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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.

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

Question Number 180207    Answers: 5   Comments: 0

(2/(15)) , ((11)/(40)) , ((26)/(75)) , ((47)/(120)) , ...

$$\:\frac{\mathrm{2}}{\mathrm{15}}\:,\:\:\frac{\mathrm{11}}{\mathrm{40}}\:\:,\:\:\frac{\mathrm{26}}{\mathrm{75}}\:\:,\:\:\frac{\mathrm{47}}{\mathrm{120}}\:\:,\:... \\ $$$$ \\ $$

Question Number 180200    Answers: 4   Comments: 2

A) How many even numbers of 3 different digits bigger than 300 can be formed from {1, 2, 3, 4} B) How many numbers bigger than 300 can be formed from the same group

$$\left.\boldsymbol{{A}}\right)\:{How}\:{many}\:{even}\:{numbers}\:{of}\:\mathrm{3}\:{different}\:{digits} \\ $$$$\:{bigger}\:{than}\:\mathrm{300}\:{can}\:{be}\:{formed}\:{from}\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right\} \\ $$$$\left.\boldsymbol{{B}}\right)\:{How}\:{many}\:{numbers}\:{bigger}\:{than}\:\mathrm{300}\:{can} \\ $$$$\:\:\:\:\:\:\:\:{be}\:{formed}\:{from}\:{the}\:{same}\:{group} \\ $$

Question Number 180197    Answers: 0   Comments: 0

Question Number 180196    Answers: 0   Comments: 0

Question Number 180192    Answers: 1   Comments: 2

With 3 blue, 2 red and 2 pink beads how many necklaces can be made if there: a• is no clasp b• is a clasp

$${With}\:\mathrm{3}\:{blue},\:\mathrm{2}\:{red}\:{and}\:\mathrm{2}\:{pink}\:{beads}\:{how}\:{many} \\ $$$$\:{necklaces}\:{can}\:{be}\:{made}\:{if}\:{there}: \\ $$$$\:{a}\bullet\:{is}\:{no}\:{clasp}\:\:\:\:\:\:{b}\bullet\:{is}\:{a}\:{clasp} \\ $$$$\: \\ $$

Question Number 180191    Answers: 1   Comments: 0

A solid cone of height 56cm and a basem diaeter of 56cm is formed from al cylindrica drum equal in height andt diameer with the cone. Find the surfacee ara of the remaining part of the drum. (if the answer is accompanied by a diagramit will be perfect and I thank youo fr it).

$$ \\ $$$$\mathrm{A}\:\mathrm{solid}\:\mathrm{cone}\:\mathrm{of}\:\mathrm{height}\:\mathrm{56cm}\:\mathrm{and}\:\mathrm{a}\:\mathrm{basem} \\ $$$$\mathrm{diaeter}\:\mathrm{of}\:\mathrm{56cm}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{al} \\ $$$$\mathrm{cylindrica}\:\mathrm{drum}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{height}\:\mathrm{andt} \\ $$$$\mathrm{diameer}\:\mathrm{with}\:\mathrm{the}\:\mathrm{cone}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{surfacee} \\ $$$$\mathrm{ara}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{drum}. \\ $$$$ \\ $$$$\left(\mathrm{if}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{accompanied}\:\mathrm{by}\:\mathrm{a}\:\right. \\ $$$$\mathrm{diagramit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{perfect}\:\mathrm{and}\:\mathrm{I}\:\mathrm{thank}\:\mathrm{youo} \\ $$$$\left.\mathrm{fr}\:\mathrm{it}\right). \\ $$

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