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AlgebraQuestion and Answers: Page 63

Question Number 196450    Answers: 3   Comments: 0

one solution of the equation (x−a)(x−b)(x−c)(x−d) = 9 is x=2. If a,b,c,d are different integers then a+b+c+d =?

$$\mathrm{one}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}−\mathrm{b}\right)\left(\mathrm{x}−\mathrm{c}\right)\left(\mathrm{x}−\mathrm{d}\right)\:=\:\mathrm{9}\: \\ $$$$\:\mathrm{is}\:\mathrm{x}=\mathrm{2}.\:\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{different}\: \\ $$$$\:\mathrm{integers}\:\mathrm{then}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:=?\: \\ $$

Question Number 196364    Answers: 1   Comments: 0

xp(x)=x^3 −2x^2 +x−a p(−1)=?

$${xp}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}−{a} \\ $$$${p}\left(−\mathrm{1}\right)=? \\ $$

Question Number 196360    Answers: 2   Comments: 1

Question Number 196355    Answers: 2   Comments: 1

log_5 (√(5(√(5(√(5(√(5.....)))))) ))= ?

$$\mathrm{log}_{\mathrm{5}} \sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}.....}}}\:}=\:? \\ $$

Question Number 196358    Answers: 0   Comments: 2

Question Number 196343    Answers: 2   Comments: 0

find the minimum value of f(x) f(x) = (√(x^2 −2x +5)) + (√(4x^2 −4x +10 ))

$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}\:+\mathrm{5}}\:\:+\:\:\sqrt{\mathrm{4x}^{\mathrm{2}} \:−\mathrm{4x}\:+\mathrm{10}\:} \\ $$

Question Number 196325    Answers: 1   Comments: 0

Σ_(n=0) ^∞ arg(n^2 +n+1+i)= π/2

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{arg}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}+{i}\right)=\:\pi/\mathrm{2}\: \\ $$

Question Number 196300    Answers: 2   Comments: 0

If → n ∈ N and n ≥ 2 Then → tan ((1/(n − 1)) Σ_(k=2) ^n arctan (1/k)) < (2/5) + (𝛄/(n − 1))

$$\mathrm{If}\:\rightarrow\:\mathrm{n}\:\in\:\mathbb{N}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{Then}\:\rightarrow\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{n}\:−\:\mathrm{1}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{k}}\right)\:<\:\frac{\mathrm{2}}{\mathrm{5}}\:+\:\frac{\boldsymbol{\gamma}}{\mathrm{n}\:−\:\mathrm{1}} \\ $$

Question Number 196299    Answers: 1   Comments: 0

If → y = x ! find → (dy/dx)

$$\mathrm{If}\:\rightarrow\:\mathrm{y}\:=\:\mathrm{x}\:!\:\:\:\:\:\mathrm{find}\:\rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$

Question Number 196288    Answers: 3   Comments: 1

4^x =(√5^y )=400 ((xy)/(2x+y))=?

$$\mathrm{4}^{{x}} =\sqrt{\mathrm{5}^{{y}} }=\mathrm{400} \\ $$$$\frac{{xy}}{\mathrm{2}{x}+{y}}=? \\ $$

Question Number 196285    Answers: 0   Comments: 1

problem 196258 (please)

$${problem}\:\mathrm{196258}\:\left({please}\right) \\ $$

Question Number 196185    Answers: 1   Comments: 0

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

Question Number 196183    Answers: 2   Comments: 0

Question Number 196165    Answers: 3   Comments: 0

{: ((x^(x+y) =y^(24) )),((y^(x+y) =x^6 )) }⇒(x,y)=(?,?)

$$\left.\begin{matrix}{{x}^{{x}+{y}} ={y}^{\mathrm{24}} }\\{{y}^{{x}+{y}} ={x}^{\mathrm{6}} }\end{matrix}\right\}\Rightarrow\left({x},{y}\right)=\left(?,?\right) \\ $$

Question Number 196121    Answers: 0   Comments: 0

In forest , a hunter obstains that Every morning a snake eats a mouse Every afternoon a scorpion kills a snake Every night a mouse corrodes a scorpion The 8^(th) day morning , there remains Only one of them , a mouse How many were they, in each species?

$${In}\:{forest}\:,\:{a}\:{hunter}\:{obstains}\:{that} \\ $$$${Every}\:{morning}\:{a}\:{snake}\:{eats}\:{a}\:{mouse} \\ $$$${Every}\:{afternoon}\:{a}\:{scorpion}\:{kills}\:{a}\:{snake} \\ $$$${Every}\:{night}\:{a}\:{mouse}\:{corrodes}\:{a}\:{scorpion} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{morning}\:,\:{there}\:{remains} \\ $$$${Only}\:{one}\:{of}\:{them}\:,\:{a}\:{mouse} \\ $$$$ \\ $$$${How}\:{many}\:{were}\:{they},\:{in}\:{each}\:{species}? \\ $$

Question Number 196119    Answers: 5   Comments: 0

solve (√(100−x^2 ))+(√(64−x^2 ))=12

$${solve} \\ $$$$\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{64}−{x}^{\mathrm{2}} }=\mathrm{12} \\ $$

Question Number 196102    Answers: 0   Comments: 0

Question Number 196086    Answers: 0   Comments: 0

Answer to the question “196008” Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1) Ans) according “de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) sinα− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 +.... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α+...] for “α_k =((kπ)/(2n+1)) ; 1≤k≤n ⇒sin(2n+1)α_k =0 ⇒∀ 1≤k≤n→ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α_k +...=0 therefore “ x_k =tan^2 α_k ” thr roots of the equation are blowe x^n − (((2n+1)),((2n−1)) ) x^n + (((2n+1)),((2n−3)) )x^(n−1) −...=0 the sume of the roots of the equation is “ s= (((2n+1)),((2n−1)) ) ” ⇒s=Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1)✓ the proof of the seconf part is done similarly

$${Answer}\:{to}\:{the}\:{question}\:``\mathrm{196008}'' \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left.{Ans}\right) \\ $$$${according}\:\:``{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} {sin}\alpha−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +.... \\ $$$$=\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)\left[\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha+...\right] \\ $$$${for}\:\:``\alpha_{{k}} =\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:\:;\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha_{{k}} =\mathrm{0} \\ $$$$\Rightarrow\forall\:\:\mathrm{1}\leqslant{k}\leqslant{n}\rightarrow\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha_{{k}} +...=\mathrm{0} \\ $$$${therefore}\:\:``\:{x}_{{k}} ={tan}^{\mathrm{2}} \alpha_{{k}} \:''\:{thr}\:{roots}\:{of}\:\:{the}\:{equation}\:{are}\:{blowe} \\ $$$${x}^{{n}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:{x}^{{n}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{3}}\end{pmatrix}{x}^{{n}−\mathrm{1}} −...=\mathrm{0} \\ $$$${the}\:{sume}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{is}\:``\:{s}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:'' \\ $$$$\Rightarrow{s}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\checkmark \\ $$$${the}\:{proof}\:{of}\:{the}\:{seconf}\:{part}\:{is}\:{done}\:{similarly} \\ $$$$ \\ $$$$ \\ $$

Question Number 196070    Answers: 1   Comments: 0

During an invasion In the amazon dominion Every evening , every invader kills an amazon warrior Every morning , every amazon kills an invader soldier The 8^(th) day evening , there remains only one amazon and no invader How many were they in each part ?

$${During}\:{an}\:{invasion}\: \\ $$$${In}\:{the}\:{amazon}\:{dominion} \\ $$$${Every}\:{evening}\:,\:{every}\:\:{invader}\: \\ $$$${kills}\:\:{an}\:\:{amazon}\:{warrior}\: \\ $$$${Every}\:{morning}\:,\:{every}\:{amazon} \\ $$$${kills}\:\:\:{an}\:{invader}\:{soldier} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{evening}\:,\:{there}\:{remains}\: \\ $$$${only}\:{one}\:{amazon}\:\:{and}\:{no}\:{invader} \\ $$$$ \\ $$$${How}\:{many}\:\:{were}\:{they}\:{in}\:{each}\:{part}\:? \\ $$

Question Number 196066    Answers: 2   Comments: 0

((( 1 −1)),((−1 2)) )^(−8) = ?

$$\begin{pmatrix}{\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}^{−\mathrm{8}} =\:\:? \\ $$

Question Number 196056    Answers: 5   Comments: 0

solve (√(2x+3))−(√(3x+8))=(√(3x+4))−(√(2x+7))

$${solve} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}−\sqrt{\mathrm{3}{x}+\mathrm{8}}=\sqrt{\mathrm{3}{x}+\mathrm{4}}−\sqrt{\mathrm{2}{x}+\mathrm{7}} \\ $$

Question Number 196024    Answers: 1   Comments: 0

lim_(x→0^+ ) [xΣ_(n=1) ^∞ ((1/n^(x+1) ))]=λ , evalute λ

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{{x}+\mathrm{1}} }\right)\right]=\lambda\:,\:{evalute}\:\lambda \\ $$

Question Number 196015    Answers: 0   Comments: 1

Question Number 195998    Answers: 0   Comments: 0

We can transform to get rid of the ((...))^(1/3) a^(1/3) +b^(1/3) =c^(1/3) (a^(1/3) +b^(1/3) )^3 =c a+b+3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )=c 3a^(1/3) b^(1/3) c^(1/3) =c−a−b 27abc=(c−a−b)^3 Is it possible to do the same for a^(1/3) +b^(1/3) =c^(1/3) +d^(1/3) I found no path yet...

$$\mathrm{We}\:\mathrm{can}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{get}\:\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\sqrt[{\mathrm{3}}]{...} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} ={c} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)={c} \\ $$$$\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} {c}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}−{a}−{b} \\ $$$$\mathrm{27}{abc}=\left({c}−{a}−{b}\right)^{\mathrm{3}} \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{do}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} +{d}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{no}\:\mathrm{path}\:\mathrm{yet}... \\ $$

Question Number 195971    Answers: 3   Comments: 0

if f′(x)=((f(x+a)−f(x))/a), find f(x).

$${if}\:{f}'\left({x}\right)=\frac{{f}\left({x}+{a}\right)−{f}\left({x}\right)}{{a}},\:{find}\:{f}\left({x}\right). \\ $$

Question Number 195911    Answers: 0   Comments: 0

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