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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.

ThreetennisplayersA,BandCplayeachotheronlyonce.TheprobabilitythatAwillbeatBis35,thatBwillbeatCis23,andthatAwillbeatCis57.Find(1)theprobabilitythatAwillnotwinbothgames(2)theprobabilitythatAwillwinnotbothgames.

Question Number 19555    Answers: 1   Comments: 0

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

Findcenterandradiusofcirclehavingequationzz¯+(1i)z+(1+i)z¯1=0.

Question Number 19391    Answers: 0   Comments: 1

Question Number 19389    Answers: 1   Comments: 0

What is the digital root of 3^(2017)

Whatisthedigitalrootof32017

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?

relatedtoQ.19333thesidelengthesofatriangleareinteger.iftheperimeterofthetriangleis100,howmanydifferenttrianglesexist?whatisthemaximumareaofthem?

Question Number 19385    Answers: 0   Comments: 5

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

tan2β=1findβ....letssolveforfun

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

Iflog2(9x1+7)log2(3x1+1)=2,thenthevaluesofxare

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

Ifα,β,γandδarefoursolutionsoftheequationtan(θ+5π4)=3tan3θ,then(1)Σtanα=0(2)Σtanαtanβ=2(3)Σtanαtanβtanγ=83(4)tanαtanβtanγtanδ=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)

TheblockQmovestotherightwithaconstantvelocityv0asshowninfigure.TherelativevelocityofbodyPwithrespecttoQis(assumeallpulleysandstringsareideal)

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

Inthearrangementshowninfiguretwobeadsslidealongasmoothhorizontalrod.Therelationbetweenvandv0ingivenpositionwillbe

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

Twoblocksareplacedonasmoothhorizontalsurfaceandconnectedbyastringpulleyarrangementasshown.IfaforceFstartsactingonblockm1,thenfindtherelationbetweenaccelerationofbothmassesandtheirvalues

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 )

Provethatz1z22=z12+z222z1z2cos(θ1θ2)

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.

Provethatz1+z22=z12+z22z1z2ispurelyimaginarynumber.

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)

Provethatz1+z2=z1z2arg(z1)arg(z2)=π2

Question Number 19349    Answers: 1   Comments: 2

Prove that ∣z_1 + z_2 + z_3 + .... + z_n ∣ ≤ ∣z_1 ∣ + ∣z_2 ∣ + ∣z_3 ∣ + .... + ∣z_n ∣

Provethatz1+z2+z3+....+znz1+z2+z3+....+zn

Question Number 19452    Answers: 1   Comments: 0

Question Number 19345    Answers: 0   Comments: 0

A thin bi − convex lens rest on a plane mirror . it is found that a point objects placed 20cm above the object coincide with it own image. Determine the position and nature of the image when the object is placed (i) 8cm and (ii) 12 from the lens mirror combinatiom

Athinbiconvexlensrestonaplanemirror.itisfoundthatapointobjectsplaced20cmabovetheobjectcoincidewithitownimage.Determinethepositionandnatureoftheimagewhentheobjectisplaced(i)8cmand(ii)12fromthelensmirrorcombinatiom

Question Number 19341    Answers: 0   Comments: 2

Question Number 19476    Answers: 1   Comments: 0

STATEMENT-1 : The graph between kinetic energy and vertical displacement is a straight line for a projectile. STATEMENT-2 : The graph between kinetic energy and horizontal displacement is a straight line for a projectile. STATEMENT-3 : The graph between kinetic energy and time is a parabola for a projectile.

STATEMENT1:Thegraphbetweenkineticenergyandverticaldisplacementisastraightlineforaprojectile.STATEMENT2:Thegraphbetweenkineticenergyandhorizontaldisplacementisastraightlineforaprojectile.STATEMENT3:Thegraphbetweenkineticenergyandtimeisaparabolaforaprojectile.

Question Number 19332    Answers: 1   Comments: 1

Let S_n = n^2 + 20n + 12, n a positive integer. What is the sum of all possible values of n for which S_n is a perfect square?

LetSn=n2+20n+12,napositiveinteger.WhatisthesumofallpossiblevaluesofnforwhichSnisaperfectsquare?

Question Number 19405    Answers: 1   Comments: 0

Convert i(√((2(√2)−1)/2)) into polarform.

Converti2212intopolarform.

Question Number 19330    Answers: 1   Comments: 0

A triangle with perimeter 7 has integer side lengths. What is the maximum possible area of such a triangle?

Atrianglewithperimeter7hasintegersidelengths.Whatisthemaximumpossibleareaofsuchatriangle?

Question Number 19329    Answers: 0   Comments: 5

Solve the equation y^3 = x^3 + 8x^2 − 6x + 8 for positive integers x and y.

Solvetheequationy3=x3+8x26x+8forpositiveintegersxandy.

Question Number 19324    Answers: 0   Comments: 0

y=tan^(−1) 3a^2 x−x^3 /a(a^2 −3x^2 )

y=tan13a2xx3/a(a23x2)

Question Number 19323    Answers: 0   Comments: 0

y=sin (2tan^(−1) (√(1−x/1+x)))

y=sin(2tan11x/1+x)

Question Number 19313    Answers: 1   Comments: 0

Prove that ∣z_1 ± z_2 ∣^2 = ∣z_2 ∣^2 + ∣z_1 ∣^2 ± 2Re(z_1 z_2 ^ ) = ∣z_1 ∣^2 + ∣z_2 ∣^2 ± 2Re(z_1 ^ .z_2 )

Provethatz1±z22=z22+z12±2Re(z1z¯2)=z12+z22±2Re(z¯1.z2)

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