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Question Number 174928 Answers: 3 Comments: 0

How many digits does 1000^(1000) have? Mastermind

$$\mathrm{How}\:\mathrm{many}\:\mathrm{digits}\:\mathrm{does}\:\mathrm{1000}^{\mathrm{1000}} \:\mathrm{have}? \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$

Question Number 174915 Answers: 1 Comments: 0

Find the value of a for which the limit lim_(x→0) ((sin (ax)−arctan x−x)/(x^3 +x^4 )) is finite and then evaluate the limit

$$\:{Find}\:{the}\:{value}\:{of}\:{a}\:{for}\:{which}\: \\ $$$$\:{the}\:{limit}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({ax}\right)−\mathrm{arctan}\:{x}−{x}}{{x}^{\mathrm{3}} +{x}^{\mathrm{4}} } \\ $$$${is}\:{finite}\:{and}\:{then}\:{evaluate}\:{the}\:{limit} \\ $$

Question Number 174939 Answers: 1 Comments: 0

sum of the all proper factors of 360 ?

$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{all}\:\mathrm{proper}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{360}\:? \\ $$

Question Number 174938 Answers: 0 Comments: 0

let u_n = ∫_0 ^1 x^n artan(nx)dx 1)lim u_n ? 2)nature of Σ u_n 3) calculate u_n 4)equivalent of u_n ?

$${let}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} {artan}\left({nx}\right){dx} \\ $$$$\left.\mathrm{1}\right){lim}\:{u}_{{n}} ? \\ $$$$\left.\mathrm{2}\right){nature}\:{of}\:\Sigma\:{u}_{{n}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{u}_{{n}} \\ $$$$\left.\mathrm{4}\right){equivalent}\:{of}\:{u}_{{n}} ? \\ $$

Question Number 174940 Answers: 0 Comments: 1

f(x)=(log3^x −2log3).(x^2 −1) let

$${f}\left({x}\right)=\left({log}\mathrm{3}^{{x}} −\mathrm{2}{log}\mathrm{3}\right).\left({x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${let} \\ $$

Question Number 174902 Answers: 1 Comments: 0

Question Number 174896 Answers: 0 Comments: 0

prove that ∫_(−∞) ^( ∞) ((x csch((x/2)))/(x^2 +π^2 ))dx=π−2

$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\int_{−\infty} ^{\:\infty} \frac{\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{csch}}\left(\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\right)}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\pi^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\pi−\mathrm{2} \\ $$

Question Number 174888 Answers: 1 Comments: 0

Question Number 174884 Answers: 1 Comments: 0

Question Number 174883 Answers: 2 Comments: 0

lim_(n→∞) ∫_0 ^1 e^x^2 sin(nx)dx

$$\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } \mathrm{sin}\left({nx}\right){dx}\: \\ $$

Question Number 174879 Answers: 0 Comments: 7

Q : (an E lementary to abstract algebra ) prove that the order of an element in” quotient group ” (Q , ⊕)/(Z , ⊕) is finite. Notice: (Q , ⊕)/(Z , ⊕) = { (a/b) + Z ∣ a,b ∈ Z , b ≠0 } ≻ Source: John B .F raleigh book ≺

$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Q}}\::\:\left(\boldsymbol{{an}}\:\:\mathscr{E}\:\boldsymbol{{lementary}}\:\boldsymbol{{to}}\:\boldsymbol{{abstract}}\:\boldsymbol{{algebra}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\:\boldsymbol{{order}}\:\:\boldsymbol{{of}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{an}}\:\:\boldsymbol{{element}}\:\:\boldsymbol{{in}}''\:\boldsymbol{{quotient}}\:\boldsymbol{{group}}\:''\:\left(\mathbb{Q}\:,\:\oplus\right)/\left(\mathbb{Z}\:,\:\oplus\right)\:\boldsymbol{{is}}\:\boldsymbol{{finite}}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Notice}}:\:\:\left(\mathbb{Q}\:,\:\oplus\right)/\left(\mathbb{Z}\:,\:\oplus\right)\:=\:\left\{\:\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\:+\:\mathbb{Z}\:\mid\:\:\boldsymbol{{a}},\boldsymbol{{b}}\:\in\:\mathbb{Z}\:,\:\boldsymbol{{b}}\:\neq\mathrm{0}\:\right\}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\succ\:\boldsymbol{{Source}}:\:\boldsymbol{{John}}\:\boldsymbol{{B}}\:.\boldsymbol{{F}}\:\boldsymbol{{raleigh}}\:\boldsymbol{{book}}\:\prec\:\:\:\: \\ $$$$ \\ $$

Question Number 174878 Answers: 0 Comments: 0

Question Number 174877 Answers: 0 Comments: 0

Question Number 174876 Answers: 1 Comments: 0

let f:[0,1]→R be a continuous function ditermine (with appropriate justification) the following limit: lim_(n→∞) ∫_0 ^1 nx^n f(x)dx

$$\:\:\mathrm{let}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:{be}\:{a}\:\mathrm{continuous} \\ $$$$\:\mathrm{function}\:\mathrm{ditermine}\:\left(\mathrm{with}\:\mathrm{appropriate}\right. \\ $$$$\left.\mathrm{justification}\right)\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\:\:\mathrm{limit}:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {nx}^{{n}} {f}\left({x}\right){dx} \\ $$

Question Number 174871 Answers: 1 Comments: 0

Question Number 174863 Answers: 1 Comments: 3

Question Number 174859 Answers: 1 Comments: 1

Question Number 174855 Answers: 1 Comments: 0

Find the number of ways a committee of 4 people can be chosen from a group of 5 men and 7 women when it contains people of both sexes and there are at least as many women as men.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\: \\ $$$$\mathrm{a}\:\mathrm{committee}\:\mathrm{of}\:\mathrm{4}\:\mathrm{people}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{from}\:\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{5}\:\mathrm{men} \\ $$$$\mathrm{and}\:\mathrm{7}\:\mathrm{women}\:\mathrm{when}\:\mathrm{it}\:\mathrm{contains} \\ $$$$\mathrm{people}\:\mathrm{of}\:\mathrm{both}\:\mathrm{sexes}\:\mathrm{and} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{as}\:\mathrm{many} \\ $$$$\mathrm{women}\:\mathrm{as}\:\mathrm{men}. \\ $$

Question Number 174853 Answers: 0 Comments: 2

By first principle, solve cos^2 x + sin^2 x=1

$$\mathrm{By}\:\mathrm{first}\:\mathrm{principle},\:\mathrm{solve}\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{x}=\mathrm{1} \\ $$

Question Number 174844 Answers: 1 Comments: 1

Question Number 174849 Answers: 0 Comments: 0

Question Number 174841 Answers: 1 Comments: 0

lim_(x→0) ((xsin(sinx)−sin^2 x )/x^6 )

$$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\mathrm{sin}\left(\mathrm{sin}{x}\right)−\mathrm{sin}^{\mathrm{2}} {x}\:\:}{{x}^{\mathrm{6}} } \\ $$

Question Number 174838 Answers: 2 Comments: 0

calculate Ω = ∫_0 ^( (π/2)) (( x^( 2) )/(( sin(x)+cos(x))^( 2) ))dx=?

$$ \\ $$$$\:\:\:\:\:{calculate} \\ $$$$ \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:{x}^{\:\mathrm{2}} }{\left(\:{sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$

Question Number 174837 Answers: 0 Comments: 0

sin(A)+ sin(B )+ sin(C)≤ ((3(√3))/2) ≻ Solution ≺ ( A+B +C =π ) l.h.s = 2sin(((A+B)/2) )cos(((A−B)/2))+2sin((C/2))cos((C/2)) = 2cos ((C/2)){2cos(A/2) .cos((B/2) )} = 4cos((A/2)).cos((B/2)).cos((C/2)) l.h.s ≤_(post) ^(previoue) 4 (((3(√3))/8) )=((3(√3))/2) note: { (( I : cos(a)+cos(b)=2cos(((a+b)/2))cos(((a−b)/2)))),(( II: sin(a)+sin(b)=2sin(((a+b)/2))cos(((a−b)/2) ))),(( { (( cos((π/2) −α)=sin(α))),(( sin((π/2) −α)= cos(α))) :})) :}

$$ \\ $$$$\:\:\:\:\:{sin}\left({A}\right)+\:{sin}\left({B}\:\right)+\:{sin}\left({C}\right)\leqslant\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\succ\:\:\:{Solution}\:\prec\:\:\:\:\:\:\left(\:{A}+{B}\:+{C}\:=\pi\:\right) \\ $$$$\:\:\:\:\:\:{l}.{h}.{s}\:=\:\mathrm{2}{sin}\left(\frac{{A}+{B}}{\mathrm{2}}\:\right){cos}\left(\frac{{A}−{B}}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{{C}}{\mathrm{2}}\right){cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left\{\mathrm{2}{cos}\frac{{A}}{\mathrm{2}}\:.{cos}\left(\frac{{B}}{\mathrm{2}}\:\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}{cos}\left(\frac{{A}}{\mathrm{2}}\right).{cos}\left(\frac{{B}}{\mathrm{2}}\right).{cos}\left(\frac{{C}}{\mathrm{2}}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:{l}.{h}.{s}\:\underset{{post}} {\overset{{previoue}} {\leqslant}}\:\mathrm{4}\:\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\right)=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\begin{cases}{\:{I}\::\:\:{cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right)}\\{\:{II}:\:{sin}\left({a}\right)+{sin}\left({b}\right)=\mathrm{2}{sin}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\:\right)}\\{\:\:\:\begin{cases}{\:{cos}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)={sin}\left(\alpha\right)}\\{\:\:{sin}\left(\frac{\pi}{\mathrm{2}}\:−\alpha\right)=\:{cos}\left(\alpha\right)}\end{cases}}\end{cases} \\ $$

Question Number 174823 Answers: 1 Comments: 1

Question Number 174816 Answers: 3 Comments: 1

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