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Question Number 151317    Answers: 2   Comments: 0

if x=(((√7) - (√6)))^(1/3) + (((√7) + (√6)))^(1/3) find E(x) = x^6 - 6x^4 + 9x^2 = ?

$$\mathrm{if}\:\:\boldsymbol{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:-\:\sqrt{\mathrm{6}}}\:+\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:+\:\sqrt{\mathrm{6}}} \\ $$$$\mathrm{find}\:\:\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{4}} \:+\:\mathrm{9x}^{\mathrm{2}} \:=\:? \\ $$

Question Number 151316    Answers: 1   Comments: 0

Find positive integers a and b such that ((a)^(1/3) +(b)^(1/3) −1)^2 = 49+ 20(6)^(1/3)

$$\:\:\:\:\:\:{Find}\:{positive}\:{integers}\: \\ $$$$\:\:\:\:\:{a}\:{and}\:{b}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\left(\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{49}+\:\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$

Question Number 151315    Answers: 1   Comments: 0

if x^2 +y^2 =16^2 ; y^2 +z^2 =24^2 z^2 +t^2 =42^2 and t^2 +x^2 =38^2 find max[(x+z)(y+t)]=?

$$\mathrm{if}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \:\:;\:\:\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{24}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} =\mathrm{42}^{\mathrm{2}} \:\:\mathrm{and}\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} =\mathrm{38}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\mathrm{max}\left[\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{y}+\mathrm{t}\right)\right]=? \\ $$

Question Number 151311    Answers: 0   Comments: 0

Question Number 151308    Answers: 0   Comments: 0

Question Number 151307    Answers: 2   Comments: 0

if x(√x) - 26(√x) = 5 find x - 5(√x) = ?

$$\mathrm{if}\:\:\mathrm{x}\sqrt{\mathrm{x}}\:-\:\mathrm{26}\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:? \\ $$

Question Number 151300    Answers: 3   Comments: 0

5 ∙ 6,02∙10^(23) = ? (solution) a)3,01∙10^(24) b)3,01∙10^(22)

$$\mathrm{5}\:\centerdot\:\mathrm{6},\mathrm{02}\centerdot\mathrm{10}^{\mathrm{23}} \:=\:?\:\left(\mathrm{solution}\right) \\ $$$$\left.\mathrm{a}\left.\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{24}} \:\:\:\mathrm{b}\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{22}} \\ $$

Question Number 151294    Answers: 1   Comments: 0

Question Number 151287    Answers: 0   Comments: 2

I = ∫_(x=a) ^( x=b) (√(u^2 + v^2 x^2 − 2uvwx)) dx = ?

$${I}\:=\:\int_{{x}={a}} ^{\:{x}={b}} \sqrt{{u}^{\mathrm{2}} \:+\:{v}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{uvwx}}\:{dx}\:=\:? \\ $$

Question Number 151284    Answers: 1   Comments: 0

if (a/3^(x-1) ) = (b/3^(y+2) ) = (c/3^(z-1) ) = (1/5) find abc = ?

$$\mathrm{if}\:\:\frac{\mathrm{a}}{\mathrm{3}^{\boldsymbol{\mathrm{x}}-\mathrm{1}} }\:=\:\frac{\mathrm{b}}{\mathrm{3}^{\boldsymbol{\mathrm{y}}+\mathrm{2}} }\:=\:\frac{\mathrm{c}}{\mathrm{3}^{\boldsymbol{\mathrm{z}}-\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{find}\:\:\mathrm{abc}\:=\:? \\ $$

Question Number 151282    Answers: 0   Comments: 0

Question Number 151281    Answers: 0   Comments: 0

Question Number 151278    Answers: 1   Comments: 0

∫ (e^(√(x - 1)) /( (√(x - 1)))) dx = ?

$$\int\:\frac{\boldsymbol{\mathrm{e}}^{\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}} }{\:\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}}\:\mathrm{dx}\:=\:? \\ $$

Question Number 151276    Answers: 2   Comments: 2

∫_0 ^(2π) cos^(−1) (cos x)dx

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 151272    Answers: 0   Comments: 1

Question Number 151269    Answers: 0   Comments: 1

Question Number 151268    Answers: 0   Comments: 0

Σ_(r=1) ^∞ (((−1)^(r−1) )/r)[ψ((r/2)+(1/4))−ψ((r/2)−(1/4))]

$$\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{r}−\mathrm{1}} }{{r}}\left[\psi\left(\frac{{r}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi\left(\frac{{r}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\right] \\ $$

Question Number 151265    Answers: 1   Comments: 3

Question Number 151256    Answers: 2   Comments: 0

A_n =2^n +3^n +4^n +5^n B_n =100^n +101^n +102^n +103^n 1)find values of n while 7∣A_n 2) show that B_n ≡A_n [7 ]

$$\:{A}_{{n}} =\mathrm{2}^{{n}} +\mathrm{3}^{{n}} +\mathrm{4}^{{n}} +\mathrm{5}^{{n}} \\ $$$${B}_{{n}} =\mathrm{100}^{{n}} +\mathrm{101}^{{n}} +\mathrm{102}^{{n}} +\mathrm{103}^{{n}} \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{find}}\:\boldsymbol{{values}}\:\boldsymbol{{of}}\:\boldsymbol{{n}}\:\boldsymbol{{while}}\:\mathrm{7}\mid\boldsymbol{{A}}_{\boldsymbol{{n}}} \\ $$$$\left.\mathrm{2}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:\boldsymbol{{B}}_{\boldsymbol{{n}}} \equiv\boldsymbol{{A}}_{\boldsymbol{{n}}} \left[\mathrm{7}\:\right] \\ $$

Question Number 151248    Answers: 1   Comments: 0

Question Number 151247    Answers: 1   Comments: 0

(1/5) + (1/(5^2 ∙2)) + (1/(5^3 ∙3)) + (1/(5^4 ∙4)) + ... = ?

$$\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} \centerdot\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{4}} \centerdot\mathrm{4}}\:+\:...\:=\:? \\ $$

Question Number 151246    Answers: 1   Comments: 0

find I=∫_0 ^(π/4) ln(cosx)dx and J=∫_0 ^(π/4) ln(sinx)dx

$$\mathrm{find}\:\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:\mathrm{and}\:\mathrm{J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sinx}\right)\mathrm{dx} \\ $$

Question Number 151241    Answers: 1   Comments: 0

Question Number 151230    Answers: 4   Comments: 2

prove: ∫_0 ^( ∞) (( ln ( 1+x^( 2) ))/(x^( 2) (1+x^( 2) )))dx= π ln((e/2) ) ..

$$ \\ $$$$\:\:\:\:{prove}: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\:\left(\:\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{{x}^{\:\mathrm{2}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)}{dx}=\:\pi\:{ln}\left(\frac{{e}}{\mathrm{2}}\:\right)\:.. \\ $$

Question Number 151226    Answers: 0   Comments: 0

Question Number 151224    Answers: 1   Comments: 0

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