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Question Number 115664 Answers: 2 Comments: 0

If lim_(x→p) ((p^x −x^p )/(x^x −p^p )) = 1 then p = ?

$${If}\:\underset{{x}\rightarrow{p}} {\mathrm{lim}}\:\frac{{p}^{{x}} −{x}^{{p}} }{{x}^{{x}} −{p}^{{p}} }\:=\:\mathrm{1}\:{then}\:{p}\:=\:? \\ $$

Question Number 115661 Answers: 2 Comments: 0

(1) ∫ sin ((√x)) dx (2) ∫ cos ((√x) ) dx (3) ∫ tan ((√x) ) dx

$$\left(\mathrm{1}\right)\:\int\:\mathrm{sin}\:\left(\sqrt{{x}}\right)\:{dx} \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{cos}\:\left(\sqrt{{x}}\:\right)\:{dx}\: \\ $$$$\left(\mathrm{3}\right)\:\int\:\mathrm{tan}\:\left(\sqrt{{x}}\:\right)\:{dx}\: \\ $$

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