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Question Number 180362    Answers: 0   Comments: 0

Question Number 180351    Answers: 0   Comments: 0

solve the differential equation y′′ +2 y′ + y = 0 using power series method that is, assume y = Σ_(n=0) ^∞ c_n x^n is a solution

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\:\:{y}''\:+\mathrm{2}\:{y}'\:+\:{y}\:=\:\mathrm{0} \\ $$$$\mathrm{using}\:\mathrm{power}\:\mathrm{series}\:\mathrm{method}\:\mathrm{that}\:\mathrm{is},\:\mathrm{assume} \\ $$$${y}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{n}} {x}^{{n}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{solution} \\ $$

Question Number 180350    Answers: 1   Comments: 0

Question Number 180348    Answers: 0   Comments: 0

Question Number 180343    Answers: 1   Comments: 1

Question Number 180340    Answers: 1   Comments: 1

please answer this question i have no idea to post an image. it is sent by mistke.

$${please}\:{answer}\:{this}\:{question} \\ $$$${i}\:{have}\:{no}\:{idea}\:{to}\:{post}\:{an}\:{image}.\:{it}\:{is}\: \\ $$$${sent}\:{by}\:{mistke}. \\ $$

Question Number 180335    Answers: 2   Comments: 1

The integers between 1 to 10^4 contain exactly one 8 and one 9 is? I got 1160 but one is arguing 1154only..kindly help me out

$${The}\:{integers}\:{between}\:\mathrm{1}\:{to}\:\mathrm{10}^{\mathrm{4}} \\ $$$${contain}\:{exactly}\:{one}\:\mathrm{8}\:\:{and}\:{one}\:\mathrm{9} \\ $$$${is}?\:{I}\:{got}\:\mathrm{1160}\:{but}\:{one}\:\:{is}\:{arguing} \\ $$$$\mathrm{1154}{only}..{kindly}\:{help}\:{me}\:{out} \\ $$

Question Number 180301    Answers: 3   Comments: 1

Question Number 180300    Answers: 3   Comments: 0

Question Number 180299    Answers: 1   Comments: 0

Question Number 180298    Answers: 1   Comments: 0

Question Number 180295    Answers: 0   Comments: 2

Express these both Cartesian and polar form (1) f(z)=3z^2 −2z+(1/z) (2) f(z)=z+(1/z) Thanks

$$\mathrm{Express}\:\mathrm{these}\:\mathrm{both}\:\mathrm{Cartesian}\:\mathrm{and}\: \\ $$$$\mathrm{polar}\:\mathrm{form} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{3z}^{\mathrm{2}} −\mathrm{2z}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thanks} \\ $$

Question Number 180285    Answers: 0   Comments: 3

Express this f(z)=((2z+i)/(z+i)) in polar form where z=re^(iθ) (polar form)

$$\mathrm{Express}\:\mathrm{this}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{2z}+\mathrm{i}}{\mathrm{z}+\mathrm{i}}\:\mathrm{in}\:\mathrm{polar}\:\mathrm{form} \\ $$$$\mathrm{where}\:\mathrm{z}=\mathrm{re}^{\mathrm{i}\theta} \:\left(\mathrm{polar}\:\mathrm{form}\right) \\ $$$$ \\ $$

Question Number 180277    Answers: 0   Comments: 4

Express the function f(z)=ze^(iz) into cartesian form and separate it into Real and Imaginary part. M.m

$$\mathrm{Express}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{ze}^{\mathrm{iz}} \:\mathrm{into} \\ $$$$\mathrm{cartesian}\:\mathrm{form}\:\mathrm{and}\:\mathrm{separate}\:\mathrm{it}\:\mathrm{into} \\ $$$$\mathrm{Real}\:\mathrm{and}\:\mathrm{Imaginary}\:\mathrm{part}. \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$

Question Number 180274    Answers: 1   Comments: 0

Solve in C the equation z^4 +(7−i)z^3 +(12−15i)z^2 +(4+4i)z+16+192i=0 Knowing that it has one real root and a purely imaginary root of equal magnitude.

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{C}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{z}^{\mathrm{4}} +\left(\mathrm{7}−{i}\right){z}^{\mathrm{3}} +\left(\mathrm{12}−\mathrm{15}{i}\right){z}^{\mathrm{2}} +\left(\mathrm{4}+\mathrm{4}{i}\right){z}+\mathrm{16}+\mathrm{192}{i}=\mathrm{0} \\ $$$$\mathrm{Knowing}\:\mathrm{that}\:\mathrm{it}\:\mathrm{has}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{and}\:\mathrm{a}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitude}. \\ $$

Question Number 180273    Answers: 0   Comments: 0

In triangle ABC with angles α , β , γ correspondently , Euler′s line interescts BC at point P. Ite′s put δ is angle between Euler′s line and BC (∠BPH). Then the following is true tan δ = ((2 cos β cos γ − cos α)/(sin (β − γ)))

$$\mathrm{In}\:\mathrm{triangle}\:\:\mathrm{ABC}\:\:\mathrm{with}\:\mathrm{angles}\:\:\alpha\:,\:\beta\:,\:\gamma \\ $$$$\mathrm{correspondently}\:,\:\mathrm{Euler}'\mathrm{s}\:\mathrm{line}\:\mathrm{interescts} \\ $$$$\mathrm{BC}\:\:\mathrm{at}\:\mathrm{point}\:\:\mathrm{P}.\:\mathrm{Ite}'\mathrm{s}\:\mathrm{put}\:\:\delta\:\:\mathrm{is}\:\mathrm{angle} \\ $$$$\mathrm{between}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{line}\:\mathrm{and}\:\:\mathrm{BC}\:\left(\angle\mathrm{BPH}\right). \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true} \\ $$$$\mathrm{tan}\:\delta\:=\:\frac{\mathrm{2}\:\mathrm{cos}\:\beta\:\mathrm{cos}\:\gamma\:−\:\mathrm{cos}\:\alpha}{\mathrm{sin}\:\left(\beta\:−\:\gamma\right)} \\ $$

Question Number 180272    Answers: 0   Comments: 0

In the triangle following identity is true: 6R^2 = a^2 + c^2 , a ≠ c. Ptove that Euler′s line is antiparallel to AC.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{following}\:\mathrm{identity} \\ $$$$\mathrm{is}\:\mathrm{true}:\:\:\:\mathrm{6R}^{\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:\:\:,\:\:\:\mathrm{a}\:\neq\:\mathrm{c}. \\ $$$$\mathrm{Ptove}\:\mathrm{that}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{line}\:\mathrm{is}\:\mathrm{antiparallel}\:\mathrm{to}\:\:\mathrm{AC}. \\ $$

Question Number 180268    Answers: 2   Comments: 0

Question Number 180260    Answers: 1   Comments: 0

x^(logx) =100x what is the value of x?

$$\mathrm{x}^{\mathrm{logx}} =\mathrm{100x} \\ $$$$ \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}? \\ $$

Question Number 180258    Answers: 2   Comments: 1

Question Number 180257    Answers: 2   Comments: 1

Question Number 180256    Answers: 3   Comments: 1

Question Number 180254    Answers: 1   Comments: 0

Question Number 180236    Answers: 0   Comments: 0

calculation Ω= Σ_(n=1) ^∞ (( ζ ( 2n ))/4^( n) ) =^? (1/2) Ω =Σ_(n=1) ^∞ {(1/2^( 2n) ) Σ_(k=1) ^∞ (1/k^( 2n) ) } = Σ_(n=1) ^∞ Σ_(k=1) ^∞ (1/((2k )^( 2n) ))=Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((2k)^( 2n) )) = Σ_(k=1) ^∞ Σ_(n=1) ^∞ (1/((4k^( 2) )^( n) )) = Σ_(k=1) ^∞ (( (1/(4k^( 2) )))/(1− (1/(4k^( 2) )))) = Σ_(k=1) ^∞ (1/((2k−1)(2k+1 ))) =(1/2) Σ_(k=1) ^∞ ((1/(2k−1)) −(1/(2k+1)) ) = (1/2) (1/(2(1)−1)) − lim_( k→∞) (1/(2k+1)) =(1/2)

$$\:\:\:\:\mathrm{calculation} \\ $$$$\:\:\:\:\Omega=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\zeta\:\left(\:\mathrm{2}{n}\:\right)}{\mathrm{4}^{\:{n}} }\:\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:\Omega\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\mathrm{2}^{\:\mathrm{2}{n}} }\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\:\mathrm{2}{n}} }\:\right\} \\ $$$$\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}\:\right)^{\:\mathrm{2}{n}} }=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)^{\:\mathrm{2}{n}} } \\ $$$$\:\:\:\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{4}{k}^{\:\mathrm{2}} \:\right)^{\:{n}} }\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\frac{\mathrm{1}}{\mathrm{4}{k}^{\:\mathrm{2}} }}{\mathrm{1}−\:\frac{\mathrm{1}}{\mathrm{4}{k}^{\:\mathrm{2}} }} \\ $$$$\:\:\:\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\:\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}\right)−\mathrm{1}}\:−\:\mathrm{lim}_{\:{k}\rightarrow\infty} \:\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 180212    Answers: 1   Comments: 0

How many polygons can be formed from a heptagon?

$${How}\:{many}\:{polygons}\:{can}\:{be}\:{formed} \\ $$$$\:{from}\:{a}\:{heptagon}?\: \\ $$

Question Number 180211    Answers: 1   Comments: 0

Find: Ω = gcd (4^(2020) −1 , 8^(2021) −1). Generalization.

$$\mathrm{Find}: \\ $$$$\Omega\:=\:\mathrm{gcd}\:\left(\mathrm{4}^{\mathrm{2020}} −\mathrm{1}\:,\:\mathrm{8}^{\mathrm{2021}} −\mathrm{1}\right).\:\mathrm{Generalization}. \\ $$

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