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Question Number 197960 Answers: 1 Comments: 0

determiner le total de nombres de 5 chiffres comprises entre 10000 et 50000 divisibles simultanement par 5 et 9 (sans utiliser les formules d arrangement et de combinaison)

$$\mathrm{determiner}\:\mathrm{le}\:\mathrm{total}\:\mathrm{de}\:\mathrm{nombres}\:\mathrm{de}\: \\ $$$$\mathrm{5}\:\mathrm{chiffres}\:\mathrm{comprises}\:\mathrm{entre}\:\mathrm{10000}\:\mathrm{et}\: \\ $$$$\mathrm{50000}\:\:\mathrm{divisibles}\:\mathrm{simultanement}\:\mathrm{par} \\ $$$$\mathrm{5}\:\mathrm{et}\:\mathrm{9}\:\:\: \\ $$$$\left(\mathrm{sans}\:\mathrm{utiliser}\:\mathrm{les}\:\mathrm{formules}\:\mathrm{d}\:\mathrm{arrangement}\right. \\ $$$$\left.\mathrm{et}\:\mathrm{de}\:\mathrm{combinaison}\right) \\ $$$$ \\ $$

Question Number 197951 Answers: 0 Comments: 0

Σ_(n=1 ) ^∞ (n/(n^4 +n^2 +1)) − Σ_(n=1) ^∞ (n^2 /(n^8 +n^4 +1)) = ?

$$\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}\:−\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{8}} +{n}^{\mathrm{4}} +\mathrm{1}}\:=\:? \\ $$

Question Number 197950 Answers: 0 Comments: 0

Let x,y,z>0 , x+y+z=3 Prove That : (1/( (√(x^2 +2x))))+(1/( (√(z^2 +2z))))+(√3)((1/(y+2))−(y/9))+((((√x)+(√y)+(√z)+24))^(1/3) /( (√3)))≥((17)/(3(√3)))

$${Let}\:{x},{y},{z}>\mathrm{0}\:,\:{x}+{y}+{z}=\mathrm{3}\:{Prove}\:{That}\:: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}+\frac{\mathrm{1}}{\:\sqrt{{z}^{\mathrm{2}} +\mathrm{2}{z}}}+\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{{y}+\mathrm{2}}−\frac{{y}}{\mathrm{9}}\right)+\frac{\sqrt[{\mathrm{3}}]{\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}+\mathrm{24}}}{\:\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{17}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

Question Number 197947 Answers: 1 Comments: 0

calculate… L = lim _(n→∞) (( (1+(1/2) )(1+(1/3))… (1+(1/n))))^(1/n) = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$

Question Number 197944 Answers: 1 Comments: 0

tan(π/(12))=((sinα−sin(π/(12)))/(cosα+cos(π/(12)))) α=?

$${tan}\frac{\pi}{\mathrm{12}}=\frac{{sin}\alpha−{sin}\frac{\pi}{\mathrm{12}}}{{cos}\alpha+{cos}\frac{\pi}{\mathrm{12}}} \\ $$$$\alpha=? \\ $$

Question Number 197937 Answers: 1 Comments: 0

(√3) sin^2 θ∙tanβ+cos^2 β=?

$$\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} \theta\centerdot{tan}\beta+{cos}^{\mathrm{2}} \beta=? \\ $$

Question Number 197935 Answers: 1 Comments: 0

Show that Σ_(n=1) ^∞ (((n!)^2 )/((2n)!)) =(1/3)+((2π(√3))/(27))

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}} \\ $$

Question Number 197922 Answers: 0 Comments: 1

Question Number 197920 Answers: 2 Comments: 1

Question Number 197919 Answers: 1 Comments: 0

I_m = ∫_0 ^1 (((⌊2^m x⌋)/3^m ) Σ_(n=m+1) ^∞ ((⌊2^n x⌋)/3^n ))dx then find the value of I = Σ_(m=1) ^∞ I_m = ?

$$\:\:\:\:\:\:\mathrm{I}_{{m}} \:\:\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\lfloor\mathrm{2}^{{m}} {x}\rfloor}{\mathrm{3}^{{m}} }\:\underset{{n}={m}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\lfloor\mathrm{2}^{{n}} {x}\rfloor}{\mathrm{3}^{{n}} }\right){dx} \\ $$$$\:\:\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\mathrm{I}\:=\:\:\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{I}_{{m}} \:\:=\:\:?\: \\ $$

Question Number 197914 Answers: 1 Comments: 3

Determiner la surface hachuree (voir figure) BC=10cm ∡B=45° ∡C=30°

$$\boldsymbol{\mathrm{Determiner}}\:\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{hachuree}} \\ $$$$\left(\boldsymbol{\mathrm{voir}}\:\:\boldsymbol{\mathrm{figure}}\right) \\ $$$$\:\boldsymbol{\mathrm{BC}}=\mathrm{10}\boldsymbol{\mathrm{cm}}\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{B}}=\mathrm{45}°\:\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{30}° \\ $$

Question Number 197908 Answers: 1 Comments: 0

sin18^° =?

$${sin}\mathrm{18}^{°} =? \\ $$

Question Number 197895 Answers: 1 Comments: 0

Solve the equation: x^4 − x^3 − 4x^2 + 3x + 2 = 0

$${Solve}\:{the}\:{equation}: \\ $$$${x}^{\mathrm{4}} \:−\:{x}^{\mathrm{3}} \:−\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$

Question Number 197891 Answers: 2 Comments: 0

Soit I=∫^( 1) _( 0) (√(t(√(t(1−t)))))dt Comment calculer I

$$\mathrm{Soit}\:\mathrm{I}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \sqrt{\mathrm{t}\sqrt{\mathrm{t}\left(\mathrm{1}−\mathrm{t}\right)}}\mathrm{dt} \\ $$$$\mathrm{Comment}\:\mathrm{calculer}\:\mathrm{I} \\ $$

Question Number 197917 Answers: 1 Comments: 1

Question Number 197916 Answers: 2 Comments: 0

Question Number 197882 Answers: 1 Comments: 0

Find the minimum value of ((5t^2 −8t+5)/((2+(√3))t^2 −2t+2−(√3))) where 2−(√3)<t<2+(√3).

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}\:\:\mathrm{where}\:\mathrm{2}−\sqrt{\mathrm{3}}<{t}<\mathrm{2}+\sqrt{\mathrm{3}}. \\ $$

Question Number 197881 Answers: 1 Comments: 1

Question Number 197880 Answers: 1 Comments: 0

find the sum of infinite series (1/2^1 )∙(1/3^2 ) + (1/2^2 )∙(1/3^4 )(1^2 +2^2 +3^2 ) + (1/2^3 )∙(1/3^6 )(1^2 +2^2 +3^2 +...+7^2 )+ (1/2^4 )∙(1/3^8 )(1^2 +2^2 +3^2 +...+15^2 )+........

$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{7}^{\mathrm{2}} \right)+ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{8}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{15}^{\mathrm{2}} \right)+........ \\ $$

Question Number 197876 Answers: 2 Comments: 1

Question Number 197874 Answers: 1 Comments: 0

Question Number 197865 Answers: 1 Comments: 0

Question Number 197860 Answers: 3 Comments: 0

((2^(17) +2^(16) +2^(15) +…+1)/(2^8 +2^7 +2^6 +…+1)) = ?

$$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$

Question Number 197851 Answers: 0 Comments: 0

If z_1 = (√((α−β)/2)) and z_2 = (√((α+β)/2)) . Show that ∣z_1 −z_2 ∣^2 + ∣z_1 +z_2 ∣ = 2∣z∣^2 +∣z_2 ∣^2 and deduce that ∣α+(√(α^2 −β^2 ))∣ + ∣α−(√(α^2 −β^2 ))∣ = ∣α+β∣ + ∣α−β∣ Thank you in advance

$$\mathrm{If}\:\mathrm{z}_{\mathrm{1}} \:=\:\sqrt{\frac{\alpha−\beta}{\mathrm{2}}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:=\:\sqrt{\frac{\alpha+\beta}{\mathrm{2}}}\:.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\:=\:\mathrm{2}\mid\mathrm{z}\mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{deduce}\:\mathrm{that}\: \\ $$$$\mid\alpha+\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:+\:\mid\alpha−\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:=\:\mid\alpha+\beta\mid\:+\:\mid\alpha−\beta\mid \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{in}\:\mathrm{advance} \\ $$

Question Number 197850 Answers: 1 Comments: 0

Solve the equation: (2sin x − 1)(2cos 2x + 2sin x +1) = 3(1−2cos 2x)

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{2}{sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{cos}\:\mathrm{2}{x}\right) \\ $$

Question Number 197848 Answers: 4 Comments: 0

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