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Question Number 156660    Answers: 0   Comments: 0

Question Number 156652    Answers: 1   Comments: 0

Sirs, please give me the general solutions to a quadratic ineqality. (1) ax^2 + bx + c > 0 (2) ax^2 + bx + c ≥ 0 (3) ax^2 + bx + c < 0 (4) ax^2 + bx + c ≤ 0

$$\mathrm{Sirs},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{ineqality}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:>\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\geqslant\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:<\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\leqslant\:\:\:\:\mathrm{0} \\ $$

Question Number 156648    Answers: 3   Comments: 0

Solve for integers: x^2 y^2 - x^2 - xy - y^2 + x + y = 0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:-\:\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{xy}\:-\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0} \\ $$

Question Number 156647    Answers: 0   Comments: 0

Question Number 156642    Answers: 1   Comments: 3

For x∈(0;∞) - Z prove that: (({x}^3 )/([x])) + (([x]^3 )/({x})) ≥ (1/8) (x^2 + [x]^2 + {x}^2 ) [∗]-GIF and {x}=x-[x]

$$\mathrm{For}\:\:\mathrm{x}\in\left(\mathrm{0};\infty\right)\:-\:\mathbb{Z}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left\{\mathrm{x}\right\}^{\mathrm{3}} }{\left[\mathrm{x}\right]}\:+\:\frac{\left[\mathrm{x}\right]^{\mathrm{3}} }{\left\{\mathrm{x}\right\}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{8}}\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\left[\mathrm{x}\right]^{\mathrm{2}} \:+\:\left\{\mathrm{x}\right\}^{\mathrm{2}} \right) \\ $$$$\left[\ast\right]-\mathrm{GIF}\:\:\mathrm{and}\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$

Question Number 156640    Answers: 1   Comments: 0

Question Number 156639    Answers: 0   Comments: 0

x^3 =x+c x^4 =x^2 +cx let cx=kx^4 +hx^2 ⇒ kx^3 +hx=c x^2 =1+c(kx^2 +h) x^2 =((1+ch)/(1−ck)) (((1+ch)/(1−ck))){((k(1+ch))/(1−ck))+h}^2 =c^2 (1+ch)(h+k)^2 =c^2 (1−ck)^3 ⇒ (1+ch)(h^2 +k^2 +2hk) =c^2 (1−c^3 k^3 +3c^2 k^2 −3ck) c^5 k^3 +(1+ch−3c^4 )k^2 +{2h(1+ch)+3c^3 }k +(1+ch)h^2 −c^2 =0 let h=3c^3 −(1/c) ⇒ c^5 k^3 +{9c^3 −(2/c)+2c(3c^3 −(1/c))^2 }k +{3c^4 (3c^3 −(1/c))^2 −c^2 }=0 ⇒ k^3 +3(6c^2 −(1/c^2 ))k +27c^5 −18c+(2/c^3 )=0 D=(((27c^5 )/2)+9c+(1/c^3 ))^2 +(6c^2 −(1/c^2 ))^3 ...

$$\mathrm{x}^{\mathrm{3}} =\mathrm{x}+\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{2}} +\mathrm{cx} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{kx}^{\mathrm{4}} +\mathrm{hx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{kx}^{\mathrm{3}} +\mathrm{hx}=\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{c}\left(\mathrm{kx}^{\mathrm{2}} +\mathrm{h}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{ch}}{\mathrm{1}−\mathrm{ck}} \\ $$$$\left(\frac{\mathrm{1}+\mathrm{ch}}{\mathrm{1}−\mathrm{ck}}\right)\left\{\frac{\mathrm{k}\left(\mathrm{1}+\mathrm{ch}\right)}{\mathrm{1}−\mathrm{ck}}+\mathrm{h}\right\}^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\mathrm{ch}\right)\left(\mathrm{h}+\mathrm{k}\right)^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{ck}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:\left(\mathrm{1}+\mathrm{ch}\right)\left(\mathrm{h}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} +\mathrm{2hk}\right) \\ $$$$\:\:\:\:=\mathrm{c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{c}^{\mathrm{3}} \mathrm{k}^{\mathrm{3}} +\mathrm{3c}^{\mathrm{2}} \mathrm{k}^{\mathrm{2}} −\mathrm{3ck}\right) \\ $$$$\mathrm{c}^{\mathrm{5}} \mathrm{k}^{\mathrm{3}} +\left(\mathrm{1}+\mathrm{ch}−\mathrm{3c}^{\mathrm{4}} \right)\mathrm{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left\{\mathrm{2h}\left(\mathrm{1}+\mathrm{ch}\right)+\mathrm{3c}^{\mathrm{3}} \right\}\mathrm{k} \\ $$$$\:\:\:\:\:+\left(\mathrm{1}+\mathrm{ch}\right)\mathrm{h}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{h}=\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\:\:\Rightarrow \\ $$$$\mathrm{c}^{\mathrm{5}} \mathrm{k}^{\mathrm{3}} +\left\{\mathrm{9c}^{\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{c}}+\mathrm{2c}\left(\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\right)^{\mathrm{2}} \right\}\mathrm{k} \\ $$$$\:\:\:\:+\left\{\mathrm{3c}^{\mathrm{4}} \left(\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{k}^{\mathrm{3}} +\mathrm{3}\left(\mathrm{6c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\right)\mathrm{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{27c}^{\mathrm{5}} −\mathrm{18c}+\frac{\mathrm{2}}{\mathrm{c}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\mathrm{D}=\left(\frac{\mathrm{27c}^{\mathrm{5}} }{\mathrm{2}}+\mathrm{9c}+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)^{\mathrm{2}} +\left(\mathrm{6c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$... \\ $$

Question Number 156617    Answers: 1   Comments: 0

Question Number 156616    Answers: 2   Comments: 0

∫_0 ^1 ((arcsin(x))/x)dx=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsin}\left({x}\right)}{{x}}{dx}=? \\ $$

Question Number 156636    Answers: 1   Comments: 0

Question Number 156611    Answers: 0   Comments: 1

lim(√(x+5))

$${lim}\sqrt{{x}+\mathrm{5}} \\ $$

Question Number 156610    Answers: 2   Comments: 0

If A and B are invertible matrices,then: (AB)^(−1) =B^(−1) A^(−1) ≠ A^(−1) B^(−1) proove.

$$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{invertible}\:\mathrm{matrices},\mathrm{then}: \\ $$$$\left(\mathrm{AB}\right)^{−\mathrm{1}} =\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \neq\:\mathrm{A}^{−\mathrm{1}} \mathrm{B}^{−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{proove}}. \\ $$

Question Number 156609    Answers: 1   Comments: 0

Question Number 156604    Answers: 1   Comments: 2

Question Number 156595    Answers: 0   Comments: 2

Mary walks 9 city blocks south, 2 blocks east, 3 blocks south and 7 blocks east. if all blocks are the same length. How far is she from her starting point. please help with the aid of a diagram

$$\mathrm{Mary}\:\mathrm{walks}\:\mathrm{9}\:\mathrm{city}\:\mathrm{blocks}\:\mathrm{south}, \\ $$$$\mathrm{2}\:\mathrm{blocks}\:\mathrm{east},\:\mathrm{3}\:\mathrm{blocks}\:\mathrm{south}\:\mathrm{and}\: \\ $$$$\mathrm{7}\:\mathrm{blocks}\:\mathrm{east}.\:\mathrm{if}\:\mathrm{all}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{length}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{is}\:\mathrm{she}\:\mathrm{from}\:\mathrm{her}\:\mathrm{starting} \\ $$$$\mathrm{point}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{the}\:\mathrm{aid}\:\mathrm{of}\:\mathrm{a}\:\mathrm{diagram} \\ $$

Question Number 156593    Answers: 1   Comments: 1

Question Number 156591    Answers: 0   Comments: 0

Find all the triplets (a, b, c) of pozitif integers such that a^3 +b^3 +c^3 =(abc)^2 ??????

$${Find}\:{all}\:{the}\:{triplets}\:\left({a},\:{b},\:{c}\right)\:{of}\:{pozitif}\:{integers} \\ $$$${such}\:{that} \\ $$$$\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\boldsymbol{\mathrm{b}}^{\mathrm{3}} +\boldsymbol{\mathrm{c}}^{\mathrm{3}} =\left(\boldsymbol{\mathrm{abc}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$?????? \\ $$

Question Number 156590    Answers: 1   Comments: 0

Question Number 156586    Answers: 0   Comments: 0

Question Number 156584    Answers: 1   Comments: 1

find p if y=1−px−3x^2 if the maximum is 13 (help pls)

$$\mathrm{find}\:\mathrm{p}\:\mathrm{if}\:\mathrm{y}=\mathrm{1}−\mathrm{px}−\mathrm{3x}^{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{13}\:\:\left(\mathrm{help}\:\mathrm{pls}\right) \\ $$

Question Number 156579    Answers: 1   Comments: 0

Find the value of x by using Cardon′s Method. (1/x) + (2/(x+1))+ (3/(x+2)) = 1

$$\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{Cardon}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{Method}}. \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}+\:\frac{\mathrm{3}}{\boldsymbol{\mathrm{x}}+\mathrm{2}}\:=\:\mathrm{1} \\ $$

Question Number 156578    Answers: 0   Comments: 0

Question Number 156577    Answers: 0   Comments: 4

Question Number 156575    Answers: 1   Comments: 1

let n≥1 and λ=2n^2 -2n+1 solve for real numbers: (√(λ + x^2 )) - (√(λ - x^2 )) = (x^2 /n)

$$\mathrm{let}\:\:\mathrm{n}\geqslant\mathrm{1}\:\:\mathrm{and}\:\:\lambda=\mathrm{2n}^{\mathrm{2}} -\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\lambda\:+\:\mathrm{x}^{\mathrm{2}} }\:-\:\sqrt{\lambda\:-\:\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}} \\ $$

Question Number 156574    Answers: 0   Comments: 0

if a;b;c>0 and (1/(a+b)) + (1/(b+c)) + (1/(c+a)) = 4 then: (1/a) + (1/b) + (1/c) + (9/(a+b+c)) ≥ 16

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\mathrm{a}+\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{b}+\mathrm{c}}\:+\:\frac{\mathrm{1}}{\mathrm{c}+\mathrm{a}}\:=\:\mathrm{4}\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:+\:\frac{\mathrm{1}}{\mathrm{c}}\:+\:\frac{\mathrm{9}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}\:\geqslant\:\mathrm{16} \\ $$

Question Number 156600    Answers: 2   Comments: 0

∫ (e^x /(e^(2x) +4)) dx =?

$$\int\:\frac{{e}^{{x}} }{{e}^{\mathrm{2}{x}} +\mathrm{4}}\:{dx}\:=?\: \\ $$

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