Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 740

Question Number 143666    Answers: 2   Comments: 1

Question Number 143653    Answers: 1   Comments: 0

Question Number 143650    Answers: 3   Comments: 0

If the function f and g are defined on the set of real numbers,are such that gof(x)=((2x−5)/(3x+7)) and g(x)=((3x+2)/(x−5)). find an expression for f(x)

$${If}\:\:{the}\:{function}\:{f}\:{and}\:{g}\:{are}\:{defined} \\ $$$${on}\:{the}\:{set}\:{of}\:{real}\:{numbers},{are}\:{such} \\ $$$${that}\:\boldsymbol{{gof}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{5}}{\mathrm{3}\boldsymbol{{x}}+\mathrm{7}}\:\:\:{and}\: \\ $$$$\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{2}}{\boldsymbol{{x}}−\mathrm{5}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{expression}}\:\boldsymbol{\mathrm{for}}\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Question Number 143889    Answers: 2   Comments: 0

A Challanging Integral: Φ = ∫_0 ^( 1) ((log(x).log(1+x))/(1−x))dx

$$ \\ $$$$\:\:\:\:\:\:{A}\:\:{Challanging}\:\:{Integral}: \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{log}\left({x}\right).{log}\left(\mathrm{1}+{x}\right)}{\mathrm{1}−{x}}{dx} \\ $$$$ \\ $$$$ \\ $$

Question Number 143641    Answers: 1   Comments: 0

sin 160−sin 20=?

$$\mathrm{sin}\:\mathrm{160}−\mathrm{sin}\:\mathrm{20}=? \\ $$

Question Number 143638    Answers: 0   Comments: 2

......Calculus.... 𝛗:=^? ∫_0 ^( 1) ((ln(1−x)ln(x)(ln(((1−x)/x))))/x) dx m.n....

$$\:\:\:\:\:\:\:\:\:\:......{Calculus}.... \\ $$$$\boldsymbol{\phi}:\overset{?} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left({x}\right)\left({ln}\left(\frac{\mathrm{1}−{x}}{{x}}\right)\right)}{{x}}\:{dx} \\ $$$$\:\:{m}.{n}.... \\ $$$$ \\ $$

Question Number 143630    Answers: 1   Comments: 1

prove that if a and c are odd integers then ab+bc is even for every integer b?

$${prove}\:{that}\:{if}\:{a}\:{and}\:{c}\:{are}\:{odd}\:{integers} \\ $$$${then}\:{ab}+{bc}\:{is}\:{even}\:{for}\:{every}\:{integer}\:{b}? \\ $$

Question Number 143628    Answers: 1   Comments: 0

∫_x ^∝ t^(α−1) e^(it) dt=??

$$\int_{{x}} ^{\propto} {t}^{\alpha−\mathrm{1}} {e}^{{it}} {dt}=?? \\ $$

Question Number 143627    Answers: 2   Comments: 0

Question Number 143635    Answers: 1   Comments: 0

Question Number 143633    Answers: 1   Comments: 0

Question Number 143624    Answers: 1   Comments: 0

∫_0 ^( ∞) z^2 e^(1/z) dz

$$\int_{\mathrm{0}} ^{\:\infty} {z}^{\mathrm{2}} {e}^{\frac{\mathrm{1}}{{z}}} {dz} \\ $$

Question Number 143622    Answers: 1   Comments: 0

∫_0 ^∝ e^(2arctg(t^2 )) dt

$$\int_{\mathrm{0}} ^{\propto} {e}^{\mathrm{2}{arctg}\left({t}^{\mathrm{2}} \right)} {dt} \\ $$

Question Number 143794    Answers: 0   Comments: 3

Prove 3^(111) +1⋮223

$$\mathrm{Prove}\:\:\:\:\:\:\:\:\:\mathrm{3}^{\mathrm{111}} +\mathrm{1}\vdots\mathrm{223} \\ $$

Question Number 143609    Answers: 1   Comments: 0

Prove that:: Ω:=∫_0 ^( 1) ((ln^2 (1−x).ln(x))/x)dx=((−1)/2) ζ (4 ) Without using the “Beta function” m.n

$$\:\:\: \\ $$$$\:{Prove}\:{that}:: \\ $$$$\:\:\Omega:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right).{ln}\left({x}\right)}{{x}}{dx}=\frac{−\mathrm{1}}{\mathrm{2}}\:\zeta\:\left(\mathrm{4}\:\right) \\ $$$${Without}\:{using}\:{the}\:``{Beta}\:{function}'' \\ $$$$\:\:{m}.{n} \\ $$

Question Number 143606    Answers: 1   Comments: 0

Question Number 143603    Answers: 2   Comments: 0

.....Calculus..... Ω:=Σ_(n=1) ^∞ (1/(n^k (1+n))) (k≥ 2) ......

$$\:\:\:\:\:\:\:\:\:\:\:\:\:.....{Calculus}..... \\ $$$$\:\:\:\:\:\:\:\:\Omega:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{k}} \left(\mathrm{1}+{n}\right)}\:\:\:\left({k}\geqslant\:\mathrm{2}\right)\:...... \\ $$

Question Number 143602    Answers: 2   Comments: 1

Question Number 143598    Answers: 3   Comments: 2

Question Number 143597    Answers: 0   Comments: 0

s = ut + (1/2)at^2 ⇒ t^2 + 2(u/a)t − 2(s/a) = 0 by the use of quadratic formula t = ((−((2u)/a) ± (√(((4u^2 )/a^2 ) + 4s)))/2) t = −(u/a) ± (√((u^2 /a^2 ) + s)) Victor Francis

$${s}\:=\:{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:\Rightarrow\:\:{t}^{\mathrm{2}} \:+\:\mathrm{2}\frac{{u}}{{a}}{t}\:−\:\mathrm{2}\frac{{s}}{{a}}\:=\:\mathrm{0} \\ $$$${by}\:{the}\:{use}\:{of}\:{quadratic}\:{formula} \\ $$$${t}\:=\:\frac{−\frac{\mathrm{2}{u}}{{a}}\:\pm\:\sqrt{\frac{\mathrm{4}{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:\mathrm{4}{s}}}{\mathrm{2}} \\ $$$${t}\:=\:−\frac{{u}}{{a}}\:\:\pm\:\:\sqrt{\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\:{s}} \\ $$$${Victor}\:\:{Francis} \\ $$$$ \\ $$$$ \\ $$

Question Number 143596    Answers: 1   Comments: 0

Given that the series (x/3^ )+(y/3^2 )+(x/3^3 )+(y/3^4 )+…+(x/3^(n−1) )+(y/3^n )=8, x,y∈R. Find the value of 3x+y.

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{{x}}{\mathrm{3}^{} }+\frac{{y}}{\mathrm{3}^{\mathrm{2}} }+\frac{{x}}{\mathrm{3}^{\mathrm{3}} }+\frac{{y}}{\mathrm{3}^{\mathrm{4}} }+\ldots+\frac{{x}}{\mathrm{3}^{{n}−\mathrm{1}} }+\frac{{y}}{\mathrm{3}^{{n}} }=\mathrm{8},\:{x},{y}\in\mathbb{R}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{3}{x}+{y}. \\ $$

Question Number 143595    Answers: 2   Comments: 0

If f(x)=((10^x −10^(−x) )/(10^x +10^(−x) )), find f^( −1) (x).

$$\mathrm{If}\:{f}\left({x}\right)=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} },\:\mathrm{find}\:{f}^{\:−\mathrm{1}} \left({x}\right). \\ $$

Question Number 143593    Answers: 0   Comments: 0

soit (X_i ),i∈{1,2,......,n} une suite de variable aleartoire independante (iid) suivant la loi binomiale B(n,p) montrer que X_n ^_ converge en loi vers E(X) 2− montrer que (1/n)Σ_(i=1) ^2 X_(i ) ^2 converge en loi vers E (X^2 ) 3−montrer que (1/n)Σ_(i=1) ^n X_i Y_i converge en loi vers E(XY)

$${soit}\:\left(\boldsymbol{{X}}_{{i}} \right),{i}\in\left\{\mathrm{1},\mathrm{2},......,{n}\right\}\:{une}\:{suite}\:{de}\:{variable}\:{a}\boldsymbol{{leartoire}}\:\boldsymbol{{independante}} \\ $$$$\left(\boldsymbol{{iid}}\right)\:\boldsymbol{{suivant}}\:\boldsymbol{{la}}\:\boldsymbol{{loi}}\:\boldsymbol{{binomiale}}\:\boldsymbol{{B}}\left(\boldsymbol{{n}},\boldsymbol{{p}}\right)\: \\ $$$$\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\overset{\_} {\boldsymbol{{X}}}_{\boldsymbol{{n}}} \boldsymbol{{converge}}\:\boldsymbol{{en}}\:\boldsymbol{{loi}}\:\boldsymbol{{vers}}\:\boldsymbol{{E}}\left(\boldsymbol{{X}}\right) \\ $$$$\mathrm{2}−\:\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}\boldsymbol{{X}}_{\boldsymbol{{i}}\:\:} ^{\mathrm{2}} \boldsymbol{{converge}}\:\boldsymbol{{en}}\:\boldsymbol{{loi}}\:\:\boldsymbol{{vers}}\:\:\:\:\:\:\:\:\boldsymbol{{E}}\:\left(\boldsymbol{{X}}^{\mathrm{2}} \right) \\ $$$$\mathrm{3}−\boldsymbol{{montrer}}\:\boldsymbol{{que}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\boldsymbol{{X}}_{{i}} \boldsymbol{{Y}}_{{i}} \:\boldsymbol{{converge}}\:\boldsymbol{{en}}\:\:\boldsymbol{{loi}}\:\boldsymbol{{vers}}\:\boldsymbol{{E}}\left(\boldsymbol{{XY}}\right) \\ $$

Question Number 143591    Answers: 1   Comments: 0

a;b;c>0 and a+b+c=k min((1/(1+a^2 )) + (1/(1+b^2 )) + (1/(1+c^2 )))=?

$${a};{b};{c}>\mathrm{0}\:\:{and}\:\:{a}+{b}+{c}={k} \\ $$$$\boldsymbol{{min}}\left(\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{c}^{\mathrm{2}} }\right)=? \\ $$

Question Number 143588    Answers: 3   Comments: 0

Question Number 143582    Answers: 0   Comments: 0

x^3 −x−c=0 let x=t+h (t−s){t^3 +3ht^2 +(3h^2 −1)t +(h^3 −h−c)}=0 let t=p+q (p+q−s){p^3 +q^3 +3pq(p+q) +3h(p^2 +q^2 )+6hpq +(3h^2 −1)(p+q)+(h^3 −h−c)} = 0 p^4 +q^4 +pq(p^2 +q^2 )+3pq(p+q)^2 +3h(p^2 +q^2 )(p+q)+6hpq(p+q) +(3h^2 −1)(p+q)^2 +(h^3 −h−c)(p+q) −s(p^3 +q^3 )−3pqs(p+q) −3hs(p^2 +q^2 )−6hspq +(3h^2 −1)s(p+q) −s(h^3 −h−c)=0 let p^4 +q^4 =p^2 +q^2 (p^2 +q^2 )^2 −(p^2 +q^2 )−2p^2 q^2 =0 ⇒ p^2 +q^2 =(1/2)±(√((1/4)+2p^2 q^2 )) say z=(1/2)+(√((1/4)+2m^2 )) (p+q)^2 =z+2m ⇒z(1+m+3m+3h^2 −1−3hs) +(z+2m)^(1/2) {3hz+6mh +h^3 −h−c−3ms −s(z−m)+(3h^2 −1)s} +6m^2 +2(3h^2 −1)m−6hsm −s(h^3 −h−c)=0 let (3h−s)z+m(6h−2s) +h^3 −h−c+3sh^2 −s=0 & 4m+3h(h−s)=0 ⇒ 2(3h−s)z−3h(h−s)(3h−s) +3sh^2 −s+h^3 −h−c=0 let s=−3h ⇒ 12hz−80h^3 −h−c=0 & m=−3h^2 ⇒ 54h^4 −6h^2 (3h^2 −1) −54h^4 +3h(h^3 −h−c)=0 ⇒ 15h^4 −3h^2 +3ch=0 And if h≠0 ⇒ 5h^3 −h+c=0 D=(c^2 /(100))−((1/(15)))^3 ⇒ D≥0 if c≥ (2/(3(√(15)))) m=−3h^2 ; s=−3h; And 12hz−80h^3 −h−c=0 ⇒ z=((20h^2 )/3)+(1/(12))+(c/(12h)) p+q=(√(z+2m)) =(√(((2h^2 )/3)+((c+h)/(12h)))) x=t+h=p+q+h x=h+(√(((2h^2 )/3)+((c+h)/(12h)))) but (c/(12h))>−(((2h^2 )/3)+(1/(12))) ⇒ c<−8h^3 −h ⇒ 8h^3 +h+c<0 5h^3 −h+c=0 ⇒ 13h^3 +2c<0 ......

$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}={t}+{h} \\ $$$$\left({t}−{s}\right)\left\{{t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\}=\mathrm{0} \\ $$$${let}\:{t}={p}+{q} \\ $$$$\left({p}+{q}−{s}\right)\left\{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right)\right. \\ $$$$\:\:+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{6}{hpq} \\ $$$$\left.\:\:+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\} \\ $$$$\:\:=\:\mathrm{0} \\ $$$${p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{pq}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{3}{pq}\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}+{q}\right)+\mathrm{6}{hpq}\left({p}+{q}\right) \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\left({h}^{\mathrm{3}} −{h}−{c}\right)\left({p}+{q}\right) \\ $$$$−{s}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)−\mathrm{3}{pqs}\left({p}+{q}\right) \\ $$$$−\mathrm{3}{hs}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{6}{hspq} \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\left({p}+{q}\right) \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:{p}^{\mathrm{4}} +{q}^{\mathrm{4}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} } \\ $$$${say}\:\:\:\:{z}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{m}^{\mathrm{2}} } \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={z}+\mathrm{2}{m} \\ $$$$\Rightarrow{z}\left(\mathrm{1}+{m}+\mathrm{3}{m}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{hs}\right) \\ $$$$+\left({z}+\mathrm{2}{m}\right)^{\mathrm{1}/\mathrm{2}} \left\{\mathrm{3}{hz}+\mathrm{6}{mh}\right. \\ $$$$\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}−\mathrm{3}{ms} \\ $$$$\left.\:\:\:\:−{s}\left({z}−{m}\right)+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\right\} \\ $$$$+\mathrm{6}{m}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){m}−\mathrm{6}{hsm} \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:\:\left(\mathrm{3}{h}−{s}\right){z}+{m}\left(\mathrm{6}{h}−\mathrm{2}{s}\right) \\ $$$$\:\:\:\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}+\mathrm{3}{sh}^{\mathrm{2}} −{s}=\mathrm{0} \\ $$$$\&\:\:\mathrm{4}{m}+\mathrm{3}{h}\left({h}−{s}\right)=\mathrm{0}\:\:\Rightarrow \\ $$$$\:\mathrm{2}\left(\mathrm{3}{h}−{s}\right){z}−\mathrm{3}{h}\left({h}−{s}\right)\left(\mathrm{3}{h}−{s}\right) \\ $$$$\:\:+\mathrm{3}{sh}^{\mathrm{2}} −{s}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{s}=−\mathrm{3}{h} \\ $$$$\Rightarrow\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\&\:\:{m}=−\mathrm{3}{h}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{54}{h}^{\mathrm{4}} −\mathrm{6}{h}^{\mathrm{2}} \left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\mathrm{54}{h}^{\mathrm{4}} +\mathrm{3}{h}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{15}{h}^{\mathrm{4}} −\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{ch}=\mathrm{0} \\ $$$${And}\:{if}\:{h}\neq\mathrm{0}\:\:\Rightarrow \\ $$$$\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$${D}=\frac{{c}^{\mathrm{2}} }{\mathrm{100}}−\left(\frac{\mathrm{1}}{\mathrm{15}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:{D}\geqslant\mathrm{0}\:\:{if}\:{c}\geqslant\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{15}}} \\ $$$${m}=−\mathrm{3}{h}^{\mathrm{2}} \:;\:{s}=−\mathrm{3}{h};\:\:{And} \\ $$$$\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:{z}=\frac{\mathrm{20}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{{c}}{\mathrm{12}{h}} \\ $$$${p}+{q}=\sqrt{{z}+\mathrm{2}{m}} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${x}={t}+{h}={p}+{q}+{h} \\ $$$$\:\:\:{x}={h}+\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${but}\:\:\:\frac{{c}}{\mathrm{12}{h}}>−\left(\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}\right) \\ $$$$\Rightarrow\:\:{c}<−\mathrm{8}{h}^{\mathrm{3}} −{h} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{3}} +{h}+{c}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{13}{h}^{\mathrm{3}} +\mathrm{2}{c}<\mathrm{0} \\ $$$$...... \\ $$

  Pg 735      Pg 736      Pg 737      Pg 738      Pg 739      Pg 740      Pg 741      Pg 742      Pg 743      Pg 744   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com