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Question Number 23687 Answers: 1 Comments: 0

Solve for x, y and z x^2 − yz = 1 .......... (i) y^2 − xz = 4 .......... (i) z^2 − xy = 9 .......... (i)

$$\mathrm{Solve}\:\mathrm{for}\:\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{yz}\:=\:\mathrm{1}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$$$\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{xz}\:=\:\mathrm{4}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$$$\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{xy}\:=\:\mathrm{9}\:\:\:\:\:\:..........\:\left(\mathrm{i}\right) \\ $$

Question Number 23686 Answers: 0 Comments: 0

A function f is defined by f : x → 3 − 2sin(x), for 0 ≤ x ≤ 360° find the range of f

$$\mathrm{A}\:\mathrm{function}\:\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:\:\mathrm{f}\::\:\mathrm{x}\:\rightarrow\:\mathrm{3}\:−\:\mathrm{2sin}\left(\mathrm{x}\right),\:\:\mathrm{for}\:\:\mathrm{0}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{360}° \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\:\mathrm{f} \\ $$

Question Number 23683 Answers: 0 Comments: 0

Question Number 23682 Answers: 0 Comments: 0

Question Number 23679 Answers: 2 Comments: 0

Question Number 23718 Answers: 0 Comments: 1

5(tan^2 x−cos^2 x)=2cosx+a then find then find the value of cos4x.???

$$\mathrm{5}\left(\mathrm{tan}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{2}} {x}\right)=\mathrm{2}{cosx}+{a}\:{then}\:{find} \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:{cos}\mathrm{4}{x}.??? \\ $$

Question Number 23739 Answers: 0 Comments: 3

Question Number 23666 Answers: 2 Comments: 1

A uniform rope of length L and mass per unit λ having one end fixed with the ceiling is released from rest. Find the tension in the fixed end as a function of the distance travelled by the movable end.

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{rope}\:\mathrm{of}\:\mathrm{length}\:{L}\:\mathrm{and}\:\mathrm{mass} \\ $$$$\mathrm{per}\:\mathrm{unit}\:\lambda\:\mathrm{having}\:\mathrm{one}\:\mathrm{end}\:\mathrm{fixed}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{ceiling}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{tension}\:\mathrm{in}\:\mathrm{the}\:\mathrm{fixed}\:\mathrm{end}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{the}\:\mathrm{movable} \\ $$$$\mathrm{end}. \\ $$

Question Number 23677 Answers: 0 Comments: 0

solve lim_(x→inf+) ∫^(2(√x)) _(2sin(1/x)) ((2t^4 +1)/((t−3)(t^3 +3))) dt

$${solve} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\underset{{x}\rightarrow{inf}+} {\mathrm{li}{m}}\:\:\underset{\mathrm{2}{sin}\frac{\mathrm{1}}{{x}}} {\int}^{\mathrm{2}\sqrt{{x}}} \frac{\mathrm{2}{t}^{\mathrm{4}} +\mathrm{1}}{\left({t}−\mathrm{3}\right)\left({t}^{\mathrm{3}} +\mathrm{3}\right)}\:{dt} \\ $$

Question Number 23663 Answers: 1 Comments: 3

solve ∫^1_ _(−1) x^2 d(lnx)

$${solve} \\ $$$$ \\ $$$$\underset{−\mathrm{1}} {\int}^{\mathrm{1}_{} } {x}^{\mathrm{2}} {d}\left({lnx}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 23657 Answers: 0 Comments: 1

Mr.Ajfour ,which app do you use for making these diagrams?

$${Mr}.{Ajfour}\:,{which}\:{app}\:{do}\:{you} \\ $$$${use}\:{for}\:{making}\:{these}\:{diagrams}? \\ $$

Question Number 23648 Answers: 1 Comments: 0

If sin(3θ + α) + sin(3θ − α) + sin(α − θ) − sin(α + θ) = cosα and cosα ≠ 0, then which of the values of θ does not satisfy the given equation? (1) nπ + (−1)^n (π/6), n ∈ I (2) nπ + (−1)^n (π/(10)), n ∈ I (3) nπ + (−1)^n (π/5), n ∈ I (4) nπ − (−1)^n ((3π)/(10)), n ∈ I

$$\mathrm{If}\:\mathrm{sin}\left(\mathrm{3}\theta\:+\:\alpha\right)\:+\:\mathrm{sin}\left(\mathrm{3}\theta\:−\:\alpha\right)\:+\:\mathrm{sin}\left(\alpha\:−\:\theta\right) \\ $$$$−\:\mathrm{sin}\left(\alpha\:+\:\theta\right)\:=\:\mathrm{cos}\alpha\:\mathrm{and}\:\mathrm{cos}\alpha\:\neq\:\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\theta\:\mathrm{does}\:\mathrm{not}\:\mathrm{satisfy} \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{equation}? \\ $$$$\left(\mathrm{1}\right)\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \:\frac{\pi}{\mathrm{6}},\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{2}\right)\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \:\frac{\pi}{\mathrm{10}},\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{3}\right)\:{n}\pi\:+\:\left(−\mathrm{1}\right)^{{n}} \:\frac{\pi}{\mathrm{5}},\:{n}\:\in\:{I} \\ $$$$\left(\mathrm{4}\right)\:{n}\pi\:−\:\left(−\mathrm{1}\right)^{{n}} \:\frac{\mathrm{3}\pi}{\mathrm{10}},\:{n}\:\in\:{I} \\ $$

Question Number 23644 Answers: 0 Comments: 0

Prove that Σ_(r=0) ^n r.^n C_r z^r = nz(1 + z)^(n−1)

$${Prove}\:{that}\:\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{r}.^{{n}} {C}_{{r}} \:{z}^{{r}} \:=\:{nz}\left(\mathrm{1}\:+\:{z}\right)^{{n}−\mathrm{1}} \\ $$

Question Number 23645 Answers: 0 Comments: 1

Ethylene on combustion gives carbon dioxide and water. Its enthalpy of combustion is 1410 kJ/mol. If the enthalpy of formation of CO_2 and H_2 O are 393.3 kJ and 286.2 kJ respectively. Calculate the enthalpy of formation of ethylene.

$$\mathrm{Ethylene}\:\mathrm{on}\:\mathrm{combustion}\:\mathrm{gives}\:\mathrm{carbon} \\ $$$$\mathrm{dioxide}\:\mathrm{and}\:\mathrm{water}.\:\mathrm{Its}\:\mathrm{enthalpy}\:\mathrm{of} \\ $$$$\mathrm{combustion}\:\mathrm{is}\:\mathrm{1410}\:\mathrm{kJ}/\mathrm{mol}.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of}\:\mathrm{CO}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\mathrm{are}\:\mathrm{393}.\mathrm{3}\:\mathrm{kJ}\:\mathrm{and}\:\mathrm{286}.\mathrm{2}\:\mathrm{kJ}\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{enthalpy}\:\mathrm{of}\:\mathrm{formation}\:\mathrm{of} \\ $$$$\mathrm{ethylene}. \\ $$

Question Number 23632 Answers: 0 Comments: 9

Question Number 23617 Answers: 2 Comments: 3

The system is released from rest. All surfaces are smooth. Find the angle θ at which the acceleration of wedge is maximum. (given (M/m) = (1/2))

$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{rest}.\:\mathrm{All} \\ $$$$\mathrm{surfaces}\:\mathrm{are}\:\mathrm{smooth}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\theta\:\mathrm{at} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{wedge}\:\mathrm{is} \\ $$$$\mathrm{maximum}.\:\left(\mathrm{given}\:\frac{{M}}{{m}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Question Number 23612 Answers: 0 Comments: 2

Find the last 2 digits from 20^(17) + 17^(20)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{digits}\:\mathrm{from} \\ $$$$\mathrm{20}^{\mathrm{17}} \:+\:\mathrm{17}^{\mathrm{20}} \\ $$

Question Number 23592 Answers: 1 Comments: 0

Let ABC be a triangle with AB = AC and ∠BAC = 30°. Let A′ be the reflection of A in the line BC; B′ be the reflection of B in the line CA; C′ be the reflection of C in the line AB. Show that A′, B′, C′ form the vertices of an equilateral triangle.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:{AB}\:=\:{AC} \\ $$$$\mathrm{and}\:\angle{BAC}\:=\:\mathrm{30}°.\:\mathrm{Let}\:{A}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{A}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{BC};\:{B}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{B}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{CA};\:{C}'\:\mathrm{be}\:\mathrm{the}\:\mathrm{reflection} \\ $$$$\mathrm{of}\:{C}\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:{AB}.\:\mathrm{Show}\:\mathrm{that}\:{A}',\:{B}',\:{C}' \\ $$$$\mathrm{form}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$

Question Number 23566 Answers: 1 Comments: 2

((a/b))^(log c) .((b/c))^(log a) .((c/a))^(log b) =1. make ((logc)/c) the subject of formula

$$\left(\frac{{a}}{{b}}\right)^{\mathrm{log}\:{c}} .\left(\frac{{b}}{{c}}\right)^{\mathrm{log}\:{a}} .\left(\frac{{c}}{{a}}\right)^{\mathrm{log}\:{b}} =\mathrm{1}. \\ $$$$ \\ $$$${make}\:\frac{{logc}}{{c}}\:{the}\:{subject}\:{of}\:{formula} \\ $$

Question Number 23564 Answers: 1 Comments: 4

The curved surface area of a cone is 21cm^2 .Calculate the curved surface area of a similar cone whose height is 4 times the other.

$${The}\:{curved}\:{surface}\:{area}\:{of}\:{a}\:{cone} \\ $$$${is}\:\mathrm{21}{cm}^{\mathrm{2}} \:.{Calculate}\:{the}\:{curved} \\ $$$${surface}\:{area}\:{of}\:{a}\:{similar}\:{cone} \\ $$$${whose}\:{height}\:{is}\:\mathrm{4}\:{times}\:{the}\:{other}. \\ $$

Question Number 23559 Answers: 1 Comments: 1

Question Number 23556 Answers: 1 Comments: 0

A ball falls vertically on to a floor, with momentum p, and then bounces repeatedly, the coefficient of restitution is e. The total momentum imparted by the ball to the floor is

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{falls}\:\mathrm{vertically}\:\mathrm{on}\:\mathrm{to}\:\mathrm{a}\:\mathrm{floor},\:\mathrm{with} \\ $$$$\mathrm{momentum}\:{p},\:\mathrm{and}\:\mathrm{then}\:\mathrm{bounces} \\ $$$$\mathrm{repeatedly},\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{restitution} \\ $$$$\mathrm{is}\:{e}.\:\mathrm{The}\:\mathrm{total}\:\mathrm{momentum}\:\mathrm{imparted}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{to}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{is} \\ $$

Question Number 23548 Answers: 0 Comments: 3

Question Number 23585 Answers: 0 Comments: 0

No work is done by a force on an object if (1) the object is stationary but the point of application of the force moves on the object (2) the object moves in such a way that the point of application of the force remains fixed (3) the force is always perpendicular to its velocity (4) the force is always perpendicular to its acceleration.

$$\mathrm{No}\:\mathrm{work}\:\mathrm{is}\:\mathrm{done}\:\mathrm{by}\:\mathrm{a}\:\mathrm{force}\:\mathrm{on}\:\mathrm{an}\:\mathrm{object}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{the}\:\mathrm{object}\:\mathrm{is}\:\mathrm{stationary}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{of}\:\mathrm{application}\:\mathrm{of}\:\mathrm{the}\:\mathrm{force}\:\mathrm{moves} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{object} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{object}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{application}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{force}\:\mathrm{remains}\:\mathrm{fixed} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{the}\:\mathrm{force}\:\mathrm{is}\:\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{its}\:\mathrm{velocity} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{the}\:\mathrm{force}\:\mathrm{is}\:\mathrm{always}\:\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{its}\:\mathrm{acceleration}. \\ $$

Question Number 23584 Answers: 1 Comments: 0

Question Number 23539 Answers: 1 Comments: 0

∫tan^6 x dx

$$\int\mathrm{tan}\:^{\mathrm{6}} \mathrm{x}\:\mathrm{dx} \\ $$

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