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Question Number 213548 Answers: 1 Comments: 0

0<c<1 such that the recursive sequence {a_n } defined by setting a_(1 ) = (c/2) , a_(n+1) = (1/2)(c+a_n ^2 ) for n∈ N monotonic and convergent

$$\mathrm{0}<{c}<\mathrm{1}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{recursive}\:\mathrm{sequence} \\ $$$$\left\{{a}_{{n}} \right\}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{setting}\: \\ $$$$\:\mathrm{a}_{\mathrm{1}\:} =\:\frac{\mathrm{c}}{\mathrm{2}}\:\:,\:{a}_{\mathrm{n}+\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{c}+\mathrm{a}_{\mathrm{n}} ^{\mathrm{2}} \right)\:\:\mathrm{for}\:\mathrm{n}\in\:\mathbb{N} \\ $$$$\mathrm{monotonic}\:\mathrm{and}\:\mathrm{convergent} \\ $$

Question Number 213530 Answers: 1 Comments: 1

Question Number 213534 Answers: 1 Comments: 1

Question Number 213522 Answers: 0 Comments: 0

The two corner points of a square lie on curve f(x)= x^2 −2x−3 and the other two corner points lie on curve g(x)= −x^2 +2x+3 . It is known that the area of a square can be expressed by p+q(√r) , for a natural number p,q ,r where r is not divisible by any perfect square number other 1. The value of p+q+r =?

$$\:\:\mathrm{The}\:\mathrm{two}\:\mathrm{corner}\:\mathrm{points}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\: \\ $$$$\:\:\mathrm{lie}\:\mathrm{on}\:\mathrm{curve}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{3}\:\mathrm{and}\: \\ $$$$\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{corner}\:\mathrm{points}\:\mathrm{lie}\:\mathrm{on}\: \\ $$$$\:\mathrm{curve}\:\mathrm{g}\left(\mathrm{x}\right)=\:−\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}\:.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known} \\ $$$$\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{can}\:\mathrm{be}\: \\ $$$$\:\:\mathrm{expressed}\:\mathrm{by}\:\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\:,\:\mathrm{for}\:\mathrm{a}\:\mathrm{natural}\: \\ $$$$\:\mathrm{number}\:\mathrm{p},\mathrm{q}\:,\mathrm{r}\:\mathrm{where}\:\mathrm{r}\:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\: \\ $$$$\:\mathrm{by}\:\mathrm{any}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{number}\:\mathrm{other}\:\mathrm{1}. \\ $$$$\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}+\mathrm{q}+\mathrm{r}\:=? \\ $$

Question Number 213518 Answers: 1 Comments: 0

∫_0 ^( 2π) ((z∙sin(z))/(1+cos^2 (z))) dz ∫_( ∣z∣=2) (1/(z^2 +1)) dz ∫_( ∣z∣=2) ((sin(z))/(z^2 +1)) dz

$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{{z}\centerdot\mathrm{sin}\left({z}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({z}\right)}\:\mathrm{d}{z} \\ $$$$\int_{\:\mid{z}\mid=\mathrm{2}} \:\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\int_{\:\mid{z}\mid=\mathrm{2}} \:\frac{\mathrm{sin}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$

Question Number 213511 Answers: 4 Comments: 0

lim_(n→∞) [Σ_(r=1) ^n (1/2^r )] where [•] greatest integer finction

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\left[\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{r}} }\right] \\ $$$$\:\:\:\mathrm{where}\:\left[\bullet\right]\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{finction} \\ $$

Question Number 213564 Answers: 1 Comments: 6

f: Z→R such that f(x).f(y)=f(x+y)+f(x−y) ⇒f(x)=¿

$${f}:\:{Z}\rightarrow{R}\:{such}\:{that} \\ $$$${f}\left({x}\right).{f}\left({y}\right)={f}\left({x}+{y}\right)+{f}\left({x}−{y}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=¿ \\ $$

Question Number 213504 Answers: 1 Comments: 0

Question Number 213503 Answers: 1 Comments: 0

Question Number 213492 Answers: 1 Comments: 2

Question Number 213520 Answers: 1 Comments: 0

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Question Number 213519 Answers: 2 Comments: 0

Question Number 213486 Answers: 1 Comments: 0

x^5 +5x−(6/x)=0 x?

$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{x}}−\frac{\mathrm{6}}{\boldsymbol{\mathrm{x}}}=\mathrm{0}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}? \\ $$

Question Number 213482 Answers: 1 Comments: 0

Question Number 213485 Answers: 2 Comments: 0

prove that (1/2^2 )+(1/3^2 )+...+(1/(2021^2 ))<((25)/(36))

$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+...+\frac{\mathrm{1}}{\mathrm{2021}^{\mathrm{2}} }<\frac{\mathrm{25}}{\mathrm{36}} \\ $$

Question Number 213468 Answers: 1 Comments: 0

show that the sequence {a_n } defined recurssively by a_1 = (3/2) a_(n ) = (√(3a_(n−1 ) −2 )) for n≥2 converges and find its limit.

$$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{defined}\: \\ $$$$\mathrm{recurssively}\:\mathrm{by}\:\mathrm{a}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}\:} =\:\sqrt{\mathrm{3a}_{\mathrm{n}−\mathrm{1}\:} \:−\mathrm{2}\:}\:\:\:\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{converges}\: \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{limit}. \\ $$

Question Number 213463 Answers: 1 Comments: 0

For p,q and r prime numbers satisfying { ((p(q+1)(r+1)=1064)),((r(p+1)(q+1)=1554)) :} find the value p(q+1)r

$$\:\:\mathrm{For}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{prime}\:\mathrm{numbers}\: \\ $$$$\:\:\mathrm{satisfying}\:\begin{cases}{\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\left(\mathrm{r}+\mathrm{1}\right)=\mathrm{1064}}\\{\mathrm{r}\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{1554}}\end{cases} \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{p}\left(\mathrm{q}+\mathrm{1}\right)\mathrm{r}\: \\ $$

Question Number 213467 Answers: 1 Comments: 3

$$\:\:\:\underbrace{\boldsymbol{{B}}} \\ $$

Question Number 213459 Answers: 0 Comments: 0

Find tupple natural numbers (a,b,c) such that { ((max{((a+b)/2)+((∣a−b∣)/2) , ((b+c)/2)+((∣b−c∣)/2) ,((c+a)/2)+((∣c−a∣)/2)}=a)),((min{((a+b)/2)−((∣a−b∣)/2) , ((b+c)/2)−((∣b−c∣)/2) , ((c+a)/2)−((∣c−a∣)/2)}=b)) :} where a+b+c = 10

$$\:\:\mathrm{Find}\:\mathrm{tupple}\:\mathrm{natural}\:\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{max}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}+\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}+\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}+\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{a}}\\{\mathrm{min}\left\{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}−\frac{\mid\mathrm{a}−\mathrm{b}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}−\frac{\mid\mathrm{b}−\mathrm{c}\mid}{\mathrm{2}}\:,\:\frac{\mathrm{c}+\mathrm{a}}{\mathrm{2}}−\frac{\mid\mathrm{c}−\mathrm{a}\mid}{\mathrm{2}}\right\}=\mathrm{b}}\end{cases} \\ $$$$\:\:\mathrm{where}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{10} \\ $$

Question Number 213451 Answers: 1 Comments: 0

∫ (1/(z^6 −1)) dz=??

$$\int\:\:\frac{\mathrm{1}}{{z}^{\mathrm{6}} −\mathrm{1}}\:\mathrm{d}{z}=?? \\ $$

Question Number 213439 Answers: 1 Comments: 5