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Question Number 119164    Answers: 0   Comments: 0

Question Number 119160    Answers: 0   Comments: 5

Question Number 119159    Answers: 2   Comments: 0

We are in C. Given Z_(0 ) =1 ; Z_(n+1 ) =(1/2)Z_(n ) +(1/2)i n ∈ N. Show that ∀ n ∈ N^(∗ ) , ∣Z_n ∣<1.

$${We}\:{are}\:{in}\:\mathbb{C}. \\ $$$${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${n}\:\in\:\mathbb{N}. \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$

Question Number 119158    Answers: 1   Comments: 0

Question Number 119157    Answers: 2   Comments: 0

Question Number 119155    Answers: 1   Comments: 5

Question Number 119151    Answers: 0   Comments: 0

Question Number 119150    Answers: 2   Comments: 0

Given that a, b and c are real numbers that stisfy the system of equation above a − (√(b^2 − (1/(16)) )) = (√(c^2 − (1/(16)) )) b − (√(c^2 − (1/(25)))) = (√(a^2 − (1/(25)) )) c − (√(a^2 − (1/(36)) )) = (√(b^2 − (1/(36)))) if a+ b + c = (x/( (√y))) where x, y are positive integers and y is square free, find the value of x + y !

$${Given}\:{that}\:{a},\:{b}\:{and}\:{c}\:{are}\:{real}\:{numbers}\: \\ $$$${that}\:{stisfy}\:{the}\:{system}\:{of}\:{equation}\:{above} \\ $$$$ \\ $$$${a}\:−\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\:=\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\: \\ $$$${b}\:−\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}}\:=\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}\:} \\ $$$${c}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}\:}\:=\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}}\: \\ $$$$ \\ $$$${if}\:\:{a}+\:{b}\:+\:{c}\:=\:\frac{{x}}{\:\sqrt{{y}}}\:{where}\:{x},\:{y}\:{are}\:{positive}\:{integers} \\ $$$${and}\:{y}\:{is}\:{square}\:{free},\:{find}\:{the}\:{value}\:\:{of}\:{x}\:+\:{y}\:! \\ $$

Question Number 119147    Answers: 0   Comments: 0

Question Number 119141    Answers: 1   Comments: 0

Question Number 119138    Answers: 1   Comments: 1

(√(2(√(4(√(8(√(16(√(32...))))))))))=? pls help

$$\sqrt{\mathrm{2}\sqrt{\mathrm{4}\sqrt{\mathrm{8}\sqrt{\mathrm{16}\sqrt{\mathrm{32}...}}}}}=? \\ $$$$\mathrm{pls}\:\mathrm{help} \\ $$

Question Number 119136    Answers: 1   Comments: 2

Question Number 119177    Answers: 1   Comments: 0

Question Number 119133    Answers: 0   Comments: 0

Informatica (11110000)_2 =0•2^0 +0•2^1 +0•2^2 +0•2^3 +1•2^4 +1•2^5 +1•2^6 +1•2^7 = =0•1+0•2+0•4+0•8+1•16+1•32+1•64+1•128= 0+0+0+0+16+32+64+128=(240)_(10) (11000101)_2 =1•2^0 +0•2^1 +1•2^2 +0•2^3 +0•2^4 +0•2^5 +1•2^6 +1•2^7 = = 1•1+0•2+1•4+0•8+0•16+0•32+1•64+1•128= (197)_(10) (01101001)_2 =

$$\mathrm{Informatica} \\ $$$$\left(\mathrm{11110000}\right)_{\mathrm{2}} =\mathrm{0}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\mathrm{0}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{0}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{1}\bullet\mathrm{16}+\mathrm{1}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}= \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}=\left(\mathrm{240}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{11000101}\right)_{\mathrm{2}} =\mathrm{1}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\:\mathrm{1}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{1}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{0}\bullet\mathrm{16}+\mathrm{0}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}=\:\left(\mathrm{197}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{01101001}\right)_{\mathrm{2}} = \\ $$

Question Number 119124    Answers: 2   Comments: 0

if matrix A^2 = (((1 3)),((0 1)) ) then matrix A=...

$$\mathrm{if}\:\mathrm{matrix}\:\mathrm{A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\mathrm{3}}\\{\mathrm{0}\:\:\mathrm{1}}\end{pmatrix}\:\:\mathrm{then}\:\mathrm{matrix}\:\mathrm{A}=... \\ $$

Question Number 119119    Answers: 2   Comments: 0

Question Number 119114    Answers: 0   Comments: 4

Question Number 119112    Answers: 0   Comments: 2

Seems my recent post was removed by Tinkutara.

$$\boldsymbol{\mathrm{Seems}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{recent}}\:\boldsymbol{\mathrm{post}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{removed}}\:\boldsymbol{\mathrm{by}} \\ $$$$\boldsymbol{\mathrm{Tinkutara}}.\: \\ $$

Question Number 119109    Answers: 1   Comments: 0

Question Number 119108    Answers: 0   Comments: 0

Question Number 119107    Answers: 1   Comments: 0

Question Number 119074    Answers: 4   Comments: 0

lim_(x→1) (x−1) tan (((πx)/2))

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}−\mathrm{1}\right)\:\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)\: \\ $$

Question Number 119070    Answers: 2   Comments: 0

∫ ((x^2 −x+6)/(x^3 +3x)) dx ∫ ((5x^2 +3x−2)/(x^3 +2x^2 )) dx

$$\int\:\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{3}} +\mathrm{3}{x}}\:{dx}\: \\ $$$$\int\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}}{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }\:{dx}\: \\ $$

Question Number 119069    Answers: 0   Comments: 0

Sorry ladies and gentlemen but who understands this forum ? Most of people post questions and and answers in the same comment. What is the point ? :−((

$$\mathrm{Sorry}\:\mathrm{ladies}\:\mathrm{and}\:\mathrm{gentlemen}\:\mathrm{but}\:\mathrm{who} \\ $$$$\mathrm{understands}\:\mathrm{this}\:\mathrm{forum}\:? \\ $$$$\mathrm{Most}\:\mathrm{of}\:\mathrm{people}\:\mathrm{post}\:\mathrm{questions}\:\mathrm{and} \\ $$$$\mathrm{and}\:\mathrm{answers}\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{comment}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:?\::−\left(\left(\right.\right. \\ $$

Question Number 119065    Answers: 3   Comments: 0

solve (D^2 −2D+1)y = e^x ln x by using the method of variation of parameters.

$${solve}\:\left({D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{1}\right){y}\:=\:{e}^{{x}} \:\mathrm{ln}\:{x}\: \\ $$$${by}\:{using}\:{the}\:{method}\:{of}\:{variation} \\ $$$${of}\:{parameters}. \\ $$

Question Number 119059    Answers: 1   Comments: 0

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