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AlgebraQuestion and Answers: Page 227

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 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 119147    Answers: 0   Comments: 0

Question Number 119141    Answers: 1   Comments: 0

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 119119    Answers: 2   Comments: 0

Question Number 119107    Answers: 1   Comments: 0

Question Number 119059    Answers: 1   Comments: 0

Question Number 119055    Answers: 2   Comments: 0

Show by recurence that: ∀ n∈N^∗ Σ_(k=1) ^n (1/(k(k+1)(k+2)))=((n(n+3))/(4(n+1)(n+2)))

$$ \\ $$$${Show}\:{by}\:{recurence}\:{that}:\:\forall\:{n}\in\mathbb{N}^{\ast} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)}=\frac{{n}\left({n}+\mathrm{3}\right)}{\mathrm{4}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)} \\ $$

Question Number 119052    Answers: 3   Comments: 0

Question Number 118967    Answers: 0   Comments: 0

lim_(x→∞) ((((x!)/x^x ))^(1/x) )^(((2∙4)/(1∙3))∙((4∙6)/(3∙5))∙∙∙) =? (2/1)∙(2/3)∙(4/3)∙(4/5)∙(6/5)∙(6/7)∙∙∙=(π/2) lim_(x→∞) ((((x!)/x^x ))^(1/x) )^((2/1)∙(4/3)∙(4/3)∙(6/5)∙∙∙) =((1/e))^(π/2) =(√(((1/e))^π ))=(1/( (√e^π )))

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}\centerdot\mathrm{4}}{\mathrm{1}\centerdot\mathrm{3}}\centerdot\frac{\mathrm{4}\centerdot\mathrm{6}}{\mathrm{3}\centerdot\mathrm{5}}\centerdot\centerdot\centerdot} =? \\ $$$$\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\frac{\mathrm{6}}{\mathrm{7}}\centerdot\centerdot\centerdot=\frac{\pi}{\mathrm{2}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \right)^{\frac{\mathrm{2}}{\mathrm{1}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{6}}{\mathrm{5}}\centerdot\centerdot\centerdot} =\left(\frac{\mathrm{1}}{{e}}\right)^{\frac{\pi}{\mathrm{2}}} =\sqrt{\left(\frac{\mathrm{1}}{{e}}\right)^{\pi} }=\frac{\mathrm{1}}{\:\sqrt{{e}^{\pi} }} \\ $$

Question Number 118937    Answers: 1   Comments: 0

What is the general rule for factorising (a+b+c)^n ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}\: \\ $$$$\mathrm{for}\:\mathrm{factorising}\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{n}} ? \\ $$

Question Number 118930    Answers: 1   Comments: 0

Let [x] denote the greatest integer ≤x. Then the number of ordered pair (x,y), where x and y are positive integers less than 30 such that [(x/2)]+[((2x)/3)]+[(y/4)]+[((4y)/5)]=((7x)/6)+((21y)/(20)) is (A) 1 (B) 2 (C) 3 (D) 4

$$\mathrm{Let}\:\left[{x}\right]\:\mathrm{denote}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\leqslant{x}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{pair}\:\left({x},\mathrm{y}\right),\:\mathrm{where}\:{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{30}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]+\left[\frac{\mathrm{y}}{\mathrm{4}}\right]+\left[\frac{\mathrm{4y}}{\mathrm{5}}\right]=\frac{\mathrm{7}{x}}{\mathrm{6}}+\frac{\mathrm{21y}}{\mathrm{20}} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$

Question Number 118936    Answers: 3   Comments: 0

show by recurrence that: for n ∈ N^∗ , 2^(6n−5) +3^(2n ) is divisible by 11.

$${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$

Question Number 118907    Answers: 0   Comments: 0

where can u fond a formula for denesting ramanujan type roots?

$${where}\:{can}\:{u}\:{fond}\:{a}\:{formula}\:{for} \\ $$$${denesting}\:{ramanujan}\:{type}\:{roots}? \\ $$

Question Number 118896    Answers: 1   Comments: 0

Question Number 118849    Answers: 1   Comments: 2

Question Number 118790    Answers: 0   Comments: 0

Show by recurence that (a+b)^n =Σ_(k=0 ) ^n C_n ^k ×a^k ×b^(n−k)

$$\mathrm{Show}\:\mathrm{by}\:\mathrm{recurence}\:\mathrm{that} \\ $$$$\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}\:} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} ×\mathrm{a}^{\mathrm{k}} ×\mathrm{b}^{\mathrm{n}−\mathrm{k}} \\ $$

Question Number 118780    Answers: 2   Comments: 0

z and z′ ∈ C . show that: 1. zz′^(−) =z^− ×z′^(−) 2. ((z/(z′)))^(−) =(z^− /(z′^(−) ))

$$\mathrm{z}\:\mathrm{and}\:\mathrm{z}'\:\in\:\mathbb{C}\:. \\ $$$$\mathrm{show}\:\mathrm{that}: \\ $$$$\mathrm{1}.\:\:\:\:\:\:\overline {\mathrm{zz}'}=\overset{−} {\mathrm{z}}×\overline {\mathrm{z}'} \\ $$$$\mathrm{2}.\:\:\:\:\:\:\:\overline {\left(\frac{\mathrm{z}}{\mathrm{z}'}\right)}=\frac{\overset{−} {\mathrm{z}}}{\overline {\mathrm{z}'}} \\ $$$$ \\ $$

Question Number 118777    Answers: 1   Comments: 0

lim_(n→∞) (((n!)/n^n ))^(1/n) =lim_(n→∞) ((((√(2nπ)) n^n )/(n^n ×e^n )))^(1/n) ⇒lim_(n→∞) (1/e)((√(2nπ)))^(1/n) =(1/e)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}{n}\pi}\:{n}^{{n}} }{{n}^{{n}} ×{e}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{e}}\left(\sqrt{\mathrm{2}{n}\pi}\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{e}} \\ $$

Question Number 118740    Answers: 1   Comments: 1

f(x+2)+f(x−1)=2x^2 +14 f(x)=?

$${f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 118712    Answers: 4   Comments: 0

Prove the following inequalities: 1)(((n+1)/2))^n >n! for ∀n∈N^∗ ,n>1 2)∣sinnx∣≤n∣sinx∣ for ∀n∈N^∗

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequalities}: \\ $$$$\left.\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} >\mathrm{n}!\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} ,\mathrm{n}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mid\mathrm{sinnx}\mid\leqslant\mathrm{n}\mid\mathrm{sinx}\mid\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \\ $$

Question Number 118651    Answers: 1   Comments: 2

Question Number 118636    Answers: 1   Comments: 2

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