Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1818

Question Number 25973    Answers: 1   Comments: 0

((k.2^k +2.2^k +2k+4)/2)=(k+3)2^k

$$\frac{\mathrm{k}.\mathrm{2}^{\mathrm{k}} +\mathrm{2}.\mathrm{2}^{\mathrm{k}} +\mathrm{2k}+\mathrm{4}}{\mathrm{2}}=\left(\mathrm{k}+\mathrm{3}\right)\mathrm{2}^{\mathrm{k}} \\ $$

Question Number 25972    Answers: 2   Comments: 0

Question Number 25971    Answers: 1   Comments: 2

In finding the equations of the bisectors of the angles between two lines a_1 x+b_1 y+c_1 =0 and a_2 x+b_2 y+c_2 =0, why we observe a_1 a_2 +b_1 b_2 >0 or <0 for obtuse and acute angle bisectors?

$${In}\:{finding}\:{the}\:{equations}\:{of}\:{the} \\ $$$${bisectors}\:{of}\:{the}\:{angles}\:{between}\:{two} \\ $$$${lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}, \\ $$$${why}\:{we}\:{observe}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} >\mathrm{0}\:{or}\:<\mathrm{0} \\ $$$${for}\:{obtuse}\:{and}\:{acute}\:{angle}\:{bisectors}? \\ $$

Question Number 25969    Answers: 0   Comments: 1

C_0 +2C_1 +3C_2 +..........+(n+1)C_n =(n+2)2^(n−1) using Bionomial teorm

$$\mathrm{C}_{\mathrm{0}} +\mathrm{2C}_{\mathrm{1}} +\mathrm{3C}_{\mathrm{2}} +..........+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{C}_{\mathrm{n}} =\left(\mathrm{n}+\mathrm{2}\right)\mathrm{2}^{\mathrm{n}−\mathrm{1}} \\ $$$$\mathrm{using}\:\:\mathrm{Bionomial}\:\mathrm{teorm} \\ $$

Question Number 25981    Answers: 1   Comments: 0

Question Number 25965    Answers: 1   Comments: 0

solve the integral ∫sin^4 θdθ

$${solve}\:{the}\:{integral} \\ $$$$\int\mathrm{sin}^{\mathrm{4}} \theta{d}\theta \\ $$

Question Number 25963    Answers: 0   Comments: 0

A bus is traveling along a straight road at 100 km/hr and the bus conductor walks at 6 km/hr on the floor of the bus and in the same direction as the bus. Find the speed of the conductor relative to the road and relative to the bus.

$$\mathrm{A}\:\mathrm{bus}\:\mathrm{is}\:\mathrm{traveling}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{at}\:\mathrm{100}\:\mathrm{km}/\mathrm{hr}\:\mathrm{and}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{conductor} \\ $$$$\mathrm{walks}\:\mathrm{at}\:\mathrm{6}\:\mathrm{km}/\mathrm{hr}\:\mathrm{on}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{the}\:\mathrm{bus}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{conductor}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{road}\:\mathrm{and}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bus}. \\ $$

Question Number 26016    Answers: 0   Comments: 0

Question Number 26013    Answers: 1   Comments: 0

Question Number 25961    Answers: 0   Comments: 0

8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?

$$\mathrm{8cos}^{\mathrm{4}} \mathrm{x}−\mathrm{8cos}^{\mathrm{2}} \mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}:\mathrm{8cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$−\mathrm{8cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sinx}=\underset{−} {+}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi+\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi−\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{5}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi+\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi−\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\mathrm{why}\:\mathrm{x}=\frac{\mathrm{k}\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}? \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{way}? \\ $$

Question Number 25960    Answers: 0   Comments: 0

answer to 25955.we introduce the parametric function F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx ∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1) let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2) and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2) and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem ∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1) =π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2) −>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx and by the changement x^(1/2) =u we find F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )

$${answer}\:{to}\:\mathrm{25955}.{we}\:{introduce}\:{the}\:{parametric}\:{function} \\ $$$${F}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}_{} } \right){t}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{after}\:{verifying}\:{that}\: \\ $$$${F}\:{is}\:{derivable}\:{on}\left[\mathrm{0}.\propto\left[\:\:{we}\:{find}\:\:\:\partial{F}/\partial{t}=\:\:\int_{\mathrm{0}} ^{\infty} \left(\:\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){t}\right)^{−\mathrm{1}} {dx}\right.\right.\right. \\ $$$$\partial{F}/\partial{t}=\mathrm{1}/\mathrm{2}\:\int_{{R}} \left({tx}^{\mathrm{2}} +{t}+\mathrm{1}\right)^{−\mathrm{1}} {dx}\:\:\:{we}\:{put}\:\:{f}\left({z}\right)\:=\left({tz}^{\mathrm{2}} +{z}+\mathrm{1}\right)^{−\mathrm{1}} \\ $$$${let}\:{find}\:{the}\:{poles}\:{of}\:{f}..{tz}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0}\:<−>\:\:{z}=+−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:{the}\:{poles}\:{are}\:\:{z}_{\mathrm{0}} ={i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \:\:{and}\:\:\:{z}_{\mathrm{1}} =−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:\:{f}\left({t}\right)\:\:=\left({t}\left({t}−{z}_{\mathrm{0}} \right)\left({t}−{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \:\:\:{by}\:{residus}\:{theorem} \\ $$$$ \\ $$$$\int_{{R}} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{R}\left({f}.{z}_{\mathrm{0}} \right)\:\:=\mathrm{2}{i}\pi\:\left({t}\left({z}_{\mathrm{0}} −{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \\ $$$$=\pi\:{t}^{−\mathrm{1}/\mathrm{2}} \left({t}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}\:} \:\:\:\:−>\partial{F}/\partial{t}\:=\pi\:\mathrm{2}^{−\mathrm{1}} \:{t}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}/\mathrm{2}} \\ $$$$−>{F}\left({t}\right)\:\:=\pi\:\mathrm{2}^{−\mathrm{1}\:} \int_{\mathrm{0}} ^{{t}} \:\:\:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx}\:+\alpha \\ $$$$\alpha={F}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:\:{F}\left({t}\right)\:=\pi\mathrm{2}^{−\mathrm{1}} \int_{\mathrm{0}} ^{{t}} \:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$${and}\:{by}\:{the}\:{changement}\:\:{x}^{\mathrm{1}/\mathrm{2}} ={u}\:\:\:{we}\:{find} \\ $$$$ \\ $$$${F}\left({t}\right)\:=\:\pi\:{ln}\left(\:{t}^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{1}+{t}\right)^{\mathrm{1}/\mathrm{2}} \right)\:{so}\:\:\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}={F}\left(\mathrm{1}\right)=\pi{ln}\left(\mathrm{1}+\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$

Question Number 25962    Answers: 0   Comments: 0

find the value of Σ_(n=1) ^(n=∝) 1/_(n^2 (n+1)) we give Σ_(n=1) ^(n=∝) 1/_n 2= π^2 /6 and H_n =1+2^(−1) +3^(−1) +...+n^(−1) = ln(n) + s + θ(1/n) s is the constant number of Euler

$${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:\:\:\mathrm{1}/_{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)} \:\:{we}\:{give}\:\:\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \mathrm{1}/_{{n}} \mathrm{2}=\:\pi^{\mathrm{2}} /\mathrm{6} \\ $$$${and}\:\:{H}_{{n}} =\mathrm{1}+\mathrm{2}^{−\mathrm{1}} +\mathrm{3}^{−\mathrm{1}} +...+{n}^{−\mathrm{1}} =\:{ln}\left({n}\right)\:+\:{s}\:+\:\theta\left(\mathrm{1}/{n}\right)\: \\ $$$${s}\:{is}\:{the}\:{constant}\:{number}\:{of}\:{Euler} \\ $$

Question Number 25955    Answers: 0   Comments: 0

find the value of ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\int_{\mathrm{0}} ^{\infty} \:\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$

Question Number 25954    Answers: 0   Comments: 0

nswer to 25929 by binome formula (1+x)^(n+m) =Σ_(k=0) ^(k=n+m) C_(n+m) ^k x^k the coefficent of x^n is C_(n+m) ^n and the coefficient of x^m is C_(n+m) ^m but we have for p<n C_n ^p = C_n ^(n−p) >>>>>C_(n+m) ^n = C_(n+m) ^m .

$${nswer}\:{to}\:\mathrm{25929}\:\:\:{by}\:{binome}\:{formula}\:\:\: \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}+{m}} \:\:=\sum_{{k}=\mathrm{0}} ^{{k}={n}+{m}} \:{C}_{{n}+{m}} ^{{k}} \:{x}^{{k}} \:\:\:\:{the}\:{coefficent}\:{of}\:{x}^{{n}} \:{is}\: \\ $$$${C}_{{n}+{m}} ^{{n}} \:{and}\:{the}\:{coefficient}\:{of}\:{x}^{{m}} \:\:{is}\:\:\:{C}_{{n}+{m}} ^{{m}} \:\:\:{but}\:{we}\:{have}\:\:{for} \\ $$$${p}<{n}\:\:\:\:{C}_{{n}} ^{{p}} \:\:\:=\:\:\:{C}_{{n}} ^{{n}−{p}} \:\:\:\:\:>>>>>{C}_{{n}+{m}} ^{{n}} \:\:=\:\:\:{C}_{{n}+{m}} ^{{m}} . \\ $$

Question Number 25949    Answers: 0   Comments: 0

Question Number 25948    Answers: 0   Comments: 1

Factorise: x^2 +y^2 −z^2 −2xy

$$\mathrm{Factorise}:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} −\mathrm{2}{xy} \\ $$

Question Number 25943    Answers: 1   Comments: 0

Question Number 26096    Answers: 0   Comments: 0

Question Number 25940    Answers: 0   Comments: 0

every extreme point is a critical point but every critical is not a extreme point justify it

$${every}\:{extreme}\:{point}\:{is}\:{a}\:{critical}\: \\ $$$${point}\:{but}\:{every}\:{critical}\:{is}\:{not}\:{a}\: \\ $$$${extreme}\:{point}\:{justify}\:{it}\: \\ $$

Question Number 25937    Answers: 1   Comments: 0

For a certain amount of work,Ade takes 6hours less than Bode.if they work together it takes them 13hours 20 minutes.How long will it take Bode alone to complete the work?

$${For}\:{a}\:{certain}\:{amount}\:{of}\:{work},{Ade}\:{takes} \\ $$$$\mathrm{6}{hours}\:{less}\:{than}\:{Bode}.{if}\:{they}\:{work}\:{together} \\ $$$${it}\:{takes}\:{them}\:\mathrm{13}{hours}\:\mathrm{20}\:{minutes}.{How} \\ $$$${long}\:{will}\:{it}\:{take}\:{Bode}\:{alone}\:{to}\:{complete} \\ $$$${the}\:{work}? \\ $$

Question Number 25934    Answers: 1   Comments: 2

Question Number 25932    Answers: 0   Comments: 0

answer to 25824 we have a^(−x^2 ) = e^(−x^2_ ln(a)) so for a>1 ln(a)=( (ln(a))^(1/2) )^2 >>>>∫_R a^(−^ x^2 ) = ∫_R e^(−(x (ln(a)^(1/2) )^2 ) dx and with the changement t=x (ln(a)^(1/2) >>>>x=t ( ln(a))^(−1/2) we have ∫_R a^(−x^2 ) dx = π^(1/2) (ln(a))^(−1/2) ...if 0<a<1 ln(a)<0 and the integrale is divergente...

$${answer}\:{to}\:\mathrm{25824}\:\:{we}\:{have}\:{a}^{−{x}^{\mathrm{2}} } \:=\:{e}^{−{x}^{\mathrm{2}_{} } {ln}\left({a}\right)} \:\:{so}\:{for}\:{a}>\mathrm{1} \\ $$$${ln}\left({a}\right)=\left(\:\left({ln}\left({a}\right)\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} >>>>\int_{{R}} {a}^{−^{} {x}^{\mathrm{2}} } \:=\:\int_{{R}} {e}^{−\left({x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} \right.} {dx} \\ $$$${and}\:{with}\:{the}\:{changement}\:\:{t}={x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \:\:>>>>{x}={t}\:\left(\:{ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right. \\ $$$$\:{we}\:{have}\:\:\int_{{R}} {a}^{−{x}^{\mathrm{2}} } {dx}\:\:=\:\pi^{\mathrm{1}/\mathrm{2}} \left({ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} ...{if}\:\mathrm{0}<{a}<\mathrm{1}\:{ln}\left({a}\right)<\mathrm{0} \\ $$$$\:{and}\:{the}\:{integrale}\:{is}\:{divergente}... \\ $$

Question Number 25930    Answers: 2   Comments: 0

A line passes through A(−3, 0) and B(0, −4). A variable line perpendicular to AB is drawn to cut x and y-axes at M and N. Find the locus of the point of intersection of the lines AN and BM.

$$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:{A}\left(−\mathrm{3},\:\mathrm{0}\right)\:\mathrm{and} \\ $$$${B}\left(\mathrm{0},\:−\mathrm{4}\right).\:\mathrm{A}\:\mathrm{variable}\:\mathrm{line}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:{AB}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{cut}\:{x}\:\mathrm{and}\:{y}-\mathrm{axes}\:\mathrm{at} \\ $$$${M}\:\mathrm{and}\:{N}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:{AN}\:\mathrm{and}\:{BM}. \\ $$

Question Number 25929    Answers: 0   Comments: 0

Show that the coefficients of x^m and x^n in (1+x)^(m+n) are equal.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{x}^{\mathrm{m}} \:\mathrm{and}\:\mathrm{x}^{\mathrm{n}} \:\mathrm{in}\:\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{m}+\mathrm{n}} \mathrm{are}\:\mathrm{equal}. \\ $$

Question Number 25928    Answers: 0   Comments: 0

Expand (1−x)^4 .Hence,find S if S=(1−x^3 )^4 −4(1−x^3 )^3 +6(1−x^3 )^2 −4(1−x^3 )^ +1.

$$\mathrm{Expand}\:\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} .\mathrm{H}\boldsymbol{\mathrm{ence}},\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{S}}\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{S}}=\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{3}} +\mathrm{6}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{} +\mathrm{1}. \\ $$

Question Number 25924    Answers: 1   Comments: 0

x+3+4=5

$${x}+\mathrm{3}+\mathrm{4}=\mathrm{5} \\ $$

  Pg 1813      Pg 1814      Pg 1815      Pg 1816      Pg 1817      Pg 1818      Pg 1819      Pg 1820      Pg 1821      Pg 1822   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com