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Two particles A and B move with constant velocities v_1 and v_2 along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0, the particles were located at distances d_1 and d_2 from O respectively. Find the time, when they are nearest and also this shortest distance.

$$\mathrm{Two}\:\mathrm{particles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{velocities}\:{v}_{\mathrm{1}} \:\mathrm{and}\:{v}_{\mathrm{2}} \:\mathrm{along}\:\mathrm{two} \\ $$$$\mathrm{mutually}\:\mathrm{perpendicular}\:\mathrm{straight}\:\mathrm{lines} \\ $$$$\mathrm{towards}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point}\:{O}.\:\mathrm{At} \\ $$$$\mathrm{moment}\:{t}\:=\:\mathrm{0},\:\mathrm{the}\:\mathrm{particles}\:\mathrm{were} \\ $$$$\mathrm{located}\:\mathrm{at}\:\mathrm{distances}\:{d}_{\mathrm{1}} \:\mathrm{and}\:{d}_{\mathrm{2}} \:\mathrm{from}\:{O} \\ $$$$\mathrm{respectively}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{time},\:\mathrm{when}\:\mathrm{they} \\ $$$$\mathrm{are}\:\mathrm{nearest}\:\mathrm{and}\:\mathrm{also}\:\mathrm{this}\:\mathrm{shortest} \\ $$$$\mathrm{distance}. \\ $$

Question Number 17169 Answers: 2 Comments: 0

The base of a pyramid is an equilateral triangle of side length 6 cm. The other edges of the pyramid are each of length (√(15)) cm. Find the volume of the pyramid.

$$\mathrm{The}\:\mathrm{base}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pyramid}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{of}\:\mathrm{side}\:\mathrm{length}\:\mathrm{6}\:\mathrm{cm}.\:\mathrm{The}\:\mathrm{other} \\ $$$$\mathrm{edges}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pyramid}\:\mathrm{are}\:\mathrm{each}\:\mathrm{of}\:\mathrm{length} \\ $$$$\sqrt{\mathrm{15}}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pyramid}. \\ $$

Question Number 16983 Answers: 1 Comments: 0

If 15 men or 24 women or 36 boys do a piece of work in 12 days, working 8 hours a day, how many men must associated with 12 women and 6 boys to do another piece of work 2(1/4) times as great in 30 days working 6 hrs a day?

$$\mathrm{If}\:\mathrm{15}\:\mathrm{men}\:\mathrm{or}\:\mathrm{24}\:\mathrm{women}\:\mathrm{or}\:\mathrm{36}\:\mathrm{boys}\:\mathrm{do}\:\mathrm{a}\: \\ $$$$\mathrm{piece}\:\mathrm{of}\:\mathrm{work}\:\mathrm{in}\:\mathrm{12}\:\mathrm{days},\:\mathrm{working}\:\mathrm{8}\:\mathrm{hours} \\ $$$$\mathrm{a}\:\mathrm{day},\:\mathrm{how}\:\mathrm{many}\:\mathrm{men}\:\mathrm{must}\:\mathrm{associated} \\ $$$$\mathrm{with}\:\mathrm{12}\:\mathrm{women}\:\mathrm{and}\:\mathrm{6}\:\mathrm{boys}\:\mathrm{to}\:\mathrm{do}\:\mathrm{another} \\ $$$$\mathrm{piece}\:\mathrm{of}\:\mathrm{work}\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{times}\:\mathrm{as}\:\mathrm{great}\:\mathrm{in}\:\mathrm{30} \\ $$$$\mathrm{days}\:\mathrm{working}\:\mathrm{6}\:\mathrm{hrs}\:\mathrm{a}\:\mathrm{day}? \\ $$

Question Number 17096 Answers: 1 Comments: 0

The total number of solutions of the equation tan 3x − tan 2x − tan 3x tan 2x = 1 in [0, 2π] is

$$\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:−\:\mathrm{tan}\:\mathrm{2}{x}\:−\:\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\mathrm{1}\:\mathrm{in} \\ $$$$\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{is} \\ $$

Question Number 16980 Answers: 0 Comments: 1

To Q16066: I have posted my solution there. Those who are intetested in this interesting question please have a critical view at it. Maybe there are alternative solutions which are easier and more direct and straight on.

$$\mathrm{To}\:\mathrm{Q16066}: \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{posted}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{there}. \\ $$$$\mathrm{Those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{intetested}\:\mathrm{in}\:\mathrm{this}\:\mathrm{interesting} \\ $$$$\mathrm{question}\:\mathrm{please}\:\mathrm{have}\:\mathrm{a}\:\mathrm{critical}\:\mathrm{view}\:\mathrm{at} \\ $$$$\mathrm{it}.\:\mathrm{Maybe}\:\mathrm{there}\:\mathrm{are}\:\mathrm{alternative}\:\mathrm{solutions} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{easier}\:\mathrm{and}\:\mathrm{more}\:\mathrm{direct}\:\mathrm{and} \\ $$$$\mathrm{straight}\:\mathrm{on}. \\ $$

Question Number 16973 Answers: 1 Comments: 0

5^(log(x)) = x^(log(2)) , find x.

$$\mathrm{5}^{\mathrm{log}\left(\mathrm{x}\right)} \:=\:\mathrm{x}^{\mathrm{log}\left(\mathrm{2}\right)} ,\:\:\:\:\:\mathrm{find}\:\:\mathrm{x}. \\ $$

Question Number 16990 Answers: 1 Comments: 4

If for positive integers a and b, a + b = (a/b) + (b/a), find a^2 + b^2 .

$$\mathrm{If}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{integers}\:{a}\:\mathrm{and}\:{b}, \\ $$$${a}\:+\:{b}\:=\:\frac{{a}}{{b}}\:+\:\frac{{b}}{{a}},\:\mathrm{find}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} . \\ $$

Question Number 16958 Answers: 0 Comments: 0

Let ABCD be a quadrilateral with an inscribed circle. Prove that the circles inscribed in triangles ABC and ADC are tangent to each other.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{with}\:\mathrm{an} \\ $$$$\mathrm{inscribed}\:\mathrm{circle}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{circles} \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{triangles}\:{ABC}\:\mathrm{and}\:{ADC} \\ $$$$\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$

Question Number 16951 Answers: 0 Comments: 0

Let M be a point in interior of ΔABC. Three lines are drawn through M, parallel to triangle′s sides, thereby producing three trapezoids. Suppose a diagonal is drawn in each trapezoid in such a way that the diagonals have no common endpoints. These three diagonals divide ABC into seven parts, four of them being triangles. Prove that the area of one of the four triangles equals the sum of the areas of the other three.

$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\Delta{ABC}. \\ $$$$\mathrm{Three}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{through}\:{M}, \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{triangle}'\mathrm{s}\:\mathrm{sides},\:\mathrm{thereby} \\ $$$$\mathrm{producing}\:\mathrm{three}\:\mathrm{trapezoids}.\:\mathrm{Suppose}\:\mathrm{a} \\ $$$$\mathrm{diagonal}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{each}\:\mathrm{trapezoid}\:\mathrm{in} \\ $$$$\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{have}\:\mathrm{no} \\ $$$$\mathrm{common}\:\mathrm{endpoints}.\:\mathrm{These}\:\mathrm{three} \\ $$$$\mathrm{diagonals}\:\mathrm{divide}\:{ABC}\:\mathrm{into}\:\mathrm{seven} \\ $$$$\mathrm{parts},\:\mathrm{four}\:\mathrm{of}\:\mathrm{them}\:\mathrm{being}\:\mathrm{triangles}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four} \\ $$$$\mathrm{triangles}\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{areas} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}. \\ $$

Question Number 16947 Answers: 0 Comments: 0

Through the vertices of the smaller base AB of the trapezoid ABCD two parallel lines are drawn, intersecting the segment CD. These lines and the trapezoid′s diagonals divide it into seven triangles and a pentagon. Show that the area of the pentagon equals the sum of the areas of the three triangles that share a common side with the trapezoid.

Question Number 16946 Answers: 0 Comments: 0

Consider the quadrilateral ABCD. The points M, N, P and Q are the midpoints of the sides AB, BC, CD and DA. Let X = AP ∩ BQ, Y = BQ ∩ CM, Q = CM ∩ DN and T= DN ∩ AP. Prove that [XYZT] = [AQX] + [BMY] + [CNZ] + [DPT].

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{quadrilateral}\:{ABCD}. \\ $$$$\mathrm{The}\:\mathrm{points}\:{M},\:{N},\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{AB},\:{BC},\:{CD} \\ $$$$\mathrm{and}\:{DA}. \\ $$$$\mathrm{Let}\:{X}\:=\:{AP}\:\cap\:{BQ},\:{Y}\:=\:{BQ}\:\cap\:{CM}, \\ $$$${Q}\:=\:{CM}\:\cap\:{DN}\:\mathrm{and}\:{T}=\:{DN}\:\cap\:{AP}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\left[{XYZT}\right]\:=\:\left[{AQX}\right]\:+\:\left[{BMY}\right] \\ $$$$+\:\left[{CNZ}\right]\:+\:\left[{DPT}\right]. \\ $$

Question Number 16944 Answers: 0 Comments: 2

Find the number of digits in the number 2^(2005) × 5^(2000) when written in full.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{2}^{\mathrm{2005}} \:×\:\mathrm{5}^{\mathrm{2000}} \:\mathrm{when}\:\mathrm{written}\:\mathrm{in} \\ $$$$\mathrm{full}. \\ $$

Question Number 16969 Answers: 0 Comments: 0

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