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Question Number 13842 Answers: 3 Comments: 0

Find the number of solutions of the equation sin 5x cos 3x = sin 6x cos 2x, x ∈ [0, π]

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{sin}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\:\mathrm{sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{2}{x}, \\ $$$${x}\:\in\:\left[\mathrm{0},\:\pi\right] \\ $$

Question Number 13840 Answers: 1 Comments: 0

Solve: (a) tanx + secx = 2cosx (b) sinθ + tanθ − sin2θ = 0

$$\mathrm{Solve}: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{tan}{x}\:+\:\mathrm{sec}{x}\:=\:\mathrm{2cos}{x} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{sin}\theta\:+\:\mathrm{tan}\theta\:−\:\mathrm{sin2}\theta\:=\:\mathrm{0} \\ $$

Question Number 13849 Answers: 0 Comments: 2

Given below is a graph between speed and time for a particle. Is the particle undergoing positive displacement or negative displacement?

$$\mathrm{Given}\:\mathrm{below}\:\mathrm{is}\:\mathrm{a}\:\mathrm{graph}\:\mathrm{between}\:\mathrm{speed} \\ $$$$\mathrm{and}\:\mathrm{time}\:\mathrm{for}\:\mathrm{a}\:\mathrm{particle}.\:\mathrm{Is}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{undergoing}\:\mathrm{positive}\:\mathrm{displacement}\:\mathrm{or} \\ $$$$\mathrm{negative}\:\mathrm{displacement}? \\ $$

Question Number 13835 Answers: 1 Comments: 1

Question Number 13830 Answers: 1 Comments: 0

Question Number 13831 Answers: 1 Comments: 0

Question Number 13832 Answers: 1 Comments: 0

Question Number 13812 Answers: 1 Comments: 0

The domain of f(x) = (√(((4/3))^(4−x) −(((27)/(64)))^(x−5) )); is?

$$\mathrm{The}\:\mathrm{domain}\:\mathrm{of}\:{f}\left({x}\right)\:= \\ $$$$\sqrt{\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{4}−{x}} −\left(\frac{\mathrm{27}}{\mathrm{64}}\right)^{{x}−\mathrm{5}} };\:\mathrm{is}? \\ $$

Question Number 13811 Answers: 1 Comments: 1

The domain of f(x) = (√(2 − 2^x − 2^(2x) )) is?

$$\mathrm{The}\:\mathrm{domain}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\sqrt{\mathrm{2}\:−\:\mathrm{2}^{{x}} \:−\:\mathrm{2}^{\mathrm{2}{x}} }\:\mathrm{is}? \\ $$

Question Number 13808 Answers: 1 Comments: 0

If f(x) defined over the domain [0, 1] then domain of function f(10^x ) is?

$$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{defined}\:\mathrm{over}\:\mathrm{the}\:\mathrm{domain}\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\mathrm{then}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{function}\:{f}\left(\mathrm{10}^{{x}} \right)\:\mathrm{is}? \\ $$

Question Number 13806 Answers: 0 Comments: 1

Prove that for −(π/2)<x<(π/2) , (1/1^3 )cos x−(1/3^3 )cos 3x+(1/5^3 )cos 5x−....to infinity =(π/8)((π^2 /4)−x^2 ) .

$${Prove}\:{that}\:{for}\:−\frac{\pi}{\mathrm{2}}<{x}<\frac{\pi}{\mathrm{2}}\:, \\ $$$$\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{3}} }\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\mathrm{cos}\:\mathrm{3}{x}+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} }\mathrm{cos}\:\mathrm{5}{x}−....{to}\:{infinity} \\ $$$$\:\:=\frac{\pi}{\mathrm{8}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{x}^{\mathrm{2}} \right)\:. \\ $$

Question Number 13778 Answers: 0 Comments: 0

Question Number 13756 Answers: 1 Comments: 5

Question Number 13751 Answers: 1 Comments: 4

(ds/dt)=v,(dv/dt)=a,(da/dt)=b,(db/dt)=e,(de/dt)=f (df/dt)=g,(dg/dt)=h,(dh/dt)=i,(di/dt)=j,(dj/dt)=k,..... now if we continue this process to infinity..and if v_0 ,v,a,b,e,f,g,h,i, j,................=1 .then calculate the formula of v and s ...

$$\frac{{ds}}{{dt}}={v},\frac{{dv}}{{dt}}={a},\frac{{da}}{{dt}}={b},\frac{{db}}{{dt}}={e},\frac{{de}}{{dt}}={f} \\ $$$$\frac{{df}}{{dt}}={g},\frac{{dg}}{{dt}}={h},\frac{{dh}}{{dt}}={i},\frac{{di}}{{dt}}={j},\frac{{dj}}{{dt}}={k},..... \\ $$$${now}\:{if}\:{we}\:{continue}\:{this}\:{process}\:{to} \\ $$$${infinity}..{and}\:{if}\:{v}_{\mathrm{0}} ,{v},{a},{b},{e},{f},{g},{h},{i}, \\ $$$${j},................=\mathrm{1}\:.{then}\:{calculate} \\ $$$${the}\:{formula}\:{of}\:{v}\:{and}\:{s}\:... \\ $$$$ \\ $$

Question Number 13745 Answers: 2 Comments: 1

The velocity of a particle moving in straight line is given by the graph shown here. Draw the acceleration position graph.

$$\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{moving}\:\mathrm{in} \\ $$$$\mathrm{straight}\:\mathrm{line}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{shown} \\ $$$$\mathrm{here}.\:\mathrm{Draw}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{position} \\ $$$$\mathrm{graph}. \\ $$

Question Number 13744 Answers: 1 Comments: 4

Solve the following 7^x ≡13(mod 18) Pl give complete process.

$${Solve}\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{7}^{{x}} \equiv\mathrm{13}\left({mod}\:\mathrm{18}\right) \\ $$$${Pl}\:{give}\:{complete}\:{process}. \\ $$

Question Number 13738 Answers: 1 Comments: 1

P,Q,R,S are four locations on the same horizontal plane.Q is on a bearing of 041° from P and the distance is 40km. S is 28km from R on a bearing 074°, R is directly due north of P and the distance between Q and R is 38km. (a)the bearing of R from Q (b)the distance between Q and S (c)the distance between P and R

$$\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}\:\mathrm{are}\:\mathrm{four}\:\mathrm{locations}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{horizontal}\:\mathrm{plane}.\mathrm{Q}\:\mathrm{is}\:\mathrm{on}\:\mathrm{a}\: \\ $$$$\mathrm{bearing}\:\mathrm{of}\:\mathrm{041}°\:\mathrm{from}\:\mathrm{P}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{is}\:\mathrm{40km}. \\ $$$$\mathrm{S}\:\mathrm{is}\:\mathrm{28km}\:\mathrm{from}\:\mathrm{R}\:\mathrm{on}\:\mathrm{a}\:\mathrm{bearing}\:\mathrm{074}°, \\ $$$$\mathrm{R}\:\mathrm{is}\:\mathrm{directly}\:\mathrm{due}\:\mathrm{north}\:\mathrm{of}\:\mathrm{P}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{R}\:\mathrm{is} \\ $$$$\mathrm{38km}. \\ $$$$\left(\mathrm{a}\right)\mathrm{the}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{R}\:\mathrm{from}\:\mathrm{Q} \\ $$$$\left(\mathrm{b}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{S} \\ $$$$\left(\mathrm{c}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{P}\:\mathrm{and}\:\mathrm{R} \\ $$

Question Number 13737 Answers: 1 Comments: 1

A platform and a building are on the same horizontal plane.The angle of depression of the bottom C of the building from the top A of the platform is 39°.The angle of elevation of the top D of the building from the top of the platform is 56°.Given that the distance between the foot of the platform and that of the building is 10m,calculate the height of the building to the nearest whole number.

Question Number 13735 Answers: 2 Comments: 6

Question Number 13731 Answers: 0 Comments: 1

Prove that if p>q>0 and x≥0 (1/p)((x^p /(p+1))−1)≥(1/q)((x^q /(q+1))−1).

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{p}>{q}>\mathrm{0}\:\mathrm{and}\:{x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}}\left(\frac{{x}^{{p}} }{{p}+\mathrm{1}}−\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{{q}}\left(\frac{{x}^{{q}} }{{q}+\mathrm{1}}−\mathrm{1}\right).\: \\ $$

Question Number 13733 Answers: 0 Comments: 4

Question continuing from mrW1 post on p^2 mod n≡1. Find a number n such that for all m<n such that HCF(m,n)=1 m^2 mod n =1 e.g. for 12 possible value of m are 1,5,7,11.

$$\mathrm{Question}\:\mathrm{continuing}\:\mathrm{from} \\ $$$$\mathrm{mrW1}\:\mathrm{post}\:\mathrm{on}\:{p}^{\mathrm{2}} \:\mathrm{mod}\:\mathrm{n}\equiv\mathrm{1}. \\ $$$$\mathrm{Find}\:\mathrm{a}\:\mathrm{number}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{for}\:\mathrm{all}\:{m}<{n}\:\mathrm{such}\:\mathrm{that}\:\mathrm{HCF}\left({m},{n}\right)=\mathrm{1} \\ $$$${m}^{\mathrm{2}} \:\mathrm{mod}\:{n}\:=\mathrm{1} \\ $$$${e}.{g}.\:\mathrm{for}\:\mathrm{12}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{m} \\ $$$$\mathrm{are}\:\mathrm{1},\mathrm{5},\mathrm{7},\mathrm{11}. \\ $$

Question Number 13728 Answers: 1 Comments: 0

Prove that n!>((n/3))^n

$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{n}!>\left(\frac{{n}}{\mathrm{3}}\right)^{{n}} \\ $$

Question Number 13727 Answers: 1 Comments: 1

A light wire which obeys hooke′s law hangs vertically on a fixed support. The wire has an unstretched lenght of 15cm. The lenght of the wire however increase to 20cm when a load of 0.5kg is attached to it lower end . What is the tension in the wire when the load is at rest ?. If the load is pulled down until the lenght of the wire is 22cm. What is the new tension in the wire (g = 9.8 m/s).

Question Number 13725 Answers: 0 Comments: 2

(1/7)=.142857^(−) (1/7) is a recurring decimal of period 6. What will be the period of (1/7^(20) )?

$$\frac{\mathrm{1}}{\mathrm{7}}=.\overline {\mathrm{142857}} \\ $$$$\frac{\mathrm{1}}{\mathrm{7}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{recurring}\:\mathrm{decimal}\:\mathrm{of}\:\mathrm{period}\:\mathrm{6}. \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\frac{\mathrm{1}}{\mathrm{7}^{\mathrm{20}} }? \\ $$

Question Number 13724 Answers: 2 Comments: 3

Expansion of 1000! has 249, 0′s at the end Find the first non−zero digit from right. 1000!=......d000...00 What is the value of d?

$$\mathrm{Expansion}\:\mathrm{of}\:\mathrm{1000}!\:\mathrm{has}\:\mathrm{249},\:\mathrm{0}'{s}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{first}\:\mathrm{non}−\mathrm{zero}\:\mathrm{digit}\:\mathrm{from} \\ $$$$\mathrm{right}. \\ $$$$\mathrm{1000}!=......{d}\mathrm{000}...\mathrm{00} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{d}? \\ $$

Question Number 13721 Answers: 0 Comments: 0

what is NBS?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{NBS}? \\ $$

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