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Question Number 111195 Answers: 2 Comments: 0

(√(bemath)) ∫ (dx/( ((x−1))^(1/(3 )) (((x+1)^2 ))^(1/(3 )) )) ?

$$\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\:\:\:\int\:\frac{\mathrm{dx}}{\:\sqrt[{\mathrm{3}\:}]{\mathrm{x}−\mathrm{1}}\:\:\sqrt[{\mathrm{3}\:}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}\:? \\ $$

Question Number 111189 Answers: 4 Comments: 0

(1) ∫ (((x+1)dx)/(x^4 (x−1))) ? (2) (dy/dx) + (y/(x−2)) = 5(x−2)(√y)

$$\:\left(\mathrm{1}\right)\:\:\:\:\:\:\:\int\:\frac{\left({x}+\mathrm{1}\right){dx}}{{x}^{\mathrm{4}} \left({x}−\mathrm{1}\right)}\:?\: \\ $$$$\:\left(\mathrm{2}\right)\:\:\:\:\:\:\frac{{dy}}{{dx}}\:+\:\frac{{y}}{{x}−\mathrm{2}}\:=\:\mathrm{5}\left({x}−\mathrm{2}\right)\sqrt{{y}}\: \\ $$

Question Number 111187 Answers: 1 Comments: 1

Question Number 111183 Answers: 2 Comments: 1

y^(old)

$${y}^{{old}} \\ $$

Question Number 111180 Answers: 0 Comments: 0

solve: x^2 y(ydx+dy)=2ydx+xdy

$${solve}:\:{x}^{\mathrm{2}} {y}\left({ydx}+{dy}\right)=\mathrm{2}{ydx}+{xdy} \\ $$

Question Number 111272 Answers: 1 Comments: 0

Tricolours flags(each flag having three different strips of non−overlapping colours) are to be designed using white,blue,red,yellow and black strips. How many of the flags have blue colour?

$$\mathrm{Tricolours}\:\mathrm{flags}\left(\mathrm{each}\:\mathrm{flag}\:\mathrm{having}\right. \\ $$$$\mathrm{three}\:\mathrm{different}\:\mathrm{strips}\:\mathrm{of} \\ $$$$\left.\mathrm{non}−\mathrm{overlapping}\:\mathrm{colours}\right)\:\mathrm{are}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{designed}\:\mathrm{using}\:\mathrm{white},\mathrm{blue},\mathrm{red},\mathrm{yellow} \\ $$$$\mathrm{and}\:\mathrm{black}\:\mathrm{strips}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{flags}\:\mathrm{have}\:\mathrm{blue}\:\mathrm{colour}? \\ $$

Question Number 111175 Answers: 1 Comments: 0

Question Number 111174 Answers: 0 Comments: 0

following the newest trend − what do I say!? − ahead of it, of course! I post this answer to one of the next questions, look out for it so you won′t miss it! I=I_1 −2I_2 =ξ(5)+Γ(7/3)−2/(√π)+C

$${following}\:{the}\:{newest}\:{trend}\:−\:{what}\:{do}\:{I} \\ $$$${say}!?\:−\:{ahead}\:{of}\:{it},\:{of}\:{course}!\:{I}\:{post}\:{this} \\ $$$${answer}\:{to}\:{one}\:{of}\:{the}\:{next}\:{questions},\:{look} \\ $$$${out}\:{for}\:{it}\:{so}\:{you}\:{won}'{t}\:{miss}\:{it}! \\ $$$$ \\ $$$${I}={I}_{\mathrm{1}} −\mathrm{2}{I}_{\mathrm{2}} =\xi\left(\mathrm{5}\right)+\Gamma\left(\mathrm{7}/\mathrm{3}\right)−\mathrm{2}/\sqrt{\pi}+{C} \\ $$

Question Number 112273 Answers: 1 Comments: 2

(1) lim_(x→∞) (e^x +e^(−x) )^(2/x) =? (2) lim_(x→(π/2)) (cos x)^(−x+(π/2)) ? (3) lim_(x→0) (√((1+tan x)/x^2 ))−(√(((1−sin x)/x^2 ) ))?

$$\:\left(\mathrm{1}\right)\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{cos}\:\mathrm{x}\right)^{−\mathrm{x}+\frac{\pi}{\mathrm{2}}} ? \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }}−\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:}? \\ $$

Question Number 111275 Answers: 1 Comments: 0

In a tennis tournament n women and 2n men played. Each player played exactly one match with every other player. If there are no ties and the number of the matches won by women to the number of matches won by men is 7:5, find n.

$$\mathrm{In}\:\mathrm{a}\:\mathrm{tennis}\:\mathrm{tournament}\:\mathrm{n}\:\mathrm{women}\:\mathrm{and} \\ $$$$\mathrm{2n}\:\mathrm{men}\:\mathrm{played}.\:\mathrm{Each}\:\mathrm{player}\:\mathrm{played} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{match}\:\mathrm{with}\:\mathrm{every}\:\mathrm{other} \\ $$$$\mathrm{player}.\:\mathrm{If}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{ties}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{matches}\:\mathrm{won}\:\mathrm{by}\:\mathrm{women} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{matches}\:\mathrm{won}\:\mathrm{by}\:\mathrm{men} \\ $$$$\mathrm{is}\:\mathrm{7}:\mathrm{5},\:\mathrm{find}\:\mathrm{n}. \\ $$$$ \\ $$

Question Number 111163 Answers: 1 Comments: 2

A bag contains 11 white balls and 9 black balls, another contains 12 white balls and 13 black balls. If two balls are drawn without replacement, from each bag find the probability that exactly one of the 4 balls is white?

$$\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{11}\:\mathrm{white}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{9} \\ $$$$\mathrm{black}\:\mathrm{balls},\:\mathrm{another}\:\mathrm{contains}\:\mathrm{12}\:\mathrm{white} \\ $$$$\mathrm{balls}\:\mathrm{and}\:\mathrm{13}\:\mathrm{black}\:\mathrm{balls}.\:\mathrm{If}\:\mathrm{two}\:\mathrm{balls} \\ $$$$\mathrm{are}\:\mathrm{drawn}\:\mathrm{without}\:\mathrm{replacement},\:\mathrm{from} \\ $$$$\mathrm{each}\:\mathrm{bag}\:\mathrm{find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{4}\:\mathrm{balls}\:\mathrm{is}\:\mathrm{white}? \\ $$

Question Number 111159 Answers: 2 Comments: 0

z is a complex number with Re(z) , Im(z)∈N. Determine z if z.z^− =1000

$${z}\:{is}\:{a}\:{complex}\:{number}\:{with}\: \\ $$$${Re}\left({z}\right)\:,\:{Im}\left({z}\right)\in\mathbb{N}. \\ $$$${Determine}\:{z}\:\:{if} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}.\overset{−} {{z}}=\mathrm{1000} \\ $$

Question Number 111157 Answers: 2 Comments: 0

question proposed MN july 1970 ∫_0 ^(π/4) tanx(ln(1+tan^2 x))dx my solution followed

$${question}\:{proposed}\:{MN}\:{july}\:\mathrm{1970} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tan}{x}\left(\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right)\right){dx} \\ $$$${my}\:{solution}\:{followed} \\ $$

Question Number 111155 Answers: 1 Comments: 0

Find the number of rational numbers r, 0<r<1, such that when r is written as a fraction in lowest term. The numerator and the denominator have a sum of 1000.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{numbers} \\ $$$$\mathrm{r},\:\mathrm{0}<\mathrm{r}<\mathrm{1},\:\mathrm{such}\:\mathrm{that}\:\mathrm{when}\:\mathrm{r}\:\mathrm{is}\:\mathrm{written} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{fraction}\:\mathrm{in}\:\mathrm{lowest}\:\mathrm{term}.\:\mathrm{The} \\ $$$$\mathrm{numerator}\:\mathrm{and}\:\mathrm{the}\:\mathrm{denominator} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{1000}. \\ $$

Question Number 111154 Answers: 0 Comments: 2

Version 2.202 adds calculator. place cursor on a line which you want to evalute. Select calculate from menu in 3 vertical dots at top. Calculator accepts complex numbers and will return result in complex when needed. principal values are return for asin, acos etc. Angle are in radian unless ° symbol is used e.g. (30+60i)° Result will be added in immediate next line can be edited to form another expression for further computation. sin^(−1) 2 (1.570796 − 1.316958i)

$$\mathrm{Version}\:\mathrm{2}.\mathrm{202}\:\mathrm{adds}\:\mathrm{calculator}. \\ $$$$\mathrm{place}\:\mathrm{cursor}\:\mathrm{on}\:\mathrm{a}\:\mathrm{line}\:\mathrm{which}\:\mathrm{you} \\ $$$$\mathrm{want}\:\mathrm{to}\:\mathrm{evalute}.\: \\ $$$$\mathrm{Select}\:\mathrm{calculate}\:\mathrm{from}\:\mathrm{menu}\:\mathrm{in}\:\mathrm{3}\:\mathrm{vertical} \\ $$$$\mathrm{dots}\:\mathrm{at}\:\mathrm{top}. \\ $$$$\mathrm{Calculator}\:\mathrm{accepts}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{and}\:\mathrm{will}\:\mathrm{return}\:\mathrm{result}\:\mathrm{in}\:\mathrm{complex} \\ $$$$\mathrm{when}\:\mathrm{needed}. \\ $$$$\mathrm{principal}\:\mathrm{values}\:\mathrm{are}\:\mathrm{return}\:\mathrm{for} \\ $$$$\mathrm{asin},\:\mathrm{acos}\:\mathrm{etc}. \\ $$$$\mathrm{Angle}\:\mathrm{are}\:\mathrm{in}\:\mathrm{radian}\:\mathrm{unless}\:°\:\mathrm{symbol} \\ $$$$\mathrm{is}\:\mathrm{used}\:\mathrm{e}.\mathrm{g}.\:\left(\mathrm{30}+\mathrm{60i}\right)° \\ $$$$\mathrm{Result}\:\mathrm{will}\:\mathrm{be}\:\mathrm{added}\:\mathrm{in}\:\mathrm{immediate} \\ $$$$\mathrm{next}\:\mathrm{line}\:\mathrm{can}\:\mathrm{be}\:\mathrm{edited}\:\mathrm{to}\:\mathrm{form} \\ $$$$\mathrm{another}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{further} \\ $$$$\mathrm{computation}. \\ $$$$\mathrm{sin}^{−\mathrm{1}} \mathrm{2} \\ $$$$\left(\mathrm{1}.\mathrm{570796}\:−\:\mathrm{1}.\mathrm{316958i}\right) \\ $$

Question Number 111276 Answers: 1 Comments: 0

A coin that comes up head with probability p and tail with probability 1−p independently of each flip is flipped five times. The probability of two heads and three tails is equal to (1/7) of the probability of three heads and two tails. Let p=(x/y), where gcd(x,y) =1 . Find x+y.

$$\mathrm{A}\:\mathrm{coin}\:\mathrm{that}\:\mathrm{comes}\:\mathrm{up}\:\mathrm{head}\:\mathrm{with} \\ $$$$\mathrm{probability}\:\mathrm{p}\:\mathrm{and}\:\mathrm{tail}\:\mathrm{with}\:\mathrm{probability} \\ $$$$\mathrm{1}−\mathrm{p}\:\mathrm{independently}\:\mathrm{of}\:\mathrm{each}\:\mathrm{flip}\:\mathrm{is}\:\mathrm{flipped}\:\mathrm{five}\:\mathrm{times}. \\ $$$$\mathrm{The}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{two}\:\mathrm{heads}\:\mathrm{and}\:\mathrm{three}\:\mathrm{tails}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{1}}{\mathrm{7}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{three} \\ $$$$\mathrm{heads}\:\mathrm{and}\:\mathrm{two}\:\mathrm{tails}.\:\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{x}}{\mathrm{y}},\:\mathrm{where} \\ $$$$\mathrm{gcd}\left(\mathrm{x},\mathrm{y}\right)\:=\mathrm{1}\:.\:\mathrm{Find}\:\mathrm{x}+\mathrm{y}. \\ $$

Question Number 111149 Answers: 0 Comments: 4

Let f_0 (x) = (1/(1−x)) and f_n (x) =f_0 (f_(n−1) (x)), n=1,2,3,... Evaluate f_(2018) (2018)

$$\mathrm{Let}\:\mathrm{f}_{\mathrm{0}} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}\:\mathrm{and}\:\mathrm{f}_{\mathrm{n}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{f}_{\mathrm{0}} \left(\mathrm{f}_{\mathrm{n}−\mathrm{1}} \left(\mathrm{x}\right)\right),\:\mathrm{n}=\mathrm{1},\mathrm{2},\mathrm{3},...\:\mathrm{Evaluate} \\ $$$$\mathrm{f}_{\mathrm{2018}} \left(\mathrm{2018}\right) \\ $$

Question Number 111279 Answers: 1 Comments: 0

Triangle ABC has AB=2∙AC. Let D and E be on AB and BC respectively such that ∠BAE =∠ACD. Let F be the intersections of segments AE and CD, and suppose that △CFE is equilateral. What is ∠ACB?

$$\mathrm{Triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{AB}=\mathrm{2}\centerdot\mathrm{AC}.\:\mathrm{Let} \\ $$$$\mathrm{D}\:\mathrm{and}\:\mathrm{E}\:\mathrm{be}\:\mathrm{on}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:\angle\mathrm{BAE} \\ $$$$=\angle\mathrm{ACD}.\:\mathrm{Let}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{intersections}\:\mathrm{of} \\ $$$$\mathrm{segments}\:\mathrm{AE}\:\mathrm{and}\:\mathrm{CD},\:\mathrm{and}\:\mathrm{suppose} \\ $$$$\mathrm{that}\:\bigtriangleup\mathrm{CFE}\:\mathrm{is}\:\mathrm{equilateral}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\angle\mathrm{ACB}? \\ $$

Question Number 111147 Answers: 0 Comments: 0