Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 1753

Question Number 25771    Answers: 1   Comments: 0

show that ((cos 50)/(sin 40))+((sin 42)/(cos 48))−((2tan 18)/(cot 72))=0

$${show}\:{that}\:\frac{\mathrm{cos}\:\mathrm{50}}{\mathrm{sin}\:\mathrm{40}}+\frac{\mathrm{sin}\:\mathrm{42}}{\mathrm{cos}\:\mathrm{48}}−\frac{\mathrm{2tan}\:\mathrm{18}}{\mathrm{cot}\:\mathrm{72}}=\mathrm{0} \\ $$

Question Number 25770    Answers: 1   Comments: 0

Question Number 25769    Answers: 0   Comments: 0

a−nser to question 25765...we put I=∫_0 ^∞ (cos(x^(2n) )(1+x^2 )^(−1) dx and J=∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx...we have 2(I+iJ)=∫_R e^(ix^(2n) ) (1+x^2 )^(−1) dx...let f(z)=e^(ix^(2n) ) (1+x^2 )^(−1) by residus therem ∫_R f(z)dz = 2iπRes(f.i) but Res(f.i)= lim_(z−i) (z−i)f(z)=e^(i(i)^(2n) ) = e^(i(−1)_ ^n ) (2i)^(−1) then ∫_R f(z)dz = πe^((−1)^n ) = π( cos(−1)^n +isin(−1)^n ) then ∫_0 ^∞ cos(x^(2n) )(1+x^2 )^(−1) dx =π2^(−1) cos(−1)^n and ∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx=π.2^(−1) sin(−1)^n_ <>....

$${a}−{nser}\:{to}\:{question}\:\mathrm{25765}...{we}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \left({cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:{J}=\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}...{we}\:{have}\:\mathrm{2}\left({I}+{iJ}\right)=\int_{{R}} {e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}...{let}\:{f}\left({z}\right)={e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {by}\:{residus}\:{therem}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi{Res}\left({f}.{i}\right)\:{but}\:{Res}\left({f}.{i}\right)=\:{lim}_{{z}−{i}} \left({z}−{i}\right){f}\left({z}\right)={e}^{{i}\left({i}\right)^{\mathrm{2}{n}} } =\:{e}^{{i}\left(−\mathrm{1}\right)_{} ^{{n}} } \left(\mathrm{2}{i}\right)^{−\mathrm{1}} {then}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\pi{e}^{\left(−\mathrm{1}\right)^{{n}} } =\:\pi\left(\:{cos}\left(−\mathrm{1}\right)^{{n}} \:+{isin}\left(−\mathrm{1}\right)^{{n}} \right)\:{then}\:\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:=\pi\mathrm{2}^{−\mathrm{1}} {cos}\left(−\mathrm{1}\right)^{{n}} {and}\:\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}=\pi.\mathrm{2}^{−\mathrm{1}} {sin}\left(−\mathrm{1}\right)^{{n}_{} } <>....\right. \\ $$

Question Number 25767    Answers: 1   Comments: 0

Question Number 25765    Answers: 0   Comments: 0

find the value of ∫_0 ^∞ cos(x^(2n) )(1+x^2 )^(−1) dx and ∫_0 ^∞ sin( x^(2n) )^ (1+x^2 )^(−1) dx

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

Question Number 25763    Answers: 1   Comments: 0

1+cos^2 2θ=2(cos^4 θ+sin^4 θ)

$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right) \\ $$

Question Number 25762    Answers: 0   Comments: 0

find the value of∫_0 ^∞ artan(2x)(1+x^2 )^(−1_ ) the key of slution put F(t)=∫_0 ^∞ artan(xt)(1+x^2 )^(−1) find ∂F/∂t first then F(t) and take t=2 you will of find find the value of integral..

$${find}\:{the}\:{value}\:{of}\int_{\mathrm{0}} ^{\infty} \:{artan}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}_{} } \:\:{the}\:{key}\:{of}\:{slution}\:{put}\:{F}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:{artan}\left({xt}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:\:{find}\:\partial{F}/\partial{t}\:{first}\:{then}\:{F}\left({t}\right)\:{and}\:{take}\:{t}=\mathrm{2}\:{you}\:{will}\:{of}\:{find}\:{find}\:{the}\:{value}\:{of}\:{integral}.. \\ $$

Question Number 25745    Answers: 1   Comments: 0

find the volume of the solid generated by thr revolution of the curve y(x^2 +a^2 )=a^3 about itd asymptote.

$${find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\: \\ $$$${generated}\:{by}\:{thr}\:{revolution}\:{of}\:{the} \\ $$$${curve}\:{y}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)={a}^{\mathrm{3}} \:{about}\:{itd}\: \\ $$$${asymptote}. \\ $$

Question Number 25741    Answers: 1   Comments: 1

If the sum of n terms f a series is A n^2 +B n , where A, B are constants, then it is an AP.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{terms}\:\mathrm{f}\:\mathrm{a}\:\mathrm{series}\:\mathrm{is}\: \\ $$$${A}\:{n}^{\mathrm{2}} +{B}\:{n}\:,\:\mathrm{where}\:{A},\:{B}\:\mathrm{are}\:\mathrm{constants}, \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}. \\ $$

Question Number 25744    Answers: 1   Comments: 0

find the surface area of the solid formed by the rotation of the arc of the cycloid x=a(t+sin t), y=a(1+cost) about x axis

$${find}\:{the}\:{surface}\:{area}\:{of}\:{the}\:{solid}\: \\ $$$${formed}\:{by}\:{the}\:{rotation}\:{of}\:{the}\:{arc}\:{of}\: \\ $$$${the}\:{cycloid}\:{x}={a}\left({t}+{sin}\:{t}\right),\: \\ $$$${y}={a}\left(\mathrm{1}+{cost}\right)\:{about}\:{x}\:{axis} \\ $$

Question Number 25712    Answers: 1   Comments: 1

Question Number 25709    Answers: 0   Comments: 1

Q. 25444 (another solution) z^ +1 =iz^2 +∣z∣^2 let z=x+iy , ⇒ x−iy+1=i(x^2 −y^2 )−2xy+x^2 +y^2 or (x−y)^2 =x+1 ,and (y^2 −x^2 )=y let y−x=u and y+x=v ⇒ 2u^2 =v−u+2 ...(i) 2uv=u+v ....(ii) ⇒ v=(u/(2u−1)) , substituting in (i) 2u^2 +u=(u/(2u−1))+2 u=1 satisfies this , so ⇒ (2u+3)(u−1)=(u/(2u−1))−1 ⇒ (2u+3)(u−1)=(((1−u))/(2u−1)) ⇒ u=1 or (2u+3)(2u−1)+1=0 or 4u^2 +4u−2=0 or 2u^2 +2u−1=0 ⇒ u=((−2±(√(4+8)))/4) =((−1±(√3))/2) for u_1 =1 , v_1 =(u_1 /(2u_1 −1))=1 so x_1 =((v_1 −u_1 )/2)=0 and y_1 =((v_1 +u_1 )/2) =1 hence z_1 =i for u_2 =(((√3)−1)/2) , v_2 =(((√3)−1)/(2((√3)−1−1))) =−((((√3)−1)(2+(√3)))/2) =−((((√3)+1))/2) x_2 =((v_2 −u_2 )/2)=−((√3)/2) y_2 =((v_2 +u_2 )/2) =−(1/2) hence z_2 =−(1/2)((√3)+i) for u_3 =−((((√3)+1))/2) v_3 =(u_3 /(2u_3 −1)) =((((√3)+1))/(2((√3)+2))) =((((√3)+1)(2−(√3)))/2) =(((√3)−1)/2) x_3 =((v_3 −u_3 )/2)=((√3)/2) and y_3 =((v_3 +u_3 )/2)=−(1/2) ⇒ z_3 =(1/2)((√3)−i) So to summarize, z_1 =i , z_2 =−(1/2)((√3)+i), and z_3 =(1/2)((√3)−i) .

$${Q}.\:\mathrm{25444}\:\:\left({another}\:{solution}\right) \\ $$$$\:\:\bar {{z}}+\mathrm{1}\:={iz}^{\mathrm{2}} +\mid{z}\mid^{\mathrm{2}} \\ $$$${let}\:\:{z}={x}+{iy}\:,\:\Rightarrow \\ $$$${x}−{iy}+\mathrm{1}={i}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${or}\:\:\left({x}−{y}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:,{and}\: \\ $$$$\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)={y} \\ $$$${let}\:\:\:{y}−{x}={u}\:\:{and}\:\:{y}+{x}={v} \\ $$$$\Rightarrow\:\:\mathrm{2}{u}^{\mathrm{2}} ={v}−{u}+\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{uv}={u}+{v}\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}\:\:,\:{substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +{u}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}+\mathrm{2} \\ $$$${u}=\mathrm{1}\:{satisfies}\:{this}\:,\:{so} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}−\mathrm{1} \\ $$$$\Rightarrow\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{\left(\mathrm{1}−{u}\right)}{\mathrm{2}{u}−\mathrm{1}} \\ $$$$\Rightarrow\:\:{u}=\mathrm{1}\:{or}\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{2}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:{u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{4}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${for}\:{u}_{\mathrm{1}} =\mathrm{1}\:\:,\:{v}_{\mathrm{1}} =\frac{{u}_{\mathrm{1}} }{\mathrm{2}{u}_{\mathrm{1}} −\mathrm{1}}=\mathrm{1} \\ $$$${so}\:\:{x}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} −{u}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{0}\:\:\:{and}\: \\ $$$$\:\:\:\:\:{y}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} +{u}_{\mathrm{1}} }{\mathrm{2}}\:=\mathrm{1} \\ $$$${hence}\:\:{z}_{\mathrm{1}} ={i} \\ $$$${for}\:{u}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\:,\:\:{v}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:{x}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} −{u}_{\mathrm{2}} }{\mathrm{2}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:{y}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} +{u}_{\mathrm{2}} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${hence}\:\:\:{z}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+{i}\right) \\ $$$${for}\:\:{u}_{\mathrm{3}} =−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:{v}_{\mathrm{3}} =\frac{{u}_{\mathrm{3}} }{\mathrm{2}{u}_{\mathrm{3}} −\mathrm{1}}\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} −{u}_{\mathrm{3}} }{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and} \\ $$$$\:{y}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} +{u}_{\mathrm{3}} }{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{z}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−{i}\right) \\ $$$${So}\:\:{to}\:{summarize}, \\ $$$$\boldsymbol{{z}}_{\mathrm{1}} =\boldsymbol{{i}}\:\:,\:\:\boldsymbol{{z}}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\boldsymbol{{i}}\right),\:{and} \\ $$$$\:\boldsymbol{{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{i}}\right)\:. \\ $$

Question Number 25706    Answers: 0   Comments: 0

Question Number 25728    Answers: 1   Comments: 3

Question Number 25726    Answers: 0   Comments: 1

x−3=11

$${x}−\mathrm{3}=\mathrm{11} \\ $$

Question Number 25702    Answers: 1   Comments: 1

Question Number 25700    Answers: 0   Comments: 1

Question Number 25689    Answers: 0   Comments: 0

Question Number 25684    Answers: 1   Comments: 0

For a geosynchronous satellite of mass m moving in a circular orbit around the earth at a constant speed v and an altitude h above the earth surface.Show the velocity v=(((GM_e )/(R_e +h)))^(1/2) . If the satellite above is synchronous how fast is it moving through space,taking the period to be 24hrs and M_e =mass of the satellite and is equal to 5.98×10^(24) kg

$${For}\:{a}\:{geosynchronous}\:{satellite}\:{of} \\ $$$${mass}\:{m}\:{moving}\:{in}\:{a}\:{circular} \\ $$$${orbit}\:{around}\:{the}\:{earth}\:{at}\:{a}\:{constant} \\ $$$${speed}\:{v}\:{and}\:{an}\:{altitude}\:{h}\:{above} \\ $$$${the}\:{earth}\:{surface}.{Show}\:{the} \\ $$$${velocity}\:{v}=\left(\frac{{GM}_{{e}} }{{R}_{{e}} +{h}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} . \\ $$$$ \\ $$$${If}\:{the}\:{satellite}\:{above}\:{is}\:{synchronous} \\ $$$${how}\:{fast}\:{is}\:{it}\:{moving}\:{through} \\ $$$${space},{taking}\:{the}\:{period}\:{to}\:{be}\:\mathrm{24}{hrs} \\ $$$${and}\:{M}_{{e}} ={mass}\:{of}\:{the}\:{satellite}\:{and} \\ $$$${is}\:{equal}\:{to}\:\mathrm{5}.\mathrm{98}×\mathrm{10}^{\mathrm{24}} {kg} \\ $$

Question Number 25683    Answers: 0   Comments: 0

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

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

Question Number 25682    Answers: 1   Comments: 0

we give ∫_0 ^∞ t^(a−1) (1 + t)^(−1) dt =π (sin(πa))^(−1) with 0<a<1 find the value of ∫_0 ^∞ (1 +x^(16) )^(−1) dx

$${we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:{t}^{{a}−\mathrm{1}} \left(\mathrm{1}\:+\:{t}\right)^{−\mathrm{1}} {dt}\:=\pi\:\left({sin}\left(\pi{a}\right)\right)^{−\mathrm{1}} \:{with}\:\mathrm{0}<{a}<\mathrm{1}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}\:+{x}^{\mathrm{16}} \right)^{−\mathrm{1}} {dx} \\ $$

Question Number 25680    Answers: 0   Comments: 0

let p(X) = (1 +iX )^(n) − ( 1 − iX )^(n) if p(X ) =∝Σ ( X −xk) find xk and α

$${let}\:{p}\left({X}\right)\:=\:\left(\mathrm{1}\:+{iX}\:\overset{{n}} {\right)}\:−\:\left(\:\mathrm{1}\:−\:{iX}\:\overset{{n}} {\right)}\:\:{if}\:\:{p}\left({X}\:\right)\:=\propto\Sigma\:\left(\:{X}\:−{xk}\right)\:\:{find}\:{xk}\:{and}\:\alpha \\ $$

Question Number 25677    Answers: 0   Comments: 0

let 0<x<1 find the value of F(x) = ∫ ln (1+x cost)dt fromt=0 to t=pi

$${let}\:\mathrm{0}<{x}<\mathrm{1}\:\:{find}\:{the}\:{value}\:{of}\:{F}\left({x}\right)\:=\:\int\:\mathrm{ln}\:\left(\mathrm{1}+{x}\:{cost}\right){dt}\:{fromt}=\mathrm{0}\:{to}\:{t}={pi} \\ $$

Question Number 25676    Answers: 0   Comments: 0

find the value of ∫_0 ^∞ sin(x^ n )/x^ 2 + 1 dx

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

Question Number 25675    Answers: 1   Comments: 1

Question Number 25667    Answers: 0   Comments: 2

Deepak srarts a work and work for 12 days and find he has ompleted 10% less work which he had to finish in order to complete the work in 24 days,^ so he ask Rahman to join him to finish the work on time . If Rahman work half day only for the remaining days. Find the efficiency of Rahman is what percent less then the efficiency of Deepak ?

$$\mathrm{Deepak}\:\mathrm{srarts}\:\mathrm{a}\:\mathrm{work}\:\mathrm{and}\:\mathrm{work}\:\mathrm{for}\:\mathrm{12} \\ $$$$\mathrm{days}\:\mathrm{and}\:\mathrm{find}\:\mathrm{he}\:\mathrm{has}\:\mathrm{ompleted}\:\mathrm{10\%}\: \\ $$$$\mathrm{less}\:\mathrm{work}\:\mathrm{which}\:\mathrm{he}\:\mathrm{had}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{in}\: \\ $$$$\mathrm{order}\:\mathrm{to}\:\mathrm{complete}\:\mathrm{the}\:\mathrm{work}\:\mathrm{in}\:\mathrm{24}\:\mathrm{days}\bar {,} \\ $$$$\mathrm{so}\:\mathrm{he}\:\mathrm{ask}\:\mathrm{Rahman}\:\mathrm{to}\:\mathrm{join}\:\mathrm{him}\:\mathrm{to}\:\mathrm{finish} \\ $$$$\mathrm{the}\:\mathrm{work}\:\mathrm{on}\:\mathrm{time}\:.\:\mathrm{If}\:\mathrm{Rahman}\:\mathrm{work}\: \\ $$$$\mathrm{half}\:\mathrm{day}\:\mathrm{only}\:\mathrm{for}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{days}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{efficiency}\:\mathrm{of}\:\mathrm{Rahman}\:\mathrm{is}\:\mathrm{what} \\ $$$$\mathrm{percent}\:\mathrm{less}\:\mathrm{then}\:\mathrm{the}\:\mathrm{efficiency}\:\mathrm{of} \\ $$$$\mathrm{Deepak}\:? \\ $$

  Pg 1748      Pg 1749      Pg 1750      Pg 1751      Pg 1752      Pg 1753      Pg 1754      Pg 1755      Pg 1756      Pg 1757   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com