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AllQuestion and Answers: Page 529

Question Number 163958 Answers: 0 Comments: 0

Question Number 163957 Answers: 1 Comments: 0

Question Number 163956 Answers: 0 Comments: 0

Question Number 163955 Answers: 0 Comments: 0

Question Number 163942 Answers: 1 Comments: 0

f(2x)−2[ f(x) ]^2 +1=0 f(x)=?

$${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left[\:{f}\left({x}\right)\:\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 163939 Answers: 0 Comments: 0

An old unsolved question#1132 let S={1,2,3,4,5}, if A,B,C is such that A∩B∩C=∅ A∩B≠∅ A∩C≠∅ how many ways can be choose A,B and C

$$\mathrm{An}\:\mathrm{old}\:\mathrm{unsolved}\:\mathrm{question}#\mathrm{1132} \\ $$$$\mathrm{let}\:\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\:\mathrm{if}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}=\varnothing \\ $$$$\mathrm{A}\cap\mathrm{B}\neq\varnothing \\ $$$$\mathrm{A}\cap\mathrm{C}\neq\varnothing \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choose}\:\mathrm{A},\mathrm{B}\:\mathrm{and} \\ $$$$\mathrm{C} \\ $$

Question Number 163938 Answers: 1 Comments: 0

lim_(x→0) ((cos (−3x)−cos (3x))/(sin^2 (x(√5) )))=?

$$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(−\mathrm{3}{x}\right)−\mathrm{cos}\:\left(\mathrm{3}{x}\right)}{\mathrm{sin}\:^{\mathrm{2}} \left({x}\sqrt{\mathrm{5}}\:\right)}=? \\ $$

Question Number 163936 Answers: 1 Comments: 0

cos^(−1) (((x^2 −1)/(x^2 +1)))+(1/2)tan^(−1) (((2x)/(1−x^2 )))=((2π)/3)

$$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$

Question Number 163934 Answers: 0 Comments: 0

Let y(x) be the solution of x^2 y′′(x)−2y(x)=0 → { ((y(1)=1)),((y(2)=1)) :} y(3)=?

$${Let}\:{y}\left({x}\right)\:{be}\:{the}\:{solution}\:{of}\: \\ $$$$\:\:{x}^{\mathrm{2}} \:{y}''\left({x}\right)−\mathrm{2}{y}\left({x}\right)=\mathrm{0}\:\rightarrow\begin{cases}{{y}\left(\mathrm{1}\right)=\mathrm{1}}\\{{y}\left(\mathrm{2}\right)=\mathrm{1}}\end{cases} \\ $$$$\:{y}\left(\mathrm{3}\right)=? \\ $$

Question Number 163931 Answers: 1 Comments: 0

lim_(x→0) (x^x^x /x)=? pleas help

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{{x}^{{x}} } }{{x}}=? \\ $$$${pleas}\:\:{help} \\ $$

Question Number 163928 Answers: 1 Comments: 0

f(x)=((2x^(100!) )/(100!))+x^(100) +1 find ((d^(100!) f(x))/dx^(100!) )=?

$${f}\left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{100}!} }{\mathrm{100}!}+{x}^{\mathrm{100}} +\mathrm{1} \\ $$$${find}\:\:\:\frac{{d}^{\mathrm{100}!} {f}\left({x}\right)}{{dx}^{\mathrm{100}!} }=? \\ $$

Question Number 163926 Answers: 2 Comments: 0

{: ((sin θ=1−sin^2 θ)),((csc^2 θ−tan^2 θ =? )) }

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left.\begin{matrix}{\mathrm{sin}\:\theta=\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta}\\{\mathrm{csc}\:^{\mathrm{2}} \theta−\mathrm{tan}\:^{\mathrm{2}} \theta\:=?\:}\end{matrix}\right\} \\ $$

Question Number 163921 Answers: 2 Comments: 0

Question Number 163914 Answers: 0 Comments: 0

Question Number 163913 Answers: 0 Comments: 0

Question Number 163912 Answers: 0 Comments: 1

Question Number 163906 Answers: 1 Comments: 0

{ ((x^3 + x + 6 = 8y)),((y^3 + y + 6 = 8z)),((z^3 + z + 6 = 8x)) :} ⇒ x;y;z = ?

$$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}\:+\:\mathrm{6}\:=\:\mathrm{8y}}\\{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{y}\:+\:\mathrm{6}\:=\:\mathrm{8z}}\\{\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\:+\:\mathrm{6}\:=\:\mathrm{8x}}\end{cases}\:\:\:\Rightarrow\:\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:=\:? \\ $$

Question Number 163905 Answers: 0 Comments: 0

Question Number 163903 Answers: 0 Comments: 0

prouve a/bc ,si (a,b)=1,alors a/c

$${prouve}\: \\ $$$${a}/{bc}\:,{si}\:\left({a},{b}\right)=\mathrm{1},{alors} \\ $$$${a}/{c} \\ $$

Question Number 163902 Answers: 1 Comments: 0

(√(x!^(x!) )) + 2^(x!) = x!^3 + 10x! + 4 find: x = ?

$$\sqrt{\mathrm{x}!^{\boldsymbol{\mathrm{x}}!} }\:\:+\:\:\mathrm{2}^{\boldsymbol{\mathrm{x}}!} \:\:=\:\mathrm{x}!^{\mathrm{3}} \:\:+\:\:\mathrm{10x}!\:\:+\:\:\mathrm{4} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 163899 Answers: 2 Comments: 0

Find: 𝛀 = ∫_( 0) ^( 1) ((cos(ax))/( (√x) ∙ (√(1 - x)))) dx

$$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{cos}\left(\mathrm{ax}\right)}{\:\sqrt{\mathrm{x}}\:\centerdot\:\sqrt{\mathrm{1}\:-\:\mathrm{x}}}\:\mathrm{dx} \\ $$

Question Number 163897 Answers: 2 Comments: 0

if x^3 = 1 and x ≠ 1 simplificar (((1/x^4 )/(1 + x^5 )))^3

$$\mathrm{if}\:\:\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{x}\:\neq\:\mathrm{1} \\ $$$$\mathrm{simplificar}\:\:\left(\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{5}} }\right)^{\mathrm{3}} \\ $$

Question Number 163891 Answers: 2 Comments: 0

if f(x^3 + 1) = x^5 + 4x + 2 find ∫_( 0) ^( 1) f(x) dx = ?

$$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}\right)\:=\:\mathrm{x}^{\mathrm{5}} \:+\:\mathrm{4x}\:+\:\mathrm{2} \\ $$$$\mathrm{find}\:\:\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 163888 Answers: 1 Comments: 1

prove∫((1−x^2 )/( (√(1−x^2 ))))dx = ∫(√(1−x^2 ))dx

$${prove}\int\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}\:=\:\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 163886 Answers: 1 Comments: 2

Evaluate the following by using integration by parts formula: ∫xsin^(−1) (x)dx

$${Evaluate}\:{the}\:{following}\:{by}\:{using}\: \\ $$$$\:{integration}\:{by}\:{parts}\:{formula}: \\ $$$$\int{xsin}^{−\mathrm{1}} \left({x}\right){dx} \\ $$

Question Number 163885 Answers: 0 Comments: 0

preuve a)HH^(−1) =H b)HH=H c)H^(−1) =H

$${preuve} \\ $$$$\left.{a}\right){HH}^{−\mathrm{1}} ={H} \\ $$$$\left.{b}\right){HH}={H} \\ $$$$\left.{c}\right){H}^{−\mathrm{1}} ={H} \\ $$

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