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Question Number 176942    Answers: 1   Comments: 0

Question Number 176936    Answers: 1   Comments: 0

Question Number 176933    Answers: 1   Comments: 0

Question Number 182216    Answers: 3   Comments: 2

Find a, b, c ∈ N ; 2^( a) + 4^( b) + 8^( c) = 328

$$\:{Find}\:{a},\:{b},\:{c}\:\in\:\mathbb{N}\:;\:\mathrm{2}^{\:{a}} +\:\mathrm{4}^{\:{b}} +\:\mathrm{8}^{\:{c}} =\:\mathrm{328} \\ $$

Question Number 176928    Answers: 1   Comments: 0

What is the smallest three digit number divisible by (2/3), (7/9), (4/(13)) without a remainder ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{three}\:\mathrm{digit}\:\mathrm{number} \\ $$$$\mathrm{divisible}\:\mathrm{by}\:\frac{\mathrm{2}}{\mathrm{3}},\:\frac{\mathrm{7}}{\mathrm{9}},\:\frac{\mathrm{4}}{\mathrm{13}}\:\mathrm{without}\:\mathrm{a}\:\mathrm{remainder}\:? \\ $$

Question Number 176925    Answers: 1   Comments: 0

Question Number 176917    Answers: 0   Comments: 0

Question Number 176915    Answers: 1   Comments: 0

Question Number 176913    Answers: 1   Comments: 1

Determiner la hauteur DE(r+x) en fonction de r r=OOC=BH BF=20 pour que distance(AB+BC+CD+DE+EF soit sgale AC+arcCDF

$${Determiner}\:{la}\:{hauteur}\:\mathrm{D}{E}\left({r}+{x}\right)\:{en}\:{fonction}\:{de}\:{r} \\ $$$${r}=\mathrm{OOC}=\mathrm{BH}\:\:\:\:\:\mathrm{BF}=\mathrm{20} \\ $$$$\mathrm{pour}\:\mathrm{que}\:\mathrm{distance}\left(\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DE}+\mathrm{EF}\:\:\mathrm{soit}\:\right. \\ $$$$\mathrm{sgale}\:\mathrm{AC}+\mathrm{arcCDF} \\ $$

Question Number 176898    Answers: 1   Comments: 0

Question Number 176746    Answers: 5   Comments: 0

a_(n+2) −5a_(n+1) +6a_n =3n+5^n a_1 =1, a_2 =0 find a_n

$${a}_{{n}+\mathrm{2}} −\mathrm{5}{a}_{{n}+\mathrm{1}} +\mathrm{6}{a}_{{n}} =\mathrm{3}{n}+\mathrm{5}^{{n}} \\ $$$${a}_{\mathrm{1}} =\mathrm{1},\:{a}_{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:{a}_{{n}} \\ $$

Question Number 176741    Answers: 1   Comments: 1

In the following figure, if ((S_1 +S_2 )/( (√S_3 ))) = (6/( (√π))) find the area of the circular crown

$$\: \\ $$$$\:{In}\:{the}\:{following}\:{figure},\:{if}\:\:\:\frac{{S}_{\mathrm{1}} \:+{S}_{\mathrm{2}} }{\:\sqrt{{S}_{\mathrm{3}} }}\:=\:\frac{\mathrm{6}}{\:\sqrt{\pi}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{the}\:{circular}\:{crown}\: \\ $$$$\: \\ $$

Question Number 176727    Answers: 0   Comments: 0

In △ABC the following relationship holds: 6r Σ_(cyc) (r_a /(s + n_a )) + Σ_(cyc) n_a ≥ 3s

$$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{the}\:\mathrm{following}\:\mathrm{relationship} \\ $$$$\mathrm{holds}: \\ $$$$\mathrm{6r}\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{r}_{\boldsymbol{\mathrm{a}}} }{\mathrm{s}\:+\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} }\:\:+\:\:\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\mathrm{n}_{\boldsymbol{\mathrm{a}}} \:\geqslant\:\mathrm{3s} \\ $$

Question Number 176723    Answers: 0   Comments: 0

donner la forme trigonometrique(i−1)^5 /(i−4)^4

$${donner}\:{la}\:{forme}\:{trigonometrique}\left({i}−\mathrm{1}\right)^{\mathrm{5}} /\left({i}−\mathrm{4}\right)^{\mathrm{4}} \\ $$

Question Number 176721    Answers: 2   Comments: 0

Question Number 176722    Answers: 0   Comments: 1

(((√(11))+1)/(2(√(11))+11)) = ((((√(11))+1)(2(√(11))−11))/((2(√(11))+11)(2(√(11))−11))) = ((2(√(11))−11+2×11+11(√(11)))/(2×11−121)) = ((13(√(11))+11)/(−99)) Ans

$$\frac{\sqrt{\mathrm{11}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{11}}\:=\:\frac{\left(\sqrt{\mathrm{11}}+\mathrm{1}\right)\left(\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}\right)}{\left(\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{11}\right)\left(\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}\right)}\:=\: \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{11}}−\mathrm{11}+\mathrm{2}×\mathrm{11}+\mathrm{11}\sqrt{\mathrm{11}}}{\mathrm{2}×\mathrm{11}−\mathrm{121}}\:=\:\frac{\mathrm{13}\sqrt{\mathrm{11}}+\mathrm{11}}{−\mathrm{99}} \\ $$$$\boldsymbol{{A}}{ns} \\ $$

Question Number 176719    Answers: 0   Comments: 0

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Question Number 176718    Answers: 1   Comments: 1

x^3 +y^3 =z^2 x^3 +z^3 =y^2 y^3 +z^3 =x^2 obviously x=y=z=0∨(1/2) trying to totally solve it let y=px∧z=qx (p^3 +1)x^3 =q^2 x^2 (q^3 +1)x^3 =p^2 x^2 (p^3 +q^3 )x^3 =x^2 ⇒ x=0 ⇒ y=0∧z=0 ★ x=(q^2 /(p^3 +1)) x=(p^2 /(q^3 +1)) x=(1/(p^3 +q^3 )) ⇒ (p^2 /(q^3 +1))=(q^2 /(p^3 +1)) (p^2 /(q^3 +1))=(1/(p^3 +q^3 )) ========== p^5 +p^2 −q^5 −q^2 =0 p^5 +p^2 q^3 −q^3 −1=0 subtracting both p^2 (q^3 −1)+q^5 −q^3 +q^2 −1=0 (q−1)(p^2 (q^2 +q+1)+q^4 +q^3 +q+1)=0 ⇒ q=1 ⇒ p^5 +p^2 −2=0 (p−1)(p^4 +p^3 +2p+2)=0 ⇒ p=1 ⇒ x=y=z=(1/2) ★ p^4 +p^3 +2p+2=0 no useable exact solutions p≈−.975564±.528237i ⇒ x≈.33635∓.515329i ∧ y≈−.0559113±.680406i ∧ z=x ★ p≈.475564±1.18273i ⇒ x≈−.586346±.562464i ∧ y≈−.944089∓.426001i ∧ z=x ★ p^2 (q^2 +q+1)+q^4 +q^3 +q+1=0 p^2 =−(((q+1)^2 (q^2 −q+1))/(q^2 +q+1)) we can be sure that (q≠−1⇒p=0) ⇒ p^2 <0 ⇒ p=±(√((q^2 −q+1)/(q^2 +q+1)))(q+1)i ...I′ll continue later

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} ={z}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{z}^{\mathrm{3}} ={y}^{\mathrm{2}} \\ $$$${y}^{\mathrm{3}} +{z}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{obviously}\:{x}={y}={z}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{totally}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left({p}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={q}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({q}^{\mathrm{3}} +\mathrm{1}\right){x}^{\mathrm{3}} ={p}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){x}^{\mathrm{3}} ={x}^{\mathrm{2}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\wedge{z}=\mathrm{0}\:\bigstar \\ $$$${x}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}} \\ $$$${x}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} } \\ $$$$========== \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} −{q}^{\mathrm{5}} −{q}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}^{\mathrm{5}} +{p}^{\mathrm{2}} {q}^{\mathrm{3}} −{q}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{subtracting}\:\mathrm{both} \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{3}} −\mathrm{1}\right)+{q}^{\mathrm{5}} −{q}^{\mathrm{3}} +{q}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({q}−\mathrm{1}\right)\left({p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{q}=\mathrm{1}\:\Rightarrow\:{p}^{\mathrm{5}} +{p}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({p}−\mathrm{1}\right)\left({p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{p}=\mathrm{1}\:\Rightarrow\:{x}={y}={z}=\frac{\mathrm{1}}{\mathrm{2}}\:\bigstar \\ $$$$\:\:\:\:\:{p}^{\mathrm{4}} +{p}^{\mathrm{3}} +\mathrm{2}{p}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$$\:\:\:\:\:{p}\approx−.\mathrm{975564}\pm.\mathrm{528237i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx.\mathrm{33635}\mp.\mathrm{515329i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{0559113}\pm.\mathrm{680406i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$$\:\:\:\:\:{p}\approx.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273i} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\approx−.\mathrm{586346}\pm.\mathrm{562464i} \\ $$$$\:\:\:\:\:\wedge\:{y}\approx−.\mathrm{944089}\mp.\mathrm{426001i} \\ $$$$\:\:\:\:\:\wedge\:{z}={x}\:\bigstar \\ $$$${p}^{\mathrm{2}} \left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)+{q}^{\mathrm{4}} +{q}^{\mathrm{3}} +{q}+\mathrm{1}=\mathrm{0} \\ $$$${p}^{\mathrm{2}} =−\frac{\left({q}+\mathrm{1}\right)^{\mathrm{2}} \left({q}^{\mathrm{2}} −{q}+\mathrm{1}\right)}{{q}^{\mathrm{2}} +{q}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{be}\:\mathrm{sure}\:\mathrm{that}\:\left({q}\neq−\mathrm{1}\Rightarrow{p}=\mathrm{0}\right) \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:{p}=\pm\sqrt{\frac{{q}^{\mathrm{2}} −{q}+\mathrm{1}}{{q}^{\mathrm{2}} +{q}+\mathrm{1}}}\left({q}+\mathrm{1}\right)\mathrm{i} \\ $$$$...\mathrm{I}'\mathrm{ll}\:\mathrm{continue}\:\mathrm{later} \\ $$

Question Number 176716    Answers: 1   Comments: 0

Question Number 176710    Answers: 3   Comments: 0

Question Number 176708    Answers: 1   Comments: 0

a_(n+2) −3a_(n+1) +2a_n =n+3^n please find a_n

$$\:{a}_{{n}+\mathrm{2}} −\mathrm{3}{a}_{{n}+\mathrm{1}} +\mathrm{2}{a}_{{n}} ={n}+\mathrm{3}^{{n}} \: \\ $$$${please}\:{find}\:{a}_{{n}} \\ $$$$ \\ $$

Question Number 176692    Answers: 1   Comments: 1

Question Number 176684    Answers: 1   Comments: 2

Question Number 176679    Answers: 2   Comments: 0

Eeasy integral.... 𝛀 = ∫_(−∫_0 ^( ∞) e^( −x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( −x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt −−−m.n−−−

$$ \\ $$$$\:\:\:\:\:\:\:\:{Eeasy}\:\:{integral}.... \\ $$$$\:\:\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{−\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:−{x}^{\:\mathrm{2}} } {dx}} {sin}^{\:\mathrm{2}} \left({t}\right).{ln}^{\:\mathrm{3}} \left(\:{t}\:+\:\sqrt{\mathrm{1}+{t}^{\:\mathrm{2}} }\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−−−{m}.{n}−−− \\ $$

Question Number 176676    Answers: 3   Comments: 0

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=−(x^4 +(1/x^4 ))

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=−\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$

Question Number 176672    Answers: 1   Comments: 0

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