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Question Number 26839 Answers: 1 Comments: 0

a=3 b=6 a−b=?

$$\mathrm{a}=\mathrm{3}\:\mathrm{b}=\mathrm{6} \\ $$$$\mathrm{a}−\mathrm{b}=? \\ $$

Question Number 26812 Answers: 1 Comments: 0

sum of infinite seris tan^(−1) (2/n^2 )

$${sum}\:{of}\:{infinite}\:{seris} \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}/{n}^{\mathrm{2}} \right) \\ $$

Question Number 26801 Answers: 0 Comments: 0

find expansion of α^3 +β^3 +γ^3

$${find}\:{expansion}\:{of}\:\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \\ $$

Question Number 26686 Answers: 1 Comments: 1

Question Number 26681 Answers: 0 Comments: 1

f:R×R→R such that f(x+iy)=(√(x^2 +y^2 .)) Then f is a) many−one and into function b) one−one and onto function c) many−one and onto function d) one−one and into function

$${f}:{R}×{R}\rightarrow{R}\:{such}\:{that}\:{f}\left({x}+{iy}\right)=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} .} \\ $$$${Then}\:{f}\:{is} \\ $$$$\left.{a}\right)\:{many}−{one}\:{and}\:{into}\:{function} \\ $$$$\left.{b}\right)\:{one}−{one}\:{and}\:{onto}\:{function} \\ $$$$\left.{c}\right)\:{many}−{one}\:{and}\:{onto}\:{function} \\ $$$$\left.{d}\right)\:{one}−{one}\:{and}\:{into}\:{function} \\ $$

Question Number 26634 Answers: 1 Comments: 1

Question Number 26625 Answers: 1 Comments: 0

3y−2x+7=0 x^2 −4y^2 −21=0

$$\mathrm{3y}−\mathrm{2x}+\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4y}^{\mathrm{2}} −\mathrm{21}=\mathrm{0} \\ $$

Question Number 26623 Answers: 2 Comments: 0

distance between 2 places A and B on road is 70 km. a car starts from A and other from B .if they travel in same direction they will meet after 7 hours. if they travel towards each other they will meet after 1 hour then find their speeds

$$\mathrm{distance}\:\mathrm{between}\:\mathrm{2}\:\mathrm{places}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{on} \\ $$$$\mathrm{road}\:\mathrm{is}\:\mathrm{70}\:\mathrm{km}.\:\mathrm{a}\:\mathrm{car}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{A}\:\mathrm{and}\:\mathrm{other}\: \\ $$$$\mathrm{from}\:\mathrm{B}\:.\mathrm{if}\:\mathrm{they}\:\mathrm{travel}\:\mathrm{in}\:\mathrm{same}\:\mathrm{direction} \\ $$$$\mathrm{they}\:\mathrm{will}\:\mathrm{meet}\:\mathrm{after}\:\mathrm{7}\:\mathrm{hours}.\:\mathrm{if}\:\mathrm{they}\:\mathrm{travel} \\ $$$$\mathrm{towards}\:\mathrm{each}\:\mathrm{other}\:\mathrm{they}\:\mathrm{will}\:\mathrm{meet}\:\mathrm{after} \\ $$$$\mathrm{1}\:\mathrm{hour}\:\mathrm{then}\:\mathrm{find}\:\mathrm{their}\:\mathrm{speeds} \\ $$

Question Number 26581 Answers: 0 Comments: 0

Question Number 26580 Answers: 0 Comments: 0

Question Number 26544 Answers: 1 Comments: 0

∫1/x^2 +y^2 dydx

$$\int\mathrm{1}/{x}^{\mathrm{2}} +{y}^{\mathrm{2}} {dydx} \\ $$

Question Number 26601 Answers: 1 Comments: 0

Use polar co-ordinates to evaluate ∫∫_R e^(−(x^2 +y^2 )) dA, where the region R is enclosed by the circle x^2 +y^2 =1.

$${Use}\:{polar}\:{co}-{ordinates}\:{to}\:{evaluate}\:\int\underset{{R}} {\int}{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dA},\:{where}\:{the}\:{region}\:{R}\:{is}\:{enclosed}\:{by}\:{the}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}. \\ $$

Question Number 26528 Answers: 1 Comments: 0

f length of a rectangle is reduced by 5 umits and its breadth is increasesd 3 units then area of rectangle is reduced by 8 sq units if lenghth is reduced 3 units and breadth is increased by 2 units then area of rectangle will increased by 67 sq units . then find length and breadth of rectangle

$${f}\:{length}\:\:{of}\:{a}\:{rectangle}\:{is}\:{reduced}\:{by}\:\mathrm{5}\: \\ $$$${umits}\:{and}\:{its}\:{breadth}\:{is}\:{increasesd} \\ $$$$\mathrm{3}\:{units}\:{then}\:{area}\:{of}\:{rectangle}\:{is}\:{reduced} \\ $$$${by}\:\mathrm{8}\:{sq}\:{units}\:{if}\:{lenghth}\:{is}\:{reduced} \\ $$$$\mathrm{3}\:{units}\:{and}\:{breadth}\:{is}\:{increased}\:{by}\: \\ $$$$\mathrm{2}\:{units}\:{then}\:{area}\:{of}\:{rectangle} \\ $$$${will}\:{increased}\:{by}\:\mathrm{67}\:{sq}\:{units}\:.\: \\ $$$${then}\:{find}\:{length}\:{and}\:{breadth}\:{of}\:{rectangle} \\ $$

Question Number 26513 Answers: 0 Comments: 2

let n be a fixed positive integer. How many ways are there to write n as a sum of positive integers, n=a_1 +a_2 +...+a_k with k arbitary positive integer and a_1 ≤a_2 ...≤a_k ≤a_1 +1. for example with n=4, there are four ways : 4, 2+2, 1+1+2,1+1+1+1

$$\mathrm{let}\:{n}\:\mathrm{be}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{positive}\:\mathrm{integer}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{are}\:\mathrm{there}\:\mathrm{to}\:\mathrm{write}\:{n}\:\mathrm{as}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{integers},\: \\ $$$${n}={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{k}} \\ $$$$\mathrm{with}\:{k}\:\mathrm{arbitary}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{and}\:{a}_{\mathrm{1}} \leqslant{a}_{\mathrm{2}} ...\leqslant{a}_{{k}} \leqslant{a}_{\mathrm{1}} +\mathrm{1}.\:\mathrm{for}\:\mathrm{example} \\ $$$$\mathrm{with}\:{n}=\mathrm{4},\:\mathrm{there}\:\mathrm{are}\:\mathrm{four}\:\mathrm{ways}\::\:\mathrm{4},\:\mathrm{2}+\mathrm{2},\:\mathrm{1}+\mathrm{1}+\mathrm{2},\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1} \\ $$

Question Number 26456 Answers: 0 Comments: 1

Question Number 26455 Answers: 0 Comments: 0

Question Number 26454 Answers: 0 Comments: 0

Question Number 26442 Answers: 0 Comments: 1

find the lateral surface area of cuboid = L=13m 12m and 10m

$${find}\:{the}\:{lateral}\:{surface}\:{area}\:{of}\:{cuboid}\:= \\ $$$${L}=\mathrm{13}{m}\:\mathrm{12}{m}\:{and}\:\mathrm{10}{m} \\ $$

Question Number 26425 Answers: 0 Comments: 0

A body is projected vertically upward with an initial velocity of u. Another Another body is projected with the same initial velocity, t seconds after the first. If T is the time when the two bodies meet, and g the acceleration due to gravity, Show that T = ((2u + gt)/(2g))

$$\mathrm{A}\:\mathrm{body}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically}\:\mathrm{upward}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{u}.\:\mathrm{Another} \\ $$$$\mathrm{Another}\:\mathrm{body}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same}\:\mathrm{initial}\:\mathrm{velocity},\:\mathrm{t}\:\mathrm{seconds}\:\mathrm{after} \\ $$$$\mathrm{the}\:\mathrm{first}.\:\mathrm{If}\:\:\mathrm{T}\:\mathrm{is}\:\mathrm{the}\:\mathrm{time}\:\mathrm{when}\:\mathrm{the}\:\mathrm{two}\:\mathrm{bodies}\:\mathrm{meet},\:\mathrm{and}\:\mathrm{g}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{gravity},\:\mathrm{Show}\:\mathrm{that}\:\:\:\boldsymbol{\mathrm{T}}\:=\:\frac{\mathrm{2u}\:+\:\mathrm{gt}}{\mathrm{2g}} \\ $$

Question Number 26424 Answers: 0 Comments: 0

The front of a train 80m long passes a signal at a speed of 72km/hr. If the rear of the train passes the signal 5s later, Find (a) The magnitude of the acceleration of the train. (b) The speed at which the rear of the train passes the signal.

$$\mathrm{The}\:\mathrm{front}\:\mathrm{of}\:\mathrm{a}\:\mathrm{train}\:\mathrm{80m}\:\mathrm{long}\:\mathrm{passes}\:\mathrm{a}\:\mathrm{signal}\:\mathrm{at}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{72km}/\mathrm{hr}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{train}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{signal}\:\mathrm{5s}\:\mathrm{later},\:\mathrm{Find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{train}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{speed}\:\mathrm{at}\:\mathrm{which}\:\mathrm{the}\:\mathrm{rear}\:\mathrm{of}\:\mathrm{the}\:\mathrm{train}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{signal}. \\ $$

Question Number 26382 Answers: 0 Comments: 1

Question Number 26381 Answers: 0 Comments: 9

Question Number 26380 Answers: 0 Comments: 0

Question Number 26352 Answers: 1 Comments: 0

x(x+9)=(x+3)(x+7)−10

$${x}\left({x}+\mathrm{9}\right)=\left({x}+\mathrm{3}\right)\left({x}+\mathrm{7}\right)−\mathrm{10} \\ $$

Question Number 26329 Answers: 1 Comments: 0

A small particle moving with a uniform acceleration a covers distances X and Y in the first two equal and consecutive intervals of time t. Show that a = ((Y − X)/t^2 )

$$\mathrm{A}\:\mathrm{small}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{a}\:\mathrm{covers}\:\mathrm{distances}\: \\ $$$$\mathrm{X}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{and}\:\mathrm{consecutive}\:\mathrm{intervals}\:\mathrm{of}\:\mathrm{time}\:\mathrm{t}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{a}\:=\:\frac{\mathrm{Y}\:−\:\mathrm{X}}{\mathrm{t}^{\mathrm{2}} } \\ $$

Question Number 26328 Answers: 1 Comments: 1

Three towns X, Y and Z are on a straight road and Y is the mid−way between X and Z. A motor cyclist moving with uniform acceleration passes X, Y and Z. The speed with which the motocyclist passes X and Z are 20m/s and 40m/s respectively. Find the speed with which the motorcyclist passes Y.

$$\mathrm{Three}\:\mathrm{towns}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{road}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}−\mathrm{way}\:\mathrm{between} \\ $$$$\mathrm{X}\:\mathrm{and}\:\mathrm{Z}.\:\mathrm{A}\:\mathrm{motor}\:\mathrm{cyclist}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{uniform}\:\mathrm{acceleration}\:\mathrm{passes}\:\mathrm{X},\:\mathrm{Y}\:\mathrm{and}\:\mathrm{Z}.\: \\ $$$$\mathrm{The}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motocyclist}\:\mathrm{passes}\:\mathrm{X}\:\mathrm{and}\:\mathrm{Z}\:\mathrm{are}\:\mathrm{20m}/\mathrm{s}\:\mathrm{and}\:\mathrm{40m}/\mathrm{s}\: \\ $$$$\mathrm{respectively}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the}\:\mathrm{motorcyclist}\:\mathrm{passes}\:\mathrm{Y}. \\ $$

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