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Question Number 208503 Answers: 0 Comments: 0

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Question Number 208693 Answers: 2 Comments: 0

Calculate the area enclosed by the curve ((1/x)−2)^2 +((1/y)−2)^2 =1

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve} \\ $$$$\left(\frac{\mathrm{1}}{{x}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{y}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 208469 Answers: 1 Comments: 0

Find: ((61^3 + 24^3 )/(61^3 + 37^3 )) = ?

$$\mathrm{Find}:\:\:\:\:\:\frac{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{24}^{\mathrm{3}} }{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{37}^{\mathrm{3}} }\:\:=\:\:? \\ $$

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Compare it: a = log_3 4 b = log_5 6 c = log_6 2

$$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\mathrm{a}\:=\:\mathrm{log}_{\mathrm{3}} \:\mathrm{4} \\ $$$$\mathrm{b}\:=\:\mathrm{log}_{\mathrm{5}} \:\mathrm{6} \\ $$$$\mathrm{c}\:=\:\mathrm{log}_{\mathrm{6}} \:\mathrm{2} \\ $$

Question Number 208455 Answers: 1 Comments: 2

Question Number 208453 Answers: 2 Comments: 0

If f(x) = (((2a + 1)∙x + 1)/(x − a)) and f(x) = f^(−1) (x) Find: a^2 + 3 = ?

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\left(\mathrm{2a}\:+\:\mathrm{1}\right)\centerdot\mathrm{x}\:+\:\mathrm{1}}{\mathrm{x}\:−\:\mathrm{a}}\:\:\:\mathrm{and}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{3}\:=\:? \\ $$

Question Number 208447 Answers: 0 Comments: 0

$$\:\:\underbrace{\:} \\ $$

Question Number 208445 Answers: 1 Comments: 0

u_(n+1) = u_n −u_n ^3 , u_0 ∈]0,1[ v_n = (1/u_(n+1) ^2 )−(1/u_n ^2 ) = f(u_n ^2 ) ; f(x) = ((2−x)/((1−x)^2 )) v_n converges to 2, v_n is decreasing . show that v_n ≥ 2 x_n =(1/(n+1))Σ_(m=0) ^m (v_m ) . show that x_0 ≥x_n ≥v_n . show that x_n is decreasing and lim_(n→∞) x_n = l ≥2 . show that 2x_(n+1) −x_n ≤v_(n+1) and deduce l . express x_(n+1) −x_n interms of u_n . deduce lim_(n→∞) nu_n ^2

$$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${v}_{{n}} \:{converges}\:{to}\:\mathrm{2},\:{v}_{{n}} \:{is}\:{decreasing} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\mathrm{2} \\ $$$${x}_{{n}} =\frac{\mathrm{1}}{{n}+\mathrm{1}}\underset{{m}=\mathrm{0}} {\overset{{m}} {\sum}}\left({v}_{{m}} \right) \\ $$$$.\:{show}\:{that}\:{x}_{\mathrm{0}} \geqslant{x}_{{n}} \geqslant{v}_{{n}} \\ $$$$.\:{show}\:{that}\:{x}_{{n}} \:{is}\:{decreasing}\:{and}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}_{{n}} =\:{l}\:\geqslant\mathrm{2} \\ $$$$.\:{show}\:{that}\:\mathrm{2}{x}_{{n}+\mathrm{1}} −{x}_{{n}} \leqslant{v}_{{n}+\mathrm{1}} \:{and}\:{deduce}\:{l} \\ $$$$.\:{express}\:{x}_{{n}+\mathrm{1}} −{x}_{{n}} \:{interms}\:{of}\:{u}_{{n}} \\ $$$$.\:{deduce}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{nu}_{{n}} ^{\mathrm{2}} \\ $$

Question Number 208441 Answers: 3 Comments: 0

Question Number 208440 Answers: 0 Comments: 0

a_1 >a_2 >a_3 >...>a_n >0 b_1 >b_2 >b_3 >...>b_n >0 prove Σ_(i=1) ^n a_i b_i ≥Σ_(i=1) ^n a_i b_(n−i+1)

$${a}_{\mathrm{1}} >{a}_{\mathrm{2}} >{a}_{\mathrm{3}} >...>{a}_{{n}} >\mathrm{0} \\ $$$${b}_{\mathrm{1}} >{b}_{\mathrm{2}} >{b}_{\mathrm{3}} >...>{b}_{{n}} >\mathrm{0} \\ $$$${prove} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{i}} \geqslant\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {b}_{{n}−{i}+\mathrm{1}} \\ $$

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h_a (x) = e^(−x) + ax^2 show that h_a admits a minimum in R

$${h}_{{a}} \left({x}\right)\:=\:{e}^{−{x}} \:+\:{ax}^{\mathrm{2}} \\ $$$${show}\:{that}\:{h}_{{a}} \:{admits}\:{a}\:{minimum}\:{in}\:\mathbb{R} \\ $$

Question Number 208423 Answers: 1 Comments: 0

calculons i=∫∫∫_([0;1]) ((dxdydz)/(1−xyz))

$$\:\:\:\boldsymbol{{calculons}}\: \\ $$$$\boldsymbol{{i}}=\int\int\int_{\left[\mathrm{0};\mathrm{1}\right]} \frac{\boldsymbol{{dxdydz}}}{\mathrm{1}−\boldsymbol{{xyz}}} \\ $$

Question Number 208421 Answers: 0 Comments: 1

Question Number 208420 Answers: 2 Comments: 1

resoudre dans R^3 { ((x+y=3)),((y+z=5)) :}x+z=4

$$\boldsymbol{{resoudre}}\:\boldsymbol{{dans}}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{\boldsymbol{{x}}+\boldsymbol{{y}}=\mathrm{3}}\\{\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{5}}\end{cases}\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{4} \\ $$

Question Number 208418 Answers: 1 Comments: 0

u_(n+1) = u_n −u_n ^3 ; u_0 ∈ ]0, 1[ . show that u_n ∈ ]0, 1[ . show that u_n converges to 0 v_n = (1/u_(n+1) ^2 ) − (1/u_n ^2 ) . express v_n interms of u_n . show that v_n converges to 2 f(x) = ((2−x)/((1−x)^2 )) . show that f is increasing and deduce that v_n is decreasing . show that v_n ≥ 2

$$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} \:;\:{u}_{\mathrm{0}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\left..\:{show}\:{that}\:{u}_{{n}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$.\:{show}\:{that}\:{u}_{{n}} \:{converges}\:{to}\:\mathrm{0} \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} } \\ $$$$.\:{express}\:{v}_{{n}} \:{interms}\:{of}\:{u}_{{n}} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:{converges}\:{to}\:\mathrm{2} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$.\:{show}\:{that}\:{f}\:{is}\:{increasing}\:{and}\:{deduce}\:{that}\: \\ $$$${v}_{{n}} \:{is}\:{decreasing} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\mathrm{2} \\ $$

Question Number 208412 Answers: 2 Comments: 0

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