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Question Number 137191 Answers: 0 Comments: 0

Some useful approximations of sine function sin((π/7))=((96)/(221)) sin((π/9))=((128)/(373)) sin((π/(11)))=((32)/(113)) ... I am counting more ..thanking you!

$${Some}\:{useful}\:{approximations}\:{of}\:{sine}\:{function} \\ $$$${sin}\left(\frac{\pi}{\mathrm{7}}\right)=\frac{\mathrm{96}}{\mathrm{221}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{9}}\right)=\frac{\mathrm{128}}{\mathrm{373}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{11}}\right)=\frac{\mathrm{32}}{\mathrm{113}} \\ $$$$... \\ $$$${I}\:{am}\:{counting}\:{more}\:..{thanking}\:{you}! \\ $$

Question Number 136995 Answers: 0 Comments: 0

1−(((1.1.3)/(2.3.4)))(1/(1!))+(((3.3.7)/(2^2 .3^2 .4^2 )))(1/(2!))−(((5.7.10)/(2^3 .3^3 .4^3 )))(1/(3!))−....

$$\mathrm{1}−\left(\frac{\mathrm{1}.\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\right)\frac{\mathrm{1}}{\mathrm{1}!}+\left(\frac{\mathrm{3}.\mathrm{3}.\mathrm{7}}{\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} }\right)\frac{\mathrm{1}}{\mathrm{2}!}−\left(\frac{\mathrm{5}.\mathrm{7}.\mathrm{10}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} .\mathrm{4}^{\mathrm{3}} }\right)\frac{\mathrm{1}}{\mathrm{3}!}−.... \\ $$

Question Number 137016 Answers: 1 Comments: 0

Question Number 136950 Answers: 2 Comments: 0

Question Number 136930 Answers: 0 Comments: 6

Question Number 136899 Answers: 2 Comments: 0

Question Number 136892 Answers: 1 Comments: 0

∫(d^2 y/dx^2 )dy

$$\int\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }{dy} \\ $$$$ \\ $$

Question Number 136885 Answers: 1 Comments: 0

Question Number 136850 Answers: 0 Comments: 1

Σ_(n=−∞) ^∞ a^((n(n+1))/2) b^((n(n−1))/2) =1+(√((2a^2 )/π))∫_0 ^∞ e^(−t^2 /2) (((1−a(√(ab)) cosh((√(log(ab))) t))/(a^3 b−2a(√(ab )) cosh((√(log(ab))) t))))dt

$$\underset{{n}=−\infty} {\overset{\infty} {\sum}}{a}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} {b}^{\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}} =\mathrm{1}+\sqrt{\frac{\mathrm{2}{a}^{\mathrm{2}} }{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} /\mathrm{2}} \left(\frac{\mathrm{1}−{a}\sqrt{{ab}}\:{cosh}\left(\sqrt{{log}\left({ab}\right)}\:{t}\right)}{{a}^{\mathrm{3}} {b}−\mathrm{2}{a}\sqrt{{ab}\:}\:{cosh}\left(\sqrt{{log}\left({ab}\right)}\:{t}\right)}\right){dt} \\ $$

Question Number 136826 Answers: 0 Comments: 0

Question Number 136823 Answers: 0 Comments: 0

Question Number 136799 Answers: 0 Comments: 0

∫_0 ^1 Π_(n=1) ^∞ (1−q^n )dq

$$\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−{q}^{{n}} \right){dq} \\ $$

Question Number 136761 Answers: 1 Comments: 1

Question Number 136760 Answers: 1 Comments: 0

Question Number 136757 Answers: 1 Comments: 0

1+((1/2))^2 (1/(2.1!))+(((1.3)/2^2 ))^2 (1/(2^2 .2!))+(((1.3.5)/2^3 )).(1/(2^3 .3!))+....=(((Γ((1/4)))/((2π^3 )^(1/4) )))^2

$$\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}.\mathrm{1}!}+\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right).\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+....=\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{2}\pi^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{4}} }\right)^{\mathrm{2}} \\ $$

Question Number 136661 Answers: 0 Comments: 0

∫_0 ^1 (1−x)^(−(7/8)) (1+x)^(−(1/(16))) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{16}}} {dx} \\ $$

Question Number 136593 Answers: 1 Comments: 0

Question Number 136520 Answers: 3 Comments: 1

Question Number 136519 Answers: 1 Comments: 0

1−((1/2))^(2k) +(((1.3)/(2.4)))^(2k) −(((1.3.5)/(2.4.6)))^(2k) +(((1.3.5.7)/(2.4.6.8)))^(2k) −.... Find the general form

$$\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}\right)^{\mathrm{2}{k}} −\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}\right)^{\mathrm{2}{k}} +\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}\right)^{\mathrm{2}{k}} −.... \\ $$$${Find}\:{the}\:{general}\:{form} \\ $$

Question Number 136420 Answers: 2 Comments: 0

Σ_(n=1) ^∞ ((sin(2n+1)θ)/((2n+1)^2 ))=(θ/2)log(2)+sinθ((log(sinθ))/4) _2 F_1 ((1/2),(1/2);(3/2);sin^2 θ)+((sinθ)/(16)) _2 F_1 ((1/2),(1/2),(1/2);(3/2);(3/2);sin^2 θ) Prove or disprove 0<θ<π

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\theta}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\theta}{\mathrm{2}}{log}\left(\mathrm{2}\right)+{sin}\theta\frac{{log}\left({sin}\theta\right)}{\mathrm{4}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \theta\right)+\frac{{sin}\theta}{\mathrm{16}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \theta\right) \\ $$$${Prove}\:{or}\:{disprove}\:\:\:\:\:\mathrm{0}<\theta<\pi \\ $$

Question Number 136105 Answers: 0 Comments: 4

Question Number 136104 Answers: 0 Comments: 0

((cos(1+(√((43)/3)))π)/1^2 )+((cos(1+(√((43)/3)))2π)/2^2 )+((cos(1+(√((43)/3)))3π)/3^2 )+...=aπ Find a

$$\frac{{cos}\left(\mathrm{1}+\sqrt{\frac{\mathrm{43}}{\mathrm{3}}}\right)\pi}{\mathrm{1}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\sqrt{\frac{\mathrm{43}}{\mathrm{3}}}\right)\mathrm{2}\pi}{\mathrm{2}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\sqrt{\frac{\mathrm{43}}{\mathrm{3}}}\right)\mathrm{3}\pi}{\mathrm{3}^{\mathrm{2}} }+...={a}\pi \\ $$$${Find}\:{a} \\ $$

Question Number 136070 Answers: 1 Comments: 0

((sin((√2)))/1^3 )+((sin(2(√2)))/2^3 )+((sin(3(√2)))/3^3 )+...=((π^b +1)/( a(√b)))−(π/b) Find a−b

$$\frac{{sin}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)}{\mathrm{3}^{\mathrm{3}} }+...=\frac{\pi^{{b}} +\mathrm{1}}{\:{a}\sqrt{{b}}}−\frac{\pi}{{b}} \\ $$$${Find}\:{a}−{b} \\ $$

Question Number 136110 Answers: 0 Comments: 1

An engine is pumping water from a well 25m deep. It discharges 0.4 m^3 of water each second with a velocity of 12 ms^(−1) . Find the power of the pump given that the density of water is 1000 kg m^(−3) . take g = 10 ms^(−2)

$$\mathrm{An}\:\mathrm{engine}\:\mathrm{is}\:\mathrm{pumping}\:\mathrm{water}\:\mathrm{from}\:\mathrm{a}\:\mathrm{well}\:\mathrm{25m}\:\mathrm{deep}.\:\mathrm{It}\:\mathrm{discharges} \\ $$$$\mathrm{0}.\mathrm{4}\:\mathrm{m}^{\mathrm{3}} \:\mathrm{of}\:\mathrm{water}\:\mathrm{each}\:\mathrm{second}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{12}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{Find}\:\mathrm{the}\: \\ $$$$\mathrm{power}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pump}\:\mathrm{given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{density}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\mathrm{1000}\:\mathrm{kg}\:\mathrm{m}^{−\mathrm{3}} . \\ $$$$\mathrm{take}\:\boldsymbol{\mathrm{g}}\:=\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{2}} \\ $$

Question Number 135976 Answers: 1 Comments: 0

Question Number 135906 Answers: 1 Comments: 0

Edit: this is already fixed. Therefore this thread is closed... There′s a glitch! Whenever you backspace from a line to delete it, you′ll end up directly in front of the first thing you typed, whatever it is. Also, is there a way for this app to run in portrait orrentation on tablets? Because I can′t seem to get it to rotate to that mode. I saw in your screenshots on the Google Play page that it was possible, but I can′t get it to work. I would perfer editing in that orrentation over the landscape rotation.

$${Edit}:\:{this}\:{is}\:{already}\:{fixed}.\:{Therefore}\:{this}\:{thread}\:{is}\:{closed}... \\ $$$$ \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{glitch}! \\ $$$$\mathrm{Whenever}\:\mathrm{you}\:\mathrm{backspace}\:\mathrm{from}\:\mathrm{a}\:\mathrm{line}\:\mathrm{to}\:\mathrm{delete}\:\mathrm{it},\:\mathrm{you}'\mathrm{ll}\:\mathrm{end}\:\mathrm{up}\:\mathrm{directly}\:\mathrm{in}\:\mathrm{front}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{thing}\:\mathrm{you}\:\mathrm{typed}, \\ $$$$\mathrm{whatever}\:\mathrm{it}\:\mathrm{is}. \\ $$$$ \\ $$$$\mathrm{Also},\:\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{for}\:\mathrm{this}\:\mathrm{app}\:\mathrm{to}\:\mathrm{run}\:\mathrm{in}\:\mathrm{portrait}\:\mathrm{orrentation}\:\mathrm{on}\:\mathrm{tablets}?\:\mathrm{Because}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}\:\mathrm{to}\:\mathrm{rotate} \\ $$$$\mathrm{to}\:\mathrm{that}\:\mathrm{mode}.\:\mathrm{I}\:\mathrm{saw}\:\mathrm{in}\:\mathrm{your}\:\mathrm{screenshots}\:\mathrm{on}\:\mathrm{the}\:\mathrm{Google}\:\mathrm{Play}\:\mathrm{page}\:\mathrm{that}\:\mathrm{it}\:\mathrm{was}\:\mathrm{possible},\:\mathrm{but}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{get}\:\mathrm{it}\:\mathrm{to}\:\mathrm{work}. \\ $$$$\mathrm{I}\:\mathrm{would}\:\mathrm{perfer}\:\mathrm{editing}\:\mathrm{in}\:\mathrm{that}\:\mathrm{orrentation}\:\mathrm{over}\:\mathrm{the}\:\mathrm{landscape}\:\mathrm{rotation}. \\ $$

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