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Isomeri di posizione C_4 H_8 → ciclobutano e metilciclopropano C_5 H_(10) → ciclopentano, metilciclobutano, dimetilciclopropano, etilciclo propano isomeri configurazionali cis−trans dell′1,2 dimetilciclobutano Guardo il legame C_1 −C_2 CH_3 e^ gruppo dominante sull′H (MM=15 contro 1) isomeri configurazionali: isomeria dovuta all′impossibilita^ di rotazione completa intorno ai legami σ. Tabella pag. 53 per la prossima verifica Capitolo 3 Elettrofili e nucleofili Tipi di reazioni Carblcationi e carbanioni Effetto induttivo → tabella Soft e hard

$$\mathrm{Isomeri}\:\mathrm{di}\:\mathrm{posizione} \\ $$$$\mathrm{C}_{\mathrm{4}} \mathrm{H}_{\mathrm{8}} \rightarrow\:\mathrm{ciclobutano}\:\mathrm{e}\:\mathrm{metilciclopropano} \\ $$$$\mathrm{C}_{\mathrm{5}} \mathrm{H}_{\mathrm{10}} \rightarrow\:\mathrm{ciclopentano},\:\mathrm{metilciclobutano},\:\mathrm{dimetilciclopropano},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{etilciclo}\:\mathrm{propano} \\ $$$$\mathrm{isomeri}\:\mathrm{configurazionali}\:\mathrm{cis}−\mathrm{trans}\:\mathrm{dell}'\mathrm{1},\mathrm{2}\:\mathrm{dimetilciclobutano} \\ $$$$\mathrm{Guardo}\:\mathrm{il}\:\mathrm{legame}\:\mathrm{C}_{\mathrm{1}} −\mathrm{C}_{\mathrm{2}} \\ $$$$\mathrm{CH}_{\mathrm{3}} \acute {\mathrm{e}}\:\mathrm{gruppo}\:\mathrm{dominante}\:\mathrm{sull}'\mathrm{H}\:\left(\mathrm{MM}=\mathrm{15}\:\mathrm{contro}\:\mathrm{1}\right) \\ $$$$\mathrm{isomeri}\:\mathrm{configurazionali}:\:\mathrm{isomeria}\:\mathrm{dovuta}\:\mathrm{all}'\mathrm{impossibilit}\acute {\mathrm{a}}\:\mathrm{di}\:\mathrm{rotazione}\: \\ $$$$\mathrm{completa}\:\mathrm{intorno}\:\mathrm{ai}\:\mathrm{legami}\:\sigma. \\ $$$$\mathrm{Tabella}\:\mathrm{pag}.\:\mathrm{53}\:\mathrm{per}\:\mathrm{la}\:\mathrm{prossima}\:\mathrm{verifica} \\ $$$$\mathrm{Capitolo}\:\mathrm{3}\: \\ $$$$\mathrm{Elettrofili}\:\mathrm{e}\:\mathrm{nucleofili} \\ $$$$\mathrm{Tipi}\:\mathrm{di}\:\mathrm{reazioni} \\ $$$$\mathrm{Carblcationi}\:\mathrm{e}\:\mathrm{carbanioni} \\ $$$$\mathrm{Effetto}\:\mathrm{induttivo}\:\rightarrow\:\mathrm{tabella} \\ $$$$\mathrm{Soft}\:\mathrm{e}\:\mathrm{hard} \\ $$

Question Number 119256 Answers: 0 Comments: 0

Determine ∫_(−(π/4)) ^( (π/4)) (cost+(√(1+t^2 sin^3 tcos^3 t))dt=?

$$\: \\ $$$$\:\:\:\:\:\:{Determine}\:\:\int_{−\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{4}}} \left({cost}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} {sin}^{\mathrm{3}} {tcos}^{\mathrm{3}} {t}}{dt}=?\right. \\ $$

Question Number 119253 Answers: 2 Comments: 0

Given three function f(x) ,g(x) and h(x). where f(x)=x^2 +x−2 and h(x)=x^2 +2x−1. If ((f(x))/(x+3)) ≤ ((g(x))/(x+3)) ≤ ((h(x))/(x+3)) , then the value of lim_(x→−1) g(x) = ?

$${Given}\:{three}\:{function}\:{f}\left({x}\right)\:,{g}\left({x}\right)\:{and}\:{h}\left({x}\right). \\ $$$${where}\:{f}\left({x}\right)={x}^{\mathrm{2}} +{x}−\mathrm{2}\:{and}\:{h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}. \\ $$$${If}\:\frac{{f}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{g}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{h}\left({x}\right)}{{x}+\mathrm{3}}\:,\:{then}\:{the}\:{value}\:{of} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\:? \\ $$

Question Number 119250 Answers: 1 Comments: 5

Question Number 119246 Answers: 2 Comments: 0

Prove the following inequalities hold true ∀x∈R a)cos(cosx)>0 b)cos(sinx)>sin(cosx)

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\:\mathrm{inequalities}\:\mathrm{hold} \\ $$$$\:\:\mathrm{true}\:\forall\mathrm{x}\in\mathrm{R} \\ $$$$\left.\mathrm{a}\right)\mathrm{cos}\left(\mathrm{cosx}\right)>\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{cos}\left(\mathrm{sinx}\right)>\mathrm{sin}\left(\mathrm{cosx}\right) \\ $$

Question Number 119241 Answers: 1 Comments: 0

675×54/100

$$\mathrm{675}×\mathrm{54}/\mathrm{100} \\ $$

Question Number 119240 Answers: 2 Comments: 0

find max and min value of f(x,y) = 4x^2 +8xy+9y^2 −8x−24y+4

$${find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${f}\left({x},{y}\right)\:=\:\mathrm{4}{x}^{\mathrm{2}} +\mathrm{8}{xy}+\mathrm{9}{y}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{24}{y}+\mathrm{4}\: \\ $$

Question Number 119235 Answers: 1 Comments: 0

(D^3 +D^2 −4D−4)y = e^(4x)

$$\:\left({D}^{\mathrm{3}} +{D}^{\mathrm{2}} −\mathrm{4}{D}−\mathrm{4}\right){y}\:=\:{e}^{\mathrm{4}{x}} \\ $$

Question Number 119233 Answers: 3 Comments: 0

(D^2 −5D+6)y = e^(3x)

$$\:\left({D}^{\mathrm{2}} −\mathrm{5}{D}+\mathrm{6}\right){y}\:=\:{e}^{\mathrm{3}{x}} \\ $$

Question Number 119229 Answers: 3 Comments: 0

sin 3A+cos 3A = (1/2)

$$\:\:\:\mathrm{sin}\:\mathrm{3}{A}+\mathrm{cos}\:\mathrm{3}{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Question Number 119220 Answers: 2 Comments: 0

2cos^2 x −25 cot x = 0 x =?

$$\:\:\:\:\:\mathrm{2cos}\:^{\mathrm{2}} {x}\:−\mathrm{25}\:\mathrm{cot}\:{x}\:=\:\mathrm{0}\: \\ $$$$\:\:\:\:\:{x}\:=? \\ $$

Question Number 119216 Answers: 4 Comments: 4

lim_(x→1) ((x^3 −3x+2)/( (x^2 )^(1/(3 )) −2 (x)^(1/(3 )) +1)) ?

$$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}}{\:\sqrt[{\mathrm{3}\:}]{{x}^{\mathrm{2}} }\:−\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{{x}}\:+\mathrm{1}}\:?\: \\ $$

Question Number 119214 Answers: 1 Comments: 0

f(x) = (∫_0 ^1 f(x)dx)x^2 + (∫_0 ^2 f(x)dx)x + (∫_0 ^3 f(x)dx)+1 the valeu f(4) = ?

$${f}\left({x}\right)\:=\:\left(\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{f}\left({x}\right){dx}\right){x}^{\mathrm{2}} \:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}\right){x}\:+\:\left(\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}\right)+\mathrm{1} \\ $$$${the}\:{valeu}\:{f}\left(\mathrm{4}\right)\:=\:? \\ $$

Question Number 119212 Answers: 1 Comments: 0

Question Number 119204 Answers: 0 Comments: 0

40−misolning yechimi: y=(x^3 /3)+2x^2 −5x+7 Kritik nuqtalarini topish uchun: 1. Funksiyadan hosila olamiz 2. Funksiya hosilasini nolga tenglab, tenglamani yechamiz. y′=x^2 +4x−5=0 ⇒ x_1 =1; x_2 =−5 x_1 +x_2 =1−5=−4 Javob: −4

$$\mathrm{40}−\mathrm{misolning}\:\:\:\mathrm{yechimi}: \\ $$$$\mathrm{y}=\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{7} \\ $$$$\mathrm{Kritik}\:\mathrm{nuqtalarini}\:\mathrm{topish}\:\mathrm{uchun}:\: \\ $$$$\mathrm{1}.\:\mathrm{Funksiyadan}\:\mathrm{hosila}\:\mathrm{olamiz} \\ $$$$\mathrm{2}.\:\mathrm{Funksiya}\:\mathrm{hosilasini}\:\mathrm{nolga}\:\mathrm{tenglab},\: \\ $$$$\mathrm{tenglamani}\:\mathrm{yechamiz}. \\ $$$$\mathrm{y}'=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{5}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\mathrm{x}_{\mathrm{1}} =\mathrm{1};\:\:\mathrm{x}_{\mathrm{2}} =−\mathrm{5} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\mathrm{1}−\mathrm{5}=−\mathrm{4}\:\:\:\:\:\mathrm{Javob}:\:\:−\mathrm{4}\:\:\:\: \\ $$

Question Number 119193 Answers: 3 Comments: 3

if x^3 +(1/x^3 )=52, find the value of x^4 +(1/x^4 )=?

$${if}\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{52},\:{find}\:{the}\:{value}\:{of} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=? \\ $$

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