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Question Number 191652 Answers: 1 Comments: 1

Question Number 191642 Answers: 0 Comments: 1

Comparer [OHDF] avec [ABHEF] (avec preuve) Sachant que: OH∣∣DF HE=EF OB=4OA CH=2BC et BC<OB.

$$\mathrm{Comparer}\:\left[\mathrm{OHDF}\right]\:\:\mathrm{avec}\:\left[\mathrm{ABHEF}\right] \\ $$$$\left({avec}\:{preuve}\right) \\ $$$$\:\:{Sa}\mathrm{c}{h}\mathrm{a}{nt}\:{que}: \\ $$$$\mathrm{OH}\mid\mid\mathrm{DF}\:\:\:\mathrm{HE}=\mathrm{EF}\:\:\:\mathrm{OB}=\mathrm{4OA} \\ $$$$\:\:\:\mathrm{CH}=\mathrm{2BC}\:\mathrm{et}\:\mathrm{BC}<\mathrm{OB}.\:\:\: \\ $$$$ \\ $$

Question Number 191640 Answers: 0 Comments: 1

x^3 −1 = (x−1) (x^2 +x+1) ⇒ x = 1 and x = ((−1±(√3) i)/2) ⇒ w = −(1/2) + (((√3)i )/2) and w^2 = −(1/2) − ((√3)/2) i ⇒ w^3 = 1 similarly x^3 + 1= 0 ⇒ x = −1 and x =−w^2 and x = −w

$${x}^{\mathrm{3}} −\mathrm{1}\:=\:\left({x}−\mathrm{1}\right)\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{1}\:{and}\:{x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}\:{i}}{\mathrm{2}} \\ $$$$\Rightarrow\:{w}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}{i}\:}{\mathrm{2}}\:{and}\:{w}^{\mathrm{2}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{i} \\ $$$$\Rightarrow\:{w}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$${similarly}\:{x}^{\mathrm{3}} \:+\:\mathrm{1}=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:−\mathrm{1}\:{and}\:{x}\:=−{w}^{\mathrm{2}} \:{and}\:{x}\:=\:−{w} \\ $$

Question Number 191639 Answers: 1 Comments: 1

Une feuille de papier format A4 est plie n fois simultanement en longueur et largeur (total plie=2n) −Quel son les nouvelles dimensions de cette feuille la feuille a la fin . −quel seront ses dimensions apres 6 pliages −Quel seront les dimensions pour une feuille care de 30 cm plie 6 fois succrssivement en long et large

$$\mathrm{Une}\:\mathrm{feuille}\:\mathrm{de}\:\mathrm{papier}\:\mathrm{format}\:\boldsymbol{\mathrm{A}}\mathrm{4}\:\mathrm{est}\:\mathrm{plie} \\ $$$$\:\boldsymbol{\mathrm{n}}\:\mathrm{fois}\:\mathrm{simultanement}\:\mathrm{en}\:\mathrm{longueur}\:\mathrm{et} \\ $$$$\mathrm{largeur}\:\left(\mathrm{total}\:\mathrm{plie}=\mathrm{2n}\right) \\ $$$$−\mathrm{Quel}\:\mathrm{son}\:\mathrm{les}\:\mathrm{nouvelles}\:\mathrm{dimensions}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{feuille} \\ $$$$\:\:\:\mathrm{la}\:\mathrm{feuille}\:\mathrm{a}\:\mathrm{la}\:\mathrm{fin}\:. \\ $$$$−\mathrm{quel}\:\mathrm{seront}\:\mathrm{ses}\:\mathrm{dimensions}\:\:\mathrm{apres}\: \\ $$$$\:\:\:\:\mathrm{6}\:\boldsymbol{\mathrm{pliages}} \\ $$$$−\mathrm{Quel}\:\mathrm{seront}\:\:\mathrm{les}\:\mathrm{dimensions}\:\mathrm{pour}\:\mathrm{une}\:\mathrm{feuille}\:\boldsymbol{\mathrm{care}}\:\mathrm{de}\:\mathrm{30}\:\boldsymbol{\mathrm{cm}}\: \\ $$$$\:\:\:\mathrm{plie}\:\mathrm{6}\:\boldsymbol{\mathrm{fois}}\:\:\mathrm{succrssivement}\:\mathrm{en}\:\mathrm{long}\:\:\mathrm{et}\:\mathrm{large} \\ $$$$ \\ $$

Question Number 191637 Answers: 0 Comments: 0

Question Number 191633 Answers: 0 Comments: 0

Question Number 191632 Answers: 0 Comments: 1

∫_0 ^∞ (√(tan θ)) dθ

$$\int_{\mathrm{0}} ^{\infty} \sqrt{\mathrm{tan}\:\theta}\:{d}\theta \\ $$

Question Number 191631 Answers: 1 Comments: 0

∫_0 ^∞ x^(1/2) e^(−x^2 ) dx

$$\int_{\mathrm{0}} ^{\infty} {x}^{\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 191626 Answers: 3 Comments: 0

Question Number 191624 Answers: 1 Comments: 0

Question Number 191623 Answers: 2 Comments: 0

Question Number 191621 Answers: 2 Comments: 0

Question Number 191615 Answers: 1 Comments: 0

a + b + c = 0. Prove that, (a/(a^2 − bc)) + (b/(b^2 − ca)) + (c/(c^2 − ab)) = 0.

$${a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that}, \\ $$$$\frac{{a}}{{a}^{\mathrm{2}} \:−\:{bc}}\:+\:\frac{{b}}{{b}^{\mathrm{2}} \:−\:{ca}}\:+\:\frac{{c}}{{c}^{\mathrm{2}} \:−\:{ab}}\:=\:\mathrm{0}. \\ $$

Question Number 191614 Answers: 0 Comments: 1

((α^(100) +β^(100) )/(α^(100) −β^(100) )) = (((−w)^(100) +(−w^2 )^(100) )/((−w)^(100) −(−w^2 )^(100) )) = ((w^(100) +w^(200) )/(w^(100) −w^(200) )) = ((1+w^(100) )/(1−w^(100 ) )) = ((1+w)/(1−w)) = (2/(2w)) = (1/w) =

$$\frac{\alpha^{\mathrm{100}} +\beta^{\mathrm{100}} }{\alpha^{\mathrm{100}} −\beta^{\mathrm{100}} }\:=\: \\ $$$$\frac{\left(−{w}\right)^{\mathrm{100}} +\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} }{\left(−{w}\right)^{\mathrm{100}} −\left(−{w}^{\mathrm{2}} \right)^{\mathrm{100}} } \\ $$$$=\:\frac{{w}^{\mathrm{100}} +{w}^{\mathrm{200}} }{{w}^{\mathrm{100}} −{w}^{\mathrm{200}} } \\ $$$$=\:\frac{\mathrm{1}+{w}^{\mathrm{100}} }{\mathrm{1}−{w}^{\mathrm{100}\:} } \\ $$$$=\:\frac{\mathrm{1}+{w}}{\mathrm{1}−{w}}\:=\:\frac{\mathrm{2}}{\mathrm{2}{w}}\:=\:\frac{\mathrm{1}}{{w}}\:=\: \\ $$

Question Number 191610 Answers: 1 Comments: 0

Q: if x+(1/x)=2cos(θ) prove it x^n +(1/x^n )=2cos(nθ)

$${Q}:\:{if}\:\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}{cos}\left(\theta\right)\:\:{prove}\:{it}\:\:{x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\mathrm{2}{cos}\left({n}\theta\right) \\ $$

Question Number 191589 Answers: 1 Comments: 1

a^x = bc, b^y = ca, c^z = ab. Prove that, (x/(1 + x)) + (y/(1 + y)) + (z/(1 + z)) = 2. (Without using log) a ≠ b ≠ c

$${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$

Question Number 191651 Answers: 2 Comments: 1