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Hello verry nice day for all of you god bless You pleas Can you use black Color shen You post Quation or Give answer is verry hard to read withe other colors

$${Hello}\: \\ $$$${verry}\:{nice}\:{day}\:{for}\:{all}\:{of}\:{you}\:{god}\:{bless}\:{You} \\ $$$${pleas}\:{Can}\:{you}\:{use}\:{black}\:{Color}\:{shen}\:{You}\:{post}\:{Quation}\: \\ $$$${or}\:{Give}\:{answer}\:{is}\:{verry}\:{hard}\:{to}\:{read}\:{withe}\:{other}\:{colors} \\ $$

Question Number 98953 Answers: 2 Comments: 0

Given f(x)=sin^2 x find the expansion of f(x) up to the n^(th) term.

$$\mathcal{G}\mathrm{iven}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{\mathrm{2}} \mathrm{x}\:\mathrm{find}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{up}\:\mathrm{to}\:\mathrm{the}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{term}. \\ $$

Question Number 98924 Answers: 0 Comments: 0

Prove that if a+ bi is a root to pz^2 + qz + r = 0 , where a,b,p,q,r ∈R then a−bi is also a root to that equation.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{a}+\:{bi}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{to} \\ $$$$\:{pz}^{\mathrm{2}} \:+\:{qz}\:+\:{r}\:=\:\mathrm{0}\:,\:\mathrm{where}\:{a},{b},{p},{q},{r}\:\in\mathbb{R} \\ $$$$\mathrm{then}\:{a}−{bi}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{root}\:\mathrm{to}\:\mathrm{that}\:\mathrm{equation}. \\ $$

Question Number 98887 Answers: 0 Comments: 0

Question Number 98673 Answers: 2 Comments: 0

find a_n in terms of n (I can′t find it...) a_1 =1; a_2 =4 a_3 =a_2 ×4×((2^2 −1)/2^2 ) a_4 =a_3 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 ) a_5 =a_4 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 )×((4^2 −1)/4^2 ) ... n≥2: a_(n+1) =4a_n Π_(k=2) ^n ((k^2 −1)/k^2 )

$$\mathrm{find}\:{a}_{{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\left(\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}...\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$... \\ $$$${n}\geqslant\mathrm{2}:\:{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 98640 Answers: 0 Comments: 1

Question Number 98638 Answers: 0 Comments: 2

Le plan complexe est rapporte^ a^ un repe^ re orthornorme directe (0,e_1 ^→ ,e_2 ^→ ). On note A et B les points d′affixes respectives i, et 2i. Soit f, l′application du plan prive^ de A dans lui-me^ me qui a^ tout point M d′affixe z distincte i associe le point M d′affixe z′ definie par: z′=((2z−i)/(iz+1)) 1\ Soit z≠i a\ On pose z−i=re^(iθ) . Interpreter ge^ ometriquement r et θ a^ l′aide des points A et M.

$$\mathrm{Le}\:\mathrm{plan}\:\mathrm{complexe}\:\mathrm{est}\:\mathrm{rapport}\acute {\mathrm{e}}\:\grave {\mathrm{a}}\:\mathrm{un}\:\mathrm{rep}\grave {\mathrm{e}re} \\ $$$$\mathrm{orthornorme}\:\mathrm{directe}\:\left(\mathrm{0},\overset{\rightarrow} {\mathrm{e}}_{\mathrm{1}} ,\overset{\rightarrow} {\mathrm{e}}_{\mathrm{2}} \right).\:\mathcal{O}\mathrm{n}\:\mathrm{note}\:\mathrm{A}\:\mathrm{et}\:\mathrm{B}\:\mathrm{les} \\ $$$$\mathrm{points}\:\mathrm{d}'\mathrm{affixes}\:\mathrm{respectives}\:\boldsymbol{\mathrm{i}},\:\mathrm{et}\:\mathrm{2}\boldsymbol{\mathrm{i}}.\:\mathrm{Soit}\:\mathrm{f},\:\mathrm{l}'\mathrm{application} \\ $$$$\mathrm{du}\:\mathrm{plan}\:\mathrm{priv}\acute {\mathrm{e}}\:\mathrm{de}\:\mathrm{A}\:\mathrm{dans}\:\mathrm{lui}-\mathrm{m}\hat {\mathrm{e}me}\:\mathrm{qui}\:\grave {\mathrm{a}}\:\mathrm{tout}\:\mathrm{point} \\ $$$$\mathrm{M}\:\mathrm{d}'\mathrm{affixe}\:\mathrm{z}\:\mathrm{distincte}\:\boldsymbol{\mathrm{i}}\:\mathrm{associe}\:\mathrm{le}\:\mathrm{point}\:\mathrm{M}\:\mathrm{d}'\mathrm{affixe} \\ $$$$\boldsymbol{\mathrm{z}}'\:\mathrm{definie}\:\mathrm{par}:\:\mathrm{z}'=\frac{\mathrm{2z}−\mathrm{i}}{\mathrm{iz}+\mathrm{1}} \\ $$$$\mathrm{1}\backslash\:\mathrm{Soit}\:\mathrm{z}\neq\mathrm{i} \\ $$$$\mathrm{a}\backslash\:\mathrm{On}\:\mathrm{pose}\:\mathrm{z}−\mathrm{i}=\mathrm{re}^{\mathrm{i}\theta} .\:\mathcal{I}\mathrm{nterpreter}\:\mathrm{g}\acute {\mathrm{e}ometriquement}\:\mathrm{r}\:\mathrm{et}\:\theta \\ $$$$\grave {\mathrm{a}}\:\mathrm{l}'\mathrm{aide}\:\mathrm{des}\:\mathrm{points}\:\mathrm{A}\:\mathrm{et}\:\mathrm{M}. \\ $$

Question Number 98620 Answers: 1 Comments: 0

how do i make use of the function gamma(n). example, gamma(n)=∫_0 ^∞ x^(n−1) e^(−x) dx? instead of typing gamma(n). i can′t find it in the app.

$$\boldsymbol{{how}}\:\boldsymbol{{do}}\:\boldsymbol{{i}}\:\boldsymbol{{make}}\:\boldsymbol{{use}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{function}} \\ $$$$\boldsymbol{{gamma}}\left(\boldsymbol{{n}}\right). \\ $$$$\boldsymbol{{example}},\:\boldsymbol{{gamma}}\left(\boldsymbol{{n}}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\boldsymbol{{x}}^{\boldsymbol{{n}}−\mathrm{1}} \boldsymbol{{e}}^{−\boldsymbol{{x}}} \boldsymbol{{dx}}? \\ $$$$\boldsymbol{{instead}}\:\boldsymbol{{of}}\:\boldsymbol{{typing}}\:\boldsymbol{{gamma}}\left(\boldsymbol{{n}}\right). \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{can}}'\boldsymbol{{t}}\:\boldsymbol{{find}}\:\boldsymbol{{it}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\:\boldsymbol{{app}}. \\ $$

Question Number 98468 Answers: 2 Comments: 10

Question Number 98443 Answers: 1 Comments: 0

Given the sequence (U_n )_(n∈N) defined by U_0 =1 and U_(n+1) =f(U_n ) where f(x)=(x/((x+1)^2 )) Show by mathematical induction that ∀n∈N^∗ 0<U_n ≤(1/n)

$$\mathcal{G}\mathrm{iven}\:\mathrm{the}\:\mathrm{sequence}\:\left(\mathrm{U}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}} \:\mathrm{defined}\:\mathrm{by}\:\mathrm{U}_{\mathrm{0}} =\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{U}_{\mathrm{n}+\mathrm{1}} =\mathrm{f}\left(\mathrm{U}_{\mathrm{n}} \right)\:\mathrm{where}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\mathcal{S}\mathrm{how}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{0}<\mathrm{U}_{\mathrm{n}} \leqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$

Question Number 98320 Answers: 1 Comments: 0

Question Number 98208 Answers: 0 Comments: 2

suppose a force given as F_1 = 24 N and F_2 = 50 N act through points AB and AC where OA = 2i +3j , OB = 5i + 6j and OC = 7i + 8j (a) find in vector notation F_1 and F_2 then find thier resultant.

$$\mathrm{suppose}\:\mathrm{a}\:\mathrm{force}\:\mathrm{given}\:\mathrm{as}\:{F}_{\mathrm{1}} \:=\:\mathrm{24}\:{N}\:\mathrm{and}\:{F}_{\mathrm{2}} \:=\:\mathrm{50}\:{N}\:\mathrm{act}\:\mathrm{through}\: \\ $$$$\mathrm{points}\:{AB}\:\mathrm{and}\:{AC}\:\mathrm{where}\:\:{OA}\:=\:\mathrm{2}{i}\:+\mathrm{3}{j}\:,\:{OB}\:=\:\mathrm{5}{i}\:+\:\mathrm{6}{j}\:\:\mathrm{and}\: \\ $$$${OC}\:=\:\mathrm{7}{i}\:+\:\mathrm{8}{j} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{find}\:\mathrm{in}\:\mathrm{vector}\:\mathrm{notation}\:{F}_{\mathrm{1}} \:\mathrm{and}\:{F}_{\mathrm{2}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{thier}\:\mathrm{resultant}. \\ $$

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