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Question Number 14977 Answers: 2 Comments: 2

Find the real roots of the equation cos^4 x + sin^7 x = 1 in the interval [−π, π].

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:+\:\mathrm{sin}^{\mathrm{7}} \:{x}\:=\:\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\left[−\pi,\:\pi\right]. \\ $$

Question Number 14999 Answers: 1 Comments: 0

A 20 kg box is released from the top of an inclined plane that is 5 m long and makes an angle of 20° to the horizontal. A 60N friction force impedes the motion of the box . How long will it take to reach the bottom of the box.

$$\mathrm{A}\:\mathrm{20}\:\mathrm{kg}\:\mathrm{box}\:\mathrm{is}\:\mathrm{released}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{an}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{is}\:\mathrm{5}\:\mathrm{m}\:\mathrm{long}\:\mathrm{and} \\ $$$$\mathrm{makes}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{20}°\:\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{A}\:\mathrm{60N}\:\mathrm{friction}\:\mathrm{force}\:\mathrm{impedes}\:\mathrm{the}\:\mathrm{motion}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{box}\:.\:\mathrm{How}\:\mathrm{long}\:\mathrm{will}\:\mathrm{it}\:\mathrm{take}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$

Question Number 14971 Answers: 0 Comments: 0

Question Number 14970 Answers: 0 Comments: 0

Question Number 14965 Answers: 2 Comments: 2

Question Number 14991 Answers: 2 Comments: 0

Question Number 14964 Answers: 0 Comments: 0

proof that ∀ x,y ∈N ∃ a,b,c ∈N ∍ (4/(x^2 +y^2 ))=(1/a) + (1/b) + (1/c)

$$\mathrm{proof}\:\mathrm{that}\: \\ $$$$\forall\:{x},{y}\:\in\mathbb{N}\:\:\exists\:{a},{b},{c}\:\in\mathbb{N}\:\backepsilon\:\frac{\mathrm{4}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}} \\ $$

Question Number 14963 Answers: 1 Comments: 0

A resistor R is connected in series with a parallel combination of two resistors of 24 and 8 ohms . The total power disipated in the circuit is 64 watt when the applied voltage is 24 volt.Find R

$$\mathrm{A}\:\mathrm{resistor}\:\mathrm{R}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{two}\:\mathrm{resistors} \\ $$$$\mathrm{of}\:\mathrm{24}\:\mathrm{and}\:\mathrm{8}\:\mathrm{ohms}\:.\:\mathrm{The}\:\mathrm{total}\:\mathrm{power}\:\mathrm{disipated}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{is}\:\mathrm{64}\:\mathrm{watt}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{applied}\:\mathrm{voltage}\:\mathrm{is}\:\mathrm{24}\:\mathrm{volt}.\mathrm{Find}\:\mathrm{R} \\ $$

Question Number 14962 Answers: 1 Comments: 0

Two 30 ohms resistor are connected in parallel, what should be the resistance to be connected in series with this parallel combination such that the power in each 30 ohms is (1/4) th of total power.

$$\mathrm{Two}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{resistor}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{parallel},\:\mathrm{what}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{resistance} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{this}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{power} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{th}\:\mathrm{of}\:\mathrm{total}\:\mathrm{power}. \\ $$

Question Number 15022 Answers: 1 Comments: 0

Evaluate: ∫_1 ^4 ((x^2 + x)/(√(2x + 1))) dx (Question ID: 53) How does the limits change in the solution of Q. No. 53?

$$\mathrm{Evaluate}:\:\int_{\mathrm{1}} ^{\mathrm{4}} \frac{{x}^{\mathrm{2}} \:+\:{x}}{\sqrt{\mathrm{2}{x}\:+\:\mathrm{1}}}\:{dx}\:\left(\mathrm{Question}\:\mathrm{ID}:\right. \\ $$$$\left.\mathrm{53}\right)\:\mathrm{How}\:\mathrm{does}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{change}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{of}\:\mathrm{Q}.\:\mathrm{No}.\:\mathrm{53}? \\ $$

Question Number 15212 Answers: 1 Comments: 1

Question Number 15208 Answers: 1 Comments: 0

Question Number 15207 Answers: 1 Comments: 0

The value of (√3) cosec 20°−sec 20° is equal to ____.

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\sqrt{\mathrm{3}}\:\mathrm{cosec}\:\mathrm{20}°−\mathrm{sec}\:\mathrm{20}°\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\_\_\_\_. \\ $$

Question Number 15206 Answers: 0 Comments: 0

If z = x + jy, detemine the cartesian equation of the locus of the point z which moves in the Argrand diagram so that ∣z + j2∣^2 + ∣z − j2∣^(2 ) = 40

$$\mathrm{If}\:\:\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy},\:\:\mathrm{detemine}\:\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\: \\ $$$$\mathrm{z}\:\mathrm{which}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Argrand}\:\mathrm{diagram}\:\mathrm{so}\:\mathrm{that} \\ $$$$\mid\mathrm{z}\:+\:\mathrm{j2}\mid^{\mathrm{2}} \:+\:\mid\mathrm{z}\:−\:\mathrm{j2}\mid^{\mathrm{2}\:} =\:\mathrm{40} \\ $$

Question Number 15202 Answers: 0 Comments: 0

if z = x + jy , find the equations of the two loci defined by: (a) ∣z − 4∣ = 3 (b) arg(z + 2) = (π/6)

$$\mathrm{if}\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{loci}\:\mathrm{defined}\:\mathrm{by}: \\ $$$$\left(\mathrm{a}\right)\:\mid\mathrm{z}\:−\:\mathrm{4}\mid\:=\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{b}\right)\:\mathrm{arg}\left(\mathrm{z}\:+\:\mathrm{2}\right)\:=\:\frac{\pi}{\mathrm{6}} \\ $$

Question Number 15201 Answers: 0 Comments: 0

If z = x + jy , where x and y are real, show that the locus ∣((z − 2)/(z + 1))∣ = 2 is a circle and determine its centre and radius.

$$\mathrm{If}\:\:\mathrm{z}\:=\:\mathrm{x}\:+\:\mathrm{jy}\:,\:\mathrm{where}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are}\:\mathrm{real},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus}\:\:\mid\frac{\mathrm{z}\:−\:\mathrm{2}}{\mathrm{z}\:+\:\mathrm{1}}\mid\:=\:\mathrm{2}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{and}\:\mathrm{determine}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{radius}. \\ $$

Question Number 15195 Answers: 0 Comments: 3

lim_(x→∞) (((2x − 5)/(2x + 1)))^(x + 3)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{2}{x}\:−\:\mathrm{5}}{\mathrm{2}{x}\:+\:\mathrm{1}}\right)^{{x}\:+\:\mathrm{3}} \\ $$

Question Number 14949 Answers: 0 Comments: 2

Find the largest prime factor of 203203. Anyone please suggest the method without calculators or log tables.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{prime}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{203203}. \\ $$$$\mathrm{Anyone}\:\mathrm{please}\:\mathrm{suggest}\:\mathrm{the}\:\mathrm{method} \\ $$$$\mathrm{without}\:\mathrm{calculators}\:\mathrm{or}\:\mathrm{log}\:\mathrm{tables}. \\ $$

Question Number 14940 Answers: 2 Comments: 12

For those who are interested in Geometry: A triangle has an area of 1 unit. Each of its sides is divided into 4 equal parts through 3 points. The first and the last point of each side will be connected with each other to form 2 inscribed triangles and these 2 triangles form a hexagon. Find the area of the hexagon. What is the result, if each side is equally divided into 5 parts, or generally into n parts?

$${For}\:{those}\:{who}\:{are}\:{interested}\:{in}\: \\ $$$${Geometry}:\: \\ $$$${A}\:{triangle}\:{has}\:{an}\:{area}\:{of}\:\mathrm{1}\:{unit}.\:{Each} \\ $$$${of}\:{its}\:{sides}\:{is}\:{divided}\:{into}\:\mathrm{4}\:{equal}\:{parts} \\ $$$${through}\:\mathrm{3}\:{points}.\:{The}\:{first}\:{and}\:{the}\:{last} \\ $$$${point}\:{of}\:{each}\:{side}\:{will}\:{be}\:{connected} \\ $$$${with}\:{each}\:{other}\:{to}\:{form}\:\mathrm{2}\:{inscribed} \\ $$$${triangles}\:{and}\:{these}\:\mathrm{2}\:{triangles}\:{form} \\ $$$${a}\:{hexagon}.\:{Find}\:{the}\:{area}\:{of}\:{the}\:{hexagon}. \\ $$$$ \\ $$$${What}\:{is}\:{the}\:{result},\:{if}\:{each}\:{side}\:{is} \\ $$$${equally}\:{divided}\:{into}\:\mathrm{5}\:{parts},\:{or} \\ $$$${generally}\:{into}\:{n}\:{parts}? \\ $$

Question Number 14939 Answers: 1 Comments: 0

A point moves in x-y plane according to the law x = 4 sin 6t and y = 4(1 − cos 6t). Find distance traversed by the particle in 5 seconds, when x and y are in metres.

$$\mathrm{A}\:\mathrm{point}\:\mathrm{moves}\:\mathrm{in}\:{x}-{y}\:\mathrm{plane}\:\mathrm{according} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{law}\:{x}\:=\:\mathrm{4}\:\mathrm{sin}\:\mathrm{6}{t}\:\mathrm{and} \\ $$$${y}\:=\:\mathrm{4}\left(\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6}{t}\right).\:\mathrm{Find}\:\mathrm{distance}\:\mathrm{traversed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{in}\:\mathrm{5}\:\mathrm{seconds},\:\mathrm{when}\:{x}\:\mathrm{and} \\ $$$${y}\:\mathrm{are}\:\mathrm{in}\:\mathrm{metres}. \\ $$

Question Number 14938 Answers: 0 Comments: 0

Question Number 14923 Answers: 1 Comments: 0

A plane is inclined at an angle of 30° with horizontal. Find the component of a force F^→ = −10k^∧ N perpendicular to the plane. Given that z-direction is vertically upwards.

$$\mathrm{A}\:\mathrm{plane}\:\mathrm{is}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{30}° \\ $$$$\mathrm{with}\:\mathrm{horizontal}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{component} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{force}\:\overset{\rightarrow} {{F}}\:=\:−\mathrm{10}\overset{\wedge} {{k}}\mathrm{N}\:\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{plane}.\:\mathrm{Given}\:\mathrm{that}\:{z}-\mathrm{direction}\:\mathrm{is} \\ $$$$\mathrm{vertically}\:\mathrm{upwards}. \\ $$

Question Number 14922 Answers: 1 Comments: 2

The velocity of particle P due East is 4 m/s, and that of Q is 3 m/s due South. What is the velocity of P w.r.t. Q?

$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{particle}\:{P}\:\mathrm{due}\:\mathrm{East}\:\mathrm{is} \\ $$$$\mathrm{4}\:\mathrm{m}/\mathrm{s},\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:{Q}\:\mathrm{is}\:\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{due} \\ $$$$\mathrm{South}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{of}\:{P}\:\mathrm{w}.\mathrm{r}.\mathrm{t}. \\ $$$${Q}? \\ $$

Question Number 14921 Answers: 1 Comments: 2

A car travelling at 36 km/h due North turns West in 5 seconds and maintains the same speed. What is the acceleration of the car?

$$\mathrm{A}\:\mathrm{car}\:\mathrm{travelling}\:\mathrm{at}\:\mathrm{36}\:\mathrm{km}/\mathrm{h}\:\mathrm{due}\:\mathrm{North} \\ $$$$\mathrm{turns}\:\mathrm{West}\:\mathrm{in}\:\mathrm{5}\:\mathrm{seconds}\:\mathrm{and}\:\mathrm{maintains} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{speed}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{car}? \\ $$

Question Number 14920 Answers: 1 Comments: 0

A swimmer crosses a flowing river of width d to and fro in time t_1 . The time taken to cover the same distance up and down the stream is t_2 . If t_3 is the time the swimmer would take to swim a distance 2d in still water, then prove that t_1 ^2 = t_2 t_3 .

$$\mathrm{A}\:\mathrm{swimmer}\:\mathrm{crosses}\:\mathrm{a}\:\mathrm{flowing}\:\mathrm{river}\:\mathrm{of} \\ $$$$\mathrm{width}\:{d}\:\mathrm{to}\:\mathrm{and}\:\mathrm{fro}\:\mathrm{in}\:\mathrm{time}\:{t}_{\mathrm{1}} .\:\mathrm{The}\:\mathrm{time} \\ $$$$\mathrm{taken}\:\mathrm{to}\:\mathrm{cover}\:\mathrm{the}\:\mathrm{same}\:\mathrm{distance}\:\mathrm{up} \\ $$$$\mathrm{and}\:\mathrm{down}\:\mathrm{the}\:\mathrm{stream}\:\mathrm{is}\:{t}_{\mathrm{2}} .\:\mathrm{If}\:{t}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{time}\:\mathrm{the}\:\mathrm{swimmer}\:\mathrm{would}\:\mathrm{take}\:\mathrm{to}\:\mathrm{swim} \\ $$$$\mathrm{a}\:\mathrm{distance}\:\mathrm{2}{d}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water},\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{t}_{\mathrm{1}} ^{\mathrm{2}} \:=\:{t}_{\mathrm{2}} {t}_{\mathrm{3}} . \\ $$

Question Number 14905 Answers: 0 Comments: 5

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