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Question Number 194853 Answers: 3 Comments: 0

Question Number 194852 Answers: 1 Comments: 0

lim_(x→0) (cosx)^(log(x)) =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\mathrm{log}\left(\mathrm{x}\right)} =? \\ $$$$ \\ $$

Question Number 194847 Answers: 0 Comments: 2

z^n = −512+512i find n.

$$\:\:\:\:\: {z}^{{n}} =\:−\mathrm{512}+\mathrm{512}{i}\: \\ $$$$\:\:\:\:{find}\:{n}. \\ $$

Question Number 194846 Answers: 1 Comments: 0

x=((((10+6(√3)))^(1/3) ((√3)−1))/( (√(6+2(√5)))−(√5))) G=(12x^3 +4x^2 −55)^(2023) =?

$$\:\:\:\:\: {x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}+\mathrm{6}\sqrt{\mathrm{3}}}\:\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}−\sqrt{\mathrm{5}}} \\ $$$$\:\:\:\: {G}=\left(\mathrm{12}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{55}\right)^{\mathrm{2023}} \:=? \\ $$

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 194836 Answers: 0 Comments: 1

Question Number 194831 Answers: 1 Comments: 1

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 194850 Answers: 0 Comments: 0

Question Number 194826 Answers: 1 Comments: 0

tan 19° = p tan 7° =?

$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{19}°\:=\:{p}\: \\ $$$$\:\:\:\:\:\:\mathrm{tan}\:\mathrm{7}°\:=? \\ $$

Question Number 194821 Answers: 0 Comments: 2

M a inside poin in ΔABC. M = bar {(A, area(MBC)), (B, area(MAC)),(C,area(MAB))}

$${M}\:{a}\:{inside}\:{poin}\:{in}\:\:\Delta{ABC}. \\ $$$${M}\:=\:{bar}\:\left\{\left({A},\:{area}\left({MBC}\right)\right),\:\left({B},\:{area}\left({MAC}\right)\right),\left({C},{area}\left({MAB}\right)\right)\right\} \\ $$

Question Number 194819 Answers: 0 Comments: 0

Question Number 194818 Answers: 1 Comments: 0

Question Number 194815 Answers: 1 Comments: 0

Question Number 194812 Answers: 2 Comments: 0

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

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

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 194796 Answers: 1 Comments: 0

A 1m^2 rectangle which length is less than 1 is a square. Why?

$${A}\:\mathrm{1}{m}^{\mathrm{2}} \:{rectangle}\:{which}\:{length}\:{is}\: \\ $$$${less}\:{than}\:\mathrm{1}\:{is}\:\:{a}\:{square}.\:{Why}? \\ $$

Question Number 194791 Answers: 1 Comments: 0

of x (((x−(√2)))^(1/7) /2) −(((x−(√2)))^(1/7) /x^2 ) = (x/2) ((x^2 /(x+(√2))))^(1/7)

$$\:\: \: \: \:{of}\:{x}\: \\ $$$$\:\:\frac{\sqrt[{\mathrm{7}}]{{x}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:−\frac{\sqrt[{\mathrm{7}}]{{x}−\sqrt{\mathrm{2}}}}{{x}^{\mathrm{2}} }\:=\:\frac{{x}}{\mathrm{2}}\:\sqrt[{\mathrm{7}}]{\frac{{x}^{\mathrm{2}} }{{x}+\sqrt{\mathrm{2}}}}\:\:\: \\ $$

Question Number 194790 Answers: 1 Comments: 0

Question Number 194786 Answers: 1 Comments: 0

x^n +y^n =¿ (n∈N^∗ )

$${x}^{{n}} +{y}^{{n}} =¿\:\left({n}\in{N}^{\ast} \right) \\ $$

Question Number 194785 Answers: 0 Comments: 0

∫∫ x^2 +y^2 dxdy (D=x^4 +y^4 ≤1)

$$\int\int\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{dxdy}\:\left({D}={x}^{\mathrm{4}} +{y}^{\mathrm{4}} \leqslant\mathrm{1}\right) \\ $$

Question Number 194781 Answers: 0 Comments: 0

f_(n ) the general sentence is seqiencee fibonacci. prove that : f_(2n−1) =f_n ^2 +f_(n−1) ^2

$${f}_{{n}\:} \:\:{the}\:{general}\:{sentence}\:{is}\:{seqiencee} \\ $$$${fibonacci}.\: \\ $$$${prove}\:{that}\::\:\:{f}_{\mathrm{2}{n}−\mathrm{1}} ={f}_{{n}} ^{\mathrm{2}} +{f}_{{n}−\mathrm{1}} ^{\mathrm{2}} \\ $$$$ \\ $$

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 194767 Answers: 2 Comments: 0

tan θ = 2 ((8sin θ+5cos θ)/(sin^3 θ+cos^3 θ+cos θ)) =?

$$\:\:\:\:\:\mathrm{tan}\:\theta\:=\:\mathrm{2}\: \\ $$$$\:\:\:\frac{\mathrm{8sin}\:\theta+\mathrm{5cos}\:\theta}{\mathrm{sin}\:^{\mathrm{3}} \theta+\mathrm{cos}\:^{\mathrm{3}} \theta+\mathrm{cos}\:\theta}\:=? \\ $$

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