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Question Number 71134 Answers: 1 Comments: 0

if a, b and c are positive, find: ((a/b))^(log_(10) c) ×((b/c))^(log_(10) a) ×((c/a))^(log_(10) b)

$${if}\:{a},\:{b}\:{and}\:{c}\:{are}\:{positive},\:{find}: \\ $$$$\left(\frac{{a}}{{b}}\right)^{{log}_{\mathrm{10}} {c}} ×\left(\frac{{b}}{{c}}\right)^{{log}_{\mathrm{10}} {a}} ×\left(\frac{{c}}{{a}}\right)^{{log}_{\mathrm{10}} {b}} \\ $$$$ \\ $$

Question Number 71117 Answers: 0 Comments: 0

Question Number 71114 Answers: 0 Comments: 0

Question Number 71096 Answers: 1 Comments: 0

Question Number 71091 Answers: 0 Comments: 3

Question Number 71115 Answers: 1 Comments: 0

Question Number 71089 Answers: 1 Comments: 0

((√((√2)−1)))^x +((√((√2)+1)))^x =4

$$\left(\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}\right)^{\mathrm{x}} +\left(\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}\right)^{\mathrm{x}} =\mathrm{4} \\ $$

Question Number 71082 Answers: 2 Comments: 0

prove that ∣ (√(∣x∣)) − (√(∣y∣)) ∣ ≤ (√(∣x−y∣))

$${prove}\:{that}\: \\ $$$$ \\ $$$$\mid\:\sqrt{\mid{x}\mid}\:−\:\sqrt{\mid{y}\mid}\:\mid\:\leqslant\:\sqrt{\mid{x}−{y}\mid}\: \\ $$$$ \\ $$

Question Number 71073 Answers: 3 Comments: 4

Question Number 71054 Answers: 0 Comments: 2

Question Number 71053 Answers: 1 Comments: 1

Question Number 71052 Answers: 0 Comments: 2

Question Number 71051 Answers: 0 Comments: 2

Question Number 71050 Answers: 1 Comments: 1

Question Number 71049 Answers: 0 Comments: 2

Question Number 71047 Answers: 0 Comments: 1

Question Number 71046 Answers: 0 Comments: 1

Question Number 71045 Answers: 0 Comments: 1

Question Number 71034 Answers: 0 Comments: 6

Question Number 71016 Answers: 2 Comments: 0

ln (e+ln (e+ln (e+...)))=?

$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+...\right)\right)\right)=? \\ $$

Question Number 71014 Answers: 0 Comments: 0

calculate ∫_0 ^∞ e^(−(x^2 +(1/x^2 ))) dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx} \\ $$

Question Number 71000 Answers: 0 Comments: 1

Where f(x)=((4x^2 −1)/(2x−1)) defined in R−{(1/2)}, determine lim_(x→(1/2)) f(x).

$${Where}\:{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\:{defined}\:{in}\: \\ $$$$\mathbb{R}−\left\{\frac{\mathrm{1}}{\mathrm{2}}\right\},\:{determine}\:\underset{{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}} {\mathrm{lim}}{f}\left({x}\right). \\ $$

Question Number 70997 Answers: 1 Comments: 0

Question Number 70996 Answers: 0 Comments: 2

Question Number 70980 Answers: 0 Comments: 0

Soit (E,A,μ) un espace mesure . On suppose qu′il existe un X∈A tel μ(X)=+∞ 1)Montrer que si μ est semi-finie alors ∀ r>0 il existe B⊆X tel que r<μ(B)< +∞

$$\:{Soit}\:\left({E},\mathcal{A},\mu\right)\:{un}\:\:{espace}\:{mesure}\:\:.\:{On}\:{suppose} \\ $$$${qu}'{il}\:{existe}\:{un}\:{X}\in\mathcal{A}\:\:{tel}\:\:\mu\left({X}\right)=+\infty \\ $$$$\left.\mathrm{1}\right){Montrer}\:{que}\:{si}\:\:\mu\:{est}\:{semi}-{finie}\:\:{alors} \\ $$$$\forall\:{r}>\mathrm{0}\:\:{il}\:{existe}\:\:{B}\subseteq{X}\:{tel}\:{que}\:\:{r}<\mu\left({B}\right)<\:+\infty \\ $$$$ \\ $$

Question Number 70974 Answers: 0 Comments: 1

Password reset changes In case you forget any password set you can simple reset by going to set/update password. Leave old password field blank. This will work only if you are trying to reset when already logged in.

$$\boldsymbol{\mathrm{Password}}\:\boldsymbol{\mathrm{reset}}\:\boldsymbol{\mathrm{changes}} \\ $$$$\mathrm{In}\:\mathrm{case}\:\mathrm{you}\:\mathrm{forget}\:\mathrm{any}\:\mathrm{password}\:\mathrm{set} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{simple}\:\mathrm{reset}\:\mathrm{by}\:\mathrm{going}\:\mathrm{to} \\ $$$$\mathrm{set}/\mathrm{update}\:\mathrm{password}. \\ $$$$ \\ $$$$\mathrm{Leave}\:\mathrm{old}\:\mathrm{password}\:\mathrm{field}\:\mathrm{blank}. \\ $$$$ \\ $$$$\mathrm{This}\:\mathrm{will}\:\mathrm{work}\:\mathrm{only}\:\mathrm{if}\:\mathrm{you}\:\mathrm{are}\:\mathrm{trying} \\ $$$$\mathrm{to}\:\mathrm{reset}\:\mathrm{when}\:\mathrm{already}\:\mathrm{logged}\:\mathrm{in}. \\ $$

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