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Question Number 175132 Answers: 0 Comments: 4

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

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

Question Number 175122 Answers: 1 Comments: 0

Question Number 175121 Answers: 2 Comments: 0

Question Number 175115 Answers: 2 Comments: 0

Question Number 175108 Answers: 1 Comments: 0

Question Number 175110 Answers: 2 Comments: 3

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

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

Question Number 175085 Answers: 0 Comments: 1

Question Number 175083 Answers: 0 Comments: 2

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)))

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

Question Number 175066 Answers: 1 Comments: 0

Question Number 175068 Answers: 0 Comments: 0

Question Number 175061 Answers: 1 Comments: 1

Question Number 175059 Answers: 1 Comments: 2

Question Number 175053 Answers: 0 Comments: 2

Question Number 175051 Answers: 0 Comments: 2

find the value of ∫_0 ^∞ ((arctanx)/((x^2 +1)^3 ))dx

$${find}\:{the}\:{value}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{arctanx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 175050 Answers: 0 Comments: 0

Question Number 175049 Answers: 1 Comments: 2

Z_( p) , is a field ...( p is prime )

$$ \\ $$$$\:\:\:\:\:\:\mathbb{Z}_{\:{p}} \:,\:{is}\:\:{a}\:\:{field}\:...\left(\:\:{p}\:{is}\:{prime}\:\right) \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 175045 Answers: 0 Comments: 5

In the figure ABCD is an square and BM=MC. If the area of PCD=14u. What is the value of: (1) Area of ABC; (2) Area of ABMP; (3) The area of ABCD

$${In}\:{the}\:{figure}\:{ABCD}\:{is}\:{an}\:{square}\:{and} \\ $$$${BM}={MC}.\:{If}\:{the}\:{area}\:{of}\:{PCD}=\mathrm{14}{u}.\: \\ $$$${What}\:{is}\:{the}\:{value}\:{of}: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{Area}\:{of}\:{ABC}; \\ $$$$\left(\mathrm{2}\right)\:{Area}\:{of}\:\:{ABMP}; \\ $$$$\left(\mathrm{3}\right)\:{The}\:{area}\:{of}\:{ABCD} \\ $$$$ \\ $$

Question Number 175042 Answers: 1 Comments: 0

Question Number 175041 Answers: 0 Comments: 0

Solve the differential (dy/dx)=((x^2 −3xy)/(x+y))

$${Solve}\:{the}\:{differential} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} −\mathrm{3}{xy}}{{x}+{y}} \\ $$

Question Number 175040 Answers: 1 Comments: 0

hello, please, someone help me to correct the equation? It's typed wrong and I can't find where Solve: 5,76[((log_a ((√(log _b ((√a))))))/(log((√(log(a)))))) + log_(log (a)) (2)]((log _2 (x)))^(1/5) + ((log_2 (x))/(25)) = [log _2 (x)]^(3/5) Answers x_1 =1 , x_2 =2^(243) , x_3 =2^(−243) , x_4 =2^(1024) , x_5 =2^(−1024)

$$ \\ $$hello, please, someone help me to correct the equation? It's typed wrong and I can't find where $${Solve}: \\ $$$$\mathrm{5},\mathrm{76}\left[\frac{\mathrm{log}_{{a}} \left(\sqrt{\mathrm{log}\:_{{b}} \left(\sqrt{{a}}\right)}\right)}{\mathrm{log}\left(\sqrt{\mathrm{log}\left({a}\right)}\right)}\:+\:\mathrm{log}_{\mathrm{log}\:\left({a}\right)} \left(\mathrm{2}\right)\right]\sqrt[{\mathrm{5}}]{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}_{\mathrm{2}} \left({x}\right)}{\mathrm{25}}\:=\:\left[\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\right]^{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$${Answers} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:,\:{x}_{\mathrm{2}} =\mathrm{2}^{\mathrm{243}} \:,\:{x}_{\mathrm{3}} =\mathrm{2}^{−\mathrm{243}} \:,\:{x}_{\mathrm{4}} =\mathrm{2}^{\mathrm{1024}} \:,\:{x}_{\mathrm{5}} =\mathrm{2}^{−\mathrm{1024}} \\ $$

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