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Question Number 16893 Answers: 0 Comments: 4

if y=u^n show that d^n y/dx^n =n!

$$\mathrm{if}\:\mathrm{y}=\mathrm{u}^{\mathrm{n}} \:\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{d}^{\mathrm{n}} \mathrm{y}/\mathrm{dx}^{\mathrm{n}} =\mathrm{n}! \\ $$

Question Number 16892 Answers: 0 Comments: 0

∫_( 0) ^( ∞) ((ln(x))/(x^2 + 2x + 4)) dx

$$\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:+\:\mathrm{4}}\:\mathrm{dx} \\ $$

Question Number 16882 Answers: 0 Comments: 2

^• In plane parallel lines are the lines which don′t meet each other. ^• What is the condition in space that two lines be parallel?

$$\:^{\bullet} \mathrm{In}\:\mathrm{plane}\:\mathrm{parallel}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\mathrm{which}\:\mathrm{don}'\mathrm{t}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}. \\ $$$$ \\ $$$$\:^{\bullet} \mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{in}\:\mathrm{space} \\ $$$$\mathrm{that}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{be}\:\mathrm{parallel}? \\ $$

Question Number 16881 Answers: 2 Comments: 0

The quadratic polynomials p(x) = a(x − 3)^2 + bx + 1 and q(x) = 2x^2 + c(x − 2) + 13 are equal for all values of x. Find the values of a, b, and c.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{polynomials} \\ $$$${p}\left({x}\right)\:=\:{a}\left({x}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:+\:{bx}\:+\:\mathrm{1}\:\mathrm{and} \\ $$$${q}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:+\:{c}\left({x}\:−\:\mathrm{2}\right)\:+\:\mathrm{13}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a}, \\ $$$${b},\:\mathrm{and}\:{c}. \\ $$

Question Number 16880 Answers: 0 Comments: 0

Find the number of positive integers less than or equal to 300 that are multiples of 3 or 5, but are not multiples of 10 or 15.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{300}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{multiples}\:\mathrm{of}\:\mathrm{3}\:\mathrm{or}\:\mathrm{5},\:\mathrm{but}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{multiples}\:\mathrm{of}\:\mathrm{10}\:\mathrm{or}\:\mathrm{15}. \\ $$

Question Number 16879 Answers: 0 Comments: 0

Let ABCD be a parallelogram. The points M, N and P are chosen on the segments BD, BC and CD, respectively, so that CNMP is a parallelogram. Let E = AN ∩ BD and F = AP ∩ BD. Prove that [AEF] = [DFP] + [BEN].

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}.\:\mathrm{The} \\ $$$$\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{segments}\:{BD},\:{BC}\:\mathrm{and}\:{CD}, \\ $$$$\mathrm{respectively},\:\mathrm{so}\:\mathrm{that}\:{CNMP}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{parallelogram}.\:\mathrm{Let}\:{E}\:=\:{AN}\:\cap\:{BD}\:\mathrm{and} \\ $$$${F}\:=\:{AP}\:\cap\:{BD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{AEF}\right]\:=\:\left[{DFP}\right]\:+\:\left[{BEN}\right]. \\ $$

Question Number 16878 Answers: 0 Comments: 2

Let P be a point on the circumcircle of the equilateral triangle ABC. Prove that the projections of any point Q onto the lines PA, PB and PC are the vertices of an equilateral triangle.

$$\mathrm{Let}\:{P}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{projections}\:\mathrm{of}\:\mathrm{any}\:\mathrm{point}\:{Q} \\ $$$$\mathrm{onto}\:\mathrm{the}\:\mathrm{lines}\:{PA},\:{PB}\:\mathrm{and}\:{PC}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16877 Answers: 0 Comments: 0

From a point on the circumcircle of an equilateral triangle ABC parallels to the sides BC, CA and AB are drawn, intersecting the sides CA, AB and BC at the points M, N, P, respectively. Prove that the points M, N and P are collinear.

$$\mathrm{From}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{parallels}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:{CA},\:{AB}\:\mathrm{and}\:{BC} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N},\:{P},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are} \\ $$$$\mathrm{collinear}. \\ $$

Question Number 16876 Answers: 1 Comments: 0

In how many ways can a family of 5 brothers be seated round a table if (i) 2 brothers must seat next to each other. (ii) 2 brothers must not seat together.

$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{5}\:\mathrm{brothers}\:\mathrm{be}\:\mathrm{seated}\:\mathrm{round}\:\mathrm{a}\:\mathrm{table} \\ $$$$\mathrm{if}\:\left(\mathrm{i}\right)\:\mathrm{2}\:\mathrm{brothers}\:\mathrm{must}\:\mathrm{seat}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{2}\:\mathrm{brothers}\:\mathrm{must}\:\mathrm{not}\:\mathrm{seat}\:\mathrm{together}. \\ $$

Question Number 16875 Answers: 0 Comments: 0

Let P_1 , P_2 , ..., P_n be a convex polygon with the following property : for any two vertices P_i and P_j , there exists a vertex P_k such that the segment P_i P_j is seen from P_k under an angle of 60°. Prove that the polygon is an equilateral triangle.

$$\mathrm{Let}\:{P}_{\mathrm{1}} ,\:{P}_{\mathrm{2}} ,\:...,\:{P}_{{n}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{polygon} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{following}\:\mathrm{property}\::\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{two}\:\mathrm{vertices}\:{P}_{{i}} \:\mathrm{and}\:{P}_{{j}} ,\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a} \\ $$$$\mathrm{vertex}\:{P}_{{k}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segment}\:{P}_{{i}} {P}_{{j}} \\ $$$$\mathrm{is}\:\mathrm{seen}\:\mathrm{from}\:{P}_{{k}} \:\mathrm{under}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}°. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16874 Answers: 0 Comments: 0

Let ABC be an acute triangle. The interior bisectors of the angles ∠B and ∠C meet the opposite sides at the points L and M, respectively. Prove that there exists a point K in the interior of the side BC such that ΔKLM is equilateral if and only if ∠A = 60°.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{triangle}.\:\mathrm{The} \\ $$$$\mathrm{interior}\:\mathrm{bisectors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angles}\:\angle{B}\:\mathrm{and} \\ $$$$\angle{C}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{L}\:\mathrm{and}\:{M},\:\mathrm{respectively}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{point}\:{K}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{BC}\:\mathrm{such}\:\mathrm{that} \\ $$$$\Delta{KLM}\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\angle{A}\:=\:\mathrm{60}°. \\ $$

Question Number 16911 Answers: 0 Comments: 1

Please solve Q. 16066 or please tell me whether to post its solution or not. I don′t understood the solution.

$$\mathrm{Please}\:\mathrm{solve}\:\mathrm{Q}.\:\mathrm{16066}\:\mathrm{or}\:\mathrm{please}\:\mathrm{tell}\:\mathrm{me} \\ $$$$\mathrm{whether}\:\mathrm{to}\:\mathrm{post}\:\mathrm{its}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{not}.\:\mathrm{I} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{understood}\:\mathrm{the}\:\mathrm{solution}. \\ $$

Question Number 16873 Answers: 0 Comments: 0

Let I be the incenter of ΔABC. It is known that for every point M ∈ (AB), one can find the points N ∈ (BC) and P ∈ (AC) such that I is the centroid of ΔMNP. Prove that ABC is an equilateral triangle.

$$\mathrm{Let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\Delta{ABC}.\:\mathrm{It}\:\mathrm{is} \\ $$$$\mathrm{known}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\:\mathrm{point}\:{M}\:\in\:\left({AB}\right), \\ $$$$\mathrm{one}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{points}\:{N}\:\in\:\left({BC}\right)\:\mathrm{and} \\ $$$${P}\:\in\:\left({AC}\right)\:\mathrm{such}\:\mathrm{that}\:{I}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of} \\ $$$$\Delta{MNP}.\:\mathrm{Prove}\:\mathrm{that}\:{ABC}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16868 Answers: 1 Comments: 0

In how many ways can the letters of the word. EVERMORE be arrange if the word must begin with (i) R (ii) E

$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{word}.\:\mathrm{EVERMORE}\:\mathrm{be}\:\mathrm{arrange} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{word}\:\mathrm{must}\:\mathrm{begin}\:\mathrm{with}\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{R} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{E} \\ $$

Question Number 16859 Answers: 0 Comments: 0

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

$$\mathrm{In}\:\mathrm{dealing}\:\mathrm{with}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile}\:\mathrm{in} \\ $$$$\mathrm{air},\:\mathrm{we}\:\mathrm{ignore}\:\mathrm{effect}\:\mathrm{of}\:\mathrm{air}\:\mathrm{resistance} \\ $$$$\mathrm{on}\:\mathrm{motion}.\:\mathrm{What}\:\mathrm{would}\:\mathrm{the}\:\mathrm{trajectory} \\ $$$$\mathrm{look}\:\mathrm{like}\:\mathrm{if}\:\mathrm{air}\:\mathrm{resistance}\:\mathrm{is}\:\mathrm{included}? \\ $$$$\mathrm{Sketch}\:\mathrm{such}\:\mathrm{a}\:\mathrm{trajectory}\:\mathrm{and}\:\mathrm{explain} \\ $$$$\mathrm{why}\:\mathrm{you}\:\mathrm{have}\:\mathrm{drawn}\:\mathrm{it}\:\mathrm{that}\:\mathrm{way}. \\ $$

Question Number 16857 Answers: 1 Comments: 0

Suppose x and y are vectors in R^n that have the same length. show that x + y bisect the angle between x and y.

$$\mathrm{Suppose}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{vectors}\:\mathrm{in}\:\mathbb{R}^{\mathrm{n}} \:\mathrm{that}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{length}.\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:\:\mathrm{bisect}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}. \\ $$

Question Number 16855 Answers: 1 Comments: 0

If sin θ = (1/2), cos φ = (1/3), then θ + φ belongs to, where 0 < θ, φ < (π/2) (1) ((π/3), (π/2)) (2) ((π/2), ((2π)/3))

$$\mathrm{If}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{cos}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{then}\:\theta\:+\:\phi \\ $$$$\mathrm{belongs}\:\mathrm{to},\:\mathrm{where}\:\mathrm{0}\:<\:\theta,\:\phi\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\left(\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$

Question Number 16845 Answers: 1 Comments: 3