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AllQuestion and Answers: Page 1080

Question Number 111213    Answers: 2   Comments: 0

Question Number 111209    Answers: 0   Comments: 6

find the ODEs of y=Ae^x +Bxe^x +Ce^(−2x) +De^(3x)

$${find}\:{the}\:{ODEs}\:{of}\: \\ $$$${y}={Ae}^{{x}} +{Bxe}^{{x}} +{Ce}^{−\mathrm{2}{x}} +{De}^{\mathrm{3}{x}} \\ $$

Question Number 111208    Answers: 1   Comments: 3

let p a prime number s.t p≥7 and a=333......3_(p−1 times) Show that 11∣a.

$$\:\mathrm{let}\:{p}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{s}.\mathrm{t}\:{p}\geqslant\mathrm{7}\:\mathrm{and}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=\underset{{p}−\mathrm{1}\:{times}} {\mathrm{333}......\mathrm{3}} \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{11}\mid{a}. \\ $$

Question Number 111235    Answers: 1   Comments: 1

Question Number 111206    Answers: 1   Comments: 0

∫ (dx/(x^3 +3x−5))

$$\int\:\frac{{dx}}{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{5}} \\ $$

Question Number 111205    Answers: 0   Comments: 1

An 800kg car is towed up an 8^0 hill by a rope attached to a truck. The tension in the rope is 2000N and there is no frictional resistance to the car motion. How much time is needed to tow the car for 50m starting from rest?

$$\mathrm{An}\:\mathrm{800kg}\:\mathrm{car}\:\mathrm{is}\:\mathrm{towed}\:\mathrm{up}\:\mathrm{an}\:\mathrm{8}^{\mathrm{0}} \:\mathrm{hill}\:\mathrm{by}\:\mathrm{a}\:\mathrm{rope}\: \\ $$$$\mathrm{attached}\:\mathrm{to}\:\mathrm{a}\:\mathrm{truck}.\:\mathrm{The}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{rope} \\ $$$$\mathrm{is}\:\mathrm{2000N}\:\mathrm{and}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{frictional}\:\mathrm{resistance} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{car}\:\mathrm{motion}.\:\mathrm{How}\:\mathrm{much}\:\mathrm{time}\:\mathrm{is}\:\mathrm{needed}\: \\ $$$$\mathrm{to}\:\mathrm{tow}\:\mathrm{the}\:\mathrm{car}\:\mathrm{for}\:\mathrm{50m}\:\mathrm{starting}\:\mathrm{from}\:\mathrm{rest}? \\ $$

Question Number 112193    Answers: 0   Comments: 1

Question Number 111196    Answers: 1   Comments: 0

two boxes one of mass 20kg and the other of mass 30kg sre sliding down on a frictionless inclined plane that makes an angle of 25^0 with the horizontal. find their redpective accelerstions.

$${two}\:{boxes}\:{one}\:{of}\:{mass}\:\mathrm{20}{kg}\:{and}\:{the}\:{other}\:{of}\: \\ $$$${mass}\:\mathrm{30}{kg}\:{sre}\:{sliding}\:{down}\:{on}\:{a}\:{frictionless} \\ $$$${inclined}\:{plane}\:{that}\:{makes}\:{an}\:{angle}\:{of}\:\mathrm{25}^{\mathrm{0}} \:{with}\:{the}\: \\ $$$${horizontal}.\:{find}\:{their}\:{redpective}\:{accelerstions}. \\ $$

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?

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

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