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Question Number 192999 Answers: 1 Comments: 0

Show that the following functions are continous on a close interval [0, 1]. f(x)={_(3 x=1) ^(((x^2 +x−2)/(x−1)) x≠1) Help!

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{functions} \\ $$$$\mathrm{are}\:\mathrm{continous}\:\mathrm{on}\:\mathrm{a}\:\mathrm{close}\:\mathrm{interval} \\ $$$$\left[\mathrm{0},\:\mathrm{1}\right]. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left\{_{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}} ^{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{2}}{\mathrm{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\neq\mathrm{1}} \right. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 192992 Answers: 1 Comments: 0

{ ((2x+3y≡1(mod26))),((7x+8y≡2(mod26))) :}

$$\begin{cases}{\mathrm{2x}+\mathrm{3y}\equiv\mathrm{1}\left(\mathrm{mod26}\right)}\\{\mathrm{7x}+\mathrm{8y}\equiv\mathrm{2}\left(\mathrm{mod26}\right)}\end{cases} \\ $$$$ \\ $$

Question Number 192991 Answers: 1 Comments: 0

the first, third and sixth terms of a linear sequence are the first three terms of an exponential sequence. find the common ratio

$${the}\:{first},\:{third}\:{and}\:{sixth}\:{terms}\:{of}\:{a} \\ $$$${linear}\:{sequence}\:{are}\:{the}\:{first}\:{three}\: \\ $$$${terms}\:{of}\:{an}\:{exponential}\:{sequence}.\: \\ $$$${find}\:{the}\:{common}\:{ratio} \\ $$

Question Number 192990 Answers: 1 Comments: 0

A fair die is tossed four times .what is the probability of obtaining a prime each time

$${A}\:{fair}\:{die}\:{is}\:{tossed}\:{four}\:{times}\:.{what} \\ $$$${is}\:{the}\:{probability}\:{of}\:{obtaining}\:{a}\: \\ $$$${prime}\:{each}\:{time} \\ $$

Question Number 192988 Answers: 2 Comments: 0

Question Number 192987 Answers: 1 Comments: 0

Question Number 192985 Answers: 1 Comments: 0

prove it : lim_(n→∞) Π_(i=1) ^n cos(θ/2^i )=((sinθ)/θ) then show : im_(n→∞) cos(π/4)cos(π/8)...cos(π/2^(n+1) ) =(2/π)

$${prove}\:{it}\:: \\ $$$${lim}_{{n}\rightarrow\infty} \:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\frac{\theta}{\mathrm{2}^{{i}} }=\frac{{sin}\theta}{\theta} \\ $$$${then}\:{show}\:: \\ $$$${im}_{{n}\rightarrow\infty} \:{cos}\frac{\pi}{\mathrm{4}}{cos}\frac{\pi}{\mathrm{8}}...{cos}\frac{\pi}{\mathrm{2}^{{n}+\mathrm{1}} }\:=\frac{\mathrm{2}}{\pi} \\ $$$$ \\ $$

Question Number 192979 Answers: 0 Comments: 0

Can this be optimized (getting the minimum) using backprobagation? α(x_i ,y_i ,h_i )=(h_i −x_i )^2 +y_i ^2 β(y_i ,h_i )=y_i h_i ^2 −2uy_i h_i γ(h_i )=h_i ^4 −4uh_i ^3 +4u^2 h_i ^2 Cost=Σ_(i=0) ^m (cα(x_i ,y_i ,h_i )−2c^2 β(y_i ,h_i )+c^3 γ(h_i )) and how?

$$\mathrm{Can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{optimized}\:\left(\mathrm{getting}\:\mathrm{the}\:\mathrm{minimum}\right)\:\mathrm{using}\:\mathrm{backprobagation}? \\ $$$$ \\ $$$$\alpha\left({x}_{{i}} ,{y}_{{i}} ,{h}_{{i}} \right)=\left({h}_{{i}} −{x}_{{i}} \right)^{\mathrm{2}} +{y}_{{i}} ^{\mathrm{2}} \\ $$$$\beta\left({y}_{{i}} ,{h}_{{i}} \right)={y}_{{i}} {h}_{{i}} ^{\mathrm{2}} −\mathrm{2}{uy}_{{i}} {h}_{{i}} \\ $$$$\gamma\left({h}_{{i}} \right)={h}_{{i}} ^{\mathrm{4}} −\mathrm{4}{uh}_{{i}} ^{\mathrm{3}} +\mathrm{4}{u}^{\mathrm{2}} {h}_{{i}} ^{\mathrm{2}} \\ $$$$\mathrm{Cost}=\underset{{i}=\mathrm{0}} {\overset{{m}} {\sum}}\left(\mathrm{c}\alpha\left({x}_{{i}} ,{y}_{{i}} ,{h}_{{i}} \right)−\mathrm{2}{c}^{\mathrm{2}} \beta\left({y}_{{i}} ,{h}_{{i}} \right)+{c}^{\mathrm{3}} \gamma\left({h}_{{i}} \right)\right) \\ $$$$\mathrm{and}\:\mathrm{how}? \\ $$

Question Number 192969 Answers: 2 Comments: 1