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Question Number 118384    Answers: 1   Comments: 0

Question Number 118376    Answers: 2   Comments: 0

If a curve y = f(x) passing through the point (1,2) is the solution of differential equation 2x^2 dy = (2xy+y^2 )dx , then the value of f(2) is equal to?

$${If}\:{a}\:{curve}\:{y}\:=\:{f}\left({x}\right)\:{passing}\:{through} \\ $$$${the}\:{point}\:\left(\mathrm{1},\mathrm{2}\right)\:{is}\:{the}\:{solution} \\ $$$${of}\:{differential}\:{equation} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:{dy}\:=\:\left(\mathrm{2}{xy}+{y}^{\mathrm{2}} \right){dx}\:,\:{then}\:{the}\: \\ $$$${value}\:{of}\:{f}\left(\mathrm{2}\right)\:{is}\:{equal}\:{to}? \\ $$

Question Number 207620    Answers: 1   Comments: 0

prove that ∫_(−a) ^a (dx/(x^n +1+(√(x^(2n) +1))))=a

$$\mathrm{prove}\:\mathrm{that}\:\underset{−{a}} {\overset{{a}} {\int}}\:\frac{{dx}}{{x}^{{n}} +\mathrm{1}+\sqrt{{x}^{\mathrm{2}{n}} +\mathrm{1}}}={a} \\ $$

Question Number 118371    Answers: 0   Comments: 1

old problem question 118120 tan tan x =tan 3x −tan 2x let t=tan x (1) tan t =((t^5 +2t^3 +t)/(3t^4 −4t^2 +1)) for t≥0 we get (approximating) t_0 =0 t_1 ≈1.28941477 t_2 ≈4.17629616 t_3 ≈7.49316173 t_4 ≈10.7303610 t_5 ≈13.9285293 ... x=nπ+arctan t let n=0 to stay in the first period 0≤t<+∞ ⇒ 0≤arctan t <(π/2) ⇒ (1) has infinite solutions for 0≤x<(π/2) graphically this is easy to see, plot these: f_1 (t)=tan t f_2 (t)=((t(t^4 +2t^2 +1))/(3t^4 −4t^2 +1))=(t/3)+((2t(5t^2 +1))/(3(3t^4 −4t^2 +1))) ⇒ g(t)=(1/3)t is asymptote of f_1 (t) and obviously tan t =at with a∈R has infinite solutions

$$\mathrm{old}\:\mathrm{problem}\:{question}\:\mathrm{118120} \\ $$$$\mathrm{tan}\:\mathrm{tan}\:{x}\:=\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{tan}\:{t}\:=\frac{{t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{for}\:{t}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{approximating}\right) \\ $$$${t}_{\mathrm{0}} =\mathrm{0} \\ $$$${t}_{\mathrm{1}} \approx\mathrm{1}.\mathrm{28941477} \\ $$$${t}_{\mathrm{2}} \approx\mathrm{4}.\mathrm{17629616} \\ $$$${t}_{\mathrm{3}} \approx\mathrm{7}.\mathrm{49316173} \\ $$$${t}_{\mathrm{4}} \approx\mathrm{10}.\mathrm{7303610} \\ $$$${t}_{\mathrm{5}} \approx\mathrm{13}.\mathrm{9285293} \\ $$$$... \\ $$$${x}={n}\pi+\mathrm{arctan}\:{t} \\ $$$$\mathrm{let}\:{n}=\mathrm{0}\:\mathrm{to}\:\mathrm{stay}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{period} \\ $$$$\mathrm{0}\leqslant{t}<+\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{arctan}\:{t}\:<\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:\mathrm{has}\:\mathrm{infinite}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{graphically}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see},\:\mathrm{plot}\:\mathrm{these}: \\ $$$${f}_{\mathrm{1}} \left({t}\right)=\mathrm{tan}\:{t} \\ $$$${f}_{\mathrm{2}} \left({t}\right)=\frac{{t}\left({t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{{t}}{\mathrm{3}}+\frac{\mathrm{2}{t}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{3}\left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\Rightarrow\:{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{3}}{t}\:\mathrm{is}\:\mathrm{asymptote}\:\mathrm{of}\:{f}_{\mathrm{1}} \left({t}\right) \\ $$$$\mathrm{and}\:\mathrm{obviously}\:\mathrm{tan}\:{t}\:={at}\:\mathrm{with}\:{a}\in\mathbb{R}\:\mathrm{has} \\ $$$$\mathrm{infinite}\:\mathrm{solutions} \\ $$

Question Number 118369    Answers: 0   Comments: 1

657×10^4 =(5/7)+

$$\:\mathrm{657}×\mathrm{10}^{\mathrm{4}} =\frac{\mathrm{5}}{\mathrm{7}}+ \\ $$

Question Number 118366    Answers: 0   Comments: 4

4. Turunan fungsi f(x)=^5 (√((10x^2 −4)^8 )) adalah f^1 (x). Nilai f^1 (1)=... a. 1 b. 8 c. 14 d. 16 e. 20

$$\mathrm{4}.\:\mathrm{Turunan}\:\mathrm{fungsi}\:\mathrm{f}\left(\mathrm{x}\right)=^{\mathrm{5}} \sqrt{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{8}} \:\:} \\ $$$$\mathrm{adalah}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right).\:\mathrm{Nilai}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{1}\right)=... \\ $$$${a}.\:\mathrm{1} \\ $$$${b}.\:\mathrm{8} \\ $$$${c}.\:\mathrm{14} \\ $$$$\mathrm{d}.\:\mathrm{16} \\ $$$$\mathrm{e}.\:\mathrm{20} \\ $$

Question Number 118364    Answers: 1   Comments: 0

...ordinary differential equation... y′′ −xy′+y=0 general solution ::=?? .m.n.1970.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{ordinary}\:{differential}\:{equation}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:{y}''\:−{xy}'+{y}=\mathrm{0} \\ $$$$\:\:\:\:{general}\:{solution}\:::=?? \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$ \\ $$

Question Number 118357    Answers: 2   Comments: 0

Find the sum of the prime factors of 20^5 +21.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{prime}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\mathrm{20}^{\mathrm{5}} +\mathrm{21}. \\ $$

Question Number 118346    Answers: 0   Comments: 0

Let 30 furniture sets arrive at two city stations A and B ,15 sets for each station All funiture sets need to be delivered to two furniture stores C and D,and 10 sets must be delivered to store C,and to store D−20.It is known that delivery one furniture set from the station A to the stores C and D costs 1 and 3 monetary units,and from station B−2 and 5 units respectively,.It is necessary to draw out such a transportation plan that the cost of transportation is the lowest

$$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\ $$$$\mathrm{units},\mathrm{and}\:\mathrm{from}\:\mathrm{station}\:\mathrm{B}−\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{units} \\ $$$$\:\mathrm{respectively},.\mathrm{It}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\mathrm{out}\:\mathrm{such}\:\mathrm{a}\:\mathrm{transportation}\:\mathrm{plan}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{cost}\:\mathrm{of}\:\mathrm{transportation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lowest} \\ $$

Question Number 118349    Answers: 1   Comments: 0

solution xdy + (3y − e^x )dx with integration factor

$${solution}\:{xdy}\:+\:\left(\mathrm{3}{y}\:−\:{e}^{{x}} \right){dx}\: \\ $$$${with}\:{integration}\:{factor} \\ $$

Question Number 118347    Answers: 1   Comments: 0

∫ (dx/( (√x) +(x)^(1/(3 )) ))

$$\:\:\:\int\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt[{\mathrm{3}\:}]{{x}}}\: \\ $$

Question Number 118362    Answers: 1   Comments: 0

Prove that 𝛑 is an irrational number

$$\mathrm{Prove}\:\mathrm{that}\:\boldsymbol{\pi}\:\mathrm{is}\:\mathrm{an}\:\mathrm{irrational}\:\mathrm{number} \\ $$

Question Number 118340    Answers: 1   Comments: 1

∫_(π/3) ^(π/2) (dx/(1+sin x−cos x))

$$\:\:\:\underset{\pi/\mathrm{3}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$

Question Number 118339    Answers: 0   Comments: 0

Question Number 118338    Answers: 3   Comments: 0

solve ∫ (dx/(3−5sin x))

$$\:\:\:\:{solve}\:\int\:\frac{{dx}}{\mathrm{3}−\mathrm{5sin}\:{x}}\: \\ $$

Question Number 118327    Answers: 4   Comments: 0

lim_(x→1) ((x−(√(2−x^2 )))/(2x−(√(2+2x^2 )))) ?

$$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}−\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\sqrt{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }}\:? \\ $$

Question Number 118325    Answers: 0   Comments: 0

Question Number 118323    Answers: 0   Comments: 0

Prove that ζ(1−s)=2^(1−s) π^(−s) cos(((sπ)/2))Γ(s)ζ(s)

$${Prove}\:{that}\: \\ $$$$\zeta\left(\mathrm{1}−{s}\right)=\mathrm{2}^{\mathrm{1}−{s}} \pi^{−{s}} {cos}\left(\frac{{s}\pi}{\mathrm{2}}\right)\Gamma\left({s}\right)\zeta\left({s}\right) \\ $$

Question Number 118322    Answers: 2   Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

Question Number 118318    Answers: 1   Comments: 0

Question Number 118308    Answers: 1   Comments: 1

Question Number 118307    Answers: 2   Comments: 1

∫^∞ _0 ((x^2 −2)/(x^4 +x^2 +1)) dx

$$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$

Question Number 118304    Answers: 0   Comments: 0

... ⧫_▼ ^▼ Advanced calculus⧫_▼ ^▼ ... prove that :: Σ_(n=1) ^∞ (1/(3^n n^3 )) =^(???) (7/8)ζ( 3 )+(1/6)ln^3 (2)−(π^2 /(12)) ln(2) ✓ ...♠M.N.July 1970♠... .Good luck.

$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\underset{\blacktrinagledown} {\overset{\blacktrinagledown} {\blacklozenge}}\mathscr{A}{dvanced}\:\:{calculus}\underset{\blacktrinagledown} {\overset{\blacktrinagledown} {\blacklozenge}}\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{3}^{{n}} {n}^{\mathrm{3}} }\:\overset{???} {=}\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\:\mathrm{3}\:\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:{ln}\left(\mathrm{2}\right)\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\spadesuit\mathscr{M}.\mathscr{N}.\mathscr{J}{uly}\:\mathrm{1970}\spadesuit... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.\mathscr{G}{ood}\:{luck}. \\ $$$$ \\ $$

Question Number 118298    Answers: 0   Comments: 1

b, x, y, c are consecutive terms of a G.P. ∴ x = br, y = br^2 , c = br^3 A.M. between b and c is a ∴ ((b+c)/2) = a ⇒ 2a = b+c 2abc = (b+c)bc = b^2 c + bc^2 = b^2 (br^3 ) + b(br^3 )^2 = b^3 r^3 + b^3 r^6 = (br)^3 + (br^2 )^3 = x^3 + y^3 ∴ x^3 + y^3 = 2abc ←

$$\:\:\:\:\:\mathrm{b},\:\mathrm{x},\:\mathrm{y},\:\mathrm{c}\:\mathrm{are}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{a}\:\mathrm{G}.\mathrm{P}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\boldsymbol{\mathrm{x}}\:=\:\boldsymbol{\mathrm{br}},\:\boldsymbol{\mathrm{y}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{2}} ,\:\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{br}}^{\mathrm{3}} \: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\mathrm{A}.\mathrm{M}.\:\mathrm{between}\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{is}\:\mathrm{a} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{2}}\:\:=\:\mathrm{a}\:\:\:\:\Rightarrow\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\:=\:\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{a}}\mathrm{bc}\:=\:\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\mathrm{bc} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{2}} \boldsymbol{\mathrm{c}}\:+\:\mathrm{b}\boldsymbol{\mathrm{c}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{2}} \left(\boldsymbol{\mathrm{br}}^{\mathrm{3}} \right)\:+\:\mathrm{b}\left(\boldsymbol{\mathrm{br}}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{b}^{\mathrm{3}} \mathrm{r}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \mathrm{r}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\boldsymbol{\mathrm{br}}\right)^{\mathrm{3}} \:+\:\left(\boldsymbol{\mathrm{br}}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{3}} \:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{2abc}\:\:\:\leftarrow \\ $$

Question Number 118292    Answers: 2   Comments: 0

Prove that: ∫_0 ^( 1) ((x^n −1)/(lnx)) = ln∣n+1∣

$$\boldsymbol{\mathrm{P}}\mathrm{rove}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{lnx}}\:=\:\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{n}}+\mathrm{1}\mid \\ $$

Question Number 118286    Answers: 2   Comments: 0

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣a−b∣ ;(b)∣ a+b∣>∣ a−b∣;(c)∣a+ b∣< ∣a−b∣

$$\mathrm{What}\:\mathrm{condition}\:\mathrm{should}\:\mathrm{be}\:\mathrm{satisfied}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{vectors}\:\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{relations}\:\mathrm{to}\:\mathrm{hold}\:\mathrm{true}\::\left(\mathrm{a}\right)\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid=\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$$$;\left(\mathrm{b}\right)\mid\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid>\mid\:\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid;\left(\mathrm{c}\right)\mid\boldsymbol{\mathrm{a}}+\:\boldsymbol{\mathrm{b}}\mid<\:\mid\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\mid \\ $$

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