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Question Number 191993 Answers: 0 Comments: 0

Question Number 191986 Answers: 1 Comments: 0

Ques. 1 Let (G,∗) be a group, then show that for each a∈G, ∃ a unique element e∈G ∣ a∗e=e∗a=a Ques. 2 If a∈G ⇒ x∈G and x is unique show that if x∗a=e, then a∗x=e. Hello!

$$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\mathrm{Let}\:\left(\mathrm{G},\ast\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{group},\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{for}\:\mathrm{each}\:\mathrm{a}\in\mathrm{G},\:\exists\:\mathrm{a}\:\mathrm{unique}\: \\ $$$$\mathrm{element}\:\mathrm{e}\in\mathrm{G}\:\mid\:\mathrm{a}\ast\mathrm{e}=\mathrm{e}\ast\mathrm{a}=\mathrm{a} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{a}\in\mathrm{G}\:\Rightarrow\:\mathrm{x}\in\mathrm{G}\:\mathrm{and}\:\mathrm{x}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{x}\ast\mathrm{a}=\mathrm{e},\:\mathrm{then}\:\mathrm{a}\ast\mathrm{x}=\mathrm{e}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Hello}! \\ $$

Question Number 191966 Answers: 1 Comments: 0

Question Number 191962 Answers: 1 Comments: 0

Question Number 191961 Answers: 0 Comments: 0

Question Number 191960 Answers: 2 Comments: 0

Question Number 191958 Answers: 1 Comments: 2

Question Number 191950 Answers: 1 Comments: 0

(√a) = 1 + (1/a) find: a^2 −a−(√a) = ?

$$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$

Question Number 191946 Answers: 1 Comments: 0

((fof^(−1) (5)+fof^(−1) (15))/(fof^(−1) (5)))=?

$$\frac{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)+\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{15}\right)}{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)}=? \\ $$

Question Number 191947 Answers: 2 Comments: 0

prove that (√(a+b(√(a−b(√(a+b(√(a−b(√(...)))))))))) = (((√(4a−3b^2 ))+b)/2)

$${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{...}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$

Question Number 191937 Answers: 1 Comments: 0

Check whether (Q, ∙) is a group or not Hello bosses!

$$\mathrm{Check}\:\mathrm{whether}\:\left(\mathrm{Q},\:\centerdot\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{or} \\ $$$$\mathrm{not} \\ $$$$ \\ $$$$\mathrm{Hello}\:\mathrm{bosses}! \\ $$

Question Number 191935 Answers: 2 Comments: 0

Question Number 191934 Answers: 1 Comments: 0

Question Number 195092 Answers: 1 Comments: 0

lim_(x→1^+ ) ((4x+5)/(x−x^2 ))=?

$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{4}{x}+\mathrm{5}}{{x}−{x}^{\mathrm{2}} }=? \\ $$

Question Number 191930 Answers: 1 Comments: 2

What is the value of inside Area of (ABCDEF)? Such that: ∡AOB=120 ∡ANB=60;°R=ON (OA=OB=32cm) ArcAE=ArcBF(r=12cm) BASE is circulare (Aider le tailleur a savoir la surface du tissu necessaire pour couvrir l ′espace indique dans la figure?)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{inside}\:\mathrm{Area}\:\mathrm{of} \\ $$$$\left(\mathrm{ABCDEF}\right)? \\ $$$$\mathrm{Such}\:\mathrm{that}:\:\measuredangle\mathrm{AOB}=\mathrm{120}\:\:\:\measuredangle\mathrm{ANB}=\mathrm{60};°\mathrm{R}=\mathrm{ON} \\ $$$$\:\left(\mathrm{OA}=\mathrm{OB}=\mathrm{32cm}\right)\:\mathrm{ArcAE}=\mathrm{ArcBF}\left(\mathrm{r}=\mathrm{12cm}\right) \\ $$$$\mathrm{BASE}\:\mathrm{is}\:\mathrm{circulare} \\ $$$$\left({Aider}\:{le}\:{tailleur}\:{a}\:{savoir}\:{la}\:{surface}\right. \\ $$$${du}\:{tissu}\:{necessaire}\:{pour}\:{couvrir}\: \\ $$$$\left.\:{l}\:'\mathrm{e}{space}\:{indique}\:{dans}\:{la}\:{figure}?\right) \\ $$

Question Number 191901 Answers: 3 Comments: 0

if 0<θ<45^0 which is bigger ? 2tanθ or tan2θ

$${if}\:\:\mathrm{0}<\theta<\mathrm{45}^{\mathrm{0}} \:\:{which}\:{is}\:{bigger}\:? \\ $$$$\mathrm{2}{tan}\theta\:\:{or}\:\:{tan}\mathrm{2}\theta \\ $$

Question Number 191897 Answers: 2 Comments: 0

if we combine 100gr Na with 180 Cl_2 then how much NaCl will produce?

$${if}\:{we}\:{combine}\:\mathrm{100}{gr}\:{Na}\:{with}\:\mathrm{180}\:{Cl}_{\mathrm{2}} \\ $$$${then}\:{how}\:{much}\:{NaCl}\:{will}\:{produce}? \\ $$

Question Number 191927 Answers: 0 Comments: 1

What is the value of laterale Shaded Area (couronne circulaire coloree?) R=10cm 20cm≤ h≤24cm ∡COD=45°

Question Number 191926 Answers: 2 Comments: 0

Question Number 191921 Answers: 1 Comments: 0

Question Number 191889 Answers: 2 Comments: 0

Σ_(n=1) ^k (1/(n^2 +2n)) =?

$$\:\:\:\:\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{k}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{2n}}\:=? \\ $$

Question Number 191887 Answers: 3 Comments: 0

Question Number 191874 Answers: 1 Comments: 0

lim_(x⇒∞) ((e^(x+1) +pi^(x−1) )/(e^(x−1) +pi^(x+1) ))

$${lim}_{{x}\Rightarrow\infty} \frac{{e}^{{x}+\mathrm{1}} +{pi}^{{x}−\mathrm{1}} }{{e}^{{x}−\mathrm{1}} +{pi}^{{x}+\mathrm{1}} } \\ $$$$ \\ $$

Question Number 191873 Answers: 4 Comments: 0

Question Number 191868 Answers: 2 Comments: 0

A particle of mass m moves under the central repulsive force ((mb)/r^3 ) and is initially moving at a distance ′a′ from the origin of a force with velocity ′v′ at right angle to ′a′. show that rcos pθ=a where p =(b/(a^2 v^2 ))+1.

$${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$ \\ $$

Question Number 191867 Answers: 1 Comments: 0

Prove that if u=f(x^3 +y^3 ),where f is arbitry function then x^2 (∂u/∂y) = y^2 (∂u/∂x)

$${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$

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