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AllQuestion and Answers: Page 327

Question Number 176692    Answers: 1   Comments: 1

Question Number 176684    Answers: 1   Comments: 2

Question Number 176679    Answers: 2   Comments: 0

Eeasy integral.... 𝛀 = ∫_(βˆ’βˆ«_0 ^( ∞) e^( βˆ’x^( 2) ) dx) ^( ∫_0 ^( ∞) e^( βˆ’x^( 2) ) dx) sin^( 2) (t).ln^( 3) ( t + (√(1+t^( 2) )))dt βˆ’βˆ’βˆ’m.nβˆ’βˆ’βˆ’

$$ \\ $$$$\:\:\:\:\:\:\:\:{Eeasy}\:\:{integral}.... \\ $$$$\:\:\:\:\:\:\boldsymbol{\Omega}\:=\:\int_{βˆ’\int_{\mathrm{0}} ^{\:\infty} {e}^{\:βˆ’{x}^{\:\mathrm{2}} } {dx}} ^{\:\int_{\mathrm{0}} ^{\:\infty} {e}^{\:βˆ’{x}^{\:\mathrm{2}} } {dx}} {sin}^{\:\mathrm{2}} \left({t}\right).{ln}^{\:\mathrm{3}} \left(\:{t}\:+\:\sqrt{\mathrm{1}+{t}^{\:\mathrm{2}} }\right){dt}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:βˆ’βˆ’βˆ’{m}.{n}βˆ’βˆ’βˆ’ \\ $$

Question Number 176676    Answers: 3   Comments: 0

If x^3 +(1/x^3 )=1, prove that x^5 +(1/x^5 )=βˆ’(x^4 +(1/x^4 ))

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{1}, \\ $$$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} }=βˆ’\left(\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }\right) \\ $$

Question Number 176672    Answers: 1   Comments: 0

Question Number 176673    Answers: 1   Comments: 0

Question Number 176668    Answers: 3   Comments: 0

Question Number 176662    Answers: 3   Comments: 0

sequence V_(n+1) βˆ’V_n =n+3^n . Find V_n .

$$\:\:{sequence}\:{V}_{{n}+\mathrm{1}} βˆ’{V}_{{n}} ={n}+\mathrm{3}^{{n}} .\:{Find}\:{V}_{{n}} . \\ $$

Question Number 176660    Answers: 1   Comments: 1

Question Number 176652    Answers: 1   Comments: 0

f(x,y)= { ((e^(1/(r^2 βˆ’1)) if r<1, where r=βˆ₯(x,y)βˆ₯)),((0 if rβ‰₯1)) :} show that f(x,y) is continuous in R^2

$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\begin{cases}{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} βˆ’\mathrm{1}}} \:\mathrm{if}\:\mathrm{r}<\mathrm{1},\:\mathrm{where}\:\mathrm{r}=\parallel\left(\mathrm{x},\mathrm{y}\right)\parallel}\\{\mathrm{0}\:\mathrm{if}\:\mathrm{r}\geqslant\mathrm{1}}\end{cases} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2}} \\ $$

Question Number 176648    Answers: 2   Comments: 1

∫ (dx/(a+bcosx)) ∫ (dx/(aβˆ’bsinx))

$$\int\:\frac{{dx}}{{a}+{bcosx}}\:\:\:\: \\ $$$$ \\ $$$$\int\:\frac{{dx}}{{a}βˆ’{bsinx}} \\ $$

Question Number 176643    Answers: 0   Comments: 2

Question Number 176638    Answers: 1   Comments: 0

lim_(xβ†’0) ((sin^2 (x)βˆ’sin (x^2 ))/(x^2 (cos^2 (x)βˆ’cos (x^2 ))))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)βˆ’\mathrm{sin}\:\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)βˆ’\mathrm{cos}\:\left(\mathrm{x}^{\mathrm{2}} \right)\right)}=? \\ $$

Question Number 176637    Answers: 0   Comments: 0

Question Number 176636    Answers: 2   Comments: 2

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} β‡’20pqr =?

$$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} }\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} }\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\:\Rightarrow\mathrm{20}{pqr}\:=? \\ $$

Question Number 176610    Answers: 1   Comments: 0

Question Number 176607    Answers: 1   Comments: 0

Question Number 176603    Answers: 2   Comments: 6

look the anser

$${look}\:{the}\:{anser} \\ $$

Question Number 176595    Answers: 2   Comments: 0

Question Number 176594    Answers: 0   Comments: 1

𝛗=∫_0 ^( 1) (( ( tanh^( βˆ’1) (x))^2 )/((1+x )^( 2) )) dx = ? β‰Ί solution ≻ note : tanh^( βˆ’1) (x)=βˆ’ (1/2) ln(((1βˆ’x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1βˆ’x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1βˆ’x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) βˆ’2∫_0 ^( 1) ln(t)dt} =βˆ’ (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) β—‚ m.n β–Ά

$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\:{tanh}^{\:βˆ’\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:{note}\::\:\:{tanh}^{\:βˆ’\mathrm{1}} \left({x}\right)=βˆ’\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\frac{\mathrm{1}βˆ’{x}}{\mathrm{1}+{x}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\frac{\mathrm{1}βˆ’{x}}{\mathrm{1}+{x}}\:\right)}{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\overset{\frac{\mathrm{1}βˆ’{x}}{\mathrm{1}+{x}}\:=\:{t}} {=}\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left({t}\:\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}_{\:} }\:\left\{\:\left[{t}.{ln}^{\:\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} βˆ’\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=βˆ’\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\blacktriangleleft\:{m}.{n}\:\blacktriangleright\: \\ $$

Question Number 176592    Answers: 2   Comments: 0

Solve the equation: 2^x + 3^x βˆ’ 4^x + 6^x βˆ’ 9^x = 1

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:βˆ’\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:βˆ’\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{1} \\ $$

Question Number 176589    Answers: 1   Comments: 0

Question Number 176598    Answers: 1   Comments: 0

x^3 +(1/x^3 )=1 (((x^5 +(1/x^5 ))^3 βˆ’1)/(x^5 +(1/x^5 )))=? Q#176387 reposted for a new answer.

$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} βˆ’\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=? \\ $$$${Q}#\mathrm{176387}\:{reposted}\:{for}\:{a}\:{new}\:{answer}. \\ $$

Question Number 176581    Answers: 1   Comments: 0

Given { ((sin a+sin b=((√2)/2))),((cos a+cos b=((√6)/2))) :} for a,b real numbers. Evaluate sin (a+b). (A)((√3)/2) (D) βˆ’((√3)/2) (B) (2/( (√3))) (E)βˆ’(2/( (√3))) (C) ((√3)/4)

$$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{a}+\mathrm{sin}\:\mathrm{b}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{a}+\mathrm{cos}\:\mathrm{b}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Evaluate} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{a}+\mathrm{b}\right). \\ $$$$\:\left(\mathrm{A}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:βˆ’\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\left(\mathrm{E}\right)βˆ’\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\left(\mathrm{C}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$

Question Number 176580    Answers: 1   Comments: 0

4^(x^2 βˆ’2x+2) βˆ’2^(x^2 βˆ’2x+3) +2=2^(x^2 βˆ’2x+2) x=?

$$\mathrm{4}^{{x}^{\mathrm{2}} βˆ’\mathrm{2}{x}+\mathrm{2}} βˆ’\mathrm{2}^{{x}^{\mathrm{2}} βˆ’\mathrm{2}{x}+\mathrm{3}} +\mathrm{2}=\mathrm{2}^{{x}^{\mathrm{2}} βˆ’\mathrm{2}{x}+\mathrm{2}} \\ $$$${x}=? \\ $$

Question Number 176571    Answers: 1   Comments: 0

lim_(xβ†’1) ((lnx)/(1+lnxβˆ’1))=?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{lnx}}{\mathrm{1}+{lnx}βˆ’\mathrm{1}}=? \\ $$

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