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Question Number 118347    Answers: 1   Comments: 0

∫ (dx/( (√x) +(x)^(1/(3 )) ))

$$\:\:\:\int\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt[{\mathrm{3}\:}]{{x}}}\: \\ $$

Question Number 118362    Answers: 1   Comments: 0

Prove that 𝛑 is an irrational number

$$\mathrm{Prove}\:\mathrm{that}\:\boldsymbol{\pi}\:\mathrm{is}\:\mathrm{an}\:\mathrm{irrational}\:\mathrm{number} \\ $$

Question Number 118340    Answers: 1   Comments: 1

∫_(Ο€/3) ^(Ο€/2) (dx/(1+sin xβˆ’cos x))

$$\:\:\:\underset{\pi/\mathrm{3}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}βˆ’\mathrm{cos}\:{x}} \\ $$

Question Number 118339    Answers: 0   Comments: 0

Question Number 118338    Answers: 3   Comments: 0

solve ∫ (dx/(3βˆ’5sin x))

$$\:\:\:\:{solve}\:\int\:\frac{{dx}}{\mathrm{3}βˆ’\mathrm{5sin}\:{x}}\: \\ $$

Question Number 118327    Answers: 4   Comments: 0

lim_(xβ†’1) ((xβˆ’(√(2βˆ’x^2 )))/(2xβˆ’(√(2+2x^2 )))) ?

$$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}βˆ’\sqrt{\mathrm{2}βˆ’{x}^{\mathrm{2}} }}{\mathrm{2}{x}βˆ’\sqrt{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }}\:? \\ $$

Question Number 118325    Answers: 0   Comments: 0

Question Number 118323    Answers: 0   Comments: 0

Prove that ΞΆ(1βˆ’s)=2^(1βˆ’s) Ο€^(βˆ’s) cos(((sΟ€)/2))Ξ“(s)ΞΆ(s)

$${Prove}\:{that}\: \\ $$$$\zeta\left(\mathrm{1}βˆ’{s}\right)=\mathrm{2}^{\mathrm{1}βˆ’{s}} \pi^{βˆ’{s}} {cos}\left(\frac{{s}\pi}{\mathrm{2}}\right)\Gamma\left({s}\right)\zeta\left({s}\right) \\ $$

Question Number 118322    Answers: 2   Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

Question Number 118318    Answers: 1   Comments: 0

Question Number 118308    Answers: 1   Comments: 1

Question Number 118307    Answers: 2   Comments: 1

∫^∞ _0 ((x^2 βˆ’2)/(x^4 +x^2 +1)) dx

$$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{x}^{\mathrm{2}} βˆ’\mathrm{2}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$

Question Number 118304    Answers: 0   Comments: 0

... ⧫_β–Ό ^β–Ό Advanced calculus⧫_β–Ό ^β–Ό ... prove that :: Ξ£_(n=1) ^∞ (1/(3^n n^3 )) =^(???) (7/8)ΞΆ( 3 )+(1/6)ln^3 (2)βˆ’(Ο€^2 /(12)) ln(2) βœ“ ...β™ M.N.July 1970β™ ... .Good luck.

$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\underset{\blacktrinagledown} {\overset{\blacktrinagledown} {\blacklozenge}}\mathscr{A}{dvanced}\:\:{calculus}\underset{\blacktrinagledown} {\overset{\blacktrinagledown} {\blacklozenge}}\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{n}} {n}^{\mathrm{3}} }\:\overset{???} {=}\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\:\mathrm{3}\:\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:{ln}\left(\mathrm{2}\right)\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\spadesuit\mathscr{M}.\mathscr{N}.\mathscr{J}{uly}\:\mathrm{1970}\spadesuit... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.\mathscr{G}{ood}\:{luck}. \\ $$$$ \\ $$

Question Number 118298    Answers: 0   Comments: 1

b, x, y, c are consecutive terms of a G.P. ∴ x = br, y = br^2 , c = br^3 A.M. between b and c is a ∴ ((b+c)/2) = a β‡’ 2a = b+c 2abc = (b+c)bc = b^2 c + bc^2 = b^2 (br^3 ) + b(br^3 )^2 = b^3 r^3 + b^3 r^6 = (br)^3 + (br^2 )^3 = x^3 + y^3 ∴ x^3 + y^3 = 2abc ←

$$\:\:\:\:\:\mathrm{b},\:\mathrm{x},\:\mathrm{y},\:\mathrm{c}\:\mathrm{are}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:\boldsymbol{\mathrm{br}},\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{2}} ,\:\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{3}} \: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathrm{A}.\mathrm{M}.\:\mathrm{between}\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}\:\:=\:\mathrm{a}\:\:\:\:\Rightarrow\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\:=\:\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\mathrm{bc}\:=\:\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\mathrm{bc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{2}} \boldsymbol{\mathrm{c}}\:+\:\mathrm{b}\boldsymbol{\mathrm{c}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{2}} \left(\boldsymbol{\mathrm{br}}^{\mathrm{3}} \right)\:+\:\mathrm{b}\left(\boldsymbol{\mathrm{br}}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{3}} \mathrm{r}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \mathrm{r}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\boldsymbol{\mathrm{br}}\right)^{\mathrm{3}} \:+\:\left(\boldsymbol{\mathrm{br}}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{3}} \:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{2abc}\:\:\:\leftarrow \\ $$

Question Number 118292    Answers: 2   Comments: 0

Prove that: ∫_0 ^( 1) ((x^n βˆ’1)/(lnx)) = ln∣n+1∣

$$\boldsymbol{\mathrm{P}}\mathrm{rove}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} βˆ’\mathrm{1}}{\mathrm{lnx}}\:=\:\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{n}}+\mathrm{1}\mid \\ $$

Question Number 118286    Answers: 2   Comments: 0

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣aβˆ’b∣ ;(b)∣ a+b∣>∣ aβˆ’b∣;(c)∣a+ b∣< ∣aβˆ’b∣

$$\mathrm{What}\:\mathrm{condition}\:\mathrm{should}\:\mathrm{be}\:\mathrm{satisfied}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{vectors}\:\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{relations}\:\mathrm{to}\:\mathrm{hold}\:\mathrm{true}\::\left(\mathrm{a}\right)\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}βˆ’\boldsymbol{\mathrm{b}}\mid \\ $$$$;\left(\mathrm{b}\right)\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid>\mid\:\boldsymbol{\mathrm{a}}βˆ’\boldsymbol{\mathrm{b}}\mid;\left(\mathrm{c}\right)\mid\boldsymbol{\mathrm{a}}+\:\boldsymbol{\mathrm{b}}\mid<\:\mid\boldsymbol{\mathrm{a}}βˆ’\boldsymbol{\mathrm{b}}\mid \\ $$

Question Number 118280    Answers: 0   Comments: 1

(26) OA^(β†’) = a^β†’ = (((4.8)),((3.6)) ) , OB^(β†’) = b^β†’ = ((( 8)),((15)) ) a^β†’ . b^β†’ = (4.8)(8) + (3.6)(15) = 92.4^(β†’ ) a = (√(4.8^2 +3.6^2 )) = (√(23.04+12.96)) = (√(36)) = 6 b = (√(8^2 +15^2 )) = (√(64+225)) = (√(289)) = 17 a.b = 6 . 17 = 102 cos AOB = ((a^β†’ .b^β†’ )/(a.b)) = ((92.4)/(102)) = 0.9059 = cos 25.06Β° ∴ ∠AOB = 25.06Β°

$$\left(\mathrm{26}\right)\:\:\:\overset{\rightarrow} {\mathrm{OA}}\:=\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\mathrm{4}.\mathrm{8}}\\{\mathrm{3}.\mathrm{6}}\end{pmatrix}\:\:,\:\:\overset{\rightarrow} {\mathrm{OB}}\:=\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\mathrm{8}}\\{\mathrm{15}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{4}.\mathrm{8}\right)\left(\mathrm{8}\right)\:+\:\left(\mathrm{3}.\mathrm{6}\right)\left(\mathrm{15}\right)\:=\:\mathrm{92}.\overset{\rightarrow\:} {\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\sqrt{\mathrm{4}.\mathrm{8}^{\mathrm{2}} +\mathrm{3}.\mathrm{6}^{\mathrm{2}} }\:=\:\:\sqrt{\mathrm{23}.\mathrm{04}+\mathrm{12}.\mathrm{96}}\:=\:\sqrt{\mathrm{36}}\:=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{64}+\mathrm{225}}\:=\:\sqrt{\mathrm{289}}\:=\:\mathrm{17} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{6}\:.\:\mathrm{17}\:=\:\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{AOB}\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{\mathrm{92}.\mathrm{4}}{\mathrm{102}}\:=\:\mathrm{0}.\mathrm{9059}\:=\:\mathrm{cos}\:\mathrm{25}.\mathrm{06}Β° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\angle\mathrm{AOB}\:=\:\mathrm{25}.\mathrm{06}Β° \\ $$

Question Number 118277    Answers: 0   Comments: 0

(23) Let a^β†’ = ((( 7)),((24)) ) and b^β†’ = ((( 3)),((βˆ’4)) ) a^β†’ .b^β†’ = (7)(3) + (24)(βˆ’4) = 21βˆ’96 = βˆ’75 a = (√(7^2 +24^2 )) = (√(49+576)) = (√(625)) = 25 b = (√(3^2 +(βˆ’4)^2 )) = (√(9+16)) = (√(25)) = 5 a.b = 25 . 5 = 125 Let ΞΈ be the angle between a^β†’ and b^β†’ . cosΞΈ = ((a^β†’ .b^β†’ )/(a.b)) = ((βˆ’75)/(125)) = βˆ’0.6 = cos126.87Β° ∴ ΞΈ = 126.87Β°

$$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{βˆ’\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(βˆ’\mathrm{4}\right)\:=\:\mathrm{21}βˆ’\mathrm{96}\:=\:βˆ’\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(βˆ’\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{25}\:.\:\mathrm{5}\:=\:\:\mathrm{125}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{a}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{βˆ’\mathrm{75}}{\mathrm{125}}\:=\:βˆ’\mathrm{0}.\mathrm{6}\:=\:\mathrm{cos126}.\mathrm{87}Β° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\theta\:=\:\mathrm{126}.\mathrm{87}Β°\:\:\:\: \\ $$

Question Number 118275    Answers: 1   Comments: 1

Question Number 118274    Answers: 2   Comments: 0

yβ€²β€²βˆ’3yβ€²+2y = (1/(1+e^(βˆ’x) ))

$$\:\:{y}''βˆ’\mathrm{3}{y}'+\mathrm{2}{y}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{βˆ’{x}} }\: \\ $$

Question Number 118272    Answers: 1   Comments: 0

lim_(xβ†’1) ((a/(1βˆ’x^a )) βˆ’ (b/(1βˆ’x^b )) ) = with (a,b) ∈(R^+ )^2

$$\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{{a}}{\mathrm{1}βˆ’{x}^{{a}} }\:βˆ’\:\frac{{b}}{\mathrm{1}βˆ’{x}^{{b}} }\:\right)\:=\: \\ $$$${with}\:\left({a},{b}\right)\:\in\left(\mathbb{R}^{+} \right)^{\mathrm{2}} \: \\ $$

Question Number 118270    Answers: 0   Comments: 1

∫x a^x (1βˆ’a)^(1βˆ’x) dx?

$$\:\:\:\int{x}\:{a}^{{x}} \:\left(\mathrm{1}βˆ’{a}\right)^{\mathrm{1}βˆ’{x}} \:{dx}? \\ $$

Question Number 118265    Answers: 1   Comments: 0

Biochimica Isomeri di posizione del C_7 H_(16) CH_3 CH_2 CH_2 CH_2 CH_2 CH_2 CH_3 CH_3 CH_2 CH(CH_3 )CH_2 CH_2 CH_3 CH_3 CH(CH_3 )CH_2 CH_2 CH_2 CH_3 CH_3 CH(CH_3 )CH(CH_3 )CH_2 CH_3 Studiare la distillazione frazionata del petrolio

$$\mathrm{Biochimica} \\ $$$$\mathrm{Isomeri}\:\mathrm{di}\:\mathrm{posizione}\:\mathrm{del}\:\mathrm{C}_{\mathrm{7}} \mathrm{H}_{\mathrm{16}} \\ $$$$\mathrm{CH}_{\mathrm{3}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{3}} \\ $$$$\mathrm{CH}_{\mathrm{3}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}\left(\mathrm{CH}_{\mathrm{3}} \right)\mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{3}} \\ $$$$\mathrm{CH}_{\mathrm{3}} \mathrm{CH}\left(\mathrm{CH}_{\mathrm{3}} \right)\mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{3}} \\ $$$$\mathrm{CH}_{\mathrm{3}} \mathrm{CH}\left(\mathrm{CH}_{\mathrm{3}} \right)\mathrm{CH}\left(\mathrm{CH}_{\mathrm{3}} \right)\mathrm{CH}_{\mathrm{2}} \mathrm{CH}_{\mathrm{3}} \\ $$$$\mathrm{Studiare}\:\mathrm{la}\:\mathrm{distillazione}\:\mathrm{frazionata}\:\mathrm{del}\:\mathrm{petrolio} \\ $$

Question Number 118263    Answers: 0   Comments: 0

Question Number 118259    Answers: 0   Comments: 0

Svolgimento prova analoga alla verifica 1C 1) Scrivere lβ€²insieme mediante elencazione A={x=8n+5,n∈N∧0<n≀3} Soluzione A={13;21;29} A={0;3;6;9;12} B={0;4;8;12} Scrivere A∩B AβˆͺB Soluzione A∩B={0;12} AβˆͺB={0;3;4;6;8;9;12} Lβ€²addizione non gode della proprieta^ invariantiva La divisione non gode della proprieta^ associativa (50:10):5β‰ 50:(10:5) Indicare la proprieta^ corrispondente 35βˆ’7=(35βˆ’2)βˆ’(7βˆ’2) proprieta^ invariantiva della sottrazione (11+7)β€’5=5β€’(11+7) proprieta^ commutativa della moltiplicazione (7βˆ’2)β€’3=21βˆ’6 proprieta^ distributiva della moltiplicazione rispetto alla sottrazione 21βˆ’6=3β€’(7βˆ’2) raccoglimento a fattor comune

$$\mathrm{Svolgimento}\:\mathrm{prova}\:\mathrm{analoga}\:\mathrm{alla}\:\mathrm{verifica}\:\mathrm{1C} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Scrivere}\:\mathrm{l}'\mathrm{insieme}\:\mathrm{mediante}\:\mathrm{elencazione} \\ $$$$\mathrm{A}=\left\{\mathrm{x}=\mathrm{8n}+\mathrm{5},\mathrm{n}\in\mathbb{N}\wedge\mathrm{0}<\mathrm{n}\leqslant\mathrm{3}\right\} \\ $$$$\mathrm{Soluzione} \\ $$$$\mathrm{A}=\left\{\mathrm{13};\mathrm{21};\mathrm{29}\right\} \\ $$$$\mathrm{A}=\left\{\mathrm{0};\mathrm{3};\mathrm{6};\mathrm{9};\mathrm{12}\right\}\:\:\:\:\:\:\mathrm{B}=\left\{\mathrm{0};\mathrm{4};\mathrm{8};\mathrm{12}\right\} \\ $$$$\mathrm{Scrivere}\:\mathrm{A}\cap\mathrm{B}\:\:\:\:\:\:\:\mathrm{A}\cup\mathrm{B} \\ $$$$\mathrm{Soluzione} \\ $$$$\mathrm{A}\cap\mathrm{B}=\left\{\mathrm{0};\mathrm{12}\right\}\:\:\:\:\:\:\mathrm{A}\cup\mathrm{B}=\left\{\mathrm{0};\mathrm{3};\mathrm{4};\mathrm{6};\mathrm{8};\mathrm{9};\mathrm{12}\right\} \\ $$$$\mathrm{L}'\mathrm{addizione}\:\mathrm{non}\:\mathrm{gode}\:\mathrm{della}\:\mathrm{propriet}\grave {\mathrm{a}}\:\mathrm{invariantiva} \\ $$$$\mathrm{La}\:\mathrm{divisione}\:\mathrm{non}\:\mathrm{gode}\:\mathrm{della}\:\mathrm{propriet}\grave {\mathrm{a}}\:\mathrm{associativa} \\ $$$$\left(\mathrm{50}:\mathrm{10}\right):\mathrm{5}\neq\mathrm{50}:\left(\mathrm{10}:\mathrm{5}\right) \\ $$$$\mathrm{Indicare}\:\mathrm{la}\:\mathrm{propriet}\acute {\mathrm{a}}\:\mathrm{corrispondente} \\ $$$$\mathrm{35}βˆ’\mathrm{7}=\left(\mathrm{35}βˆ’\mathrm{2}\right)βˆ’\left(\mathrm{7}βˆ’\mathrm{2}\right)\:\:\:\:\:\:\mathrm{propriet}\acute {\mathrm{a}}\:\mathrm{invariantiva}\:\mathrm{della}\:\mathrm{sottrazione} \\ $$$$\left(\mathrm{11}+\mathrm{7}\right)\bullet\mathrm{5}=\mathrm{5}\bullet\left(\mathrm{11}+\mathrm{7}\right)\:\:\:\:\:\:\:\:\:\:\:\mathrm{propriet}\acute {\mathrm{a}}\:\mathrm{commutativa}\:\mathrm{della}\:\mathrm{moltiplicazione} \\ $$$$\left(\mathrm{7}βˆ’\mathrm{2}\right)\bullet\mathrm{3}=\mathrm{21}βˆ’\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{propriet}\acute {\mathrm{a}}\:\mathrm{distributiva}\:\mathrm{della}\:\mathrm{moltiplicazione}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{rispetto}\:\mathrm{alla}\:\mathrm{sottrazione} \\ $$$$\mathrm{21}βˆ’\mathrm{6}=\mathrm{3}\bullet\left(\mathrm{7}βˆ’\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{raccoglimento}\:\mathrm{a}\:\mathrm{fattor}\:\mathrm{comune} \\ $$

Question Number 118251    Answers: 1   Comments: 0

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