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Question Number 26486 Answers: 1 Comments: 5

∫_( −1) ^( 1) ((xdx)/((x^2 + 1)^2 ))

$$\int_{\:−\mathrm{1}} ^{\:\:\mathrm{1}} \frac{\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 26495 Answers: 0 Comments: 0

Question Number 26480 Answers: 1 Comments: 0

Find the number of all possible different words into which the word INTERFERE can be converted by change of place of letters,if no two consonants must be together.

$${Find}\:{the}\:{number}\:{of}\:{all}\:{possible} \\ $$$${different}\:{words}\:{into}\:{which}\:{the} \\ $$$${word}\:{INTERFERE}\:{can}\:{be}\: \\ $$$${converted}\:{by}\:{change}\:{of}\:{place}\:{of}\: \\ $$$${letters},{if}\:{no}\:{two}\:{consonants} \\ $$$${must}\:{be}\:{together}. \\ $$

Question Number 26477 Answers: 1 Comments: 0

A number of four different digits is formed by using the digits 1,2,3,4,5,6,7,in all possible ways each digit occuring once only.find how many of them are greater than 3400

$${A}\:{number}\:{of}\:{four}\:{different} \\ $$$${digits}\:{is}\:{formed}\:{by}\:{using}\:{the}\: \\ $$$${digits}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},{in}\:{all}\:{possible} \\ $$$${ways}\:{each}\:{digit}\:{occuring}\:{once} \\ $$$${only}.{find}\:{how}\:{many}\:{of}\:{them}\:{are} \\ $$$${greater}\:{than}\:\mathrm{3400} \\ $$

Question Number 26473 Answers: 1 Comments: 0

How many numbers less than 1000 and divisible by 5 can be formed with the digits 0, 1, 2 ,3 ,4 ,5 6 ,7 ,8 ,9,each digit not occuring more than once in each number?

$${How}\:{many}\:{numbers}\:{less}\:{than} \\ $$$$\mathrm{1000}\:{and}\:{divisible}\:{by}\:\mathrm{5}\:{can}\:{be} \\ $$$${formed}\:{with}\:{the}\:{digits}\:\mathrm{0},\:\mathrm{1},\:\mathrm{2}\:,\mathrm{3}\:,\mathrm{4}\:,\mathrm{5}\: \\ $$$$\mathrm{6}\:,\mathrm{7}\:,\mathrm{8}\:,\mathrm{9},{each}\:{digit}\:{not}\:{occuring} \\ $$$${more}\:{than}\:{once}\:{in}\:{each}\:{number}? \\ $$

Question Number 26463 Answers: 0 Comments: 1

∫_0 ^(2π) cos^2 θsin θdθ

$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}^{\mathrm{2}} \theta\mathrm{sin}\:\theta{d}\theta \\ $$

Question Number 26461 Answers: 0 Comments: 1

Question Number 26456 Answers: 0 Comments: 1

Question Number 26455 Answers: 0 Comments: 0

Question Number 26454 Answers: 0 Comments: 0

Question Number 26452 Answers: 1 Comments: 0

∫(√(tanx))

$$\int\sqrt{{tanx}} \\ $$

Question Number 26450 Answers: 2 Comments: 5

A man of mass M is standing on a platform of mass m_1 holding a string passing over a system of ideal pulleys. Another mass m_2 is hanging as shown (m_2 = 20 kg, m_1 = 10 kg, g = 10 m/s^2 ) Force exerted by man on string to accelerate upwards

$$\mathrm{A}\:\mathrm{man}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{is}\:\mathrm{standing}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{platform}\:\mathrm{of}\:\mathrm{mass}\:{m}_{\mathrm{1}} \:\mathrm{holding}\:\mathrm{a}\:\mathrm{string} \\ $$$$\mathrm{passing}\:\mathrm{over}\:\mathrm{a}\:\mathrm{system}\:\mathrm{of}\:\mathrm{ideal}\:\mathrm{pulleys}. \\ $$$$\mathrm{Another}\:\mathrm{mass}\:{m}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{hanging}\:\mathrm{as}\:\mathrm{shown} \\ $$$$\left({m}_{\mathrm{2}} \:=\:\mathrm{20}\:\mathrm{kg},\:{m}_{\mathrm{1}} \:=\:\mathrm{10}\:\mathrm{kg},\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$$$\mathrm{Force}\:\mathrm{exerted}\:\mathrm{by}\:\mathrm{man}\:\mathrm{on}\:\mathrm{string}\:\mathrm{to} \\ $$$$\mathrm{accelerate}\:\mathrm{upwards} \\ $$

Question Number 26449 Answers: 1 Comments: 0

Transform the equation 5x^2 + 4xy + 2y^2 − 2x + 4y + 4 = 0 into one without xy, x and y terms.

$${Transform}\:{the}\:{equation}\:\mathrm{5}{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy} \\ $$$$+\:\mathrm{2}{y}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{4}{y}\:+\:\mathrm{4}\:=\:\mathrm{0}\:{into}\:{one} \\ $$$${without}\:{xy},\:{x}\:{and}\:{y}\:{terms}. \\ $$

Question Number 26447 Answers: 0 Comments: 1

Question Number 26446 Answers: 1 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 26438 Answers: 1 Comments: 0

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 26405 Answers: 1 Comments: 0