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Question Number 195092    Answers: 1   Comments: 0

lim_(x→1^+ ) ((4x+5)/(x−x^2 ))=?

$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{4}{x}+\mathrm{5}}{{x}−{x}^{\mathrm{2}} }=? \\ $$

Question Number 191930    Answers: 1   Comments: 2

What is the value of inside Area of (ABCDEF)? Such that: ∡AOB=120 ∡ANB=60;°R=ON (OA=OB=32cm) ArcAE=ArcBF(r=12cm) BASE is circulare (Aider le tailleur a savoir la surface du tissu necessaire pour couvrir l ′espace indique dans la figure?)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{inside}\:\mathrm{Area}\:\mathrm{of} \\ $$$$\left(\mathrm{ABCDEF}\right)? \\ $$$$\mathrm{Such}\:\mathrm{that}:\:\measuredangle\mathrm{AOB}=\mathrm{120}\:\:\:\measuredangle\mathrm{ANB}=\mathrm{60};°\mathrm{R}=\mathrm{ON} \\ $$$$\:\left(\mathrm{OA}=\mathrm{OB}=\mathrm{32cm}\right)\:\mathrm{ArcAE}=\mathrm{ArcBF}\left(\mathrm{r}=\mathrm{12cm}\right) \\ $$$$\mathrm{BASE}\:\mathrm{is}\:\mathrm{circulare} \\ $$$$\left({Aider}\:{le}\:{tailleur}\:{a}\:{savoir}\:{la}\:{surface}\right. \\ $$$${du}\:{tissu}\:{necessaire}\:{pour}\:{couvrir}\: \\ $$$$\left.\:{l}\:'\mathrm{e}{space}\:{indique}\:{dans}\:{la}\:{figure}?\right) \\ $$

Question Number 191901    Answers: 3   Comments: 0

if 0<θ<45^0 which is bigger ? 2tanθ or tan2θ

$${if}\:\:\mathrm{0}<\theta<\mathrm{45}^{\mathrm{0}} \:\:{which}\:{is}\:{bigger}\:? \\ $$$$\mathrm{2}{tan}\theta\:\:{or}\:\:{tan}\mathrm{2}\theta \\ $$

Question Number 191897    Answers: 2   Comments: 0

if we combine 100gr Na with 180 Cl_2 then how much NaCl will produce?

$${if}\:{we}\:{combine}\:\mathrm{100}{gr}\:{Na}\:{with}\:\mathrm{180}\:{Cl}_{\mathrm{2}} \\ $$$${then}\:{how}\:{much}\:{NaCl}\:{will}\:{produce}? \\ $$

Question Number 191927    Answers: 0   Comments: 1

What is the value of laterale Shaded Area (couronne circulaire coloree?) R=10cm 20cm≤ h≤24cm ∡COD=45°

$$\boldsymbol{\mathrm{W}}\mathrm{hat}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{laterale}\:\mathrm{Shaded} \\ $$$$\:\mathrm{Area}\:\:\left(\boldsymbol{\mathrm{couronne}}\:\boldsymbol{\mathrm{circulaire}}\:\:\boldsymbol{\mathrm{coloree}}?\right) \\ $$$$\:\:\boldsymbol{\mathrm{R}}=\mathrm{10}\boldsymbol{\mathrm{cm}}\:\:\:\mathrm{20}\boldsymbol{\mathrm{cm}}\leqslant\:\boldsymbol{\mathrm{h}}\leqslant\mathrm{24}\boldsymbol{\mathrm{cm}}\:\:\measuredangle\mathrm{COD}=\mathrm{45}° \\ $$$$\:\: \\ $$

Question Number 191926    Answers: 2   Comments: 0

Question Number 191921    Answers: 1   Comments: 0

Question Number 191889    Answers: 2   Comments: 0

Σ_(n=1) ^k (1/(n^2 +2n)) =?

$$\:\:\:\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\:=? \\ $$

Question Number 191887    Answers: 3   Comments: 0

Question Number 191874    Answers: 1   Comments: 0

lim_(x⇒∞) ((e^(x+1) +pi^(x−1) )/(e^(x−1) +pi^(x+1) ))

$${lim}_{{x}\Rightarrow\infty} \frac{{e}^{{x}+\mathrm{1}} +{pi}^{{x}−\mathrm{1}} }{{e}^{{x}−\mathrm{1}} +{pi}^{{x}+\mathrm{1}} } \\ $$$$ \\ $$

Question Number 191873    Answers: 4   Comments: 0

Question Number 191868    Answers: 2   Comments: 0

A particle of mass m moves under the central repulsive force ((mb)/r^3 ) and is initially moving at a distance ′a′ from the origin of a force with velocity ′v′ at right angle to ′a′. show that rcos pθ=a where p =(b/(a^2 v^2 ))+1.

$${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$ \\ $$

Question Number 191867    Answers: 1   Comments: 0

Prove that if u=f(x^3 +y^3 ),where f is arbitry function then x^2 (∂u/∂y) = y^2 (∂u/∂x)

$${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$

Question Number 191862    Answers: 1   Comments: 1

Find the last digit from (2^(400) −2^(320) )(2^(200) +2^(160) )(2^(200) −2^(160) )

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{from}\: \\ $$$$\:\left(\mathrm{2}^{\mathrm{400}} −\mathrm{2}^{\mathrm{320}} \right)\left(\mathrm{2}^{\mathrm{200}} +\mathrm{2}^{\mathrm{160}} \right)\left(\mathrm{2}^{\mathrm{200}} −\mathrm{2}^{\mathrm{160}} \right) \\ $$

Question Number 191859    Answers: 1   Comments: 1

Question Number 191856    Answers: 1   Comments: 0

Question Number 191855    Answers: 0   Comments: 0

Question Number 191854    Answers: 1   Comments: 0

Question Number 191846    Answers: 1   Comments: 0

find the last three digits of 4^2^(42) Mohammed Alwan

$${find}\:{the}\:{last}\:{three}\:{digits} \\ $$$${of}\:\mathrm{4}^{\mathrm{2}^{\mathrm{42}} } \\ $$$${Mohammed}\:{Alwan} \\ $$

Question Number 191841    Answers: 2   Comments: 0

Question Number 191840    Answers: 0   Comments: 0

calcul ∫_0 ^1 2∣cosu∣(√(1+3u^2 ))du

$${calcul} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}\mid{cosu}\mid\sqrt{\mathrm{1}+\mathrm{3}{u}^{\mathrm{2}} \:}{du} \\ $$

Question Number 191839    Answers: 1   Comments: 0

2^a = 3^b = 36^c then prove that ab = 2c(a + b).

$$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$

Question Number 191833    Answers: 3   Comments: 0

Question Number 191832    Answers: 0   Comments: 0

Question Number 191831    Answers: 1   Comments: 0

Question Number 191830    Answers: 0   Comments: 0

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