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Question Number 41722 Answers: 0 Comments: 0

sir Tinkutara is the anyway i could get to talk to you guys(Equation Editor group) i′ve got some ideas i wanna share with you.just comment the messenger media below and the name i′ll get you soon= thanks

$${sir}\:{Tinkutara}\:{is}\:{the}\:{anyway}\:{i}\:{could}\:{get}\:{to}\:{talk}\:{to}\:{you}\:{guys}\left({Equation}\:{Editor}\:{group}\right) \\ $$$${i}'{ve}\:{got}\:{some}\:{ideas}\:{i}\:{wanna}\:{share}\:{with}\:{you}.{just}\:{comment} \\ $$$${the}\:{messenger}\:{media}\:{below}\:{and}\:{the}\:{name}\:{i}'{ll}\:{get}\:{you}\:{soon}= \\ $$$$\:\:{thanks} \\ $$

Question Number 41721 Answers: 0 Comments: 1

show that U_n = 1+ nx + ((n(n−1))/(2!)) x^(2 ) + ((n(n−1)(n−2))/(3!))x^3 + .... for which U_n = (1 + x)^n .

$${show}\:{that} \\ $$$${U}_{{n}} =\:\mathrm{1}+\:{nx}\:+\:\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}\:{x}^{\mathrm{2}\:\:} +\:\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +\:.... \\ $$$${for}\:{which}\:{U}_{{n}} =\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} . \\ $$

Question Number 41720 Answers: 0 Comments: 2

Difference of last two digits of 2019^(2019^(2019) ) is ?

$$\mathrm{Difference}\:\mathrm{of}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{of}\:\:\:\:\mathrm{2019}^{\mathrm{2019}^{\mathrm{2019}} } \:\:\:\mathrm{is}\:\:? \\ $$

Question Number 41710 Answers: 2 Comments: 0

Factorize: 60x^3 y + 115x^2 y^2 − 120xy^3

$$\mathrm{Factorize}:\:\:\mathrm{60x}^{\mathrm{3}} \mathrm{y}\:+\:\mathrm{115x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{120xy}^{\mathrm{3}} \\ $$

Question Number 41707 Answers: 2 Comments: 0

Here′s a question that has troubled me for months now.Please help 5^(√x) − 5^(x−7) =100 find the possible value(s) of x.

$${Here}'{s}\:{a}\:{question}\:{that}\:{has} \\ $$$${troubled}\:{me}\:{for}\:{months}\:{now}.{Please} \\ $$$${help} \\ $$$$ \\ $$$$\mathrm{5}^{\sqrt{{x}}} \:−\:\mathrm{5}^{{x}−\mathrm{7}} \:=\mathrm{100} \\ $$$$ \\ $$$${find}\:{the}\:{possible}\:{value}\left({s}\right)\:{of}\:{x}. \\ $$

Question Number 41706 Answers: 1 Comments: 1

study the convergence of u_n =Σ_(k=2) ^n (1/(kln(k))) −ln(ln(n)

$${study}\:{the}\:{convergence}\:{of}\: \\ $$$${u}_{{n}} =\sum_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{\mathrm{1}}{{kln}\left({k}\right)}\:−{ln}\left({ln}\left({n}\right)\right. \\ $$

Question Number 41705 Answers: 0 Comments: 1

find nsture of the serie Σ u_n with u_n =^n (√(n/(n+1))) −1

$${find}\:{nsture}\:{of}\:{the}\:{serie}\:\Sigma\:{u}_{{n}} \\ $$$${with}\:{u}_{{n}} =^{{n}} \sqrt{\frac{{n}}{{n}+\mathrm{1}}}\:\:\:−\mathrm{1} \\ $$

Question Number 41704 Answers: 0 Comments: 0

let u_n = (1/(√(n−1))) +(2/(√n)) +(1/(√(n+1))) calculate Σ_(n=2) ^∞ u_n

$${let}\:{u}_{{n}} =\:\frac{\mathrm{1}}{\sqrt{{n}−\mathrm{1}}}\:+\frac{\mathrm{2}}{\sqrt{{n}}}\:+\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}} \\ $$$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:{u}_{{n}} \\ $$

Question Number 41703 Answers: 0 Comments: 1

calculate I = ∫_0 ^(π/4) ((cosx)/(cos^3 x +sin^3 x))dx

$${calculate}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{cosx}}{{cos}^{\mathrm{3}} {x}\:+{sin}^{\mathrm{3}} {x}}{dx} \\ $$

Question Number 41702 Answers: 1 Comments: 3

find the value of ∫_0 ^(√3) arcsin(((2t)/(1+t^2 )))dt

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:{arcsin}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$

Question Number 41699 Answers: 0 Comments: 0

find the perpendicular distance btn two lines y=2x−3 and y=3x−1

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{perpendicular}}\:\boldsymbol{\mathrm{distance}}\:\boldsymbol{\mathrm{btn}} \\ $$$$\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{lines}}\:\boldsymbol{{y}}=\mathrm{2}\boldsymbol{{x}}−\mathrm{3} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{x}}−\mathrm{1} \\ $$

Question Number 44610 Answers: 0 Comments: 1

Let A be 2×3 matrix , whereas B be 3×2 matrix. If determinant(AB)=4, then the value of determinant (BA) ?

$${Let}\:{A}\:{be}\:\mathrm{2}×\mathrm{3}\:{matrix}\:,\:{whereas}\:{B}\:{be} \\ $$$$\mathrm{3}×\mathrm{2}\:{matrix}.\:{If}\:{determinant}\left({AB}\right)=\mathrm{4}, \\ $$$${then}\:{the}\:{value}\:{of}\:{determinant}\:\left({BA}\right)\:? \\ $$

Question Number 41691 Answers: 2 Comments: 0

tan^2 20+tan^2 40+tan^2 80=33

$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33} \\ $$

Question Number 41686 Answers: 1 Comments: 2

Question Number 41682 Answers: 2 Comments: 0

find radius of curvature to y=sin x at x=π/6 .

$${find}\:{radius}\:{of}\:{curvature}\:{to} \\ $$$${y}=\mathrm{sin}\:{x}\:\:{at}\:\:{x}=\pi/\mathrm{6}\:. \\ $$

Question Number 41679 Answers: 1 Comments: 5

let f(x) = ∫_0 ^1 ln(1+t +xt^2 )dt 1) calculate f^′ (x) then find a simple form of f(x) 2) calculate ∫_0 ^1 ln(1+t +t^2 )dt 3) calculate ∫_0 ^1 ln(1−t^3 )dt .

$${let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\:+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}\:+{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{ln}\left(\mathrm{1}−{t}^{\mathrm{3}} \right){dt}\:. \\ $$

Question Number 41678 Answers: 1 Comments: 0

prove that ∫_0 ^∞ cos(x^2 )dx=∫_0 ^∞ sin(x^2 )dx by using only series.

$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\:{by}\:{using} \\ $$$${only}\:{series}. \\ $$

Question Number 41677 Answers: 2 Comments: 2

calculate A = ∫_0 ^(π/4) cos^8 xdx and B= ∫_0 ^(π/4) sin^8 xdx 2) calculate A +B and A−B 3) calculate A^2 −B^2

$${calculate}\:{A}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{8}} {xdx}\:{and}\: \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sin}^{\mathrm{8}} {xdx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}\:+{B}\:{and}\:{A}−{B} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:{A}^{\mathrm{2}} \:−{B}^{\mathrm{2}} \\ $$

Question Number 41675 Answers: 1 Comments: 1

Question Number 41672 Answers: 0 Comments: 1

find Σ_(k=0) ^n (1/(3k+1)) interms of H_n

$${find}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{3}{k}+\mathrm{1}}\:{interms}\:{of}\:{H}_{{n}} \\ $$

Question Number 41671 Answers: 0 Comments: 0

if p=6.4×10^4 and q=3.2×10^5 find the values of (i)p×q (ii)p+q write the answers in standard form

Question Number 41651 Answers: 2 Comments: 1

∫( 1+2x+3x^2 +4x^3 +.........) dx , (0<∣x∣<1)

$$\int\left(\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{3}} +.........\right)\:{dx}\:,\:\:\: \\ $$$$\left(\mathrm{0}<\mid{x}\mid<\mathrm{1}\right) \\ $$

Question Number 41642 Answers: 1 Comments: 0

n(n − 1)(n − 2)(n − 3) .... (n − r + 1) = ??

$$\mathrm{n}\left(\mathrm{n}\:−\:\mathrm{1}\right)\left(\mathrm{n}\:−\:\mathrm{2}\right)\left(\mathrm{n}\:−\:\mathrm{3}\right)\:....\:\left(\mathrm{n}\:−\:\mathrm{r}\:+\:\mathrm{1}\right)\:=\:?? \\ $$

Question Number 41634 Answers: 2 Comments: 1

Question Number 41622 Answers: 4 Comments: 9

let z_1 and z_2 the roots of x^2 −2x+2=0 1) calculate z_1 ^3 +z_2 ^3 then (1/z_1 ^3 ) +(1/z_2 ^3 ) 2) calculate z_1 ^4 +z_2 ^4 then (1/z_1 ^4 ) +(1/z_2 ^4 ) 3) let n from N simplify A_n = z_1 ^n +z_2 ^n and B_n = z_1 ^n −z_2 ^n 4) simplify S_n =Σ_(k=0) ^(n−1) (z_1 ^k +z_2 ^k )

$${let}\:{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{z}_{\mathrm{1}} ^{\mathrm{3}} \:+{z}_{\mathrm{2}} ^{\mathrm{3}} \:\:\:{then}\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{\mathrm{3}} }\:+\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{z}_{\mathrm{1}} ^{\mathrm{4}} \:+{z}_{\mathrm{2}} ^{\mathrm{4}} \:\:{then}\:\:\frac{\mathrm{1}}{{z}_{\mathrm{1}} ^{\mathrm{4}} }\:+\frac{\mathrm{1}}{{z}_{\mathrm{2}} ^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right)\:{let}\:{n}\:{from}\:{N}\:\:{simplify} \\ $$$${A}_{{n}} =\:{z}_{\mathrm{1}} ^{{n}} \:+{z}_{\mathrm{2}} ^{{n}} \:\:\:\:\:\:\:{and}\:\:{B}_{{n}} =\:{z}_{\mathrm{1}} ^{{n}} \:−{z}_{\mathrm{2}} ^{{n}} \\ $$$$\left.\mathrm{4}\right)\:{simplify}\:\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({z}_{\mathrm{1}} ^{{k}} \:\:\:+{z}_{\mathrm{2}} ^{{k}} \right) \\ $$

Question Number 41620 Answers: 2 Comments: 1

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