Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 999

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

Question Number 119057    Answers: 3   Comments: 0

show that ∣x+y∣≤∣x∣+∣y∣

$${show}\:{that} \\ $$$$\mid{x}+{y}\mid\leqslant\mid{x}\mid+\mid{y}\mid \\ $$$$ \\ $$

Question Number 119055    Answers: 2   Comments: 0

Show by recurence that: ∀ n∈N^∗ Σ_(k=1) ^n (1/(k(k+1)(k+2)))=((n(n+3))/(4(n+1)(n+2)))

$$ \\ $$$${Show}\:{by}\:{recurence}\:{that}:\:\forall\:{n}\in\mathbb{N}^{\ast} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{4}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$

Question Number 119054    Answers: 2   Comments: 0

Show by recurence that: 5^n ≥1+4n ; n∈N

$${Show}\:{by}\:{recurence}\:{that}: \\ $$$$\mathrm{5}^{{n}} \geqslant\mathrm{1}+\mathrm{4}{n}\:;\:{n}\in\mathbb{N} \\ $$

Question Number 119052    Answers: 3   Comments: 0

Question Number 119048    Answers: 1   Comments: 0

Show that : ⌊x⌋ = ⌊((⌊nx⌋)/n)⌋ where n∈N^∗ x∈R

$$\mathrm{Show}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\lfloor{x}\rfloor\:=\:\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:\:\:\:\mathrm{where}\:{n}\in\mathbb{N}^{\ast} \:\:\:{x}\in\mathbb{R} \\ $$$$ \\ $$

Question Number 119038    Answers: 2   Comments: 0

∫_(−π/4) ^(+π/4) ((√(1+tan x))/( (√(1−tan x))))dx

$$\underset{−\pi/\mathrm{4}} {\overset{+\pi/\mathrm{4}} {\int}}\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:{x}}}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}}{dx} \\ $$

Question Number 119035    Answers: 1   Comments: 0

Among the positive integers less than 1200, how many of them are relatively prime to 60?

$$\mathrm{Among}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{1200}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\ $$$$\mathrm{to}\:\mathrm{60}? \\ $$

Question Number 119032    Answers: 1   Comments: 0

Question Number 119025    Answers: 0   Comments: 0

Question Number 119094    Answers: 1   Comments: 0

Why the Fermat formula about prime numbers is for n = 5 and n = 6 fail? f(n) = 2^2^n + 1

$${Why}\:{the}\:{Fermat}\:{formula}\:{about}\:{prime}\:{numbers}\:{is}\:{for}\:{n}\:=\:\mathrm{5}\:{and}\:{n}\:=\:\mathrm{6}\:{fail}? \\ $$$$ \\ $$$${f}\left({n}\right)\:=\:\mathrm{2}^{\mathrm{2}^{{n}} } \:+\:\mathrm{1} \\ $$

Question Number 119021    Answers: 1   Comments: 0

  Pg 994      Pg 995      Pg 996      Pg 997      Pg 998      Pg 999      Pg 1000      Pg 1001      Pg 1002      Pg 1003   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com