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Question Number 93933 Answers: 0 Comments: 5

Let ∗′ be the binary operation on the set {1,2,3,4,5} defined by a∗′b=H.C.Fof a and b. Is the operation ∗′ same as the operation ∗ defined above? justify your answer.

$${Let}\:\ast'\:{be}\:{the}\:{binary}\:{operation}\:{on}\:{the}\:{set}\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\}\:{defined}\:{by}\:\mathrm{a}\ast'\mathrm{b}=\mathrm{H}.\mathrm{C}.\mathrm{F}{of}\:{a}\:{and}\:{b}.\: \\ $$$${Is}\:{the}\:{operation}\:\ast'\:{same}\:{as}\:{the}\:{operation}\:\ast\:{defined}\:{above}?\:{justify}\:{your}\:{answer}. \\ $$

Question Number 93567 Answers: 1 Comments: 1

Given that A={0,1,3,5} B={1,2,4,7} and C={1,2,3,5,8} prove that (i) (A∩B)∩C = A∩(B∩C) (ii) (A∪B)∪C = A∪(B∪C) (iii) (A∪B)∩C = (A∪C)∪(B∩C) (iv) (A∩C)∪B = (A∪B)∩(C∪B)

$${Given}\:{that}\:{A}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{5}\right\}\:{B}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{7}\right\}\:{and}\:{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{8}\right\}\:{prove}\:{that} \\ $$$$\left(\mathrm{i}\right)\:\left(\mathrm{A}\cap\mathrm{B}\right)\cap\mathrm{C}\:=\:\mathrm{A}\cap\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{ii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C}\:=\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right) \\ $$$$\left(\mathrm{iii}\right)\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\mathrm{C}\:=\:\left(\mathrm{A}\cup\mathrm{C}\right)\cup\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\left(\mathrm{iv}\right)\:\left(\mathrm{A}\cap\mathrm{C}\right)\cup\mathrm{B}\:=\:\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{C}\cup\mathrm{B}\right) \\ $$

Question Number 88761 Answers: 0 Comments: 4

if a_n =((n!)/(n^n e^(−n) (√(2πn)))) and b_n =(((2n)!(√n))/(4^n (n!)^2 )) lim_(n→∞) a_n =1 find lim_(n→∞) b_n =?

$${if}\:{a}_{{n}} =\frac{{n}!}{{n}^{{n}} \:{e}^{−{n}} \sqrt{\mathrm{2}\pi{n}}} \\ $$$${and}\:{b}_{{n}} =\frac{\left(\mathrm{2}{n}\right)!\sqrt{{n}}}{\mathrm{4}^{{n}} \:\left({n}!\right)^{\mathrm{2}} } \\ $$$$\underset{{n}\rightarrow\infty} {{lim}a}_{{n}} =\mathrm{1} \\ $$$${find}\:\underset{{n}\rightarrow\infty} {{lim}b}_{{n}} =? \\ $$

Question Number 88473 Answers: 0 Comments: 2

Question Number 82792 Answers: 1 Comments: 1

show that if A⊂R^m and B⊂R^n are compact sets. then A×B={(a,b)∈R^(m+n) :a∈A and b∈B}

$${show}\:{that}\:{if}\:{A}\subset\mathbb{R}^{{m}} \:{and}\:{B}\subset\mathbb{R}^{{n}} \:{are}\: \\ $$$${compact}\:{sets}.\: \\ $$$${then}\:{A}×{B}=\left\{\left({a},{b}\right)\in\mathbb{R}^{{m}+{n}} :{a}\in{A}\:{and}\:{b}\in{B}\right\} \\ $$

Question Number 77746 Answers: 0 Comments: 4

Master comes after Chess disappeared. Boyka comes after Master disappeared. BK comes after Boyka disappeared. What comes after BK disappeared? 1. A−Team 2. Girlka 3. Bezirksschornsteinfegermeister 4. none from above [A question from NMO 2019 in Madagascar]

$${Master}\:{comes}\:{after}\:{Chess}\:{disappeared}. \\ $$$${Boyka}\:{comes}\:{after}\:{Master}\:{disappeared}. \\ $$$${BK}\:{comes}\:{after}\:{Boyka}\:{disappeared}. \\ $$$${What}\:{comes}\:{after}\:{BK}\:{disappeared}? \\ $$$$\mathrm{1}.\:\:{A}−{Team} \\ $$$$\mathrm{2}.\:\:{Girlka} \\ $$$$\mathrm{3}.\:\:{Bezirksschornsteinfegermeister} \\ $$$$\mathrm{4}.\:\:{none}\:{from}\:{above} \\ $$$$ \\ $$$$\left[{A}\:{question}\:{from}\:{NMO}\:\mathrm{2019}\:{in}\:{Madagascar}\right] \\ $$

Question Number 74639 Answers: 0 Comments: 4

$${If} \\ $$

Question Number 74590 Answers: 1 Comments: 0

Question Number 74234 Answers: 0 Comments: 0

f(x)=(1/((4x^3 ))^(1/5) ) find f^ ′(x) with using lim_(h→0) ((f(x+h)−f(x))/h)

$${f}\left({x}\right)=\frac{\mathrm{1}}{\sqrt[{\mathrm{5}}]{\mathrm{4}{x}^{\mathrm{3}} }} \\ $$$$ \\ $$$${find}\:{f}^{\:} '\left({x}\right) \\ $$$$ \\ $$$${with}\:{using}\:\underset{{h}\rightarrow\mathrm{0}} {{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$

Question Number 71420 Answers: 0 Comments: 0

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Question Number 70704 Answers: 1 Comments: 2

z^4 −2z^2 +2=0

$$\mathrm{z}^{\mathrm{4}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$

Question Number 69568 Answers: 1 Comments: 3

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Question Number 56744 Answers: 1 Comments: 0

If R be a relation on a set of real number defined by R={(x,y): x^2 +y^2 =0}, find i− R in roster form ii−Domain of R iii−Range of R

$$\mathrm{If}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{on}\:\mathrm{a}\:\mathrm{set}\:\mathrm{of}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{defined}\:\mathrm{by}\:\mathrm{R}=\left\{\left(\mathrm{x},\mathrm{y}\right):\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{0}\right\}, \\ $$$$\mathrm{find}\: \\ $$$$\:\:\mathrm{i}−\:\mathrm{R}\:\mathrm{in}\:\mathrm{roster}\:\mathrm{form} \\ $$$$\:\:\mathrm{ii}−\mathrm{Domain}\:\mathrm{of}\:\mathrm{R} \\ $$$$\:\:\mathrm{iii}−\mathrm{Range}\:\mathrm{of}\:\mathrm{R}\: \\ $$

Question Number 56203 Answers: 0 Comments: 0

Question Number 55633 Answers: 1 Comments: 0

Known set A⊆R not empty, If Sup A=Inf A, then set A is..

$$\mathrm{Known}\:\mathrm{set}\:{A}\subseteq\mathbb{R}\:\mathrm{not}\:\mathrm{empty}, \\ $$$$\mathrm{If}\:\mathrm{Sup}\:{A}=\mathrm{Inf}\:{A},\:\mathrm{then}\:\mathrm{set}\:{A}\:\mathrm{is}.. \\ $$

Question Number 47354 Answers: 0 Comments: 4

Question Number 45840 Answers: 0 Comments: 0

a≦7⇒P(!∃x_a )=0, b≦9⇒Q(!∃y_b )=0 for a, b∈N And A⊋A′: A={(x, y)∣P(x)∙Q(y)=0}=A′, B_(∈A) ={(x, y)∈A∣x=y} Then ∀t∈N: ∣B∣=n(t)=f(P(x), Q(y)), also only t can be in [N, M]. find M. :(

$${a}\leqq\mathrm{7}\Rightarrow\mathrm{P}\left(!\exists{x}_{{a}} \right)=\mathrm{0}, \\ $$$${b}\leqq\mathrm{9}\Rightarrow\mathrm{Q}\left(!\exists{y}_{{b}} \right)=\mathrm{0}\:\mathrm{for}\:{a},\:{b}\in\mathbb{N} \\ $$$$\mathrm{And}\:{A}\supsetneq{A}':\:{A}=\left\{\left({x},\:{y}\right)\mid\mathrm{P}\left({x}\right)\centerdot\mathrm{Q}\left({y}\right)=\mathrm{0}\right\}={A}', \\ $$$${B}_{\in{A}} =\left\{\left({x},\:{y}\right)\in{A}\mid{x}={y}\right\} \\ $$$$\mathrm{Then}\:\forall{t}\in\mathbb{N}:\:\mid{B}\mid={n}\left({t}\right)={f}\left(\mathrm{P}\left({x}\right),\:\mathrm{Q}\left({y}\right)\right), \\ $$$$\mathrm{also}\:\mathrm{only}\:{t}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\left[{N},\:{M}\right]. \\ $$$$\mathrm{find}\:{M}. \\ $$$$:\left(\right. \\ $$

Question Number 44826 Answers: 2 Comments: 0

Let A and B be sets. Prove that A = B if and only if A ∪ B = A ∩ B

$$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{be}\:\mathrm{sets}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{A}\:=\:{B}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if}\:{A}\:\cup\:{B}\:=\:{A}\:\cap\:{B} \\ $$

Question Number 42763 Answers: 0 Comments: 1

For A = {1, 2, 3}, let B be the set of 2−element sets belonging to P(A) and let C be the set consisting of the sets that are intersections of two distinct elements of B. Determine C P(A) = power set of A

$$\mathrm{For}\:{A}\:=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3}\right\},\:\mathrm{let}\:{B}\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{2}−\mathrm{element}\:\mathrm{sets} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:{P}\left({A}\right)\:\mathrm{and}\:\mathrm{let}\:{C}\:\mathrm{be}\:\mathrm{the}\:\mathrm{set}\:\mathrm{consisting}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{sets}\:\mathrm{that}\:\mathrm{are}\:\mathrm{intersections}\:\mathrm{of}\:\mathrm{two}\:\mathrm{distinct}\:\mathrm{elements} \\ $$$$\mathrm{of}\:{B}.\:\mathrm{Determine}\:{C} \\ $$$$ \\ $$$${P}\left({A}\right)\:=\:\mathrm{power}\:\mathrm{set}\:\mathrm{of}\:{A} \\ $$

Question Number 36061 Answers: 1 Comments: 2

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Question Number 33551 Answers: 1 Comments: 1

Question Number 31711 Answers: 0 Comments: 0

Given p is primes and A={−(m/n)−p(n/m) ∣ m , n ∈N} find sup A

$$\mathrm{Given}\:{p}\:\mathrm{is}\:\mathrm{primes}\:\mathrm{and}\:\mathrm{A}=\left\{−\frac{{m}}{{n}}−{p}\frac{{n}}{{m}}\:\mid\:{m}\:,\:{n}\:\in\mathbb{N}\right\} \\ $$$$\mathrm{find}\:\mathrm{sup}\:\mathrm{A} \\ $$

Question Number 31669 Answers: 0 Comments: 0

A={(m/n)+((8n)/m) : m, n ∈ N} N= natural numbers supremum ? infimum?

$$\mathrm{A}=\left\{\frac{{m}}{{n}}+\frac{\mathrm{8}{n}}{{m}}\::\:{m},\:{n}\:\in\:\mathrm{N}\right\}\:\mathrm{N}=\:\mathrm{natural}\:\mathrm{numbers} \\ $$$$\mathrm{supremum}\:? \\ $$$$\mathrm{infimum}? \\ $$

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