Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1039

Question Number 115683    Answers: 1   Comments: 0

Question Number 115656    Answers: 3   Comments: 1

∫ (√((x−a)/(b−x))) dx = ? where a <x < b

$$\int\:\sqrt{\frac{{x}−{a}}{{b}−{x}}}\:{dx}\:=\:? \\ $$$${where}\:{a}\:<{x}\:<\:{b} \\ $$

Question Number 115653    Answers: 0   Comments: 1

Question Number 115646    Answers: 1   Comments: 3

Question Number 115645    Answers: 1   Comments: 2

Question Number 115643    Answers: 0   Comments: 0

Question Number 115642    Answers: 0   Comments: 0

A vertical post of height h m rises from a plane which slopes down towards the South at an angle α to the horizontal. Prove that the length of its shadow when the sun is S𝛉W at an elevation β is ((h(√((1+tan^2 α cos^2 θ) )))/(tanβ + tanα cos θ))m

$${A}\:{vertical}\:{post}\:{of}\:{height}\:{h}\:{m}\:{rises}\:{from}\:{a}\:{plane}\:{which}\: \\ $$$${slopes}\:{down}\:{towards}\:{the}\:{South}\:{at}\:{an}\:{angle} \\ $$$$\alpha\:{to}\:{the}\:{horizontal}.\:{Prove}\:{that}\:{the}\:{length} \\ $$$${of}\:{its}\:{shadow}\:{when}\:{the}\:{sun}\:{is}\:\boldsymbol{{S}\theta{W}}\:\: \\ $$$${at}\:{an}\:{elevation}\:\beta\:{is} \\ $$$$ \\ $$$$\frac{{h}\sqrt{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}} \theta\right)\:}}{{tan}\beta\:+\:{tan}\alpha\:\mathrm{cos}\:\theta}{m} \\ $$

Question Number 115632    Answers: 1   Comments: 6

Question Number 115629    Answers: 0   Comments: 2

Question Number 115628    Answers: 0   Comments: 0

Question Number 115627    Answers: 0   Comments: 1

Question Number 115621    Answers: 2   Comments: 0

... advanced calculus... evaluate :: show that lim_(n→∞) (1/n)[cos^(2p) (π/(2n))+cos^(2p) ((2π)/(2n))+cos^(2p) ((3π)/(2n))......cos^(2p) (π/2)] =Π_(r=1) ^p ((p+r)/(4r))

$$ \\ $$$$\:\:\:\:\:\:\:...\:{advanced}\:\:\:{calculus}...\: \\ $$$$\:\:\:\:\:\:\:{evaluate}\::: \\ $$$${show}\:{that}\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left[{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{2}\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{3}\pi}{\mathrm{2}{n}}......{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}}\right]\:=\underset{{r}=\mathrm{1}} {\overset{{p}} {\prod}}\frac{{p}+{r}}{\mathrm{4}{r}} \\ $$

Question Number 115616    Answers: 2   Comments: 0

lim_(x→0) ((((256+tan x))^(1/(8 )) −((1+sin x))^(1/(3 )) −1)/( ((1+tan x))^(1/(6 )) −1)) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{8}\:}]{\mathrm{256}+\mathrm{tan}\:\:{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:? \\ $$

Question Number 115604    Answers: 6   Comments: 1

(1)lim_(x→(π/2)) ((cos 2x)/(tan x)) =? (2) lim_(x→π) (((√(2 +cos x)) −1)/((π−x)^2 )) = ?

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

Question Number 115597    Answers: 4   Comments: 0

Question Number 115595    Answers: 1   Comments: 0

Prove that, for all primes p>3, 13∣10^(2p) −10^p +1

$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{all}\:\mathrm{primes}\:\mathrm{p}>\mathrm{3}, \\ $$$$\mathrm{13}\mid\mathrm{10}^{\mathrm{2p}} −\mathrm{10}^{\mathrm{p}} +\mathrm{1} \\ $$

Question Number 115594    Answers: 1   Comments: 0

old question, I couldn′t find it: ∫(√(x−(√x)))dx=?

$$\mathrm{old}\:\mathrm{question},\:\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}: \\ $$$$\int\sqrt{{x}−\sqrt{{x}}}{dx}=? \\ $$

Question Number 115592    Answers: 0   Comments: 1

Question Number 115625    Answers: 0   Comments: 1

Question Number 115575    Answers: 3   Comments: 2

lim_(n→∞) Π_(k=1) ^n (1−(1/(k+1)))=?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)=? \\ $$

Question Number 115564    Answers: 1   Comments: 0

Question Number 115558    Answers: 2   Comments: 2

... advanced calculus... evaluate :: ∫_0 ^( ∞) ln(1+ax^2 )ln(1+(b/x^2 ))dx m.n.july

$$\:\:\:\:\:\:\:...\:{advanced}\:\:\:{calculus}...\: \\ $$$$\:\:\:\:\:\:\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} {ln}\left(\mathrm{1}+{ax}^{\mathrm{2}} \right){ln}\left(\mathrm{1}+\frac{{b}}{{x}^{\mathrm{2}} }\right){dx} \\ $$$$\:\:\:\:\:\:\:{m}.{n}.{july} \\ $$$$ \\ $$

Question Number 115555    Answers: 3   Comments: 1

Given that x,y∈R ∀ x^2 −y^2 =32, (x+y)^4 +(x−y)^4 =4352, Find the value of x^2 +y^2 .

$$\mathrm{Given}\:\mathrm{that}\:{x},{y}\in\mathbb{R}\:\forall\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{32}, \\ $$$$\left({x}+{y}\right)^{\mathrm{4}} +\left({x}−{y}\right)^{\mathrm{4}} =\mathrm{4352},\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} . \\ $$

Question Number 115551    Answers: 2   Comments: 0

lim_(x→2) (((x^4 −4x^3 +5x^2 −4x+4))^(1/(4 )) /( (√(x^2 −3x+2)))) = ?

$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}}\:=\:? \\ $$

Question Number 115544    Answers: 1   Comments: 1

Given x^2 +12(√x) = 5 then x+2(√x) ?

$${Given}\:{x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}\:=\:\mathrm{5} \\ $$$${then}\:{x}+\mathrm{2}\sqrt{{x}}\:? \\ $$

Question Number 115541    Answers: 3   Comments: 0

lim_(x→a) (2−(x/a))^(tan (((πx)/(2a)))) =?

$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\left(\mathrm{2}−\frac{{x}}{{a}}\right)^{\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}{a}}\right)} =? \\ $$

  Pg 1034      Pg 1035      Pg 1036      Pg 1037      Pg 1038      Pg 1039      Pg 1040      Pg 1041      Pg 1042      Pg 1043   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com