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Question Number 87751 Answers: 0 Comments: 2

find in the form y= f(x) the general solution of the differentail equation (d^2 y/dx^2 ) −(dy/dx)−6y = e^(3x)

$$\mathrm{find}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{y}=\:{f}\left({x}\right)\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{differentail}\:\mathrm{equation} \\ $$$$\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\frac{{dy}}{{dx}}−\mathrm{6}{y}\:=\:{e}^{\mathrm{3}{x}} \\ $$$$ \\ $$

Question Number 87550 Answers: 0 Comments: 2

(1/(2e^(−x) −1)) > (2/(e^(−x) −2))

$$\frac{\mathrm{1}}{\mathrm{2e}^{−\mathrm{x}} −\mathrm{1}}\:>\:\frac{\mathrm{2}}{\mathrm{e}^{−\mathrm{x}} −\mathrm{2}} \\ $$

Question Number 87536 Answers: 1 Comments: 0

solve ∣2x−1∣=3⌊x⌋+2{x}

$${solve}\: \\ $$$$\mid\mathrm{2}{x}−\mathrm{1}\mid=\mathrm{3}\lfloor{x}\rfloor+\mathrm{2}\left\{{x}\right\} \\ $$$$ \\ $$

Question Number 87497 Answers: 0 Comments: 0

A complex number z is defined by z = (1/2)(cos θ + isin θ),such that z^n = (1/2^n ) (cos nθ + isin nθ) Using De Moivre′s theorem,or otherwise, show that (i) Σ_(r=0) ^∞ (1/4^r ) sin 2rθ is a convergent geometic progression. (ii) Σ_(r=0) ^∞ (1/4^r ) sin 2r = ((14 sin 2θ)/(17−16cos 2θ))

$$\mathrm{A}\:\mathrm{complex}\:\mathrm{number}\:{z}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:\theta\:+\:{i}\mathrm{sin}\:\theta\right),\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\left(\mathrm{cos}\:{n}\theta\:+\:{i}\mathrm{sin}\:{n}\theta\right) \\ $$$$\mathrm{Using}\:\mathrm{De}\:\mathrm{Moivre}'\mathrm{s}\:\mathrm{theorem},\mathrm{or}\:\mathrm{otherwise},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\left(\mathrm{i}\right)\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{4}^{{r}} }\:\mathrm{sin}\:\mathrm{2r}\theta\:\mathrm{is}\:\mathrm{a}\:\mathrm{convergent}\:\mathrm{geometic}\:\mathrm{progression}. \\ $$$$\left(\mathrm{ii}\right)\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}^{{r}} }\:\mathrm{sin}\:\mathrm{2}{r}\:=\:\frac{\mathrm{14}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{17}−\mathrm{16cos}\:\mathrm{2}\theta} \\ $$

Question Number 87308 Answers: 0 Comments: 0

A sequence (U_n ) is defined reculsively as U_o = (1/2) and U_(n+1) = (2/(1 + U_n )) for n ∈ N a) Show by mathematical induction that all terms in the sequence are positive. b) Given that the sequence (U_n ) is convergent, show that the limit,l, is a solution to the equation x^2 + x−2 = 0. Hence find l c) Given that (V_n ) is a sequence of general term such that V_n = ((U_n −1)/(U_n +2)) , ∀ n ∈ N. show that (V_n ) is convergent and determine its limit. hence deduce the convergence of the sequence (U_n ). Please recommend me textbooks for this topic even youtube vids please

$$\:\mathrm{A}\:\mathrm{sequence}\:\left({U}_{{n}} \right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{reculsively}\:\mathrm{as}\: \\ $$$$\:{U}_{{o}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:{U}_{{n}+\mathrm{1}} \:=\:\frac{\mathrm{2}}{\mathrm{1}\:+\:{U}_{{n}} }\:\mathrm{for}\:\mathrm{n}\:\in\:\mathbb{N} \\ $$$$\left.\:\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\:\mathrm{all}\:\mathrm{terms}\:\mathrm{in}\:\mathrm{the}\:\mathrm{sequence} \\ $$$$\:\:\:\:\:\mathrm{are}\:\mathrm{positive}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left({U}_{{n}} \right)\:\mathrm{is}\:\mathrm{convergent},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{limit},{l},\:\mathrm{is} \\ $$$$\:\:\:\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{x}−\mathrm{2}\:=\:\mathrm{0}.\:\mathrm{Hence}\:\mathrm{find}\:{l} \\ $$$$\left.\:\mathrm{c}\right)\:\:\mathrm{Given}\:\mathrm{that}\:\left({V}_{{n}} \right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{general}\:\mathrm{term}\:\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:{V}_{{n}} \:=\:\frac{{U}_{{n}} −\mathrm{1}}{{U}_{{n}} +\mathrm{2}}\:,\:\forall\:{n}\:\in\:\mathbb{N}. \\ $$$$\:\:\mathrm{show}\:\mathrm{that}\:\left({V}_{{n}} \right)\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{and}\:\mathrm{determine}\:\mathrm{its}\:\:\mathrm{limit}. \\ $$$$\mathrm{hence}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence}\:\left({U}_{{n}} \right). \\ $$$$\:\:{Please}\:{recommend}\:{me}\:{textbooks}\:{for}\:{this}\:{topic}\:{even}\:{youtube}\:{vids} \\ $$$${please} \\ $$$$ \\ $$

Question Number 87069 Answers: 0 Comments: 1

((cos x−sin x)/(√(1+sin 2x))) = sec 2x−tan 2x prove it

$$\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{2x}}}\:=\:\mathrm{sec}\:\mathrm{2x}−\mathrm{tan}\:\mathrm{2x} \\ $$$$\mathrm{prove}\:\mathrm{it}\: \\ $$

Question Number 87031 Answers: 1 Comments: 1

Question Number 86886 Answers: 0 Comments: 1

Question Number 86873 Answers: 0 Comments: 0

A company paid a total dividend of K12 600.00 at the end of 2018 on 6000 shares. If Freddy owned 200 shares in the company, how much was paid out in dividents to him?

$${A}\:{company}\:{paid}\:{a}\:{total}\:{dividend}\:{of}\:\boldsymbol{\mathrm{K}}\mathrm{12}\:\mathrm{600}.\mathrm{00}\:{at}\:{the}\:{end}\:{of}\:\mathrm{2018}\:{on}\:\mathrm{6000}\:{shares}.\:{If}\:{Freddy}\:{owned}\:\mathrm{200}\:{shares}\:{in}\:{the}\:{company},\:{how}\:{much}\:{was}\:{paid}\:{out}\:{in}\:{dividents}\:{to}\:{him}? \\ $$

Question Number 87017 Answers: 1 Comments: 0

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Question Number 86849 Answers: 0 Comments: 0

If z,w ε C and ∣z∣>1, ∣w∣<1 so ∣((z−w)/(1−z^− w))∣>1, demostrate thr veracity of the statment. (V or F)

$${If}\:\:{z},{w}\:\epsilon\:\mathbb{C}\:{and}\:\mid{z}\mid>\mathrm{1},\:\mid{w}\mid<\mathrm{1} \\ $$$${so}\:\mid\frac{{z}−{w}}{\mathrm{1}−\overset{−} {{z}w}}\mid>\mathrm{1},\:{demostrate} \\ $$$${thr}\:{veracity}\:{of}\:{the} \\ $$$$\:{statment}.\:\left({V}\:{or}\:{F}\right) \\ $$

Question Number 86708 Answers: 1 Comments: 0

∫x (√((√2) x−(√(2x^2 −1)))) dx

$$\int\mathrm{x}\:\sqrt{\sqrt{\mathrm{2}}\:\mathrm{x}−\sqrt{\mathrm{2x}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$

Question Number 86598 Answers: 0 Comments: 1

write out the general summation formula for the maclaurin series expansion for (1/2) (cos x + cosh x)

$$\:\mathrm{write}\:\mathrm{out}\:\mathrm{the}\:\mathrm{general}\:\mathrm{summation}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\:\mathrm{the}\:\mathrm{maclaurin}\:\mathrm{series}\:\mathrm{expansion}\:\mathrm{for}\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{cos}\:{x}\:+\:\mathrm{cosh}\:{x}\right) \\ $$

Question Number 86461 Answers: 3 Comments: 0

Use exponential representation of sin θ and cos θ to show that a) sin^2 θ + cos^2 θ = 1 b) cos^2 θ − sin^2 θ = cos2θ c) 2 sinθ cosθ = 2sin2θ.

$$\mathrm{Use}\:\mathrm{exponential}\:\mathrm{representation}\:\mathrm{of}\:\mathrm{sin}\:\theta\:\mathrm{and}\:\mathrm{cos}\:\theta\:\mathrm{to}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left.\mathrm{a}\left.\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:\mathrm{cos}^{\mathrm{2}} \:\theta\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\right)\:\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{sin}^{\mathrm{2}} \theta\:=\:\mathrm{cos2}\theta \\ $$$$\left.\mathrm{c}\right)\:\mathrm{2}\:\mathrm{sin}\theta\:\mathrm{cos}\theta\:=\:\mathrm{2sin2}\theta. \\ $$

Question Number 86365 Answers: 0 Comments: 0

I think it will be ∫_0 ^(π/4) (dx/(√(1+tanx))) ≈∫_0 ^(π/4) (dx/(√(1+x))) =(𝛑/4)−(1/2).(1/2).((𝛑/4))^2 +((1.3)/(2.4)).(1/3).((𝛑/4))^3 −((1.3.5)/(2.4.6)).(1/4)((𝛑/4))^4 +....

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{tanx}}}\:\approx\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{x}}}\: \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{3}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{3}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}.\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{4}} +.... \\ $$

Question Number 86356 Answers: 1 Comments: 1

Question Number 86198 Answers: 0 Comments: 7

any methods to sketch these curves r = a(1−cosθ) r= a + b cosθ a>b r= a + bcosθ a<b

$$\mathrm{any}\:\mathrm{methods}\:\mathrm{to}\:\mathrm{sketch}\:\mathrm{these}\:\mathrm{curves} \\ $$$$\:\mathrm{r}\:=\:\mathrm{a}\left(\mathrm{1}−\mathrm{cos}\theta\right) \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{b}\:\mathrm{cos}\theta\:\:{a}>{b} \\ $$$$\:\mathrm{r}=\:\mathrm{a}\:+\:\mathrm{bcos}\theta\:\:\:{a}<{b} \\ $$

Question Number 86140 Answers: 0 Comments: 1

Question Number 86132 Answers: 1 Comments: 0

∫x^3 sin(2x^2 +6)^5 dx

$$\int{x}^{\mathrm{3}} \:{sin}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}\right)^{\mathrm{5}} \:{dx} \\ $$

Question Number 86128 Answers: 1 Comments: 0

⌊2x−(1/2)⌋=⌊∣x∣−(1/2)⌋=2x−2

$$\lfloor\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\lfloor\mid{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\rfloor=\mathrm{2}{x}−\mathrm{2} \\ $$

Question Number 86116 Answers: 0 Comments: 1

lim_(x→∞) ((tanx)/x)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{tanx}}{{x}} \\ $$

Question Number 86062 Answers: 2 Comments: 1

∫((√(x^2 −25))/x)dx

$$\int\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{25}}}{{x}}{dx} \\ $$

Question Number 85982 Answers: 0 Comments: 2

A primitive of the function defned by f(x) = x −1 + (1/(x+1)) is A. F(x) = (x^2 /2) −x + ln(x + 1) B. F(x) = (x^2 /2) + ln(x−1) C. F(x) = (x^2 /2)−x + ln(1−x) D. F(x) = −x + ln(x−1)

$$\mathrm{A}\:\mathrm{primitive}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{defned}\:\mathrm{by}\:\mathrm{f}\left({x}\right)\:=\:{x}\:−\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:−{x}\:+\:\mathrm{ln}\left({x}\:+\:\mathrm{1}\right)\:\:\:\:\mathrm{B}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\:\mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{C}.\:\mathrm{F}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\:+\:\mathrm{ln}\left(\mathrm{1}−{x}\right)\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{F}\left({x}\right)\:=\:−{x}\:+\:\mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 85888 Answers: 0 Comments: 5

i−∫_0 ^5 (x+[2x])^([(x/3)]) dx ii−∫_0 ^3 (z−{z})^([z]) dz

$${i}−\int_{\mathrm{0}} ^{\mathrm{5}} \left({x}+\left[\mathrm{2}{x}\right]\right)^{\left[\frac{{x}}{\mathrm{3}}\right]} {dx} \\ $$$$ \\ $$$${ii}−\int_{\mathrm{0}} ^{\mathrm{3}} \left({z}−\left\{{z}\right\}\right)^{\left[{z}\right]} \:{dz} \\ $$

Question Number 85872 Answers: 1 Comments: 1

∫cos^(2020) x dx = ?

$$\int\mathrm{cos}^{\mathrm{2020}} \mathrm{x}\:\mathrm{dx}\:=\:? \\ $$

Question Number 85858 Answers: 0 Comments: 0

Is a matrix A^T A always positive definite?

$$\mathrm{Is}\:\mathrm{a}\:\mathrm{matrix} \\ $$$$\mathrm{A}^{\mathrm{T}} \mathrm{A}\:\mathrm{always}\:\mathrm{positive}\:\mathrm{definite}? \\ $$

Pg 62 Pg 63 Pg 64 Pg 65 Pg 66 Pg 67 Pg 68 Pg 69 Pg 70 Pg 71

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