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Question Number 98806 Answers: 0 Comments: 1

for a is integer number such that ∣∣x−1∣ −2∣ ≤ a exactly has 2013 solution

$$\mathrm{for}\:{a}\:\mathrm{is}\:\mathrm{integer}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mid\mid{x}−\mathrm{1}\mid\:−\mathrm{2}\mid\:\leqslant\:{a}\:\:\mathrm{exactly}\:\mathrm{has}\:\mathrm{2013} \\ $$$$\mathrm{solution} \\ $$

Question Number 98788 Answers: 2 Comments: 0

lim_(n→∞) (1/n)[ (n+1)(n+2)...(n+n)_ ^ ]^(1/n)

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\left[\:\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)...\left(\mathrm{n}+\mathrm{n}\right)_{} ^{} \right]^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$

Question Number 98776 Answers: 0 Comments: 0

∫_0 ^∞ (((x−1))/(ln(F(x)(√5)+cos(πx)(ϕ)^(−x) −1)(√(F(x)(√5)+cos(πx)(ϕ)^(−x) −1))))dx F(x)=Fib(x)=xth Extended fibonacci number f:R→R ϕ=((1+(√5))/2)

$$\int_{\mathrm{0}} ^{\infty} \frac{\left({x}−\mathrm{1}\right)}{{ln}\left({F}\left({x}\right)\sqrt{\mathrm{5}}+{cos}\left(\pi{x}\right)\left(\varphi\right)^{−{x}} −\mathrm{1}\right)\sqrt{{F}\left({x}\right)\sqrt{\mathrm{5}}+{cos}\left(\pi{x}\right)\left(\varphi\right)^{−{x}} −\mathrm{1}}}{dx} \\ $$$$ \\ $$$${F}\left({x}\right)={Fib}\left({x}\right)={xth}\:{Extended}\:{fibonacci}\:{number} \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Question Number 98773 Answers: 2 Comments: 0

if x is a selected number of the number from 20−99, then what is probalility x^3 −x is divided by 12?

$$\mathrm{if}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{selected}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{from}\:\mathrm{20}−\mathrm{99},\:\mathrm{then}\:\mathrm{what} \\ $$$$\mathrm{is}\:\mathrm{probalility}\:\mathrm{x}^{\mathrm{3}} −\mathrm{x}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by} \\ $$$$\mathrm{12}?\: \\ $$

Question Number 98770 Answers: 0 Comments: 2

Question Number 98768 Answers: 2 Comments: 0

lim_(x→0) (((√(x+1)) sin x+ln(1+x^2 )−x)/(((1+x^2 ))^(1/(3 )) −1))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}+\mathrm{1}}\:\mathrm{sin}\:\mathrm{x}+\mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{x}}{\sqrt[{\mathrm{3}\:\:}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }−\mathrm{1}} \\ $$

Question Number 98761 Answers: 0 Comments: 5

(√(x+(√x) )) −(√(x−(√x))) = m(√(x/(x+(√x)))) m is a real parameter

$$\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:}\:−\sqrt{\mathrm{x}−\sqrt{\mathrm{x}}}\:=\:\mathrm{m}\sqrt{\frac{\mathrm{x}}{\mathrm{x}+\sqrt{\mathrm{x}}}} \\ $$$$\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{parameter} \\ $$

Question Number 98750 Answers: 0 Comments: 0

Question Number 98744 Answers: 0 Comments: 2

∫_0 ^π ∫_0 ^(2sinθ) (1+rsinθ)r dr dθ

$$\int_{\mathrm{0}} ^{\pi} \int_{\mathrm{0}} ^{\mathrm{2}{sin}\theta} \left(\mathrm{1}+{rsin}\theta\right){r}\:{dr}\:{d}\theta \\ $$

Question Number 98728 Answers: 0 Comments: 5

Currently working on enhancing this app to draw shapes. So posting a math problem realted to drawing.^ Ref. Frame1 X-Y Frame 2: Axis translated by (h,k) and rotated about point (u,v). Consider a point (x_1 ,y_1 ) on X−Y axis. 1. What will be the postion of the point on X−Y axis after it is translated and plotted in frame 2. 2. A point is moved by distance dx,dy in X−Y. How much distane will it moved in the new frame.

$$\mathrm{Currently}\:\mathrm{working}\:\mathrm{on}\:\mathrm{enhancing} \\ $$$$\mathrm{this}\:\mathrm{app}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{shapes}. \\ $$$$\mathrm{So}\:\mathrm{posting}\:\mathrm{a}\:\:\mathrm{math}\:\mathrm{problem}\:\mathrm{realted} \\ $$$$\mathrm{to}\:\mathrm{drawing}\bar {.} \\ $$$$\mathrm{Ref}.\:\mathrm{Frame1}\:\mathrm{X}-\mathrm{Y} \\ $$$$\mathrm{Frame}\:\mathrm{2}: \\ $$$$\mathrm{Axis}\:\mathrm{translated}\:\mathrm{by}\:\left({h},{k}\right)\:\mathrm{and} \\ $$$$\mathrm{rotated}\:\mathrm{about}\:\mathrm{point}\:\left({u},{v}\right). \\ $$$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{point}\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:\mathrm{on}\:\mathrm{X}−\mathrm{Y}\:\mathrm{axis}. \\ $$$$\mathrm{1}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{postion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{X}−\mathrm{Y}\:\mathrm{axis}\:\mathrm{after}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{translated}\:\mathrm{and}\:\mathrm{plotted}\:\mathrm{in}\:\mathrm{frame}\:\mathrm{2}. \\ $$$$\mathrm{2}.\:\mathrm{A}\:\mathrm{point}\:\mathrm{is}\:\mathrm{moved}\:\mathrm{by}\:\mathrm{distance} \\ $$$${dx},{dy}\:\mathrm{in}\:\mathrm{X}−\mathrm{Y}.\:\mathrm{How}\:\mathrm{much}\:\mathrm{distane} \\ $$$$\mathrm{will}\:\mathrm{it}\:\mathrm{moved}\:\mathrm{in}\:\mathrm{the}\:\mathrm{new}\:\mathrm{frame}. \\ $$

Question Number 98723 Answers: 0 Comments: 4

Version 2.084 has fixes for all crashes which were reported on playstore in last week. If anyone is facing crashes please update to v2.084 and see if the problem is solved. If the problem is still present, please send us an email as the problem may be specifuc to device model.

$$\mathrm{Version}\:\mathrm{2}.\mathrm{084}\:\mathrm{has}\:\mathrm{fixes}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{crashes}\:\mathrm{which}\:\mathrm{were}\:\mathrm{reported}\:\mathrm{on} \\ $$$$\mathrm{playstore}\:\mathrm{in}\:\mathrm{last}\:\mathrm{week}. \\ $$$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{is}\:\mathrm{facing}\:\mathrm{crashes}\:\mathrm{please} \\ $$$$\mathrm{update}\:\mathrm{to}\:\mathrm{v2}.\mathrm{084}\:\mathrm{and}\:\mathrm{see}\:\mathrm{if}\:\mathrm{the}\: \\ $$$$\mathrm{problem}\:\mathrm{is}\:\mathrm{solved}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{is}\:\mathrm{still}\:\mathrm{present},\:\mathrm{please} \\ $$$$\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email}\:\mathrm{as}\:\mathrm{the}\:\mathrm{problem} \\ $$$$\mathrm{may}\:\mathrm{be}\:\mathrm{specifuc}\:\mathrm{to}\:\mathrm{device}\:\mathrm{model}. \\ $$

Question Number 98722 Answers: 3 Comments: 0

let f(x) =arctan((3/x)) 1) calculste f^((n)) (x) and f^((n)) (1) 2) developp f at integr seri at point x_0 =1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculste}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{seri}\:\mathrm{at}\:\mathrm{point}\:\mathrm{x}_{\mathrm{0}} =\mathrm{1} \\ $$

Question Number 98721 Answers: 2 Comments: 0

calculate ∫_0 ^∞ (dx/(x^4 +x^2 +1)) 1) by using residue theorem 2) by using complex decomposition

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{residue}\:\mathrm{theorem} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{by}\:\mathrm{using}\:\mathrm{complex}\:\mathrm{decomposition} \\ $$

Question Number 98713 Answers: 2 Comments: 1

∫((sin(x))/x)dx

$$\int\frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$ \\ $$

Question Number 98690 Answers: 1 Comments: 3

Question Number 98687 Answers: 1 Comments: 4

Question Number 98679 Answers: 1 Comments: 2

prove that ∫_0 ^∞ ((3+2(√x))/(x^2 +2x+5))dx=4.13049

$${prove}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}+\mathrm{2}\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{dx}=\mathrm{4}.\mathrm{13049}\: \\ $$

Question Number 98678 Answers: 2 Comments: 9

∫_0 ^∞ e^(−ax) ((sin mx)/x) dx = tan^(−1) ((m/a)), a>0

$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ax}}} \frac{\mathrm{sin}\:\boldsymbol{\mathrm{mx}}}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{a}}}\right),\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98677 Answers: 1 Comments: 0

prove ∫_0 ^a ((ln(1+ax))/(1+x^2 ))dx=(1/2)ln(1+a^2 )tan^(−1) a, a>0

$$\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{ax}}\right)}{\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\mathrm{tan}^{−\mathrm{1}} \boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{a}}>\mathrm{0} \\ $$

Question Number 98675 Answers: 0 Comments: 10

Question Number 98673 Answers: 2 Comments: 0

find a_n in terms of n (I can′t find it...) a_1 =1; a_2 =4 a_3 =a_2 ×4×((2^2 −1)/2^2 ) a_4 =a_3 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 ) a_5 =a_4 ×4×((2^2 −1)/2^2 )×((3^2 −1)/3^2 )×((4^2 −1)/4^2 ) ... n≥2: a_(n+1) =4a_n Π_(k=2) ^n ((k^2 −1)/k^2 )

$$\mathrm{find}\:{a}_{{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n} \\ $$$$\left(\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}...\right) \\ $$$${a}_{\mathrm{1}} =\mathrm{1};\:{a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{3}} ={a}_{\mathrm{2}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{4}} ={a}_{\mathrm{3}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} ×\mathrm{4}×\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\frac{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }×\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$... \\ $$$${n}\geqslant\mathrm{2}:\:{a}_{{n}+\mathrm{1}} =\mathrm{4}{a}_{{n}} \underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 98672 Answers: 1 Comments: 0

∫_0 ^4 ∫_0 ^(x/4) e^x^2 dx dy

$$\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\mathrm{0}} ^{\frac{{x}}{\mathrm{4}}} {e}^{{x}^{\mathrm{2}} } \:{dx}\:{dy} \\ $$

Question Number 98661 Answers: 0 Comments: 1

using cayley − hamilton theorem what is the inverse of matrix A= [((0 1 −1)),((1 2 2)),((0 1 −1)) ]

$$\mathrm{using}\:\mathrm{cayley}\:−\:\mathrm{hamilton} \\ $$$$\mathrm{theorem}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\mathrm{matrix}\:\mathrm{A}=\:\begin{bmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{bmatrix}\: \\ $$

Question Number 98796 Answers: 2 Comments: 0

Question Number 98657 Answers: 2 Comments: 0

solve y^(′′) −3y^′ +2y =((sinx)/x)

$$\mathrm{solve}\:\:\mathrm{y}^{''} \:−\mathrm{3y}^{'} \:\:+\mathrm{2y}\:=\frac{\mathrm{sinx}}{\mathrm{x}} \\ $$

Question Number 98656 Answers: 2 Comments: 0

solve xy^(′′) +(2+x^2 )y^′ =xe^(−x^2 )

$$\mathrm{solve}\:\:\:\mathrm{xy}^{''} \:+\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}^{'} \:\:=\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \\ $$

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