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AlgebraQuestion and Answers: Page 66

Question Number 195124    Answers: 1   Comments: 0

Question Number 195122    Answers: 0   Comments: 0

Question Number 195116    Answers: 0   Comments: 2

hi. please represntation (2)^(1/3) on number′s axis with ruler and compass and pen. thank you

$${hi}.\:{please}\:{represntation}\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:{on}\:{number}'{s}\:{axis} \\ $$$${with}\:{ruler}\:{and}\:{compass}\:{and}\:{pen}. \\ $$$${thank}\:{you} \\ $$

Question Number 195121    Answers: 2   Comments: 0

x^2 −x−1=0 x^8 +2x^7 −47x=?

$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{7}} −\mathrm{47}{x}=? \\ $$

Question Number 195107    Answers: 1   Comments: 0

Question Number 195101    Answers: 2   Comments: 0

(√(ln2)) >^? ln2

$$\sqrt{{ln}\mathrm{2}}\:\:\overset{?} {>}{ln}\mathrm{2} \\ $$

Question Number 195027    Answers: 1   Comments: 2

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) find (a_1 )

$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 195017    Answers: 3   Comments: 0

(x+1)^3 =1 x=?

$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1}\:\:\:\:\:\:\:\:\:\:{x}=? \\ $$

Question Number 195002    Answers: 1   Comments: 0

x^(√x) =((√x))^x x=?

$${x}^{\sqrt{{x}}} =\left(\sqrt{{x}}\right)^{{x}} \:\:\:\:\:{x}=? \\ $$

Question Number 194991    Answers: 0   Comments: 4

a_3 x^3 −x^2 +a_1 x−7=0 is a cubic polynomial in x whose Roots are α , β , γ positive real numbers satisfying ((225α^2 )/(α^2 +7))=((144β^2 )/(β^2 +7))=((100γ^2 )/(γ^2 +7)) Find (a_1 )

$$ \\ $$$${a}_{\mathrm{3}} {x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}−\mathrm{7}=\mathrm{0}\:{is}\:{a}\:{cubic}\:{polynomial}\:{in}\:{x} \\ $$$${whose}\:{Roots}\:{are}\:\alpha\:,\:\beta\:,\:\gamma\:{positive}\:{real}\:{numbers} \\ $$$${satisfying} \\ $$$$\frac{\mathrm{225}\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{144}\beta^{\mathrm{2}} }{\beta^{\mathrm{2}} +\mathrm{7}}=\frac{\mathrm{100}\gamma^{\mathrm{2}} }{\gamma^{\mathrm{2}} +\mathrm{7}} \\ $$$${Find}\:\left({a}_{\mathrm{1}} \right) \\ $$

Question Number 194953    Answers: 1   Comments: 0

Let P(x)= x^2 +(x/2)+b and Q(x)=x^2 +cx+d be two polynomial with real coefficients such that P(x)Q(x)= Q(P(x)) for all real x . Find all the real roots of P(Q(x))=0

$$\:{Let}\:{P}\left({x}\right)=\:{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}+{b}\:{and} \\ $$$$\:\:{Q}\left({x}\right)={x}^{\mathrm{2}} +{cx}+{d}\:{be}\:{two}\: \\ $$$$\:\:{polynomial}\:{with}\:{real}\:{coefficients} \\ $$$$\:\:{such}\:{that}\:{P}\left({x}\right){Q}\left({x}\right)=\:{Q}\left({P}\left({x}\right)\right) \\ $$$$\:{for}\:{all}\:{real}\:{x}\:. \\ $$$$\:\:{Find}\:{all}\:{the}\:{real}\:{roots}\:{of}\: \\ $$$$\:\:{P}\left({Q}\left({x}\right)\right)=\mathrm{0}\: \\ $$

Question Number 194952    Answers: 1   Comments: 0

x + (√(17−x^2 )) + x(√(17−x^2 )) =9 find the possible value of X

$$\boldsymbol{\mathrm{x}}\:+\:\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\:+\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{17}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\:=\mathrm{9} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{X}} \\ $$

Question Number 194942    Answers: 2   Comments: 2

$$\:\:\:\:\: \\ $$

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)

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

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