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Question Number 175174 Answers: 0 Comments: 0

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if x is a real number in [0,1] then the value of lim_(m→∞ ) lim_(n→∞) [1+cos^(2m) (n!πx)]

$$\:\:\:\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\:\:\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\underset{{m}\rightarrow\infty\:} {\mathrm{lim}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{1}+\mathrm{cos}^{\mathrm{2}{m}} \left({n}!\pi{x}\right)\right]\: \\ $$

Question Number 175137 Answers: 1 Comments: 0

lim_(x→2) ((((5x−9))^(1/5) ((4x−7))^(1/4) ((3x−5))^(1/3) (√(2x−3))−1)/(x−2))=?

$$\:\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{5}}]{\mathrm{5}{x}−\mathrm{9}}\:\sqrt[{\mathrm{4}}]{\mathrm{4}{x}−\mathrm{7}}\:\sqrt[{\mathrm{3}}]{\mathrm{3}{x}−\mathrm{5}}\:\sqrt{\mathrm{2}{x}−\mathrm{3}}−\mathrm{1}}{{x}−\mathrm{2}}=? \\ $$

Question Number 175132 Answers: 0 Comments: 4

≺ Question− algebra ≻ Count the number of ” zero divisor ” of ring , ( Z_( 45) , , ⊕ ) ■ m.n −−−−−−−−

$$ \\ $$$$\:\:\:\:\:\:\prec\:\:\boldsymbol{{Question}}−\:\boldsymbol{{algebra}}\:\succ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{Count}}\:\boldsymbol{{the}}\:\boldsymbol{{number}}\:\boldsymbol{{of}}\:\:\:''\:\boldsymbol{{zero}}\:\boldsymbol{{divisor}}\:'' \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{of}}\:\:\boldsymbol{{ring}}\:,\:\:\:\left(\:\mathbb{Z}_{\:\mathrm{45}} \:,\: \:,\:\oplus\:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:\boldsymbol{{m}}.\boldsymbol{{n}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−\:\:\:\: \\ $$

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Prove that determinant ((1,1,1),(a,b,c),(a^2 ,b^2 ,c^2 ))= (a−b)(b−c)(c−a)

$$\mathrm{Prove}\:\mathrm{that}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{{a}}&{{b}}&{{c}}\\{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\end{vmatrix}=\:\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\ $$

Question Number 175104 Answers: 1 Comments: 0

log _3 (a+1)=log _4 (a+8) a=?

$$\:\:\mathrm{log}\:_{\mathrm{3}} \left({a}+\mathrm{1}\right)=\mathrm{log}\:_{\mathrm{4}} \left({a}+\mathrm{8}\right) \\ $$$$\:{a}=? \\ $$

Question Number 175103 Answers: 1 Comments: 0

If 20 (mod 9) is equivalent to y (mod 6). Find y

$$\mathrm{If}\:\:\mathrm{20}\:\left(\mathrm{mod}\:\mathrm{9}\right)\:\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\:\:\mathrm{y}\:\left(\mathrm{mod}\:\mathrm{6}\right).\:\:\mathrm{Find}\:\:\mathrm{y} \\ $$

Question Number 175095 Answers: 1 Comments: 0

x^3 +x^2 +x+c=0 find x

$$\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$

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how the next number 1,2,5,9,6,13,8,17,....

$${how}\:{the}\:{next}\:{number}\:\mathrm{1},\mathrm{2},\mathrm{5},\mathrm{9},\mathrm{6},\mathrm{13},\mathrm{8},\mathrm{17},.... \\ $$

Question Number 175081 Answers: 0 Comments: 0

If in △ABC , m(∢B)=3m(∢C) then: a = (b − c) (√((b/c) + 1)) , b > a + (c/2) R = (√(c^3 /(3c − b)))

$$\mathrm{If}\:\mathrm{in}\:\bigtriangleup\mathrm{ABC}\:,\:\mathrm{m}\left(\sphericalangle\mathrm{B}\right)=\mathrm{3m}\left(\sphericalangle\mathrm{C}\right)\:\mathrm{then}: \\ $$$$\mathrm{a}\:=\:\left(\mathrm{b}\:−\:\mathrm{c}\right)\:\sqrt{\frac{\mathrm{b}}{\mathrm{c}}\:+\:\mathrm{1}}\:\:,\:\:\mathrm{b}\:>\:\mathrm{a}\:+\:\frac{\mathrm{c}}{\mathrm{2}} \\ $$$$\mathrm{R}\:=\:\sqrt{\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{3c}\:−\:\mathrm{b}}} \\ $$

Question Number 175080 Answers: 0 Comments: 0

calculate .. lim_( x→ 0^( +) ) ∫_0 ^( (𝛑/2)) J_0 ( x.cos(∅ )).cos(∅)d∅ = ? where : J_𝛎 ( x )= x^( v) .Σ_(n=0) ^∞ (((−1 )^( n) .x^( 2n) )/(2^( 2n+v) .𝚪 (n + 𝛎 +1 )))

$$ \\ $$$$\:\:\:\boldsymbol{{calculate}}\:.. \\ $$$$\:\boldsymbol{{lim}}_{\:\boldsymbol{{x}}\rightarrow\:\mathrm{0}^{\:+} } \int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \boldsymbol{{J}}_{\mathrm{0}} \:\left(\:\boldsymbol{{x}}.\boldsymbol{{cos}}\left(\boldsymbol{\emptyset}\:\right)\right).\boldsymbol{{cos}}\left(\boldsymbol{\emptyset}\right)\boldsymbol{{d}\emptyset}\:=\:? \\ $$$$\:\:\:\boldsymbol{{where}}\:: \\ $$$$\:\:\:\:\:\boldsymbol{{J}}_{\boldsymbol{\nu}} \:\left(\:\boldsymbol{{x}}\:\right)=\:\boldsymbol{{x}}^{\:\boldsymbol{{v}}} .\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\:\right)^{\:\boldsymbol{{n}}} .\boldsymbol{{x}}^{\:\mathrm{2}\boldsymbol{{n}}} }{\mathrm{2}^{\:\mathrm{2}\boldsymbol{{n}}+\boldsymbol{{v}}} .\boldsymbol{\Gamma}\:\left(\boldsymbol{{n}}\:+\:\boldsymbol{\nu}\:+\mathrm{1}\:\right)} \\ $$

Question Number 175077 Answers: 1 Comments: 0

Show that ∫_0 ^1 ((x^b −x^a )/(lnx))=ln(((b+1)/(a+1)))

$${Show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{{lnx}}={ln}\left(\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\right) \\ $$

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