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AllQuestion and Answers: Page 317

Question Number 189013    Answers: 2   Comments: 0

Suppose (G, ∙ ) and (H, ∗ ) are groups. Take homomorphism φ : G → H. Suppose ∃g∈G : ∣g∣ = n, then ∣φ(g)∣ ≤ n. Does ∀g∈G, ∣g∣ = ∣φ(g)∣ ⇒ G ≅ H ?

$$\mathrm{Suppose}\:\left({G},\:\centerdot\:\right)\:\mathrm{and}\:\left({H},\:\ast\:\right)\:\mathrm{are}\:\mathrm{groups}. \\ $$$$\mathrm{Take}\:\mathrm{homomorphism}\:\phi\::\:{G}\:\rightarrow\:{H}. \\ $$$$\mathrm{Suppose}\:\exists{g}\in{G}\::\:\mid{g}\mid\:=\:{n},\:\mathrm{then}\:\mid\phi\left({g}\right)\mid\:\leqslant\:{n}. \\ $$$$\: \\ $$$$\mathrm{Does}\:\forall{g}\in{G},\:\mid{g}\mid\:=\:\mid\phi\left({g}\right)\mid\:\Rightarrow\:{G}\:\cong\:{H}\:? \\ $$

Question Number 189012    Answers: 0   Comments: 0

Triangle ABC have: sin2A+sin2B+sin2C=(√3)(cosA+cosB+cosC) => Prove that ABC is equilateral triangle

$${Triangle}\:{ABC}\:{have}:\: \\ $$$${sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}=\sqrt{\mathrm{3}}\left({cosA}+{cosB}+{cosC}\right) \\ $$$$=>\:{Prove}\:{that}\:{ABC}\:{is}\:{equilateral}\:{triangle} \\ $$

Question Number 189000    Answers: 1   Comments: 0

Question Number 188998    Answers: 2   Comments: 3

Question Number 188985    Answers: 1   Comments: 1

Lim_(x→∼) ((4x^3 −2x^2 −5x+4)/(9x^3 −4x^2 +9)) = ??

$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{4}}{\mathrm{9}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}}\:=\:??\: \\ $$

Question Number 188984    Answers: 1   Comments: 4

Lim_(x→∼) (√(16x^2 −2x−1))−4x−5 = ??

$$ \\ $$$$\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\sqrt{\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{1}}−\mathrm{4}\boldsymbol{{x}}−\mathrm{5}\:=\:??\:\:\:\: \\ $$$$ \\ $$

Question Number 188982    Answers: 0   Comments: 2

Question Number 188981    Answers: 1   Comments: 0

Question Number 188980    Answers: 1   Comments: 0

Question Number 189091    Answers: 1   Comments: 0

1 : Ω = Σ_(n=1) ^∞ (( (− 1 )^( n) H_( n) )/n^( 2) ) = ? 2 : η (−1 )= ?

$$ \\ $$$$\:\:\:\:\mathrm{1}\::\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\:\mathrm{1}\:\right)^{\:{n}} \mathrm{H}_{\:{n}} }{{n}^{\:\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\mathrm{2}\::\:\:\:\:\:\eta\:\left(−\mathrm{1}\:\right)=\:? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 189090    Answers: 2   Comments: 0

Question Number 188972    Answers: 0   Comments: 4

what is the answer A ? B ? C?.

$${what}\:{is}\:{the}\:{answer} \\ $$$${A}\:\:?\:{B}\:\:\:?\:{C}?. \\ $$

Question Number 188970    Answers: 1   Comments: 0

LCM(x, 144, 150) = 10800 how many value of x.

$$\mathrm{LCM}\left({x},\:\mathrm{144},\:\mathrm{150}\right)\:=\:\mathrm{10800}\:{how}\:{many}\:{value}\:{of}\:\:{x}. \\ $$$$ \\ $$

Question Number 188968    Answers: 1   Comments: 0

Question Number 188966    Answers: 0   Comments: 0

hey everyone i changed accounts my new one is gatocomcirrose i will not use this one anymore

$$ \\ $$$$\mathrm{hey}\:\mathrm{everyone}\:\mathrm{i}\:\mathrm{changed}\:\mathrm{accounts} \\ $$$$\mathrm{my}\:\mathrm{new}\:\mathrm{one}\:\mathrm{is}\:\mathrm{gatocomcirrose} \\ $$$$\mathrm{i}\:\mathrm{will}\:\mathrm{not}\:\mathrm{use}\:\mathrm{this}\:\mathrm{one}\:\mathrm{anymore} \\ $$

Question Number 188965    Answers: 2   Comments: 0

Question Number 188954    Answers: 0   Comments: 1

Solve : sin(x) + x^2 + tan(x) = 3x Hey!

$$\mathrm{Solve}\:: \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{tan}\left(\mathrm{x}\right)\:=\:\mathrm{3x} \\ $$$$ \\ $$$$\mathrm{Hey}! \\ $$

Question Number 188948    Answers: 2   Comments: 0

Question Number 188937    Answers: 0   Comments: 2

A body of mass 20kg lies on a horizontal floor and is acted on by constant forces of magnitude 100N and mN in the directions 180° and 060° respectively, where m is a constant. The body begins to move in the direction 090°. If there's also a resistance of 1N/5 per kilogram to the motion of the body; (a) calculate the value of m and the magnitude of the resultant force acting on the body (b) find the acceleration of the body.

A body of mass 20kg lies on a horizontal floor and is acted on by constant forces of magnitude 100N and mN in the directions 180° and 060° respectively, where m is a constant. The body begins to move in the direction 090°. If there's also a resistance of 1N/5 per kilogram to the motion of the body; (a) calculate the value of m and the magnitude of the resultant force acting on the body (b) find the acceleration of the body.

Question Number 188933    Answers: 2   Comments: 1

lim_(x→1) (((√(4x−3))+(√(2x−1))−3x+1)/(x^2 −2x+1))=?

$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}{x}−\mathrm{3}}+\sqrt{\mathrm{2}{x}−\mathrm{1}}−\mathrm{3}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}=? \\ $$

Question Number 188928    Answers: 0   Comments: 0

Question Number 188927    Answers: 1   Comments: 0

Question Number 188926    Answers: 1   Comments: 0

Question Number 188924    Answers: 1   Comments: 0

Question Number 188912    Answers: 4   Comments: 0

Question Number 188908    Answers: 1   Comments: 0

the radius of a circle is 12cmunits find the perimeter of a regular inscribed a. triangle b.heptagon c. decagon

$${the}\:{radius}\:{of}\:{a}\:{circle}\:{is}\:\mathrm{12}{cmunits} \\ $$$${find}\:{the}\:{perimeter}\:{of}\:{a}\:{regular}\: \\ $$$${inscribed}\: \\ $$$${a}.\:{triangle} \\ $$$${b}.{heptagon} \\ $$$${c}.\:{decagon} \\ $$

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