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Question Number 161102 Answers: 1 Comments: 1

Question Number 161101 Answers: 1 Comments: 0

solve: ∫((x+1)/(x^2 −7x−3))dx

$${solve}: \\ $$$$\:\:\:\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{7}{x}−\mathrm{3}}{dx} \\ $$

Question Number 161100 Answers: 0 Comments: 0

f(x^2 )= 2+∫_( 0) ^( x^2 ) f(y) (1−tan y)dy , ∀x∈R f(−π)=?

$$\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)=\:\mathrm{2}+\int_{\:\mathrm{0}} ^{\:\mathrm{x}^{\mathrm{2}} } \mathrm{f}\left(\mathrm{y}\right)\:\left(\mathrm{1}−\mathrm{tan}\:\mathrm{y}\right)\mathrm{dy}\:,\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\:\mathrm{f}\left(−\pi\right)=? \\ $$

Question Number 161096 Answers: 0 Comments: 0

if x;y;z>0 and a;b;c>0 different in pairs and n;k∈N^∗ ((log x^n )/(b^k - c^k )) = ((log y^n )/(c^k - a^k )) = ((log z^n )/(a^k - b^k )) then find (√(xyz))

Question Number 161091 Answers: 1 Comments: 0

Solve for real numbers: (√(1 - x)) = 2x^2 - 1 - 2x (√(1 - x^2 ))

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\mathrm{1}\:-\:\mathrm{x}}\:=\:\mathrm{2x}^{\mathrm{2}} \:-\:\mathrm{1}\:-\:\mathrm{2x}\:\sqrt{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 161114 Answers: 0 Comments: 0

Let f(x)= sin^3 (2x) for −(π/4)≤x≤(π/4) then Df^(−1) ((1/8))=(a/(b(√b))) so { ((a=?)),((b=?)) :}

$$\:\:{Let}\:{f}\left({x}\right)=\:\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{2}{x}\right)\:{for}\:−\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\:{then}\:{Df}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{{a}}{{b}\sqrt{{b}}}\:{so}\:\begin{cases}{{a}=?}\\{{b}=?}\end{cases} \\ $$

Question Number 161089 Answers: 3 Comments: 0

prove that I= ∫_0 ^( (π/2)) ln ( 1+ sin (2 α )) dα = 2G − π ln ((√2) ) G: catalan constant

$$ \\ $$$$\:\:{prove}\:{that} \\ $$$$\:\:\:\mathrm{I}=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \mathrm{ln}\:\left(\:\mathrm{1}+\:{sin}\:\left(\mathrm{2}\:\alpha\:\right)\right)\:{d}\alpha\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{2G}\:−\:\pi\:\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{G}:\:\:{catalan}\:{constant} \\ $$

Question Number 161084 Answers: 0 Comments: 0

Question Number 161079 Answers: 1 Comments: 3

Question Number 161076 Answers: 1 Comments: 0

Ω = ∫_0 ^( ∞) ((ln (1+ x ))/((1+ x^( 2) )^( 2) )) dx = ? −−−−−−−−−−−−

$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\:\left(\mathrm{1}+\:{x}\:\right)}{\left(\mathrm{1}+\:{x}^{\:\mathrm{2}} \right)^{\:\mathrm{2}} }\:{dx}\:=\:? \\ $$$$\:\:\:\:\:−−−−−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 161075 Answers: 0 Comments: 0

simplify Σ_(n=1) ^∞ (( n)/(( n^( 2) −(( 1)/4) )^( 3) )) = ?

$$ \\ $$$$\:\:\:\:{simplify} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{n}}{\left(\:{n}^{\:\mathrm{2}} −\frac{\:\mathrm{1}}{\mathrm{4}}\:\right)^{\:\mathrm{3}} }\:=\:? \\ $$$$ \\ $$

Question Number 161071 Answers: 2 Comments: 0

For a,b,c > 0 . Find (x,y,z) that satisfy this equation system ax + by = (x−y)^2 by + cz = (y−z)^2 cz + ax = (z−x)^2

$${For}\:\:{a},{b},{c}\:>\:\mathrm{0}\:. \\ $$$${Find}\:\:\left({x},{y},{z}\right)\:\:{that}\:\:{satisfy}\:\:{this}\:\:{equation}\:\:{system}\: \\ $$$$\:\:\:{ax}\:+\:{by}\:=\:\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{by}\:+\:{cz}\:=\:\left({y}−{z}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{cz}\:+\:{ax}\:=\:\left({z}−{x}\right)^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 161068 Answers: 2 Comments: 0

∫ ((2x)/((1−x^2 )(√(x^4 −1)))) dx =?

$$\:\:\:\:\:\int\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}\:{dx}\:=? \\ $$

Question Number 161066 Answers: 2 Comments: 0

x_1 ,x_2 be the roots of the equation x^2 +x+m=0 & x_1 ^5 +x_2 ^5 = 2021. Find the sum of the possible values of m.

$$\:{x}_{\mathrm{1}} \:,{x}_{\mathrm{2}} \:{be}\:{the}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} +{x}+{m}=\mathrm{0}\:\&\:{x}_{\mathrm{1}} ^{\mathrm{5}} +{x}_{\mathrm{2}} ^{\mathrm{5}} \:=\:\mathrm{2021}. \\ $$$$\:{Find}\:{the}\:{sum}\:{of}\:{the}\:{possible}\:{values} \\ $$$$\:\:{of}\:{m}. \\ $$

Question Number 161065 Answers: 1 Comments: 0

{ ((((x+abc))^(1/4) +((x−abc))^(1/8) = a)),((((x+abc))^(1/4) −((x−abc))^(1/8) = b)),((((x+abc))^(1/4) −((x−abc))^(1/4) = c)) :} find (√(x+abc)) +(√(x−abc))

$$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}}\end{cases} \\ $$$$\:{find}\:\sqrt{{x}+{abc}}\:+\sqrt{{x}−{abc}} \\ $$

Question Number 161061 Answers: 1 Comments: 0

Question Number 161060 Answers: 1 Comments: 2

Given sin(5x−38)=cos(2x+16), 0°≤x≤90°, find the value of x

$$\mathrm{Given}\:\mathrm{sin}\left(\mathrm{5x}−\mathrm{38}\right)=\mathrm{cos}\left(\mathrm{2x}+\mathrm{16}\right),\:\mathrm{0}°\leqslant\mathrm{x}\leqslant\mathrm{90}°, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$

Question Number 161059 Answers: 0 Comments: 0

Find: 𝛀 =∫_( 0) ^( 1) ∫_( 0) ^( 1) (x^2 +2xy+x)ln(1 + (1/(x+y)))dxdy

$$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right)\mathrm{dxdy} \\ $$

Question Number 161058 Answers: 1 Comments: 0

Solve the differential equation: x(y-1)dx + (x+1)dy = 0

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\mathrm{x}\left(\mathrm{y}-\mathrm{1}\right)\mathrm{dx}\:+\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{dy}\:=\:\mathrm{0} \\ $$$$ \\ $$

Question Number 161039 Answers: 0 Comments: 0

let the differential equation: (1 + x) y^(′′) (x) + (1 - x) y^′ (x) = ((1-x)/(1+x)) y(x) y(0) = 1 , y^′ (0) = 0 then prove that: ∫_( 0) ^( ∞) (y^(′′) (x) + y^′ (x) + y(x)) e^(-x) dx = (3/2)

$$\mathrm{let}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\left(\mathrm{1}\:+\:\mathrm{x}\right)\:\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}-\mathrm{x}}{\mathrm{1}+\mathrm{x}}\:\mathrm{y}\left(\mathrm{x}\right) \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:,\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\left(\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:+\:\mathrm{y}\left(\mathrm{x}\right)\right)\:\mathrm{e}^{-\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 161026 Answers: 0 Comments: 0

Question Number 161025 Answers: 1 Comments: 0

Two commodities A and B cost $70 and $80 per kg respectively. If 34.5kg of A is mixed with 26kg of B and the mixture is sold at $85 per kg, calculate the percentage profit. please help me out, I′m somehow confused

$${Two}\:{commodities}\:{A}\:{and}\:{B}\:{cost} \\ $$$$\$\mathrm{70}\:{and}\:\$\mathrm{80}\:{per}\:{kg}\:{respectively}. \\ $$$${If}\:\mathrm{34}.\mathrm{5}{kg}\:{of}\:{A}\:{is}\:{mixed}\:{with}\:\mathrm{26}{kg} \\ $$$${of}\:{B}\:{and}\:{the}\:{mixture}\:{is}\:{sold}\:{at} \\ $$$$\$\mathrm{85}\:{per}\:{kg},\:{calculate}\:{the}\:{percentage} \\ $$$${profit}. \\ $$$${please}\:{help}\:{me}\:{out},\:{I}'{m}\:{somehow} \\ $$$${confused} \\ $$

Question Number 161023 Answers: 1 Comments: 2

Question Number 161020 Answers: 1 Comments: 0

etudier la convergence ∫_0 ^(+oo) (1/( (√(x(1+x^2 )))))dx

$${etudier}\:{la}\:{convergence} \\ $$$$\int_{\mathrm{0}} ^{+{oo}} \frac{\mathrm{1}}{\:\sqrt{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}}{dx} \\ $$

Question Number 161035 Answers: 2 Comments: 0

Solve for x ε R (√((√3)−x)) = x(√((√3)+x))

$${Solve}\:{for}\:{x}\:\epsilon\:\mathbb{R}\: \\ $$$$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$

Question Number 161033 Answers: 1 Comments: 0

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