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Question Number 137652 Answers: 1 Comments: 0

Given that α,β,γ are the three roots of the equation (x−59)^3 +(2x−64)^3 =(3x−123)^3 , find the value of α+β+γ.

$$\mathrm{Given}\:\mathrm{that}\:\alpha,\beta,\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{three}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left({x}−\mathrm{59}\right)^{\mathrm{3}} +\left(\mathrm{2}{x}−\mathrm{64}\right)^{\mathrm{3}} =\left(\mathrm{3}{x}−\mathrm{123}\right)^{\mathrm{3}} , \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha+\beta+\gamma. \\ $$

Question Number 137650 Answers: 2 Comments: 0

Question Number 137648 Answers: 1 Comments: 1

Question Number 137645 Answers: 1 Comments: 0

The sum of the squares of two consecutive odd numbers is 130. Find the larger number.

$${The}\:{sum}\:{of}\:{the}\:{squares}\:{of}\:{two}\:{consecutive} \\ $$$${odd}\:{numbers}\:{is}\:\mathrm{130}.\:{Find}\:{the}\:{larger}\:{number}. \\ $$

Question Number 137686 Answers: 0 Comments: 0

Question Number 137694 Answers: 1 Comments: 3

find ∫ (x^3 /((x−3)^3 (x(√2)+7)^4 ))dx

$${find}\:\int\:\:\:\frac{{x}^{\mathrm{3}} }{\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}\sqrt{\mathrm{2}}+\mathrm{7}\right)^{\mathrm{4}} }{dx} \\ $$

Question Number 137641 Answers: 1 Comments: 0

Question Number 137637 Answers: 0 Comments: 0

Prove or disprove ((cos(1+(1/( (√3))))𝛑)/1^2 )+((cos(1+(1/( (√3))))2π)/2^2 )+((cos(1+(1/( (√3))))3π)/3^2 )+...=0

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$$$\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\boldsymbol{\pi}}{\mathrm{1}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{2}\pi}{\mathrm{2}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{3}\pi}{\mathrm{3}^{\mathrm{2}} }+...=\mathrm{0} \\ $$

Question Number 137635 Answers: 0 Comments: 1

Question Number 137634 Answers: 1 Comments: 0

x=2^(p ) and y=2^(q ) . Evaluate in terms of x and/ or y (i)2^(p+q) (ii) 2^(2q ) (iii) 2^(p−1)

$${x}=\mathrm{2}^{{p}\:} {and}\:{y}=\mathrm{2}^{{q}\:} .\:{Evaluate}\:{in}\:{terms}\:{of}\:{x}\:{and}/\:{or}\:{y}\: \\ $$$$\left({i}\right)\mathrm{2}^{{p}+{q}} \:\:\left({ii}\right)\:\mathrm{2}^{\mathrm{2}{q}\:} \:\:\:\left({iii}\right)\:\mathrm{2}^{{p}−\mathrm{1}} \\ $$

Question Number 137627 Answers: 1 Comments: 0

Question Number 137626 Answers: 0 Comments: 0

Question Number 137625 Answers: 0 Comments: 0

Question Number 137624 Answers: 0 Comments: 0

(dy/dx) = ((6xy+y^2 )/(x+y))

$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:=\:\frac{\mathrm{6}\boldsymbol{\mathrm{xy}}+\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}} \\ $$

Question Number 137618 Answers: 1 Comments: 2

Question Number 137615 Answers: 0 Comments: 1

∫_0 ^( (π/2)) ln(((1+tanx)/(1−tanx)))^(xcos(8x)) dx

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}\right)^{{xcos}\left(\mathrm{8}{x}\right)} {dx} \\ $$

Question Number 137613 Answers: 1 Comments: 0

Two particles of mass m and 4m are connected by a light inelastic string of length 3a, which passes through a small smooth fixed ring. The heavier particle hangs at rest at a distance 2a beneath the ring, while the lighter particle describes a horizontal circle with constant speed. Find (a) The distance of the plane of the circle below the ring. (b) The angular speed of the lighter particle.

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{and}\:\mathrm{4}{m}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light}\:\mathrm{inelastic} \\ $$$$\mathrm{string}\:\mathrm{of}\:\mathrm{length}\:\mathrm{3}{a},\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{a}\:\mathrm{small}\:\mathrm{smooth}\:\mathrm{fixed}\:\mathrm{ring}. \\ $$$$\mathrm{The}\:\mathrm{heavier}\:\mathrm{particle}\:\mathrm{hangs}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{2}{a}\:\mathrm{beneath}\:\mathrm{the}\:\mathrm{ring}, \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{speed}. \\ $$$$\:\mathrm{Find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{below}\:\mathrm{the}\:\mathrm{ring}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{angular}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}. \\ $$

Question Number 137611 Answers: 1 Comments: 0

Question Number 137610 Answers: 0 Comments: 0

........ mathematical analysis (II).... prove that :: 𝛀=∫_0 ^( 1) (1/(1+x))ln(((x^2 +2x+1)/(1+x+x^2 )))=Σ_(n=1) ^∞ (1/(n^2 (((2n)),(( n)) )))=(π^2 /(18))..

$$\:\:\:\:\:\:\:\:\:\:........\:{mathematical}\:\:\:{analysis}\:\left({II}\right).... \\ $$$$\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}}=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}.. \\ $$

Question Number 137605 Answers: 0 Comments: 0

.....nice calculus... prove that:: Σ_(n=0) ^∞ tan^(−1) ((1/F_n )).tan^(−1) ((1/F_(n+1) ))=(π^2 /4) F_n is fibonacci sequence....

$$\:\:\:\:\:\:\:\:\:\:\:\:.....{nice}\:\:\:\:{calculus}... \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}} }\right).{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}+\mathrm{1}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{F}_{{n}} \:{is}\:{fibonacci}\:{sequence}.... \\ $$

Question Number 137598 Answers: 0 Comments: 0

Question Number 137597 Answers: 2 Comments: 0

Find the minimum value of x^(2) +y^(2) +z^(2) , subject to the condition 2x+3y+5z=30?

$$ \\ $$Find the minimum value of x^(2) +y^(2) +z^(2) , subject to the condition 2x+3y+5z=30?

Question Number 137594 Answers: 1 Comments: 0

(x+(√(x^2 +1)))(y+(√(y^4 +4)))=9 x(√(y^4 +4))+y(√(x^2 +1))=?

$$\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{4}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${x}\sqrt{{y}^{\mathrm{4}} +\mathrm{4}}+{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=? \\ $$

Question Number 137592 Answers: 3 Comments: 0

......advanced.....calculus.... 𝛀=Σ_(n=1) ^∞ ((ψ′′(n))/n)=??? I havefound :: Ω=−(π^4 /(36)) ... !

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......{advanced}.....{calculus}.... \\ $$$$\:\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi''\left({n}\right)}{{n}}=??? \\ $$$$\:{I}\:{havefound}\:::\:\:\Omega=−\frac{\pi^{\mathrm{4}} }{\mathrm{36}}\:\:...\:! \\ $$

Question Number 137591 Answers: 0 Comments: 1

a=(4)^(1/3) +(2)^(1/3) +(1)^(1/3) (3/a)+(3/a^2 )+(1/a^3 )=?

$${a}=\sqrt[{\mathrm{3}}]{\mathrm{4}}+\sqrt[{\mathrm{3}}]{\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{1}} \\ $$$$\frac{\mathrm{3}}{{a}}+\frac{\mathrm{3}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=? \\ $$

Question Number 137590 Answers: 0 Comments: 1

x=1+((π+(π+1)^2 +(π+2)^3 +(π+3)^4 )/(4+5^2 +6^3 +7^4 )) (√(x+2(√(x−1))))+(√(x−2(√(x−1))))=?

$${x}=\mathrm{1}+\frac{\pi+\left(\pi+\mathrm{1}\right)^{\mathrm{2}} +\left(\pi+\mathrm{2}\right)^{\mathrm{3}} +\left(\pi+\mathrm{3}\right)^{\mathrm{4}} }{\mathrm{4}+\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{3}} +\mathrm{7}^{\mathrm{4}} } \\ $$$$\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}=? \\ $$

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