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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 Solve for value of x AS SIMPLE AS POSSIBLE.

$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{1}}−\mathrm{5}=\mathrm{0} \\ $$$${Solve}\:{for}\:{value}\:{of}\:{x}\:{AS}\:{SIMPLE}\:{AS}\:{POSSIBLE}. \\ $$

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

_1 : _2 = .... (A) 1 : (2)^(1/3) (B) (2)^(1/3) : 1 (C) 1 : (√2) (D) (√2) : 1 (E) 1 : (√3)

$$\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ _{\mathrm{1}} \::\: _{\mathrm{2}} \:=\:.... \\ $$$$\:\:\left({A}\right)\:\mathrm{1}\::\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\:\:\:\left({B}\right)\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\::\:\mathrm{1}\:\:\:\:\:\left({C}\right)\:\mathrm{1}\::\:\sqrt{\mathrm{2}} \\ $$$$\:\:\left({D}\right)\:\sqrt{\mathrm{2}}\::\:\mathrm{1}\:\:\:\:\left({E}\right)\:\mathrm{1}\::\:\sqrt{\mathrm{3}} \\ $$

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