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Question Number 19472 Answers: 0 Comments: 3

A fishing boat is anchored 9 km away from the nearest point on the shore. A messanger must be sent from the fishing boat to a camp, 15 km from the point on shore closest to boat. If the messanger can walk at a speed of 5 km per hour and can row at 4 km/h, determine the distance of that point (in km) from the shore, where he must land so as to reach the shore in least possible time.

$$\mathrm{A}\:\mathrm{fishing}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{anchored}\:\mathrm{9}\:\mathrm{km}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{shore}.\:\mathrm{A} \\ $$$$\mathrm{messanger}\:\mathrm{must}\:\mathrm{be}\:\mathrm{sent}\:\mathrm{from}\:\mathrm{the}\:\mathrm{fishing} \\ $$$$\mathrm{boat}\:\mathrm{to}\:\mathrm{a}\:\mathrm{camp},\:\mathrm{15}\:\mathrm{km}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{shore}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{boat}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{messanger} \\ $$$$\mathrm{can}\:\mathrm{walk}\:\mathrm{at}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{5}\:\mathrm{km}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{and}\:\mathrm{can}\:\mathrm{row}\:\mathrm{at}\:\mathrm{4}\:\mathrm{km}/\mathrm{h},\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{of}\:\mathrm{that}\:\mathrm{point}\:\left(\mathrm{in}\:\mathrm{km}\right)\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{shore},\:\mathrm{where}\:\mathrm{he}\:\mathrm{must}\:\mathrm{land}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to} \\ $$$$\mathrm{reach}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{in}\:\mathrm{least}\:\mathrm{possible}\:\mathrm{time}. \\ $$

Question Number 19455 Answers: 1 Comments: 0

Prove that if z = cos 6° + i sin 6°, then (1/(z^2 + 1)) − ((iz)/(z^4 − 1)) + ((iz^3 )/(z^8 − 1)) + ((iz^7 )/(z^(16) − 1)) = 0.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{z}\:=\:\mathrm{cos}\:\mathrm{6}°\:+\:{i}\:\mathrm{sin}\:\mathrm{6}°,\:\mathrm{then} \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{iz}}{{z}^{\mathrm{4}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{3}} }{{z}^{\mathrm{8}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{7}} }{{z}^{\mathrm{16}} \:−\:\mathrm{1}}\:=\:\mathrm{0}. \\ $$

Question Number 19435 Answers: 1 Comments: 0

If α = cos ((2π)/5) + i sin ((2π)/5) , then find the value of α + α^2 + α^3 + α^4 .

$$\mathrm{If}\:\alpha\:=\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:+\:{i}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\alpha\:+\:\alpha^{\mathrm{2}} \:+\:\alpha^{\mathrm{3}} \:+\:\alpha^{\mathrm{4}} . \\ $$

Question Number 19434 Answers: 1 Comments: 0

If (((1 + i(√3))/(1 − i(√3))))^n is an integer, then n is

$$\mathrm{If}\:\left(\frac{\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}{\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\right)^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{n}\:\mathrm{is} \\ $$

Question Number 19433 Answers: 1 Comments: 0

(((√(5 + 12i)) + (√(5 − 12i)))/((√(5 + 12i)) − (√(5 − 12i)))) =

$$\frac{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:+\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:−\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}\:= \\ $$

Question Number 19419 Answers: 1 Comments: 1

sin z=200 find z

$$\mathrm{sin}\:\boldsymbol{{z}}=\mathrm{200} \\ $$$$ \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{z}} \\ $$

Question Number 19415 Answers: 1 Comments: 0

PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. What is the area of the region QXROQ bounded by the two semicircular arcs?

$${PS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{of}\:\mathrm{length}\:\mathrm{4}\:\mathrm{and}\:{O} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PS}.\:\mathrm{A}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:{PS}\:\mathrm{as}\:\mathrm{diameter}.\:\mathrm{Let} \\ $$$${X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{this}\:\mathrm{arc}.\:{Q}\:\mathrm{and}\:{R} \\ $$$$\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{arc}\:{PXS}\:\mathrm{such}\:\mathrm{that}\:{QR} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{PS}\:\mathrm{and}\:\mathrm{the}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{drawn}\:\mathrm{with}\:{QR}\:\mathrm{as}\:\mathrm{diameter}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:{PS}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{region}\:{QXROQ}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{semicircular}\:\mathrm{arcs}? \\ $$

Question Number 19413 Answers: 1 Comments: 1

How many integer pairs (x, y) satisfy x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{integer}\:\mathrm{pairs}\:\left({x},\:{y}\right)\:\mathrm{satisfy} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:−\:\mathrm{2}{x}\:−\:\mathrm{4}{y}\:−\:\mathrm{8}\:=\:\mathrm{0}? \\ $$

Question Number 19409 Answers: 2 Comments: 1

What is the sum of the squares of the roots of the equation x^2 − 7[x] + 5 = 0? (Here [x] denotes the greatest integer less than or equal to x. For example [3.4] = 3 and [−2.3] = −3.)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:\mathrm{7}\left[{x}\right]\:+\:\mathrm{5}\:=\:\mathrm{0}? \\ $$$$\left(\mathrm{Here}\:\left[{x}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\right. \\ $$$$\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x}.\:\mathrm{For}\:\mathrm{example} \\ $$$$\left.\left[\mathrm{3}.\mathrm{4}\right]\:=\:\mathrm{3}\:\mathrm{and}\:\left[−\mathrm{2}.\mathrm{3}\right]\:=\:−\mathrm{3}.\right) \\ $$

Question Number 19403 Answers: 1 Comments: 1

Let P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9). What is the largest integer that is a divisor of P(n) for all positive even integers n?

$$\mathrm{Let}\:{P}\left({n}\right)\:=\:\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{3}\right)\left({n}\:+\:\mathrm{5}\right)\left({n}\:+\:\mathrm{7}\right)\left({n}\:+\:\mathrm{9}\right). \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{divisor}\:\mathrm{of}\:{P}\left({n}\right)\:\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{even} \\ $$$$\mathrm{integers}\:{n}? \\ $$

Question Number 19457 Answers: 1 Comments: 0

prove that ∫tan^2 xdx =tan x−x

$${prove}\:{that}\:\int\mathrm{tan}\:^{\mathrm{2}} {xdx} \\ $$$$ \\ $$$$=\mathrm{tan}\:{x}−{x} \\ $$

Question Number 19394 Answers: 1 Comments: 0

Question Number 19393 Answers: 1 Comments: 0

Three tennis players A, B and C play each other only once. The probability that A will beat B is (3/5), that B will beat C is (2/3), and that A will beat C is (5/7). Find (1) the probability that A will not win both games (2) the probability that A will win not both games.

$$\mathrm{Three}\:\mathrm{tennis}\:\mathrm{players}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{play}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{only}\:\mathrm{once}.\:\mathrm{The}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{A}\:\mathrm{will} \\ $$$$\mathrm{beat}\:\mathrm{B}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{5}},\:\mathrm{that}\:\mathrm{B}\:\mathrm{will}\:\mathrm{beat}\:\mathrm{C}\:\mathrm{is}\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{A}\:\mathrm{will}\:\mathrm{beat}\:\mathrm{C}\:\mathrm{is}\:\frac{\mathrm{5}}{\mathrm{7}}.\:\mathrm{Find}\:\left(\mathrm{1}\right)\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{A}\:\mathrm{will}\:\mathrm{not}\:\mathrm{win}\:\mathrm{both}\:\mathrm{games} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{A}\:\mathrm{will}\:\mathrm{win}\:\mathrm{not}\:\mathrm{both}\:\mathrm{games}. \\ $$

Question Number 19555 Answers: 1 Comments: 0

Find center and radius of circle having equation zz^ + (1 − i)z + (1 + i)z^ − 1 = 0.

$$\mathrm{Find}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\mathrm{equation}\:{z}\bar {{z}}\:+\:\left(\mathrm{1}\:−\:{i}\right){z}\:+\:\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:−\:\mathrm{1}\:=\:\mathrm{0}. \\ $$

Question Number 19391 Answers: 0 Comments: 1

Question Number 19389 Answers: 1 Comments: 0

What is the digital root of 3^(2017)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{digital}\:\mathrm{root}\:\mathrm{of}\:\:\mathrm{3}^{\mathrm{2017}} \\ $$

Question Number 19388 Answers: 0 Comments: 1

related to Q.19333 the side lengthes of a triangle are integer. if the perimeter of the triangle is 100, how many different triangles exist? what is the maximum area of them?

Question Number 19385 Answers: 0 Comments: 5

tan^2 β=−1 find β.... lets solve for fun

$${tan}^{\mathrm{2}} \beta=−\mathrm{1} \\ $$$$ \\ $$$${find}\:\beta....\:{lets}\:{solve}\:{for}\:{fun} \\ $$

Question Number 19382 Answers: 2 Comments: 0

If log_2 (9^(x−1) +7)−log_2 (3^(x−1) +1)=2, then the values of x are

$$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{9}^{{x}−\mathrm{1}} +\mathrm{7}\right)−\mathrm{log}_{\mathrm{2}} \left(\mathrm{3}^{{x}−\mathrm{1}} +\mathrm{1}\right)=\mathrm{2},\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{are} \\ $$

Question Number 19367 Answers: 1 Comments: 0

If α, β, γ and δ are four solutions of the equation tan (θ + ((5π)/4)) = 3 tan 3θ, then (1) Σtan α = 0 (2) Σtan α tan β = −2 (3) Σtan α tan β tan γ = −(8/3) (4) tan α tan β tan γ tan δ = −3

$$\mathrm{If}\:\alpha,\:\beta,\:\gamma\:\mathrm{and}\:\delta\:\mathrm{are}\:\mathrm{four}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:\left(\theta\:+\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:=\:\mathrm{3}\:\mathrm{tan}\:\mathrm{3}\theta,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\Sigma\mathrm{tan}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:=\:−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\delta\:=\:−\mathrm{3} \\ $$

Question Number 19362 Answers: 0 Comments: 5

The block Q moves to the right with a constant velocity v_0 as shown in figure. The relative velocity of body P with respect to Q is (assume all pulleys and strings are ideal)

$$\mathrm{The}\:\mathrm{block}\:{Q}\:\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}. \\ $$$$\mathrm{The}\:\mathrm{relative}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{body}\:{P}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:{Q}\:\mathrm{is}\:\left(\mathrm{assume}\:\mathrm{all}\:\mathrm{pulleys}\:\mathrm{and}\right. \\ $$$$\left.\mathrm{strings}\:\mathrm{are}\:\mathrm{ideal}\right) \\ $$

Question Number 19358 Answers: 1 Comments: 1

In the arrangement shown in figure two beads slide along a smooth horizontal rod. The relation between v and v_0 in given position will be

$$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure} \\ $$$$\mathrm{two}\:\mathrm{beads}\:\mathrm{slide}\:\mathrm{along}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{rod}.\:\mathrm{The}\:\mathrm{relation}\:\mathrm{between} \\ $$$${v}\:\mathrm{and}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{given}\:\mathrm{position}\:\mathrm{will}\:\mathrm{be} \\ $$

Question Number 19355 Answers: 0 Comments: 3

Two blocks are placed on a smooth horizontal surface and connected by a string pulley arrangement as shown. If a force F starts acting on block m_1 , then find the relation between acceleration of both masses and their values

$$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{and}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{string}\:\mathrm{pulley}\:\mathrm{arrangement}\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{force}\:{F}\:\mathrm{starts}\:\mathrm{acting}\:\mathrm{on}\:\mathrm{block}\:{m}_{\mathrm{1}} , \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{both}\:\mathrm{masses}\:\mathrm{and}\:\mathrm{their}\:\mathrm{values} \\ $$

Question Number 19352 Answers: 1 Comments: 0

Prove that ∣z_1 − z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 − 2∣z_1 ∣ ∣z_2 ∣ cos (θ_1 − θ_2 )

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \\ $$$$−\:\mathrm{2}\mid{z}_{\mathrm{1}} \mid\:\mid{z}_{\mathrm{2}} \mid\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \:−\:\theta_{\mathrm{2}} \right) \\ $$

Question Number 19351 Answers: 1 Comments: 2

Prove that ∣z_1 + z_2 ∣^2 = ∣z_1 ∣^2 + ∣z_2 ∣^2 ⇔ (z_1 /z_2 ) is purely imaginary number.

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:=\:\mid{z}_{\mathrm{1}} \mid^{\mathrm{2}} \:+\:\mid{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\Leftrightarrow \\ $$$$\frac{{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} }\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{number}. \\ $$

Question Number 19350 Answers: 1 Comments: 0

Prove that ∣z_1 + z_2 ∣ = ∣z_1 − z_2 ∣ ⇔ arg(z_1 ) − arg(z_2 ) = (π/2)

$$\mathrm{Prove}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \mid\:=\:\mid{z}_{\mathrm{1}} \:−\:{z}_{\mathrm{2}} \mid\:\Leftrightarrow \\ $$$$\mathrm{arg}\left({z}_{\mathrm{1}} \right)\:−\:\mathrm{arg}\left({z}_{\mathrm{2}} \right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

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