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Question Number 156458 Answers: 0 Comments: 0

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Solve for real numbers: 9 + log_2 (x/(x^4 + 48)) = 2 + (2/( (√(x - 1))))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{9}\:+\:\mathrm{log}_{\mathrm{2}} \:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{48}}\:=\:\mathrm{2}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{x}\:-\:\mathrm{1}}} \\ $$

Question Number 156452 Answers: 0 Comments: 0

Find x;y;z≥0 such that: { ((x-y-z = sinx-siny-sinz)),((x^2 -y^2 -z^2 = sin^2 x-sin^2 y-sin^2 z)),((x^3 -y^3 -z^3 = sin^3 x-sin^3 y-sin^3 z)) :}

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If 0<a≤b<π then prove that: ((sin(√(ab)))/(sin(((a+b)/2)))) ≥ ((32a^2 b^2 (√(ab)))/((a+b)^5 ))

$$\mathrm{If}\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}<\pi\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{sin}\sqrt{\mathrm{ab}}}{\mathrm{sin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)}\:\geqslant\:\frac{\mathrm{32a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \sqrt{\mathrm{ab}}}{\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{5}} } \\ $$

Question Number 156447 Answers: 1 Comments: 0

Find all functions f : Z → R such that f(n+m)=nf(n)+mf(m)+nm-n-m ∀n;m∈Z

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}\::\:\mathbb{Z}\:\rightarrow\:\mathbb{R}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{n}+\mathrm{m}\right)=\mathrm{nf}\left(\mathrm{n}\right)+\mathrm{mf}\left(\mathrm{m}\right)+\mathrm{nm}-\mathrm{n}-\mathrm{m} \\ $$$$\forall\mathrm{n};\mathrm{m}\in\mathbb{Z} \\ $$

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in a second order differenti equation, a 4 henry inductor,an 8 ohm resistor and o.2 farad capacito are connected in series with the temperature of the battery with ggl. E=80 sin 3t. solid=0 the charge in the capacitor and the current in the circuit is zero. a.charge b.current at t>0

$$\mathrm{in}\:\mathrm{a}\:\mathrm{second}\:\mathrm{order}\:\mathrm{differenti}\:\mathrm{equation},\:\mathrm{a}\:\mathrm{4}\:\mathrm{henry} \\ $$$$\mathrm{inductor},\mathrm{an}\:\mathrm{8}\:\mathrm{ohm}\:\mathrm{resistor}\:\mathrm{and}\:\mathrm{o}.\mathrm{2}\:\mathrm{farad}\:\mathrm{capacito} \\ $$$$\mathrm{are}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{the}\:\mathrm{temperature}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{battery}\:\mathrm{with}\:\mathrm{ggl}.\:\mathrm{E}=\mathrm{80}\:\mathrm{sin}\:\mathrm{3t}.\:\mathrm{solid}=\mathrm{0}\:\mathrm{the} \\ $$$$\mathrm{charge}\:\mathrm{in}\:\mathrm{the}\:\mathrm{capacitor}\:\mathrm{and}\:\mathrm{the}\:\mathrm{current}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{circuit}\:\mathrm{is}\:\mathrm{zero}. \\ $$$$\mathrm{a}.\mathrm{charge} \\ $$$$\mathrm{b}.\mathrm{current}\:\mathrm{at}\:\mathrm{t}>\mathrm{0} \\ $$

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

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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