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Question Number 157660 Answers: 0 Comments: 0

Given x_1 = 1, x_2 , x_3 , …, is a real numbers sequence for n ≥ 1 with recurrence relation x_(n+1) − x_n = (1/(2x_n )) . [x] is expressed as the largest integer of x . [25x_(625) ] = ?

$${Given}\:\:{x}_{\mathrm{1}} \:=\:\mathrm{1},\:{x}_{\mathrm{2}} \:,\:{x}_{\mathrm{3}} \:,\:\ldots,\:{is}\:\:{a}\:\:{real}\:\:{numbers}\:\:{sequence}\:\:{for}\:\:{n}\:\geqslant\:\mathrm{1}\:\:{with}\:\: \\ $$$${recurrence}\:\:{relation}\:\:{x}_{{n}+\mathrm{1}} \:−\:{x}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{x}_{{n}} }\:\:. \\ $$$$\left[{x}\right]\:\:{is}\:\:{expressed}\:\:{as}\:\:{the}\:\:{largest}\:\:{integer}\:\:{of}\:\:{x}\:\:. \\ $$$$\left[\mathrm{25}{x}_{\mathrm{625}} \right]\:\:=\:\:? \\ $$

Question Number 157655 Answers: 1 Comments: 1

x^2 f(x^3 )+(1/((1+x)^2 )) f(((1−x)/(1+x)))=4x^3 (1+x^4 )^5 ∫_( 0) ^( 1) f(x) dx =?

$$\:\:{x}^{\mathrm{2}} \:{f}\left({x}^{\mathrm{3}} \right)+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{f}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=\mathrm{4}{x}^{\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{5}} \\ $$$$\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right)\:{dx}\:=? \\ $$

Question Number 157647 Answers: 0 Comments: 3

bonjour ,calculer la limite suivante en utilisant les developpements limites: lim_(x→0) ((1/x^2 ) − (1/(sin^2 x))).

$${bonjour}\:,{calculer}\:{la}\:{limite}\:{suivante}\:{en}\:{utilisant}\:{les}\:{developpements}\:{limites}: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}}\right). \\ $$

Question Number 157645 Answers: 1 Comments: 0

what is the latest version of this app available i m having 2.265

$${what}\:{is}\:{the}\:{latest}\:{version} \\ $$$${of}\:{this}\:{app}\:{available} \\ $$$${i}\:{m}\:{having}\:\:\:\mathrm{2}.\mathrm{265} \\ $$

Question Number 157644 Answers: 2 Comments: 0

Prove (1/2)(√(2−(√3)))=(((√6)−(√2))/4)

$$\mathrm{Prove}\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Question Number 157637 Answers: 2 Comments: 0

3^x =2^x y+1 {x:y} εN.

$$\:\:\:\:\:\:\mathrm{3}^{{x}} =\mathrm{2}^{{x}} {y}+\mathrm{1} \\ $$$$\:\:\:\:\:\left\{\boldsymbol{{x}}:\boldsymbol{{y}}\right\}\:\varepsilon\mathbb{N}.\: \\ $$

Question Number 157635 Answers: 0 Comments: 0

Question Number 157631 Answers: 0 Comments: 0

what is higher order derivatives? discuss its importants.

$${what}\:{is}\:{higher}\:{order}\:{derivatives}? \\ $$$${discuss}\:{its}\:{importants}. \\ $$

Question Number 157630 Answers: 1 Comments: 1

Question Number 157652 Answers: 0 Comments: 4

Question Number 157628 Answers: 1 Comments: 0

find (C_0 ^(100) )^2 +(C_2 ^(100) )^2 +(C_4 ^(100) )^2 +(C_6 ^(100) )^2 +...+(C_(100) ^(100) )^2 =?

$${find} \\ $$$$\left({C}_{\mathrm{0}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{2}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{4}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{6}} ^{\mathrm{100}} \right)^{\mathrm{2}} +...+\left({C}_{\mathrm{100}} ^{\mathrm{100}} \right)^{\mathrm{2}} =? \\ $$

Question Number 157611 Answers: 1 Comments: 0

Given g(x) = (1/(1 + 3^((1/2) − x) )) g((1/(2017))) + g((2/(2017))) + g((3/(2017))) + … + g(((2016)/(2017))) = ?

$${Given}\:\:{g}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}\:−\:{x}} } \\ $$$${g}\left(\frac{\mathrm{1}}{\mathrm{2017}}\right)\:+\:{g}\left(\frac{\mathrm{2}}{\mathrm{2017}}\right)\:+\:{g}\left(\frac{\mathrm{3}}{\mathrm{2017}}\right)\:+\:\ldots\:+\:{g}\left(\frac{\mathrm{2016}}{\mathrm{2017}}\right)\:\:=\:\:? \\ $$

Question Number 157604 Answers: 1 Comments: 3

Question find the” minimum” value of: f (x):= ∣1+x∣+∣ 2+x∣ + ∣4 +2x∣

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Q}{uestion}\: \\ $$$$\:{find}\:{the}''\:{minimum}''\:{value}\:{of}: \\ $$$$ \\ $$$${f}\:\left({x}\right):=\:\mid\mathrm{1}+{x}\mid+\mid\:\mathrm{2}+{x}\mid\:+\:\mid\mathrm{4}\:+\mathrm{2}{x}\mid \\ $$$$ \\ $$

Question Number 157598 Answers: 1 Comments: 0

Given a,b,c nonnegative numbers which satisfy a+b+c = 3. Prove that (1/(2ab^2 + 1)) + (1/(2bc^2 +1)) + (1/(2ac^2 + 1)) ≥ 1 .

$${Given}\:\:{a},{b},{c}\:\:{nonnegative}\:\:{numbers}\:\:{which}\:\:{satisfy}\:\:\:{a}+{b}+{c}\:=\:\mathrm{3}. \\ $$$${Prove}\:\:{that}\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{ab}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{bc}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{ac}^{\mathrm{2}} \:+\:\mathrm{1}}\:\geqslant\:\mathrm{1}\:. \\ $$

Question Number 157599 Answers: 1 Comments: 0

Find the number of x ∈ [1, 2016 ] , x ∈ N which making the expression 4x^6 + x^3 + 5 is divided by 11 .

$${Find}\:\:{the}\:\:{number}\:\:{of}\:\:{x}\:\in\:\left[\mathrm{1},\:\mathrm{2016}\:\right]\:\:,\:\:{x}\:\in\:\mathbb{N} \\ $$$${which}\:\:{making}\:\:{the}\:\:{expression}\:\:\mathrm{4}{x}^{\mathrm{6}} \:+\:\:{x}^{\mathrm{3}} \:+\:\mathrm{5}\:\:\:{is}\:\:{divided}\:\:\:{by}\:\:\mathrm{11}\:. \\ $$

Question Number 157593 Answers: 0 Comments: 0

let f:C→R z→min(y−[y],[y+1]−y) , y=Im(z) let w=e^(i((2π)/n)) , n∈N^∗ evaluate S_n =Σ_(0≤k<n) f(w^k )

$${let}\:{f}:{C}\rightarrow{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:{z}\rightarrow{min}\left({y}−\left[{y}\right],\left[{y}+\mathrm{1}\right]−{y}\right)\:,\:{y}={Im}\left({z}\right) \\ $$$${let}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} ,\:{n}\in{N}^{\ast} \\ $$$${evaluate}\:{S}_{{n}} =\underset{\mathrm{0}\leqslant{k}<{n}} {\sum}{f}\left({w}^{{k}} \right) \\ $$

Question Number 157592 Answers: 0 Comments: 0

let A={0^. ,1^. ,2^. } prove that every application from A to A is a 2nd degree polynom

$${let}\:{A}=\left\{\overset{.} {\mathrm{0}},\overset{.} {\mathrm{1}},\overset{.} {\mathrm{2}}\right\} \\ $$$${prove}\:{that}\:{every}\:{application}\:{from}\:{A}\: \\ $$$${to}\:{A}\:{is}\:{a}\:\mathrm{2}{nd}\:{degree}\:{polynom} \\ $$

Question Number 157591 Answers: 1 Comments: 0

lim_(x→π) ((x−π)^2 cos ((1/(x−π)))+x^4 +sin^3 x)=?

$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\left(\left({x}−\pi\right)^{\mathrm{2}} \mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}−\pi}\right)+{x}^{\mathrm{4}} +\mathrm{sin}\:^{\mathrm{3}} {x}\right)=? \\ $$

Question Number 157585 Answers: 1 Comments: 0

if 0<a≤b≤c<(π/2) then: (5/(tana)) + (3/(tanb)) + (1/(tanc)) ≥ ((27)/(tana + tanb + tanc))

$$\mathrm{if}\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}\leqslant\mathrm{c}<\frac{\pi}{\mathrm{2}}\:\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{5}}{\mathrm{tan}\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{3}}{\mathrm{tan}\boldsymbol{\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\boldsymbol{\mathrm{c}}}\:\geqslant\:\frac{\mathrm{27}}{\mathrm{tan}\boldsymbol{\mathrm{a}}\:+\:\mathrm{tan}\boldsymbol{\mathrm{b}}\:+\:\mathrm{tan}\boldsymbol{\mathrm{c}}} \\ $$

Question Number 157584 Answers: 3 Comments: 0

suppose the ratio of Jim to Rohn is 2:1 and the ratio of Rohn to Bill is 3:4, how do i find for the ratio of Jim to Bill.....???

$${suppose}\: \\ $$$${the}\:{ratio}\:{of}\:{Jim}\:{to}\:{Rohn}\:{is}\:\mathrm{2}:\mathrm{1} \\ $$$$\:{and}\:{the}\:{ratio}\:{of}\:{Rohn}\:{to}\:{Bill}\:{is} \\ $$$$\:\mathrm{3}:\mathrm{4},\:{how}\:{do}\:{i}\:{find}\:{for}\:{the}\:{ratio} \\ $$$${of}\:{Jim}\:{to}\:{Bill}.....??? \\ $$

Question Number 157576 Answers: 1 Comments: 2

Question Number 157575 Answers: 1 Comments: 3

(√((1/2)−(1/2)(√((1/2)+(1/2)cos 𝛂)))) (Π<𝛂<2Π)

$$ \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\alpha}}} \\ $$$$\left(\Pi<\boldsymbol{\alpha}<\mathrm{2}\Pi\right) \\ $$

Question Number 157565 Answers: 0 Comments: 0

Question Number 157562 Answers: 2 Comments: 0

Question Number 157560 Answers: 0 Comments: 0

Question Number 157561 Answers: 0 Comments: 0

if a;b;c and a+b+c≥3 prove that: Σ (a^3 /(b + kbc)) ≥ (3/(1 + k)) ; k>0

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}\geqslant\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Sigma\:\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}\:+\:\mathrm{kbc}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{1}\:+\:\mathrm{k}}\:\:;\:\:\mathrm{k}>\mathrm{0}\: \\ $$

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