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Question Number 29321 Answers: 0 Comments: 1

is it possible to divide an angle into three equal parts?

$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{divide}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{into}\:\mathrm{three}\:\mathrm{equal}\:\mathrm{parts}? \\ $$

Question Number 29320 Answers: 0 Comments: 1

Find the principal argument of ((−1+2i)/(1−3i)) .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{principal}\:\mathrm{argument}\:\mathrm{of}\:\frac{−\mathrm{1}+\mathrm{2i}}{\mathrm{1}−\mathrm{3i}}\:. \\ $$

Question Number 29319 Answers: 0 Comments: 0

mxmz]33]2n3xmxksnd

$$\left.{m}\left.{xmz}\right]\mathrm{33}\right]\mathrm{2}{n}\mathrm{3}{xmxksnd} \\ $$

Question Number 29313 Answers: 0 Comments: 0

Question Number 29311 Answers: 0 Comments: 0

∫(2x^3 −3x^2 +3x−1)^(1/5) dx and limit is from 0 to 1

$$\int\left(\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} {dx}\:{and}\:{limit}\:{is}\:{from}\:\mathrm{0}\:{to}\:\mathrm{1} \\ $$

Question Number 29310 Answers: 0 Comments: 0

Show that the sequence: {1 + (1/2) + (1/4) + ... + (1/2^n )}_(n = 0) ^∞ is convergent.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}:\:\:\:\:\:\left\{\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\:+\:...\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\right\}_{\mathrm{n}\:\:=\:\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\:\mathrm{is}\:\mathrm{convergent}. \\ $$

Question Number 29314 Answers: 0 Comments: 3

Question Number 29287 Answers: 0 Comments: 0

plz is there any app for medical forum just like this one

$${plz}\:{is}\:{there}\:{any}\:{app}\:{for}\:{medical} \\ $$$${forum}\:{just}\:{like}\:{this}\:{one} \\ $$

Question Number 29286 Answers: 0 Comments: 1

To Developer Tinku Tara: Since the recent updates, it is not more possible to change an existing image via Edit Post”. This was possible in earlier versions of the app. Can you please fix this problem? Thank you!

$${To}\:{Developer}\:{Tinku}\:{Tara}: \\ $$$${Since}\:{the}\:{recent}\:{updates},\:{it}\:{is}\:{not}\:{more} \\ $$$${possible}\:{to}\:{change}\:{an}\:{existing}\:{image}\:{via} \\ $$$${Edit}\:{Post}''.\:{This}\:{was}\:{possible}\:{in}\:{earlier} \\ $$$${versions}\:{of}\:{the}\:{app}.\:{Can}\:{you}\:{please} \\ $$$${fix}\:{this}\:{problem}?\:{Thank}\:{you}! \\ $$

Question Number 29276 Answers: 1 Comments: 0

if the (x/a)+(y/b)=1 passes through the point lf intersection of the lines x+y=3 and 2x−3y=1 and is parallel to the line y=x−6 then find the value of a and b.

$${if}\:{the}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{passes}\:{through}\:{the}\:{point}\:{lf}\:{intersection}\:{of}\:{the}\:{lines}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1}\:{and}\:{is}\:{parallel}\:{to}\:{the}\:{line}\:{y}={x}−\mathrm{6}\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{a}\:\:{and}\:{b}.\: \\ $$

Question Number 29275 Answers: 0 Comments: 0

x(1−2xln x)y^((2)) +(1−4x^2 ln x)y^((1)) −(2+4x)y=0 y_1 =ln x the public answer?

$${x}\left(\mathrm{1}−\mathrm{2}{x}\mathrm{ln}\:{x}\right){y}^{\left(\mathrm{2}\right)} +\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \mathrm{ln}\:{x}\right){y}^{\left(\mathrm{1}\right)} −\left(\mathrm{2}+\mathrm{4}{x}\right){y}=\mathrm{0} \\ $$$${y}_{\mathrm{1}} =\mathrm{ln}\:{x} \\ $$$${the}\:{public}\:{answer}? \\ $$

Question Number 29268 Answers: 0 Comments: 2

A man moves 20m North , then 12m East and finally 15m South.His displacement from the starting point is now (a) 13m (b) 27m (c) 47m (d) 23m

$$\mathrm{A}\:\mathrm{man}\:\mathrm{moves}\:\:\mathrm{20m}\:\:\mathrm{North}\:,\:\:\mathrm{then}\:\:\mathrm{12m}\:\mathrm{East}\:\:\mathrm{and}\:\:\mathrm{finally}\:\:\mathrm{15m}\:\:\mathrm{South}.\mathrm{His} \\ $$$$\mathrm{displacement}\:\mathrm{from}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{is}\:\mathrm{now} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{13m}\:\:\left(\mathrm{b}\right)\:\:\mathrm{27m}\:\:\left(\mathrm{c}\right)\:\:\mathrm{47m}\:\:\left(\mathrm{d}\right)\:\:\mathrm{23m} \\ $$

Question Number 29272 Answers: 0 Comments: 0

(2x−3x^3 )y^((2)) +4y^((1)) +6xy=0 this equation has an answer in the form of several sentences. get the public answer.

$$\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{3}} \right){y}^{\left(\mathrm{2}\right)} +\mathrm{4}{y}^{\left(\mathrm{1}\right)} +\mathrm{6}{xy}=\mathrm{0} \\ $$$${this}\:{equation}\:{has}\:{an}\:{answer}\:{in}\:{the}\:{form}\:{of}\:{several}\:{sentences}. \\ $$$${get}\:{the}\:{public}\:{answer}. \\ $$

Question Number 29265 Answers: 0 Comments: 0

An electric pump with efficiency of 70% raises water to a height of 15m . If water is delivered at the rate of 350 dm^3 per minute. (i) what is the power rating of the pump ? (mass of 1 dm^3 = 1 kg) (ii) what is the energy lost by the pump ? (g = 10 m/s^2 ) (Answer: 1250W. 22.5 KJ)

$$\mathrm{An}\:\mathrm{electric}\:\mathrm{pump}\:\mathrm{with}\:\mathrm{efficiency}\:\mathrm{of}\:\:\:\mathrm{70\%}\:\:\mathrm{raises}\:\mathrm{water}\:\mathrm{to}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of}\:\:\mathrm{15m} \\ $$$$.\:\mathrm{If}\:\mathrm{water}\:\mathrm{is}\:\mathrm{delivered}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\:\mathrm{350}\:\mathrm{dm}^{\mathrm{3}} \:\mathrm{per}\:\mathrm{minute}.\:\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{power}\:\mathrm{rating}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pump}\:?\:\:\:\left(\mathrm{mass}\:\mathrm{of}\:\mathrm{1}\:\mathrm{dm}^{\mathrm{3}} \:=\:\mathrm{1}\:\mathrm{kg}\right) \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{energy}\:\mathrm{lost}\:\mathrm{by}\:\mathrm{the}\:\mathrm{pump}\:?\:\:\:\left(\mathrm{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{Answer}:\:\:\:\mathrm{1250W}.\:\:\:\:\mathrm{22}.\mathrm{5}\:\mathrm{KJ}\right) \\ $$

Question Number 29264 Answers: 1 Comments: 0

there are 25 persons in a conical tent every person needs an area of 4 sq m on the ground under the tent. if height if the tent is 18m.find the volume of the tent.

$$\mathrm{there}\:\mathrm{are}\:\mathrm{25}\:\mathrm{persons}\:\mathrm{in}\:\mathrm{a}\:\mathrm{conical}\:\mathrm{tent} \\ $$$$\mathrm{every}\:\mathrm{person}\:\mathrm{needs}\:\mathrm{an}\:\mathrm{area}\:\mathrm{of}\:\mathrm{4}\:\mathrm{sq}\:\mathrm{m}\: \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{under}\:\mathrm{the}\:\mathrm{tent}.\:\mathrm{if} \\ $$$$\mathrm{height}\:\mathrm{if}\:\mathrm{the}\:\mathrm{tent}\:\mathrm{is}\:\mathrm{18m}.\mathrm{find}\:\mathrm{the}\:\mathrm{volume} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{tent}. \\ $$

Question Number 29273 Answers: 0 Comments: 4

Question Number 29260 Answers: 0 Comments: 0

The eriodic time of a simple pendulum is given by T=2π(√(L/g)). if L=100±0.1cm(s.e) and the time s for 10 ossilations are: 48.2,48.2,48.7,48.5,48.4,48.2,48.4, 48.6,48.5,48.2. calcukate g and the standard error involved.

Question Number 29259 Answers: 1 Comments: 0

solve the equation 2x+3y=5 3x+4y=4

$${solve}\:{the}\:{equation} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{5} \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{4}\: \\ $$

Question Number 29308 Answers: 2 Comments: 0

Solve: w^3 = − 16

$$\mathrm{Solve}:\:\:\:\mathrm{w}^{\mathrm{3}} \:=\:−\:\mathrm{16} \\ $$

Question Number 29249 Answers: 1 Comments: 1

Question Number 29241 Answers: 1 Comments: 0

Question Number 29242 Answers: 0 Comments: 0

Question Number 29254 Answers: 1 Comments: 1

Question Number 29222 Answers: 1 Comments: 0

3x−4y=12, xy=2

$$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{12},\:{xy}=\mathrm{2} \\ $$

Question Number 29217 Answers: 1 Comments: 0

The velocity of a physical system is given by V={(√((p+1/n)))}/x where p=pressure. Find the dimensions of n and x

$${The}\:{velocity}\:{of}\:{a}\:{physical}\:{system} \\ $$$${is}\:{given}\:{by}\:{V}=\left\{\sqrt{\left({p}+\mathrm{1}/{n}\right)}\right\}/{x} \\ $$$${where}\:{p}={pressure}.\:{Find}\:{the} \\ $$$${dimensions}\:{of}\:{n}\:{and}\:{x} \\ $$

Question Number 29216 Answers: 1 Comments: 0

A horizontal force of 0.8N is required to pull a 5kg block across a table top at a constant speed. With the block initially at rest,a 20g bullet fired horizontally into the block to slide 1.5m before coming to rest again.Determine the speed v of the bullet,where the bullet is assumed to be embedded in the block.

$${A}\:{horizontal}\:{force}\:{of}\:\mathrm{0}.\mathrm{8}{N}\:{is} \\ $$$${required}\:{to}\:{pull}\:{a}\:\mathrm{5}{kg}\:{block}\:{across} \\ $$$${a}\:{table}\:{top}\:{at}\:{a}\:{constant}\:{speed}. \\ $$$${With}\:{the}\:{block}\:{initially}\:{at}\:{rest},{a} \\ $$$$\mathrm{20}{g}\:{bullet}\:{fired}\:{horizontally}\:{into} \\ $$$${the}\:{block}\:{to}\:{slide}\:\mathrm{1}.\mathrm{5}{m}\:{before} \\ $$$${coming}\:{to}\:{rest}\:{again}.{Determine} \\ $$$${the}\:{speed}\:{v}\:{of}\:{the}\:{bullet},{where} \\ $$$${the}\:{bullet}\:{is}\:{assumed}\:{to}\:{be} \\ $$$${embedded}\:{in}\:{the}\:{block}. \\ $$

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