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Question Number 156426 Answers: 1 Comments: 0

Given that tan α and tan β are the roots of the equation x^2 +3ax+4a+1=0, where a>1 and α,β∈(−(π/2),(π/2)). Evaluate tan(((α+β)/2)).

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{tan}\:\alpha\:\mathrm{and}\:\mathrm{tan}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} +\mathrm{3}{ax}+\mathrm{4}{a}+\mathrm{1}=\mathrm{0},\: \\ $$$$\mathrm{where}\:{a}>\mathrm{1}\:\mathrm{and}\:\alpha,\beta\in\left(−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right).\: \\ $$$$\mathrm{Evaluate}\:\mathrm{tan}\left(\frac{\alpha+\beta}{\mathrm{2}}\right). \\ $$

Question Number 156423 Answers: 1 Comments: 2

Question Number 156422 Answers: 1 Comments: 2

Question Number 156421 Answers: 0 Comments: 2

Question Number 156419 Answers: 1 Comments: 1

x is positive integer number can you check if Q=((((x+2)^4 −x^4 ))^(1/3) is a natural number

$$\mathrm{x}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{number}\: \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{check}\:\mathrm{if}\:\mathrm{Q}=\sqrt[{\mathrm{3}}]{\left(\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{x}^{\mathrm{4}} \right.} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{natural}\:\mathrm{number} \\ $$

Question Number 156413 Answers: 2 Comments: 0

Question Number 156404 Answers: 0 Comments: 3

x^4 +bx^2 +cx+d=0 let cx=m+px^2 +x^4 ⇒ 2x^4 +(b+p)x^2 +m+d=0 x^2 =−(((b+p)/4))±(√((((b+p)/4))^2 −(((m+d)/2)))) (p−b)x^2 −2cx+m−d=0 x=(c/(p−b))±(√(((c/(p−b)))^2 −(((m−d)/(p−b))))) ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −8(p−b)^2 (m+d))/(16))) +((2c^2 )/(b−p))∓(√((4c^4 −4c^2 (p−b)(m−d))/((p−b)^2 ))) +m−d=0 Just if m=d ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −16d(p−b))/(16))) +((4c^2 )/(b−p))=0 ⇒ (((b^2 −p^2 )/4)+((4c^2 )/(b−p)))^2 +d(p−b)=(((b^2 −p^2 )^2 )/(16)) ⇒ ((16c^4 )/((b−p)^2 ))+(d+2c^2 )(b+p)=0 let b−p=z z^2 (2b−z)=−((16c^4 )/(2c^2 +d)) z^3 −2bz^2 −k=0 (k>0) if b=−1 , d=0, c→−c z^3 +2z^2 −8c^2 =0 let z=((2c)/t) 8c^3 +8c^2 t−8c^2 t^3 =0 ⇒ t^3 −t=c back to square one.

$$\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{m}+\mathrm{px}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2x}^{\mathrm{4}} +\left(\mathrm{b}+\mathrm{p}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{m}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} =−\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)\pm\sqrt{\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}+\mathrm{d}}{\mathrm{2}}\right)} \\ $$$$\left(\mathrm{p}−\mathrm{b}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2cx}+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\pm\sqrt{\left(\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}−\mathrm{d}}{\mathrm{p}−\mathrm{b}}\right)} \\ $$$$\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} \left(\mathrm{m}+\mathrm{d}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{2c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\mp\sqrt{\frac{\mathrm{4c}^{\mathrm{4}} −\mathrm{4c}^{\mathrm{2}} \left(\mathrm{p}−\mathrm{b}\right)\left(\mathrm{m}−\mathrm{d}\right)}{\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$${Just}\:{if}\:\:{m}={d}\:\:\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16d}\left(\mathrm{p}−\mathrm{b}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\right)^{\mathrm{2}} +\mathrm{d}\left(\mathrm{p}−\mathrm{b}\right)=\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow\:\frac{\mathrm{16c}^{\mathrm{4}} }{\left(\mathrm{b}−\mathrm{p}\right)^{\mathrm{2}} }+\left(\mathrm{d}+\mathrm{2c}^{\mathrm{2}} \right)\left(\mathrm{b}+\mathrm{p}\right)=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{b}−\mathrm{p}=\mathrm{z} \\ $$$$\mathrm{z}^{\mathrm{2}} \left(\mathrm{2b}−\mathrm{z}\right)=−\frac{\mathrm{16c}^{\mathrm{4}} }{\mathrm{2c}^{\mathrm{2}} +\mathrm{d}} \\ $$$$\mathrm{z}^{\mathrm{3}} −\mathrm{2bz}^{\mathrm{2}} −\mathrm{k}=\mathrm{0}\:\:\:\:\left(\mathrm{k}>\mathrm{0}\right) \\ $$$$\mathrm{if}\:\mathrm{b}=−\mathrm{1}\:,\:\mathrm{d}=\mathrm{0},\:\mathrm{c}\rightarrow−\mathrm{c} \\ $$$$\mathrm{z}^{\mathrm{3}} +\mathrm{2z}^{\mathrm{2}} −\mathrm{8c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{z}=\frac{\mathrm{2c}}{\mathrm{t}} \\ $$$$\mathrm{8c}^{\mathrm{3}} +\mathrm{8c}^{\mathrm{2}} \mathrm{t}−\mathrm{8c}^{\mathrm{2}} \mathrm{t}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{t}^{\mathrm{3}} −\mathrm{t}=\mathrm{c} \\ $$$$\mathrm{back}\:\mathrm{to}\:\mathrm{square}\:\mathrm{one}. \\ $$$$ \\ $$$$ \\ $$

Question Number 156403 Answers: 0 Comments: 1

find two numbers between (1/3)and (3/4)

$$\mathrm{find}\:\mathrm{two}\:\mathrm{numbers}\:\mathrm{between}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{and}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Question Number 156400 Answers: 0 Comments: 1

Question Number 156384 Answers: 0 Comments: 0

Question Number 156381 Answers: 0 Comments: 1

∫_0 ^( (𝛑/2)) (x^2 /(sin(x))) dx

$$\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}\:\boldsymbol{{dx}} \\ $$

Question Number 156386 Answers: 1 Comments: 0

φ (n )= Σ_(k=1) ^n (−1 )^( k−1) ((( n)),(( k)) ) H_( k) Find the value of : Σ_(n=1) ^∞ (−1)^( n−1) φ ( n^( 2) ) =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\phi\:\left({n}\:\right)=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\:\right)^{\:{k}−\mathrm{1}} \begin{pmatrix}{\:{n}}\\{\:{k}}\end{pmatrix}\:\mathrm{H}_{\:{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Find}\:\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} \:\phi\:\left(\:{n}^{\:\mathrm{2}} \right)\:=? \\ $$

Question Number 156373 Answers: 1 Comments: 0

A. lim_(x→+∞) Σ_(k=1) ^n [tan((kπ)/(2n))]

$$ \\ $$$$\:\:\mathrm{A}.\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\mathrm{tan}\frac{{k}\pi}{\mathrm{2n}}\right] \\ $$

Question Number 156369 Answers: 1 Comments: 2

solve by betta function ∫_0 ^(2π) sin^5 (z) dz

$${solve}\:{by}\:{betta}\:{function}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{5}} \left({z}\right)\:{dz} \\ $$

Question Number 156367 Answers: 0 Comments: 0

lim→oo nΣ_(k=0) ^(2n−1) (1/k^2 )

$$\:{lim}\rightarrow{oo}\:{n}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 156363 Answers: 0 Comments: 0

∫_0 ^(π/2) t^(2n) sin^(2n) tdt

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}^{\mathrm{2}{n}} {sin}^{\mathrm{2}{n}} {tdt} \\ $$

Question Number 156362 Answers: 1 Comments: 0

Question Number 156361 Answers: 1 Comments: 2

Solve in R (1/( (√(x^2 - 1)))) =1− (2/x)

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\mathrm{1}−\:\frac{\mathrm{2}}{\mathrm{x}} \\ $$

Question Number 156359 Answers: 0 Comments: 0

if a;b;c>0 and ab+bc+ca=3 then prove that: (((a^3 + 1)(b^3 + 1)(c^3 + 1)))^(1/3) ≥ 2

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\mathrm{3} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{1}\right)\left(\mathrm{c}^{\mathrm{3}} \:+\:\mathrm{1}\right)}\:\geqslant\:\mathrm{2} \\ $$

Question Number 156339 Answers: 1 Comments: 0

∫_0 ^1 (x^(2n) /(arcsinx))dx=...???

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{arcsinx}}{dx}=...??? \\ $$

Question Number 156338 Answers: 1 Comments: 2

Solve in R (1/( (√(x^2 - 1)))) = (2/x) - 1

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:-\:\mathrm{1} \\ $$$$ \\ $$

Question Number 156335 Answers: 1 Comments: 1

Question Number 156333 Answers: 1 Comments: 0

Question Number 156395 Answers: 0 Comments: 0

A car is moving along a straight road level when the driver see a boy crossing the road 1.5m when the driver immediately applies the brakes which produces constant retardation in the first second. After applying the brakes the car travels 25m and in the next second it traveles 15m. i) find the retardation in m/s^2 ii) show that the car comes to rest at the point where it started

$${A}\:{car}\:{is}\:{moving}\:{along}\:{a}\:{straight}\:{road}\:{level}\:{when}\: \\ $$$${the}\:{driver}\:{see}\:{a}\:{boy}\:{crossing}\:{the}\:{road}\:\mathrm{1}.\mathrm{5}{m}\:{when}\:{the}\:{driver} \\ $$$${immediately}\:{applies}\:{the}\:{brakes}\:{which}\:{produces} \\ $$$$\:{constant}\:{retardation}\:{in}\:{the}\:{first}\:{second}.\:{After}\:{applying}\:{the}\:{brakes} \\ $$$${the}\:{car}\:{travels}\:\mathrm{25}{m}\:{and}\:{in}\:{the}\:{next}\:{second}\:{it} \\ $$$${traveles}\:\mathrm{15}{m}.\: \\ $$$$\left.{i}\right)\:{find}\:{the}\:{retardation}\:{in}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left.{ii}\right)\:{show}\:{that}\:{the}\:{car}\:{comes}\:{to}\:{rest}\:{at}\: \\ $$$${the}\:{point}\:{where}\:{it}\:{started} \\ $$

Question Number 156330 Answers: 0 Comments: 0

Question Number 156328 Answers: 1 Comments: 4

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