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Question Number 201897    Answers: 0   Comments: 0

Question: The congruence equation ′′ ( 5a +3 )x ≡^(3a + 4) 19 ′′ is given. Find the sum of digits of the smallest three −digit natural number ” a ” such that the assumed equation has no solution in ” Z ”.

$$ \\ $$$$\:\:\:\:\:\mathrm{Q}{uestion}:\:\:{The}\:{congruence} \\ $$$$\:\:\:\:\:\:\:{equation}\:\:\:''\:\:\:\:\left(\:\mathrm{5}{a}\:+\mathrm{3}\:\right){x}\:\:\overset{\mathrm{3}{a}\:+\:\mathrm{4}} {\equiv}\:\mathrm{19}\:\:\:''\:\:{is}\:{given}. \\ $$$$\:\:\:\:\:\:\:{Find}\:{the}\:{sum}\:{of}\:{digits}\:{of}\:\: \\ $$$$\:\:\:\:\:\:\:{the}\:{smallest}\:\:{three}\:−{digit}\:{natural}\:{number}\:\:''\:{a}\:''\: \\ $$$$\:\:\:\:\:\:\:{such}\:{that}\:{the}\:{assumed}\:{equation}\:{has}\: \\ $$$$\:\:\:\:\:\:\:\:\:{no}\:\:{solution}\:\:{in}\:\:\:''\:\:\:\mathbb{Z}\:\:''. \\ $$$$ \\ $$

Question Number 201848    Answers: 1   Comments: 0

Question Number 201846    Answers: 0   Comments: 1

$$\:\:\: \\ $$

Question Number 201839    Answers: 0   Comments: 1

Question Number 201837    Answers: 0   Comments: 0

y′′′−y′′+y′=sec(t),−(π/2)<t<(π/2)

$${y}'''−{y}''+{y}'={sec}\left({t}\right),−\frac{\pi}{\mathrm{2}}<{t}<\frac{\pi}{\mathrm{2}} \\ $$

Question Number 201982    Answers: 0   Comments: 0

Question Number 201977    Answers: 0   Comments: 1

Question Number 201980    Answers: 1   Comments: 0

Question Number 201979    Answers: 2   Comments: 1

Question Number 201972    Answers: 1   Comments: 0

Question Number 201969    Answers: 2   Comments: 0

A dice is cast twice, and the sum of the appearing numbers is 10. The probability that the number 5 has appeared at least once is.

$$\mathrm{A}\:\mathrm{dice}\:\mathrm{is}\:\mathrm{cast}\:\mathrm{twice},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{appearing}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{10}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{5}\:\mathrm{has}\: \\ $$$$\mathrm{appeared}\:\mathrm{at}\:\mathrm{least}\:\mathrm{once}\:\mathrm{is}. \\ $$

Question Number 201966    Answers: 2   Comments: 0

solve ((x^2 +3x+2))^(1/3) (((x+1))^(1/3) −((x+2))^(1/3) )= 1

$$ \\ $$$$\:\:\:\:\:\:\:{solve}\: \\ $$$$\:\: \\ $$$$\:\:\:\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\right)=\:\mathrm{1} \\ $$$$ \\ $$

Question Number 201965    Answers: 1   Comments: 0

m−h=2p p(m−h)=k−q mk−qh=(1/3) k−2q=1+ph Assume one find the rest! ✓

$${m}−{h}=\mathrm{2}{p} \\ $$$${p}\left({m}−{h}\right)={k}−{q} \\ $$$${mk}−{qh}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${k}−\mathrm{2}{q}=\mathrm{1}+{ph} \\ $$$${Assume}\:{one}\:{find}\:{the}\:{rest}! \\ $$$$\checkmark \\ $$

Question Number 201832    Answers: 1   Comments: 0

Question Number 201831    Answers: 0   Comments: 0

Question Number 201829    Answers: 2   Comments: 0

shortest distance from (−6,0)to x^2 −y^2 +16=0

$${shortest}\:{distance}\:{from}\:\left(−\mathrm{6},\mathrm{0}\right){to}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +\mathrm{16}=\mathrm{0} \\ $$

Question Number 201822    Answers: 0   Comments: 4

Do Not Use sin(θ)∼θ (θ is small Enough) θ^ +(g/ℓ)sin(θ)=0 y′′(t)+(g/ℓ) sin(y(t))=0 y′′(t)y′(t)+(g/ℓ)sin(y(t))y′(t)=0 y′(t)y′′(t)=(1/2)∙((d )/dt)(y′(t))^2 (g/ℓ)sin(y(t))y′(t)=−(g/ℓ)∙((d )/dt)cos(y(t)) ∴ ((d )/dt)[(1/2)(y′(t))^2 −(g/ℓ)cos(y(t))]=0 ∴(1/2)(y′(t))^2 −(g/ℓ)cos(y(t))=c_1 Const y′′(t)+(g/ℓ)sin(y(t))=0→(y′(t))^2 −((2g)/ℓ)cos(y(t))=c_1 and... I can′t Sove Diff Equa..

$$\mathrm{Do}\:\mathrm{Not}\:\mathrm{Use}\:\mathrm{sin}\left(\theta\right)\sim\theta\:\left(\theta\:\:\mathrm{is}\:\mathrm{small}\:\mathrm{Enough}\right) \\ $$$$\ddot {\theta}+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left(\theta\right)=\mathrm{0} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\:\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0} \\ $$$${y}''\left({t}\right){y}'\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=\mathrm{0} \\ $$$${y}'\left({t}\right){y}''\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=−\frac{\mathrm{g}}{\ell}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{cos}\left({y}\left({t}\right)\right) \\ $$$$\therefore\:\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)\right]=\mathrm{0} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)={c}_{\mathrm{1}} \:\:\boldsymbol{\mathrm{Const}} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0}\rightarrow\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{2g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)={c}_{\mathrm{1}} \\ $$$$\mathrm{and}...\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{Sove}\:\mathrm{Diff}\:\:\mathrm{Equa}.. \\ $$

Question Number 201820    Answers: 1   Comments: 0

((∣3x+1∣−∣x+2∣)/(3−∣2x∣)) ≥ 0 find the solution set.

$$\:\:\:\frac{\mid\mathrm{3x}+\mathrm{1}\mid−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{3}−\mid\mathrm{2x}\mid}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}. \\ $$

Question Number 201819    Answers: 0   Comments: 0

If xyz ∈R^+ , xyz=1 , prove that the following inequality holds: (x/(2x^5 +x+4))+(y/(2y^5 +y+4))+(z/(2z^5 +z+4))≥(3/7). Solution please with an advice to get better at inequalities and which book would you recommend. Thanks in advance!

$$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\ $$

Question Number 201817    Answers: 1   Comments: 0

Question Number 201808    Answers: 1   Comments: 1

Question Number 201806    Answers: 1   Comments: 2

Question Number 201804    Answers: 0   Comments: 0

If xyz ∈R^+ , xyz=1 , prove that the following inequality holds: (x/(2x^5 +x+4))+(y/(2y^5 +y+4))+(z/(2z^5 +z+4))≥(3/7). Solution please with an advice to get better at inequalities and which book would you recommend. Thanks in advance!

$$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\ $$

Question Number 201802    Answers: 0   Comments: 1

Solution of equations like this: ((f(x)))^(1/n) +((g(x)))^(1/n) =((h(x)))^(1/n) If the solution is not obvious we must get rid of the radicals. In the following cases this is easy: (√a)+(√b)=(√c) ⇒ a+2(√(ab))+b=c ⇒ 4ab=(c−a−b)^2 (a)^(1/3) +(b)^(1/3) =(c)^(1/3) ⇒ a+3((ab))^(1/3) ((a)^(1/3) +(b)^(1/3) )+b=c ⇒ a+3((abc))^(1/3) +b=c ⇒ 27abc=(c−a−b)^3 But how to solve for n≥4? I found this formula to get an equation without radicals: a^(1/n) +b^(1/n) =c^(1/n) ⇒ Π_(k=0) ^(n−1) (c−(a^(1/n) +b^(1/n) e^(i((2πk)/n)) )^n ) =0 For n=2, 3 we get above equations. For n=4: c^4 −4(a+b)c^3 +2(3a^2 −62ab+3b^2 )c^2 −4(a+b)(a^2 +30ab+b^2 )c+(a−b)^4 =0 ⇔ 8ab(17c^2 +14(a+b)c+a^2 +b^2 )=(c−a−b)^4 For n=5: 625abc(c^2 +3(a+b)c+a^2 −3ab+b^2 )=(c−a−b)^5 I hope this helps...

$$\mathrm{Solution}\:\mathrm{of}\:\mathrm{equations}\:\mathrm{like}\:\mathrm{this}: \\ $$$$\sqrt[{{n}}]{{f}\left({x}\right)}+\sqrt[{{n}}]{{g}\left({x}\right)}=\sqrt[{{n}}]{{h}\left({x}\right)} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{obvious}\:\mathrm{we}\:\mathrm{must}\:\mathrm{get} \\ $$$$\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radicals}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{easy}: \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\sqrt{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{2}\sqrt{{ab}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4}{ab}=\left({c}−{a}−{b}\right)^{\mathrm{2}} \\ $$$$\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}=\sqrt[{\mathrm{3}}]{{c}} \\ $$$$\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{ab}}\left(\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}\right)+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{a}+\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}+{b}={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{27}{abc}=\left({c}−{a}−{b}\right)^{\mathrm{3}} \\ $$$$\mathrm{But}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{n}\geqslant\mathrm{4}? \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{get}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{without}\:\mathrm{radicals}: \\ $$$${a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} ={c}^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\:\left({c}−\left({a}^{\frac{\mathrm{1}}{{n}}} +{b}^{\frac{\mathrm{1}}{{n}}} \mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi{k}}{{n}}} \right)^{{n}} \right)\:=\mathrm{0} \\ $$$$\mathrm{For}\:{n}=\mathrm{2},\:\mathrm{3}\:\mathrm{we}\:\mathrm{get}\:\mathrm{above}\:\mathrm{equations}. \\ $$$$\mathrm{For}\:{n}=\mathrm{4}: \\ $$$${c}^{\mathrm{4}} −\mathrm{4}\left({a}+{b}\right){c}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −\mathrm{62}{ab}+\mathrm{3}{b}^{\mathrm{2}} \right){c}^{\mathrm{2}} −\mathrm{4}\left({a}+{b}\right)\left({a}^{\mathrm{2}} +\mathrm{30}{ab}+{b}^{\mathrm{2}} \right){c}+\left({a}−{b}\right)^{\mathrm{4}} =\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{8}{ab}\left(\mathrm{17}{c}^{\mathrm{2}} +\mathrm{14}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{4}} \\ $$$$\mathrm{For}\:{n}=\mathrm{5}: \\ $$$$\mathrm{625}{abc}\left({c}^{\mathrm{2}} +\mathrm{3}\left({a}+{b}\right){c}+{a}^{\mathrm{2}} −\mathrm{3}{ab}+{b}^{\mathrm{2}} \right)=\left({c}−{a}−{b}\right)^{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{this}\:\mathrm{helps}... \\ $$

Question Number 201798    Answers: 1   Comments: 0

Find the differential of the function: y = (√(x^2 − 1))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}: \\ $$$$\boldsymbol{\mathrm{y}}\:=\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:−\:\mathrm{1}} \\ $$

Question Number 201796    Answers: 0   Comments: 2

x^4 −15x^2 −30x+104=0 for x∈R x=2, 4 I want to know the best way to arrive at these answers (without guessing). I found one new way. Shall post later.

$${x}^{\mathrm{4}} −\mathrm{15}{x}^{\mathrm{2}} −\mathrm{30}{x}+\mathrm{104}=\mathrm{0} \\ $$$${for}\:{x}\in\mathbb{R}\:\:\:\:\:{x}=\mathrm{2},\:\mathrm{4} \\ $$$${I}\:{want}\:{to}\:{know}\:{the}\:{best}\:{way}\:{to} \\ $$$${arrive}\:{at}\:{these}\:{answers}\:\left({without}\right. \\ $$$$\left.{guessing}\right).\:{I}\:{found}\:{one}\:{new}\:{way}. \\ $$$$\:{Shall}\:{post}\:{later}. \\ $$

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