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Question Number 132250 Answers: 0 Comments: 0

Question Number 132173 Answers: 0 Comments: 2

If v = ((√(p + (1/n)))/x), where p = pressure. find the dimension of n and x

$$\mathrm{If}\:\:\:\:\mathrm{v}\:\:\:=\:\:\:\frac{\sqrt{\mathrm{p}\:\:\:+\:\:\:\frac{\mathrm{1}}{\mathrm{n}}}}{\mathrm{x}},\:\:\:\:\:\:\:\:\:\mathrm{where}\:\:\:\:\mathrm{p}\:\:=\:\:\:\mathrm{pressure}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{dimension}\:\mathrm{of}\:\:\:\:\:\mathrm{n}\:\:\:\mathrm{and}\:\:\:\mathrm{x} \\ $$

Question Number 132162 Answers: 1 Comments: 0

Σ_(n=1) ^∞ ((sin(n))/n^2 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}^{\mathrm{2}} } \\ $$

Question Number 132124 Answers: 1 Comments: 2

Calculate ∫_0 ^( (π/4)) (√((tan(x)+tan^2 (x))/(tan(x)−tan^2 (x)))) cos(x)dx

$$ \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{tan}\left(\mathrm{x}\right)+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{tan}\left(\mathrm{x}\right)−\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}}\:\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\: \\ $$

Question Number 132064 Answers: 0 Comments: 2

Question Number 131935 Answers: 1 Comments: 0

Question Number 131877 Answers: 0 Comments: 4

Σ_(n=1) ^∞ (1/(n(e^(2πn) −1)))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({e}^{\mathrm{2}\pi{n}} −\mathrm{1}\right)} \\ $$

Question Number 131795 Answers: 1 Comments: 0

(1/1^3 )+(1/2^3 )+(1/5^3 )+(1/(10^3 ))+(1/(17^3 ))+(1/(26^3 ))+(1/(37^3 ))+(1/(50^3 ))+(1/(65^3 ))+(1/(82^3 ))+(1/(101^3 ))+...

Question Number 131686 Answers: 1 Comments: 6

Σ_(n=1) ^∞ ((coth(nπ))/n^3 )

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)}{{n}^{\mathrm{3}} } \\ $$

Question Number 131580 Answers: 0 Comments: 2

Prove or disprove Σ_(n=0) ^∞ (1/((n^2 +97)^2 ))=(𝛑^2 /(97(e^(𝛑(√(97))) −e^(−𝛑(√(97))) )^2 ))+(𝛑/(388)).((e^(2𝛑(√(97))) +1)/(e^(2𝛑(√(97))) −1))+((37635)/(37636))−(1/( 388(√(97))))

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$$$\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{97}\right)^{\mathrm{2}} }=\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{97}\left(\boldsymbol{{e}}^{\boldsymbol{\pi}\sqrt{\mathrm{97}}} −{e}^{−\boldsymbol{\pi}\sqrt{\mathrm{97}}} \right)^{\mathrm{2}} }+\frac{\boldsymbol{\pi}}{\mathrm{388}}.\frac{{e}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} +\mathrm{1}}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi}\sqrt{\mathrm{97}}} −\mathrm{1}}+\frac{\mathrm{37635}}{\mathrm{37636}}−\frac{\mathrm{1}}{\:\mathrm{388}\sqrt{\mathrm{97}}} \\ $$

Question Number 131507 Answers: 1 Comments: 0

1/Show that (2019)^(2021) +(2021)^(2019) divided by 2020 2/Show that 2222^(5555) +5555^(2222) divided by 7

$$ \\ $$$$\:\:\:\mathrm{1}/\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{2019}\right)^{\mathrm{2021}} +\left(\mathrm{2021}\right)^{\mathrm{2019}} \:\mathrm{divided}\:\:\mathrm{by}\:\mathrm{2020} \\ $$$$\:\:\:\mathrm{2}/\mathrm{Show}\:\mathrm{that}\:\mathrm{2222}^{\mathrm{5555}} +\mathrm{5555}^{\mathrm{2222}} \:\mathrm{divided}\:\:\mathrm{by}\:\mathrm{7} \\ $$$$ \\ $$

Question Number 131346 Answers: 0 Comments: 0

Question Number 131094 Answers: 1 Comments: 1

Calculate 1/ I = ∮_c^+ ((zdz)/((z−1)^2 (z^2 −2z+1−2i))) ,C={z/∣z∣=2} 2/ J =∮_c^+ ((ch(z)dz)/(z(e^z −1))) , C={z/∣z−3i∣=4} 3/ K=∮_c^+ ((sin(z)dz)/(z^3 (z+1)^2 )) , C={z/∣z∣=2}

$$\:\: \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\mathrm{1}/\:\mathrm{I}\:=\:\oint_{\mathrm{c}^{+} } \frac{\mathrm{zdz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{1}−\mathrm{2i}\right)}\:\:,\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\}\: \\ $$$$\:\:\mathrm{2}/\:\mathrm{J}\:=\oint_{\mathrm{c}^{+} } \frac{\mathrm{ch}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}\left(\mathrm{e}^{\mathrm{z}} −\mathrm{1}\right)}\:\:,\:\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}−\mathrm{3i}\mid=\mathrm{4}\right\} \\ $$$$\:\:\mathrm{3}/\:\mathrm{K}=\oint_{\mathrm{c}^{+} } \frac{\mathrm{sin}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}^{\mathrm{3}} \left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} }\:\:,\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\} \\ $$

Question Number 131083 Answers: 0 Comments: 0

1−(1/(5−((16)/(13−((81)/(25−((256)/(41−((625)/(61−((1296)/(85−((2401)/(113−...))))))))))))))=(6/𝛑^2 )

$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}−\frac{\mathrm{16}}{\mathrm{13}−\frac{\mathrm{81}}{\mathrm{25}−\frac{\mathrm{256}}{\mathrm{41}−\frac{\mathrm{625}}{\mathrm{61}−\frac{\mathrm{1296}}{\mathrm{85}−\frac{\mathrm{2401}}{\mathrm{113}−...}}}}}}}=\frac{\mathrm{6}}{\boldsymbol{\pi}^{\mathrm{2}} } \\ $$

Question Number 130992 Answers: 1 Comments: 0

If I = (V/R) and V=250 volts and R=50 ohms Find the change in I resulting from an increase of 1 volt in V and increase of 0.5 ohm in R.

$$\mathrm{If}\:\mathrm{I}\:=\:\frac{\mathrm{V}}{\mathrm{R}}\:\mathrm{and}\:\mathrm{V}=\mathrm{250}\:\mathrm{volts}\:\mathrm{and}\:\mathrm{R}=\mathrm{50}\:\mathrm{ohms} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{change}\:\mathrm{in}\:\mathrm{I}\:\mathrm{resulting}\:\mathrm{from}\:\mathrm{an}\: \\ $$$$\mathrm{increase}\:\mathrm{of}\:\mathrm{1}\:\mathrm{volt}\:\mathrm{in}\:\mathrm{V}\:\mathrm{and}\:\mathrm{increase}\:\mathrm{of}\:\mathrm{0}.\mathrm{5}\:\mathrm{ohm} \\ $$$$\mathrm{in}\:\mathrm{R}. \\ $$

Question Number 131078 Answers: 0 Comments: 0

Σ_(n=1) ^∞ (1/(n^8 +1))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{8}} +\mathrm{1}} \\ $$

Question Number 130978 Answers: 1 Comments: 5