Question and Answers Forum

All Questions   Topic List

AllQuestion and Answers: Page 756

Question Number 140831    Answers: 1   Comments: 0

......nice .... calculus...... prove that:: ξ := ∫_(−∞) ^( ∞) ((cos (πx^2 ))/(cosh(πx)))dx=(1/( (√2))) .... .......

$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:......{nice}\:....\:{calculus}...... \\ $$$$\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\xi\::=\:\int_{−\infty} ^{\:\infty} \frac{{cos}\:\left(\pi{x}^{\mathrm{2}} \right)}{{cosh}\left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.... \\ $$$$\:\:\:\:....... \\ $$

Question Number 140829    Answers: 1   Comments: 0

factorise 5x^(2m) 8x^m +8

$${factorise}\:\mathrm{5}{x}^{\mathrm{2}{m}} \mathrm{8}{x}^{{m}} +\mathrm{8} \\ $$

Question Number 140828    Answers: 2   Comments: 0

factorise 3k^2 +2kh−8h^2

$$\mathrm{factorise}\:\mathrm{3k}^{\mathrm{2}} \:+\mathrm{2kh}−\mathrm{8h}^{\mathrm{2}} \\ $$

Question Number 140827    Answers: 1   Comments: 0

Find 1. ∫(dx/([x]^2 )) , x≥1 2. ∫(([x]^λ )/x^(λ+1) ) , x≥1 Where [∗] denote the integer part

$${Find} \\ $$$$\mathrm{1}.\:\int\frac{{dx}}{\left[{x}\right]^{\mathrm{2}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$$\mathrm{2}.\:\int\frac{\left[{x}\right]^{\lambda} }{{x}^{\lambda+\mathrm{1}} }\:,\:{x}\geqslant\mathrm{1} \\ $$$${Where}\:\left[\ast\right]\:{denote}\:{the}\:{integer}\:{part} \\ $$

Question Number 140826    Answers: 0   Comments: 0

Question Number 140825    Answers: 1   Comments: 0

Use the limit comparison test to determine if the series converges or diverges Σ_(n=2) ^∞ (1/(7+8n ln (ln n))).

$$\mathrm{Use}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{comparison}\:\mathrm{test} \\ $$$$\mathrm{to}\:\mathrm{determine}\:\mathrm{if}\:\mathrm{the}\:\mathrm{series}\:\mathrm{converges} \\ $$$$\mathrm{or}\:\mathrm{diverges}\: \\ $$$$\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{7}+\mathrm{8n}\:\mathrm{ln}\:\left(\mathrm{ln}\:\mathrm{n}\right)}.\: \\ $$

Question Number 140823    Answers: 1   Comments: 1

Determine whether the improper integral converges or diverges ∫_1 ^( ∞) ((2x+7)/(7x^3 +5x^2 +1)) dx

$$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{improper} \\ $$$$\mathrm{integral}\:\mathrm{converges}\:\mathrm{or}\:\mathrm{diverges}\: \\ $$$$\int_{\mathrm{1}} ^{\:\infty} \:\frac{\mathrm{2x}+\mathrm{7}}{\mathrm{7x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{dx}\: \\ $$

Question Number 140821    Answers: 0   Comments: 1

O is centre of circle and square. find yellow area in terms of radius of circle

$${O}\:{is}\:{centre}\:{of}\:{circle}\:{and}\:{square}. \\ $$$${find}\:{yellow}\:{area}\:{in}\:{terms}\:{of}\:{radius}\: \\ $$$${of}\:{circle} \\ $$

Question Number 140816    Answers: 0   Comments: 0

Question Number 140815    Answers: 0   Comments: 0

Question Number 140813    Answers: 0   Comments: 0

Question Number 142208    Answers: 1   Comments: 0

Find the equation of the circle which is orthogonal to the circles x^2 +y^2 −7x−y=0 and x^2 +y^2 +3x−6y+5=0 and which passes through the point (−3,0).

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{7}{x}−{y}=\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}{y}+\mathrm{5}=\mathrm{0}\:\mathrm{and}\:\mathrm{which}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{3},\mathrm{0}\right). \\ $$

Question Number 142207    Answers: 2   Comments: 0

Question Number 142205    Answers: 0   Comments: 5

Question Number 142204    Answers: 1   Comments: 0

Question Number 140810    Answers: 0   Comments: 0

Find the natural value of x that satisfies the equation: (1/(6∙7)) + (1/(7∙8)) + (1/(8∙9)) +...+ (1/(x∙(x+1))) = (1/(12))

$${Find}\:{the}\:{natural}\:{value}\:{of}\:\boldsymbol{{x}}\:{that} \\ $$$${satisfies}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{7}}\:+\:\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{8}\centerdot\mathrm{9}}\:+...+\:\frac{\mathrm{1}}{{x}\centerdot\left({x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Question Number 140809    Answers: 1   Comments: 0

z^5 + (1/z^5 ) = ((205)/(16)) ∙ (z + (1/z))

$$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:=\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right) \\ $$

Question Number 140805    Answers: 1   Comments: 0

find x 2cosh 2x+10sinh 2x=5

$${find}\:{x}\:\mathrm{2cosh}\:\mathrm{2}{x}+\mathrm{10sinh}\:\mathrm{2}{x}=\mathrm{5} \\ $$

Question Number 140803    Answers: 1   Comments: 0

Question Number 140798    Answers: 0   Comments: 1

Σ_(n=1) ^∞ (1/(n^3 (((4n)),(n) )))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{4}{n}}\\{{n}}\end{pmatrix}} \\ $$

Question Number 140793    Answers: 0   Comments: 0

∫(√((x+2)/e^x ))dx=...?

$$\int\sqrt{\frac{{x}+\mathrm{2}}{{e}^{{x}} }}{dx}=...? \\ $$

Question Number 140786    Answers: 0   Comments: 0

Question Number 140785    Answers: 0   Comments: 2

P-intersection point of bimedians in ABCD-convexe quadrilateral with a;b;c;d-sides, e;f-diagonals, E-point in plane, x;y;z;t-distances from, E-to A;B;C;D. Prove that... (1/4)(a^2 +b^2 +c^2 +e^2 +f^2 )+4PE^2 =x^2 +y^2 +z^2 +t^2

$${P}-{intersection}\:{point}\:{of}\:{bimedians}\:{in} \\ $$$${ABCD}-{convexe}\:{quadrilateral}\:{with} \\ $$$${a};{b};{c};{d}-{sides},\:{e};{f}-{diagonals}, \\ $$$${E}-{point}\:{in}\:{plane},\:{x};{y};{z};{t}-{distances} \\ $$$${from},\:{E}-{to}\:{A};{B};{C};{D}.\:{Prove}\:{that}... \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)+\mathrm{4}{PE}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{t}^{\mathrm{2}} \\ $$

Question Number 140784    Answers: 0   Comments: 0

Let the i-j plane be the complex plane, with basic operations ij=−i ji=−j i^2 =−1 j^2 =−1 z=r+xi+yj w=s+pi+qj zw=rs+pri+qrj+sxi−px−qxi +syj−pyj−qy = (rs−px−qy)+(pr+sx−qx)i +(qr+sy−py)j wz=rs+sxi+syj+pri−px−pyi +qrj−qxj−qy = (rs−px−qy)+(pr+sx−py)i +(qr+sy−qx)j (little difference..) ⇒ zw−wz= (py−qx)(i−j) = 0 if py−qx= 0 z^2 = (r^2 −x^2 −y^2 )+(2rx−xy)i +(2ry−xy)j And if y=0 ⇒ z=r+xi z^2 =(r^2 −x^2 )+2rxi either way! (z^2 )z=r(r^2 −x^2 )+2r^2 xi +x(r^2 −x^2 )i−2rx^2 ⇒ (z^2 )z= r(r^2 −3x^2 )+x(3r^2 −x^2 )i z(z^2 )= r(r^2 −x^2 )+x(r^2 −x^2 )i +2r^2 xi−2rx^2 = r(r^2 −3x^2 )+x(3r^2 −x^2 )i so z(z^2 )=(z^2 )z = z^3 (so far so good) .....

$${Let}\:{the}\:{i}-{j}\:{plane}\:{be}\:{the}\:{complex} \\ $$$$\:{plane},\:{with}\:{basic}\:{operations} \\ $$$$\:\:{ij}=−{i} \\ $$$$\:\:{ji}=−{j} \\ $$$$\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{j}^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}={r}+{xi}+{yj}\:\:\:\:{w}={s}+{pi}+{qj} \\ $$$${zw}={rs}+{pri}+{qrj}+{sxi}−{px}−{qxi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{syj}−{pyj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{qx}\right){i} \\ $$$$\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{py}\right){j} \\ $$$${wz}={rs}+{sxi}+{syj}+{pri}−{px}−{pyi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{qrj}−{qxj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{py}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{qx}\right){j} \\ $$$$\left({little}\:{difference}..\right) \\ $$$$\Rightarrow\:\:{zw}−{wz}=\:\left({py}−{qx}\right)\left({i}−{j}\right) \\ $$$$\:\:\:=\:\mathrm{0}\:\:{if}\:\:\:{py}−{qx}=\:\mathrm{0} \\ $$$${z}^{\mathrm{2}} =\:\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+\left(\mathrm{2}{rx}−{xy}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\mathrm{2}{ry}−{xy}\right){j} \\ $$$${And}\:{if}\:\:{y}=\mathrm{0}\:\:\Rightarrow\:\:{z}={r}+{xi} \\ $$$${z}^{\mathrm{2}} =\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{rxi} \\ $$$$\:\:{either}\:{way}! \\ $$$$\left({z}^{\mathrm{2}} \right){z}={r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{r}^{\mathrm{2}} {xi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:\left({z}^{\mathrm{2}} \right){z}=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$${z}\left({z}^{\mathrm{2}} \right)=\:{r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{r}^{\mathrm{2}} {xi}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:{so}\:\:\:{z}\left({z}^{\mathrm{2}} \right)=\left({z}^{\mathrm{2}} \right){z}\:=\:{z}^{\mathrm{3}} \:\:\left({so}\:{far}\:{so}\:{good}\right) \\ $$$$..... \\ $$$$ \\ $$

Question Number 142200    Answers: 2   Comments: 0

Let p & q real positive number what the minimum of ((p^2 /q^2 ) +(q/p))^3 .

$$\:{Let}\:{p}\:\&\:{q}\:{real}\:{positive}\:{number} \\ $$$$\:\:{what}\:{the}\:{minimum}\:{of}\:\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:+\frac{{q}}{{p}}\right)^{\mathrm{3}} . \\ $$

Question Number 140779    Answers: 1   Comments: 1

{ ((7x≡3(mod5))),((5x≡3(mod9))) :} solve for x

$$\begin{cases}{\mathrm{7}{x}\equiv\mathrm{3}\left({mod}\mathrm{5}\right)}\\{\mathrm{5}{x}\equiv\mathrm{3}\left({mod}\mathrm{9}\right)}\end{cases} \\ $$$${solve}\:{for}\:{x} \\ $$

  Pg 751      Pg 752      Pg 753      Pg 754      Pg 755      Pg 756      Pg 757      Pg 758      Pg 759      Pg 760   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com