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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 )=?

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

Question Number 119192    Answers: 1   Comments: 0

Question Number 119187    Answers: 0   Comments: 3

Question Number 119184    Answers: 0   Comments: 0

Nama : Attila Abiem Farhan Kelas : XI BKP Tugas KD 3.18 MTK 1. jawab : u^→ = (((−3)),((−4)),(( 12)) ) ∣u^→ ∣=(√((−3^2 )+(−4)^2 +12^2 )) =(√(9+16+144)) =(√(169))=13 2. jawab : a^→ = ((2),((−1)),(3) ) b^→ = ((7),((13)),(5) ) a^→ +b^→ = ((9),((12)),(8) ) ∣a^→ +b^→ ∣=(√(9^2 +12^2 +8^2 )) =(√(81+144+64)) =(√(289)) 3. jawab u^→ = ((7),(9),((−17)) ) v^→ = (((−2)),((−3)),((19)) ) u^→ −v^→ = ((9),((12)),((36)) ) ∣u^→ +v^→ ∣=(√(9^2 +12^2 +36^2 )) =(√(81+144+1296)) =(√(1521)) 4. jawab : A. 2a^→ = (((−6)),((−8)),((−24)) ) ∣2a^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676)) =26 (1/2)b^→ = ((3),(4),((12)) ) ∣(1/2)b^→ ∣=(√(3^2 +4^2 +12^2 ))=(√(9+16+144))=(√(169))=13 B. 4a^→ +b^→ = (((−12)),((−16)),((−48)) ) + ((6),(8),((24)) ) = (((−6)),((−8)),((−24)) ) ∣4a^→ +b^→ ∣=(√((−6)^2 +(−8)^2 +(−24)^2 )) =(√(36+64+576)) =(√(676))=26 5. jawab : u^→ =8i+3k v^→ =5j−9k u^→ .v^→ =(8.0)i+(0.5)j+(3.−9)k =0+0−27k u^→ .v^→ =−27k

$${Nama}\::\:{Attila}\:{Abiem}\:{Farhan} \\ $$$${Kelas}\::\:{XI}\:{BKP} \\ $$$${Tugas}\:{KD}\:\mathrm{3}.\mathrm{18}\:{MTK} \\ $$$$\mathrm{1}.\:{jawab}\:: \\ $$$$\:\overset{\rightarrow} {{u}}=\begin{pmatrix}{−\mathrm{3}}\\{−\mathrm{4}}\\{\:\:\mathrm{12}}\end{pmatrix}\: \\ $$$$\mid\overset{\rightarrow} {{u}}\mid=\sqrt{\left(−\mathrm{3}^{\mathrm{2}} \right)+\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{9}+\mathrm{16}+\mathrm{144}} \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{169}}=\mathrm{13} \\ $$$$\mathrm{2}.\:{jawab}\:: \\ $$$$\overset{\rightarrow} {{a}}=\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{13}}\\{\mathrm{5}}\end{pmatrix}\: \\ $$$$\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{12}}\\{\mathrm{8}}\end{pmatrix} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{81}+\mathrm{144}+\mathrm{64}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{289}} \\ $$$$\mathrm{3}.\:{jawab} \\ $$$$\overset{\rightarrow} {{u}}=\begin{pmatrix}{\mathrm{7}}\\{\mathrm{9}}\\{−\mathrm{17}}\end{pmatrix}\:\overset{\rightarrow} {{v}}=\begin{pmatrix}{−\mathrm{2}}\\{−\mathrm{3}}\\{\mathrm{19}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {{u}}−\overset{\rightarrow} {{v}}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{12}}\\{\mathrm{36}}\end{pmatrix} \\ $$$$\mid\overset{\rightarrow} {{u}}+\overset{\rightarrow} {{v}}\mid=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} +\mathrm{36}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{81}+\mathrm{144}+\mathrm{1296}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1521}} \\ $$$$\mathrm{4}.\:{jawab}\::\: \\ $$$${A}.\:\:\:\mathrm{2}\overset{\rightarrow} {{a}}=\begin{pmatrix}{−\mathrm{6}}\\{−\mathrm{8}}\\{−\mathrm{24}}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\mid\mathrm{2}\overset{\rightarrow} {{a}}\mid=\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{8}\right)^{\mathrm{2}} +\left(−\mathrm{24}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{36}+\mathrm{64}+\mathrm{576}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{676}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{26} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{b}}=\begin{pmatrix}{\mathrm{3}}\\{\mathrm{4}}\\{\mathrm{12}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\mid\frac{\mathrm{1}}{\mathrm{2}}\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\sqrt{\mathrm{9}+\mathrm{16}+\mathrm{144}}=\sqrt{\mathrm{169}}=\mathrm{13} \\ $$$$ \\ $$$${B}.\:\:\:\:\mathrm{4}\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}=\begin{pmatrix}{−\mathrm{12}}\\{−\mathrm{16}}\\{−\mathrm{48}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{6}}\\{\mathrm{8}}\\{\mathrm{24}}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{6}}\\{−\mathrm{8}}\\{−\mathrm{24}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\mid\mathrm{4}\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{8}\right)^{\mathrm{2}} +\left(−\mathrm{24}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{36}+\mathrm{64}+\mathrm{576}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{676}}=\mathrm{26} \\ $$$$\mathrm{5}.\:{jawab}\::\: \\ $$$$\overset{\rightarrow} {{u}}=\mathrm{8}{i}+\mathrm{3}{k} \\ $$$$\overset{\rightarrow} {{v}}=\mathrm{5}{j}−\mathrm{9}{k} \\ $$$$\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}=\left(\mathrm{8}.\mathrm{0}\right){i}+\left(\mathrm{0}.\mathrm{5}\right){j}+\left(\mathrm{3}.−\mathrm{9}\right)\mathrm{k} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{0}+\mathrm{0}−\mathrm{27}{k} \\ $$$$\overset{\rightarrow} {{u}}.\overset{\rightarrow} {{v}}=−\mathrm{27}{k} \\ $$

Question Number 119173    Answers: 3   Comments: 0

Question Number 119164    Answers: 0   Comments: 0

Question Number 119160    Answers: 0   Comments: 5

Question Number 119159    Answers: 2   Comments: 0

We are in C. Given Z_(0 ) =1 ; Z_(n+1 ) =(1/2)Z_(n ) +(1/2)i n ∈ N. Show that ∀ n ∈ N^(∗ ) , ∣Z_n ∣<1.

$${We}\:{are}\:{in}\:\mathbb{C}. \\ $$$${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${n}\:\in\:\mathbb{N}. \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$

Question Number 119158    Answers: 1   Comments: 0

Question Number 119157    Answers: 2   Comments: 0

Question Number 119155    Answers: 1   Comments: 5

Question Number 119151    Answers: 0   Comments: 0

Question Number 119150    Answers: 2   Comments: 0

Given that a, b and c are real numbers that stisfy the system of equation above a − (√(b^2 − (1/(16)) )) = (√(c^2 − (1/(16)) )) b − (√(c^2 − (1/(25)))) = (√(a^2 − (1/(25)) )) c − (√(a^2 − (1/(36)) )) = (√(b^2 − (1/(36)))) if a+ b + c = (x/( (√y))) where x, y are positive integers and y is square free, find the value of x + y !

$${Given}\:{that}\:{a},\:{b}\:{and}\:{c}\:{are}\:{real}\:{numbers}\: \\ $$$${that}\:{stisfy}\:{the}\:{system}\:{of}\:{equation}\:{above} \\ $$$$ \\ $$$${a}\:−\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\:=\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{16}}\:}\: \\ $$$${b}\:−\:\sqrt{{c}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}}\:=\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{25}}\:} \\ $$$${c}\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}\:}\:=\:\sqrt{{b}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{36}}}\: \\ $$$$ \\ $$$${if}\:\:{a}+\:{b}\:+\:{c}\:=\:\frac{{x}}{\:\sqrt{{y}}}\:{where}\:{x},\:{y}\:{are}\:{positive}\:{integers} \\ $$$${and}\:{y}\:{is}\:{square}\:{free},\:{find}\:{the}\:{value}\:\:{of}\:{x}\:+\:{y}\:! \\ $$

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