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AlgebraQuestion and Answers: Page 64

Question Number 194914    Answers: 2   Comments: 0

Question Number 194899    Answers: 1   Comments: 0

Question Number 194900    Answers: 1   Comments: 0

Given d = (((2+ (√5)))^(1/3) /(1+(√5))) then d^3 −4d^2 +8d −2 =?

$$\:\:\:\:\:\:{Given}\:\:\:{d}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}+\:\sqrt{\mathrm{5}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\:\: \\ $$$$\:\:\:\:\:\:{then}\:\:{d}^{\mathrm{3}} −\mathrm{4}{d}^{\mathrm{2}} \:+\mathrm{8}{d}\:−\mathrm{2}\:=?\: \\ $$$$\:\:\:\:\: \\ $$

Question Number 194888    Answers: 1   Comments: 0

Question Number 194884    Answers: 1   Comments: 0

x! = 6!. 7! x^2 =?

$$\:\:\:\:\:\:{x}!\:=\:\mathrm{6}!.\:\mathrm{7}!\: \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:=?\: \\ $$

Question Number 194877    Answers: 0   Comments: 1

without calculator Prove: ((π^5 +π^4 ))^(1/6) <e

$${without}\:{calculator}\:{Prove}:\:\sqrt[{\mathrm{6}}]{\pi^{\mathrm{5}} +\pi^{\mathrm{4}} }<{e} \\ $$

Question Number 194853    Answers: 3   Comments: 0

Question Number 194846    Answers: 1   Comments: 0

$$\:\:\:\:\:\:\underbrace{\:} \\ $$

Question Number 194844    Answers: 1   Comments: 0

(x^2 +1)^2 +(x+3)^2 =(x^2 +ax+b)(x^2 +cx+d) Find a,b,c,d.

$$\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\left({x}^{\mathrm{2}} +{ax}+{b}\right)\left({x}^{\mathrm{2}} +{cx}+{d}\right) \\ $$$${Find}\:{a},{b},{c},{d}. \\ $$

Question Number 194837    Answers: 2   Comments: 0

for x>0 find the minimum of the function f(x)=x^3 +(5/x).

$${for}\:{x}>\mathrm{0}\:{find}\:{the}\:{minimum}\:{of}\:{the} \\ $$$${function}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\frac{\mathrm{5}}{{x}}. \\ $$

Question Number 194851    Answers: 0   Comments: 1

Name Zainab Bibi BC200400692 Assignmeng No#2 Mth 621 solution.. a_n =(((n− )!)/((n+ )^2 )) a_(n+ ) =(((n+ − )!)/((n+ + )^2 ))=(((n)!)/((n+2)^2 )) by ratio test (a_(n+ ) /a_n )=((((n)!)/((n+2)))/(((n− )!)/((n+ )^2 )))=(((n)!)/((n+2)^2 ))×(((n+ )^2 )/((n+ )!)) lim_(n→∞) (a_(n+ ) /a_n )=lim_(n→∞) ((n(n− )!)/((n+2)^2 ))×(((n+ )^2 )/((n− )!)) ∵(n)!=n(n− )! lim_(n→∞) (a_(n+ ) /a_n )=lim_(n→∞) ((n(n+ )^2 )/((n+2)^2 ))=lim_(n→∞) ((n(n^2 +1+2n))/(n^2 +4+4n)) lim_(n→∞) (((n^3 +n+2n^2 ))/(n^2 +4+4n))=lim_(n→∞) (((1+(1/n^2 )+(2/n)))/((1/n)+(1/n^3 )+(4/n^2 ))) Divided by n^3 to numerator and denominator Now by Applying limit (((1+(1/∞^(2 ) )+(2/∞)))/((1/∞)+(4/∞^3 )+(4/∞^2 )))=((1+0+0)/(0+0+0))=(1/0)=∞ lim_(n→∞) (a_(n+1) /a_n )=∞

$$\boldsymbol{\mathrm{Name}}\:\:\:\boldsymbol{\mathrm{Zainab}}\:\boldsymbol{\mathrm{Bibi}} \\ $$$$\boldsymbol{\mathrm{BC}}\mathrm{200400692} \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ssignmeng}\:\mathrm{No}#\mathrm{2}\: \\ $$$$\boldsymbol{\mathrm{Mth}}\:\mathrm{621} \\ $$$$\mathrm{solution}.. \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} } \\ $$$$\mathrm{a}_{\mathrm{n}+ } =\frac{\left(\mathrm{n}+ − \right)!}{\left(\mathrm{n}+ + \right)^{\mathrm{2}} }=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{by}\:\mathrm{ratio}\:\mathrm{test} \\ $$$$\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\frac{\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)}}{\frac{\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+ \right)^{\mathrm{2}} }}=\frac{\left(\mathrm{n}\right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+ \right)!} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\:\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}− \right)!}{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}− \right)!}\:\:\:\:\:\because\left(\mathrm{n}\right)!=\mathrm{n}\left(\mathrm{n}− \right)! \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+ } }{\mathrm{a}_{\mathrm{n}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}+ \right)^{\mathrm{2}} }{\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} }=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}+\mathrm{2n}\right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}} \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{n}^{\mathrm{3}} +\mathrm{n}+\mathrm{2n}^{\mathrm{2}} \right)}{\mathrm{n}^{\mathrm{2}} +\mathrm{4}+\mathrm{4n}}=\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{n}}\right)}{\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{n}^{\mathrm{2}} }} \\ $$$$\mathrm{Divided}\:\mathrm{by}\:\mathrm{n}^{\mathrm{3}} \:\mathrm{to}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator}\: \\ $$$$\mathrm{Now}\:\mathrm{by}\:\mathrm{Applying}\:\mathrm{limit} \\ $$$$\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\infty^{\mathrm{2}\:} }+\frac{\mathrm{2}}{\infty}\right)}{\frac{\mathrm{1}}{\infty}+\frac{\mathrm{4}}{\infty^{\mathrm{3}} }+\frac{\mathrm{4}}{\infty^{\mathrm{2}} }}=\frac{\mathrm{1}+\mathrm{0}+\mathrm{0}}{\mathrm{0}+\mathrm{0}+\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{0}}=\infty \\ $$$$\mathrm{li}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }=\infty \\ $$

Question Number 194809    Answers: 2   Comments: 0

If f(x)=ax^2 −5x+3 and g(x)=3x−3 intersection at points (1,h) and (3,t). Find

$$\:\:\:\:{If}\:{f}\left({x}\right)={ax}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\:{and}\: \\ $$$$\:\:\:{g}\left({x}\right)=\mathrm{3}{x}−\mathrm{3}\:{intersection}\:{at} \\ $$$$\:{points}\:\left(\mathrm{1},{h}\right)\:{and}\:\left(\mathrm{3},{t}\right). \\ $$$$\:\:{Find}\:\:\underline{ } \\ $$

Question Number 194808    Answers: 0   Comments: 4

suppose a,b,c are positive real numbers prove the inequality (((a+b)/2))(((b+c)/2))(((c+a)/2))≥(((a+b+c)/3))(((abc)^2 ))^(1/3)

$$ \\ $$$${suppose}\:{a},{b},{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${prove}\:{the}\:{inequality} \\ $$$$\left(\frac{{a}+{b}}{\mathrm{2}}\right)\left(\frac{{b}+{c}}{\mathrm{2}}\right)\left(\frac{{c}+{a}}{\mathrm{2}}\right)\geqslant\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)\sqrt[{\mathrm{3}}]{\left({abc}\right)^{\mathrm{2}} } \\ $$

Question Number 194791    Answers: 1   Comments: 0

x

$$\:\:\:\underline{\underbrace{\boldsymbol{{x}}}} \\ $$

Question Number 194779    Answers: 1   Comments: 4

If a divided by b gives q remaining r Then (a/b) = q,rrr... in base b+1

$${If}\:\:{a}\:\:{divided}\:{by}\:{b}\:{gives}\:{q}\:\:{remaining}\:{r} \\ $$$${Then}\:\:\frac{{a}}{{b}}\:=\:{q},{rrr}...\:\:{in}\:{base}\:{b}+\mathrm{1} \\ $$

Question Number 194756    Answers: 3   Comments: 0

b

$$\:\:\:\:\:\:\underbrace{\boldsymbol{{b}}} \\ $$

Question Number 194710    Answers: 0   Comments: 21

let p be a prime number & let a_1 ,a_2 ,a_3 ,...,a_(p ) be integers show that , there exists an integer k such that the numbers a_1 +k, a_2 +k,a_3 +k,....,a_p +k produce at least (1/2)p distinct remainders when divided by p.

$${let}\:{p}\:{be}\:{a}\:{prime}\:{number} \\ $$$$\&\:{let}\:{a}_{\mathrm{1}} \:,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,...,{a}_{{p}\:} {be}\:{integers} \\ $$$${show}\:{that}\:,\:{there}\:{exists}\:{an}\:{integer}\:{k}\:{such}\:{that}\:{the}\:{numbers} \\ $$$${a}_{\mathrm{1}} +{k},\:{a}_{\mathrm{2}} +{k},{a}_{\mathrm{3}} +{k},....,{a}_{{p}} +{k} \\ $$$${produce}\:{at}\:{least}\:\frac{\mathrm{1}}{\mathrm{2}}{p}\:{distinct}\:{remainders} \\ $$$${when}\:{divided}\:{by}\:{p}. \\ $$

Question Number 194695    Answers: 1   Comments: 0

⇃

$$\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$

Question Number 194648    Answers: 3   Comments: 3

Question Number 194637    Answers: 4   Comments: 1

x+y=1 x^2 +y^2 =2 x^(11) +y^(11) =?

$$ \\ $$$${x}+{y}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{11}} +{y}^{\mathrm{11}} =? \\ $$$$ \\ $$$$ \\ $$

Question Number 194636    Answers: 0   Comments: 3

Question Number 194634    Answers: 1   Comments: 0

a_1 ,a_2 ,a_3 ,....,a_n >0 such that a_i ∈[0,i] ∀ i∈{1,2,3,4,...,n} prove that 2^n .a_1 (a_1 +a_2 )...(a_1 +a_2 +...+a_n )≥(n+1)(a_1 ^2 .a_2 ^2 ...a_n ^2 )

$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,....,{a}_{{n}} >\mathrm{0}\:{such}\:{that}\:{a}_{{i}} \in\left[\mathrm{0},{i}\right]\: \\ $$$$\forall\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},...,{n}\right\}\:{prove}\:{that} \\ $$$$\mathrm{2}^{{n}} .{a}_{\mathrm{1}} \left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)...\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} \right)\geqslant\left({n}+\mathrm{1}\right)\left({a}_{\mathrm{1}} ^{\mathrm{2}} .{a}_{\mathrm{2}} ^{\mathrm{2}} ...{a}_{{n}} ^{\mathrm{2}} \right) \\ $$

Question Number 194619    Answers: 1   Comments: 0

Find the sum of the roots of the equation: −3x^3 + 8x^2 − 6x − 7 = 0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}: \\ $$$$−\mathrm{3x}^{\mathrm{3}} \:+\:\mathrm{8x}^{\mathrm{2}} \:−\:\mathrm{6x}\:−\:\mathrm{7}\:=\:\mathrm{0} \\ $$

Question Number 194612    Answers: 1   Comments: 2

Question Number 194610    Answers: 1   Comments: 0

where can I learn about multiple sigma notaions of dependent and independent variables something like this Σ_(1≤i) Σ_(<j) Σ_(<k≤1) (i+j+k)=λ find λ I want to know what to study

$${where}\:{can}\:{I}\:{learn}\:{about}\:{multiple}\:{sigma}\:{notaions} \\ $$$${of}\:{dependent}\:{and}\:{independent}\:{variables} \\ $$$$ \\ $$$${something}\:{like}\:{this} \\ $$$$\underset{\mathrm{1}\leqslant{i}} {\sum}\underset{<{j}} {\sum}\underset{<{k}\leqslant\mathrm{1}} {\sum}\left({i}+{j}+{k}\right)=\lambda \\ $$$${find}\:\lambda \\ $$$${I}\:{want}\:{to}\:{know}\:{what}\:{to}\:{study} \\ $$

Question Number 194586    Answers: 1   Comments: 2

abc = e^3 + d^3 + f^3 edf = a^3 + b^3 + c^3 find: abc and edf

$$\mathrm{abc}\:=\:\mathrm{e}^{\mathrm{3}} \:+\:\mathrm{d}^{\mathrm{3}} \:+\:\mathrm{f}^{\mathrm{3}} \\ $$$$\mathrm{edf}\:=\:\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \\ $$$$\mathrm{find}:\:\mathrm{abc}\:\:\mathrm{and}\:\:\mathrm{edf}\: \\ $$

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