Question and Answers Forum

All Questions   Topic List

OthersQuestion and Answers: Page 70

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

$$ \\ $$

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?

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

Question Number 85698    Answers: 0   Comments: 0

please any recommendation of a youtube video on General conics??

$$\mathrm{please}\:\mathrm{any}\:\mathrm{recommendation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{youtube}\:\mathrm{video} \\ $$$$\mathrm{on}\:\mathrm{General}\:\mathrm{conics}?? \\ $$

Question Number 85676    Answers: 0   Comments: 15

∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)

$$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{sin}\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{t}\right)^{\mathrm{4}} } \\ $$$$=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\mathrm{4cos}^{\mathrm{4}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)\:{dt} \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}=\:{u}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} {u}\:×\mathrm{2}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\:}}\left(\mathrm{tan}\:^{\mathrm{2}} {u}+\mathrm{1}\right)\:{d}\left(\mathrm{tan}\:{u}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} {u}\:+\:\mathrm{tan}\:{u}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{0}−\left(−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\right]=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Question Number 85580    Answers: 0   Comments: 0

Solve: (D^2 +2D+1)y= x cos x

$$\boldsymbol{\mathrm{Solve}}: \\ $$$$\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{2D}+\mathrm{1}\right)\mathrm{y}=\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x} \\ $$$$ \\ $$

Question Number 85354    Answers: 0   Comments: 0

∫x(sin(cos(x)))^(x−1) dx

$$\int{x}\left({sin}\left({cos}\left({x}\right)\right)\right)^{{x}−\mathrm{1}} \:\:{dx} \\ $$

Question Number 85286    Answers: 0   Comments: 2

  Pg 65      Pg 66      Pg 67      Pg 68      Pg 69      Pg 70      Pg 71      Pg 72      Pg 73      Pg 74   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com