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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 !

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

Question Number 119147    Answers: 0   Comments: 0

Question Number 119141    Answers: 1   Comments: 0

Question Number 119138    Answers: 1   Comments: 1

(√(2(√(4(√(8(√(16(√(32...))))))))))=? pls help

$$\sqrt{\mathrm{2}\sqrt{\mathrm{4}\sqrt{\mathrm{8}\sqrt{\mathrm{16}\sqrt{\mathrm{32}...}}}}}=? \\ $$$$\mathrm{pls}\:\mathrm{help} \\ $$

Question Number 119136    Answers: 1   Comments: 2

Question Number 119177    Answers: 1   Comments: 0

Question Number 119133    Answers: 0   Comments: 0

Informatica (11110000)_2 =0•2^0 +0•2^1 +0•2^2 +0•2^3 +1•2^4 +1•2^5 +1•2^6 +1•2^7 = =0•1+0•2+0•4+0•8+1•16+1•32+1•64+1•128= 0+0+0+0+16+32+64+128=(240)_(10) (11000101)_2 =1•2^0 +0•2^1 +1•2^2 +0•2^3 +0•2^4 +0•2^5 +1•2^6 +1•2^7 = = 1•1+0•2+1•4+0•8+0•16+0•32+1•64+1•128= (197)_(10) (01101001)_2 =

$$\mathrm{Informatica} \\ $$$$\left(\mathrm{11110000}\right)_{\mathrm{2}} =\mathrm{0}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\mathrm{0}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{0}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{1}\bullet\mathrm{16}+\mathrm{1}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}= \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}=\left(\mathrm{240}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{11000101}\right)_{\mathrm{2}} =\mathrm{1}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\:\mathrm{1}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{1}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{0}\bullet\mathrm{16}+\mathrm{0}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}=\:\left(\mathrm{197}\right)_{\mathrm{10}} \\ $$$$\left(\mathrm{01101001}\right)_{\mathrm{2}} = \\ $$

Question Number 119124    Answers: 2   Comments: 0

if matrix A^2 = (((1 3)),((0 1)) ) then matrix A=...

$$\mathrm{if}\:\mathrm{matrix}\:\mathrm{A}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{1}\:\:\mathrm{3}}\\{\mathrm{0}\:\:\mathrm{1}}\end{pmatrix}\:\:\mathrm{then}\:\mathrm{matrix}\:\mathrm{A}=... \\ $$

Question Number 119119    Answers: 2   Comments: 0

Question Number 119114    Answers: 0   Comments: 4

Question Number 119112    Answers: 0   Comments: 2

Seems my recent post was removed by Tinkutara.

$$\boldsymbol{\mathrm{Seems}}\:\boldsymbol{\mathrm{my}}\:\boldsymbol{\mathrm{recent}}\:\boldsymbol{\mathrm{post}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{removed}}\:\boldsymbol{\mathrm{by}} \\ $$$$\boldsymbol{\mathrm{Tinkutara}}.\: \\ $$

Question Number 119109    Answers: 1   Comments: 0

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