrm{Equation}: \\ $$$$\left(\mathrm{7x}+\mathrm{1}\right)\left(\mathrm{9x}+\mathrm{1}\right)\left(\mathrm{21x}+\mathrm{1}\right)\left(\mathrm{63x}+\mathrm{1}\right)=\:\frac{\mathrm{160}}{\mathrm{189}} \\ $$

Question Number 210843 Answers: 1 Comments: 0

Question Number 210842 Answers: 1 Comments: 0

Question Number 210851 Answers: 1 Comments: 1

Question Number 210840 Answers: 1 Comments: 0

Prove that if x, y are rational numbers satisfying the equation x^5 + y^5 = 2(x^2)(y^2) then 1 - xy is the square of rational number

$$ \\ $$Prove that if x, y are rational numbers satisfying the equation x^5 + y^5 = 2(x^2)(y^2) then 1 - xy is the square of rational number

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