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Question Number 119564 Answers: 0 Comments: 0

((3/x) − ((15)/(2y)) ) ÷ (6/(xy)) = ((6y − 15x)/(2xy)) × ((xy)/6).

$$\left(\frac{\mathrm{3}}{{x}}\:−\:\frac{\mathrm{15}}{\mathrm{2}{y}}\:\right)\:\boldsymbol{\div}\:\frac{\mathrm{6}}{{xy}}\:=\:\frac{\mathrm{6}{y}\:−\:\mathrm{15}{x}}{\mathrm{2}{xy}}\:×\:\frac{{xy}}{\mathrm{6}}. \\ $$

Question Number 119548 Answers: 1 Comments: 2

Question Number 119540 Answers: 1 Comments: 0

Question Number 119538 Answers: 1 Comments: 1

Question Number 119529 Answers: 3 Comments: 7

Question Number 119524 Answers: 1 Comments: 0

Question Number 119517 Answers: 2 Comments: 0

lim_(x→∞) [ sin (x+(1/x))−sin x ] =?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\mathrm{sin}\:\left({x}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{sin}\:{x}\:\right]\:=? \\ $$

Question Number 119527 Answers: 2 Comments: 2

Question Number 119510 Answers: 0 Comments: 0

Differential Equation (1 − x)(d^2 y/dx^2 ) + x(dy/dx) − xy = (1/(1 − x)) , x ≠ 1 has the power series solution for ∣x∣<1

$${Differential}\:{Equation}\: \\ $$$$\left(\mathrm{1}\:−\:{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:{x}\frac{{dy}}{{dx}}\:−\:{xy}\:=\:\frac{\mathrm{1}}{\mathrm{1}\:−\:{x}}\:,\:{x}\:\neq\:\mathrm{1} \\ $$$${has}\:{the}\:{power}\:{series}\:{solution}\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$

Question Number 119506 Answers: 1 Comments: 0

cos (π/5)cos x+sin (π/5)sin x ≤ ((√2)/2)

$$\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{5}}\mathrm{cos}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{5}}\mathrm{sin}\:{x}\:\leqslant\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Question Number 119503 Answers: 2 Comments: 0

solve ⌊ (√x) ⌋ = ⌊ (x)^(1/(3 )) ⌋

$${solve}\:\lfloor\:\sqrt{{x}}\:\rfloor\:=\:\lfloor\:\sqrt[{\mathrm{3}\:}]{{x}}\:\rfloor\: \\ $$

Question Number 119495 Answers: 0 Comments: 0

Question Number 119494 Answers: 3 Comments: 0

Russian olympiad find real solution of the system { ((sin x+2sin (x+y+z)=0)),((sin y+3sin (x+y+z)=0)),((sin z+4sin (x+y+z)=0)) :}

$${Russian}\:{olympiad}\: \\ $$$${find}\:{real}\:{solution}\:{of}\:{the}\:{system}\: \\ $$$$\begin{cases}{\mathrm{sin}\:{x}+\mathrm{2sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{y}+\mathrm{3sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\\{\mathrm{sin}\:{z}+\mathrm{4sin}\:\left({x}+{y}+{z}\right)=\mathrm{0}}\end{cases} \\ $$$$ \\ $$

Question Number 119490 Answers: 0 Comments: 0

Let a<c<b such that c−a=b−c. If f:R→R is a function satisfying the relation f(x+a)+f(x+b)=f(x+c) for all x∈R then a period of f is (A) (b−a) (B) 2(b−a) (C) 3(b−a) (D) 4(b−a)

$$\mathrm{Let}\:{a}<\mathrm{c}<\mathrm{b}\:\mathrm{such}\:\mathrm{that}\:\mathrm{c}−{a}=\mathrm{b}−\mathrm{c}.\:\mathrm{If}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{function}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\:\:{f}\left(\mathrm{x}+{a}\right)+{f}\left(\mathrm{x}+\mathrm{b}\right)={f}\left(\mathrm{x}+\mathrm{c}\right)\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\left(\mathrm{b}−{a}\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{3}\left(\mathrm{b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4}\left(\mathrm{b}−{a}\right) \\ $$

Question Number 119487 Answers: 2 Comments: 0

find element of set S = { ((x^3 −3x^2 +2)/(2x+1)) ∈ Z for x∈Z }

$${find}\:{element}\:{of}\:{set}\:{S}\:=\:\left\{\:\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}}\:\in\:\mathbb{Z}\:{for}\:{x}\in\mathbb{Z}\:\right\} \\ $$

Question Number 119483 Answers: 1 Comments: 0

Let a>0 and f:R→R a function satisfying f(x+a)=1+[2−3f(x)+3f(x)^2 −f(x)^3 ]^(1/3) for all x∈R. Then a period of f(x) is ka where k is a positive integer whose value is (A)1 (B)2 (C)3 (D)4

$$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{a}\:\mathrm{function}\:\mathrm{satisfying} \\ $$$$\:\:\:\:\:{f}\left(\mathrm{x}+{a}\right)=\mathrm{1}+\left[\mathrm{2}−\mathrm{3}{f}\left(\mathrm{x}\right)+\mathrm{3}{f}\left(\mathrm{x}\right)^{\mathrm{2}} −{f}\left(\mathrm{x}\right)^{\mathrm{3}} \right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}.\:\mathrm{Then}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:{f}\left(\mathrm{x}\right)\:\mathrm{is}\:{ka}\:\mathrm{where}\:{k}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{whose}\:\mathrm{value}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\mathrm{4} \\ $$

Question Number 119482 Answers: 1 Comments: 0

lim_(x→0) x^2 cos ((1/x)) ?

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{\mathrm{2}} \:\mathrm{cos}\:\left(\frac{\mathrm{1}}{{x}}\right)\:? \\ $$

Question Number 119479 Answers: 3 Comments: 0

Π_(k=1) ^∞ cos((x/2^k )) = ?

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}^{\mathrm{k}} }\right)\:=\:? \\ $$

Question Number 119474 Answers: 0 Comments: 0

Please help create an algorithm when the userviews in website around 0− 11:59 it will alert “Good Morning” and from 12− 15:59 it will alert “Goodafternoon” and lastly from 16− 23:59 it will alert “Good evening”.

$$ \\ $$$$\boldsymbol{{P}\mathrm{lease}}\:\boldsymbol{\mathrm{help}} \\ $$$$\mathrm{create}\:\mathrm{an}\:\mathrm{algorithm}\:\mathrm{when}\:\mathrm{the}\: \\ $$$$\mathrm{userviews}\:\mathrm{in}\:\mathrm{website}\:\mathrm{around}\:\mathrm{0}− \\ $$$$\mathrm{11}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\:``\mathrm{Good}\:\mathrm{Morning}''\: \\ $$$$\mathrm{and}\:\mathrm{from}\:\mathrm{12}−\:\mathrm{15}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\: \\ $$$$``\mathrm{Goodafternoon}''\:\mathrm{and}\:\mathrm{lastly}\:\mathrm{from}\: \\ $$$$\mathrm{16}−\:\mathrm{23}:\mathrm{59}\:\mathrm{it}\:\mathrm{will}\:\mathrm{alert}\:``\mathrm{Good}\:\mathrm{evening}''. \\ $$

Question Number 119473 Answers: 4 Comments: 1

Question Number 119472 Answers: 1 Comments: 0

Question Number 122799 Answers: 1 Comments: 0

Question Number 122798 Answers: 2 Comments: 1

Question Number 119464 Answers: 2 Comments: 0

Expand as far as the term in x^3 (1) (3−2x−x^2 )^9 (2) (1−x+x^2 )^8

$$\mathrm{Expand}\:\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:\mathrm{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{3}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{9}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{8}} \\ $$

Question Number 119463 Answers: 1 Comments: 2

one cube of side length 3cm is painted by three colour say (red blue and black) opposte face same colour Now cube is cut into 27 small cube i)How many cube has three face coloured ii)How many cube two face coloured iii)How many cube has no colour in its face

$${one}\:{cube}\:{of}\:{side}\:{length}\:\mathrm{3}{cm}\:{is}\:{painted}\:{by}\:{three}\:{colour} \\ $$$${say}\:\left({red}\:{blue}\:{and}\:{black}\right)\:{opposte}\:{face}\:{same}\:{colour} \\ $$$${Now}\:{cube}\:{is}\:{cut}\:{into}\:\mathrm{27}\:{small}\:{cube} \\ $$$$\left.{i}\right){How}\:{many}\:{cube}\:{has}\:{three}\:{face}\:{coloured} \\ $$$$\left.{ii}\right){How}\:{many}\:{cube}\:{two}\:{face}\:{coloured} \\ $$$$\left.{iii}\right){How}\:{many}\:{cube}\:\:{has}\:{no}\:{colour}\:{in}\:{its}\:{face} \\ $$

Question Number 119462 Answers: 2 Comments: 0

... advanced calculus... evaluate :: Ω=∫_0 ^( ∞) ((tan^(−1) (x))/(e^(2πx) −1))dx =? m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

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