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Question Number 55258    Answers: 0   Comments: 0

3x+5y=?_

$$\mathrm{3}{x}+\mathrm{5}{y}=?_{} \\ $$

Question Number 55267    Answers: 0   Comments: 0

1) calculate f(x)=∫_0 ^(π/4) ln(1+xtanθ)dθ 2) find the values of integrals ∫_0 ^(π/4) ln(1+tanθ) and ∫_0 ^(π/4) ln(1+2tanθ)dθ . 1) we have f^′ (x)=∫_0 ^(π/4) ((tanθ)/(1+xtanθ)) dθ =∫_0 ^(π/4) (((sinθ)/(cosθ))/(1+x((sinθ)/(cosθ))))dθ =∫_0 ^(π/4) ((sinθ)/(cosθ +xsinθ)) dθ =_(tan((θ/2))=t) ∫_0 ^((√2)−1) (((2t)/(1+t^2 ))/(((1−t^2 )/(1+t^2 )) +((2xt)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((4t)/((1+t^2 )(1−t^2 +2xt)))dt =−∫_0 ^((√2)−1) ((4t)/((t^2 +1)(t^2 −2xt −1)))dt let decompose F(t) = ((4t)/((t^2 +1)(t^2 −2xt −1))) roots of t^2 −2xt −1 Δ^′ =x^2 +1 ⇒t_1 =x+(√(x^2 +1)) and t_2 =x−(√(x^2 +1)) F(t)=(a/(t−t_1 )) +(b/(t−t_2 )) +((ct +d)/(t^2 +1)) a =lim_(t→t_1 ) (t−t_1 )F(t)=((4t_1 )/((t_1 ^2 +1)(t_1 −t_2 ))) =α b =lim_(t→t_2 ) (t−t_2 )F(t) =((4t_2 )/((t_2 ^2 +1)(t_2 −t_1 ))) =β ⇒F(t)=(α/(t−t_1 )) +(β/(t−t_2 )) +((ct +d)/(t^2 +1)) F(0) =0=−(α/t_1 ) −(β/t_2 ) +d ⇒d =(α/t_1 ) +(β/t_2 ) F(1)=(2/(−2x)) =−(1/x)=(α/(1−t_1 )) +(β/(1−t_2 )) +((c+d)/2) ⇒(1/x) =(α/(t_1 −1)) +(β/(t_2 −1)) −(c/2) −(d/2) ⇒(c/2) =(α/(t_1 −1)) +(β/(t_2 −1)) −(d/2) −(1/x) ⇒c =((2α)/(t_1 −1)) +((2β)/(t_2 −1)) −d−(2/x) ∫ F(t)dt =αln∣t−t_1 ∣ +βln∣t−t_2 ∣ +(c/2)ln(t^2 +1) +d arctan(t) ⇒ ∫_0 ^((√2)−1) F(t)dt =[αln∣t−t_1 ∣+βln∣t−t_2 ∣ +(c/2)ln(t^2 +1)]_0 ^((√2)−1) =αln∣(√2)−1−t_1 ∣ +βln∣(√2)−1−t_2 ∣ +(c/2)ln(4−2(√2)) =αln∣(√2)−1−x−(√(1+x^2 )))+βln∣(√2)−1−x+(√(1+x^2 ))) +((ln(4−2(√2)))/2)c =f^′ (x) ⇒ f(x)=∫ αln∣(√2)−1−x−(√(1+x^2 ))∣)dx+β∫ ln∣(√2)−1+(√(1+x^2 ))∣dx +((cx)/2)ln(4−2(√2)) +C ....be continued...

$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{xtan}\theta\right){d}\theta \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:{integrals}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{ln}\left(\mathrm{1}+{tan}\theta\right)\:\:{and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\mathrm{2}{tan}\theta\right){d}\theta\:. \\ $$$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{tan}\theta}{\mathrm{1}+{xtan}\theta}\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\frac{{sin}\theta}{{cos}\theta}}{\mathrm{1}+{x}\frac{{sin}\theta}{{cos}\theta}}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{sin}\theta}{{cos}\theta\:+{xsin}\theta}\:{d}\theta\:\:=_{{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{xt}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{xt}\right)}{dt}\:=−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1}\right)}\:\:{roots}\:{of}\:\:{t}^{\mathrm{2}} −\mathrm{2}{xt}\:−\mathrm{1} \\ $$$$\Delta^{'} ={x}^{\mathrm{2}} +\mathrm{1}\:\Rightarrow{t}_{\mathrm{1}} ={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:{and}\:{t}_{\mathrm{2}} ={x}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−{t}_{\mathrm{1}} }\:+\frac{{b}}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow{t}_{\mathrm{1}} } \left({t}−{t}_{\mathrm{1}} \right){F}\left({t}\right)=\frac{\mathrm{4}{t}_{\mathrm{1}} }{\left({t}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\:=\alpha \\ $$$${b}\:={lim}_{{t}\rightarrow{t}_{\mathrm{2}} } \left({t}−{t}_{\mathrm{2}} \right){F}\left({t}\right)\:=\frac{\mathrm{4}{t}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}_{\mathrm{2}} −{t}_{\mathrm{1}} \right)}\:=\beta\:\Rightarrow{F}\left({t}\right)=\frac{\alpha}{{t}−{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}−{t}_{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}=−\frac{\alpha}{{t}_{\mathrm{1}} }\:−\frac{\beta}{{t}_{\mathrm{2}} }\:+{d}\:\:\Rightarrow{d}\:=\frac{\alpha}{{t}_{\mathrm{1}} }\:+\frac{\beta}{{t}_{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{−\mathrm{2}{x}}\:=−\frac{\mathrm{1}}{{x}}=\frac{\alpha}{\mathrm{1}−{t}_{\mathrm{1}} }\:+\frac{\beta}{\mathrm{1}−{t}_{\mathrm{2}} }\:+\frac{{c}+{d}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{1}}{{x}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{c}}{\mathrm{2}}\:−\frac{{d}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{c}}{\mathrm{2}}\:=\frac{\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−\frac{{d}}{\mathrm{2}}\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow{c}\:=\frac{\mathrm{2}\alpha}{{t}_{\mathrm{1}} −\mathrm{1}}\:+\frac{\mathrm{2}\beta}{{t}_{\mathrm{2}} −\mathrm{1}}\:−{d}−\frac{\mathrm{2}}{{x}} \\ $$$$\int\:{F}\left({t}\right){dt}\:=\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\:+{d}\:{arctan}\left({t}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {F}\left({t}\right){dt}\:=\left[\alpha{ln}\mid{t}−{t}_{\mathrm{1}} \mid+\beta{ln}\mid{t}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$=\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{1}} \mid\:+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{t}_{\mathrm{2}} \mid\:+\frac{{c}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\: \\ $$$$\left.=\left.\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\beta{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}}{c}\:={f}^{'} \left({x}\right)\:\Rightarrow \\ $$$$\left.{f}\left({x}\right)=\int\:\alpha{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}−{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid\right){dx}+\beta\int\:{ln}\mid\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\mid{dx} \\ $$$$+\frac{{cx}}{\mathrm{2}}{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:+{C}\:....{be}\:{continued}... \\ $$

Question Number 55296    Answers: 1   Comments: 0

simplify the following expression if x<0 ∣4x−(√((3x−1)^2 ))∣

$${simplify}\:{the}\:{following}\:{expression}\:{if} \\ $$$${x}<\mathrm{0} \\ $$$$\mid\mathrm{4}{x}−\sqrt{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }\mid \\ $$

Question Number 55295    Answers: 0   Comments: 0

Question Number 55245    Answers: 1   Comments: 0

Question Number 55240    Answers: 1   Comments: 0

make r the subject of the relation m=((4(√(u+r)))/(v−r))

$${make}\:{r}\:{the}\:{subject}\:{of}\:{the}\:{relation} \\ $$$${m}=\frac{\mathrm{4}\sqrt{{u}+{r}}}{{v}−{r}} \\ $$

Question Number 55237    Answers: 1   Comments: 1

∫_0 ^3 ∫_1 ^2 (x^3 +y^2 )dxdy

$$\int_{\mathrm{0}} ^{\mathrm{3}} \int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dxdy} \\ $$

Question Number 55230    Answers: 0   Comments: 1

calculate lim_(ξ→0) ∫_1 ^(1+ξ) ((arctan(ξt))/t) dt .

$${calculate}\:{lim}_{\xi\rightarrow\mathrm{0}} \:\:\:\:\:\int_{\mathrm{1}} ^{\mathrm{1}+\xi} \:\:\:\:\frac{{arctan}\left(\xi{t}\right)}{{t}}\:{dt}\:. \\ $$

Question Number 55229    Answers: 0   Comments: 1

calculate lim_(n→+∞) ∫_0 ^n (e^(nx) /(1+nx^2 )) dx .

$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\int_{\mathrm{0}} ^{{n}} \:\:\frac{{e}^{{nx}} }{\mathrm{1}+{nx}^{\mathrm{2}} }\:{dx}\:\:. \\ $$

Question Number 55224    Answers: 3   Comments: 0

log_2 (x^2 +7x−2)=log_2 (x^2 +3x−6)+log_4 8 find x

$${log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}\right)={log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}\right)+{log}_{\mathrm{4}} \mathrm{8} \\ $$$${find}\:{x} \\ $$

Question Number 55223    Answers: 0   Comments: 5

Question Number 55217    Answers: 1   Comments: 0

Question Number 55214    Answers: 1   Comments: 1

let U_n =∫_(1/n) ^1 (√(x^2 +(3/n)))dx .calculate lim_(n→+∞) U_n

$${let}\:{U}_{{n}} =\int_{\frac{\mathrm{1}}{{n}}} ^{\mathrm{1}} \sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{3}}{{n}}}{dx}\:\:\:.{calculate}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\ $$

Question Number 55211    Answers: 1   Comments: 0

Question Number 55205    Answers: 0   Comments: 0

Please, can you help me with the following? A boat B is found at 40km at at the east of a boat A. A is moving at 30° east at 20km/h, B is moving qt 10km/h at 30°west. After how many hours will the two boats be the closest to each other? What will therefore be the position of the boat B and the boat A? Thank you

$$\mathrm{Please},\:\mathrm{can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{the}\:\mathrm{following}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{A}\:\mathrm{boat}\:\boldsymbol{\mathrm{B}}\:\mathrm{is}\:\mathrm{found}\:\mathrm{at}\:\mathrm{40km}\:\mathrm{at}\:\mathrm{at}\:\mathrm{the}\:\mathrm{east}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{boat}\:\boldsymbol{\mathrm{A}}.\:\boldsymbol{\mathrm{A}}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{at}\:\mathrm{30}°\:\mathrm{east}\:\mathrm{at}\:\mathrm{20km}/\mathrm{h}, \\ $$$$\boldsymbol{\mathrm{B}}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{qt}\:\mathrm{10km}/\mathrm{h}\:\mathrm{at}\:\mathrm{30}°\mathrm{west}. \\ $$$$ \\ $$$$\mathrm{After}\:\mathrm{how}\:\mathrm{many}\:\mathrm{hours}\:\mathrm{will}\:\mathrm{the}\:\mathrm{two}\:\mathrm{boats} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}? \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{therefore}\:\mathrm{be}\:\mathrm{the}\:\mathrm{position}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{boat}\:\boldsymbol{\mathrm{B}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{boat}\:\boldsymbol{\mathrm{A}}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$$$ \\ $$

Question Number 55198    Answers: 2   Comments: 1

x^5 −4x^4 +6x^3 +8x−32=0 Find at least one root.

$${x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}−\mathrm{32}=\mathrm{0} \\ $$$${Find}\:{at}\:{least}\:{one}\:{root}. \\ $$

Question Number 55197    Answers: 0   Comments: 4

∫(dα/(1−sin^3 α))=? ∫(dβ/(1−cos^3 β))=? ∫(dγ/(1−tan^3 γ))=?

$$\int\frac{{d}\alpha}{\mathrm{1}−\mathrm{sin}^{\mathrm{3}} \:\alpha}=? \\ $$$$\int\frac{{d}\beta}{\mathrm{1}−\mathrm{cos}^{\mathrm{3}} \:\beta}=? \\ $$$$\int\frac{{d}\gamma}{\mathrm{1}−\mathrm{tan}^{\mathrm{3}} \:\gamma}=? \\ $$

Question Number 55195    Answers: 1   Comments: 0

x^2 +ax+(√2)b=0,has 2 roots:c and d,also x^2 +cx+(√2)d=0,has 2 roots:a and b.such that:a, b, c, d,are defferent non zero numbers. find possible value(s) for:a^2 +b^2 +c^2 +d^2 .

$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{b}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{c}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{also}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{cx}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{d}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{such}} \\ $$$$\boldsymbol{\mathrm{that}}:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}},\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{defferent}}\:\boldsymbol{\mathrm{non}}\:\boldsymbol{\mathrm{zero}}\: \\ $$$$\boldsymbol{\mathrm{numbers}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\left(\boldsymbol{\mathrm{s}}\right)\:\boldsymbol{\mathrm{for}}:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\boldsymbol{\mathrm{d}}^{\mathrm{2}} . \\ $$

Question Number 55193    Answers: 1   Comments: 0

Question Number 55192    Answers: 1   Comments: 1

what will be the numbers of zeroes in the expension of a) 100!×25! b) 100!+25! please help

$${what}\:{will}\:{be}\:{the}\:{numbers}\:{of}\:{zeroes}\:{in}\:{the}\:{expension}\:{of} \\ $$$$\left.{a}\right)\:\mathrm{100}!×\mathrm{25}! \\ $$$$\left.{b}\right)\:\mathrm{100}!+\mathrm{25}! \\ $$$$ \\ $$$${please}\:{help} \\ $$

Question Number 55191    Answers: 2   Comments: 0

Question Number 55185    Answers: 0   Comments: 3

Question Number 55174    Answers: 0   Comments: 3

center and radius convergence of series Σ_(n=0) ^(∝) (((4−2i)/(1+5i)))^n z^n is...

$$\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{convergence} \\ $$$$\:\mathrm{of}\:\mathrm{series}\:\underset{{n}=\mathrm{0}} {\overset{\propto} {\Sigma}}\:\left(\frac{\mathrm{4}−\mathrm{2}{i}}{\mathrm{1}+\mathrm{5}{i}}\right)^{{n}} {z}^{{n}} \:\mathrm{is}... \\ $$$$ \\ $$

Question Number 55172    Answers: 1   Comments: 0

Question Number 55171    Answers: 0   Comments: 3

Question Number 55166    Answers: 1   Comments: 0

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