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AlgebraQuestion and Answers: Page 65

Question Number 188248 Answers: 0 Comments: 0

(x^3 −y−3x)[(x^3 −3x)^2 −y^2 ]=200 (x^3 +y−3x)[(x^3 −3x)^2 +y^2 ]=600 solved in R

$$\left({x}^{\mathrm{3}} −{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} \right]=\mathrm{200} \\ $$$$\left({x}^{\mathrm{3}} +{y}−\mathrm{3}{x}\right)\left[\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \right]=\mathrm{600} \\ $$$${solved}\:{in}\:{R} \\ $$

Question Number 188247 Answers: 1 Comments: 0

Prove that (1) 5555^(2222) +2222^(5555) divisible by 7 (2) 3^(105) +4^(105) divisible by 7

$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5555}^{\mathrm{2222}} +\mathrm{2222}^{\mathrm{5555}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{105}} +\mathrm{4}^{\mathrm{105}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$

Question Number 188226 Answers: 1 Comments: 0

If Ω = Σ_(cyc) ((sin(A − (π/6)))/(cos(B − (π/6))cos(C − (π/6)))) in △ABC Solve for real numbers: x^4 − 4Ωx^3 + 6Ωx^2 − 4Ωx + 1 = 0

$$\mathrm{If}\:\:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{sin}\left(\mathrm{A}\:−\:\frac{\pi}{\mathrm{6}}\right)}{\mathrm{cos}\left(\mathrm{B}\:−\:\frac{\pi}{\mathrm{6}}\right)\mathrm{cos}\left(\mathrm{C}\:−\:\frac{\pi}{\mathrm{6}}\right)}\:\:\:\mathrm{in}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{x}^{\mathrm{4}} \:−\:\mathrm{4}\Omega\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{6}\Omega\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\Omega\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 188224 Answers: 1 Comments: 0

If Ω = Σ_(n=1) ^∞ (Π_(k=2) ^∞ ((k^3 − 1)/(k^3 + 1)))^n Solve for complex numbees: z^4 + 3z^3 + Ωz^2 + 3z + 1 = 0

$$\mathrm{If}\:\:\:\Omega\:=\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\frac{\mathrm{k}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{k}^{\mathrm{3}} \:+\:\mathrm{1}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{complex}\:\mathrm{numbees}: \\ $$$$\mathrm{z}^{\mathrm{4}} \:+\:\mathrm{3z}^{\mathrm{3}} \:+\:\Omega\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{3z}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Question Number 188170 Answers: 1 Comments: 0

solve ⌊ cos (x )⌋ + ⌊ cos(x) +(1/2) ⌋+ ⌊ −2cosx ⌋ =0

$$ \\ $$$$\:\:\:\:\:\:{solve} \\ $$$$ \\ $$$$\:\lfloor\:{cos}\:\left({x}\:\right)\rfloor\:+\:\lfloor\:{cos}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\rfloor+\:\lfloor\:−\mathrm{2}{cosx}\:\rfloor\:=\mathrm{0} \\ $$$$ \\ $$

Question Number 188156 Answers: 1 Comments: 0

a,b,c∈N 5a = 6b = 9c (abc)_(min) = ?

$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{5a}\:=\:\mathrm{6b}\:=\:\mathrm{9c} \\ $$$$\left(\mathrm{abc}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$

Question Number 188151 Answers: 1 Comments: 0

Simplify: (√(1+(1/1^2 )+(1/2^2 ))) +(√(1+(1/2^2 )+(1/3^2 ))) +...+(√(1+(1/(2022^2 ))+(1/(2023^2 ))))

$$\mathrm{Simplify}: \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}\:+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }}\:+...+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2022}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2023}^{\mathrm{2}} }} \\ $$

Question Number 188126 Answers: 0 Comments: 0

In convex polygon ABCD AB = 10 (√6) , CD = 18 ∠ ABD = 60° , ∠ BDC = 45° and BD = 13 (√6) + 9 (√2) find AC = ?

$$\mathrm{In}\:\mathrm{convex}\:\mathrm{polygon}\:\:\mathrm{ABCD} \\ $$$$\mathrm{AB}\:=\:\mathrm{10}\:\sqrt{\mathrm{6}}\:\:,\:\:\mathrm{CD}\:=\:\mathrm{18} \\ $$$$\angle\:\mathrm{ABD}\:=\:\mathrm{60}°\:\:,\:\:\angle\:\mathrm{BDC}\:=\:\mathrm{45}° \\ $$$$\mathrm{and}\:\:\mathrm{BD}\:=\:\mathrm{13}\:\sqrt{\mathrm{6}}\:+\:\mathrm{9}\:\sqrt{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\mathrm{AC}\:=\:? \\ $$

Question Number 188100 Answers: 2 Comments: 0

Question Number 188082 Answers: 1 Comments: 0

P(x) is a polynomial If P(x^2 + 1) = 6x^4 − x^2 + 5 Find P(x^2 − 1) = ?

$$\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial} \\ $$$$\mathrm{If}\:\:\:\mathrm{P}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:=\:\mathrm{6x}^{\mathrm{4}} \:−\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{5} \\ $$$$\mathrm{Find}\:\:\:\mathrm{P}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{1}\right)\:=\:? \\ $$

Question Number 188073 Answers: 2 Comments: 0

If x = (√((1 + (√5))/( (√5) − 1))) find 5x^2 −5x−1=?

$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\sqrt{\frac{\mathrm{1}\:+\:\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:−\:\mathrm{1}}}\:\:\:\:\:\mathrm{find}\:\:\:\:\mathrm{5x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{1}=? \\ $$

Question Number 188072 Answers: 0 Comments: 0

If x_1 =−1 and x_(n+1) = (1 + (2/n))x_n + (4/n) Find x_(2023) = ?

$$\mathrm{If}\:\:\:\mathrm{x}_{\mathrm{1}} =−\mathrm{1}\:\:\:\mathrm{and}\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} =\:\left(\mathrm{1}\:+\:\frac{\mathrm{2}}{\mathrm{n}}\right)\mathrm{x}_{\boldsymbol{\mathrm{n}}} +\:\frac{\mathrm{4}}{\mathrm{n}} \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{x}_{\mathrm{2023}} \:=\:? \\ $$

Question Number 188071 Answers: 0 Comments: 1

If ((sin^4 x)/5) + ((cos^4 x)/7) = (1/(12)) Find ((sin^2 2x)/5) + ((cos^2 2x)/7) = ?

$$\mathrm{If}\:\:\:\:\:\frac{\mathrm{sin}^{\mathrm{4}} \mathrm{x}}{\mathrm{5}}\:+\:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\mathrm{7}}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2x}}{\mathrm{5}}\:+\:\frac{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2x}}{\mathrm{7}}\:=\:? \\ $$

Question Number 188069 Answers: 0 Comments: 1

how is solution ∫(√e^x )ln (√e^x )dx=?

$${how}\:{is}\:{solution} \\ $$$$\int\sqrt{{e}^{{x}} }\mathrm{ln}\:\sqrt{{e}^{{x}} }{dx}=? \\ $$

Question Number 188016 Answers: 1 Comments: 0

Question Number 187998 Answers: 1 Comments: 0

Question Number 187989 Answers: 2 Comments: 0

Question Number 187980 Answers: 1 Comments: 0

Question Number 187971 Answers: 2 Comments: 0

If , x^( 5) = 1 ∧ x≠1 ( (( 1)/(x^( 2) −x +1)) + (1/(x^( 2) + x +1)) )^( 10) = ?

$$ \\ $$$$\:\:\mathrm{If}\:\:\:,\:\:{x}^{\:\mathrm{5}} \:=\:\mathrm{1}\:\:\:\wedge\:\:{x}\neq\mathrm{1} \\ $$$$ \\ $$$$\:\:\:\:\left(\:\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2}} \:−{x}\:+\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} \:+\:{x}\:+\mathrm{1}}\:\right)^{\:\mathrm{10}} =\:? \\ $$$$ \\ $$

Question Number 187916 Answers: 1 Comments: 0

how is solution y=(cosx)^((3x^2 −1)^e^x ) (dy/dx)=?

$${how}\:{is}\:{solution} \\ $$$${y}=\left({cosx}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \\ $$$$\frac{{dy}}{{dx}}=? \\ $$

Question Number 187908 Answers: 0 Comments: 0

Question Number 187905 Answers: 0 Comments: 1

Question Number 187877 Answers: 2 Comments: 0

Simplify completely ((256^(−(7/(16))) × 128^(9/(28)) )/(512^((17)/(36)) × 64^(−((11)/(12))) ))

$$\mathrm{Simplify}\:\mathrm{completely} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{256}^{−\frac{\mathrm{7}}{\mathrm{16}}} \:×\:\mathrm{128}^{\frac{\mathrm{9}}{\mathrm{28}}} }{\mathrm{512}^{\frac{\mathrm{17}}{\mathrm{36}}} \:×\:\mathrm{64}^{−\frac{\mathrm{11}}{\mathrm{12}}} } \\ $$

Question Number 187874 Answers: 2 Comments: 0

how is solution ((√2)−1)^(13) =x ((√2)+1)^(221) =? 1)x^(−16) 2)x^(−17) 3)x^(221) 4)x^(21)

$${how}\:{is}\:{solution} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{13}} =\mathrm{x}\:\:\:\:\:\:\:\:\:\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{221}} =? \\ $$$$\left.\mathrm{1}\left.\right)\left.\mathrm{x}^{−\mathrm{16}} \left.\:\:\:\:\:\:\:\:\:\:\mathrm{2}\right)\mathrm{x}^{−\mathrm{17}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\right)\mathrm{x}^{\mathrm{221}} \:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\right)\mathrm{x}^{\mathrm{21}} \\ $$

Question Number 187871 Answers: 1 Comments: 0

how is solution sgn(cos((21)/(10)))=?

$${how}\:{is}\:{solution} \\ $$$$\mathrm{sgn}\left(\mathrm{cos}\frac{\mathrm{21}}{\mathrm{10}}\right)=? \\ $$

Question Number 187856 Answers: 1 Comments: 0

find x 2^(√x) =8x

$${find}\:{x} \\ $$$$\mathrm{2}^{\sqrt{{x}}} =\mathrm{8}{x} \\ $$

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