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Question Number 98290 Answers: 0 Comments: 1

Question Number 98022 Answers: 1 Comments: 0

Derive the relation between an Arithmetic Mean and a Geometric Mean ((x_1 x_2 ...x_n ))^(1/n) ≤((x_1 +x_2 +∙∙∙+x_n )/n) ∀n∈N^∗ , ∀(x_1 ,x_2 ,...x_n )∈(R_+ ^∗ )^n

$$\mathcal{D}\mathrm{erive}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{an}\:\mathcal{A}\mathrm{rithmetic}\:\mathcal{M}\mathrm{ean} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathcal{G}\mathrm{eometric}\:\mathcal{M}\mathrm{ean} \\ $$$$\sqrt[{\mathrm{n}}]{\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} ...\mathrm{x}_{\mathrm{n}} }\leqslant\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{x}_{\mathrm{n}} }{\mathrm{n}}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} ,\:\forall\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,...\mathrm{x}_{\mathrm{n}} \right)\in\left(\mathbb{R}_{+} ^{\ast} \right)^{\mathrm{n}} \\ $$

Question Number 97953 Answers: 1 Comments: 3

Question Number 97833 Answers: 0 Comments: 10

Hello members of maths editor i want to say something concerning this forum. Well it is my own point of view: This forum is a place where people from all over the world come to interact together. A place where people from different back grounds, class and qualification speak a common language − mathematics. I in particular thank Tinkutara′s team for such a great platform. But i notice some people literally don′t show respect to others, like persuading others to answer thier questions, others show no appreciation for the given answers while others give rude comments with no reason behind them. please i just want to urge the Maths editor users to show more respect for others since we don′t know the identity or qualification of people who post and solve questions here. we remain one family as God guides us through our endervous.

$$\:\mathrm{Hello}\:\mathrm{members}\:\mathrm{of}\:\mathrm{maths}\:\mathrm{editor}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{say} \\ $$$$\mathrm{something}\:\mathrm{concerning}\:\mathrm{this}\:\mathrm{forum}.\:\mathrm{Well}\:\mathrm{it} \\ $$$$\mathrm{is}\:\mathrm{my}\:\mathrm{own}\:\mathrm{point}\:\mathrm{of}\:\mathrm{view}:\:\mathrm{This}\:\mathrm{forum}\:\mathrm{is}\:\mathrm{a}\:\mathrm{place}\:\mathrm{where} \\ $$$$\mathrm{people}\:\mathrm{from}\:\mathrm{all}\:\mathrm{over}\:\mathrm{the}\:\mathrm{world}\:\mathrm{come}\:\mathrm{to}\:\mathrm{interact}\:\mathrm{together}. \\ $$$$\mathrm{A}\:\mathrm{place}\:\mathrm{where}\:\mathrm{people}\:\mathrm{from}\:\mathrm{different}\:\mathrm{back}\:\mathrm{grounds}, \\ $$$$\mathrm{class}\:\mathrm{and}\:\mathrm{qualification}\:\mathrm{speak}\:\mathrm{a}\:\mathrm{common}\:\mathrm{language}\:−\: \\ $$$$\mathrm{mathematics}.\:\mathrm{I}\:\mathrm{in}\:\mathrm{particular}\:\mathrm{thank}\:\mathrm{Tinkutara}'\mathrm{s}\:\mathrm{team}\:\mathrm{for} \\ $$$$\mathrm{such}\:\mathrm{a}\:\mathrm{great}\:\mathrm{platform}.\:\mathrm{But}\:\mathrm{i}\:\mathrm{notice}\:\mathrm{some}\:\mathrm{people}\:\mathrm{literally} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{show}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{others},\:\mathrm{like}\:\mathrm{persuading}\:\mathrm{others} \\ $$$$\mathrm{to}\:\mathrm{answer}\:\mathrm{thier}\:\mathrm{questions},\:\mathrm{others}\:\mathrm{show}\:\mathrm{no}\: \\ $$$$\mathrm{appreciation}\:\mathrm{for}\:\mathrm{the}\:\mathrm{given}\:\mathrm{answers}\:\mathrm{while}\:\mathrm{others}\:\mathrm{give}\:\mathrm{rude} \\ $$$$\mathrm{comments}\:\mathrm{with}\:\mathrm{no}\:\mathrm{reason}\:\mathrm{behind}\:\mathrm{them}. \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{urge}\:\mathrm{the}\:\mathrm{Maths}\:\mathrm{editor}\:\mathrm{users}\:\mathrm{to}\:\mathrm{show} \\ $$$$\mathrm{more}\:\mathrm{respect}\:\mathrm{for}\:\mathrm{others}\:\mathrm{since}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{or} \\ $$$$\mathrm{qualification}\:\mathrm{of}\:\mathrm{people}\:\mathrm{who}\:\mathrm{post}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{questions}\:\mathrm{here}. \\ $$$$\mathrm{we}\:\mathrm{remain}\:\mathrm{one}\:\mathrm{family}\:\mathrm{as}\:\mathrm{God}\:\mathrm{guides}\:\mathrm{us}\:\mathrm{through}\:\mathrm{our}\:\mathrm{endervous}. \\ $$

Question Number 97803 Answers: 1 Comments: 0

The annual salaries of employees in a large company are approximately normally disributed with a mean of $50,000 and a standard deviation of $20,000. a. what percent of people earn less than $40,000? b. what percent of people earn between $45,000 and $65,000? c. what percent of people earn more than $70,000?

$$\mathrm{The}\:\mathrm{annual}\:\mathrm{salaries}\:\mathrm{of}\:\mathrm{employees}\:\mathrm{in}\:\mathrm{a}\:\mathrm{large} \\ $$$$\mathrm{company}\:\mathrm{are}\:\mathrm{approximately}\:\mathrm{normally}\: \\ $$$$\mathrm{disributed}\:\mathrm{with}\:\mathrm{a}\:\mathrm{mean}\:\mathrm{of}\:\$\mathrm{50},\mathrm{000}\:\mathrm{and}\:\mathrm{a}\:\mathrm{standard} \\ $$$$\mathrm{deviation}\:\mathrm{of}\:\$\mathrm{20},\mathrm{000}. \\ $$$$\mathrm{a}.\:\mathrm{what}\:\mathrm{percent}\:\mathrm{of}\:\mathrm{people}\:\mathrm{earn}\:\mathrm{less}\:\mathrm{than} \\ $$$$\$\mathrm{40},\mathrm{000}? \\ $$$$\mathrm{b}.\:\mathrm{what}\:\mathrm{percent}\:\mathrm{of}\:\mathrm{people}\:\mathrm{earn}\:\mathrm{between} \\ $$$$\$\mathrm{45},\mathrm{000}\:\mathrm{and}\:\$\mathrm{65},\mathrm{000}? \\ $$$$\mathrm{c}.\:\mathrm{what}\:\mathrm{percent}\:\mathrm{of}\:\mathrm{people}\:\mathrm{earn}\:\mathrm{more}\:\mathrm{than} \\ $$$$\$\mathrm{70},\mathrm{000}? \\ $$

Question Number 97606 Answers: 1 Comments: 0

given that the polynomial p(x)=(3x+2)(x−1)q(x)−2x−4 of degree 3 is exactly divisible by x−2 and that p(−1)=−12. find q(x).

$${given}\:{that}\:{the}\:{polynomial}\:{p}\left({x}\right)=\left(\mathrm{3}{x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right){q}\left({x}\right)−\mathrm{2}{x}−\mathrm{4} \\ $$$${of}\:{degree}\:\mathrm{3}\:{is}\:{exactly}\:{divisible}\:{by}\:{x}−\mathrm{2}\:{and}\: \\ $$$${that}\:{p}\left(−\mathrm{1}\right)=−\mathrm{12}.\:{find}\:{q}\left({x}\right). \\ $$

Question Number 97478 Answers: 0 Comments: 0

Find the global parametrization of the curve { x^2 +y^2 +z^2 =1; x+y−z=0 }

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{global}\:\mathrm{parametrization} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\left\{\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{1};\:\mathrm{x}+\mathrm{y}−\mathrm{z}=\mathrm{0}\:\right\}\: \\ $$

Question Number 97129 Answers: 0 Comments: 0

Given z=x+iy z∈C z≠0 1\ A, B, and C are the images of z, iz, and (2−i)+z a\ Calculate the lengths AB, AC, and BC. b\ Deduce that the triangle ABC is isosceles and not equilateral. 2\Find z, such that ∣z∣=∣((2+i)/z)∣=∣z−1∣ 3\Given Z, Z∈C such that ((Z−1)/(Z+1))=(((z−1)/(z+1)))^2 a\Express Z in terms of z b\What can we say of the images of Z, z, and (1/z) ?

$$\mathrm{Given}\:\:\mathrm{z}=\mathrm{x}+\mathrm{iy}\:\:\mathrm{z}\in\mathbb{C}\:\:\mathrm{z}\neq\mathrm{0} \\ $$$$\mathrm{1}\backslash\:\mathrm{A},\:\mathrm{B},\:\mathrm{and}\:\mathrm{C}\:\mathrm{are}\:\mathrm{the}\:\mathrm{images}\:\mathrm{of}\:\mathrm{z},\:\mathrm{iz},\:\mathrm{and}\:\left(\mathrm{2}−\mathrm{i}\right)+\mathrm{z} \\ $$$$\mathrm{a}\backslash\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{lengths}\:\mathrm{AB},\:\mathrm{AC},\:\mathrm{and}\:\mathrm{BC}. \\ $$$$\mathrm{b}\backslash\:\mathrm{Deduce}\:\mathrm{that}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{and}\:\mathrm{not} \\ $$$$\mathrm{equilateral}. \\ $$$$\mathrm{2}\backslash\mathrm{Find}\:\mathrm{z},\:\mathrm{such}\:\mathrm{that}\:\mid\mathrm{z}\mid=\mid\frac{\mathrm{2}+\mathrm{i}}{\mathrm{z}}\mid=\mid\mathrm{z}−\mathrm{1}\mid \\ $$$$\mathrm{3}\backslash\mathrm{Given}\:\mathrm{Z},\:\mathrm{Z}\in\mathbb{C}\:\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{Z}−\mathrm{1}}{\mathrm{Z}+\mathrm{1}}=\left(\frac{\mathrm{z}−\mathrm{1}}{\mathrm{z}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$\mathrm{a}\backslash\mathrm{Express}\:\mathrm{Z}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{z} \\ $$$$\mathrm{b}\backslash\mathrm{What}\:\mathrm{can}\:\mathrm{we}\:\mathrm{say}\:\mathrm{of}\:\mathrm{the}\:\mathrm{images}\:\mathrm{of}\:\mathrm{Z},\:\mathrm{z},\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{z}}\:? \\ $$

Question Number 96764 Answers: 1 Comments: 5

A particle P of mass m, is projected vertically upward with a speed u from a point A, on horizontal ground. When P is at x above its initial position, its speed is v. The only forces acting on P is its weight and resistance mgkv^2 . where k is a positive constant. (a) Show that the greatest height reached is (1/(2gk)) ln(1 +ku^2 ). (b) show that the speed with which P returns to A is (u/(√(1+ ku^2 ))) .

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\mathrm{of}\:\mathrm{mass}\:{m},\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with} \\ $$$$\mathrm{a}\:\mathrm{speed}\:{u}\:\mathrm{from}\:\mathrm{a}\:\mathrm{point}\:{A},\:\mathrm{on}\:\mathrm{horizontal}\:\mathrm{ground}.\:\mathrm{When}\:\mathrm{P}\:\mathrm{is}\:\mathrm{at}\:{x}\:\mathrm{above} \\ $$$$\mathrm{its}\:\mathrm{initial}\:\mathrm{position},\:\mathrm{its}\:\mathrm{speed}\:\mathrm{is}\:{v}.\:\mathrm{The}\:\mathrm{only}\:\mathrm{forces}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{P}\:\mathrm{is} \\ $$$$\mathrm{its}\:\mathrm{weight}\:\mathrm{and}\:\mathrm{resistance}\:{m}\mathrm{g}{kv}^{\mathrm{2}} .\:\mathrm{where}\:{k}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{constant}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{height}\:\mathrm{reached}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2g}{k}}\:\mathrm{ln}\left(\mathrm{1}\:+{ku}^{\mathrm{2}} \right). \\ $$$$\left(\mathrm{b}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{P}\:\mathrm{returns}\:\mathrm{to}\:\mathrm{A}\:\mathrm{is}\:\frac{{u}}{\sqrt{\mathrm{1}+\:{ku}^{\mathrm{2}} }}\:. \\ $$

Question Number 96761 Answers: 0 Comments: 0

A particle P moving at constant angular velocity describes a part y = f(θ). At time t = 0, the particle is at the point with coordinate (a,(π/2)) and moving with a transverse acceleration of −2aω^2 sinθ. find the polar equation of the curve described by this particle.Show that the radial component of the acceleration of P is −aω^2 (1 + cos θ).

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{P}\:\:\:\mathrm{moving}\:\mathrm{at}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{velocity} \\ $$$$\mathrm{describes}\:\mathrm{a}\:\mathrm{part}\:{y}\:=\:{f}\left(\theta\right).\:\mathrm{At}\:\mathrm{time}\:{t}\:=\:\mathrm{0},\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{with}\:\mathrm{coordinate}\:\left({a},\frac{\pi}{\mathrm{2}}\right)\:\mathrm{and}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\: \\ $$$$\mathrm{transverse}\:\mathrm{acceleration}\:\mathrm{of}\:−\mathrm{2}{a}\omega^{\mathrm{2}} \:\mathrm{sin}\theta.\:\mathrm{find}\:\mathrm{the}\:\mathrm{polar}\:\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{described}\:\mathrm{by}\:\mathrm{this}\:\mathrm{particle}.\mathrm{Show}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{radial}\:\mathrm{component}\:\mathrm{of}\:\mathrm{the}\:\:\mathrm{acceleration}\:\:\mathrm{of}\:\mathrm{P}\:\mathrm{is}\:−{a}\omega^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{cos}\:\theta\right). \\ $$

Question Number 96671 Answers: 1 Comments: 0