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Question Number 181484 Answers: 0 Comments: 6

If a hen and a half lay an egg and a half in a day and a half how many eggs would one hen lay in one day?

$${If}\:{a}\:{hen}\:{and}\:{a}\:{half} \\ $$$${lay}\:{an}\:{egg}\:{and}\:{a}\:{half} \\ $$$${in}\:{a}\:{day}\:{and}\:{a}\:{half} \\ $$$${how}\:{many}\:{eggs}\:{would} \\ $$$${one}\:{hen}\:{lay}\:{in}\:{one} \\ $$$${day}? \\ $$

Question Number 175240 Answers: 1 Comments: 1

Question Number 175230 Answers: 1 Comments: 0

Question Number 175221 Answers: 2 Comments: 0

Question Number 175218 Answers: 1 Comments: 0

⌊ ( 1 + (√5) )^( 8) ⌋ = ?

$$ \\ $$$$\:\:\lfloor\:\:\left(\:\mathrm{1}\:+\:\sqrt{\mathrm{5}}\:\right)^{\:\mathrm{8}} \:\rfloor\:=\:? \\ $$$$\:\:\: \\ $$

Question Number 175217 Answers: 0 Comments: 1

lim_( x→0) { ⌊ ((cos(x))/x) ⌋ −⌊((1+cos(x))/(1−cos(x))) ⌋}

$$ \\ $$$$\:\:{lim}_{\:{x}\rightarrow\mathrm{0}} \left\{\:\lfloor\:\frac{{cos}\left({x}\right)}{{x}}\:\rfloor\:−\lfloor\frac{\mathrm{1}+{cos}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\:\rfloor\right\} \\ $$

Question Number 175213 Answers: 2 Comments: 0

Solve by Method of variation parameter (d^2 y/dx^2 )−3(dy/dx)+2y=sinx M.m

$$\mathrm{Solve}\:\mathrm{by}\:\mathrm{Method}\:\mathrm{of}\:\mathrm{variation}\:\mathrm{parameter} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }−\mathrm{3}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{2y}=\mathrm{sinx} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 175211 Answers: 1 Comments: 0

Assume that the sequence terms tend to the constant value u, so that as n→∞, u_(n−1) →u and u_n →u. (i) show that u^2 +u−1=0 (ii) show that (1/(1+(1/(1+(1/(1+(1/(1+.....))))))))=((−1+(√5))/2)

$$\boldsymbol{\mathrm{Assume}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{tend}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{constant}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{{u}},\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{as}} \\ $$$$\boldsymbol{\mathrm{n}}\rightarrow\infty,\:\boldsymbol{{u}}_{\boldsymbol{{n}}−\mathrm{1}} \rightarrow\boldsymbol{{u}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{u}}_{\boldsymbol{{n}}} \rightarrow\boldsymbol{{u}}. \\ $$$$\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\boldsymbol{{u}}^{\mathrm{2}} +\boldsymbol{{u}}−\mathrm{1}=\mathrm{0} \\ $$$$\:\left(\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+.....}}}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Question Number 175210 Answers: 0 Comments: 0

y′x + y = y^( 2) ln(x) u=y^( −1) ⇒ u′ =−y′y^( −2) −y′y^( −2) x −y^(−1) = −ln(x) u′x −u = −ln(x) u′−(1/x) u =((−ln(x))/x) u = e^( −∫−(1/x)dx) ( ∫−((ln(x))/x)e^( −∫(1/x)dx) dx +C) = x (−∫ ((ln(x))/x^( 2) )dx +C) ln(x)=t ∫te^( −t) dt= [ −e^( −t) .t +∫e^(−t) dt] = −(1/x)ln(x) −(1/x) u= −ln(x) −Cx −1 y = (1/(−ln(x)−Cx−1)) ✓

$$ \\ $$$$\:\:\:{y}'{x}\:+\:{y}\:=\:{y}^{\:\mathrm{2}} {ln}\left({x}\right) \\ $$$$\:\:\:\:{u}={y}^{\:−\mathrm{1}} \:\Rightarrow\:{u}'\:=−{y}'{y}^{\:−\mathrm{2}} \\ $$$$\:\:\:\:\:−{y}'{y}^{\:−\mathrm{2}} {x}\:−{y}^{−\mathrm{1}} =\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:\:\:\:{u}'{x}\:−{u}\:=\:−{ln}\left({x}\right) \\ $$$$\:\:\:\:{u}'−\frac{\mathrm{1}}{{x}}\:{u}\:=\frac{−{ln}\left({x}\right)}{{x}}\: \\ $$$$\:\:\:\:\:{u}\:=\:{e}^{\:−\int−\frac{\mathrm{1}}{{x}}{dx}} \left(\:\int−\frac{{ln}\left({x}\right)}{{x}}{e}^{\:−\int\frac{\mathrm{1}}{{x}}{dx}} {dx}\:+{C}\right) \\ $$$$\:\:=\:\:{x}\:\left(−\int\:\frac{{ln}\left({x}\right)}{{x}^{\:\mathrm{2}} }{dx}\:+{C}\right) \\ $$$$\:\:\:\:{ln}\left({x}\right)={t} \\ $$$$\:\:\:\:\:\:\int{te}^{\:−{t}} {dt}=\:\left[\:−{e}^{\:−{t}} .{t}\:+\int{e}^{−{t}} {dt}\right] \\ $$$$\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{{x}}{ln}\left({x}\right)\:−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:\:\:\:\:{u}=\:−{ln}\left({x}\right)\:−{Cx}\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{1}}{−{ln}\left({x}\right)−{Cx}−\mathrm{1}}\:\checkmark \\ $$

Question Number 175199 Answers: 1 Comments: 0

Question Number 175193 Answers: 2 Comments: 0

Question Number 175191 Answers: 2 Comments: 1

∫(√(1−(1/x)))dx=?

$$\int\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}{dx}=? \\ $$

Question Number 175188 Answers: 1 Comments: 0

Question Number 175180 Answers: 3 Comments: 0

a_n is an AP and S_n is sum of n terms of this AP. Given that S_(11) −S_7 =72, determine a_6 +a_(13) .

$${a}_{{n}} \:{is}\:{an}\:{AP}\:{and}\:{S}_{{n}} \:{is}\:{sum}\:{of}\:{n}\:{terms} \\ $$$${of}\:{this}\:{AP}. \\ $$$${Given}\:{that}\:{S}_{\mathrm{11}} −{S}_{\mathrm{7}} =\mathrm{72},\:{determine} \\ $$$${a}_{\mathrm{6}} +{a}_{\mathrm{13}} . \\ $$

Question Number 175176 Answers: 2 Comments: 0

Find the equation of the locus of points equidistant from the point A(4, −1) and the line x−y+2=0.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\: \\ $$$$\mathrm{points}\:\mathrm{equidistant}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\left(\mathrm{4},\:−\mathrm{1}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{2}=\mathrm{0}. \\ $$

Question Number 175175 Answers: 1 Comments: 0

Question Number 175174 Answers: 0 Comments: 0

Question Number 175173 Answers: 0 Comments: 0

Question Number 175172 Answers: 0 Comments: 0

Question Number 175166 Answers: 1 Comments: 0

Question Number 175160 Answers: 1 Comments: 1

Question Number 175154 Answers: 1 Comments: 1

Question Number 175149 Answers: 1 Comments: 0

Question Number 175148 Answers: 0 Comments: 1

Question Number 175147 Answers: 1 Comments: 0

if x is a real number in [0,1] then the value of lim_(m→∞ ) lim_(n→∞) [1+cos^(2m) (n!πx)]

$$\:\:\:\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\:\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\underset{{m}\rightarrow\infty\:} {\mathrm{lim}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2}{m}} \left({n}!\pi{x}\right)\right]\: \\ $$

Question Number 175137 Answers: 1 Comments: 0

lim_(x→2) ((((5x−9))^(1/5) ((4x−7))^(1/4) ((3x−5))^(1/3) (√(2x−3))−1)/(x−2))=?

$$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\:\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\:\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\:\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}}=? \\ $$

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