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AlgebraQuestion and Answers: Page 46

Question Number 204129 Answers: 2 Comments: 0

a , b , x , y ∈ R a + b = 23 ax + by = 79 ax^2 + by^2 = 217 ax^3 + by^3 = 661 Find: ax^4 + by^4 = ?

$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{x}\:,\:\mathrm{y}\:\in\:\mathbb{R} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{23} \\ $$$$\mathrm{ax}\:+\:\mathrm{by}\:=\:\mathrm{79} \\ $$$$\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{by}^{\mathrm{2}} \:=\:\mathrm{217} \\ $$$$\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{by}^{\mathrm{3}} \:=\:\mathrm{661} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{by}^{\mathrm{4}} \:=\:? \\ $$

Question Number 204123 Answers: 2 Comments: 1

Question Number 204082 Answers: 1 Comments: 3

(√(((√(x^2 +66^2 ))+x)/x))−(√(x(√(x^2 +66^2 ))−x^2 ))=5

$$\sqrt{\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }+{x}}{{x}}}−\sqrt{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{66}^{\mathrm{2}} }−{x}^{\mathrm{2}} }=\mathrm{5} \\ $$

Question Number 204056 Answers: 2 Comments: 0

{ ((x + 2y + z = 8)),((3x + 2y + z = 10)),((4x + 3y − 2z = 4)) :} Solve with the help of matrix

$$\begin{cases}{\mathrm{x}\:+\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{8}}\\{\mathrm{3x}\:+\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{10}}\\{\mathrm{4x}\:+\:\mathrm{3y}\:−\:\mathrm{2z}\:=\:\mathrm{4}}\end{cases} \\ $$$$\mathrm{Solve}\:\mathrm{with}\:\mathrm{the}\:\mathrm{help}\:\mathrm{of}\:\mathrm{matrix} \\ $$

Question Number 204055 Answers: 2 Comments: 0

y = (2/3) arctg (x^4 ) find: y^′ = ?

$$\mathrm{y}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{arctg}\:\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$

Question Number 204054 Answers: 2 Comments: 0

y = (√(sinx)) + cos^3 x find: y^′ = ?

$$\mathrm{y}\:=\:\sqrt{\mathrm{sinx}}\:+\:\mathrm{cos}^{\mathrm{3}} \mathrm{x} \\ $$$$\mathrm{find}:\:\:\mathrm{y}^{'} \:=\:? \\ $$

Question Number 204041 Answers: 2 Comments: 0

Find: determinant ((1,7,(−1)),(9,(−3),5),((−1),5,3))= ?

$$\mathrm{Find}:\:\:\:\begin{vmatrix}{\mathrm{1}}&{\mathrm{7}}&{−\mathrm{1}}\\{\mathrm{9}}&{−\mathrm{3}}&{\mathrm{5}}\\{−\mathrm{1}}&{\mathrm{5}}&{\mathrm{3}}\end{vmatrix}=\:? \\ $$

Question Number 204039 Answers: 2 Comments: 0

determinant ((1,7,(−1)),(9,(−3),x),((−1),5,3))= 0 ⇒ x = ?

$$\begin{vmatrix}{\mathrm{1}}&{\mathrm{7}}&{−\mathrm{1}}\\{\mathrm{9}}&{−\mathrm{3}}&{\boldsymbol{\mathrm{x}}}\\{−\mathrm{1}}&{\mathrm{5}}&{\mathrm{3}}\end{vmatrix}=\:\mathrm{0}\:\:\:\Rightarrow\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 204038 Answers: 1 Comments: 0

y = (x^3 + 1) ∙ 3^x ⇒ y^′ = ?

$$\mathrm{y}\:=\:\left(\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}\right)\:\centerdot\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \\ $$$$\Rightarrow\:\mathrm{y}^{'} \:=\:? \\ $$

Question Number 204019 Answers: 1 Comments: 2

G is a group : prove that : (G/(Z (G ))) ≅ Inn(G ) Where , Inn(G)= {f ∣ f: G →^(f is an Automorphism) G}

$$ \\ $$$$\:\:\:\:\:{G}\:{is}\:{a}\:{group}\:: \\ $$$$\:\:\:\:\:{prove}\:{that}\::\:\:\frac{{G}}{{Z}\:\left({G}\:\right)}\:\cong\:{Inn}\left({G}\:\right) \\ $$$$\:\:\:\:{Where}\:,\:{Inn}\left({G}\right)=\:\left\{{f}\:\mid\:{f}:\:{G}\:\overset{{f}\:{is}\:{an}\:{Automorphism}} {\rightarrow}\:{G}\right\} \\ $$$$ \\ $$

Question Number 204018 Answers: 1 Comments: 0

Aut (Z )= ? where , Aut (G )= { f ∣ f :G →_(G is a group) ^(f is a isomorphism) G}

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{Aut}\:\left(\mathbb{Z}\:\right)=\:? \\ $$$$\:\:\:\:\:\:\:{where}\:,\:{Aut}\:\left({G}\:\right)=\:\left\{\:{f}\:\mid\:{f}\::{G}\:\underset{{G}\:{is}\:{a}\:{group}} {\overset{{f}\:{is}\:{a}\:{isomorphism}} {\rightarrow}}\:{G}\right\} \\ $$

Question Number 204005 Answers: 2 Comments: 1

Find: cos44° − cos84° + ctg45° = ?

$$\mathrm{Find}: \\ $$$$\mathrm{cos44}°\:−\:\mathrm{cos84}°\:+\:\mathrm{ctg45}°\:=\:? \\ $$

Question Number 203995 Answers: 3 Comments: 0

3x+4x=5

$$\mathrm{3}{x}+\mathrm{4}{x}=\mathrm{5} \\ $$

Question Number 203975 Answers: 0 Comments: 0

Let P∈C[X] p(x^2 +1)=p^2 (x)+1 p(0)=0 determin all polynom

$$\mathrm{Let}\:\mathrm{P}\in\mathbb{C}\left[\mathrm{X}\right] \\ $$$$\mathrm{p}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{p}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1} \\ $$$$\mathrm{p}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{determin}\:\mathrm{all}\:\mathrm{polynom}\: \\ $$

Question Number 203919 Answers: 1 Comments: 0

Question Number 203891 Answers: 1 Comments: 0

Question Number 203881 Answers: 2 Comments: 1

Question Number 203846 Answers: 1 Comments: 0

lim_(n→∞) Π_(i=2) ^n (((i^2 −1)/i^2 ))

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\frac{{i}^{\mathrm{2}} −\mathrm{1}}{{i}^{\mathrm{2}} }\right) \\ $$

Question Number 203826 Answers: 3 Comments: 0

Evaluate (26 + 15(√3))^(1/3) + (26 − 15(√3))^(1/3)

$${Evaluate} \\ $$$$\left(\mathrm{26}\:+\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} +\:\left(\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$

Question Number 203809 Answers: 1 Comments: 0

x^3 +y^3 +z^3 =33

$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{33} \\ $$

Question Number 203807 Answers: 2 Comments: 0

proof : ∫_0 ^1 f^2 (t)dt=0 ⇒ f=0

$$\mathrm{proof}\::\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{f}^{\mathrm{2}} \left(\mathrm{t}\right)\mathrm{dt}=\mathrm{0}\:\Rightarrow\:\mathrm{f}=\mathrm{0} \\ $$

Question Number 203774 Answers: 2 Comments: 0

determine whether the series is convergent or divergent 𝚺_(n=1) ^∞ (n/( (√(4n^2 +1))))

$$\boldsymbol{{determine}}\:\boldsymbol{{whether}}\:\boldsymbol{{the}}\:\boldsymbol{{series}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{convergent}}\:\boldsymbol{{or}}\:\boldsymbol{{divergent}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{{n}}}{\:\sqrt{\mathrm{4}\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}} \\ $$

Question Number 203771 Answers: 1 Comments: 0

Question Number 203742 Answers: 3 Comments: 0

Question Number 203638 Answers: 2 Comments: 2

Question Number 203570 Answers: 0 Comments: 2

Suggested solution method to question 203502 ze^z =1 Obviously the only real solution is z=W(1)≈.567143290 z=a+bi∧b≠0 (a+bi)e^(a+bi) =1 (a+bi)(cos b +isin b)e^a =1 e^a (acos b −bsin b)+e^a (asin b +bcos b)i=1 { ((e^a (acos b −bsin b)=1)),((e^a (asin b +bcos b)=0 ⇒ a=−bcot b)) :} Inserting & transforming leaves us with: { ((be^(−bcot b) +sin b =0)),((a=−bcot b)) :} We can only approximate. The first solutions are: b≈±4.37518515 ⇒ a≈−1.53391332 ⇒ z≈−1.53391332±4.37518515i b≈±10.7762995 ⇒ a≈−2.40158510 ⇒ z≈−2.40158510±10.7762995i These are the values of the complex LambertW−function W_n (1); n∈Z Following this path we can solve ze^z =w with z, w∈Z I hope this is helpful!

$$\mathrm{Suggested}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{to} \\ $$$$\mathrm{question}\:\mathrm{203502} \\ $$$$ \\ $$$${z}\mathrm{e}^{{z}} =\mathrm{1} \\ $$$$\mathrm{Obviously}\:\mathrm{the}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{is} \\ $$$${z}={W}\left(\mathrm{1}\right)\approx.\mathrm{567143290} \\ $$$$ \\ $$$${z}={a}+{b}\mathrm{i}\wedge{b}\neq\mathrm{0} \\ $$$$\left({a}+{b}\mathrm{i}\right)\mathrm{e}^{{a}+{b}\mathrm{i}} =\mathrm{1} \\ $$$$\left({a}+{b}\mathrm{i}\right)\left(\mathrm{cos}\:{b}\:+\mathrm{isin}\:{b}\right)\mathrm{e}^{{a}} =\mathrm{1} \\ $$$$\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)+\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)\mathrm{i}=\mathrm{1} \\ $$$$\begin{cases}{\mathrm{e}^{{a}} \left({a}\mathrm{cos}\:{b}\:−{b}\mathrm{sin}\:{b}\right)=\mathrm{1}}\\{\mathrm{e}^{{a}} \left({a}\mathrm{sin}\:{b}\:+{b}\mathrm{cos}\:{b}\right)=\mathrm{0}\:\Rightarrow\:{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{Inserting}\:\&\:\mathrm{transforming}\:\mathrm{leaves}\:\mathrm{us}\:\mathrm{with}: \\ $$$$\begin{cases}{{b}\mathrm{e}^{−{b}\mathrm{cot}\:{b}} +\mathrm{sin}\:{b}\:=\mathrm{0}}\\{{a}=−{b}\mathrm{cot}\:{b}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}. \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{solutions}\:\mathrm{are}: \\ $$$${b}\approx\pm\mathrm{4}.\mathrm{37518515}\:\Rightarrow\:{a}\approx−\mathrm{1}.\mathrm{53391332} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{1}.\mathrm{53391332}\pm\mathrm{4}.\mathrm{37518515i} \\ $$$${b}\approx\pm\mathrm{10}.\mathrm{7762995}\:\Rightarrow\:{a}\approx−\mathrm{2}.\mathrm{40158510} \\ $$$$\Rightarrow\:{z}\approx−\mathrm{2}.\mathrm{40158510}\pm\mathrm{10}.\mathrm{7762995i} \\ $$$$\mathrm{These}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex} \\ $$$$\mathrm{LambertW}−\mathrm{function}\:{W}_{{n}} \left(\mathrm{1}\right);\:{n}\in\mathbb{Z} \\ $$$$\mathrm{Following}\:\mathrm{this}\:\mathrm{path}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$${z}\mathrm{e}^{{z}} ={w}\:\mathrm{with}\:{z},\:{w}\in\mathbb{Z} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{is}\:\mathrm{helpful}! \\ $$

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