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Question Number 34169    Answers: 1   Comments: 0

if the sum S_4 of the first 4 terms of a G P is 24 and the sum S_6 of the first 6 terms is 30 find the first term and common ratio..

$$\:{if}\:{the}\:{sum}\:{S}_{\mathrm{4}} \:{of}\:{the}\:{first}\:\mathrm{4}\:{terms} \\ $$$${of}\:{a}\:{G}\:{P}\:{is}\:\mathrm{24}\:{and}\:{the}\:{sum}\:{S}_{\mathrm{6}} \:{of}\: \\ $$$${the}\:{first}\:\mathrm{6}\:{terms}\:{is}\:\mathrm{30}\:{find}\:{the}\: \\ $$$${first}\:{term}\:{and}\:{common}\:{ratio}.. \\ $$

Question Number 34160    Answers: 1   Comments: 3

prove that lim_(x→0^+ ) ln x∙ln (1+x)=lim_(x→1) ln x∙ln (1+x)

$${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$

Question Number 34159    Answers: 1   Comments: 1

let S=1+2+...+2018 compute S(mod 2)+S(mod8)+S(mod2018)

$${let}\:{S}=\mathrm{1}+\mathrm{2}+...+\mathrm{2018} \\ $$$${compute}\:{S}\left(\mathrm{mod}\:\mathrm{2}\right)+{S}\left(\mathrm{mod8}\right)+{S}\left(\mathrm{mod2018}\right) \\ $$

Question Number 34142    Answers: 0   Comments: 1

what is the derivative of (x+3)(x^2 + 5) and find the n sequence of Σ_(r=n+1 ) ^(2n) (4r^3 −3)

$${what}\:{is}\:{the}\:{derivative}\:{of}\: \\ $$$$\:\:\:\:\:\left(\boldsymbol{{x}}+\mathrm{3}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\:\mathrm{5}\right) \\ $$$${and}\:{find}\:{the}\:{n}\:{sequence}\:{of}\: \\ $$$$\:\:\:\underset{{r}={n}+\mathrm{1}\:} {\overset{\mathrm{2}{n}} {\sum}}\:\:\left(\mathrm{4}{r}^{\mathrm{3}} −\mathrm{3}\right) \\ $$

Question Number 34140    Answers: 2   Comments: 0

what is the remainder when 767^(1009) is divided by 25

$${what}\:{is}\:{the}\:{remainder}\:{when} \\ $$$$\mathrm{767}^{\mathrm{1009}} \:{is}\:{divided}\:{by}\:\mathrm{25} \\ $$

Question Number 34139    Answers: 2   Comments: 0

What is the remainder when 17^(200) is divided by 18

$${What}\:{is}\:{the}\:{remainder}\:{when} \\ $$$$\mathrm{17}^{\mathrm{200}} \:{is}\:{divided}\:{by}\:\mathrm{18} \\ $$

Question Number 34129    Answers: 2   Comments: 2

find the value of ∫_0 ^1 ln(x)ln(1+x)dx .

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}+{x}\right){dx}\:. \\ $$

Question Number 34126    Answers: 0   Comments: 0

let give n natural integr not o calculate A_n = ∫_0 ^∞ (dx/(Π_(k=1) ^n (x^2 +k))) .

$${let}\:{give}\:{n}\:{natural}\:{integr}\:{not}\:{o} \\ $$$${calculate}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\prod_{{k}=\mathrm{1}} ^{{n}} \left({x}^{\mathrm{2}} \:+{k}\right)}\:. \\ $$

Question Number 34119    Answers: 3   Comments: 0

if u and v are real valued functions of x, then (d/dx)((u/v))= A. ((v(du/dx)− u(dv/dx))/u^2 ) B. ((v(du/dx)− u(dv/dx))/u^2 ) C. ((v(du/dx)−u (dv/dx))/v^2 ) D. ((u(dv/dx) − v(du/dx))/v^2 )

$${if}\:{u}\:{and}\:{v}\:{are}\:{real}\:{valued}\:{functions} \\ $$$${of}\:{x},\:{then}\:\:\frac{{d}}{{dx}}\left(\frac{{u}}{{v}}\right)= \\ $$$${A}.\:\frac{{v}\frac{{du}}{{dx}}−\:{u}\frac{{dv}}{{dx}}}{{u}^{\mathrm{2}} }\:\:\:\:\:\:\:\:{B}.\:\:\frac{{v}\frac{{du}}{{dx}}−\:{u}\frac{{dv}}{{dx}}}{{u}^{\mathrm{2}} } \\ $$$${C}.\:\frac{{v}\frac{{du}}{{dx}}−{u}\:\frac{{dv}}{{dx}}}{{v}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:{D}.\:\frac{{u}\frac{{dv}}{{dx}}\:−\:{v}\frac{{du}}{{dx}}}{{v}^{\mathrm{2}} } \\ $$

Question Number 34117    Answers: 0   Comments: 1

Question Number 34116    Answers: 0   Comments: 1

Question Number 34110    Answers: 0   Comments: 2

Question Number 34109    Answers: 0   Comments: 0

Question Number 34107    Answers: 0   Comments: 6

Question Number 34114    Answers: 0   Comments: 0

Question Number 34113    Answers: 0   Comments: 0

Question Number 34115    Answers: 0   Comments: 1

Question Number 34095    Answers: 0   Comments: 0

Question Number 34094    Answers: 0   Comments: 2

Question Number 34092    Answers: 2   Comments: 9

l_n i_ m_∞ ((((n!))/((nm)^n )))^(1/n)

$$\underset{{n}} {{l}}\underset{} {{i}}\underset{\infty} {{m}}\:\left(\frac{\left({n}!\right)}{\left({nm}\right)^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$

Question Number 34098    Answers: 0   Comments: 13

Since the beginning several persons with their experties had left the forum... Newcommers don′t know them,but they can explore the past of the forum and recognise their precious work! Think the forum as cricket. Some were ′bowlers′ and others were ′batsmen′. Bowlers: questioners and batsmen: answerers. In my period(When I was somewhat more active) some good ′bowlers′ were sanusihammed,tawakalitu,tawa.Some good ′batsmen′ were prakash jain, Yozzi. 123556 was ′all rounder′.An other energetic ′all-rounder player′ was filups. ′Bowlers′ also play vital role... ... Speaking of present period. Recently our ′Captain′ has left the forum.An expert,an honest and a hardworker ′player′ and leader. Also a good ′bowler′ of the present period has left us. I request mrW and Mr tinkutara to come back please! The forum needs you! Also request Ajeet bhaya(a lovely name of Mr Ajfour) to become more active.He is, like mrW, an important figure. Request by a 12th player.

$$\mathrm{Since}\:\mathrm{the}\:\mathrm{beginning}\: \\ $$$$\mathrm{several}\:\mathrm{persons}\:\mathrm{with}\:\mathrm{their}\:\mathrm{experties} \\ $$$$\mathrm{had}\:\mathrm{left}\:\mathrm{the}\:\mathrm{forum}... \\ $$$$\mathrm{Newcommers}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{them},\mathrm{but} \\ $$$$\mathrm{they}\:\mathrm{can}\:\mathrm{explore}\:\mathrm{the}\:\mathrm{past}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forum} \\ $$$$\mathrm{and}\:\mathrm{recognise}\:\mathrm{their}\:\mathrm{precious}\:\mathrm{work}! \\ $$$$\mathrm{Think}\:\mathrm{the}\:\mathrm{forum}\:\mathrm{as}\:\mathrm{cricket}. \\ $$$$\mathrm{Some}\:\mathrm{were}\:'\mathrm{bowlers}'\:\mathrm{and}\:\mathrm{others}\:\mathrm{were} \\ $$$$'\mathrm{batsmen}'.\:\mathrm{Bowlers}:\:\mathrm{questioners}\:\:\mathrm{and} \\ $$$$\mathrm{batsmen}:\:\mathrm{answerers}. \\ $$$$\mathrm{In}\:\mathrm{my}\:\mathrm{period}\left(\mathrm{When}\:\mathrm{I}\:\mathrm{was}\:\mathrm{somewhat}\right. \\ $$$$\left.\mathrm{more}\:\mathrm{active}\right)\:\mathrm{some}\:\mathrm{good}\:'\mathrm{bowlers}'\:\mathrm{were} \\ $$$$\boldsymbol{\mathrm{sanusihammed}},\boldsymbol{\mathrm{tawakalitu}},\boldsymbol{\mathrm{tawa}}.\mathrm{Some} \\ $$$$\mathrm{good}\:'\mathrm{batsmen}'\:\mathrm{were}\:\boldsymbol{\mathrm{prakash}}\:\boldsymbol{\mathrm{jain}}, \\ $$$$\boldsymbol{\mathrm{Yozzi}}.\:\mathrm{123556}\:\mathrm{was}\:'\mathrm{all}\:\mathrm{rounder}'.\mathrm{An}\:\mathrm{other} \\ $$$$\mathrm{energetic}\:\:'\mathrm{all}-\mathrm{rounder}\:\mathrm{player}'\:\mathrm{was}\:\boldsymbol{\mathrm{filups}}. \\ $$$$'\mathrm{Bowlers}'\:\mathrm{also}\:\mathrm{play}\:\mathrm{vital}\:\mathrm{role}... \\ $$$$... \\ $$$$\mathrm{Speaking}\:\mathrm{of}\:\mathrm{present}\:\mathrm{period}. \\ $$$$\mathrm{Recently}\:\mathrm{our}\:'\mathrm{Captain}'\:\mathrm{has}\:\mathrm{left}\:\mathrm{the} \\ $$$$\mathrm{forum}.\mathrm{An}\:\boldsymbol{\mathrm{expert}},\mathrm{an}\:\boldsymbol{\mathrm{honest}}\:\mathrm{and}\:\mathrm{a} \\ $$$$\boldsymbol{\mathrm{hardworker}}\:\:'\mathrm{player}'\:\mathrm{and}\:\mathrm{leader}.\: \\ $$$$\mathrm{Also}\:\mathrm{a}\:\mathrm{good}\:'\mathrm{bowler}'\:\mathrm{of}\:\mathrm{the}\:\mathrm{present} \\ $$$$\mathrm{period}\:\mathrm{has}\:\mathrm{left}\:\mathrm{us}. \\ $$$$\mathrm{I}\:\mathrm{request}\:\boldsymbol{\mathrm{mrW}}\:\mathrm{and}\:\boldsymbol{\mathrm{Mr}}\:\boldsymbol{\mathrm{tinkutara}} \\ $$$$\mathrm{to}\:\mathrm{come}\:\mathrm{back}\:\mathrm{please}!\:\mathrm{The}\:\mathrm{forum} \\ $$$$\boldsymbol{\mathrm{needs}}\:\mathrm{you}! \\ $$$$\mathrm{Also}\:\mathrm{request}\:{Ajeet}\:{bhaya}\left(\mathrm{a}\:\mathrm{lovely}\right. \\ $$$$\left.\mathrm{name}\:\mathrm{of}\:\boldsymbol{\mathrm{Mr}}\:\boldsymbol{\mathrm{Ajfour}}\right)\:\mathrm{to}\:\mathrm{become}\:\mathrm{more} \\ $$$$\mathrm{active}.\mathrm{He}\:\mathrm{is},\:\mathrm{like}\:\mathrm{mrW},\:\mathrm{an}\:\mathrm{important} \\ $$$$\mathrm{figure}. \\ $$$$\mathrm{Request}\:\mathrm{by}\:\mathrm{a}\:\mathrm{12th}\:\mathrm{player}. \\ $$

Question Number 34081    Answers: 1   Comments: 0

2^n −2^(n−1) =4 .find n^(n.)

$$\mathrm{2}^{{n}} −\mathrm{2}^{{n}−\mathrm{1}} =\mathrm{4}\:.{find}\:{n}^{{n}.} \\ $$

Question Number 34066    Answers: 1   Comments: 0

show that− sin 10−(√3)sec10=4.

$$\mathrm{show}\:\mathrm{that}− \\ $$$$\mathrm{sin}\:\mathrm{10}−\sqrt{\mathrm{3}}\mathrm{sec10}=\mathrm{4}. \\ $$

Question Number 34064    Answers: 1   Comments: 0

Question Number 34056    Answers: 2   Comments: 3

x^(3z) =1 x^2 =y z=y^n FIND THE VALUE OF n please i need your help ASAP. thanks

$${x}^{\mathrm{3}{z}} =\mathrm{1}\: \\ $$$${x}^{\mathrm{2}} ={y} \\ $$$${z}={y}^{{n}} \\ $$$${FIND}\:{THE}\:{VALUE}\:{OF}\:{n} \\ $$$${please}\:{i}\:{need}\:{your}\:{help}\:{ASAP}.\:{thanks} \\ $$

Question Number 34051    Answers: 1   Comments: 0

2dy/dx+y=0 y(0)=−3

$$\mathrm{2}{dy}/{dx}+{y}=\mathrm{0}\:\:{y}\left(\mathrm{0}\right)=−\mathrm{3} \\ $$

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