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Question Number 174059    Answers: 2   Comments: 3

if x^3 +y^3 +3xy=1, then x+y=?

$${if}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}=\mathrm{1},\:{then}\:{x}+{y}=? \\ $$

Question Number 174057    Answers: 0   Comments: 1

Question Number 174056    Answers: 0   Comments: 0

Question Number 174036    Answers: 1   Comments: 0

Factorize x^2 + (√(2x ))+ x + 2

$$\:\:\mathrm{Factorize}\:{x}^{\mathrm{2}} \:+\:\sqrt{\mathrm{2}{x}\:}+\:{x}\:+\:\mathrm{2} \\ $$

Question Number 174029    Answers: 1   Comments: 1

x+y=1 x^5 +xy+y^5 =x^2 y^(2 ) faind the volue of x^(2022) +xy+y^(2022) =?

$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{5}} +{xy}+{y}^{\mathrm{5}} ={x}^{\mathrm{2}} {y}^{\mathrm{2}\:\:\:\:\:\:} \\ $$$${faind}\:{the}\:{volue}\:{of}\:\:\:{x}^{\mathrm{2022}} +{xy}+{y}^{\mathrm{2022}} =? \\ $$

Question Number 174023    Answers: 2   Comments: 1

Question Number 173997    Answers: 2   Comments: 0

If (x)^(1/3) − 3 = ((x−36))^(1/3) then ((x^( 2) −1)/x) = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{If}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\mathrm{3}\:=\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{36}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{then}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{x}^{\:\mathrm{2}} −\mathrm{1}}{{x}}\:=\:?\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 173993    Answers: 1   Comments: 0

Question Number 173992    Answers: 1   Comments: 1

Question Number 173984    Answers: 0   Comments: 0

Question Number 173983    Answers: 1   Comments: 0

in AB^Δ C prove : (( cos(A ))/a^( 3) ) +((cos(B))/b^( 3) ) +((cos(C))/c^( 3) ) ≥((81)/(16p^( 3) )) where : p= (a+b +c )/2

$$ \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\:\:{prove}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\frac{\:{cos}\left({A}\:\right)}{{a}^{\:\mathrm{3}} }\:+\frac{{cos}\left({B}\right)}{{b}^{\:\mathrm{3}} }\:+\frac{{cos}\left({C}\right)}{{c}^{\:\mathrm{3}} }\:\geqslant\frac{\mathrm{81}}{\mathrm{16}{p}^{\:\mathrm{3}} } \\ $$$$\:\:\:{where}\::\:\:{p}=\:\left({a}+{b}\:+{c}\:\right)/\mathrm{2} \\ $$$$ \\ $$

Question Number 173981    Answers: 2   Comments: 0

Question Number 173978    Answers: 1   Comments: 0

Solve system of equations: x+((3x−y)/(x^2 +y^2 ))=3 y−((x+3y)/(x^2 +y^2 ))=0

$$\mathrm{Solve}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\mathrm{x}+\frac{\mathrm{3x}−\mathrm{y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\mathrm{3} \\ $$$$\mathrm{y}−\frac{\mathrm{x}+\mathrm{3y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\mathrm{0} \\ $$$$ \\ $$

Question Number 173976    Answers: 1   Comments: 0

B(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx Γ(s)= ∫_0 ^∞ t^(s−1) e^(−t) dt Why B(a,b)= ((Γ(a)Γ(b))/(Γ(a+b))) ?

$$ \\ $$$$\:\:\:\:{B}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{a}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{b}−\mathrm{1}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\Gamma\left({s}\right)=\:\int_{\mathrm{0}} ^{\infty} {t}^{{s}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$ \\ $$$$\:\:{Why}\:\:\:\:{B}\left({a},{b}\right)=\:\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)}\:? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173972    Answers: 0   Comments: 0

Question Number 173971    Answers: 0   Comments: 2

Question Number 173970    Answers: 1   Comments: 3

Question Number 173975    Answers: 0   Comments: 0

∫_0 ^∞ (y^(a−1) /((1+y)^b )) dy =^(u=(1/(1+y))) ∫_0 ^1 (((1−u)^(a−1) )/u^(a−1) ) u^b (du/u^2 ) = ∫_0 ^1 u^(b−a−1) (1−u)^(a−1) du = B(b−a,a)=((Γ(b−a)Γ(a))/(Γ(b)))

$$ \\ $$$$\: \\ $$$$ \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{y}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{y}\right)^{{b}} }\:{dy}\:\overset{{u}=\frac{\mathrm{1}}{\mathrm{1}+{y}}} {=}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{u}\right)^{{a}−\mathrm{1}} }{{u}^{{a}−\mathrm{1}} }\:{u}^{{b}} \:\:\frac{{du}}{{u}^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{b}−{a}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{a}−\mathrm{1}} {du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{B}\left({b}−{a},{a}\right)=\frac{\Gamma\left({b}−{a}\right)\Gamma\left({a}\right)}{\Gamma\left({b}\right)} \\ $$$$\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 173948    Answers: 1   Comments: 4

Question Number 173946    Answers: 1   Comments: 0

If 3sin(x)+4cos(y)−3sin(y)=8 then .. sin(x) + sin(y)=?

$$ \\ $$$$\:\:\:{If}\:\mathrm{3}{sin}\left({x}\right)+\mathrm{4}{cos}\left({y}\right)−\mathrm{3}{sin}\left({y}\right)=\mathrm{8} \\ $$$$\:\:\:{then}\:..\:\:\:\:{sin}\left({x}\right)\:+\:{sin}\left({y}\right)=? \\ $$$$ \\ $$$$ \\ $$

Question Number 173933    Answers: 1   Comments: 1

Question Number 173958    Answers: 0   Comments: 0

Question Number 173926    Answers: 2   Comments: 3

Q: a_( n ) is an arithmatic sequence. a ( first term) and d (difference ) such that , a_( a) + a_( d) = a_( ad) find : a_( n) =? note: a , d ∈ N

$$ \\ $$$$\:\:\:\:\:{Q}: \\ $$$$\:\:\:\:\:\:\:{a}_{\:{n}\:} \:\:{is}\:{an}\:{arithmatic}\:{sequence}. \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}\:\left(\:{first}\:{term}\right)\:{and}\:\:{d}\:\left({difference}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:,\:\:{a}_{\:{a}} \:+\:{a}_{\:{d}} \:=\:{a}_{\:{ad}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{find}\:\:\::\:\:\:\:{a}_{\:{n}} \:\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:{note}:\:\:{a}\:\:,\:\:{d}\:\:\:\in\:\mathbb{N}\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 173923    Answers: 1   Comments: 1

Question Number 173917    Answers: 0   Comments: 2

prove: (−1)!

$${prove}: \\ $$$$\left(−\mathrm{1}\right)! \\ $$

Question Number 173908    Answers: 0   Comments: 6

find the value of b so that the line y=b divides the region bound by the graphs of the two functinos , into two regions of equal area. f(x)=9−x^2 and g(x)=0

$${find}\:{the}\:{value}\:{of}\:{b}\:{so}\:{that}\:{the}\:{line}\:{y}={b} \\ $$$${divides}\:{the}\:{region}\:{bound}\:{by}\:{the}\:{graphs}\:{of} \\ $$$${the}\:{two}\:{functinos}\:,\:{into}\:{two}\:{regions}\:{of}\:{equal} \\ $$$${area}. \\ $$$${f}\left({x}\right)=\mathrm{9}−{x}^{\mathrm{2}} \:{and}\:{g}\left({x}\right)=\mathrm{0} \\ $$

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