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Question Number 212445 Answers: 1 Comments: 1

Question Number 212444 Answers: 0 Comments: 0

Question Number 212435 Answers: 1 Comments: 0

Question Number 212432 Answers: 2 Comments: 0

(9/(2∙4)) + (9/(4∙6)) +...+ (9/(4n∙(n + 1))) = ((15)/8) Find: n = ?

$$\frac{\mathrm{9}}{\mathrm{2}\centerdot\mathrm{4}}\:+\:\frac{\mathrm{9}}{\mathrm{4}\centerdot\mathrm{6}}\:+...+\:\frac{\mathrm{9}}{\mathrm{4n}\centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{15}}{\mathrm{8}} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{n}}\:=\:? \\ $$

Question Number 212430 Answers: 0 Comments: 0

Question Number 212428 Answers: 3 Comments: 0

Question Number 212427 Answers: 0 Comments: 1

Question Number 212424 Answers: 1 Comments: 0

{ ((x + 2y − 3z = 1)),((2x − y + z = 4)),((3x + y + 2z = 7)) :} find: x,y,z = ?

$$\begin{cases}{\mathrm{x}\:+\:\mathrm{2y}\:−\:\mathrm{3z}\:=\:\mathrm{1}}\\{\mathrm{2x}\:−\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{4}}\\{\mathrm{3x}\:+\:\mathrm{y}\:+\:\mathrm{2z}\:=\:\mathrm{7}}\end{cases}\:\:\:\:\:\mathrm{find}:\:\mathrm{x},\mathrm{y},\mathrm{z}\:=\:? \\ $$

Question Number 212417 Answers: 1 Comments: 1

Question Number 212416 Answers: 0 Comments: 0

((((∫_0 ^(+∞) e^(−s) s^5 ds)/2)+((∫_(−∞) ^(+∞) e^(−(t^2 /2)) dt)/(∫_0 ^(+∞) sint^2 dt))(((Σ_(n=0) ^∞ (((−1)^n )/(2n+1)))/(∫_0 ^(+∞) ((sin x)/x)dx))+((Σ_(n=1) ^∞ arctan(2/n^2 ))/(lim_(t→0^+ ) ∫_(−2020) ^(2020) ((tcos x)/(x^2 +t^2 ))dx))))/(lim_(n→∞) {[(∫_0 ^1 (x^(n−1) /(1+x))dx)n−(1/2)](n/2)}))

$$\frac{\frac{\int_{\mathrm{0}} ^{+\infty} {e}^{−{s}} {s}^{\mathrm{5}} {ds}}{\mathrm{2}}+\frac{\int_{−\infty} ^{+\infty} {e}^{−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} {dt}}{\int_{\mathrm{0}} ^{+\infty} \mathrm{sin}{t}^{\mathrm{2}} {dt}}\left(\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}}{\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{sin}\:{x}}{{x}}{dx}}+\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{2}}{{n}^{\mathrm{2}} }}{\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\int_{−\mathrm{2020}} ^{\mathrm{2020}} \frac{{t}\mathrm{cos}\:{x}}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dx}}\right)}{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\left[\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}\right){n}−\frac{\mathrm{1}}{\mathrm{2}}\right]\frac{{n}}{\mathrm{2}}\right\}} \\ $$

Question Number 212415 Answers: 1 Comments: 0

lim_(n→∞) Σ_(k=1) ^n (2^(k/n) /(n+(1/k)))

$$\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}^{\frac{{k}}{{n}}} }{{n}+\frac{\mathrm{1}}{{k}}} \\ $$$$ \\ $$

Question Number 212377 Answers: 0 Comments: 0

Question Number 212384 Answers: 0 Comments: 1

Reponse a la question N° Q212291 S(ABCD)= 66,69

$$\mathrm{Reponse}\:\mathrm{a}\:\mathrm{la}\:\mathrm{question}\:\mathrm{N}° \\ $$$$\mathrm{Q212291}\:\:\:\:\:\:\mathrm{S}\left(\mathrm{ABCD}\right)=\:\:\mathrm{66},\mathrm{69} \\ $$

Question Number 212368 Answers: 1 Comments: 0

Question Number 212405 Answers: 1 Comments: 0

lim_(x→3) ((e^x −e^3 )/(x−3))=?

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{{e}^{{x}} −{e}^{\mathrm{3}} }{{x}−\mathrm{3}}=? \\ $$

Question Number 212361 Answers: 1 Comments: 2

Question Number 212360 Answers: 1 Comments: 0

T=Σ_(n=0) ^∞ ((1/((4n+1)^2 ))−(1/((4n+3)^2 ))

$$ \\ $$$$\:\:\:\:\:{T}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{4}{n}+\mathrm{3}\right)^{\mathrm{2}} }\right. \\ $$

Question Number 212359 Answers: 2 Comments: 0

Σ_(k=0) ^∞ ((1/( (√(4k+1))))−(1/( (√(4k+3)))))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{k}+\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{k}+\mathrm{3}}}\right) \\ $$

Question Number 212358 Answers: 0 Comments: 0

∫∫∫_R^3 (e^(−(√(x^2 +y^2 +z^2 ))) /((1+x^2 +y^2 +z^2 )))dx dy dz

$$ \\ $$$$\:\:\:\:\:\int\int\int_{\mathbb{R}^{\mathrm{3}} } \frac{{e}^{−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }} }{\left(\mathrm{1}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{dx}\:{dy}\:{dz} \\ $$

Question Number 212354 Answers: 2 Comments: 0

ax^2 +bx+c=0 a,b,c and two roots of the eqn. are 5 consecutive integers in some order. Find their values.

$$\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\:{a},{b},{c}\:{and}\:{two}\:{roots}\:{of}\:{the}\:{eqn}. \\ $$$$\:{are}\:\mathrm{5}\:{consecutive}\:{integers}\:{in}\: \\ $$$$\:{some}\:{order}.\:{Find}\:{their}\:{values}. \\ $$

Question Number 212347 Answers: 2 Comments: 1

Question Number 212342 Answers: 1 Comments: 0

The number abc^(−) is divisible by 37. Prove that bca^(−) + cab^(−) is divisible by 37.

$$\mathrm{The}\:\mathrm{number}\:\:\overline {\mathrm{abc}}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\overline {\mathrm{bca}}\:+\:\overline {\mathrm{cab}}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}.\: \\ $$

Question Number 212341 Answers: 2 Comments: 0

Question Number 212339 Answers: 3 Comments: 0

Question Number 212335 Answers: 0 Comments: 0

lim Σ_(k=1) ^n (1−(k/n))ln(1+(k/n^2 ))_(n→∞)

$$ \\ $$$$\underset{{n}\rightarrow\infty} {\:\:\:\:\:\:\:\:\mathrm{lim}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{{k}}{{n}}\right)\mathrm{ln}\left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right)} \\ $$

Question Number 212332 Answers: 0 Comments: 4

Please help 1.1.Let XUY=X for all sets X. Prove that Y=0(empty set). From Singler book "Excercises in set theory". I think this task is totaly wrong and cannot be proved. I would ask someone to provide me valid proof of that. I have sets X and Y such as Y is subset of X. For example. If Y={1} and X={1,2} then XUY=X is correct but that doesn't imply Y is empty. Another example when X=Y since X is any set. I can choose X=Y. Why not? Then YUY=Y is always true, but again, that doesnt imply Y is empty set Proof in book claim that is correct if we suppose Y is not empty and if we choose for instance X is empty set. Then 0UY=0 but this is wrong since 0UY=Y. Therefore, Y must be empty?

$$ \\ $$Please help 1.1.Let XUY=X for all sets X. Prove that Y=0(empty set). From Singler book "Excercises in set theory". I think this task is totaly wrong and cannot be proved. I would ask someone to provide me valid proof of that. I have sets X and Y such as Y is subset of X. For example. If Y={1} and X={1,2} then XUY=X is correct but that doesn't imply Y is empty. Another example when X=Y since X is any set. I can choose X=Y. Why not? Then YUY=Y is always true, but again, that doesnt imply Y is empty set Proof in book claim that is correct if we suppose Y is not empty and if we choose for instance X is empty set. Then 0UY=0 but this is wrong since 0UY=Y. Therefore, Y must be empty?

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