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AlgebraQuestion and Answers: Page 266

Question Number 91377 Answers: 0 Comments: 1

Question Number 91362 Answers: 1 Comments: 3

Question Number 91277 Answers: 2 Comments: 0

p=1−(1/2)+(1/3)−(1/4)+...+(1/(2003))−(1/(2004)) q=(1/(1003))+(1/(1004))+...+(1/(2004)) p^2 +q^2 =

$${p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+...+\frac{\mathrm{1}}{\mathrm{2003}}−\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{1003}}+\frac{\mathrm{1}}{\mathrm{1004}}+...+\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} \:=\: \\ $$

Question Number 91258 Answers: 0 Comments: 0

prove that _2 F_1 (α,β,β−a+1,−1)=((Γ(β−a+1)Γ((β/2)+1))/(Γ(β+1)Γ((β/2)−α+1)))

$${prove}\:{that} \\ $$$$\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\alpha,\beta,\beta−{a}+\mathrm{1},−\mathrm{1}\right)=\frac{\Gamma\left(\beta−{a}+\mathrm{1}\right)\Gamma\left(\frac{\beta}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma\left(\beta+\mathrm{1}\right)\Gamma\left(\frac{\beta}{\mathrm{2}}−\alpha+\mathrm{1}\right)} \\ $$

Question Number 91195 Answers: 1 Comments: 0

what is the duble fictorial furmolla?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{duble}\:\mathrm{fictorial}\:\mathrm{furmolla}? \\ $$

Question Number 91178 Answers: 0 Comments: 2

Question Number 91166 Answers: 0 Comments: 7

Question Number 91047 Answers: 1 Comments: 6

Find the square root of: (√7) + (√5)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}:\:\:\:\:\sqrt{\mathrm{7}}\:\:+\:\:\sqrt{\mathrm{5}} \\ $$

Question Number 91038 Answers: 1 Comments: 1

if sin((α/2))=(4/5) and cos((β/2))=(3/5) prove sin(α)=cos(β)

$${if}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${and}\:{cos}\left(\frac{\beta}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${prove} \\ $$$${sin}\left(\alpha\right)={cos}\left(\beta\right) \\ $$

Question Number 90946 Answers: 0 Comments: 0

determine x,y,z ∈ R such that 2x^2 +y^2 +2z^2 −8x+2y−2xy+2xz−16z+35=0

$${determine}\:{x},{y},{z}\:\in\:\mathbb{R}\:{such}\:{that}\: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{2}{y}−\mathrm{2}{xy}+\mathrm{2}{xz}−\mathrm{16}{z}+\mathrm{35}=\mathrm{0} \\ $$

Question Number 90916 Answers: 0 Comments: 0

Question Number 90842 Answers: 3 Comments: 4

x^2 −(y−z)^2 = 3 y^2 − (z−x)^2 = 5 z^2 − (x−y)^2 = 12

$${x}^{\mathrm{2}} −\left({y}−{z}\right)^{\mathrm{2}} \:=\:\mathrm{3} \\ $$$${y}^{\mathrm{2}} \:−\:\left({z}−{x}\right)^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$${z}^{\mathrm{2}} \:−\:\left({x}−{y}\right)^{\mathrm{2}} \:=\:\mathrm{12} \\ $$

Question Number 90793 Answers: 1 Comments: 4

a+b+c+d=4 a^2 +b^2 +c^2 +d^2 =10 a^3 +b^3 +c^3 +d^3 =22 a^4 +b^4 +c^4 +d^4 = ?

$${a}+{b}+{c}+{d}=\mathrm{4} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{10} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} =\mathrm{22} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} =\:? \\ $$

Question Number 90772 Answers: 0 Comments: 2

show that the roots of the equation x^2 −2x=(b−c)^2 −1 are rational if b and c are rational numbers.

$${show}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}=\left({b}−{c}\right)^{\mathrm{2}} −\mathrm{1}\:{are}\:{rational}\:{if} \\ $$$${b}\:{and}\:{c}\:{are}\:{rational}\:{numbers}. \\ $$

Question Number 90747 Answers: 1 Comments: 0

solve: t^(1/3) + t^(1/2) = 12

$$\mathrm{solve}:\:\:\:\mathrm{t}^{\mathrm{1}/\mathrm{3}} \:\:\:+\:\:\:\mathrm{t}^{\mathrm{1}/\mathrm{2}} \:\:\:=\:\:\:\mathrm{12} \\ $$

Question Number 90716 Answers: 0 Comments: 1

If x + (1/x) = 4 , what the value of ((x^6 −1)/x^3 )

$${If}\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{4}\:,\:{what}\:{the}\: \\ $$$${value}\:{of}\:\frac{{x}^{\mathrm{6}} −\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$

Question Number 90661 Answers: 2 Comments: 0

show that (n^4 −n^2 ) is divisible by 12

$${show}\:{that}\:\left({n}^{\mathrm{4}} −{n}^{\mathrm{2}} \right)\:{is}\:{divisible}\:{by}\:\mathrm{12} \\ $$

Question Number 90557 Answers: 0 Comments: 7

Find the area enclose by the line y = x − 1 and the parabola y^2 = 2x + 6

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{enclose}\:\mathrm{by}\:\mathrm{the}\:\mathrm{line}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}\:\:−\:\:\mathrm{1}\:\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:\:\:\mathrm{y}^{\mathrm{2}} \:\:=\:\:\mathrm{2x}\:\:+\:\:\mathrm{6} \\ $$

Question Number 90550 Answers: 0 Comments: 3

x^4 + (1/x^4 ) = 527 (x−1)(x−2)(x−3)(x−4) ?

$${x}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\mathrm{527}\: \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\:?\: \\ $$

Question Number 90483 Answers: 0 Comments: 0

prove that/ ((sin^3 a)/(sin b))+((cos^3 a)/(cos b))≥sec(a−b) for all a,b∈ (0,(π/2))

$${prove}\:{that}/\:\frac{{sin}^{\mathrm{3}} {a}}{{sin}\:{b}}+\frac{{cos}^{\mathrm{3}} {a}}{{cos}\:{b}}\geqslant{sec}\left({a}−{b}\right) \\ $$$${for}\:{all}\:{a},{b}\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right) \\ $$

Question Number 90473 Answers: 1 Comments: 2

Question Number 90406 Answers: 2 Comments: 0

If x − (1/x) = 3 x^4 − (1/x^4 ) = ???

$$\mathrm{If}\:\:\:\:\:\:\mathrm{x}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}}\:\:\:=\:\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\:\:\:=\:\:\:??? \\ $$

Question Number 90379 Answers: 0 Comments: 2

Question Number 90360 Answers: 1 Comments: 1

Question Number 90350 Answers: 1 Comments: 0

n^2 x−5a^2 y^2 −n^2 y^2 +5a^2 x

$${n}^{\mathrm{2}} {x}−\mathrm{5}{a}^{\mathrm{2}} {y}^{\mathrm{2}} −{n}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} {x} \\ $$

Question Number 90318 Answers: 0 Comments: 0

Please can this be resolve in partial fraction? ((sec^2 x − (2/x^2 ))/((tan x + (1/x))^2 ))

$$\mathrm{Please}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{resolve}\:\mathrm{in}\:\mathrm{partial}\:\mathrm{fraction}? \\ $$$$\:\:\:\:\:\:\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}\:\:−\:\:\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}{\left(\mathrm{tan}\:\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} } \\ $$

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