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Question Number 204568 Answers: 1 Comments: 0

how to convert 31230 in base 60? pls help

$$\boldsymbol{{how}}\:\boldsymbol{{to}}\:\boldsymbol{{convert}}\:\mathrm{31230}\:\boldsymbol{{in}}\:\boldsymbol{{base}}\:\mathrm{60}? \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{help}} \\ $$

Question Number 204574 Answers: 2 Comments: 0

lim_(n→∞) n^(−3/2) [(n+1)^((n+1)) (n+2)^((n+2)) ...(2n)^(2n) ]^(1/n^2 ) = ?

$$ \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}^{−\mathrm{3}/\mathrm{2}} \left[\left(\mathrm{n}+\mathrm{1}\right)^{\left(\mathrm{n}+\mathrm{1}\right)} \left(\mathrm{n}+\mathrm{2}\right)^{\left(\mathrm{n}+\mathrm{2}\right)} ...\left(\mathrm{2n}\right)^{\mathrm{2n}} \right]^{\mathrm{1}/\mathrm{n}^{\mathrm{2}} } \:=\:? \\ $$$$ \\ $$

Question Number 204560 Answers: 2 Comments: 1

Question Number 204558 Answers: 0 Comments: 0

lim_(n→∞) n^(−3/2) [(n+1)^((n+1)) (n+2)^((n+2)) ...(2n)^(2n) ]^(1/n^2 ) = ?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}^{−\mathrm{3}/\mathrm{2}} \left[\left(\mathrm{n}+\mathrm{1}\right)^{\left(\mathrm{n}+\mathrm{1}\right)} \left(\mathrm{n}+\mathrm{2}\right)^{\left(\mathrm{n}+\mathrm{2}\right)} ...\left(\mathrm{2n}\right)^{\mathrm{2n}} \right]^{\mathrm{1}/\mathrm{n}^{\mathrm{2}} } \:=\:? \\ $$

Question Number 204545 Answers: 2 Comments: 0

If a = (1/2^2 ) + (1/3^2 ) + ... + (1/(100^2 )) b = 0,99 Prove that: a < b

$$\mathrm{If} \\ $$$$\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:\:+\:\:...\:\:+\:\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} } \\ $$$$\mathrm{b}\:=\:\mathrm{0},\mathrm{99} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{a}\:<\:\mathrm{b} \\ $$

Question Number 204541 Answers: 0 Comments: 4

In a regular pentagon PQRST , PR intersects QS at O. Calculate ROS?

$${In}\:{a}\:{regular}\:{pentagon}\:{PQRST}\:,\:{PR} \\ $$$${intersects}\:{QS}\:{at}\:{O}.\:{Calculate}\:{ROS}? \\ $$

Question Number 204533 Answers: 1 Comments: 0

Question Number 204522 Answers: 1 Comments: 3

Question Number 204521 Answers: 1 Comments: 1

Question Number 204517 Answers: 2 Comments: 0

Question Number 204518 Answers: 1 Comments: 0

Question Number 204512 Answers: 2 Comments: 0

solve for x∈C 3^(2ix) −3^(ix) 2+5=0

$$\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{3}^{\mathrm{2i}{x}} −\mathrm{3}^{\mathrm{i}{x}} \mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$

Question Number 204511 Answers: 1 Comments: 0

solve for x x^2 −10⌊x⌋+((57)/4)=0

$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}\lfloor{x}\rfloor+\frac{\mathrm{57}}{\mathrm{4}}=\mathrm{0} \\ $$

Question Number 204510 Answers: 1 Comments: 0

solve for x≠y∧y≠z∧z≠x (exact solutions required) (√((−3+4i)x))=y (√((−3+4i)y))=z (√((−3+4i)z))=x

$$\mathrm{solve}\:\mathrm{for}\:{x}\neq{y}\wedge{y}\neq{z}\wedge{z}\neq{x} \\ $$$$\left(\mathrm{exact}\:\mathrm{solutions}\:\mathrm{required}\right) \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){x}}={y} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){y}}={z} \\ $$$$\sqrt{\left(−\mathrm{3}+\mathrm{4i}\right){z}}={x} \\ $$

Question Number 204509 Answers: 0 Comments: 0

Question Number 204544 Answers: 1 Comments: 0

Prove that: (1/3^3 ) + (1/4^3 ) + ... + (1/n^3 ) < (1/(12))

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:\:+\:\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }\:\:+\:\:...\:\:+\:\:\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{3}} }\:\:<\:\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Question Number 204505 Answers: 1 Comments: 0

Question Number 204500 Answers: 1 Comments: 0

Given that I = ∫∫_R (x^2 +y^2 )^(5/2) dxdy where R is the region x^2 +y^2 ≤ a^2 use a suitable transformation to evaluate I

$$\mathrm{Given}\:\mathrm{that}\:{I}\:=\:\int\int_{{R}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} {dxdy}\:\mathrm{where}\:{R} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{region}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:{a}^{\mathrm{2}} \\ $$$$\mathrm{use}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{transformation}\:\mathrm{to}\:\mathrm{evaluate}\:{I} \\ $$

Question Number 204499 Answers: 1 Comments: 0

1. Find the directional derivative of F(x,y,z) = 2xy−z^2 at the point (2,−1,1) in a direction towards (3,1,−1) in what direction is the directional derivative maximum? what is the value of this maximum?

$$\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{derivative}\:\mathrm{of}\: \\ $$$${F}\left({x},{y},{z}\right)\:=\:\mathrm{2}{xy}−{z}^{\mathrm{2}} \:\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},−\mathrm{1},\mathrm{1}\right)\:\mathrm{in} \\ $$$$\mathrm{a}\:\mathrm{direction}\:\mathrm{towards}\:\left(\mathrm{3},\mathrm{1},−\mathrm{1}\right)\: \\ $$$$\mathrm{in}\:\mathrm{what}\:\mathrm{direction}\:\mathrm{is}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{derivative} \\ $$$$\mathrm{maximum}?\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this}\:\mathrm{maximum}? \\ $$

Question Number 204480 Answers: 2 Comments: 1

x + (1/x) = 2.05 x = ?

$${x}\:+\:\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}.\mathrm{05} \\ $$$${x}\:=\:? \\ $$

Question Number 204478 Answers: 1 Comments: 0

soit f: R^3 →R^3 f(x,y,z)=(x+y,2x−y,x+z) •1 Ecrire la matrice M de cette application dans la base canonique B de R^3 •2 Calculer f(1,2,3)de 2 manieres differentes −en utilisant la definition de f −en utilisant la matrice M •3 determiner bsse de Ker( f) et de Im(f) •4 soient v_1 =(1,1,0) v_2 =(1,2,1) v_3 =(1,3,1) Montrer que la famille E=(v_1 , v_2 , v_3 )est une base de R^3 •5Calculer f(v_1 ) donner ses coordonnes(locus) dans bass E avec f(v_2 )=v_1 +6v_2 −4v_3 f(v_3 )=2v_1 +8v_2 −6v_3 •6 Ecrire la matrice N de f dans base F •7 Retrouver cette matrice a partir de M en utilisant la formule de changement de base

$$\mathrm{soit}\:\boldsymbol{\mathrm{f}}:\:\:\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)=\left(\mathrm{x}+\mathrm{y},\mathrm{2x}−\mathrm{y},\mathrm{x}+\mathrm{z}\right) \\ $$$$\bullet\mathrm{1}\:\:\mathrm{Ecrire}\:\mathrm{la}\:\mathrm{matrice}\:\mathrm{M}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{application} \\ $$$$\:\:\:\mathrm{dans}\:\mathrm{la}\:\mathrm{base}\:\mathrm{canonique}\:{B}\:\mathrm{de}\:\:\mathbb{R}^{\mathrm{3}} \: \\ $$$$\bullet\mathrm{2}\:\:\mathrm{Calculer}\:\boldsymbol{\mathrm{f}}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\mathrm{de}\:\mathrm{2}\:\mathrm{manieres}\:\mathrm{differentes} \\ $$$$\:−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{definition}\:\mathrm{de}\:\mathrm{f} \\ $$$$−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{matrice}\:{M}\: \\ $$$$\bullet\mathrm{3}\:\:\mathrm{determiner}\:\mathrm{bsse}\:\mathrm{de}\:\mathrm{Ker}\left(\:\boldsymbol{\mathrm{f}}\right)\:\mathrm{et}\:\mathrm{de}\:{I}\mathrm{m}\left(\boldsymbol{\mathrm{f}}\right) \\ $$$$\bullet\mathrm{4}\:\:\mathrm{soient}\:\mathrm{v}_{\mathrm{1}} =\left(\mathrm{1},\mathrm{1},\mathrm{0}\right)\:\mathrm{v}_{\mathrm{2}} =\left(\mathrm{1},\mathrm{2},\mathrm{1}\right)\:\:\mathrm{v}_{\mathrm{3}} =\left(\mathrm{1},\mathrm{3},\mathrm{1}\right) \\ $$$$\mathrm{Montrer}\:\mathrm{que}\:\mathrm{la}\:\mathrm{famille}\:{E}=\left(\mathrm{v}_{\mathrm{1}} ,\:\mathrm{v}_{\mathrm{2}} ,\:\mathrm{v}_{\mathrm{3}} \right)\mathrm{est} \\ $$$$\mathrm{une}\:\mathrm{base}\:\mathrm{de}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\bullet\mathrm{5Calculer}\:\mathrm{f}\left(\mathrm{v}_{\mathrm{1}} \right)\:\mathrm{donner}\:\mathrm{ses}\:\mathrm{coordonnes}\left(\boldsymbol{\mathrm{locus}}\right) \\ $$$$\:\mathrm{dans}\:\mathrm{bass}\:{E} \\ $$$$\:\:\mathrm{avec}\:\mathrm{f}\left(\mathrm{v}_{\mathrm{2}} \right)=\mathrm{v}_{\mathrm{1}} +\mathrm{6v}_{\mathrm{2}} −\mathrm{4v}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{v}_{\mathrm{3}} \right)=\mathrm{2v}_{\mathrm{1}} +\mathrm{8v}_{\mathrm{2}} −\mathrm{6v}_{\mathrm{3}} \\ $$$$\bullet\mathrm{6}\:\:\mathrm{Ecrire}\:\mathrm{la}\:\mathrm{matrice}\:{N}\:\mathrm{de}\:\boldsymbol{\mathrm{f}}\:\:\mathrm{dans}\:\mathrm{base}\:{F} \\ $$$$\bullet\mathrm{7}\:\:\mathrm{Retrouver}\:\mathrm{cette}\:\mathrm{matrice}\:\mathrm{a}\:\mathrm{partir}\:\mathrm{de}\:{M} \\ $$$$\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{formule}\:\mathrm{de}\:\mathrm{changement}\:\mathrm{de}\:\mathrm{base} \\ $$$$ \\ $$

Question Number 204477 Answers: 1 Comments: 0

lim_(n→∞) (2n∫_0 ^1 (x^n /(1+x^2 ))dx)^n =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right)^{\mathrm{n}} =? \\ $$

Question Number 204472 Answers: 2 Comments: 0

Calculate ... Ω=Σ_(k=1) ^n ⌊(( 1)/( (e)^(1/k) −1)) ⌋ =?

$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Calculate}\:... \\ $$$$\:\:\:\:\:\:\:\Omega=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\lfloor\frac{\:\mathrm{1}}{\:\sqrt[{{k}}]{{e}}\:−\mathrm{1}}\:\rfloor\:=? \\ $$$$ \\ $$

Question Number 204471 Answers: 2 Comments: 0

Question Number 204469 Answers: 1 Comments: 0

Given the function f(x) = { ((2, 0< x <2)),((−2, −2 <x < 0)) :} of period 4 (a) sketch the graph of y = f(x) , for −6 < x < 6 (b) Find the Fourier coefficient a_0 , a_n , and b_n (c) write down the Fourier series. (d) hence show that Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) = (π/4)

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:=\:\begin{cases}{\mathrm{2},\:\mathrm{0}<\:{x}\:<\mathrm{2}}\\{−\mathrm{2},\:−\mathrm{2}\:<{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{of}\:\mathrm{period}\:\mathrm{4} \\ $$$$\left(\mathrm{a}\right)\:\:\mathrm{sketch}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:{y}\:=\:{f}\left({x}\right)\:,\:\mathrm{for}\:−\mathrm{6}\:<\:{x}\:<\:\mathrm{6} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{coefficient}\:{a}_{\mathrm{0}} ,\:{a}_{{n}} ,\:\mathrm{and}\:{b}_{{n}} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{write}\:\mathrm{down}\:\mathrm{the}\:\mathrm{Fourier}\:\mathrm{series}.\: \\ $$$$\left(\mathrm{d}\right)\:\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}−\mathrm{1}}\:=\:\frac{\pi}{\mathrm{4}} \\ $$

Question Number 204468 Answers: 1 Comments: 0

How Can we prove Σ_(h=−∞) ^∞ J_h (z)=1

$$\mathrm{How}\:\mathrm{Can}\:\mathrm{we}\:\mathrm{prove}\:\underset{{h}=−\infty} {\overset{\infty} {\sum}}\:{J}_{{h}} \left({z}\right)=\mathrm{1} \\ $$

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