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Question Number 17782 Answers: 0 Comments: 1

let a,b,c,x,y and z be complex numbers such that : a=((b+c)/(x−2)), b=((c+a)/(y−2)), c=((a+b)/(z−2)) if xy+yz+zx=1000 and x+y+z=2016, find the value of xyz

$${let}\:{a},{b},{c},{x},{y}\:{and}\:{z}\:{be}\:{complex}\:{numbers} \\ $$$${such}\:{that}\:: \\ $$$${a}=\frac{{b}+{c}}{{x}−\mathrm{2}},\:{b}=\frac{{c}+{a}}{{y}−\mathrm{2}},\:{c}=\frac{{a}+{b}}{{z}−\mathrm{2}} \\ $$$${if}\:{xy}+{yz}+{zx}=\mathrm{1000}\:{and}\:{x}+{y}+{z}=\mathrm{2016}, \\ $$$${find}\:{the}\:{value}\:{of}\:{xyz} \\ $$

Question Number 17779 Answers: 1 Comments: 0

show that {log_a ab}{log_b ab}=logab_a +logab_b

$${show}\:{that}\:\left\{{lo}\underset{{a}} {{g}ab}\right\}\left\{{lo}\underset{{b}} {{g}ab}\right\}={loga}\underset{{a}} {{b}}+{loga}\underset{{b}} {{b}} \\ $$

Question Number 17775 Answers: 2 Comments: 0

Σ_(m=1) ^(10) (((sin 2πm)/(11)) − i((cos 2πm)/(11))) =????? Solve..it

$$\underset{{m}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\frac{\mathrm{sin}\:\mathrm{2}\pi{m}}{\mathrm{11}}\:−\:{i}\frac{\mathrm{cos}\:\mathrm{2}\pi{m}}{\mathrm{11}}\right)\:=????? \\ $$$${Solve}..{it} \\ $$$$ \\ $$

Question Number 17771 Answers: 1 Comments: 1

Question Number 17770 Answers: 1 Comments: 0

Find the number of ways the digits 0,1,2 and 3 can be permuted to give rise to a number greater than 2000.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{the} \\ $$$$\mathrm{digits}\:\mathrm{0},\mathrm{1},\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{can}\:\mathrm{be}\:\mathrm{permuted} \\ $$$$\mathrm{to}\:\mathrm{give}\:\mathrm{rise}\:\mathrm{to}\:\mathrm{a}\:\mathrm{number}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{2000}. \\ $$

Question Number 17767 Answers: 1 Comments: 0

Σ_(n = 1) ^(30) (n^2 + 1) = (A) Σ_(n = 1) ^(15) (2n^2 + 30n + 224) (B) Σ_(n = 1) ^(15) (2n^2 + 30n + 225) (C) Σ_(n = 1) ^(15) (2n^2 + 30n + 226) (D) Σ_(n = 1) ^(15) (2n^2 + 30n + 227) (E) Σ_(n = 1) ^(15) (2n^2 + 30n + 228)

$$\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{30}} {\sum}}\left({n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:=\: \\ $$$$\left(\mathrm{A}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{224}\right) \\ $$$$\left(\mathrm{B}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{225}\right) \\ $$$$\left(\mathrm{C}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{226}\right) \\ $$$$\left(\mathrm{D}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{227}\right) \\ $$$$\left(\mathrm{E}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{228}\right) \\ $$

Question Number 17759 Answers: 1 Comments: 0

2^(4(y^2 −2)) =y^((y^2 −8)) please help me find y... pls!

$$\mathrm{2}^{\mathrm{4}\left(\mathrm{y}^{\mathrm{2}} −\mathrm{2}\right)} =\mathrm{y}^{\left(\mathrm{y}^{\mathrm{2}} −\mathrm{8}\right)} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{find}\:\mathrm{y}...\:\mathrm{pls}! \\ $$

Question Number 17748 Answers: 0 Comments: 1

f p^2 =qr, prove thathat log_r ^p +log_(q=) ^p 2log_q ^p log_r ^p

$${f}\:{p}^{\mathrm{2}} ={qr},\:\:{prove}\:{thathat}\:{log}_{{r}} ^{{p}} +{log}_{{q}=} ^{{p}} \\ $$$$\mathrm{2}{log}_{{q}} ^{{p}} {log}_{{r}} ^{{p}} \\ $$

Question Number 17742 Answers: 1 Comments: 1

x^3 = 3^x , find x.

$$\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3}^{\mathrm{x}} ,\:\:\:\mathrm{find}\:\mathrm{x}. \\ $$

Question Number 17740 Answers: 0 Comments: 1

Question Number 17743 Answers: 2 Comments: 0

Question Number 17736 Answers: 0 Comments: 1

evaluate; ∫ln (sin 2x)dx

$${evaluate};\:\int\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{2}{x}\right){dx} \\ $$

Question Number 17735 Answers: 1 Comments: 0

Solve: (dy/dx) + ysec(x) = tan(x)

$$\mathrm{Solve}:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{ysec}\left(\mathrm{x}\right)\:=\:\mathrm{tan}\left(\mathrm{x}\right) \\ $$

Question Number 17734 Answers: 0 Comments: 0

If x^2 + y^3 − 3xy = 0, Show that, (d^2 y/dx^2 ) = ((− 2xy)/(y^2 − x^2 ))

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{3}} \:−\:\mathrm{3xy}\:=\:\mathrm{0}, \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\frac{−\:\mathrm{2xy}}{\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{x}^{\mathrm{2}} } \\ $$

Question Number 17729 Answers: 1 Comments: 3

A lotus plant in a pool of water is (1/2) cubit above water level. When propelled by air, the lotus sinks in the pool 2 cubits away from its position. Find the depth of water in the pool.

$$\mathrm{A}\:\mathrm{lotus}\:\mathrm{plant}\:\mathrm{in}\:\mathrm{a}\:\mathrm{pool}\:\mathrm{of}\:\mathrm{water}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cubit}\:\mathrm{above}\:\mathrm{water}\:\mathrm{level}.\:\mathrm{When} \\ $$$$\mathrm{propelled}\:\mathrm{by}\:\mathrm{air},\:\mathrm{the}\:\mathrm{lotus}\:\mathrm{sinks}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{pool}\:\mathrm{2}\:\mathrm{cubits}\:\mathrm{away}\:\mathrm{from}\:\mathrm{its}\:\mathrm{position}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{depth}\:\mathrm{of}\:\mathrm{water}\:\mathrm{in}\:\mathrm{the}\:\mathrm{pool}. \\ $$

Question Number 17704 Answers: 2 Comments: 1

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v (t) = 2t(3 − t) ; 0 < t < 3 and v (t) = − (t − 3)(6 − t) for 3 < t < 6 s in m/s. It repeats this cycle till it reaches the height of 20 m. At what time is its average velocity maximum?

$$\mathrm{A}\:\mathrm{monkey}\:\mathrm{climbs}\:\mathrm{up}\:\mathrm{a}\:\mathrm{slippery}\:\mathrm{pole}\:\mathrm{for} \\ $$$$\mathrm{3}\:\mathrm{seconds}\:\mathrm{and}\:\mathrm{subsequently}\:\mathrm{slips}\:\mathrm{for}\:\mathrm{3} \\ $$$$\mathrm{seconds}.\:\mathrm{Its}\:\mathrm{velocity}\:\mathrm{at}\:\mathrm{time}\:{t}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:{v}\:\left({t}\right)\:=\:\mathrm{2}{t}\left(\mathrm{3}\:−\:{t}\right)\:;\:\mathrm{0}\:<\:{t}\:<\:\mathrm{3}\:\mathrm{and} \\ $$$${v}\:\left({t}\right)\:=\:−\:\left({t}\:−\:\mathrm{3}\right)\left(\mathrm{6}\:−\:{t}\right)\:\mathrm{for}\:\mathrm{3}\:<\:{t}\:<\:\mathrm{6}\:\mathrm{s} \\ $$$$\mathrm{in}\:\mathrm{m}/\mathrm{s}.\:\mathrm{It}\:\mathrm{repeats}\:\mathrm{this}\:\mathrm{cycle}\:\mathrm{till}\:\mathrm{it} \\ $$$$\mathrm{reaches}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{20}\:\mathrm{m}.\:\mathrm{At}\:\mathrm{what} \\ $$$$\mathrm{time}\:\mathrm{is}\:\mathrm{its}\:\mathrm{average}\:\mathrm{velocity}\:\mathrm{maximum}? \\ $$

Question Number 17692 Answers: 1 Comments: 0

Question Number 17689 Answers: 1 Comments: 0

A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 km/h while the other has the speed of 27 km/h. The bird starts moving from first car towards the other and is moving with the speed of 36 km/h and when the two cars were separated by 36 km. What is the total displacement of the bird?

$$\mathrm{A}\:\mathrm{bird}\:\mathrm{is}\:\mathrm{tossing}\:\left(\mathrm{flying}\:\mathrm{to}\:\mathrm{and}\:\mathrm{fro}\right) \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{cars}\:\mathrm{moving}\:\mathrm{towards} \\ $$$$\mathrm{each}\:\mathrm{other}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}.\:\mathrm{One}\:\mathrm{car} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{18}\:\mathrm{km}/\mathrm{h}\:\mathrm{while}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{has}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{27}\:\mathrm{km}/\mathrm{h}.\:\mathrm{The}\:\mathrm{bird} \\ $$$$\mathrm{starts}\:\mathrm{moving}\:\mathrm{from}\:\mathrm{first}\:\mathrm{car}\:\mathrm{towards} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{and}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{the}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{36}\:\mathrm{km}/\mathrm{h}\:\mathrm{and}\:\mathrm{when}\:\mathrm{the}\:\mathrm{two}\:\mathrm{cars}\:\mathrm{were} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{36}\:\mathrm{km}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total} \\ $$$$\mathrm{displacement}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bird}? \\ $$

Question Number 17676 Answers: 2 Comments: 0

Question Number 17675 Answers: 1 Comments: 0

Question Number 17653 Answers: 0 Comments: 6

A line segment moves in the plane with its end points on the coordinate axes so that the sum of the length of its intersect on the coordinate axes is a constant C . Find the locus of the mid points of this segment . Ans. is 8(∣x∣^3 +∣y∣^3 )=C . Λ means power . pls. solve it.

$$\mathrm{A}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{with}\:\mathrm{its}\:\mathrm{end}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{coordinate} \\ $$$$\mathrm{axes}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{intersect}\:\mathrm{on}\:\mathrm{the}\:\mathrm{coordinate}\: \\ $$$$\mathrm{axes}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{C}\:. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{points}\:\mathrm{of} \\ $$$$\mathrm{this}\:\mathrm{segment}\:. \\ $$$$\mathrm{Ans}.\:\mathrm{is}\:\:\:\mathrm{8}\left(\mid\mathrm{x}\mid^{\mathrm{3}} +\mid\mathrm{y}\mid^{\mathrm{3}} \right)=\mathrm{C}\:. \\ $$$$\Lambda\:\:\mathrm{means}\:\mathrm{power}\:.\:\mathrm{pls}.\:\mathrm{solve}\:\mathrm{it}. \\ $$

Question Number 17638 Answers: 0 Comments: 4

please help me with this confusing question x^(2x/y) ×y^(y/x) =4......(1) (xy)^(xy+yx) =16.....(2) solve for x and y

$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{this} \\ $$$$\mathrm{confusing}\:\mathrm{question} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2x}/\mathrm{y}} ×\mathrm{y}^{\mathrm{y}/\mathrm{x}} =\mathrm{4}......\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\left(\mathrm{xy}\right)^{\mathrm{xy}+\mathrm{yx}} =\mathrm{16}.....\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$

Question Number 17634 Answers: 1 Comments: 0

y = x! , Find y′

$$\mathrm{y}\:=\:\mathrm{x}!\:\:,\:\:\:\:\:\mathrm{Find}\:\:\:\mathrm{y}' \\ $$

Question Number 17625 Answers: 1 Comments: 1

Find the fourier series of : f(x) = x, from 0 < x < π

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{fourier}\:\mathrm{series}\:\mathrm{of}\::\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x},\:\:\mathrm{from}\:\:\:\mathrm{0}\:<\:\mathrm{x}\:<\:\pi \\ $$

Question Number 17617 Answers: 0 Comments: 5

A string is stretched and fastened to two points l apart. Motion is started by displacing the string into the form y = (lx − x^2 ) from which it is release at time t = 0. Find the displacement of any point on the spring at a distance x from one end at time t.

$$\mathrm{A}\:\mathrm{string}\:\mathrm{is}\:\mathrm{stretched}\:\mathrm{and}\:\mathrm{fastened}\:\mathrm{to}\:\mathrm{two}\:\mathrm{points}\:\:\mathrm{l}\:\:\mathrm{apart}.\:\mathrm{Motion}\:\mathrm{is}\:\mathrm{started} \\ $$$$\mathrm{by}\:\mathrm{displacing}\:\mathrm{the}\:\mathrm{string}\:\mathrm{into}\:\mathrm{the}\:\mathrm{form}\:\:\mathrm{y}\:=\:\left(\mathrm{lx}\:−\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{from}\:\mathrm{which}\:\mathrm{it}\:\mathrm{is}\:\mathrm{release} \\ $$$$\mathrm{at}\:\mathrm{time}\:\mathrm{t}\:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{displacement}\:\mathrm{of}\:\mathrm{any}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance} \\ $$$$\mathrm{x}\:\mathrm{from}\:\mathrm{one}\:\mathrm{end}\:\mathrm{at}\:\mathrm{time}\:\:\mathrm{t}.\: \\ $$

Question Number 17614 Answers: 0 Comments: 3

The triangle ABC has CA = CB. P is a point on the circumcircle between A and B (and on the opposite side of the line AB to C). D is the foot of the perpendicular from C to PB. Show that PA + PB = 2∙PD.

$$\mathrm{The}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{CA}\:=\:\mathrm{CB}.\:\mathrm{P}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{between}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{B}\:\left(\mathrm{and}\:\mathrm{on}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{line}\:\mathrm{AB}\:\mathrm{to}\:\mathrm{C}\right).\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{perpendicular}\:\mathrm{from}\:\mathrm{C}\:\mathrm{to}\:\mathrm{PB}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{PA}\:+\:\mathrm{PB}\:=\:\mathrm{2}\centerdot\mathrm{PD}. \\ $$

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