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AllQuestion and Answers: Page 130

Question Number 206321    Answers: 2   Comments: 0

Question Number 206319    Answers: 0   Comments: 0

Question Number 206318    Answers: 0   Comments: 0

∫_0 ^1 ∫_0 ^1 ((xln(1+y)ln(1−x))/(1+x^2 y))dxdy

$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\mathrm{1}+{y}\right){ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} {y}}{dxdy} \\ $$

Question Number 206362    Answers: 2   Comments: 0

sin(π/7) × sin((2π)/7) × sin((3π)/7) = ?

$$\mathrm{sin}\frac{\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}\:×\:\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\:? \\ $$

Question Number 206311    Answers: 0   Comments: 0

Hi everyone... This is not a question per say (sorry ′bout that). It′s just that for past 4 years, to type math messages to my friends (text messages) I′ve been painfully using the two most famous android ones (matheditor and dxmath). And I was thinking, I realised how typing on this app on android (matheditor) is so much more practical. So if any one here knows how to code, I would love to see an android−keyboard adaptation (only with the unicode charters on this keyboard that is... and with the extra ^(1234567890−+×()inx) _(1234567890inx) unicode ones that might come in handy once we can′t use proper indexation) I would personnaly be ready to pay for it. And I would be intersted to know if anyone of you would be too. I don′t think it would be very hard to do... I am ready to help if I can. I don′t know how to make an android app, but being a mathematician and a physicist, I′m sure I can make myself usefull...

$$\mathrm{Hi}\:\mathrm{everyone}... \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{question}\:\mathrm{per}\:\mathrm{say}\:\left(\mathrm{sorry}\:'\mathrm{bout}\right. \\ $$$$\left.\mathrm{that}\right).\:\mathrm{It}'\mathrm{s}\:\mathrm{just}\:\mathrm{that}\:\mathrm{for}\:\mathrm{past}\:\mathrm{4}\:\mathrm{years},\:\mathrm{to}\:\mathrm{type} \\ $$$$\mathrm{math}\:\mathrm{messages}\:\mathrm{to}\:\mathrm{my}\:\mathrm{friends}\:\left(\mathrm{text}\:\mathrm{messages}\right) \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{painfully}\:\mathrm{using}\:\mathrm{the}\:\mathrm{two}\:\mathrm{most}\:\mathrm{famous} \\ $$$$\mathrm{android}\:\mathrm{ones}\:\left({matheditor}\:\mathrm{and}\:{dxmath}\right). \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{was}\:\mathrm{thinking},\:\mathrm{I}\:\mathrm{realised}\:\mathrm{how}\:\mathrm{typing} \\ $$$$\mathrm{on}\:\mathrm{this}\:\mathrm{app}\:\mathrm{on}\:\mathrm{android}\:\left({matheditor}\right)\:\mathrm{is}\:\mathrm{so} \\ $$$$\mathrm{much}\:\mathrm{more}\:\mathrm{practical}. \\ $$$$ \\ $$$$\mathrm{So}\:\mathrm{if}\:\mathrm{any}\:\mathrm{one}\:\mathrm{here}\:\mathrm{knows}\:\mathrm{how}\:\mathrm{to}\:\mathrm{code},\:\mathrm{I}\:\mathrm{would} \\ $$$$\mathrm{love}\:\mathrm{to}\:\mathrm{see}\:\mathrm{an}\:\mathrm{android}−\mathrm{keyboard}\:\mathrm{adaptation} \\ $$$$\left(\mathrm{only}\:\mathrm{with}\:\mathrm{the}\:\mathrm{unicode}\:\mathrm{charters}\:\mathrm{on}\:\mathrm{this}\right. \\ $$$$\mathrm{keyboard}\:\mathrm{that}\:\mathrm{is}...\:\mathrm{and}\:\mathrm{with}\:\mathrm{the}\:\mathrm{extra} \\ $$$$\:^{\mathrm{1234567890}−+×\left(\right)\mathrm{inx}} \:_{\mathrm{1234567890inx}} \:\mathrm{unicode}\:\mathrm{ones} \\ $$$$\mathrm{that}\:\mathrm{might}\:\mathrm{come}\:\mathrm{in}\:\mathrm{handy}\:\mathrm{once}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{use} \\ $$$$\left.\mathrm{proper}\:\mathrm{indexation}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{would}\:\mathrm{personnaly}\:\mathrm{be}\:\mathrm{ready}\:\mathrm{to}\:\mathrm{pay}\:\mathrm{for}\:\mathrm{it}. \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{would}\:\mathrm{be}\:\mathrm{intersted}\:\mathrm{to}\:\mathrm{know}\:\mathrm{if}\:\mathrm{anyone} \\ $$$$\mathrm{of}\:\mathrm{you}\:\mathrm{would}\:\mathrm{be}\:\mathrm{too}. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{very}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{do}... \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{ready}\:\mathrm{to}\:\mathrm{help}\:\mathrm{if}\:\mathrm{I}\:\mathrm{can}.\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{how} \\ $$$$\mathrm{to}\:\mathrm{make}\:\mathrm{an}\:\mathrm{android}\:\mathrm{app},\:\mathrm{but}\:\mathrm{being}\:\mathrm{a} \\ $$$$\mathrm{mathematician}\:\mathrm{and}\:\mathrm{a}\:\mathrm{physicist},\:\mathrm{I}'\mathrm{m}\:\mathrm{sure} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{make}\:\mathrm{myself}\:\mathrm{usefull}... \\ $$

Question Number 206309    Answers: 2   Comments: 0

Find the ways to express 11025 as product of two factors. (a) 13 (b) 14 (c) 26 (d) 27

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{express}\:\mathrm{11025}\: \\ $$$$\mathrm{as}\:\mathrm{product}\:\mathrm{of}\:\mathrm{two}\:\mathrm{factors}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{14}\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{26}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{27} \\ $$

Question Number 206303    Answers: 1   Comments: 0

log_2 4

$${log}_{\mathrm{2}} \mathrm{4} \\ $$

Question Number 206302    Answers: 2   Comments: 0

Question Number 206300    Answers: 0   Comments: 4

Find the area of shaded region.

$${Find}\:{the}\:{area}\:{of}\:{shaded}\:{region}. \\ $$

Question Number 206294    Answers: 2   Comments: 2

Solve the system (a+b)^(−1) +c^(−1) =2^(−1) (c+b)^(−1) +a^(−1) =3^(−1) (a+c)^(−1) +b^(−1) =4^(−1)

$${Solve}\:{the}\:{system} \\ $$$$\left({a}+{b}\right)^{−\mathrm{1}} +{c}^{−\mathrm{1}} =\mathrm{2}^{−\mathrm{1}} \\ $$$$\left({c}+{b}\right)^{−\mathrm{1}} +{a}^{−\mathrm{1}} =\mathrm{3}^{−\mathrm{1}} \\ $$$$\left({a}+{c}\right)^{−\mathrm{1}} +{b}^{−\mathrm{1}} =\mathrm{4}^{−\mathrm{1}} \\ $$

Question Number 206292    Answers: 2   Comments: 0

Question Number 206275    Answers: 3   Comments: 0

Question Number 206273    Answers: 1   Comments: 0

Does anyone know how this works ? I have dψ = (x^2 -cy^2 )dy And my physics teacher says it is (or can be) a harmonic function (Δψ = 0) Can anyone explain ?

$$\mathrm{Does}\:\mathrm{anyone}\:\mathrm{know}\:\mathrm{how}\:\mathrm{this}\:\mathrm{works}\:? \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{have}\:{d}\psi\:=\:\left({x}^{\mathrm{2}} -{cy}^{\mathrm{2}} \right){dy} \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{my}\:\mathrm{physics}\:\mathrm{teacher}\:\mathrm{says}\:\mathrm{it}\:\mathrm{is}\:\left(\mathrm{or}\:\mathrm{can}\right. \\ $$$$\left.\mathrm{be}\right)\:\mathrm{a}\:\mathrm{harmonic}\:\mathrm{function}\:\left(\Delta\psi\:=\:\mathrm{0}\right) \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{explain}\:? \\ $$

Question Number 206269    Answers: 3   Comments: 2

Question Number 206267    Answers: 0   Comments: 1

Question Number 206253    Answers: 2   Comments: 0

f(x)= log_( 2) ( x + 2(√x) +4 ) ⇒ f^( −1) ( 13 −4(√3) ) = ? −−−−−

$$\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:{log}_{\:\mathrm{2}} \:\left(\:{x}\:+\:\mathrm{2}\sqrt{{x}}\:+\mathrm{4}\:\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:{f}^{\:−\mathrm{1}} \left(\:\mathrm{13}\:−\mathrm{4}\sqrt{\mathrm{3}}\:\right)\:=\:? \\ $$$$\:\:\:\:\:\:\:−−−−− \\ $$$$\:\:\:\: \\ $$

Question Number 206251    Answers: 1   Comments: 0

3x^2 −12x−5(√(x^2 −4x−1))−5=0

$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}−\mathrm{5}=\mathrm{0} \\ $$

Question Number 206248    Answers: 1   Comments: 0

∫(1/( (√((1−t)(2−t)))))dt=...?

$$\int\frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{t}\right)\left(\mathrm{2}−{t}\right)}}{dt}=...? \\ $$

Question Number 206244    Answers: 1   Comments: 0

ζ

$$\:\:\:\zeta \\ $$

Question Number 206232    Answers: 4   Comments: 0

Question Number 206230    Answers: 1   Comments: 0

Expand x^2 + 2x + 3 respect to x = −2. (a) (x − 2)^2 −2(x + 2) + 3 (b) (x + 2)^2 −2(x + 2) + 3 (c) (x + 2)^2 + 2(x + 2) + 3 (d) (x − 2)^2 −2(x − 2) − 3 is it taylors theorem?

$$\mathrm{Expand}\:\:\:\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{3}\:\:\:{respect}\:{to}\:{x}\:=\:−\mathrm{2}. \\ $$$$\left(\mathrm{a}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{b}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{c}\right)\:\left({x}\:+\:\mathrm{2}\right)^{\mathrm{2}} +\:\mathrm{2}\left({x}\:+\:\mathrm{2}\right)\:+\:\mathrm{3} \\ $$$$\left(\mathrm{d}\right)\:\left({x}\:−\:\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\left({x}\:−\:\mathrm{2}\right)\:−\:\mathrm{3} \\ $$$$ \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{taylors}\:\mathrm{theorem}? \\ $$

Question Number 206227    Answers: 1   Comments: 0

OA=(4^x ) OB=_7 ^5 and AB=5 units

$${OA}=\left(\overset{{x}} {\mathrm{4}}\right)\:{OB}=_{\mathrm{7}} ^{\mathrm{5}} \:{and}\:{AB}=\mathrm{5}\:{units} \\ $$

Question Number 206224    Answers: 3   Comments: 0

find ∫_0 ^1 arctan(x^5 )dx

$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{5}} \right){dx} \\ $$

Question Number 206212    Answers: 2   Comments: 4

Question Number 206216    Answers: 3   Comments: 0

Question Number 206198    Answers: 2   Comments: 0

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